Yearly Archives: 2022

Page No 196:

Exercise 16A

Question 1:

Take three noncollinear points A, B and C on a page of your notebook. Join AB, BC and CA. What figure do you get?
Name: (i) the side opposite to ∠C
           (ii) the angle opposite to the side BC
           (iii) the vertex opposite to the side CA
           (iv) the side opposite to the vertex B
Figure

ANSWER:

We get a triangle by joining the three non-collinear points A, B, and C.
(i) The side opposite to ∠C is AB.
(ii) The angle opposite to the side BC is ∠A.
(iii) The vertex opposite to the side CA is B.
(iv) The side opposite to the vertex B is AC.

Page No 196:

Question 2:

The measures of two angles of a triangle are 72° and 58°. Find the measure of the third angle.

ANSWER:

The measures of two angles of a triangle are 72° and 58°. 
Let the third angle be x. 
Now, the sum of the measures of all the angles of a triangle is 180^o​.
 ∴∴    x + 72^o + 58^o = 180^o
    ⇒ x + 130^o = 180^o                   
    ⇒ x = 180^o​ −- 130^o
    ⇒ x = 50^o
​The measure of the third angle of the triangle is 50^o​.

Page No 196:

Question 3:

The angles of a triangle are in the ratio 1 : 3 : 5. Find the measure of each of the angles.

ANSWER:

The angles of a triangle are in the ratio 1:3:5. 
Let the measures of the angles of the triangle be (1x), (3x) and (5x)
Sum of the measures of the angles of the triangle = 180^o
      ∴ 1x + 3x + 5x = 180^o
        ⇒ 9x = 180^o
        ⇒ x = 20^o
 1x = 20^o
3x = 60^o
​5x = 100^o
The measures of the angles are 20^o, 60^o and 100^o. 

Page No 196:

Question 4:

One of the acute angles of a right triangle is 50°. Find the other acute angle.

ANSWER:

In a right angle triangle, one of the angles is 90^o.
It is given that one of the acute angled of the right angled triangle is 50^o.
We know that the sum of the measures of all the angles of a triangle is 180^o.
Now, let the third angle be x.
​Therefore, we have:
            90^o​ + 50^o + x = 180^o
   ⇒        140^o + x = 180^o
​   ⇒                    x = 180^o −- 140^o
  ⇒                     x =  40^o
​ The third acute angle is 40^o​.

Page No 196:

Question 5:

One of the angles of a triangle is 110° and the other two angles are equal. What is the measure of each of these equal angles?

ANSWER:

Given:
∠A = 110^o and ​∠B = ∠C
Now, the sum of the measures of all the angles of a traingle is 180^o .
              ∠A + ∠B + ∠C = 180^o
      ⇒    110^o + ​∠B + ∠B = 180^o
​      ⇒    110^o  + 2​∠B = 180^o
​       ⇒                2​∠B = 180^o −- 110^o
       ⇒                2∠B =  70^o
      ⇒                  ∠B = 70^o / 2
      ⇒                  ∠B = 35^o

      ∴ ​∠C = 35^o
The measures of the three angles:
∠A = 110^o, ∠B = 35^o, ​∠C = 35^o

Page No 196:

Question 6:

If one angle of a triangle is equal to the sum of other two, show that the triangle is a right triangle.

ANSWER:

Given:
 ∠A = ∠B + ∠C
​We know:
       ∠A + ∠B + ∠C = 180^o
    ⇒ ∠B +∠C + ∠B + ∠C = 180^o
​    ⇒ 2∠B + 2∠C = 180^o
​    ⇒ 2(∠B +∠C) = 180^o
​    ⇒ ∠B + ∠C = 180/2
    ⇒ ​∠B + ∠C = 90^o
∴∴ ∠A = 90^o
This shows that the triangle is a right angled triangle.

Page No 196:

Question 7:

In a ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate the angles.

ANSWER:

Let 3∠A = 4 ∠B = 6 ∠C = x
Then, we have:
 ∠A = x3, ∠B = x4, ∠C = x6But, ∠A + ∠B + ∠C = 180°∴ x3 + x4 + x6 = 180°or 4x + 3x + 2×12 = 180°or 9x = 180° × 12 = 2160°or x = 240° ∴ ∠A = 2403 = 80°, ∠B = 2404 = 60°, ∠C = 2406 = 40°∠A = x3, ∠B = x4, ∠C = x6But, ∠A + ∠B + ∠C = 180°∴ x3 + x4 + x6 = 180°or 4x + 3x + 2×12 = 180°or 9x = 180° × 12 = 2160°or x = 240° ∴ ∠A = 2403 = 80°, ∠B = 2404 = 60°, ∠C = 2406 = 40°

Page No 196:

Question 8:

Look at the figures given below. State for each triangle whether it is acute, right or obtuse.
Figure

ANSWER:

(i) It is an obtuse angle triangle as one angle is 130^o, which is greater than 90^o.

(ii) It is an acute angle triangle as all the angles in it are less than 90^o.

(iii) It is a right angle triangle as one angle is 90^o.

(iv) It is an obtuse angle triangle as one angle is 92^o, which is greater than 90^o.

Page No 197:

Question 9:

In the given figure some triangles have been given. State for each triangle whether it is scalene, isosceles or equilateral.
Figure

ANSWER:

Equilateral Triangle: A triangle whose all three sides are equal in length and each of the three angles measures 60^o.
Isosceles Triangle: A triangle whose two sides are equal in length and the angles opposite them are equal to each other.
Scalene Triangle: A triangle whose all three sides and angles are unequal in measure.

(i) Isosceles
   AC = CB = 2 cm
(ii) Isosceles
   DE = EF = 2.4 cm
(iii) Scalene
   All the sides are unequal.
(iv) Equilateral
    XY = YZ = ZX = 3 cm
(v) Equilateral
     All three angles are 60^o.
(vi) Isosceles
     Two angles are equal in measure.
(vii) Scalene
    All the angles are unequal.

Page No 197:

Question 10:

Draw a ∆ABC. Take a point D on BC. Join AD. How many triangles do you get? Name them.
Figure

ANSWER:

In ∆ABC, if we take a point D on BC, then we get three triangles, namely ∆ADB, ∆ADC and ∆ABC.

Page No 197:

Question 11:

Can a triangle have
(i) two right angles?
(ii) two obtuse angles?
(iii) two acute angles?
(iv) each angle more than 60°?
(v) each angle less than 60°?
(vi) each angle equal to 60°?

ANSWER:

(i) No
     If the two angles are 90^o each, then the sum of two angles of a triangle will be 180^o​, which is not possible.
(ii) No
      For example, let the two angles be 120^o and 150^o. Then, their sum will be 270^o​, which cannot form a triangle.
(iii) Yes
       For example, let the two angles be 50^o and 60^o​, which on adding, gives 110^o. They can easily form a triangle whose third angle is 180^o −- 110^o = 70^o​.
(iv) No
      For example, let the two angles be 70^o​ and 80^o, which on adding, gives 150^o. They cannot form a triangle whose third angle is 180^o​ −- 150^o = 30^o, which is less than 60^o.
(v) No
      For example, let the two angles be 50^o and 40^o, which on adding, gives 90^o . Thus, they cannot form a triangle whose third angle is 180^o −- 90^o = 90^o​, which is greater than 60^o.
(vi) Yes
      Sum of all angles = 60^o + 60^o + 60^o​ = 180^o

Page No 197:

Question 12:

Fill in the blanks.
(i) A triangle has …… sides, …… angles and …… vertices.
(ii) The sum of the angles of a triangle is …… .
(iii) The sides of a scalene triangle are of ……. lengths.
(iv) Each angle of an equilateral triangle measures …… .
(v) The angles opposite to equal sides of an isosceles triangle are ……. .
(vi) The sum of the lengths of the sides of a triangle is called its ………. .

ANSWER:

(i) A triangle has 3 sides 3 angles and 3 vertices.
(ii) The sum of the angles of a triangle is 180^o. 
(iii) The sides of a scalene triangle are of different lengths.
(iv) Each angle of an equilateral triangle measures 60^o.
(v) The angles opposite to equal sides of an isosceles triangle are equal.
(vi) The sum of the lengths of the sides of a triangle is called its perimeter.

Page No 197:

Exercise 16B

Question 1:

How many parts does a triangle have?
(a) 2
(b) 3
(c) 6
(d) 9

ANSWER:

Correct option: (c)
A triangle has 6 parts: three sides and three angles.

Page No 197:

Question 2:

With the angles given below, in which case the construction of triangle is possible?
(a) 30°, 60°, 70°
(b) 50°, 70°, 60°
(c) 40°, 80°, 65°
(d) 72°, 28°, 90°

ANSWER:

Correct option: (b) 
(a) Sum = 30° + 60° + 70° = 160^o
     This is not equal to the sum of all the angles of a triangle. 
(b) Sum = 50° + 70° + 60° = 180^o
     Hence, it is possible to construct a triangle with these angles.
(c) Sum = 40° + 80° + 65° = 185^o
      This is not equal to the sum of all the angles of a triangle.
(d) Sum = 72° + 28° + 90° = 190^o​
     This is not equal to the sum of all the angles of a triangle.

Page No 197:

Question 3:

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is
(a) 60°
(b) 80°
(c) 76°
(d) 84°

ANSWER:

(b) 80^o
Let the measures of the given angles be (2x)^o, (3x)^o​ and (4x)^o.
∴∴ (2x)^o + (3x)^o + (4x)^o = 180^o
  ⇒ (9x)^o = 180^o
​  ⇒ x = 180 / 9
  ⇒ x = 20^o
∴∴ ​2x =  40^o, 3x = 60^o, 4x = 80^o

Hence, the measures of the angles of the triangle are 40^o​, 60^o, 80^o.
Thus, the largest angle is 80^o.

Page No 198:

Question 4:

The two angles of a triangle are complementary. The third angle is
(a) 60°
(b) 45°
(c) 36°
(d) 90°

ANSWER:

Correct option: (d)
The measure of two angles are complimentary if their sum is 90^o degrees. 
Let the two angles be x and y, such that x + y = 90^o .
Let the third angle be z.
Now, we know that the sum of all the angles of a triangle is 180^o​.
   x + y + z​ = 180^o
 ⇒ 90^o + z = 180^o
 ⇒ z  = 180^o −- 90^o 
        = 90^o
​The third angle is 90^o.

Page No 198:

Question 5:

One of the base angles of an isosceles triangle is 70°. The vertical angle is
(a) 60°
(b) 80°
(c) 40°
(d) 35°

ANSWER:

Correct option: (c)
Let ∠A = 70^o
The triangle is an isosceles triangle.
We know that the angles opposite to the equal sides of an isosceles triangle are equal.
∴∴ ​∠B = 70^o
​We need to find the vertical angle ​∠C.
Now, sum of all the angles of a triangle is 180^o.
    ∠A + ∠B + ∠C = 180^o
⇒​ 70^o + 70^o​ + ∠C = 180^o
⇒ 140^o + ​∠C = 180^o
⇒ ∠C = 180^o −- 140^o
⇒ ∠C = 40^o

Page No 198:

Question 6:

A triangle having sides of different lengths is called
(a) an isosceles triangle
(b) an equilateral triangle
(c) a scalene triangle
(d) a right triangle

ANSWER:

Correct option: (c)
A triangle having sides of different lengths is called a scalene triangle.

Page No 198:

Question 7:

In an isosceles ∆ABC, the bisectors of ∠B and ∠C meet at a point O. If ∠A = 40°, then ∠BOC =
?
(a) 110°
(b) 70°
(c) 130°
(d) 150°

ANSWER:

Correct option: (a)

In the isosceles ABC, ​the bisectors of ∠B and ∠C meet at point O.
Since the triangle is isosceles, the angles opposite to the equal sides are equal.
∠B = ∠C
∴∴ ∠A + ∠B + ∠C = 180^o
  ⇒  40^o + 2∠B = 180^o
  ⇒ 2∠B = 140^o
  ⇒ ∠B = 70^o
Bisectors of an angle divide the angle into two equal angles.
So, in  ∆BOC:
∠OBC = 35^o and ∠OCB = 35^o
∠BOC + ∠OBC + ∠OCB = 180​^o
  ⇒ ∠BOC + 35^o + 35^o = 180^o
  ⇒ ∠BOC = 180^o​ – 70^o
  ⇒ ∠BOC = 110^o

Page No 198:

Question 8:

The sides of a triangle are in the ratio 3 : 2 : 5 and its perimeter is 30 cm. The length of the longest side is
(a) 20 cm
(b) 15 cm
(c) 10 cm
(d) 12 cm

ANSWER:

Correct option: (b)
The sides of a triangle are in the ratio 3:2:5.
Let the lengths of the sides of the triangle be (3x), (2x), (5x).
We know:
 Sum of the lengths of the sides of a triangle = Perimeter
   (3x) + (2x) + (5x) = 30
  ⇒ 10x = 30
  ⇒ x =  30 
            10
  ⇒ x = 3
First side = 3x = 9 cm
Second side = 2x = 6 cm
Third side = 5x = 15 cm
The length of the longest side is 15 cm.

Page No 198:

Question 9:

Two angles of a triangle measure 30° and 25° respectively. The measure of the third angle is
(a) 35°
(b) 45°
(c) 65°
(d) 125°

ANSWER:

Correct option: (d)
Two angles of a triangle measure 30° and 25°, respectively.
  Let the third angle be x.
   x + 30^o + 25^o = 180^o
​                    x = 180^o −- 55^o
                     x = 125^o

Page No 198:

Question 10:

Each angle of an equilateral triangle measures
(a) 30°
(b) 45°
(c) 60°
(d) 80°

ANSWER:

Correct option: (c)
Each angle of an equilateral triangle measures 60^o.

Page No 198:

Question 11:

In the adjoining figure, the point P lies
(a) in the interior of ∆ABC
(b) in the exterior of ∆ABC
(c) on ∆ABC
(d) outside ∆ABC
Figure

ANSWER:

Correct option: (c)
Point P lies on ∆ABC.

Exercise 17 A

Page No 202:

Question 1:

In the adjacent figure, a quadrilateral has been shown.
Name: (i) its diagonals,
           (ii) two pairs of opposite sides,
           (iii) two pairs of opposite angles,
           (iv) two pairs of adjacent sides,
           (v) two pairs of adjacent angles.
Figure

ANSWER:

(i) The diagonals are AC, and BD.
(ii) AB  and CD, and AD and BC are the two pairs of opposite sides.
(iii) ∠A and ∠C, and ∠B and ∠D are the two pairs of opposite angles.
(iv) AB and BC, and AD and DC are the two pairs of adjacent sides.
(v) ​∠A and ∠B, and ∠C and ∠D are the two pairs of adjacent angles.

Page No 202:

Question 2:

Draw a parallelogram ABCD in which AB = 6.5 cm, AD = 4.8 cm and ∠BAD = 70°. Measure its diagonals.

ANSWER:

Since ABCD is a parallelogram, AB = DC = 6.5 cm and AD = BC = 4.8 cm.
Given:
∠A = 70°∠A = 70°

Steps of construction :
1) Draw AD equal to 4.8 cm.
2) Make an angle of 70^° at A and cut an arc of 6.5 cm. Name it B.
3) Cut an arc of 4.8 cm from B and 6.5 cm from D. Name it C.
4) Join AB, BC and CD.
5) Measuring the diagonals AC and BD, we get AC equal to 9.2 cm and BD equal to 6.6 cm.

Page No 202:

Question 3:

Two sides of a parallelogram are in the ratio 4 : 3. If its perimeter is 56 cm, find the lengths of its sides.

ANSWER:

Two sides of a parallelogram are in the ratio 4:3.
Let the two sides be 4x and 3x.
In a parallelogram, opposite sides are equal and parallel. So, they are also in the ratio of 4:3, i.e. 4x and 3x.
Perimeter = 4x + 3x + 4x +3x 
           56 = 14x
             x = 56145614
                        
             x = 4

∴ 4x = 16
3x = 12
Length of its sides are 16cm, 12 cm, 16cm and 12cm.

          

Page No 202:

Question 4:

Name each of the following parallelograms:
(i) The diagonals are equal and the adjacent sides are unequal.
(ii) The diagonals are equal and the adjacent sides are equal.
(iii) The diagonals are unequal and the adjacent sides are equal.

ANSWER:

(i) Rectangle
(ii) Square
(iii) Rhombus

Page No 203:

Question 5:

What is a trapezium? When do you call a trapezium an isosceles trapezium?
Draw an isosceles trapezium. Measure its sides and angles.

ANSWER:

A trapezium has only one pair of parallel sides.
A trapezium is said to be an isosceles trapezium if its non-parallel sides are equal.
Following are the measures of the isosceles trapezium:
AB = 5.4 cm
BC = 3 cm
DC = 7.4 cm
AD = 3 cm

∠A =∠B =110°∠D = ∠C = 70°∠A =∠B =110°∠D = ∠C = 70°

Page No 203:

Question 6:

Which of the following statements are true and which are false?
(a) The diagonals of a parallelogram are equal.
(b) The diagonals of a rectangle are perpendicular to each other.
(c) The diagonals of a rhombus are equal.

ANSWER:

(a) False
(b) False
(c) False

Page No 203:

Question 7:

Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A square can be thought of as a special rhombus.
(c) A rectangle can be thought of as a special parallelogram.
(d) a square is also a parallelogram.

ANSWER:

(a) This is because a rectangle with equal sides becomes a square.
(b) This is because a rhombus with each angle a right angle becomes a square.
(c) This is because a parallelogram with each angle a right angle becomes a rectangle.
(d) This is because in a square opposite sides are parallel.

Page No 203:

Question 8:

A figure is said to be regular if its sides are equal in length and angles are equal in measure. What do you mean by a regular quadrilateral?

ANSWER:

A square is a regular quadrilateral all of whose sides are equal in length and all of whose angles are equal in measure.

Page No 203:

Exercise 17 B

Question 1:

The sum of all the angles of a quadrilateral is
(a) 180°
(b) 270°
(c) 360°
(d) 400°

ANSWER:

(c) 360°
The sum of all the angles of a quadrilateral is  360^°​.

Page No 203:

Question 2:

The three angles of a quadrilateral are 80°, 70° and 120°. The fourth angle is
(a) 110°
(b) 100°
(c) 90°
(d) 80°

ANSWER:

(c) 90°

The three angles of a quadrilateral are 80°, 70° and 120°.
Let the fourth angle be x.
We know that the sum of all the angles of a quadrilateral is 360°.
                  80° + 70°​ + 120° + x = 360°
​                         ⇒  270°​ + x = 360°
^​                                 ⇒              x = 360° − 270°
^​                                   ⇒             x = 90°
Thus, the fourth angle is 90°.

Page No 203:

Question 3:

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The largest of these angles is
(a) 90°
(b) 120°
(c) 150°
(d) 102°

ANSWER:

Let the angles of a quadrilateral be (3x)^°, (4x)^°​, (5x)^° and (6x)^°.
Sum of all the angles of a quadrilateral is 360^°.

∴  3x + 4x + 5x + 6x = 360°
         ⇒              18x  = 360°
         ⇒                  x =  3601836018
          ⇒                x = 20°
So,
       3x = 60°
      4x =  80°
​      5x = 100°
​      6x = 120°
The largest of these angles is 120°​.
So, the correct answer is given in option (b).

Page No 203:

Question 4:

A quadrilateral having one and only one pair of parallel sides is called
(a) a parallelogram
(b) a kite
(c) a rhombus
(d) a trapezium

ANSWER:

(d) a trapezium
A trapezium is a quadrilateral that has only one pair of parallel sides.

Page No 203:

Question 5:

A quadrilateral whose opposite sides are parallel is called
(a) rhombus
(b) a kite
(c) a trapezium
(d) a parallelogram

ANSWER:

(d) a parallelogram
A parallelogram is a quadrilateral whose opposite sides are parallel.

Page No 203:

Question 6:

An isosceles trapezium has
(a) equal parallel sides
(b) equal nonparallel sides
(c) equal opposite sides
(d) none of these

ANSWER:

(b) equal nonparallel sides
The non-parallel sides of an isosceles trapezium are equal.

Page No 203:

Question 7:

If the diagonals of a quadrilateral bisect each other at right angles, then this quadrilateral is
(a) a rectangle
(b) a rhombus
(c) a kite
(d) none of these

ANSWER:

(b) a rhombus
The diagonals of a rhombus bisect each other at right angle.

Page No 203:

Question 8:

A square has
(a) all sides equal and diagonals unequal
(b) all sides equal and diagonals equal
(c) all sides unequal and diagonals equal
(d) none of these

ANSWER:

(b) all sides equal and diagonals equal
In a square, all the sides are equal. All of its diagonals are also equal. 

Page No 203:

Question 9:

A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides, is called a
(a) trapezium
(b) parallelogram
(c) kite
(d) rectangle

ANSWER:

(c) kite

A kite has two pairs of equal adjacent sides, but unequal opposite sides.

Page No 203:

Question 10:

What do you mean by a regular quadrilateral?
(a) A rectangle
(b) A rhombus
(c) A square
(d) A trapezium

ANSWER:

(c) A square
The only regular quadrilateral is a square. This is because all of its sides and angles are equal.

Page No 207:

Exercise 18A

Question 1:

Take a point O on your notebook and draw circles of radii 4 cm, 5.3 cm and 6.2 cm, each having the same centre O.

ANSWER:

This is the required diagram as asked in the question.

 

Page No 207:

Question 2:

Draw a circle with centre C and radius 4.5 cm. Mark points P, Q, R such taht P lies in the interior of the circle, Q lies on the circle, and R lies in the exterior of the circle.

ANSWER:

This is the required diagram as asked in the question.

Page No 207:

Question 3:

Draw a circle with centre O and radius 4 cm. Draw a chord AB of the circle. Indicate by marking points X and Y, the minor arc AXB and the major arc AYB of the circle.

ANSWER:

This is the required diagram as asked in the question.

Page No 207:

Question 4:

Which of the following statements are true and which are false?
(i) Each radius of a circle is also a chord of the circle.
(ii) Each diameter of a circle is also a chord of the circle.
(iii) The centre of a circle bisects each chord of the circle.
(iv) A secant of a circle is a segment having its end points on the circle.
(v) A chord of a circle is a segment having its end points on teh circle.

ANSWER:

(i)  False
Diameter of a circle is a chord of the circle, not radius.
(ii) True
It is the longest chord of the circle.
(iii) False
A perpendicular drawn from the centre of the circle to the chord, bisects the chord.
(iv) False
It is a line passing through the circle that intersects the circle at two points.
(v) True.

Page No 207:

Question 5:

Draw a circle with centre O and radius 3.7 cm. Draw a sector having the angle 72°.

ANSWER:

Therefore, the required arc is arc OACB.

Page No 207:

Question 6:

Fill in the blanks by using <, >, = or ≤.
(i) OP …… OQ, where O is the centre of the circle, P lies on the circle and Q is in the interior of the circle.
(ii) OP …… OR, where O is the centre of the circle, P lies on the circle and R lies in the exterior of the circle.
(iii) Major arc …… minor arc of the circle.
(iv) Major arc …… semicircumference of the circle.

ANSWER:

(i) >
(ii) <
(iii) >
(iv) >
This is because the major arc covers more than half of the circumference of the circle.

Page No 207:

Question 7:

Fill in the blanks:
(i) A diameter of a circle is a chord that ………. the centre.
(ii) A radius of a circle is a line segment with one end point ……….. and the other end point …… .
(iii) If we join any two points of a circle by a line segment, we obtain a ……….. of the circle.
(iv) Any part of a circle is called an ……….. of the circle.
(v) The figure bounded by an arc and the two radii joining the end points of the arc with the centre is called a ……….. of the circle.

ANSWER:

(i)  passes through
(ii)  on the circle, at the centre of the circle
(iii)  chord
(iv)  arc
(v)  sector

Page No 209:

Exercise 18B

Question 1:

Define each of the following:
(a) Closed figures
(b) Open figures
(c) Polygons

ANSWER:

(i) A closed figure is a figure that can be traced with the same starting and ending points, and that too without crossing or retracing any part of the figure.
For example: Polygon, circle, etc.

(ii) A figure having no boundary and no starting or ending points is called as an open figure.

(iii)  A polygon is a plane shape with 3 or more straight sides. It is a 2 dimensional closed figure.

Page No 209:

Question 2:

Define each of the following:
(a) A scalene triangle
(b) An isosceles triangle
(c) An obtuse triangle

ANSWER:

(a) A triangle having no sides or angles equal is called a scalene triangle.

(b) A triangle having two sides and the corresponding opposite angles equal is called an isosceles triangle.

(c) A triangle having one of the angles more than 90°° is called an obtuse triangle.

Page No 209:

Question 3:

(i) What do you mean by a convex quadrilateral?
(ii) Define a regular polygon.

ANSWER:

(i)  A quadrilateral with no interior angles greater than 180° is known as a convex quadrilateral.

(ii)  A regular polygon is a polygon all of whose sides are of the same lengths and all of whose interior angles are of the same measures.

Page No 209:

Question 4:

The angles of a triangle are in the ratio 3 : 5 : 7. Find the measures of these angles.

ANSWER:

The angles are in ratio 3:5:7.
Suppose the angles are 3x, 5x and 7x.

∴ 3x + 5x + 7x = 180                (angle sum property of a triangle)
                  15x =180
                     x = 12^o

Therefore, the angles are of the measures 36°, 60° and 84°.

Page No 209:

Question 5:

The angles of a quadrilateral are in the ratio 2 : 3 : 4 : 6. Find the measures of these angles.

ANSWER:

Suppose the angles are 2x, 3x, 4x, and 6x.
We know that the sum of the angles of a quadrilateral is 360°.
∴ 2x + 3x + 4x + 6x = 360
                         15x = 360
                            x = 24

Therefore, the measures of the angles are 48°, 72°, 96° and 144°.

Page No 209:

Question 6:

State the properties of a rhombus.

ANSWER:

1. A rhombus is a parallelogram whose opposite sides are parallel.
2. All four of its sides are equal in length. Also, the opposite angles are equal.
3. The diagonals bisect each other at right angles.

Page No 209:

Question 7:

Define (i) a trapezium (ii) a kite.

ANSWER:

(i) A trapezium is a quadrilateral with only one pair of parallel sides.   

(ii) A kite is a quadrilateral that has two pairs of equal adjacent sides, but unequal opposite sides.

Page No 209:

Question 8:

Draw a circle with centre O and radius 3 cm. Draw a sector having an angle of 54°.

ANSWER:

Page No 209:

Question 9:

A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides is called a
(a) parallelogram
(b) rectangle
(c) trapezium
(d) kite

ANSWER:

(d) kite

Page No 209:

Question 10:

If the diagonals of a quadrilateral bisect each other at right angles, then this quadrilateral is a
(a) rectangle
(b) parallelogram
(c) rhombus
(d) kite

ANSWER:

(c) rhombus

Page No 209:

Question 11:

A quadrilateral having one and only one pair of parallel sides is called a
(a) parallelogram
(b) a kite
(c) a trapezium
(d) a rhombus

ANSWER:

(c) a trapezium

Page No 209:

Question 12:

One of the base angles of an isosceles triangle is 70°. The vertical angle is
(a) 35°
(b) 40°
(c) 70°
(d) 80°

ANSWER:

(b) 40°

Since one base angle is 70°, the other base angle will also be 70° because the triangle is isosceles.
Vertical angle:
180 − 70 − 70 = 40°°                 (angle sum property of a triangle)

Page No 209:

Question 13:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) The diagonals of a rhombus are equal.
(ii) The diagonals of a parallelogram bisect each other.
(iii) The centre of a circle bisects each chord of a circle.
(iv) Each diameter of a circle is a chord of the circle.
(v) The diagonals of a rhombus bisect each other at right angles.

ANSWER:

(i) False
Diagonals are perpendicular and bisect each other.

(ii) True
The diagonals of a parallelogram bisect each other.

(iii) False
A perpendicular drawn from the centre of a circle to the chord, bisects the chord.

(iv)  True
It divides the circle in two equal parts.

(v) True
The diagonals of a rhombus bisect each other at right angles.

Exercise 14A

Page No 186:

Question 1:

Draw a line segment PQ = 6.2 cm. Draw the perpendicular bisector of PQ.

ANSWER:

Steps for construction:
1. Draw a line segment PQ, which is equal 6.2 cm.
2. With P as the centre and radius more than half of PQ, draw arcs, one on each side of PQ.
3. With Q as the centre and the same radius as before, draw arcs cutting the previously drawn arcs at A and B, respectively.
4. Draw AB, meeting PQ at R.

Page No 186:

Question 2:

Draw a line segment AB = 5.6 cm. Draw the perpendicular bisector of AB.

ANSWER:

Steps for construction:
1. Draw a line segment AB = 5.6 cm.
2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.
3. With B as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at P and Q, respectively.
4. Draw PQ, meeting AB at R.

Page No 186:

Question 3:

Draw an angle equal to ∠AOB, given in the adjoining figure.

ANSWER:

Here ∠∠AOB is given.
Steps for construction:
1. Draw a ray QP.
2. With O as the centre and any suitable radius, draw an arc cutting OA and OB at C and E, respectively..
3. With Q as the centre and the same radius as in step (2), draw an arc cutting QP at D.
4. With D as the centre and radius equal to CE, cut the arc through D at F.
5. Draw QF and produce it to point R.
∴∴ ∠∠PQR = ∠∠AOB

Page No 186:

Question 4:

Draw an angle of 50° with the help of a protractor. Draw a ray bisecting this angle.

ANSWER:

Steps for construction:

1. Draw ∠∠BAC = 50°° with the help of protractor.
2. With A as the centre and any convenient radius, draw an arc cutting AB and AC at Q and P, respectively.
3. With P as the centre and radius more than half of PQ, draw an arc.
4. With Q as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point S.
5. Draw SA and produce it to point R.
Then, ray AR bisects ∠∠BAC.

Page No 186:

Question 5:

Construct ∠AOB = 85° with the help of a protractor. Draw a ray OX bisecting ∠AOB.

ANSWER:

Steps for construction:

1. Draw ∠∠AOB = 85°° with the help of a protractor.
2. With O as the centre and any convenient radius, draw an arc cutting OA and OB  at P and Q, respectively.
3. With P as the centre and radius more than half of PQ, draw an arc.
4. With Q as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point R.
5. Draw RO and produce it to point X.
Then, ray OX bisects ∠∠AOB.

Page No 186:

Question 6:

Draw a line AB. Take a point P on it. Draw a line passing through P and perpendicular to AB.

ANSWER:

Steps for construction:

1. Draw a line AB.
2. Take a point P on line AB.
3. With P as the centre, draw an arc of any radius, which intersects line AB at M and N, respectively.
4. With M as the centre and radius more than half of MN, draw an arc.
5. With N as the centre and the same radius as in step (4), draw an arc that cuts the previously drawn arc at R.
6. Draw PR.
PR is the required line, which is perpendicular to AB.

Page No 186:

Question 7:

Draw a line AB. Take a point P outside it. Draw a line passing through P and perpendicular to AB.

ANSWER:

Steps for construction:

1. Draw a line AB.
2. Take a point P outside AB.
3. With P as the centre and a convenient radius, draw an arc intersecting AB at M and N, respectively.
4. With M as the centre and radius more than half of MN, draw an arc.
5. With N as the centre and the same radius, draw an arc cutting the previously drawn arc at Q.
6. Draw PQ meeting AB at S.
PQ is the required line that passes through P and is perpendicular to AB.

Page No 186:

Question 8:

Draw a line AB. Take a point P outside it. Draw a line passing through P and parallel to AB.

ANSWER:

Steps for construction:

1. Draw a line AB.
2. Take a point P outside AB and another point O on AB.
3. Draw PO.
4. Draw ∠∠FPO such that ∠∠FPO is equal to AOP.
5. Extend FP to E.
Then, the line EF passes through the point P and EF||AB.

Page No 186:

Question 9:

Draw ∠ABC of measure 60° such that AB = 4.5 cm and AC = 5 cm. Through C draw a line parallel to AB and through B draw a line parallel to AC, intersecting each other at D. Measure BD and CD.

ANSWER:

Steps for construction:

1. Draw a line BX and take a point A, such that AB is equal to 4.5 cm.
2. Draw ∠∠ABP = 60°° with the help of protractor.
3. With A as the centre and a radius of 5 cm, draw an arc cutting PB at C.
4. Draw AC.
5. Now, draw ∠∠BCY = 60°°.
6. Then, draw ∠∠ABW, such that ∠∠ABW is equal to∠∠CAX, which cut the ray CY at D.
7. Draw BD.

When we measure BD and CD, we have:
BD = 5 cm and CD = 4.5 cm

Page No 186:

Question 10:

Draw a line segment AB = 6 cm. Take a point C on AB such taht AC = 2.5 cm. Draw CD perpendicular to AB.

ANSWER:

Steps of constructions

1. Draw a line segment AB, which is equal to 6 cm.
2. Take a point C on AB such that AC is equal to 2.5 cm.
3. With C as the centre, draw an arc cutting AB at M and N.
4. With M as the centre and radius more than half of MN, draw an arc.
5. With N as the centre and the same radius as before, draw another arc cutting the perviously drawn arc at S.
6. Draw SC and produce it to D.

Page No 186:

Question 11:

Draw a line segment AB = 5.6 cm. Draw the right bisector of AB.

ANSWER:

Steps for construction:

1. Draw a line segment AB, which is equal to 5.6 cm.
2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.
3. With B as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at M and N, respectively.
4. Draw MN, meeting AB at R.

Page No 188:

Exercise 14B

Question 1:

Using a pair of compasses construct the following angles:
(i) 60°
(ii) 120°
(iii) 90°

ANSWER:

(i) 
Steps for construction:
1. Draw a ray QP.
2. With Q as the centre and any convenient radius, draw an arc cutting QP at N.
3. With N as the centre and the same radius as before, draw another arc to cut the previous arc at M.
4. Draw QM and produce it to R.
^∠∠ PQR is the required angle of 60^o.
(ii)
Steps for construction:
1. Draw a ray QP.
2. With Q as the centre and any convenient radius, draw an arc cutting QP at N.
3. With N as the centre and the same radius, cut the arc at A. Again, with A as the centre and the same radius, cut the arc at M.
4. Draw QM and produce it to R.
^{∠PQR is the required angle of 120°.∠PQR is the required angle of 120°.}

(iii)
Steps for construction:
1. Draw a line PX.
2. Take a point Q on AC. With Q as the centre and any convenient radius, draw an arc cutting AX at M and N.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at W.
5. Draw QW produce it to R.
∠PQR is required angle of 90°.∠PQR is required angle of 90°.

Page No 188:

Question 2:

Draw an angle of 60°, using a pair of compasses. Bisect it to make an angle of 30°.

ANSWER:

Constructions steps:

1. Draw a ray QP.
2. Wth Q as the centre and any convenient radius,draw an arc cutting QP at N.
3. With N as the centre and radius same as before, draw another arc to cut the previous arc at M.
4. Draw QM and produce it to R.
∠PQR is an angle of 60°.∠PQR is an angle of 60°.
5. With M as the centre and radius more than half of MN, draw an arc.
6. With N as the centre and radius same as in step (5), draw another arc, cutting the previously drawn arc at point X.
7. Draw QX and produce it to point S.
Ray QS is the bisector of ∠PQR∠PQR.

Page No 188:

Question 3:

Draw an angle of 45°, using a pair of compasses.

ANSWER:

Construction steps:

1. Draw a line PR.
2. Take a point Q on PR. With Q as the centre and any convenient radius, draw an arc cutting AC at M and N.
3.With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at X.
5. Draw QX, meeting the arc at Z. Produce it to W.
6. With Z as the centre and radius more than half of ZN, draw an arc.
7. With N as the centre and the same radius as in step (6), draw another arc, cutting the previously drawn arc at a point Y.
8. Draw QY and produce it to point S.
∠PQS is the required angle of 45°.∠PQS is the required angle of 45°.

Page No 188:

Question 4:

Use a pair of compasses and construct the following angles:
(i) 150°
(ii) 15°
(iii) 135°
(iv) 2212°2212°
(v) 105°
(vi) 75°
(vii) 6712°6712°
(viii) 45°

ANSWER:

(i)

Steps for construction:
1. Draw a line XY and take a point O.
2. With O as the centre and any suitable radius, draw an arc cutting XY at M and N.
3.With N as the centre and the same radius,draw an arc cutting MN at R.
4.With R as the centre and the same radius as before, draw another arc cutting MN at Q .
5. With Q as the centre and radius less than MQ draw an arc.
6. With M as the centre and the same radius draw another arc cutting the previously drawn arc at P
5. Join PO.
 ∴∴ ∠∠XOP = 150°°

(ii)

Steps for construction:
1. Draw a ray XY.
2. With X as the centre and any convenient radius, draw an arc cutting XY at M.
3. With M as the centre and the same radius, draw an arc cutting the previously drawn arc at N.
4. Draw YN and produce it to B.
4. Draw the bisector AY of ∠∠XYB.
5. Again, draw the bisector YZ of ∠∠XYA.
 ∴∴ ∠∠XYZ = 15°°

(iii)

Steps for construction:
1. Draw a line XY and take a point A.
2. With A as the centre and any convenient radius, draw an arc cutting XY at M and N.
3. With N as the centre and the same radius, draw an arc.
4. With M as the centre and the same radius as before, draw another arc cutting the previously drawn arc at R.
5. Draw RA.
6. Draw draw the bisector ZA of ∠∠YAR.
∴∴ ∠∠XAZ = 135°°

(iv)

Steps for construction:
1. Draw a line XY.
2. Take a point A on XY. With A as the centre and any convenient radius, draw an arc cutting XY at M and N.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at P.
5. Draw PA meeting the arc at C. Produce it to E.
6. With C as the centre and radius more than half of CN, draw an arc.
7. With N as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point Q.
8. Draw AQ and produce it to point F.
9. Draw the bisector ZA of ∠∠XAF.
 ∴∴ ∠∠XAZ = 22.5°°

(v)


Steps for construction:
1. Draw a line XY.
2. Take a point O on XY. With O as the centre and any convenient radius, draw an arc cutting XY at M and N. Draw arcs with the same radius cutting MN at P and Q.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at B.
5. Draw BO meeting the arc at E.
6. With Q as the centre and radius more than half of PE, draw an arc.
7. With E as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point A.
8. Draw AO and produce it to point Z.

 ∴∴ ∠∠XOZ = 105°°
(vi)

Steps for construction:
1. Draw a line XY.
2. Take a point O on XY. With O as the centre and any convenient radius, draw an arc cutting XY at M and N. Draw arcs with the same radius cutting MN at P.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at R.
5. Draw RO meeting the arc at E. Produce it to A.
6. With P as the centre and radius more than half of PE, draw an arc.
7. With E as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point S.
8. Draw OS and produce it to point Z.

 ∴∴ ∠∠XOZ = 75°°

(vii)

Steps for construction:
1. Draw a line XY and take a point O.
2. With O as the centre and any convinient radius, draw an arc cutting XY at M and N.
3. With N as the centre and the same radius, draw an arc.
4. With M as the centre and the same radius as before, draw another arc cutting the previously drawn arc at Q.
5. Draw QO.
6. Draw PO bisector of ∠∠YOA.
7. Draw ZO bisector of ∠∠POX.
 ∴∴ ∠∠XAZ = 67.5°°

(viii)

Steps for construction:
1. Draw a line PR.
2. Take a point Q on PR. With Q as the centre and any convenient radius, draw an arc cutting AC at M and N.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at X.
5. Draw QX, meeting the arc at Z. Produce it to W.
6. With Z as the centre and radius more than half of ZN, draw an arc.
7. With N as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point Y.
8. Draw QY and produce it to point S.

Page No 188:

Question 5:

Draw a rectangle whose two adjacent sides are 5 cm and 3.5 cm. Make use of a pair of compasses and a ruler only.

ANSWER:

Construction steps:

1. Draw a ray AX.
2. With A as the centre, cut the ray AX at B such that AB is equal to 5 cm.
3. With B as the centre and any convenient radius, draw an arc cutting AX at M and N.
4. With N as the centre and radius more than half of MN, draw an arc.
5. With M as the centre and the same radius as before, draw another arc to cut the previous arc at Y.
6. Draw BY and produce it to W.
7. With B as the centre and a radius of 3.5 cm, cut ray BW at point C.
8. With C as the centre and a radius of 5 cm, draw an arc on the right side of BC.
9. With A as the centre and a radius of 3.5 cm, draw an arc cutting the previous arc at D.
10. Join CD and AD.
ABCD is the required rectangle.

Page No 188:

Question 6:

Draw a square, each of whose sides is 5 cm. Use a pair of compasses and a ruler in your construction.

ANSWER:

Construction steps:

1. Draw a ray AX.
2. With A as centre cut the ray AX at B such that AB=5 cm
3. With B as centre and any convenient radius,draw an arc cutting AX at M and N.
4. With N as centre and radius more than half of MN draw an arc.
5. With M as centre and the same radius as before,draw another arc to cut the previous arc at Y.
6. Join BY and produced it to W.
7. With B as centre and radius 5 cm cut ray BW at point C.
8.With C as centre and  radius  5 cm draw an arc on right side of BC.
9. With A as centre and  radius 5 cm draw an arc cutting the previous arc at D.
10.Join CD and AD.
ABCD is required square.

Page No 189:

Exercise 14C

Question 1:

How many lines can be drawn to pass through
(i) a given point
(ii) two given points
(iii) three given points

ANSWER:

(i) We can draw infinite number of lines passing through a given point.


(ii) Only one line can be drawn with two given points.


(iii) We can draw one line with three given points if all the three point are collinear.
 But, if the points are not collinear, then we cannot draw any line passing through the points.
 

Page No 189:

Question 2:

Classify the angles whose magnitudes are given below.
(i) 50°
(ii) 92°
(iii) 185°
(iv) 90°
(v) 180°

ANSWER:

(i) It is an acute angle because it is more than 0°° and less than 90°°.
(ii) It is an obtuse angle because it is more than 90°° and less than 180°°.
(iii) It is a reflex angle because it is more than 180°° and less than 360°°.
(iv) It is a right angle because it is 90°°.
(v) It is an straight angle because it is 180°°.

Page No 189:

Question 3:

Draw the perpendicular bisector of a given line segment AB of length 6 cm.

ANSWER:

Steps for construction:

1. Draw a line segment AB, which is equal to 6 cm.
2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.
3. With B as the centre and radius same as before, draw arcs, cutting the perviously drawn arcs at M and N, respectively.
4. Draw MN meeting AB at D.
MN is the required perpendicular bisector of AB.

Page No 189:

Question 4:

Construct an angle of 120° and bisect it.

ANSWER:

Steps of construction:

1. Draw a ray QP.
2.With Q as the centre and any convenient radius, draw an arc cutting QP at N.
3.With N as the centre and the same radius, cut the arc at A. Again, with A as the centre and the same radius, cut the arc at M.
4. Draw QM and produce it to R.
∠PQR is 120°.∠PQR is 120°.
5. With M as the centre and radius more than half of MN, draw an arc.
6. With N as the centre and the same radius mentioned in step(5), draw another arc, cutting the previously drawn arc at point X.
7. Draw QX and produce it to point S.
Ray QS is a bisector of ∠PQR.Ray QS is a bisector of ∠PQR.

Page No 189:

Question 5:

Construct an angle of 90° and bisect it.

ANSWER:

Construction steps:

1. Draw a line OA.
2. Take a point B on OA. With B as the centre and any convenient radius, draw an arc cutting OA at M and N.
3. With N as the centre and radius more than half of MN, draw an arc.
4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at W.
5. Draw WB, meeting the arc at S. Produce it to C.
∠ABC is the required angle of 90°∠ABC is the required angle of 90°.
6. With S as the centre and radius more than half of SN, draw an arc.
7. With N as centre and the same radius as in step (6), draw another arc, cutting the previously drawn arc at point X.
8. Draw BX and produce it to point D.
Ray BD is the angle bisctor of ∠ABC.Ray BD is the angle bisctor of ∠ABC. Ray BD is the angle bisctor of ∠ABC.Ray BD is the angle bisctor of ∠ABC.

Page No 189:

Question 6:

Draw a rectangle whose two adjacent sides are 5.4 cm and 3.5 cm.

ANSWER:

Steps of construction:

1. Draw a ray AX.
2. With A as the centre, cut the ray XA at B, such that AB is equal to 3.5 cm.
3. With B as the centre and with any convenient radius, draw an arc cutting AX at M and N.
4. With N as the centre and with radius more than half of MN, draw an arc.
5. With M as the centre and with the radius same as before, draw another arc to cut the previous arc at Y.
6. Draw BY and produced it to W.
7. With B as the centre and a radius of 5.4 cm, cut ray BW at point C.
8. With C as the centre and a radius 3.5 cm, draw an arc on the right side of BC.
9. With A as the centre and a radius 5.4 cm, draw an arc cutting the previous arc at D.
10. Join CD and AD.
ABCD is the required rectangle.

Page No 189:

Question 7:

Which of the following has no end points?
(a) A line segment
(b) A ray
(c) A line
(d) none of these

ANSWER:

(c) A line
 A line has no end points. We can produce it infinitely in both directions.

Page No 189:

Question 8:

Which of the following has one end point?
(a) A line
(b) A ray
(c) A line segment
(d) none of these

ANSWER:

(b) A ray
 A ray has one end point. We can produce a ray infinitely in one direction.

Page No 189:

Question 9:

Which of the following has two end points?
(a) A line segment
(b) A ray
(c) A line
(d) none of these

ANSWER:

(a) A line segment
 A line segment has two end points and both of them are fixed. Thus, a line segment has a fixed length.

Page No 189:

Question 10:

Two planes intersect
(a) at a point
(b) in a line
(c) in a plane
(d) none of these

ANSWER:

(b) in a line
When the common points of two planes intersect, they form a line.

Page No 189:

Question 11:

3232 right angles = …….
(a) 115°
(b) 135°
(c) 230°
(d) 270°

ANSWER:

(b) 135°°

32 right angles = 32 × 90° = 135°32 right angles = 32 × 90° = 135°

Page No 189:

Question 12:

Where does the vertex of an angle lie?
(a) in its interior
(b) in its exterior
(c) on the angle
(d) none of these

ANSWER:

(c) on the angle

Page No 189:

Question 13:

An angle measuring 270° is
(a) an obtuse angle
(b) an acute angle
(c) a straight line
(d) a reflex angle

ANSWER:

(d) a reflex angle
This is because it is more than 180°° and less than 360°°.

Page No 189:

Question 14:

Fill in the blanks.
(i) A line has …… end point.
(ii) A ray has …… end point
(iii) A line …… be drawn on a paper.
(iv) 0° …… acute angle …… 90° < obtuse angle < 180°.

ANSWER:

(i) A line has no end point.
(ii) A ray has one end point
(iii) A line cannot be drawn on a paper.
(iv) 0° < acute angle < 90° < obtuse angle < 180°.
(v) The standard unit of measuring an angle is degree(°).

Page No 189:

Question 15:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) If two line segments do not intersect, they are parallel.
(ii) If two rays do not intersect, they are parallel.
(iii) If two lines do not meet even when produced, they are called parallel lines.
(iv) Two parallel lines are everywhere the same distance apart.
(v) A ray has a finite length.
(vi) Ray AB−→−AB→ is the same as ray BA−→−BA→.

ANSWER:

(i) False
     If two line segments do not intersect, they may or may not be parallel.

(ii)False
If two rays do not intersect, they may or may not be parallel.

(iii) True

(iv)True

(v)False
    We can produce a ray in one direction.

(vi)False
     AB−→−AB→ means A is fixed and B is not fixed. In other words, we can produce AB towards B.
     BA−→−BA→ means B is fixed and A is not fixed. In other words, we can produce B towards A.

Page No 176:

Exercise 13A

Question 1:

Name three examples of angles from your daily life.

ANSWER:

Three examples from our daily life are as follows:

1)  Angle formed at the vertex of our elbow with the upper arm and the lower arm as the two rays. This angle will vary as per the position of our arm.

2)  Angle formed between the two hands of the clock that are hinged at a point.

3)  Angle formed between the two hands of a windmill. They are also hinged at a point, which is called the vertex of that angle.

Page No 176:

Question 2:

Name the vertex and the arms of ∠ABC, given in the figure below.
Figure

ANSWER:

The vertex is B.

Arms of ∠ABC are rays BA−→− and BC−→−∠ABC are rays BA→ and BC→.

Page No 176:

Question 3:

How many angles are formed in each of the figures given below? Name them.
Figure

ANSWER:

(i) Here, three angles are formed. They are ∠ABC, ∠ACB and ∠BAC.∠ABC, ∠ACB and ∠BAC.
(ii) Here, four angles are formed. They are ∠ABC, ∠BCD, ∠CDA and ∠DAB∠ABC, ∠BCD, ∠CDA and ∠DAB.
(iii) Here, eight angles are formed. They are ∠ABC, ∠BCD,∠CDA,∠DAB, ∠ABD, ∠ADB, ∠CDB, ∠CBD∠ABC, ∠BCD,∠CDA,∠DAB, ∠ABD, ∠ADB, ∠CDB, ∠CBD.

Page No 176:

Question 4:

In the given figure, list the points which
(i) are in the interior of ∠AOB
(ii) are in the interior of ∠AOB
(iii) lie on ∠AOB
Figure

ANSWER:

(i) Q and S are in the interior of ∠∠AOB.

(ii) P and R are in the exterior of ∠∠AOB.

(iii) A, O, B, N and T lie on the angle ∠∠AOB.

Page No 176:

Question 5:

See the adjacent figure and state which of the following statements are true and which are false.
(i) Point C is in the interior of ∠AOC.
(ii) Point C is in the interior of ∠AOD.
(iii) Point D is in the interior of ∠AOC.
(iv) Point B is in the interior of ∠AOD.
(v) Point C lies on ∠AOB.
Figure

ANSWER:

(i)False
Point C is on the angle ∠∠AOC.

(ii)True
Point C lies in the interior of ∠∠AOD.

(iii) False
Point D lies in the exterior of ∠∠AOC.

(iv) True
Point B lies in the exterior of ∠∠AOD.

(v) False
Point C lies in the interior of ∠∠AOB.

Page No 177:

Question 6:

In the adjoining figure, write another name for:
(i) ∠1
(ii) ∠2
(iii) ∠3
Figure

ANSWER:

(i) ∠∠EPB
(ii) ∠∠PQC
(iii) ∠∠FQD

Page No 179:

Exercise 13B

Question 1:

State the type of each of the following angles:
Figure

ANSWER:

(i) ∠∠AOB is an obtuse angle since its measure is more than 90°°.

(ii) ∠∠COD is a right angle since its measure is 90°°.

(iii) ∠∠FOE is a straight angle since its measure is 180°°.

(iv) ∠∠POQ is a reflex angle since its measure is more than 180°° but less than 360°°.

(v)   ∠∠HOG is an acute angle since its measure is more than 0 but less than 90°°.

(vi)  ∠∠POP is a complete angle since its measure is 360°°.

Page No 179:

Question 2:

Classify the angles whose magnitudes are given below:
(i) 30°
(ii) 91°
(iii) 179°
(iv) 90°
(v) 181°
(vi) 360°
(vii) 128°
(viii) (90.5)°
(ix) (38.3)°
(x) 80°
(xi) 0°
(xii) 15°

ANSWER:

(i) Acute angle
This is because its measure is less than 90°° but more than 0°°.
(ii) Obtuse angle
This is because its measure is more than 90°° but less than 180°°
(iii) Obtuse angle
This is because its measure is more than 90°° but less than 180°°.
(iv)Right angle
This is because its measure is 90°°.
(v) Reflex angle
This is because its measure is more than 180°° but less than 360°°.
(vi) Complete angle
This is because its measure is 360°°.
(vii)  Obtuse angle
This is because its measure is more than 90°° but less than 180°°.
(viii) Obtuse angle
This is because its measure is more than 90°° but less than 180°°.
(ix) Acute angle
This is because its measure is more than 0°° but less than 90°°.
(x)  Acute angle
This is because its measure is more than 0°° but less than 90°°.
(xi)  Zero angle
This is because its measure is zero.
(xii)  Acute angle
This is because its measure is more than 0°° but less than 90°°.

Page No 179:

Question 3:

How many degrees are there in
(i) one right angle?
(ii) two right angles?
(iii) three right angles?
(iv) four right angles?
(v) 2323 right angle?
(vi) 112112 right angles?

ANSWER:

(i) One right angle has 90°°.
(ii)  Two right angles have 90°° + 90°° = 180°°.
(iii) Three right angles have 90°° + 90°° + 90°° = 270°°.
(iv)  Four right angles have 90°° + 90°° + 90°° + 90°° = 360°°.
(v) 23×90=60°23×90=60°
(vi) (1+12)right angles =32×90=135°1+12right angles =32×90=135°

Page No 179:

Question 4:

How many degrees are there in the angle between the hour hand and the minute hand of a clock, when it is
(i) 3 o’clock?
(ii) 6 o’clock?
(iii) 12 o’clock?
(iv) 9 o’clock?

ANSWER:

(i) At 3 o’clock the angle formed between the hour hand and the minute hand is right angle, i.e. 90°°.
(ii) At 6 o’clock the angle formed between the hour hand and the minute hand is a straight angle, i.e. 180°°.
(iii) At 12 o’clock the angle formed between the hour hand and the minute hand is a complete angle, i.e. 0°°.
       This is because the hour hand and minute hand coincides to each other at 12 o’clock.
(iv) At 9 o’clock the angle formed between the hour hand and the minute hand is a right angle, i.e. 90°°.

Page No 179:

Question 5:

Using only a ruler, draw an acute angle, an obtuse angle and a straight angle.

ANSWER:

(i) Acute angle

(ii) Obtuse angle
 
(iii) Straight angle

Page No 182:

Exercise 13C

Question 1:

Measure each of the following angles with the help of a protractor and write the measure in degrees:
Figure

ANSWER:

(i) ∠AOB = 45°∠AOB = 45°
(ii) ∠PQR = 75°∠PQR = 75°
(iii)  ∠DEF = 135°∠DEF = 135°
(iv)  ∠LMN = 55°∠LMN = 55°
(v) ∠TSR = 135°∠TSR = 135°
(vi) ∠GHI = 75°∠GHI = 75°

We have measured all the above angles by placing the protractor on one of the arms of the angle and measuring the angle through the other arm that coincides with the angle on the protractor.

Page No 182:

Question 2:

Construct each of the following angles with the help of a protractor:
(i) 25°
(ii) 72°
(iii) 90°
(iv) 117°
(v) 165°
(vi) 23°
(vii) 180°
(viii) 48°

ANSWER:

Steps to follow:

1. Draw a ray QP with Q as the initial point.
2. Place the protractor on QP. With its centre on Q, mark a point R against the given angle mark of the protractor.
3. Join RQ. Now, PQR is the required angle.

(i)  

(ii) 

(iii) 

(iv) 

(v) 

(vi)

(vii) 

(viii) 

Page No 182:

Question 3:

Measure ∠ABC given in the adjoining figure and construct an angle DEF equal to ∠ABC.

ANSWER:

We can see that ∠ABC = 47°∠ABC = 47°.
Steps to follow to construct angle ∠∠DEF equal to ∠∠ABC:

• Draw a ray EF with E as the initial point.
• Place the protractor on EF. With its centre at E, mark a point D against the angle 47°° of the protractor.
• Join DE.  ∠∠DEF = 47°° = ∠∠ABC is the required angle.

Page No 182:

Question 4:

Draw a line segment AB = 6 cm. Take a point C on AB such that AC = 4 cm. From C, draw CD ⊥ AB.

ANSWER:

1. Draw a line segment AB of length 6 cm.
2. Mark point C on AB such that AC is equal to 4 cm.
3. Place the protractor on AB such that the centre of the protractor is on C and its base lies along AB.
4. Holding the protractor, mark a point D on the paper against the 90°° mark of the protractor.
5. Remove the protractor and draw a ray CD with C as the initial point.

Now, CD ⊥⊥ AB

Page No 182:

Exercise 13C

Question 1:

Where does the vertex of an angle lie?
(a) In its interior
(b) In its exterior
(c) On the angle
(d) None of these

ANSWER:

(c) On the angle
Vertex is the initial point of two rays between which the angle is formed. Therefore, it lies on the angle.

Page No 182:

Question 2:

The figure formed by two rays with the same initial point is called
(a) a ray
(b) a line
(c) an angle
(d) none of these

ANSWER:

(c) an angle
The initial point is called the vertex.

Page No 182:

Question 3:

An angle measuring 180° is called
(a) a complete angle
(b) a reflex angle
(c) a straight angle
(d) none of these

ANSWER:

(c) straight angle

Page No 182:

Question 4:

An angle measuring 90° is called
(a) a straight angle
(b) a right angle
(c) a complete angle
(d) a reflex angle

ANSWER:

(b) right angle

Page No 182:

Question 5:

An angle measuring 91° is
(a) an acute angle
(b) an obtuse angle
(c) a reflex angle
(d) none of these

ANSWER:

(b) an obtuse angle
This is because it is more than 90°° but less than 180°°.

Page No 182:

Question 6:

An angle measuring 270° is
(a) an obtuse angle
(b) an  acute angle
(c) a straight angle
(d) a reflex angle

ANSWER:

(d) a reflex angle
This is because it is more than 180°° but less than 360°°.

Page No 182:

Question 7:

The measure of a straight angle is
(a) 90°
(b) 150°
(c) 180°
(d) 360°

ANSWER:

(c) 180°°

Page No 183:

Question 8:

An angle measuring 200° is
(a) an obtuse angle
(b) an acute angle
(c) a reflex angle
(d) none of these

ANSWER:

(c) a reflex angle
This is because it is more than 180°° but less than 360°°.

Page No 183:

Question 9:

An angle measuring 360° is
(a) a reflex angle
(b) an obtuse angle
(c) a straight angle
(d) a complete angle

ANSWER:

(d) a complete angle
This is because it completes the rotation of 360°°.

Page No 183:

Question 10:

A reflex angle measures
(a) more than 180° but less than 270°
(b) more than 180° but less than 360°
(c) more than 90° but less than 180°
(d) none of these

ANSWER:

(b) more than 180° but less than 360°° but less than 360°

Page No 183:

Question 11:

2 right angles = ?
(a) 90°
(b) 180°
(c) 270°
(d) 360°

ANSWER:

(b)
2 right angles =  2×90°= 180°2×90°= 180° (straight angle)

Page No 183:

Question 12:

3232 right angles = ?
(a) 115°
(b) 135°
(c) 270°
(d) 230°

ANSWER:

(b) 135°°

32 right angle =32× 90°=135°32 right angle =32× 90°=135°

Page No 183:

Question 13:

If there are 36 spokes in a bicycle wheel, then the angle between a pair of adjacent spokes is
(a) 15°
(b) 12°
(c) 10°
(d) 18°

ANSWER:

( c)  10°°

Number of spokes = 36
Measure of the angle of the wheel = Complete angle = 360°°
Angle between a pair of adjacent spokes=Measure of angleNumber of spokes=360°36=10°Measure of angleNumber of spokes=360°36=10°

Page No 172:

Exercise 12A

Question 1:

In the figure of a table given below, name the pairs of parallel edges of the top.
Figure

ANSWER:

Following are the parallel edges of the top:

AD∥BCThis is because AD and BC will not intersect even if both these line segments are produced indefinitely in both the directions.AB∥DCThis is because AB and DC will not intersect even if both these line segments are produced indefinitely in both the directions.AD∥BCThis is because AD and BC will not intersect even if both these line segments are produced indefinitely in both the directions.AB∥DCThis is because AB and DC will not intersect even if both these line segments are produced indefinitely in both the directions.

Page No 172:

Question 2:

Name the groups of all possible parallel edges of the box whose figure is shown below.
Figure

ANSWER:

The groups of parallel edges are (AD∥GH∥BC∥FE), (AB∥DC∥GF∥HE) and (AH∥BE∥CF∥DG).The above mentioned groups of edges are parallel because they will not meet each other if produced infinitely to both sides.The groups of parallel edges are (AD∥GH∥BC∥FE), (AB∥DC∥GF∥HE) and (AH∥BE∥CF∥DG).The above mentioned groups of edges are parallel because they will not meet each other if produced infinitely to both sides.

Page No 173:

Question 3:

Identify parallel line segments in each of the figures given below:
(i) Figure
(ii) Figure
(iii) Figure
(iv) Figure
(v) Figure

ANSWER:

(i)
DE∥BCThis is because they do not intersect each other.DE∥BCThis is because they do not intersect each other.
(ii)
AB∥DC and AD∥BC This is because these pairs of line segments do not intersect each other.AB∥DC and AD∥BC This is because these pairs of line segments do not intersect each other.
(iii)
  AB∥DC and AD∥BC This is because these pairs of line segments do not intersect each other.AB does not intersect DC and AD does not intersect BC.AB∥DC and AD∥BC This is because these pairs of line segments do not intersect each other.AB does not intersect DC and AD does not intersect BC.

(iv)
LM∥RQ, RS∥PM and LS∥PQLM∥RQ, RS∥PM and LS∥PQ
These pairs of line segments are non-intersecting.
So, these pairs of lines are parallel.

(v)
AB∥DC, AB∥EF. DC∥EFAC∥BD, CE∥DFAB∥DC, AB∥EF. DC∥EFAC∥BD, CE∥DF
These pairs of line segments are non-intersecting.
So, these pairs of lines are parallel.

Page No 173:

Question 4:

Find the distance between the parallel lines l and m, using a set square.
(i) Figure
(ii) Figure

ANSWER:

(i) Distance between l and m is 1.3 cm.

Place the ruler so that one of its measuring edges lies along the line l. Hold it with one hand. Now place a set square with one arm of the right angle coinciding with the edge of the ruler. Draw the line segment PM along the edge of the set square, as shown in the figure. Then, measure the distance (PM) between l and m, which will be equal to 1.3 cm.


(ii) Distance between l and m is 1 cm.

Place the ruler so that one of its measuring edges lies along the line l. Hold it with one hand. Now place a set square with one arm of the right angle coinciding with the edge of the ruler. Draw the line segment PM along the edge of the set square, as shown in figure. Then, measure the distance (PQ) between l and m as 1 cm.

Page No 173:

Question 5:

In the figure, l||m. If AB ⊥ l, CD ⊥ l and AB = 2.3 cm, find CD.
Figure

ANSWER:

At point A, AB is the perpendicular distance between l and m.
At point C, CD is the perpendicular distance between l and m.
The perpendicular distance between two parallel lines is same at all points.
∴ CD = AB = 2.3 cm

Page No 173:

Question 6:

In the figure, do the segments AB and CD intersect? Are they parallel? Give reasons for your answer.
Figure

ANSWER:

Line segments AB and CD will intersect if they are produced endlessly towards the ends A and C, respectively.

Therefore, they are not parallel to each other.

Page No 173:

Question 7:

Using a set square and a ruler, test whether l || m in each of the following cases:
(i) Figure
(ii) Figure

ANSWER:

(i) Place the ruler so that one of its measuring edges lies along the line l. Hold it firmly with one hand. Now place a set square with one arm of the right angle coinciding with the edge of the ruler. Draw line segments between l and m (say PM, RS, AB) with the set square.

Now, we see that PM = AB = RS.
Thus, we can say that l ∥∥m.



(ii) In this case, we see that when we draw line segments between l and m, they are unequal, i.e. PM≠RSPM≠RS.
Therefore, l is not parallel to m.

Page No 173:

Question 8:

Which of the following statements are true and which are false?
(i) Two lines are parallel if they do not meet, even when produced.
(ii) Two parallel lines are everywhere the same distance apart.
(iii) If two line segments do not intersect, they are parallel.
(iv) If two rays do not intersect, they are parallel.

ANSWER:

(i) True
The statement is true because such lines do not intersect even when produced.

(ii) True
Perpendicular distance between two parallel lines is same at all points on the lines.

(iii) True
If the corresponding lines are produced infinitely, they will not intersect. Hence, they are parallel.

(iv) True
The corresponding lines determined by them will not intersect. Hence, they are parallel to each other.

Exercise 11A

Page No 166:

Question 1:

Name all the line segments in each of the following figures:
(i) Figure
(ii) Figure
(iii) Figure

ANSWER:

(i)  The line segments are
YX ¯¯¯¯¯¯This is because it has two end points Y and X.YZ ¯¯¯¯¯¯ This is because it has two end points Y an Z.YX ¯This is because it has two end points Y and X.YZ ¯ This is because it has two end points Y an Z.

(ii)  
AD¯¯¯¯¯This is because it has two end points A and D.AB¯¯¯¯¯ This is because it has two end points A and B.AC¯¯¯¯¯ This is because it has two end points A and C.AE¯¯¯¯¯ This is because it has two end points A and E.DB¯¯¯¯¯ This is because it has two end points B and D.BC¯¯¯¯¯ This is because it has two end points B and C.CE¯¯¯¯¯ This is because it has two end points C and E.AD¯This is because it has two end points A and D.AB¯ This is because it has two end points A and B.AC¯ This is because it has two end points A and C.AE¯ This is because it has two end points A and E.DB¯ This is because it has two end points B and D.BC¯ This is because it has two end points B and C.CE¯ This is because it has two end points C and E.

(iii)
PS¯¯¯¯¯ This is because it has two end points P and S.PQ¯¯¯¯¯ This is because it has two end points P and Q.QR¯¯¯¯¯ This is because it has two end points Q and R.RS¯¯¯¯¯ This is because it has two end points R and S.PR¯¯¯¯¯ This is because it has two end points P and R.QS¯¯¯¯¯ This is because it has two end points Q and S.PS¯ This is because it has two end points P and S.PQ¯ This is because it has two end points P and Q.QR¯ This is because it has two end points Q and R.RS¯ This is because it has two end points R and S.PR¯ This is because it has two end points P and R.QS¯ This is because it has two end points Q and S.

Page No 166:

Question 2:

Identify and name the line segments and rays in each of the following figures:
(i) Figure
(ii) Figure
(iii) Figure

ANSWER:

(i)  Line segment is AB¯¯¯¯¯. This is because it has two end points A and B.AB¯. This is because it has two end points A and B.
Rays are:
−→AC This is because it has only one end point A.−→BD This is because it has only one end point B.→AC This is because it has only one end point A.→BD This is because it has only one end point B.


(ii) Line segments are:
EP¯¯¯¯¯  This is because it has two end points Eand P.EG¯¯¯¯¯ This is because it has two end points E and G.GP¯¯¯¯¯ This is because it has two end points G and P.EP¯  This is because it has two end points Eand P.EG¯ This is because it has two end points E and G.GP¯ This is because it has two end points G and P.

Rays are:
EF−→− This is because it has only one end point, i.e. E.GH−→− This is because it has only one end point, i.e. G.PQ−→− This is because it has only one end point, i.e. P.EF→ This is because it has only one end point, i.e. E.GH→ This is because it has only one end point, i.e. G.PQ→ This is because it has only one end point, i.e. P.

(iii) Line segments are:
OL¯¯¯¯¯ This is because it has two end points O and L.OP¯¯¯¯¯ This is because it has two end points O and P.OL¯ This is because it has two end points O and L.OP¯ This is because it has two end points O and P.

Rays are:
LM−→− This is because it has only one end point, i.e. L.PQ−→−  This is because it has only one end point, i.e. P.LM→ This is because it has only one end point, i.e. L.PQ→  This is because it has only one end point, i.e. P.

Page No 167:

Question 3:

In the adjoining figure, name
(i) four line segments;
(ii) four rays;
(iii) two non-intersecting line segments.
Figure

ANSWER:

(i)
PR¯¯¯¯¯ This is because it has two end points P and R.QS¯¯¯¯¯ This is because it has two end points Q and S.PQ¯¯¯¯¯ This is because it has two end points P and Q.RS¯¯¯¯¯ This is because it has two end points R and S.PR¯ This is because it has two end points P and R.QS¯ This is because it has two end points Q and S.PQ¯ This is because it has two end points P and Q.RS¯ This is because it has two end points R and S.


(ii)
PA−→− This is because it has only one end point, i.e. P.RB−→− This is because it has only one end point, i.e. R.QC−→− This is because it has only one end point, i.e. Q.SD−→− This is because it has only one end point, i.e. S.PA→ This is because it has only one end point, i.e. P.RB→ This is because it has only one end point, i.e. R.QC→ This is because it has only one end point, i.e. Q.SD→ This is because it has only one end point, i.e. S.

(iii)
PR ¯¯¯¯¯¯and QS¯¯¯¯¯ are the two non−intersecting line segments as they do not have any point in common.PR ¯and QS¯ are the two non-intersecting line segments as they do not have any point in common.

Page No 167:

Question 4:

What do you mean by collinear points?
(i) How many lines can you draw passing through three collinear points?
(ii) Given three collinear points A, B, C. How many line segments do they determine? Name them.
Figure

ANSWER:

 COLLINEAR POINTS :
 Three or more points in a plane are said to be collinear if they all lie in the same line. This line is called the line of collinearity for the given points.

(i) We can draw only one line passing through three collinear points.

(ii) 3 Line segments are:
AB¯¯¯¯¯ This is because it has two end points A and B.BC¯¯¯¯¯ This is because it has two end points B and C.AC¯¯¯¯¯ This is because it has two end points A and C.AB¯ This is because it has two end points A and B.BC¯ This is because it has two end points B and C.AC¯ This is because it has two end points A and C.

Page No 167:

Question 5:

In the adjoining figure, name:
(i) four pairs of intersecting lines
(ii) four collinear points
(iii) three noncollinear points
(iv) three concurrent lines
(v) three lines whose point of intersection is P
Figure

ANSWER:

(i)
PS←→ and AB ←→− intersecting at S.CD←→and RS←→ intersecting at R.PS←→ and CD←→ intersecting at P.AB←→ and RS←→ intersecting at S.PS↔ and AB ↔ intersecting at S.CD↔and RS↔ intersecting at R.PS↔ and CD↔ intersecting at P.AB↔ and RS↔ intersecting at S.

(ii) A, Q, S and B are four collinear points as they all lie on the same line AB ←→−AB ↔.

(iii) A, C and B are non-collinear points as they do not lie on the same line. 

(iv)

PS←→ , RS←→ and AB←→ are three concurrent lines passing through the same point SPS↔ , RS↔ and AB↔ are three concurrent lines passing through the same point S.

(v)

PS←→ , PQ←→ and CD←→ have common point of intersection PPS↔ , PQ↔ and CD↔ have common point of intersection P.

Page No 167:

Question 6:

Mark three noncollinear points A, B, C, as shown. Draw lines through these points taking two at a time. Name the lines. How many such different lines can be drawn?
Figure

ANSWER:

Taking points A and B, we can draw only one line AB ←→−AB ↔.
Taking points B and C, we can draw only one line BC←→ BC↔ .
Taking points A and C, we can draw only one line AC ←→−AC ↔.
We can draw only three lines through these non-collinear points A ,B and C.

Page No 167:

Question 7:

Count the number of line segments drawn in each of the following figures and name them.
(i) Figure
(ii) Figure
(iii) Figure
(iv) Figure

ANSWER:

(i) There are 6 line segments. These are:
AB¯¯¯¯¯ (with end points A and B)AC¯¯¯¯¯ (with end points A and C)AD¯¯¯¯¯ (with end points A and D)BC¯¯¯¯¯ (with end points B and C)BD¯¯¯¯¯ (with end points B and D)CD¯¯¯¯¯ (with end points C and D)AB¯ (with end points A and B)AC¯ (with end points A and C)AD¯ (with end points A and D)BC¯ (with end points B and C)BD¯ (with end points B and D)CD¯ (with end points C and D)

(ii) There are 10 line segments. These are:
AB¯¯¯¯¯  (with end points A and B)BC ¯¯¯¯¯¯ (with end points B and C)CD¯¯¯¯¯  (with end points C and D)AD¯¯¯¯¯  (with end points A and D)AC¯¯¯¯¯  (with end points A anc C)BD ¯¯¯¯¯¯ (with end points B and D)AO ¯¯¯¯¯¯ (with end points A and O)CO¯¯¯¯¯  (with end points C and O)BO¯¯¯¯¯  (with end points B and O)DO¯¯¯¯¯  (with end points D and O)AB¯  (with end points A and B)BC ¯ (with end points B and C)CD¯  (with end points C and D)AD¯  (with end points A and D)AC¯  (with end points A anc C)BD ¯ (with end points B and D)AO ¯ (with end points A and O)CO¯  (with end points C and O)BO¯  (with end points B and O)DO¯  (with end points D and O)

(iii) There are 6 line segments. They are:
AB¯¯¯¯¯, AF¯¯¯¯¯, FB¯¯¯¯¯, EC¯¯¯¯¯, ED¯¯¯¯¯, DC¯¯¯¯¯AB¯, AF¯, FB¯, EC¯, ED¯, DC¯

(iv) There are 12 line segments. They are:
AB¯¯¯¯¯, AD¯¯¯¯¯, AE¯¯¯¯¯BC¯¯¯¯¯, BF¯¯¯¯¯ CG¯¯¯¯¯, CD¯¯¯¯¯HG¯¯¯¯¯¯, HE¯¯¯¯¯¯ , DH¯¯¯¯¯¯EF¯¯¯¯¯, GF¯¯¯¯¯AB¯, AD¯, AE¯BC¯, BF¯ CG¯, CD¯HG¯, HE¯ , DH¯EF¯, GF¯

Page No 167:

Question 8:

Consider the line PQ←→PQ↔ given below and find whether the given statements are true or false:
(i) M Is a point on ray NQ−→−NQ→.
(ii) L is a point on ray MP−→−MP→.
(iii) Ray MQ−→−MQ→ is different from ray NQ−→−NQ→.
(iv) L, M, N are points on line segment LN¯¯¯¯¯LN¯.
(v) Ray LP−→LP→ is different from ray LQ−→−LQ→.
Figure

ANSWER:

(i) False
M is outside ray NQ.

(ii) True
L is placed between M and P.

(iii) True
Ray MQ is extended endlessly from M to Q and ray NQ is extended endlessly from N to Q.

(iv) True

(v) True
  LP −→−is extended endlessly from L to P.LQ−→− is extended endlessly from L to Q.LP →is extended endlessly from L to P.LQ→ is extended endlessly from L to Q.

Page No 168:

Question 9:

Write ‘T’ for true and ‘F’ for false in case of each of the following statements:
(i) Every point has a size.
(ii) A line segment has no length.
(iii) Every ray has a finite length.
(iv) The ray AB−→−AB→ is the same as the ray BA−→−BA→.
(v) The line segment AB¯¯¯¯¯AB¯ is the same as the line segment BA¯¯¯¯¯BA¯.
(vi) The line AB←→AB↔ is the same as the line BA←→BA↔.
(vii) Two points A and B in a plane determine a unique line segment.
(viii) Two intersecting lines intersect at a point.
(ix) Two intersecting planes intersect at a point.
(x) If points A, B, C are collinear and points C, D, E are collineaer then the pints A,B, C, D, E are collinear.
(xi) One and only one ray can be drawn with a given end point.
(xii) One and only one line can be drawn to pass through two given points.
(xiii) An unlimited number of lines can be drawn to pass through a given point.

ANSWER:

(i)  False
A point does not have any length, breadth or thickness.

(ii)   False
A line segment has a definite length.

(iii) False
A ray has no definite length.

(iv) False
Ray AB has initial point A and is extended endlessly towards B, while ray BA has initial point B and is extended endlessly towards A.

(v) True
This is because both the line segments have definite length with end points A and B.

(vi)  True
This is because it neither has a definite length nor any end point.

(vii) True
Only one line segment can pass through the two given points.

(viii) True

(ix) False
Two intersecting planes intersect at a line.

(x) False
Different set of collinear points need not be collinear.


(xi) False
    With point P, endless rays (like PA, PB, PC, PD, PE, PF) can be drawn.


(xii) True
Two points define one unique line.
(xiii) True

Page No 168:

Question 10:

Fill in the blanks:
(i) A line segment has a ………….. length.
(ii) A ray has ………….. end point.
(iii) A line has ………….. end point.
(iv) A ray has no ………….. length.
(v) A line ………….. be drawn on a paper.

ANSWER:

 (i) definite
(ii) one
(iii) no
(iv) definite
(v) cannot

Page No 168:

Exercise 11B

Question 1:

Which of the following has no end points?
(a) A line segment
(b) A ray
(c) A line
(d) None of these

ANSWER:

(c) A line does not have any end point. It is a line segment that is extended endlessly on both sides.

Page No 168:

Question 2:

Which of the following has one end point?
(a) A line
(b) A ray
(c) A line segment
(d) None of these

ANSWER:

(b) A ray has one end point, which is called the initial point. It is extended endlessly towards the other direction.

Page No 168:

Question 3:

Which of the following has two end points?
(a) A line segment
(b) A ray
(c) A line
(d) None of these

ANSWER:

(a) A line segment has two end points and a definite length that can be measured.

Page No 168:

Question 4:

Which of the following has definite length?
(a) A line
(b) A line segment
(c) A ray
(d) None of these

ANSWER:

(b) A line segment has a definite length that can be measured by a ruler and, therefore, it can be drawn on a paper.

Page No 168:

Question 5:

Which of the following can be drawn on a piece of paper?
(a) A line
(b) A line segment
(c) A ray
(d) A plane

ANSWER:

(b) A line segment has a definite length that can be measured by a ruler. So, it can be drawn on a paper.

Page No 168:

Question 6:

How many lines can be drawn passing through a given point?
(a) One only
(b) Two
(c) Three
(d) Unlimited number

ANSWER:

(d) Unlimited number of lines can be drawn.

Page No 168:

Question 7:

How many lines can be drawn passing through two given point?
(a) One only
(b) Two
(c) Three
(d) Unlimited number

ANSWER:

(a) Only one line can be drawn that passes through two given points.

Page No 168:

Question 8:

Two planes intersect
(a) at a point
(b) in a plane
(c) in a line
(d) none of these

ANSWER:

(c)  Two intersecting planes intersect in a line.

Page No 168:

Question 9:

Two lines intersect
(a) at a point
(b) at two points
(c) at an infinite number of points
(d) in a line

ANSWER:

(a) Two lines intersect at a point.

Page No 169:

Question 10:

Two points in a plane determine
(a) exactly one line segment
(b) exactly two line segments
(c) an infinite number of line segments
(d) none of these

ANSWER:

(a) exactly one line segment

Two points in a plane determine exactly one line segment with those two points as its end points.

Page No 169:

Question 11:

The minimum number of points of intersection of three lines in a plane is
(a) 1
(b) 2
(c) 3
(d) 0

ANSWER:

(d) 0
Three lines will not necessarily intersect in a plane. Thus, the minimum point of intersection will be 0.

Page No 169:

Question 12:

The maximum number of points of intersection of three lines in a plane is
(a) 0
(b) 1
(c) 2
(d) 3

ANSWER:

(d) 3

The maximum number of points of intersection of three lines that intersect in a plane are three.

Page No 169:

Question 13:

Choose the correct statement:
(a) every line has a definite length
(b) every ray has a definite length
(c) every line segment has a definite length
(d) none of these

ANSWER:

(c) Every line segment has a definite length.

Every line segment has a definite length, which can be measured using a ruler.

Page No 169:

Question 14:

Choose the false statement:
(a) Line AB←→AB↔ is the same as line BA←→BA↔
(b) Ray AB−→−AB→ is the same as ray BA−→−BA→
(c) Line segment AB¯¯¯¯¯AB¯ is the same as teh line segment BA¯¯¯¯¯BA¯
(d) None of these

ANSWER:

(b) Ray AB−→− is same as ray BA−→− AB→ is same as ray BA→ 
This is because the initial points in these rays are A and B, respectively, and are extended endlessly towards B and A, respectively.

Page No 169:

Question 15:

How many rays can be drawn with a given point as the initial point?
(a) One
(b) Two
(c) An unlimited number
(d) A limited number only

ANSWER:

(c) An unlimited number of rays can be drawn with a given point as the initial point. For example:

Exercise 9A

Page No 139:

Question 1:

Write each of the following statements as an equation:
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice  a number is 16.
(vi) If 12 is subtracted from a number, the result is 24.
(vii) Twice a number subtracted from 19 is 11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the number.

ANSWER:

(i) Let the required number be x.
  So, five times the number will be 5x.
  ∴ 5x = 40

(ii) Let the required number be x.
     So, when it is increased by 8, we get x + 8.
     ∴ x + 8 = 15

(iii) Let the required number be x.
     So, when 25 exceeds the number, we get 25 −- x.
 ∴ 25 −- x  = 7

(iv) Let the required number be x.
    So, when the number exceeds 5, we get x −- 5.
   ∴ x −- 5  = 3

(v) Let the required number be x.
     So, thrice the number will be 3x.
  ∴ 3x −- 5 = 16

(vi) Let the required number be x.
     So, 12 subtracted from the number will be x −- 12.
   ∴ x −- 12 = 24

(vii) Let the required number be x.
     So, twice the number will be 2x.
   ∴ 19 −- 2x = 11

(viii) Let the required number be x.
     So, the number when divided by 8 will be x8x8.
   ∴ x8x8 = 7

(ix) Let the required number be x.
     So, four times the number will be 4x.
 ∴ 4x −- 3 = 17

(x) Let the required number be x.
     So, 6 times the number will be 6x.
    ∴ 6x = x + 5

Page No 140:

Question 2:

Write a statement for each of the equations, given below:
(i) x − 7 = 14
(ii) 2y = 18
(iii) 11 + 3x = 17
(iv) 2x − 3 = 13
(v) 12y − 30 = 6
(vi) 2z3=82z3=8

ANSWER:

(i) 7 less than the number x equals 14.
(ii) Twice the number y equals 18.
(iii) 11 more than thrice the number x equals 17.
(iv) 3 less than twice the number x equals 13.
(v) 30 less than 12 times the number y equals 6.
(vi) When twice the number z is divided by 3, it equals 8.

Page No 140:

Question 3:

Verify by substitution that
(i) the root of 3x − 5 = 7 is x = 4
(ii) the root of 3 + 2x = 9 is x = 3
(iii) the root of 5x − 8 = 2x − 2 is x = 2
(iv) the root of 8 − 7y = 1 is y = 1
(v) the root of z7=8z7=8 is z = 56

ANSWER:

(i)

   3x − 5 = 7Substituting x = 4  in the given equation:L.H.S. : 3×4 −5or, 12 − 5 = 7 = R.H.S.L.H.S. = R.H.S. Hence,  x = 4  is the root of the given equation. 3x – 5 = 7Substituting x = 4  in the given equation:L.H.S. : 3×4 -5or, 12 – 5 = 7 = R.H.S.L.H.S. = R.H.S. Hence,  x = 4  is the root of the given equation. 

(ii)

    3 + 2x= 9Substituting x = 3 in the given equation:L.H.S. : 3 + 2×3or,  3 + 6 = 9 = R.H.S. L.H.S. = R.H.S. Hence,  x = 3  is the root of the given equation. 3 + 2x= 9Substituting x = 3 in the given equation:L.H.S. : 3 + 2×3or,  3 + 6 = 9 = R.H.S. L.H.S. = R.H.S. Hence,  x = 3  is the root of the given equation. 

(iii)

    5x − 8 = 2x −2 Substituting x = 2 in the given equation:L.H.S. :                                   R.H.S. :5×2− 8                           = 2×2−2or, 10 − 8 = 2                      = 4 −2 = 2     L.H.S. = R.H.S. Hence, x =2  is the root of the given equation. 5x – 8 = 2x -2 Substituting x = 2 in the given equation:L.H.S. :                                   R.H.S. :5×2- 8                           = 2×2-2or, 10 – 8 = 2                      = 4 -2 = 2     L.H.S. = R.H.S. Hence, x =2  is the root of the given equation. 

(iv)

     8 − 7y = 1 Substituting y = 1  in the given equation:L.H.S. : 8 −7×1or, 8 − 7 = 1 = R.H.S. L.H.S. = R.H.S. Hence, y =1  is the root of the given equation. 8 – 7y = 1 Substituting y = 1  in the given equation:L.H.S. : 8 -7×1or, 8 – 7 = 1 = R.H.S. L.H.S. = R.H.S. Hence, y =1  is the root of the given equation. 

(v)

     z7 = 8 Substituting z = 56 in the given equation:L.H.S. : 567 = 8 =R.H.S.L.H.S. = R.H.S. Hence, z =56  is the root of the given equation.z7 = 8 Substituting z = 56 in the given equation:L.H.S. : 567 = 8 =R.H.S.L.H.S. = R.H.S. Hence, z =56  is the root of the given equation.

Page No 140:

Question 4:

Solve each of the following equations by the trial-and-error method:
(i) y + 9 = 13
(ii) x − 7 = 10
(iii) 4x = 28
(iv) 3y = 36
(v) 11 + x = 19
(vi) x3=4×3=4
(vii) 2x − 3 = 9
(viii) 12x + 7 = 1112x + 7 = 11
(ix) 2y + 4 = 3y
(x) z − 3 = 2z − 5

ANSWER:

(i) y + 9 = 13
    We try several values of y until we get the  L.H.S. equal to the R.H.S.

y	L.H.S.	R.H.S.	Is LHS =RHS ?
1	1 + 9 = 10	13	No
2	2 + 9 = 11	13	No
3	3 + 9 = 12	13	No
4	4 + 9 = 13	13	Yes

 ∴ y = 4

(ii) x − 7= 10
We try several values of x until we get the  L.H.S. equal to the R.H.S.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
10	10 − 7 = 3	10	No
11	11 − 7 = 4	10	No
12	12 − 7 = 5	10	No
13	13 − 7 = 6	10	No
14	14 − 7 = 7	10	No
15	15 − 7 = 8	10	No
16	16 − 7 = 9	10	No
17	17 − 7 = 10	10	Yes

∴ x = 17

(iii) 4x = 28
     We try several values of x until we get the  L.H.S. equal to the R.H.S.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	4 ×× 1 = 4	28	No
2	4 ×× 2 = 8	28	No
3	4 ×× 3 = 12	28	No
4	4 ×× 4 = 16	28	No
5	4 ×× 5 = 20	28	No
6	4 ×× 6 = 24	28	No
7	4 ×× 7 = 28	28	Yes

  ∴ x = 7

(iv) 3y = 36
    We try several values of x until we get the L.H.S. equal to the R.H.S.

y	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
6	3 ×× 6 = 18	36	No
7	3 ×× 7 = 21	36	No
8	3 ×× 8 = 24	36	No
9	3 ×× 9 = 27	36	No
10	3 ×× 10 = 30	36	No
11	3 ××11 = 33	36	No
12	3 ×× 12 = 36	36	Yes

 ∴ y = 12

(v) 11 + x = 19
     We try several values of x until we get the L.H.S. equal to the R.H.S.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	11 + 1 = 12	19	No
2	11 + 2 = 13	19	No
3	11 + 3 = 14	19	No
4	11 + 4 = 15	19	No
5	11 + 5 = 16	19	No
6	11 + 6 = 17	19	No
7	11 + 7 = 18	19	No
8	11 + 8 = 19	19	Yes

   ∴ x = 8

(vi) x3 = 4×3 = 4
    Since R.H.S. is an natural number so L.H.S. must also be a natural number. Thus, x has to be a multiple of 3.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
3	33=133=1	4	No
6	63=263=2	4	No
9	93=393=3	4	No
12	123=4123=4	4	Yes

  ∴ x = 12

(vii) 2x − 3 = 9

  We try several values of x until we get the L.H.S. equal to the R.H.S.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	2 ×× 1 − 3 = −1	9	No
2	2 ×× 2 − 3 = 1	9	No
3	2 ×× 3 − 3 = 3	9	No
4	2 ×× 4 − 3 = 5	9	No
5	2 ×× 5 − 3 = 7	9	No
6	2 ×× 6 − 3 = 9	9	Yes

  ∴  x = 6

(viii) 12x + 7 = 1112x + 7 = 11
     Since, R.H.S. is a natural number so L.H.S. must be a natural number Thus, we will try values if x which are multiples of ‘x’

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
2	2/2 + 7 = 8	11	No
4	4/2 + 7 = 9	11	No
6	6/2 + 7 = 10	11	No
8	8/2 + 7 = 11	11	Yes

    ∴ x = 8

(ix) 2y + 4 = 3y
      We try several values of y until we get the L.H.S. equal to the R.H.S.

y	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	2 ×× 1 + 4 = 6	3 ×× 1 = 3	No
2	2 ×× 2 + 4 = 8	3 ×× 2 = 6	No
3	2 ×× 3 + 4 = 10	3 ×× 3 = 9	No
4	2 ×× 4 + 4 = 12	3 ×× 4 = 12	Yes

  ∴ y = 4

(x) z − 3 = 2z − 5
 We try several values of z till we get the L.H.S. equal to the R.H.S.

z	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	1 − 3 = −2	2 ×× 1 − 5 = −3	No
2	2 − 3 = −1	2 ×× 2 − 5 = −1	Yes

∴ z = 2

Page No 143:

Exercise 9B

Question 1:

Solve each of the following equations and verify the answer in each case:
x + 5 = 12

ANSWER:

 x + 5 = 12

Subtracting 5 from both the sides:
⇒ x + 5 − 5 = 12 − 5              
⇒ x = 7
Verification:
Substituting x = 7 in the L.H.S.:
⇒ 7 + 5 = 12 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 2:

Solve each of the following equations and verify the answer in each case:
x + 3 = −2

ANSWER:

 x + 3 = −2

Subtracting 3 from both the sides:
⇒ x + 3 − 3 = −2 − 3             
⇒ x = −5

Verification:
Substituting x = −5 in the L.H.S.:
⇒ −5 + 3 =  −2 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 3:

Solve each of the following equations and verify the answer in each case:
x − 7 = 6

ANSWER:

 x − 7 = 6
Adding 7 on both the sides:
⇒ x − 7 + 7 = 6 + 7               
⇒ x = 13

Verification:
Substituting x = 13 in the L.H.S.:
⇒ 13 − 7 =  6 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 4:

Solve each of the following equations and verify the answer in each case:
x − 2 = −5

ANSWER:

 x − 2 = −5

Adding 2 on both sides:
⇒ x − 2 + 2 = −5 + 2            
⇒ x = −3
Verification:
Substituting x = −3 in the L.H.S.:
⇒  −3 − 2 = −5 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 5:

Solve each of the following equations and verify the answer in each case:
3x − 5 = 13

ANSWER:

3x − 5 = 13
⇒ 3x − 5 + 5 = 13 + 5             [Adding 5 on both the sides]
⇒ 3x = 18
⇒ 3×3 = 1833×3 = 183                        [Dividing both the sides by 3]
⇒ x = 6
Verification:
Substituting x = 6 in the L.H.S.:
⇒  3 ×× 6 − 5 = 18 − 5 = 13 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 6:

Solve each of the following equations and verify the answer in each case:
4x + 7 = 15

ANSWER:

4x + 7 = 15
⇒ 4x + 7 − 7 = 15 − 7              [Subtracting 7 from both the sides]
⇒ 4x = 8
⇒ 4×4 = 844×4 = 84                        [Dividing both the sides by 4]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.:
⇒  4××2 + 7 = 8 + 7 = 15 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 7:

Solve each of the following equations and verify the answer in each case:
x5=12×5=12

ANSWER:

x5 = 12×5 = 12
⇒ x5×5  = 12×5×5×5  = 12×5                                  [Multiplying both the sides by 5]
⇒ x = 60
Verification:
Substituting x = 60 in the L.H.S.:
⇒ 605605 = 12 = R.H.S.
⇒ L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 8:

Solve each of the following equations and verify the answer in each case:
3×5=153×5=15

ANSWER:

     3×5 = 153×5 = 15
⇒ 3×5× 5  = 15 × 53×5× 5  = 15 × 5                                  [Multiplying both the sides by 5]
⇒ 3x = 75
⇒ 3×3 = 7533×3 = 753
⇒ x = 25
Verification:
Substituting x = 25 in the L.H.S.:
⇒ 3 × 2553 × 255 = 15 = R.H.S.
⇒ L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 9:

Solve each of the following equations and verify the answer in each case:
5x − 3 = x + 17

ANSWER:

5x − 3 = x + 17
⇒ 5x − x = 17 + 3                  [Transposing x to the L.H.S. and 3 to the R.H.S.]
⇒ 4x = 20
⇒ 4×4 = 2044×4 = 204                        [Dividing both the sides by 4]
⇒ x = 5
Verification:
Substituting x = 5 on both the sides:
L.H.S.:  5(5) − 3
⇒ 25 − 3
⇒ 22

R.H.S.:  5 + 17 = 22
⇒ L.H.S. = R.H.S.
 Hence, verified.

Page No 143:

Question 10:

Solve each of the following equations and verify the answer in each case:
2x−12=32x-12=3

ANSWER:

2x−12 = 32x-12 = 3
⇒ 2x −-1212 + 1212 = 3 + 1212                              [Adding 1212 on both the sides]
⇒ 2x  = 6 + 126 + 12
⇒ 2x = 7272
⇒ 2×2 = 72 × 22×2 = 72 × 2                                            [Dividing both the sides by 3]
⇒ x = 7474
Verification:
Substituting  x = 7474 in the  L.H.S.:
2(74) − 12= 72 − 12 = 62 = 3 = R.H.S.274 – 12= 72 – 12 = 62 = 3 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 11:

Solve each of the following equations and verify the answer in each case:
3(x + 6) = 24

ANSWER:

3(x + 6) = 24
⇒ 3×x + 3×6 = 243×x + 3×6 = 24                     [On expanding the brackets]
⇒  3x + 18 = 24
⇒ 3x + 18 −- 18 = 24 −- 18            [Subtracting 18 from both the sides]
⇒ 3x = 6
⇒ 3×3 = 633×3 = 63                                [Dividing both the sides by 3]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.:
 3(2 + 6) = 3 ××8 = 24  = R.H.S.
 L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 12:

Solve each of the following equations and verify the answer in each case:
6x + 5 = 2x + 17

ANSWER:

6x + 5 = 2x + 17
⇒⇒6x  −- 2x = 17 −- 5                          [Transposing 2x to the L.H.S. and 5 to the R.H.S.]
⇒⇒4x = 12
⇒⇒4×4= 1244×4= 124                                      [Dividing both the sides by 4]
⇒⇒x = 3
Verification:
Substituting x = 3 on both the sides:
L.H.S.: 6(3) + 5
=18 + 5
=23
R.H.S.:   2(3) + 17
= 6 + 17
= 23
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 13:

Solve each of the following equations and verify the answer in each case:
x4−8=1×4-8=1

ANSWER:

x4− 8 = 1×4- 8 = 1
⇒x4− 8 + 8 = 1 + 8⇒x4- 8 + 8 = 1 + 8                  [Adding 8 on both the sides]
⇒x4 = 9⇒x4 = 9
⇒x4 × 4 = 9 × 4⇒x4 × 4 = 9 × 4                        [Multiplying both the sides by 4]
or, x = 36
Verification:
Substituting x = 36 in the L.H.S.:
or, 364 − 8364 – 8 = 9 − 8 = 1 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 14:

Solve each of the following equations and verify the answer in each case:
x2=x3+1×2=x3+1

ANSWER:

x2 = x3 + 1×2 = x3 + 1
⇒x2 − x3 = 1⇒x2 – x3 = 1                                      [Transposing x3x3 to the L.H.S.]
⇒3x − 2×6 = 1⇒3x – 2×6 = 1
⇒x6 = 1⇒x6 = 1
⇒x6 × 6 = 1 × 6⇒x6 × 6 = 1 × 6                                    [Multiplying both the sides by 6]
or, x = 6
Verification:
Substituting x = 6 on both the sides:
L.H.S.: 62 62 = 3
R.H.S.: 63 + 163 + 1 =  2 + 1 =  3
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 15:

Solve each of the following equations and verify the answer in each case:
3(x + 2) − 2(x − 1) = 7

ANSWER:

3(x + 2) − 2(x − 1) = 7
⇒3×x + 3×2 − 2×x −2×(−1) = 7⇒3×x + 3×2 – 2×x -2×(-1) = 7              [On expanding the brackets]
or, 3x + 6 −-2x + 2 = 7
or, x + 8 = 7
or, x + 8 −- 8 = 7 −- 8                                        [Subtracting 8 from both the sides]
or, x = −-1
Verification:
Substituting x = −-1 in the L.H.S.:
3(−1+2) −2(−1−1)or, 3(1) −2(−2)or, 3 + 4 = 7 = R.H.S.3(-1+2) -2(-1-1)or, 3(1) -2(-2)or, 3 + 4 = 7 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 16:

Solve each of the following equations and verify the answer in each case:

ANSWER:

5(x-1) +2(x+3) + 6 = 0
⇒⇒5x -5 +2x +6 +6 = 0        (Expanding within the brackets)
⇒⇒7x +7 = 0
⇒⇒x +1 = 0       (Dividing by 7)
⇒⇒x = -1

Verification:
Putting x = -1 in the L.H.S.:
L.H.S.: 5(-1 -1) + 2(-1 + 3) + 6
          = 5(-2) + 2(2) + 6
          = -10 + 4 + 6  = 0 = R.H.S.

Hence, verified.

Page No 143:

Question 17:

Solve each of the following equations and verify the answer in each case:
6(1 − 4x) + 7(2 + 5x) = 53

ANSWER:

6(1 − 4x) + 7(2 + 5x) = 53
or, 6 × 1 − 6 × 4x + 7 × 2 + 7 × 5x = 536 × 1 – 6 × 4x + 7 × 2 + 7 × 5x = 53              [On expanding the brackets]
or, 6 −- 24x + 14 + 35x = 53
or, 11x + 20 = 53
or, 11x + 20 −- 20 = 53 −- 20                                        [Subtracting 20 from both the sides]
or, 11x = 33
or, 11×11= 331111×11= 3311                                                      [Dividing both the sides by 11]
or, x = 3
Verification:
Substituting x = 3 in the L.H.S.:
6(1 − 4 × 3) + 7(2 + 5 × 3)⇒6(1 − 12) + 7(2 + 15)⇒6(−11) + 7(17)⇒−66 + 119  = 53 = R.H.S.6(1 – 4 × 3) + 7(2 + 5 × 3)⇒6(1 – 12) + 7(2 + 15)⇒6(-11) + 7(17)⇒-66 + 119  = 53 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 18:

Solve each of the following equations and verify the answer in each case:
16(3x − 5) − 10(4x − 8) = 40

ANSWER:

16(3x − 5) − 10(4x − 8) = 40
or, 16 × 3x − 16 × 5 −10 × 4x − 10 × (−8) = 4016 × 3x – 16 × 5 -10 × 4x – 10 × (-8) = 40              [On expanding the brackets]
or, 48x −- 80 −- 40x + 80 = 40
or, 8x  = 40
or, 8×8=4088×8=408                                                      [Dividing both the sides by 8]
or, x = 5
Verification:
Substituting x = 5 in the L.H.S.:

16(3 × 5 − 5) − 10( 4 × 5 − 8)⇒16(15 − 5) − 10(20 − 8)⇒16(10) −10(12)⇒160 − 120  = 40 = R.H.S.16(3 × 5 – 5) – 10( 4 × 5 – 8)⇒16(15 – 5) – 10(20 – 8)⇒16(10) -10(12)⇒160 – 120  = 40 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 19:

Solve each of the following equations and verify the answer in each case:
3(x + 6) + 2(x + 3) = 64

ANSWER:

3(x + 6) + 2(x + 3) = 64
⇒⇒3 × x   +   3 × 6 + 2 × x  + 2 × 3   = 64            [On expanding the brackets]
⇒⇒3x + 18 +  2x + 6 = 64
⇒5x + 24 = 64
⇒5x + 24 −- 24 = 64 −- 24                                       [Subtracting 24 from both the sides]
⇒5x = 40
⇒5×5 = 4055×5 = 405                                                           [Dividing both the sides by 5]
⇒x = 8
Verification:
Substituting x = 8 in the L.H.S.:
3(8 + 6) + 2(8 + 3)3(14) + 2(11)42 + 22 = 64 = R.H.S.3(8 + 6) + 2(8 + 3)3(14) + 2(11)42 + 22 = 64 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 20:

Solve each of the following equations and verify the answer in each case:
3(2 − 5x) − 2(1 − 6x) = 1

ANSWER:

3(2 − 5x) − 2(1 − 6x) = 1
or, 3 × 2  + 3 × (−5x) − 2 × 1 − 2 × (−6x) = 1           [On expanding the brackets]
or, 6 − 15x −  2 + 12x = 1
or, 4 – 3x = 1
or,  3  =3x                                                        
or, x = 1

Verification:
Substituting x = 1 in the L.H.S.:
3(2 − 5 × 1) − 2(1 − 6 × 1)⇒3(2 − 5) − 2(1− 6)⇒3(−3) −2(−5)⇒−9 + 10 = 1 = R.H.S.3(2 – 5 × 1) – 2(1 – 6 × 1)⇒3(2 – 5) – 2(1- 6)⇒3(-3) -2(-5)⇒-9 + 10 = 1 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 21:

Solve each of the following equations and verify the answer in each case:
n4−5=n6+12n4-5=n6+12

ANSWER:

n4−5 = n6 + 12n4-5 = n6 + 12
or, n4 − n6 = 12 + 5   n4 – n6 = 12 + 5                                     [Transposing n/6 to the L.H.S. and 5 to the R.H.S.]
or, 3n−2n12 = 1+1023n-2n12 = 1+102
or, n12 = 112n12 = 112
or, n12×12 = 112×12n12×12 = 112×12                                    [Dividing both the sides by 12]
or, n = 66
Verification:
Substituting n = 66 on both the sides:

L.H.S.:
664−5 =332 − 5  =33 − 102 =232 = 232 R.H.S.: 666+12= 11 + 12 = 22+12= 232 664-5 =332 – 5  =33 – 102 =232 = 232 R.H.S.: 666+12= 11 + 12 = 22+12= 232 
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 22:

Solve each of the following equations and verify the answer in each case:
2m3+ 8=m2−12m3+ 8=m2-1

ANSWER:

2m3 + 8 = m2− 12m3 + 8 = m2- 1
or, 2m3 − m2 = −1 −82m3 – m2 = -1 -8                      [Transposing m/2 to the L.H.S. and 8 to the R.H.S.]
or, 4m−3m6 = −9or, m6 = −9or, 4m-3m6 = -9or, m6 = -9
or, m6×6 = −9×6m6×6 = -9×6                            [Multiplying both the sides by 6]
or, m = −-54
Verification:
Substituting x = −54 on both the sides:

L.H.S.: 2(−54)3 +  8 = −542−1= −1083 + 8 = −36+ 8 = −28  R.H.S.:−542−1 = −27 − 1= −28L.H.S.: 2(-54)3 +  8 = -542-1= -1083 + 8 = -36+ 8 = -28  R.H.S.:-542-1 = -27 – 1= -28
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 23:

Solve each of the following equations and verify the answer in each case:
2×5−32=x2+12×5-32=x2+1

ANSWER:

2×5 −32 = x2 + 12×5 -32 = x2 + 1
or, 2×5− x2 = 1+ 322×5- x2 = 1+ 32                        [Transposing x/2 to the L.H.S. and 3/2 to R.H.S.]
or, 4x−5×10= 2+32or, −x10 = 52or, 4x-5×10= 2+32or, -x10 = 52
or, −x10(−10) =52 ×(−10)or, -x10(-10) =52 ×(-10)               [Multiplying both the sides by −10]
or, x = −25
Verification:
Substituting x = −25 on both the sides:
L.H.S.: 2(−25)5 − 32 = −505 − 32  = −10 − 32 = −232R.H.S.: −252+ 1= −25 + 22 = −232L.H.S.: 2(-25)5 – 32 = -505 – 32  = -10 – 32 = -232R.H.S.: -252+ 1= -25 + 22 = -232
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 24:

Solve each of the following equations and verify the answer in each case:
x−35−2=2x5x-35-2=2×5

ANSWER:

x−35 − 2 = 2×5 x-35 – 2 = 2×5 
or, x5− 35 −2 = 2x5x5- 35 -2 = 2×5
or, − 35− 2 = 2×5−x5- 35- 2 = 2×5-x5                        [Transposing x/5 to the R.H.S.]
or, −3− 105 = x5-3- 105 = x5
or, −135 = x5-135 = x5
or, −135(5) =x5 ×(5)-135(5) =x5 ×(5)               [Multiplying both the sides by 5]
or, x = −13
Verification:
Substituting x = −13 on both the sides:
L.H.S.: −13 − 35 − 2 =−165 − 2= −16 − 105 = −265R.H.S.: 2×(−13)5  = −265  L.H.S.: -13 – 35 – 2 =-165 – 2= -16 – 105 = -265R.H.S.: 2×(-13)5  = -265  

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 25:

Solve each of the following equations and verify the answer in each case:
3×10−4=143×10-4=14

ANSWER:

3×10 − 4 = 14 3×10 – 4 = 14 
or, 3×10− 4  + 4= 14 + 43×10- 4  + 4= 14 + 4                        [Adding 4 on both the sides]
or, 3×10 = 183×10 = 18
or, 3×10×10 = 18×103×10×10 = 18×10                              [Multiplying both the sides by 10]
or, 3x = 1803x = 180              
or, 3×3 = 18033×3 = 1803                  [Dividing both the sides by 3]
or, x = 60
Verification:
Substituting x = 60 on both the sides:
3×6010 − 4 =18010 − 4 = 18 − 4 = 14 = R.H.S.3×6010 – 4 =18010 – 4 = 18 – 4 = 14 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 26:

Solve each of the following equations and verify the answer in each case:
34 (x − 1) = x − 334 (x – 1) = x – 3

ANSWER:

34(x−1) = x − 334x-1 = x – 3
 ⇒34×x  − 34 × 1= x − 3 ⇒34×x  – 34 × 1= x – 3                   [On expanding the brackets]
⇒3×4− 34  = x − 3⇒3×4- 34  = x – 3                      
⇒3×4− x = −3 + 34⇒3×4- x = -3 + 34                           [Transposing x to the L.H.S. and −34-34 to the R.H.S.]
⇒3x−4×4=  −12+34⇒3x-4×4=  -12+34
⇒−x4 =  −94⇒-x4 =  -94
⇒−x4×(−4) =  −94×(−4)⇒-x4×-4 =  -94×-4                              [Multiplying both the sides by -4]
or, x = 9             

Verification:
Substituting x = 9 on both the sides:
L.H.S. : 34(9−1) = 34(8) = 6  R.H.S.: 9 − 3 = 6L.H.S. : 349-1 = 34(8) = 6  R.H.S.: 9 – 3 = 6

L.H.S. = R.H.S.
Hence, verified.

Page No 144:

Exercise 9C

Question 1:

If 9 is added to a certain number, the result is 36. Find the number.

ANSWER:

Let the required number be x.
According to the question:
9 + x = 36
or, x + 9 −- 9 = 36 −- 9                        [Subtracting 9 from both the sides]
or, x = 27
Thus, the required number is 27.

Page No 144:

Question 2:

If 11 is subtracted from 4 times a number, the result is 89. Find the number.

ANSWER:

Let the required number be x.
According to the question:
4x −-11 = 89
or, 4x −- 11 +11 = 89 + 11                        [Adding 11 on both the sides]
or, 4x = 100
or, 4×4 = 10044×4 = 1004                                       [Dividing both the sides by 4]
or, x = 25
Thus, the required number is 25.

Page No 144:

Question 3:

Find a number which when multiplied by 5 is increased by 80.

ANSWER:

Let the required number be x.
According to the question:
or, 5x = x + 80
or, 5x −- x = 80                       [Transposing x to the L.H.S.]
or, 4x = 80
or, 4×4 = 8044×4 = 804                                        [Dividing both the sides by 4]
or, x = 20
Thus, the required number is 20.

Page No 144:

Question 4:

The sum of three consecutive natural numbers is 114. Find the numbers.

ANSWER:

Let the three consecutive natural numbers be x, (x+1), (x+2).
According to the question:
x + (x + 1) + (x + 2) = 114
or, x + x + 1 + x + 2 = 114
or, 3x + 3 = 114
or, 3x + 3 −- 3 = 114 −- 3                     [Subtracting 3 from both the sides]
or, 3x = 111
or, 3×3 = 11133×3 = 1113                                   [Dividing both the sides by 3]
or, x = 37
Required numbers are:
x = 37
or, x + 1 = 37 + 1 = 38
or ,x + 2 = 37 + 2 = 39
Thus, the required numbers are 37, 38 and 39.

Page No 144:

Question 5:

When Raju multiplies a certain number by 17 and adds 4 to the product, he gets 225. Find that number.

ANSWER:

Let the required number be x.
When Raju multiplies it with 17, the number becomes 17x.
According to the question :
17x + 4 = 225
or, 17x + 4 −- 4 = 225 −- 4                          [Subtracting 4 from both the sides]
or, 17x = 221
or, 17×17 = 2211717×17 = 22117                                        [Dividing both the sides by 17]
or, x = 13
Thus, the required number is 13.

Page No 144:

Question 6:

If a number is tripled and the result is increased by 5, we get 50. Find the number.

ANSWER:

Let the required number be x.
According to the question, the number is tripled and 5 is added to it
∴ 3x + 5
or, 3x  + 5 = 50
or, 3x + 5 −- 5 = 50 −- 5                         [Subtracting 5 from both the sides]
or, 3x = 45
or, 3×3 =4533×3 =453                              [Dividing both the sides by 3]
or, x = 15
Thus, the required number is 15.

Page No 144:

Question 7:

Find two numbers such that one of them exceeds the other by 18 and their sum is 92.

ANSWER:

Let one of the number be x.
∴ The other number = (x + 18)
According to the question:
x + (x + 18) = 92
or, 2x + 18 −- 18 = 92 −- 18                         [Subtracting 18 from both the sides]
or, 2x =74
or, 2×2 = 7422×2 = 742                                           [Dividing both the sides by 2]
or, x = 37
Required numbers are:
x = 37
or, x + 18 = 37 + 18 = 55

Page No 144:

Question 8:

One out of two numbers is thrice the other. If their sum is 124, find the numbers.

ANSWER:

Let one of the number be ‘x’
∴ Second number = 3x
According to the question:
x + 3x = 124
or, 4x  = 124                         
or, 4×4 = 12444×4 = 1244                      [Dividing both the sides by 4]
or, x = 31
Thus, the required number is x = 31 and 3x = 3××31 = 93.

Page No 144:

Question 9:

Find two numbers such that one of them is five times the other and their difference is 132.

ANSWER:

Let one of the number be x.
∴ Second number = 5x
According to the question:
5x −- x = 132
or, 4x = 132
or, 4×4 = 13244×4 = 1324                        [Dividing both the sides by 4]
or, x = 33
Thus, the required numbers are x = 33 and 5x = 5××33 = 165.

Page No 144:

Question 10:

The sum of two consecutive even numbers is 74. Find the numbers.

ANSWER:

Let one of the even number be x.
Then, the other consecutive even number is (x + 2).
According to the question:
x + (x + 2)  = 74
or, 2x + 2  = 74
or, 2x + 2 −- 2 = 74 −- 2           [Subtracting 2 from both the sides]
or, 2x = 72
or, 2×2 = 7222×2 = 722                         [Dividing both the sides by 2]
or, x = 36
Thus, the required numbers are x = 36 and x+ 2 = 38.

Page No 144:

Question 11:

The sum of three consecutive odd numbers is 21. Find the numbers.

ANSWER:

Let the first odd number be x.
Then, the next consecutive odd numbers will be (x + 2) and (x + 4).
According to the question:
x + (x + 2) + (x + 4)  = 21
or, 3x + 6  = 21
or, 3x + 6 −- 6 = 21 −- 6           [Subtracting 6 from both the sides]
or, 3x = 15
or, 3×3 = 1533×3 = 153                         [Dividing both the sides by 3]
or, x = 5
∴ Required numbers are:
x = 5
x + 2 = 5 + 2 = 7
x + 4 = 5 + 4 = 9

Page No 144:

Question 12:

Reena is 6 years older than her brother Ajay. If the sum of their ages is 28 years, what are their present ages?

ANSWER:

Let the present age of Ajay be x years.
Since Reena is 6 years older than Ajay, the present age of Reena will be (x+ 6) years.
According to the question:
x + (x + 6) = 28
or, 2x + 6 = 28
or, 2x + 6 −- 6 = 28 −- 6                 [Subtracting 6 from both the sides]
or, 2x = 22
or, 2×2 = 2222×2 = 222                              [Dividing both the sides by 2]
or, x = 11
∴ Present age of Ajay = 11 years
Present age of Reena  = x +6 = 11 + 6
                                 = 17 years

Page No 144:

Question 13:

Deepak is twice as old as his brother Vikas. If the difference of their ages be 11 years, find their present ages.

ANSWER:

Let the present age of Vikas be x years.
Since Deepak is twice as old as Vikas, the present age of Deepak will be 2x years.
According to the question:
2x −- x = 11
x = 11
∴ Present age of Vikas = 11 years
Present age of Deepak  = 2x = 2××11
                                   = 22 years

Page No 144:

Question 14:

Mrs Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old as Rekha. Find their present ages.

ANSWER:

Let the present age of Rekha be x years.
As Mrs. Goel is 27 years older than Rekha, the present age of Mrs. Goel will be (x + 27) years.
After 8 years:
Rekha’s age = (x + 8) years
Mrs. Goel’s age = (x + 27 + 8)
                      = (x + 35) years

According to the question:
(x + 35) = 2(x + 8)
or, x + 35 = 2××x + 2××8                [On expanding the brackets]
or, x + 35 = 2x + 16
or, 35 −- 16 = 2x −- x                [Transposing 16 to the L.H.S. and x to the R.H.S.]
or, x = 19
∴ Present age of Rekha = 19 years
Present age of Mrs. Goel = x + 27
                                     = 19 + 27
                                     = 46 years

Page No 145:

Question 15:

A man is 4 times as old as his son. After 16 years he will be only twice as old as his son. Find their present ages.

ANSWER:

Let the present age of the son be x years.
As the man is 4 times as old as his son, the present age of the man will be (4x) years.
After 16 years:
Son’s age  = (x + 16) years
Man’s age = (4x + 16) years

According to the question:
(4x + 16) = 2(x + 16)
or, 4x + 16 = 2××x + 2××16                [On expanding the brackets]
or, 4x + 16 = 2x + 32
or, 4x −- 2x = 32 −- 16                [Transposing 16 to the R.H.S. and 2x to the L.H.S.]
or, 2x = 16
or, 2×2 = 1622×2 = 162                           [Dividing both the sides by 2]
or, x = 8
∴ Present age of the son = 8 years
Present age of the man  = 4x = 4××8
                                    = 32 years

Page No 145:

Question 16:

A man is thrice as old as his son. Five years ago the man was four times as old as his son. Find their present ages.

ANSWER:

Let the present age of the son be x years.
As the man is 3 times as old as his son, the present age of the man will be (3x) years.

5 years ago:
Son’s age = (x −- 5) years
Man’s age = (3x −- 5) years

According to the question:
(3x −- 5) = 4(x −- 5)
or, 3x −- 5 = 4××x −- 4××5                [On expanding the brackets]
or, 3x −- 5 = 4x −- 20
or, 20 −- 5 = 4x −- 3x                [Transposing 3x to the R.H.S. and 20 to the L.H.S.]
or, x = 15
∴ Present age of the son = 15 years
Present age of the man  = 3x = 3××15
                                    = 45 years

Page No 145:

Question 17:

After 16 years, Fatima will be three times as old as she is now. Find her present age.

ANSWER:

Let the present age of Fatima be x years.

After 16 years:
Fatima’s age = (x + 16) years

According to the question:
x + 16 = 3(x)
or, 16 = 3x −- x               [Transposing x to the R.H.S.]
or, 16 = 2x
or, 2×2 = 1622×2 = 162                [Dividing both the sides by 2]
or, x = 8
∴ Present age of Fatima = 8 years

Page No 145:

Question 18:

After 32 years, Rahim will be 5 times as old as he was 8 years ago. How old is Rahim today?

ANSWER:

Let the present age of Rahim be x years.
After 32 years:
Rahim’s age = (x + 32) years
8 years ago:
Rahim’s age = (x −- 8) years
According to the question:
x + 32 = 5(x −- 8)
or, x + 32  = 5x −- 5××8              
or, x + 32 = 5x −- 40
or, 40 + 32 = 5x −- x                      [Transposing ‘x’ to the R.H.S. and 40 to the L.H.S.]
or, 72 = 4x
or, 4×4 = 7244×4 = 724                               [Dividing both the sides by 4]
or, x = 18
Thus, the present age of Rahim is 18 years.

Page No 145:

Question 19:

A bag contains 25-paisa and 50-paisa coins whose total value is Rs 30. If the number of 25-paisa coins is four times that of 50-paisa coins, find the number of each type of coins.

ANSWER:

Let the number of 50 paisa coins be x.
Then, the number of 25 paisa coins will be 4x.
According to the question:
0.50(x) + 0.25(4x) = 30
or, 0.5x + x = 30
or, 1.5x = 30
or, 1.5×1.5 = 301.51.5×1.5 = 301.5          [Dividing both the sides by 1.5]
or, x = 20
Thus, the number of 50 paisa coins is 20.
Number of 25 paisa coins = 4x = 4××20 = 80

Page No 145:

Question 20:

Five times the price of a pen is Rs 17 more than three times its price. Find the price of the pen.

ANSWER:

Let the price of one pen be Rs x.
According to the question:
5x = 3x + 17
or, 5x −- 3x = 17                    [Transposing 3x to the L.H.S.]
or, 2x = 17
or, 2×2 = 1722×2 = 172                        [Dividing both the sides by 2]
or, x = 8.50
∴ Price of one pen = Rs 8.50

Page No 145:

Question 21:

The number of boys in a school is 334 more than the number of girls. If the total strength of the school is 572, find the number of girls in the school.

ANSWER:

Let the number of girls in the school be x.
Then, the number of boys in the school will be (x + 334).
Total strength of the school = 572

∴ x + (x + 334) = 572
or, 2x + 334 = 572
or, 2x + 334 −- 334 = 572 −- 334                {Subtracting 334 from both the sides]
or, 2x = 238
or, 2×2 = 23822×2 = 2382                                          [Dividing both the sides by 2]
or, x = 119
∴ Number of girls in the school = 119

Page No 145:

Question 22:

The length of a rectangular park is thrice its breadth. If the perimeter of the park is 168 metres, fund its dimensions.

ANSWER:

Let the breadth of the park be x metres.
Then, the length of the park will be 3x metres.
Perimeter of the park = 2 (Length + Breadth) = 2 ( 3x + x ) m
Given perimeter = 168 m

∴ 2(3x + x) = 168                           
or, 2 ( 4x ) = 168
or, 8x = 168                                      [On expanding the brackets]
or, 8×8 = 16888×8 = 1688                                [Dividing both the sides by 8]
or, x = 21 m
∴ Breadth of the park = x = 21 m
Length of the park = 3x = 3××21 = 63 m

Page No 145:

Question 23:

The length of a rectangular hall is 5 metres more than its breadth. If the perimeter of the hall is 74 metres, find its length and breadth.

ANSWER:

Let the breadth of the hall be x metres.
Then, the length of the hall will be (x + 5) metres.
Perimeter of the hall = 2(Length + Breadth) = 2( x + 5 + x) metres
Given perimeter of the rectangular hall = 74 metres

∴ 2( x + 5 + x) = 74                             
or, 2 ( 2x + 5 ) = 74
or, 2 ×2x + 2 ×5 = 74                              [On expanding the brackets]
or, 4x + 10 = 74
or, 4x + 10 −- 10 = 74 −- 10                     [Subtracting 10 from both the sides]
or, 4x = 64
or, 4×4 = 6444×4 = 644                                        [Dividing both the sides by 4]
or, x = 16 metres
∴ Breadth of the park = x
                                  = 16 metres
Length of the park = x + 5 = 16 + 5
                            = 21 metres

Page No 145:

Question 24:

A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm more than its breadth. Find the length and the breadth of the rectangle so formed.

ANSWER:

Let the breadth of the rectangle be x cm.
Then, the length of the rectangle will be (x + 7) cm.
Perimeter of the rectangle = 2(Length + Breadth) = 2( x + 7 + x) cm
Given perimeter of the rectangle = Length of the wire = 86 cm

∴ 2( x + 7 + x) = 86                            
or, 2 ( 2x + 7 ) = 86
or, 2 ×2x + 2 × 7 = 86                              [On expanding the brackets]
or, 4x + 14 = 86
or, 4x + 14 – 14 = 86 – 14                   [Subtracting 14 from both the sides]
or, 4x = 72
or, 4×4 = 7244×4 = 724                                   [Dividing by 4 on both the sides]
or, x = 18 metres
Breadth of the hall = x
                             = 18 metres
Length of the hall = x + 7
                           = 18 + 7
                           = 25 metres

Page No 146:

Exercise 9D

Question 1:

A man earns Rs 25 per hour. How much does he earn in x hours?

ANSWER:

Earning of the man per hour = Rs 25

Earning of the man in x hours = Rs (25××x)
                                             = Rs 25x

Page No 146:

Question 2:

The cost of 1 pen is Rs 16 and the cost of 1 pencil is Rs 5. What is the total cost of x pens and y pencils.

ANSWER:

Cost of 1 pen = Rs 16
∴ Cost of ‘x‘ pens = Rs 16 ×× x
                           = Rs 16x
Similarly, cost of 1 pencil = Rs 5
∴ Cost of ‘y’ pencils = Rs 5××y
                               = Rs 5y
∴ Total cost of x pens and y pencils = Rs (16x + 5y)

Page No 146:

Question 3:

Lalit earns Rs x per day and spends Rs y per day. How much does he save in 30 days?

ANSWER:

Lalit’s earning per day = Rs x
∴ Lalit’s earning in 30 days = Rs 30 ××x
                                          = Rs 30x

Similarly, Lalit’s expenditure per day = Rs y
∴ Lalit’s expenditure in 30 days = Rs 30 ×× y
                                                = Rs 30y
∴ In 30 days, Lalit saves = (Total earnings −- Total expenditure)
                                      = Rs (30x −- 30y)
                                      = Rs 30(x – y)

Page No 146:

Question 4:

Three times a number added to 8 gives 20. Find the number.

ANSWER:

Let the required number be x.
Three times this number is 3x.
On adding 8, the number becomes 3x + 8.
3x + 8 = 20
or, 3x + 8 −- 8 = 20 −- 8                [Subtracting 8 from both the sides]
or, 3x = 12
or, 3×3 = 1233×3 = 123                             [Dividing both the sides by 3]
or, x = 4
∴ Required number = 4

Page No 146:

Question 5:

If x = 1, y = 2 and z = 3, find the value of x² + y² + 2xyz.

ANSWER:

Given:
x =1
y = 2
z = 3

Substituting x = 1, y = 2 and z = 3 in the given equation (x² + y² + 2xyz):

(1)2 +( 2)2  + 2(1)(2)(3)⇒1 + 4 + 12 = 1712 + 22  + 2123⇒1 + 4 + 12 = 17

Page No 146:

Question 6:

Solve: 4x + 9 = 17.

ANSWER:

4x + 9 = 17
or, 4x + 9 −- 9 = 17 −- 9                          [Subtracting 9 from both the sides]
or, 4x = 8
or, 4×4 = 844×4 = 84                                          [Dividing both the sides with 4]
or, x = 2

Page No 146:

Question 7:

Solve: 3(x + 2) − 2(x − 1) = 7.

ANSWER:

3(x + 2) − 2(x − 1) = 7.
or, 3 × x + 3 × 2 − 2 × x − 2 × (−1) = 73 × x + 3 × 2 – 2 × x – 2 × (-1) = 7                    [On expanding the brackets]
or, 3x + 6 − 2x + 2 = 7
or, x + 8 = 7
or, x + 8 − 8 = 7 − 8                                            [Subtracting 8 from both the sides]
or, x = −1

Page No 146:

Question 8:

Solve: 2×5−x2=522×5-x2=52.

ANSWER:

2×5 −x2 = 522×5 -x2 = 52
or, 4x − 5×10 = 524x – 5×10 = 52                               [Taking the L.C.M. as 10]
or, −x10 = 52-x10 = 52
or, −x10×(−10) = 52×(−10)-x10×-10 = 52×-10                          [Multiplying both the sides by (−10)]
or, x = −25

Page No 146:

Question 9:

The sum of three consecutive natural numbers is 51. Find the numbers.

ANSWER:

Let the three consecutive natural numbers be x, (x + 1) and (x + 2).

∴ x + (x + 1) + (x + 2) = 51
3x + 3 = 51
3x + 3 −- 3 = 51 −- 3                   [Subtracting 3 from both the sides]
3x = 48
3×3 = 4833×3 = 483                                [Dividing both the sides by 3]
x = 16
Thus, the three natural numbers are x = 16, x+1 = 17 and x+2 = 18.

Page No 146:

Question 10:

After 16 years, Seema will be three times as old as she is now. Find her present age.

ANSWER:

Let the present age of Seema be x years.
After 16 years:
Seema’s age = x + 16

After 16 years, her age becomes thrice of her age now
∴ x + 16 = 3x
or, 16 = 3x −- x                     [Transposing x to the R.H.S.]
or, 2x = 16
or, 2×2 = 1622×2 = 162                      [Dividing both the sides by 2]
or, x = 8 years

Page No 146:

Question 11:

By how much does I exceed 2x − 3y − 4?
(a) 2x − 3y − 5
(b) 2x − 3y − 3
(c) 5 − 2x + 3y
(d) none of these

ANSWER:

(c) 5 − 2x + 3y

1 exceeds 2x − 3y − 4.

∴1 −- (2x − 3y − 4) = 1 −-2x + 3y + 4
                              = 5 −- 2x + 3y

∴ 1 exceeds 2x − 3y − 4 by 5 −- 2x + 3y.

Page No 146:

Question 12:

What must be added to 5x³ − 2x² + 6x + 7 to make the sum x³ + 3x² − x + 1?
(a) 4x³ − 5x² + 7x + 6
(b) −4x³ + 5x² − 7x − 6
(c) 4x³ + 5x² − 7x + 6
(d) none of these

ANSWER:

  (b) −4x³ + 5x² − 7x − 6
In order to find what must be added, we subtract (5x³ − 2x² + 6x + 7) from (x³ + 3x² − x + 1).

    (x³ + 3x² − x + 1)  − ( 5x³ − 2x² + 6x + 7)
or,  x³ + 3x² − x + 1 − 5x³  + 2x² − 6x − 7
or,  x³ − 5x³+ 3x²+ 2x²− x − 6x+ 1 − 7
or, −4x³ + 5x² − 7x − 6

Page No 146:

Question 13:

2x − [3y − {2x − (y − x)}] = ?
(a) 5x − 4y
(b) 4y − 5x
(c) 5y − 4x
(d) 4x − 5y

ANSWER:

(a) 5x − 4y

2x − [3y − {2x − (y − x)}]
= 2x − [3y − {2x −  y + x}]
= 2x − [3y − {3x −  y}]
= 2x −  [3y −  3x + y]
= 2x −  [4y −  3x]
= 2x −  4y + 3x
= 5x −  4y

Page No 146:

Question 14:

The coefficient of x in −5xyz is
(a) −5
(b) 5yz
(c) −5yz
(d) yz

ANSWER:

(c) −5yz

All the terms in the expression −5xyz barring x will be the coefficient of x, i.e. −5yz.

Page No 146:

Question 15:

13(x + 7 + z)13(x + 7 + z) is a
(a) monomial
(b) binomial
(c) trinomial
(d) quadrinomial

ANSWER:

(b) trinomial
Since it contains three variables, i.e. ‘x’, ‘y’ and ‘z’, it is a trinomial.

Page No 146:

Question 16:

If x5=1×5=1, then
(a) x=15x=15
(b) x = 5
(c) x = (5 + 1)
(d) none of these

ANSWER:

(b) x = 5

x5 = 1or, x5 × 5= 1 × 5   [Multiplying both the sides by 5]x5 = 1or, x5 × 5= 1 × 5   [Multiplying both the sides by 5]
or, x = 5

Page No 146:

Question 17:

If x = 1, y = 2 and z = 3 then (x² + y² + z²) = ?
(a) 6
(b) 12
(c) 14
(d) 15

ANSWER:

(c) 14
Substituting x = 1, y = 2 and z = 3 in (x² + y² + z²):
(1)2  + (2)2 + (3)212  + 22 + 32
or, 1 + 4 + 9 = 14

Page No 146:

Question 18:

If 13 x + 5 = 813 x + 5 = 8, then x = ?
(a) 3
(b) 6
(c) 9
(d) 12

ANSWER:

(c) 9

13x + 5 = 813x + 5 = 8
or, 13x + 5 − 5 = 8−513x + 5 – 5 = 8-5                                  [Subtracting 5 from both the sides]
or, 13x = 313x = 3
or, 13x × 3 = 3×313x × 3 = 3×3                                        [Multiplying both the sides by 3]
or, x = 9

Page No 146:

Question 19:

Fill in the blanks.
(i) An expression having one term is called a …… .
(ii) An expression having two term is called a …… .
(i) An expression having three term is called a …… .
(iv) 3x − 5 = 7 − x ⇒ x = …… .3x – 5 = 7 – x ⇒ x = …… .
(v) (b2 − a2) − (a2 − b2) = …… .(b2 – a2) – (a2 – b2) = …… .

ANSWER:

(i) monomial
(ii) binomial
(iii) trinomial
(iv) x = 3
3x −- 5 = 7 −- x
or, 3x + x = 7 + 5                   [Transposing x to the L.H.S. and 5 to the R.H.S.]
or, 4x = 12
or, 4×4 = 1244×4 = 124
or, x = 3
(v) 2b² −- 2a²
or, b² −- a² −- a² + b²
or, 2b² −- 2a²
( b2 − a2) − (a2 − b2)
(b2 − a2) − (a2 − b2
(b2 − a2) − (a2 − b2)
(b2 − a2) − (a2 − b2)

Page No 147:

Question 20:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) −3xy²z is a monomial.
(ii) x=23x=23 is solution of 2x + 5 = 8.
(iii) 2x + 3 = 5 is a linear equation.
(iv) The coefficient of x in 5xy is 5.
(v) 8 − x = 5 ⇒ x = 38 – x = 5 ⇒ x = 3.

ANSWER:

(i) True
Since it has one term, it is a monomial.

(ii) False
2x + 5 = 8
or, 2x + 5 −- 5 = 8 −- 5               [Subtracting 5 from both the sides]
or, 2x = 3
or, x = 3/2 and not x = 2/3

(iii) True
This is because the maximum power of the variable x is 1.

(iv) False
The coefficient of x in 5xy would be 5y.

(v) True
    8 − x = 5
  or, 8 − 5 = x
    or, 3 = x

Exercise 10A

Page No 152:

Question 1:

Find each of the following ratios in the simplest form:
(i) 24 to 56
(ii) 84 paise to Rs 3
(iii) 4 kg to 750 g
(iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m

ANSWER:

(i) 24:56 =     24  =     24 ​÷ 8     =    3     
                       56       56 ​÷ 8            7
As the H.C.F. of 3 and 7 is 1, the simplest form of 24:56 is 3:7.

(ii) 84 paise to Rs 3 = Rs 0.84 to R. 3 =  0.84  =  0.84​ ÷ 3    =  0.28   =   28    =  28 ​÷  4   =   7  
                                                                   3            3 ​÷ 3             1          100      100 ​÷ 4        25
As the H.C.F. of 7 and 25 is 1, the simplest form of  0.84:3 is 7:25.

(iii) 4 kg:750 g = 4000 g:750 g =   4000 ​÷ 250   =    16   
                                                            750 ​÷ 250            3
As the H.C.F. of 16 and 3 is 1, the simplest form of 4000:750 is 16:3.

(iv) 1.8 kg:6 kg  =   1.8   =   18   =  18 ​÷ 6  =     3  
                               6           60      60 ​÷ 6        10 
As the H.C.F. of 3 and 10 is 1, the simplest form of 1.8:6 is 3:1.

(v) 48 minutes to 1 hour = 48 minutes to 60 minutes = 48:60 =  48 ​÷ 12   =  4   
                                                                                                 60 ​÷ 12        5
As the H.C.F. of 4 and 5 is 1, the simplest form of 48:60 is 4:5.
 
(vi) 2.4 km to 900 m = 2400m:900m =    2400   =   24    =   24 ​÷ 3   =    8  
                                                               900          9           9 ​÷ 3          3
 As the H.C.F. of 8 and 3 is 1, the simplest form of 2400:900 is 8:3.

                    

Page No 152:

Question 2:

Express each of the following ratios in the simplest form:
(i) 36 : 90
(ii) 324 : 144
(iii) 85 : 561
(iv) 480 : 384
(v) 186 : 403
(vi) 777 : 1147

ANSWER:

(i) 36:90 =   36    =    36 ​÷ 18    =     2          (As the H.C.F. of 36 and 90 is 18.)
                   90          90 ​÷ 18            5
Since the H.C.F. of 2 and 5 is 1, the simplest form of 36:90 is 2:5.

(ii) 324:144 =   324   =    324 ​÷ 36    =    9       (As the H.C.F. of  324 and 144 is 36.)
                          144          144 ​÷ 36           4
 Since the H.C.F. of 9 and 4 is 1, the simplest form of 324:144 is 9:4.

(iii) 85:561 =  85   =   85 ​÷ 17   =      5         (As the H.C.F. of 85 and 561 is 17.)
                        561      561 ​​÷ 17           33
Since the H.C.F. of 5 and 33 is 1, the simplest form of 85:561 is 5:33.

(iv) 480:384 =    480    =    480 ​÷ 96    =   5       (As the H.C.F. of 480 and 384 is 96.)
                            384          384 ​​÷ 96          4
Since the H.C.F. of 5 and 4 is 1, the simplest form of 480:384 is 5:4.

(v) 186:403 =     186   =   186 ÷ 31    =   6        (As the H.C.F. of 186 and 403 is 31.)
                             403        403 ÷ 31        13
Since the H.C.F. of 6 and 13 is 1, the simplest form of 186:403 is 6:13.

(vi) 777:1147 =  777  ​÷ 37      =   21           (As the H.C.F. of 777 and 1147 is 37.)
                            1147  ​÷  37         31
Since the H.C.F. of 21 and 31 is 1, the simplest form of 777:1147 is 21:31.

Page No 152:

Question 3:

Write each of the following ratios in the simplest form:
(i) Rs 6.30 : Rs 16.80
(ii) 3 weeks : 30 days
(iii) 3 m 5 cm : 35 cm
(iv) 48 min : 2 hours 40 min
(v) 1 L 35 mL : 270 mL
(vi) 4 kg : 2 kg 500 g

ANSWER:

(i) Rs 6.30:Rs 16.80
             6.30     =    63    =   63 ​÷ 21     =   3        (H.C.F. of 63 and 168 is 21.)
            16.80         168        168  ​÷ 21        8
    Ratio = 3 : 8
(ii)3 weeks:30 days = 21days:30 days          (1 week = 7 days)
          21     =   21 ​÷ 3   =   7          (H.C.F. of 21 and 30 is 3.)
          30          30 ​ ​÷ 3       10
     Ratio = 7 : 10
 (iii) 3 m 5 cm:35 cm = 305 cm:35 cm      (1 m = 100 cm)
            305  =   305  ​÷ 5   =   61        (H.C.F. of 305 and 35 is 5.)
             35         35  ​÷ 5          7
      Ratio = 61:7
(iv) 48 min:2 hours 40 min = 48 min:160 min        (1 hour = 60 mins)
             48   =    48  ​÷ 16    =    3         (H.C.F. of 48 and 160 is 16.)
            160        160  ​÷ 16       10
      Ratio = 3:10
(v) 1 L 35 mL:270 mL = 1035 mL:270 mL         (1 L = 1000 mL)
             1035   =   1035  ​÷ 45    =   23        (H.C.F. of 1035 and 270 is 45.)
              270           270  ​÷ 45          6
       Ratio = 23:6
(vi) 4 kg:2 kg 500 g = 4000 g:2500 g        (1 kg= 1000 g)
            4000   =   40  =  40  ​÷ 5   =   8     (H.C.F. of 40 and 25 is 5.)
            2500         25      25  ​÷ 5        5
      Ratio = 8:5

Page No 152:

Question 4:

Mr Sahai and his wife are both school teachers and earn Rs 16800 and Rs 10500 per month respectively. Find the ratio of
(i) Mr Sahai’s income to his wife’s income;
(ii) Mrs Sahai’s income to her husband’s income;
(iii) Mr Sahai’s income to the total income of the two.

ANSWER:

Mr Sahai’s earning = Rs 16800
Mrs Sahai’s earning = Rs 10500
(i) Ratio = 16800:10500 = 168:105 =  168  ​÷ 21  =   8          (H.C.F. of 168 and 105 is 21.)
                                                           105 ​ ​÷ 21           5
Mr Sahai’s income:Mrs Sahai’s income = 8:5
(ii)Ratio = 10500:16800 = 105:168 =  105  ​÷ 21   =    5        (H.C.F. of 168 and 105 is 21.)
                                                         168 ​ ​÷ 21          8
Mrs Sahai’s income:Mr Sahai’s income = 5:8

(iii) Total income = 16800 + 10500 = Rs  27300
   Ratio = 16800:27300 = 168:273 =  168   =  168  ​÷ 21   =  8  (H.C.F. of 168 and 273 is 21.)
                                                        273        273  ​÷ 21     13
Mrs Sahai’s income:Total income = 8:13

Page No 152:

Question 5:

Rohit earns Rs 15300 and saves Rs 1224 per month. Find the ratio of
(i) his income and savings;
(ii) his income and expenditure;
(iii) his expenditure and savings.

ANSWER:

Rohit’s income = Rs 15300
Rohit’s savings = Rs 1224
(i) Income:Savings = 15300:1224 = 15300 ​÷ 612   =  25        (H.C.F. of 15300 and 1224 is 612.)
                                                                  1224  ​÷ 612         2 
    Income:Savings = 25:2
(ii) Monthly expenditure = Rs (15300 −- 1224) = Rs 14076
    Income:Expenditure = 15300:14076 =  15300  ÷ 612  =   25       (H.C.F. of 15300 and 14076 is 612.)
                                                                       14076  ​÷ 612        23
     Income:Expenditure = 25:23
(iii) Expenditure : Savings = 14076:1224 =  14076  ÷ 612  =  23       (H.C.F. of 14076 and 1224 is 612.)
                                                                           1224  ​÷ 612         2
      Expenditure:Savings = 23:2
                    

Page No 152:

Question 6:

The ratio of the number of male and female workers in a textile mill is 5 : 3. If there are 115 male workers, what is the numkber of female workers in the mill?

ANSWER:

Number of male:Number of female = 5:3
Let the number be x.
Number of male = 5x
​Number of female = 3x
Number of male workers = 115
   Now, 5x = 115
         ⇒  x =   115   = 23
                       5
Number of female workers in the mill = 3x = 3 × 23 = 69

Page No 152:

Question 7:

The bosys and the girls in a school are in the ratio 9 : 5. If the total strength of the school is 448, find the number of girls.

ANSWER:

Boys:Girls = 9:5
Let the number of boys = 9x 
Let the number of girls = 5x
Total strength of the school = 448
According to given condition, we have:
                                                      9x + 5x = 448
                                              ⇒         14x = 448
                                               ⇒            x =   448   = 32
                                                                      14
Number of boys = 9x = 9 × 32 = 288
Number of girls = 5x = 5 ​× 32 = 160

Page No 152:

Question 8:

Divide Rs 1575 between Kamal and Madhu in the ratio 7 : 2.

ANSWER:

Kamal:Madhu = 7:2
Sum of the ratio terms = 7 + 2 = 9
Kamal’s share =   7   × 1575 =   11025   = Rs 1225
                            9                         9
Madhu’s share =   2   × 1575 =   3150   = Rs 350
                            9                       9

Page No 152:

Question 9:

Divide Rs 3450 among A, B and C in the ratio 3 : 5 : 7.

ANSWER:

A:B:C = 3:5:7
Sum of the ratio terms = 3 + 5 +7 = 15
A’s share =   3    × 3450 =   10350   = Rs 690
                   15                       15

B’s share =   5   × 3450 =   17250   = Rs 1150
                   15                        15

 C’s share =     7   × 3450 =   24150  = Rs 1610      
                      15                       15

Page No 152:

Question 10:

Two numbers are in the ratio 11 : 12 and their sum is 460. Find the numbers.

ANSWER:

Two number are in the ratio 11:12.
Let the numbers be 11x and 12x.
Given:     11x + 12x = 460
         ⇒          23x = 460
        ⇒               x =   460   =  20
                                    23
First number = 11x = 11 × 20 = 220
Second number = 12x = 12 × 20 = 240
Hence, the numbers are 220 and 240.

Page No 152:

Question 11:

A 35-cm line segment is divided into two parts in the ratio 4 : 3. Find the length of each part.

ANSWER:

Ratio of the two parts of line segment = 4:3
Sum of the ratio terms = 4 + 3 = 7
First part =   4   × 35 cm = 4 × 5 cm = 20 cm
                    7
Second part =    3   × 35 cm = 3 × 5 cm = 15 cm
                         7

Page No 152:

Question 12:

A factory produces electric bulbs. If 1 out of every 10 bulbs is defective and the factory produces 630 bulbs per day, find the number of defective bulbs produced each day.

ANSWER:

Number of bulbs produced each day = 630
Out of 10 bulbs, 1 is defective.
Number of defective bulbs =  630  = 63
                                             10

∴∴ Number of defective bulbs produced each day = 63

Page No 152:

Question 13:

Find the ratio of the price of a pencil to that of a ball pen if pencils cost Rs 96 per score and ball pens cost Rs 50.40 per dozen.

ANSWER:

Price of pencil = Rs 96 per score 
Price of ball pen = Rs 50.40 per dozen
Price per unit of pencil =  96  = 4.8
                                       20
Price per unit of ball pen =  50.40  =  4.2
                                            12
   Ratio =    4.8   =   48   =   48  ​÷ 6    =    8  
                   4.2        42        42  ​÷  6         7
Price of a pencil:Price of a ball pen = 8:7

Page No 152:

Question 14:

The ratio of the length of a field to its width is 5 : 3. Find its length if the width is 42 metres.

ANSWER:

Length:Width = 5:3
Let the length and the width of the field be 5x m and 3x m, respectively.
Width = 42 m
3x = 42
 x =   42   = 14
          3
∴∴ Length = 5x = 5 × 14 = 70 metres

Page No 152:

Question 15:

The ratio of income to savings of a family is 11 : 2. Find the expenditure if the savings is Rs 1520.

ANSWER:

Income:Savings = 11:2
Let the income and the saving be Rs 11x and Rs 2x, respectively.
Saving = Rs 1520
  2x = 1520
   x =   1520   = 760
              2
∴∴ Income = Rs 11x =Rs (11 × 760) = Rs 8360
   Expenditure = Income −- Saving
                      = Rs (8360 −- 1520 )
                      = Rs 6840

Page No 152:

Question 16:

The ratio of income to expenditure of a family is 7 : 6. Find the savings if the income is Rs 14000.

ANSWER:

Income:Expenditure = 7:6
Let the income and the expenditure be Rs 7x and Rs 6x, respectively.
Income = Rs 14000
7x = 14000
x =    14000  =  2000
             7
Expenditure = Rs 6x = Rs 6 × 2000 = Rs 12000
∴∴ Saving = Income −- Expenditure
          = Rs (14000 −- 12000)
         = Rs 2000

Page No 152:

Question 17:

The ratio of zinc and copper in an alloy is 7 : 9. If the weight of copper in the alloy is 11.7 kg find the weight of zinc in it.

ANSWER:

Let the weight of zinc be x kg.
Ratio of zinc and copper = 7:9
Weight of copper in the alloy = 11.7 kg
     7   =    x     
     9       11.7
⇒  x =  11.7 × 7   =  81.9   = 9.1
                9                9
Weight of zinc = 9.1 kg

Page No 152:

Question 18:

A bus covers 128 km in 2 hours and a train covers 240 km in 3 hours. Find the ratio of their speeds.

ANSWER:

A bus covers 128 km in 2 hours.
Speed of the bus =   Distance  =  128 km   = 64 km/ hr
                               Time               2 hr

A train covers 240 km in 3 hours.
Speed of the train =  Distance   =  240   = 80 km /hr
                                 Time             3

Ratio of their speeds = 64:80 =  64  =   64 ÷ 16   =   4  
                                                 80       80 ÷ 16        5
∴∴ Ratio of the speeds of the bus and the train = 4:5

Page No 153:

Question 19:

From each of the given pairs, find which ratio is larger:
(i) (3 : 4) or (9 : 16)
(ii) (5 : 12) or (17 : 30)
(iii) (3 : 7) or (4 : 9)
(iv) (1 : 2) or (13 : 27)

ANSWER:

(i) (3:4) or (9:16) 

Making the denominator equal:

        3 × 4   =  12  and 12   >   9 
        4 × 4       16         16       16

 ∴∴ (3:4) > (9:16)

(ii) (5:12) or (17:30) 

Making the denominator equal:

        5 × 5   =    25   and   17 × 2    =    34     
      12 × 5         60            30 × 2          60
⇒    25   <    34         
      60         60
 ∴∴ (5:12) < (17:30)

(iii) (3:7) or (4:9) 

Making the denominator equal:

          3 × 9  =   27   and   4 × 7   =  28  
          7 × 9       63            9 ​× 7       63
⇒       27  <   28      
         63       63

(3:7) < (4:9)

(iv) (1:2) or (13:27)

Making the denominator equal:

        1× 27   =   27   and   13 × 2   =   26  
        2 × 27       54            27 ​× 2        54

⇒   27  >  26          
     54       54

(1:2) > (13:27)

Page No 153:

Question 20:

Fill in the place holders:
(i) 2440=   5=12   2440=   5=12   
(ii) 3663=4   =   213663=4   =   21
(iii) 57=   28=35   57=   28=35   

ANSWER:

(i)   24   =   24 ​÷ 8  =   3   =    3 × 4  =  12      
      40         40 ​÷ 8      5          5  × 4      20

(ii)    36  =   36  ​÷ 9  =  4   =   4 × 3  =   12       
        63        63  ​÷ 9      7        7 × 3        21

(iii)   5   =   5 × 4   =  20  =   5 × 7   =   35  
        7        7 × 4       28        7 × 7         49

Page No 155:

Exercise 10B

Question 1:

Determine if the following numbers are in proportion:
(i) 4, 6, 8, 12
(ii) 7, 42, 13, 78
(iii) 33, 121, 9, 96
(iv) 22, 33, 42, 63
(v) 32, 48, 70, 210
(vi) 150, 200, 250, 300

ANSWER:

(i) 4, 6, 8, 12
    4  =    4 ​÷ 2   =   2 ;     8  =    8  ​÷ 4   =   2  
    6        6  ​÷ 2       3       12       12  ​÷ 4       3
Hence, 4:9::8:12 are in proportion.

(ii) 7, 42, 13, 78
      7    =  7  ​÷ 7   =   1 ;     13   =  13  ​÷ 13     =    1  
      42      42  ​÷ 7      6        78        78  ​÷ 13           6
Hence, 7:42::13:78 are in proportion.

(iii) 33, 121, 9, 96
      33   =   33  ​÷ 11   =   3   ;     9   =   9  ​÷ 3   =   3  
     121      121  ​÷ 11      11         96       96  ​÷ 3      32
 Hence, 33:121::9:96 are not in proportion.

(iv) 22, 33, 42, 63
     
2233=22÷1133÷11=23 and 4263=42÷2163÷21=232233=22÷1133÷11=23 and 4263=42÷2163÷21=23

Hence, 22:33 :: 42 : 63 are not in proportion.
 
(v) 32, 48, 70, 210
       32   =   32  ​÷ 6  =   7 ;    70   =   70  ​÷ 70   =  1 
       48        48  ​÷ 6       8     210      210  ​÷ 70      3
    Hence, 32:48::70:210 are not in proportion.


 (vi) 150, 200, 250, 300
       150   =   150  ​÷ 50   =  3;   250  =  250  ​÷ 50  =   5  
      200         200  ​÷ 50        4   300      300  ​÷ 50       6
    Hence, 150:200::250:300 are not in proportion.

Page No 155:

Question 2:

Verify the following:
(i) 60 : 105 : : 84 : 147
(ii) 91 : 104 : : 119 : 136
(iii) 108 : 72 : : 129 : 86
(iv) 39 : 65 : : 141 : 235

ANSWER:

(i) 60:105::84:147
      60   =   60  ​÷ 15   =   4         (H.C.F. of 60 and 105 is 15.)
     105       105  ​÷ 15       7
       84   =   84  ​÷ 21  =   4         (H.C.F. of 84 and 147 is 21.)
      147     147  ​÷ 21        7
    Hence, 60:105::84:147 are in proportion.
(ii) 91:104::119:136
      91   =  91  ​÷ 13   =   7         (H.C.F. of 91 and 104 is 13.)
     104       104  ​÷ 13      8  
      119  =   119  ​÷ 17  =   7      (H.C.F. of 11 and 136 is 17.)
     136        136  ​÷ 17       8
  Hence, 91:104::119:136 are in proportion.
(iii) 108:72::129:86
      108  =   108  ​÷ 36   =   3        (H.C.F. of 108 and 72 is 36.)
      72           72 ​ ​÷ 36        2
       129   =   ​129  ​÷  43  =  3       (H.C.F. of 129 and 86 is 43.)
        86          86   ​÷ 43        2
   Hence, 108:72::129:86 are in proportion.
(iv) 39:65::141:235
       39   =   39  ​÷ 13  =   3        (H.C.F. of 39 and 65 is 13.)
       65         65  ​÷ 13      5
      141   =   141  ​÷  47   =   3     (H.C.F. of 141 and 235 is 47.)
      235         235  ​÷ 47        5
   Hence, 39:65::141:235 are in proportion.

Page No 155:

Question 3:

Find the value of x in each of the following proportions:
(i) 55 : 11 : : x : 6
(ii) 27 : x : : 63 : 84
(iii) 51 : 85 : : 57 : x
(iv) x : 92 : : 87 : 116

ANSWER:

(i) 55:11::x:6
      Product of extremes = Product of means
                          55 × 6 = 11 × x
⇒                            11x = 330
⇒                               x =  330   = 30
                                          11
(ii) 27:x::63:84
      Product of extremes = Product of means
                      27 ​× 84 = x ​× 63
 ⇒                         63x = 2268
 ⇒                             x =  2268  = 36
                                           63
(iii) 51:85::57:x 
      Product of extremes = Product of means
                           51 × x = 85 × 57
 ⇒                           51x = 4845
 ⇒                               x =   4845   = 95
                                             51
(iv) x:92::87:116
     Product of extremes = Product of means
                       x ×  116 = 92 ​× 87
 ⇒                      116x = 8004
 ⇒                           x  =   8004  = 69
                                       116

Page No 155:

Question 4:

Write (T) for true and (F) for false in case of each of the following:
(i) 51 : 68 : : 85 : 102
(ii) 36 : 45 : : 80 : 100
(iii) 30 bags : 18 bags : : Rs 450 : Rs 270
(iv) 81 kg : 45 kg : : 18 men : 10 men
(v) 45 km : 60 km : : 12 h : 15 h
(vi) 32 kg : Rs 36 : : 8 kg : Rs 9

ANSWER:

(i) 51:68::85:102
   Product of means = 68 × 85 = 5780
   Product of extremes = 51 × 102 = 5202
   Product of means ≠ Product of extremes
    Hence, (F).
(ii) 36:45::80:100
  Product of means = 45 ​× 80 = 3600
  Product of extremes = 36 × 100 = 3600
  Product of means = Product of extremes 
   Hence, (T).
(iii) 30 bags:18 bags::Rs 450:Rs 270
       or 30:18::450:270
     Product of means = 18 × 450 = 8100
     Product of extremes = 30 ​× 270 = 8100
     Product of means = Product of extremes 
     Hence, (T).
(iv) 81 kg:45 kg::18 men:10 men
      or 81:45::18:10
     Product of means = 45 × 18 = 810
     Product of extremes = 81 × 10 = 810
     Product of means = Product of extremes
      Hence, (T).
(v) 45 km:60 km::12 h:15 h
     or,45:60::12:15
     Product of means = 60 × 12 = 720
     Product of extremes = 45 × 15 = 675
     Product of means ≠ Product of extremes 
      Hence, (F).
(vi) 32 kg:Rs 36::8 kg:Rs 9
     Product of means = 36 × 8 = 288
     Product of extremes = 32 × 9 = 288
     Product of means = Product of extremes
     Hence, (T).
    

Page No 155:

Question 5:

Determine if the following ratios form a proportion:
(i) 25 cm : 1 m and Rs 40 : Rs 160
(ii) 39 litres : 65 litres and 6 bottles : 10 bottles
(iii) 200 mL : 2.5 L and Rs 4 : Rs 50
(iv) 2 kg : 80 kg and 25 g : 625 kg

ANSWER:

(i) 25 cm:1 m and Rs 40:Rs 160 (or) 25 cm:100 cm and Rs 40:Rs 160
      25  =  25 ​÷ 25  =  1  and  40  =  40 ÷ 40  =  1
    100      100 ​​÷ 25     4         160    160 ​÷ 40      4 
       Hence, they are in proportion.

(ii) 39 litres:65 litres and 6 bottles:10 bottles
       39   =  39 ​÷ 13   =   3    and   6   =   6 ​÷ 2   =   3 
       65        65 ​​÷ 13       5            10      10 ​÷ 2        5
      Hence they are  in proportion.

(iii) 200 mL:2.5 L and Rs 4:Rs 50 (or) 200 mL:2500 mL and Rs 4:Rs 50
        200   =   2   and   4   =   4 ​÷ 2   =    2  
       2500      25           50       50 ÷ 2       25
     Hence, they are in proportion.

(iv) 2 kg:80 kg and 25 g:625 kg  (or)  2 kg:80 kg and 25 g:625000 g
        2   =  2 ​÷ 2    =   1    and   25      =   25 ​÷ 25     =     1   
       80      80 ​÷ 2       40         625000    625000 ​​÷ 25     25000
        Hence, they are not in proportion.

Page No 155:

Question 6:

In a proportion, the 1st, 2nd and 4th terms are 51, 68 and 108 respectively. Find the 3rd term.

ANSWER:

Let the 3rd term be x.
Thus, 51:68::x:108
 We know:
                      Product of extremes = Product of means
                               51 × 108 = 68 × x
             ⇒                     5508 = 68x
             ⇒                          x =  5508  = 81
                                                   68
Hence, the third term is 81.

Page No 155:

Question 7:

The 1st, 3rd and 4th terms of a proportion are 12, 8 and 14 respectively. Find the 2nd term.

ANSWER:

Let the second term be x.
Then. 12:x::8:14
We know:
  Product of extremes = Product of means
                                 12 × 14 = 8x
            ⇒                       168 = 8x
​            ⇒                           x =   168  = 21
                                                    8
 Hence, the second term is 21.

Page No 155:

Question 8:

Show that the following numbers are in continued proportion:
(i) 48, 60, 75
(ii) 36, 90, 225
(iii) 16, 84, 441

ANSWER:

(i) 48:60, 60:75
      Product of means = 60 × 60 = 3600
      Product of extremes = 48 × 75 = 3600
Product of means = Product of extremes
       Hence, 48:60::60:75 are in continued proportion.

(ii) 36:90, 90:225
     Product of means = 90 × 90 = 8100
     Product of extremes = 36 × 225 = 8100
Product of means = Product of extremes
      Hence, 36:90::90:225 are in continued proportion.

(iii) 16:84, 84:441
    Product of means = 84 × 84 = 7056
    Product of extremes = 16 × 441 = 7056
Product of means = Product of extremes
    Hence, 16:84::84:441 are in continued proportion.
                             

Page No 155:

Question 9:

If 9, x, x 49 are in proportion, find the value of x.

ANSWER:

Given: 9:x::x:49
We know:
  Product of means = Product of extremes
                                     x × x = 9 × 49
                       ⇒               x² = 441
                       ⇒               x² = (21)²
                       ⇒                x = 21

Page No 155:

Question 10:

An electric pole casts a shadow of length 20 m at a time when a tree 6 m high casts a shadow of length 8 m. Find the height of the pole.

ANSWER:

Let the height of the pole = x m
Then, we have:
      x:20::6:8
Now, we know:
        Product of extremes = Product of means
                                   8x = 20​ × 6
                                     x =  120  = 15
                                              8
​Hence, the height of the pole is 15 m.

Page No 155:

Question 11:

Find the value of x if 5 : 3 : : x : 6.

ANSWER:

5:3::x:6
We know:
   Product of means = Product of extremes
                               3x = 5 ​× 6
                            ⇒ x =  30  = 10
                                        3
∴∴ x = 10

Page No 157:

Exercise 10C

Question 1:

If the cost of 14 m of cloth is Rs 1890, find the cost of 6 m of cloth.

ANSWER:

Cost of 14 m of cloth = Rs 1890
Cost of 1 m of cloth =  1890  = Rs 135
                                     14
Cost of 6 m of cloth = 6​ × 135 = Rs 810

Page No 157:

Question 2:

If the cost of a dozen soaps is Rs 285.60, what wil be the cost of 15 such soaps?

ANSWER:

Cost of dozen soaps = Rs 285.60
Cost of 1 soap =  285.60 
                            12
Cost of 15 soaps = 15​ ×  285.60  =  4284  = Rs 357
                                         12            12

Page No 157:

Question 3:

If 9 kg of rice costs Rs 327.60, what will be the cost of 50 kg of rice?

ANSWER:

Cost of 9 kg of rice = Rs 327.60
Cost of 1 kg of rice =  327.60  
                                     9 
Cost of 50 kg of rice = 50​ ×  327.60  =  16380  = Rs 1820
                                               9              9
Hence, the cost of 50 kg of rice is Rs 1820.

Page No 157:

Question 4:

If 22.5 m of a uniform iron rod weighs 85.5 kg, what will be the weight of 5 m of the  same rod?

ANSWER:

Weight of 22.5 m of uniform iron rod = 85.5 kg
Weight of 1 m of uniform iron rod =  85.5  kg
                                                        22.5
Weight of 5 m of uniform iron rod = 5​ ×  85.5  =  427.5  = 19 kg
                                                              22.5       22.5
Thus, the weight of 5 m of iron rod is 19 kg.

Page No 157:

Question 5:

If 15 tins of the same size contain 234 kg of oil, how much oil will there be in 10 such tins?

ANSWER:

Oil contained by 15 tins = 234 kg
Oil contained by 1 tin =  234  kg
                                       15
Oil contained by 10 tins = 10 ×  234  =  2340  = 156 kg
                                                  15         15

Page No 157:

Question 6:

If 12 L of diesel is consumed by a car in covering a distance of 222 km, how many kilometres will it go in 22 L of diesel?

ANSWER:

Distance covered by a car in 12 L diesel = 222 km
Distance covered by it in 1 L diesel =  222  km
                                                          12
Distance covered by it in 22 L diesel = 22 ×  222  =  4884  = 407 km
                                                                   12          12

Page No 157:

Question 7:

A transport company charges Rs 540 to carry 25 tonnes of weight. What will it charge to carry 35 tonnes?

ANSWER:

Cost of transporting 25 tonnes of weight = Rs 540
Cost of transporting 1 tone of weight =  540 
                                                             25
Cost of transporting 35 tonnes of weight = 35​ ×  540  =  18900  = Rs 756
                                                                          25          25

Page No 158:

Question 8:

4.5 g of an alloy of copper and zinc contains 3.5 g of copper. What weight of copper will there be in 18.9 g of the alloy?

ANSWER:

Let the weight of copper be x g.
​Then, 4.5:3.5::18.9:x
        Product of extremes = Product of means
              4.5 × x  = 3.5 × 18.9
          ⇒ x =  66.15  = 14.7
                       4.5
So, the weight of copper is 14.7 g.

Page No 158:

Question 9:

35 inland letters cost Rs 87.50. How many such letters can we buy for 315?

ANSWER:

Number of inland letters whose total cost is Rs 87.50 = 35
Number of inland letters of whose cost is Re 1 =   35    
                                                                         87.50
Number of inland letters whose cost is Rs 315 = 315​ ×   35    =  11025  = 126
                                                                                  87.50      87.50
Hence, we can buy 126 inland letters for Rs 315.

Page No 158:

Question 10:

Cost of 4 dozen bananas is Rs 104. How many bananas can be purchased for Rs 6.50?

ANSWER:

Number of bananas that can be purchased for Rs 104 = 48 (4 dozen)
Number of bananas that can be purchased for Re 1 =  48 
                                                                               104
Number of bananas that can be purchased for Rs 6.50 = 6.50 ×  48   =  312  = 3
                                                                                               104       104
Hence, 3 bananas can be purchased for Rs 6.50.

Page No 158:

Question 11:

The cost of 18 chairs is Rs 22770. How many such chairs can be bought for Rs 10120?

ANSWER:

Number of chairs that can be bought for Rs 22770 = 18
Number of chairs that can be bought for Re 1 =    18   
                                                                        22770
Number of chairs that can be bought for Rs 10120 = 10120 ×    18     =  182160  = 8
                                                                                             22770       22770

Page No 158:

Question 12:

A car travels 195 km in 3 hours.
(i) How long will it take to travel 520 km?
(ii) How far will it travel in 7 hours with the same speed?

ANSWER:

(i) Time taken by the car to travel 195 km = 3 hours
   Time taken by it to travel 1 km =   3   hours
                                                      195
  Time taken by it to travel 520 km = 520 ×   3   =  1560  = 8 hours
                                                                  195      195

(ii) Distance covered by the car in 3 hours = 195 km
    Distance covered by it in 1 hour =  195  = 65 km
                                                          3
   Distance covered by it in 7 hours = 7 × 65 = 455 km

Page No 158:

Question 13:

A labourer earns Rs 1980 in 12 days.
(i) How much does he earn in 7 days?
(ii) In how many days will he earn Rs 2640?

ANSWER:

(i) Earning of a labourer in 12 days = Rs 1980
    Earning of the labourer in 1 day =  1980  = Rs 165
                                                        12
    Earning of the labourer in 7 days = 7​ × 165 = Rs 1155
(ii) Number of days taken by the labourer to earn Rs 1980 = 12 days
     Number of days taken by him to earn Re 1 =  12  days
                                                                       1980
     Number of days taken by him to earn Rs 2640 = 2640 ×  12   =  31680  = 16 days
                                                                                         1980      1980

Page No 158:

Question 14:

The weight of 65 books is 13 kg.
(i) What is the weight of 80 such books?
(ii) How many such books weigh 6.4 kg?

ANSWER:

Weight of 65 books = 13 kg
(i) Weight of 1 book =  13  kg
                                   65
    Weight of 80 books = 80 ×  13   =  1040   = 16 kg
                                              65         65

(ii) Number of books weighing 13 kg = 65
    Number of books weighing 1 kg =  65  = 5 
                                                        13
   Number of books weighing 6.4 kg = 6.4 × 5 = 32

Page No 158:

Question 15:

If 48 boxes contain 6000 pens, how many such boxes will be needed for 1875 pens?

ANSWER:

Number of boxes containing 6000 pens = 48
Number of boxes containing 1 pen =   48  
                                                       6000
Number of boxes containing 1875 pens = 1875 ×   48    =   90000  = 15
                                                                          6000        6000
15 boxes are needed for 1875 pens.

Page No 158:

Question 16:

24 workers can build a wall in 15 days. How many days will 9 workers take to build a similar wall?

ANSWER:

Number of days taken by 24 workers to build a wall = 15 days
Number of days taken by 1 worker to build the wall = 15 × 24 = 360 days         (less worker means more days)
Number of days taken by 9 workers to build the wall =  360  = 40 days
                                                                                     9

Page No 158:

Question 17:

40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?

ANSWER:

Number of men required to complete the work in 26 days = 40
Number of men required to complete the work in 1 day = 40 × 26 = 1040 men  (less men more days)
Number of men required to complete the work in 16 days =  1040  = 65
                                                                                                   16

Page No 158:

Question 18:

In an army capm, there were provisions for 550 men for 28 days. But, 700 men attended the camp. How long did the provisions last?

ANSWER:

Number of days the provisions will last for 550 men = 28 days
Number of days the provisions will last for 1 man = 28 × 550 = 15400 days  (less men means more days)
Number of days the provisions will last for 700 men =  15400  = 22 days
                                                                                    700
The provision will last for 22 days. 

Page No 158:

Question 19:

A given quantity of rice is sufficient for 60 persons for 3 days. How many days would the rice last for 18 persons?

ANSWER:

Number of days for which the given quantity of rice is sufficient for 60 persons = 3 days
Number of days for which it is sufficient for 1 person = 3 × 60 = 180 days      (less men means more days )
Number of days for which it is sufficient for 18 persons =  180  = 10 days
                                                                                       18

Page No 158:

Exercise 10D

Question 1:

The ratio 92 : 115 in its simplest for is
(a) 23 : 25
(b) 18 : 23
(c) 3 : 5
(d) 4 : 5

ANSWER:

(d) 4 : 5
92:115 =   92 ​÷ 23   =  4  (As H.C.F. of 92 and 115 is 23.)
                115 ​÷ 23       5

Page No 158:

Question 2:

If 57 : x : : 51 : 85, then the value of x is
(a) 95
(b) 76
(c) 114
(d) none of these

ANSWER:

(a) 95
57:x::51:85
    57  =   51 
     x        85
⇒ x =  57 × 85  
               51
⇒ x =  4845  = 95
            51

Page No 158:

Question 3:

If 25 : 35 : : 45 : x, then the value of x is
(a) 63
(b) 72
(c) 54
(d) none of these

ANSWER:

(a) 63
25:35::45:x
          25  =  45 
          35       x
⇒ x =  35 × 45  =  1575  = 63
              25             25

Page No 158:

Question 4:

If 4 : 5 : : x : 35, then the value of x is
(a) 42
(b) 32
(c) 28
(d) none of these

ANSWER:

(c) 28
4:5::x:35
⇒  4  =  x  
     5      35
⇒ x =  4 × 35  = 4 × 7 = 28
              5

Page No 158:

Question 5:

If a, b, c, d are in proportion, then
(a) ac = bd
(b) ad = bc
(c) ab = cd
(d) none of these

ANSWER:

(b) ad = bc
Given:
a, b, c, d are in proportion.
a:b::c:d
     a  =  c  
     b      d
⇒ ad = bc

Page No 158:

Question 6:

If a, b, c are in proportion, then
(a) a² = bc
(b) b² = ac
(c) c² = ab
(d) none of these

ANSWER:

(b) b² = ac
Given:
a, b, c are in proportion.
a:b::b:c
    Product of means = Product of extremes
⇒​ b^​2 = ac

Page No 158:

Question 7:

Choose the correct statement:
(a) (5 : 8) > (3 : 4)
(b) (5 : 8) < (3 : 4)
(c) two ratios cannot be compared

ANSWER:

(b) (5 : 8) < (3 : 4)

We can write
(5:8) = 58 and (3:4) = 34(5:8) = 58 and (3:4) = 34
Making the denominator equal:
 5   and   3 × 2  =   6       
 8           4 × 2        8
As 6 > 5,   5   <   3    
                 8        4

Page No 159:

Question 8:

If Rs 760 is divided between A and B in the ratio 8 : 11, then B’s share is
(a) Rs 440
(b) Rs 320
(c) Rs 430
(d) Rs 330

ANSWER:

(a) Rs 440
A:B = 8:11
Sum of ratio terms = 8 + 11 = 19
B’s share =  11  × 760 =  8360  = Rs 440
                   19                  19

Page No 159:

Question 9:

Two numbers are in the ratio 5 : 7 and the sum of these numbers is 252. The larger of these numbers is
(a) 85
(b) 119
(c) 105
(d) 147

ANSWER:

(d) 147
Ratio = 5:7
Let x be any number such that we have:
   5x + 7x = 252
⇒ 12x = 252
⇒ x =  252  = 21
           12
Now, 5x = 5 × 21= 105
7x = 7 × 21 = 147

The largest number is 147.

Page No 159:

Question 10:

The sides of a triangle are in the ratio 1 : 3 : 5 and its perimeter is 90 cm. The length of its largest side is
(a) 40 cm
(b) 50 cm
(c) 36 cm
(d) 54 cm

ANSWER:

(b) 50 cm
The sides of the triangle are in the ratio 1:3:5.
Let x be any number such that the sides are 1x cm, 3x cm and 5x cm.
          1x + 3x + 5x = 90
       ⇒ 9x = 90
       ​⇒ x =  90  = 10
                   9
First side = 1x = 1 ​× 10 = 10 cm
Second side = 3x = 3 ​× 10 = 30 cm
Third side = 5x = 5 × 10 = 50 cm
The length of the largest side is 50 cm.

Page No 159:

Question 11:

The ratio of boys and girls in a school is 12 : 5. If the number of girls is 840, the total strength of the school is
(a) 1190
(b) 2380
(c) 2856
(d) 2142

ANSWER:

(c) 2856
Ratio of boys and girls = 12:5
Let x be any number such that the number of boys and girls are 12x and 5x, respectively.
Number of girls = 840
      5x = 840
 ⇒ x =  840  = 168
             5
Number of boys = 12x = 12 × 168 = 2016
Number of girls = 840
Total strength of the school = 2016 + 840 = 2856

Page No 159:

Question 12:

If the cost of 12 pens is Rs 138, then the cost of 14 such pens is
(a) Rs 164
(b) Rs 161
(c) Rs 118.30
(d) Rs 123.50

ANSWER:

(b) Rs 161
Cost of 12 pens = Rs 138
Cost of 1 pen = Rs  138  
                              12
Cost of 14 pens = Rs  138  × 14 = Rs 1932  = Rs 161
                                     12                     12

Page No 159:

Question 13:

If 24 workers can build a wall in 15 days, how many days will 8 workers take to build a similar wall?
(a) 42 days
(b) 45 days
(c) 48 days
(d) none of these

ANSWER:

(b) 45 days
Time taken by 24 workers to build a wall = 15 days
Time taken by 1 worker to build a wall = 24 × 15 = 360 days        (clearly less workers will take more time to build a wall)
Time taken by 8 workers to build a wall =  360  = 45 days
                                                                       8

Page No 159:

Question 14:

If 40 men can finish a piece of work in 26 days, how many men will be required to finish it in 20 days?
(a) 52
(b) 31
(c) 13
(d) 65

ANSWER:

(a) 52
Number of men required to finish the work in 26 days = 40
Number of men required to finish it in 1 day = 40 × 26 = 1040 men          (More men means less days)
Number of men required to finish it in 20 days =  1040  = 52
                                                                                  20

Page No 159:

Question 15:

In covering 111 km, a car consumes 6 L of petrol. How many kilometres will it go in 10 L of petrol?
(a) 172 km
(b) 185 km
(c) 205 km
(d) 266.4 km

ANSWER:

(b) 185 km
Distance covered in 6 L of petrol = 111 km
Distance covered in 1 L of  petrol =  111 km
                                                       6
Distance covered in 10 L of petrol =  111  × 10 =  1110  = 185 km
                                                              6                    6

Page No 159:

Question 16:

In a fort, 550 men had provisions for 28 days. How many days will it last for 700 men?
(a) 22 days
(b) 3571135711 days
(c) 34 days
(d) none of these

ANSWER:

(a) 22 days
Number of days for which 550 men had provisions = 28 days
Number of days for which 1 man had provisions = 28 × 550 = 15400 days (more men means less days)
Number of days for which 700 men had provisions =  15400  = 22 days
                                                                                        700

Page No 159:

Question 17:

The angles of a triangle are in the ratio 3 : 1 : 2. The measure of the largest angle is
(a) 30°
(b) 60°
(c) 90°
(d) 120°

ANSWER:

(c) 90°
Ratio of the angles of a triangle is 3:1: 2
Let x be any number such that the three angles are (3x)°°, (1x)°° and (2x)°°.
We know, the sum of the angles of a triangle is 180°°.
        3x + 1x + 2x = 180
      ⇒ 6x = 180
​      ⇒ x =  180  = 30
                  6
∴∴  (3x )°° = (3 ​× 30)°° = 90^o
​     (1x)°° = (1​ × 30)°° = 30^o
​     (2x)°° = (2 × 30)°° = 60^o
The measure of the largest angle is 90^o​.

Page No 159:

Question 18:

Length and breadth of a rectangular field are in the ratio 5 : 4. If the width of the field is 36 m, what is its length?
(a) 40 m
(b) 45 m
(c) 54 m
(d) 50 m

ANSWER:

(b) 45 m
Length:Breadth = 5:4
Let x be any number such that the length and the breadth are 5x and 4x, respectively.
Now , 4x = 36
            x =  36  = 9
                    4
Length = 5x = 5 × 9 = 45 m

Page No 159:

Question 19:

If a bus covers 195 km in 3 hours and a train covers 300 km in 4 hours, then the ratio of their speeds is
(a) 13 : 15
(b) 15 : 13
(c) 13 : 12
(d) 12 : 13

ANSWER:

(a) 13 : 15

Speed =  Distance  
                Time 
Speed of the bus =  195 km = 65 km/hr
                                  3 hr
Speed of the train =  300 km  = 75 km/hr
                                    4 hr
Ratio =  65  =  65 ÷ 5  =  13  = 13:15
             75       75 ÷ 5      15

Page No 159:

Question 20:

If the cost of 5 bars of soap is Rs 82.50, then the cost of one dozen such bars is
(a) Rs 208
(b) Rs 192
(c) Rs 198
(d) Rs 204

ANSWER:

(c) Rs 198
Cost of 5 bars of soap = Rs 82.50
Cost of 1 bar of soap =  82.50  = Rs 16.5
                                         5
Cost of 12 (1 dozen) bars of soap = 16.5 × 12 = Rs 198

Page No 159:

Question 21:

If the cost of 30 packets of 8 pencils each is Rs 600, what is the cost of 25 packets of 12 pencils each?
(a) Rs 725
(b) Rs 750
(c) Rs 480
(d) Rs 720

ANSWER:

(b) Rs 750
Cost of 30 packets of 8 pencils each = Rs 600
Cost of 1 packet of 8 pencils =  600   = Rs 20
                                                 30
Cost of  1 pencil = Rs  20   
                                      8
Cost of 1 packet of 12 pencils = 12​ ×  20  =  240  = Rs 30
                                                              8         8
Cost of 25 packets of 12 pencils each = 25 × 30 = Rs 750

Page No 159:

Question 22:

A rail journey of 75 km costs Rs 215. How much will a journey of 120 km cost?
(a) Rs 344
(b) Rs 324
(c) Rs 268.75
(d) none of these

ANSWER:

(a) Rs 344
Cost of rail journey of 75 km = Rs 215
Cost of rail journey of 1 km = Rs  215 
                                                   75
Cost of rail journey of 120 km = 120​ ×  215   = 25800 = Rs 344
                                                              75          75

Page No 159:

Question 23:

The 1st, 2nd and 4th terms of a proportion are 12, 21 and 14 respectively. Its third term is
(a) 16
(b) 18
(c) 21
(d) 8

ANSWER:

(d) 8
Let the third term be x.
Then, we have:
12:21::x:14
We know:
    Product of means = Product of extremes
      21x = 12 × 14
  ⇒ 21x = 168
  ⇒ x =  168  = 8
              21
The third term is 8

Page No 159:

Question 24:

10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?
(a) 9 h 36 min
(b) 15 h
(c) 6 h 40 min
(d) 13 h 20 min

ANSWER:

(b) 15 h
Time taken by 10 boys to dig a pitch = 12 hours
Time taken by 1 boy to dig a pitch = 12 × 10 = 120 hours        (less boys means more time)
Time taken by 8 boys to dig a pitch =  120 = 15 hours
                                                                8

Page No 161:

Exercise 10E

Question 1:

Find the ratio of:
(a) 90 cm to 1.05 m
(b) 35 minutes to an hour
(c) 150 mL to 2 L
(d) 2 dozens to a score

ANSWER:

(a) 90 cm:1.05 m (or) 90 cm:105 cm             (1 m = 100 cm)
        90   =  90 ​÷ 15  =   6         (H.C.F. of 90 and 105 is 15.) 
       105      105 ​÷ 15      7
     ∴∴ 6:7

(b) 35 minutes to an hour (or) 35 minutes:60 minutes      (1 hour = 60 minutes)
        35  =  35 ​÷ 5  =  7       (H.C.F. of 35 and 60 is 5.)
        60       60 ​÷ 5     12  
   ∴∴  7:12

(c) 150 mL to 2 L (or) 150 L:2000 L                (1 L= 1000 mL)
         150  =   150 ​÷ 50  =   3    (HCF of 150 and 2000 is 50)
        2000      2000​ ​÷50       40
    ∴∴ 3:40

(d) 2 dozens to a score (or) 24:20       (1 dozen = 12 and 1 score = 20)
        24  =   24 ​÷ 4  =  6     (H.C.F. of 24 and 20 is 4)
        20        20​ ​÷ 4      5
   ∴∴ 6:5

Page No 161:

Question 2:

The ratio of zinc and copper in an alloy is 7 : 9. If the weight of copper in the alloy is 12.6 kg, find the weight of zinc in it.

ANSWER:

Ratio of zinc and copper in an alloy is = 7:9
Let the weight of zinc and copper in it be (7x) and (9x), respectively.
Now, the weight of a copper = 12.6 kg   (given)
∴ 9x = 12.6
⇒  x = 12.6  = 1.4
             9
∴ Weight of zinc = 7x = 7​ × 1.4 = 9.8 kg

Page No 161:

Question 3:

Divide Rs 1400 among A. B and C in the ratio 2 : 3 : 5.

ANSWER:

Given:
  A:B:C = 2:3:5
Sum of ratio = 2 + 3 + 5 = 10
Total money = Rs 1400
Then, share of A =   2   × Rs 1400 = Rs  2800  = Rs 280
                             10                             10
Share of B =  3  ​× Rs 1400 = Rs  4200  = Rs 420
                    10                             10
Share of C =  5  ​× Rs 1400 = Rs  7000  = Rs 700
                    10                            10

Page No 161:

Question 4:

Prove that (5 : 6) > (3 : 4).

ANSWER:

We can write:
(5:6) = 56 and (3:4) = 34(5:6) = 56 and (3:4) = 34
By making their denominators same: (Taking the L.C.M. of 6 and 4, which is 24.)
Consider, 5:6
         5 ​× 4  =  20 
         6 ​× 4      24
              
And,  3 ​× 6  =  18 
         4 ​× 6       24
As 20 > 18
Clearly, (5:6) > (3:4)

Page No 161:

Question 5:

40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?

ANSWER:

Number of men needed to finish a piece of work in 26 days = 40
Number of men needed to finish it in 1 day = 26 × 40 = 1040    (less days means more men)
Number of men needed to finish it in 16 days =  1040  = 65
                                                                          16 

Page No 161:

Question 6:

In an army camp, there were provisions for 425 men for 30 days. How long did the provisions last for 375 men?

ANSWER:

Number of days for which provisions last for 425 men = 30 days
Number of days for which provisions last for 1 men = 30 × 425 = 12750 days. (less men means more days)
Number of days for which provisions last for 375 men = 12750  = 34 days
                                                                                   375
Hence, provisions will last for 34 days for 375 men.

Page No 161:

Question 7:

Find the value of x when 36 : x : : x : 16.

ANSWER:

Given:
36:x::x:16
We know:
Product of means = Product of extremes 
  x × x = 36 × 16
⇒ x² = 576
⇒ x² = 24²
⇒ x = 24

Page No 161:

Question 8:

Show that 48, 60, 75 are in continued proportion.

ANSWER:

Consider 48:60::60:75

Product of means = 60 × 60 = 3600
Product of extremes = 48 × 75 = 3600
So product of means = Product of extremes
Hence, 48, 60, 75 are in continued proportion.

Page No 161:

Question 9:

Two numbers are in the ratio 3 : 5 and their sum is 96. The larger number is
(a) 36
(b) 42
(c) 60
(d) 70

ANSWER:

(c) 60
Ratio = 3:5
Let x be any number such that we have:
    3x + 5x = 96
 ⇒ 8x = 96
 ⇒ x =  96  = 12
             8
The numbers are:
     3x = 3 ​× 12 = 36
     5x = 5 ​× 12 = 60
The largest number = 60

Page No 161:

Question 10:

A car travels 288 km is 4 hours and a train travels 540 km in 6 hours. The ratio of their speeds is
(a) 5 : 4
(b) 4 : 5
(c) 5 : 6
(d) 3 : 5

ANSWER:

(b) 4 : 5
Speed of the car =  Distance  =  288 km  = 72 km/hr
                                 Time            4 hr
 
Speed of the train =  Distance  =  540 km  = 90 km/hr
                                   Time             6 hr
 
Ratio of their speeds = 72:90
where, 72  = 72 ​÷ 18  =  4      (H.C.F. of 72 and 90 is 18.)
           90      90 ​÷ 18      5

Page No 161:

Question 11:

The first three terms of a proportion are 12, 21 and 8 respectively. The 4th term is
(a) 18
(b) 16
(c) 14
(d) 20

ANSWER:

(c) 14
Let the 4th term be x, such that we have:
     12:21::8:x
 Now, we know:
     Product of extremes = Product of means
                                    12x = 21 × 8 
                                       x =  168  = 14
                                               12

Page No 161:

Question 12:

The ratio 92 : 115 in simplest form is
(a) 23 : 25
(b) 18 : 23
(c) 3 : 5
(d) 4 : 5

ANSWER:

(d) 4 : 5
92:115
 92  =  92 ​÷ 23   =  4         (H.C.F. of 92 and 115 is 23)
115     115 ​÷ 23      5

Page No 161:

Question 13:

If 57 : x : : 51 : 85, then the value of x is
(a) 95
(b) 76
(c) 114
(d) none of these

ANSWER:

(a) 95
Given :  
57:x::51:85
We know:
Product of means = Product of extremes
                                      51x = 57 × 85
                                          x =   4845  = 95
                                                     51

Page No 161:

Question 14:

If 4 : 5 : : x : 45, then the value of x is
(a) 54
(b) 60
(c) 36
(d) 30

ANSWER:

(c) 36
Given:
4:5::x:45
We know:
Product of mean = Product of extremes
                                      5x = 4 ​× 45
                                       x =    180   = 36
                                                  5

Page No 161:

Question 15:

If a, b, c are in proportion, then
(a) a² = bc
(b) b² = ac
(c) c² = ab
(d) none of these

ANSWER:

(b) b² = ac
Given:
a, b, c are in proportion, such that we have:
  a:b::b:c
Now, we know:
  Product of means = Product of extremes
                         b ​× b = a ​× c
                              b² = ac

Page No 161:

Question 16:

10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?
(a) 9 hrs 36 min
(b) 15 hrs
(c) 6 hrs 40 min
(d) 13 hrs 10 min

ANSWER:

(b) 15 hrs
Time taken by 10 boys to dig a pitch = 12 hours
Time taken by 1 boy to dig a pitch = 12 × 10 = 120 hours   (Less boys would take more hours.)
Time taken by 8 boys to dig a pitch =  120  = 15 hours
                                                               8

Page No 161:

Question 17:

In covering 148 km, a car consumes 8 litres of petrol. How many kilometres will it go in 10 litres of petrol?
(a) 172 km
(b) 185 km
(c) 205 km
(d) 266.4 km

ANSWER:

 (b) 185 km
Distance covered by a car in 8 litres of petrol = 148 km
Distance covered by it in 1 litre of petrol =  148  km
                                                                    8 
Distance covered by it in 10 litres of petrol = 10 × 148 = 1480 = 185 km
                                                                              8         8

Page No 161:

Question 18:

Fill in the blanks.
(i) 1421=   3=6   1421=   3=6   
(ii) 90 cm : 1.5 m = …… .
(iii) If 36 : 81 : : x : 63, then x = …… .
(iv) If 25, 35, x are in proportion, then x = …… .
(v) If 9, x, x, 49 are in proportion, then x = …… .

ANSWER:

(i)
           Let 1421 = x3Thus, we have: 21x = 14 × 3 ⇒ x = 14 × 321 = 2∴ 1421 = 23Again,  let 23=6yThus, we have: 2y = 6 × 3 ⇒ y = 6 × 32 = 9∴ 23 = 69∴  1421 = 23 = 69Let 1421 = x3Thus, we have: 21x = 14 × 3 ⇒ x = 14 × 321 = 2∴ 1421 = 23Again,  let 23=6yThus, we have: 2y = 6 × 3 ⇒ y = 6 × 32 = 9∴ 23 = 69∴  1421 = 23 = 69

(ii) 90 cm:1.5 m (or) 90 cm:150 cm          (1 m = 100 cm)
      90  =  9  =  9 ÷ 3  =  3              (H.C.F. of 9 and 15 is 3.)
     150     15    15 ​​÷ 3      5
  
(iii) If 36:81::x:63
      Product of means = Product of extremes
                          81x = 36 × 63
                             x =  2268 
                                      81
                             x = 28

(iv) Given:
      25, 35, x are in proportion.
       25:35::35:x
     Now, we know:
     Product of extremes = Product of means
                          25 × x = 35 ​× 35
                            25x = 1225
                              x =  1225  = 49
                                       25

(v) Given:
     9, x, x, 49 are in proportion.
         9:x::x:49
     Now, we know:
     Product of means = Product of extremes
                             x ​× x = 9 ​× 49
                                   x² = 441
                                   x² = 21²
                                    x = 21

Page No 162:

Question 19:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) 30, 40, 45, 60 are in proportion.
(ii) 6 : 8 and 9 : 12 are equivalent ratios of 3 : 4.
(iii) a dozen : a score = 5 : 3.
(iv) 60 p : Rs 3 = 1 : 5.

ANSWER:

(i) 30, 40, 45, 60 
      30  =    3 ,    45  =   45 ​÷ 15  =     3      They are in proportion.
      40        4      60        60 ​÷ 15         4
  Hence, true.

(ii)  6  =  6 ​÷ 2  =  3 ,  9  =  9 ​÷ 3  =  3     Hence, they are equivalent to 3:4.
      8       8 ​÷ 2      4    2      12 ​÷ 3     4
  Hence, true.
(iii) 1 dozen:1 score = 12:20
      12  =  12 ​÷ 4  =  3       
      20      20 ​÷ 4       5
Hence, false.
(iv) 60p:Rs 3 = 60p:300p                        (1 Re = 100 p)
       60  =  6  =  6 ​÷ 6  =  1 
      300     30    30 ​÷ 6      5

 Hence, true.

Exercise 8A

Page No 130:

Question 1:

Write the following using literals, numbers and signs of basic operations:
(i) x increased by 12
(ii) y decreased by 7
(iii) The difference of a and b, when a > b
(iv) The product of x and y added to their sum
(v) One-third of x multiplied by the sum of a and b
(vi) 5 times x added to 7 times y
(vii) Sum of x and the quotient of y by 5
(viii) x taken away from 4
(ix) 2 less than the quotient of x by y
(x) x multiplied by itself
(xi) Twice x increased by y
(xii) Thrice x added to y squared
(xiii) x minus twice y
(xiv) x cubed less than y cubed
(xv) The quotient of x by 8 is multiplied by y

ANSWER:

(i) x increased by 12 is (x + 12).
(ii) y decreased by 7 is (y – 7).
(iii) The difference of a and b, when a>b is (a – b).
(iv) The product of x and y is xy.
The sum of x and y is (x + y).
      So, product of x and y added to their sum is xy + (x + y).
(v) One third of x is x3x3.
The sum of a and b is (a + b).
      ∴∴ One-third of x multiplied by the sum of a and b = x3×(a+b)=x(a+b)3 x3×(a+b)=x(a+b)3 
(vi) 5 times x added to 7 times y = (5×x)+(7×y),  which is equal to 5x+7y.(5×x)+(7×y),  which is equal to 5x+7y.

(vii) Sum of x and the quotient of y by 5 is x+y5x+y5.
(viii) x taken away from 4 is (4-x).
(ix) 2 less than the quotient of x by y is xy−2xy-2.
(x) x multiplied by itself is x×x=x2x×x=x2.
(xi) Twice x increased by y is (2×x)+y = 2x+y(2×x)+y = 2x+y.
(xii) Thrice x added to y squared is (3×x)+(y×y)=3x+y2(3×x)+(y×y)=3x+y2.
(xiii) x minus twice y is x−(2×y)=x−2yx-(2×y)=x-2y.
(xiv) x cubed less than y cubed is (y×y×y)−(x×x×x)=y3−x3.(y×y×y)-(x×x×x)=y3-x3.
(xv) The quotient of x by 8 is multiplied by y is x8×y=xy8x8×y=xy8.

Page No 130:

Question 2:

Ranjit scores 80 marks in English and x marks in Hindi. What is his total score in the two subjects?

ANSWER:

Ranjit’s score in English = 80 marks
Ranjit’s score in Hindi = x marks
Total score in the two subjects = (Ranjit’s score in English + Ranjit’s score in Hindi)
∴ Total score in the two subjects = (80 + x) marks

Page No 130:

Question 3:

Write the following in the exponential form:
(i) b × b × b × … 15 times
(ii) y × y × y × … 20 times
(iii) 14 × a × a × a × a × b × b × b
(iv) 6 × x × x × y × y
(v) 3 × z × z × z × y × y × x

ANSWER:

(i) b × b × b × … 15 times = b15b15
(ii) y × y × y × … 20 times = y20y20
(iii) 14 × a × a × a × a × b × b × b = 14×(a×a×a×a)×(b×b×b) =14a4b314×(a×a×a×a)×(b×b×b) =14a4b3
(iv) 6 × x × x × y × y = 6×(x×x)×(y×y)=6x2y26×(x×x)×(y×y)=6x2y2
(v) 3 × z × z × z × y × y × x = 3×(z×z×z)×(y×y)×x=3z3y2x3×(z×z×z)×(y×y)×x=3z3y2x

Page No 130:

Question 4:

Write down the following in the product form:
(i) x²y⁴
(ii) 6y⁵
(iii) 9xy²z
(iv) 10a³b³c³

ANSWER:

(i) x2y4=(x×x)×(y×y×y×y)=x×x×y×y×y×yx2y4=(x×x)×(y×y×y×y)=x×x×y×y×y×y
(ii) 6y5=6×(y×y×y×y×y)=6×y×y×y×y×y6y5=6×(y×y×y×y×y)=6×y×y×y×y×y
(iii) 9xy2z=9×x×(y×y)×z=9×x×y×y×z9xy2z=9×x×(y×y)×z=9×x×y×y×z
(iv) 10a3b3c3=10×(a×a×a)×(b×b×b)×(c×c×c)=10×a×a×a×b×b×b×c×c×c10a3b3c3=10×(a×a×a)×(b×b×b)×(c×c×c)=10×a×a×a×b×b×b×c×c×c

Page No 132:

Question 1:

If a = 2 and b = 3, find the value of
(i) a + b
(ii) a² + ab
(ii) ab − a²
(iv) 2a − 3b
(v) 5a² − 2ab
(vi) a³ − b³

ANSWER:

(i) a+b
    Substituting a = 2 and b = 3 in the given expression:
    2+3 = 5

(ii) a2+aba2+ab
    Substituting a = 2 and b = 3 in the given expression:
    (2)2+(2×3)=4+6=10(2)2+(2×3)=4+6=10

(iii) ab−a2ab-a2
    Substituting a = 2 and b = 3 in the given expression:
     (2×3)−(2)2=6−4=2(2×3)-(2)2=6-4=2

(iv) 2a-3b
    Substituting a = 2 and b = 3 in the given expression:
    (2×2)−(3×3)=4−9=−5(2×2)-(3×3)=4-9=-5

(v) 5a2−2ab5a2-2ab
    Substituting a=2 and b=3 in the given expression:
    5×(2)2−2×2×3=5×4−12=20−12=85×(2)2-2×2×3=5×4-12=20-12=8

(vi) a3−b3a3-b3
    Substituting a=2 and b=3 in the given expression:
    23−33=2×2×2−3×3×3=8−27=−1923-33=2×2×2-3×3×3=8-27=-19

Page No 132:

Question 2:

If x = 1, y = 2 and z = 5, find the value of
(i) 3x − 2y + 4z
(ii) x² + y² + z²
(iii) 2x² − 3y² + z²
(iv) xy + yz − zx
(v) 2x² y − 5yz + xy²
(vi) x³ − y³ − z³

ANSWER:

(i) 3x-2y+4z
    Substituting x = 1, y = 2 and z = 5 in the given expression:
    3×(1)−2×(2)+4×(5)=3−4+20=193×(1)-2×(2)+4×(5)=3-4+20=19

(ii) x2+y2+z2 x2+y2+z2 
    Substituting x = 1, y = 2 and  z = 5 in the given expression:
    12+22+52=(1×1)+(2×2)+(5×5)=1+4+25=3012+22+52=(1×1)+(2×2)+(5×5)=1+4+25=30

(iii) 2×2−3y2+z22x2-3y2+z2
      Substituting x = 1, y = 2 and z = 5 in the given expression:
     2×(1)2−3×(2)2+52=2×(1×1)−3×(2×2)+(5×5)=2−12+25=152×(1)2-3×(2)2+52=2×(1×1)-3×(2×2)+(5×5)=2-12+25=15

(iv) xy+yz−zxxy+yz-zx
     Substituting x = 1, y = 2 and z = 5 in the given expression:
      (1×2)+(2×5)−(5×1)=2+10−5=7(1×2)+(2×5)-(5×1)=2+10-5=7

(v) 2x2y−5yz+xy22x2y-5yz+xy2
      Substituting x = 1, y = 2 and z = 5 in the given expression:
     2×(1)2×2−5×2×5+1×(2)2=4−50+4=−422×(1)2×2-5×2×5+1×(2)2=4-50+4=-42

(vi) x3−y3−z3x3-y3-z3
    Substituting x = 1, y = 2 and z = 5 in the given expression:
    13−23−53=(1×1×1)−(2×2×2)−(5×5×5)=1−8−125=−13213-23-53=(1×1×1)-(2×2×2)-(5×5×5)=1-8-125=-132

Page No 132:

Question 3:

If p = −2, q = −1 and r = 3, find the value of
(i) p² + q² − r²
(ii) 2p² − q² + 3r²
(iii) p − q − r
(iv) p³ + q³ + r³ + 3pqr
(v) 3p²q + 5pq² + 2pqr
(vi) p⁴ + q⁴ − r⁴

ANSWER:

(i) p2+q2−r2p2+q2-r2
     Substituting p = -2, q = -1 and r = 3 in the given expression:
    (−2)2+(−1)2−(3)2=(−2×−2)+(−1×−1)−(3×3)⇒4+1−9=−4(-2)2+(-1)2-(3)2=(-2×-2)+(-1×-1)-(3×3)⇒4+1-9=-4

(ii) 2p2−q2+3r22p2-q2+3r2
     Substituting p = -2, q = -1 and r = 3 in the given expression:
   
2×(−2)2−(−1)2+3×(3)2=2×(−2×−2)−(−1×−1)+3×(3×3)⇒8−1+27=342×(-2)2-(-1)2+3×(3)2=2×(-2×-2)-(-1×-1)+3×(3×3)⇒8-1+27=34

(iii) p−q−rp-q-r
      Substituting p = -2, q = -1 and r = 3 in the given expression:
     (−2)−(−1)−(3)=−2+1−3=−4(-2)-(-1)-(3)=-2+1-3=-4

(iv) p3+q3+r3+3pqrp3+q3+r3+3pqr
    Substituting p = -2, q = -1 and r = 3 in the given expression:

   (−2)3+(−1)3+(3)3+3×(−2×−1×3)=(−2×−2×−2)+(−1×−1×−1)+(3×3×3)+3×(6)=(−8)+(−1)+(27)+18=36(-2)3+(-1)3+(3)3+3×(-2×-1×3)=(-2×-2×-2)+(-1×-1×-1)+(3×3×3)+3×(6)=(-8)+(-1)+(27)+18=36
   
(v) 3p2q+5pq2+2pqr3p2q+5pq2+2pqr
     Substituting p = -2, q = -1 and r = 3 in the given expression:

     3×(−2)2×(−1)+5×(−2)×(−1)2+2×(−2×−1×3)=3×(−2×−2)×(−1)+5×(−2)×(−1×−1)+2×(−2×−1×3)=−12−10+12=−103×(-2)2×(-1)+5×(-2)×(-1)2+2×(-2×-1×3)=3×(-2×-2)×(-1)+5×(-2)×(-1×-1)+2×(-2×-1×3)=-12-10+12=-10

(vi) p4+q4−r4p4+q4-r4
     Substituting p = -2, q = -1 and r = 3 in the given expression:
    (−2)4+(−1)4−(3)4=(−2×−2×−2×−2)+(−1×−1×−1×−1)−(3×3×3×3)=16+1−81=−64(-2)4+(-1)4-(3)4=(-2×-2×-2×-2)+(-1×-1×-1×-1)-(3×3×3×3)=16+1-81=-64
  

Page No 132:

Question 4:

Write the coefficient of
(i) x in 13x
(ii) y in −5y
(iii) a in 6ab
(iv) z in −7xz
(v) p in −2pqr
(vi) y² in 8xy²z
(vii) x³ in x³
(viii) x² in −x²

ANSWER:

(i) Coefficient of x in 13x is 13.
(ii) Coefficient of y in -5y is -5.
(iii) Coefficient of a in 6ab is 6b.
(iv) Coefficient of z in -7xz is -7x.
(v) Coefficient of p in -2pqr is -2qr.
(vi) Coefficient of y² in 8xy²z is 8xz.
(vii) Coefficient of x³ in  x³ is 1.
(viii) Coefficient of x² in -x² is -1.

Page No 132:

Question 5:

Write the numerical coefficient of
(i) ab
(ii) −6bc
(iii) 7xyz
(iv) −2x³y²z

ANSWER:

(i) Numerical coefficient of ab is 1.
(ii) Numerical coefficient of -6bc is -6.
(iii) Numerical coefficient of 7xyz is 7.
(iv) Numerical coefficient of −2x³y²z is -2.

Page No 132:

Question 6:

Write the constant term of
(i) 3x² + 5x + 8
(ii) 2x² − 9
(iii) 4y2−5y+354y2-5y+35
(iv) z3−2z2+z−83z3-2z2+z-83

ANSWER:

A term of expression having no literal factors is called a constant term.
(i) In the expression 3x² + 5x + 8, the constant term is 8.
(ii) In the expression 2x² − 9, the constant term is -9.
(iii) In the expression  4y2−5y+354y2−5y+35, the constant term is 3535.
(iv) In the expression z3−2z2+z−83z3−2z2+z−83 , the constant term is −83-83.

Page No 132:

Question 7:

Identify the monomials, binomials and trinomials in the following:
(i) −2xyz
(ii) 5 + 7x³y³z³
(iii) −5x³
(iv) a + b − 2c
(v) xy + yz − zx
(vi) x⁵
(vii) ax³ + bx² + cx + d
(viii) −14
(ix) 2x + 1

ANSWER:

The expressions given in (i), (iii), (vi) and (viii) contain only one term. So, each one of them is monomial.
The expressions given in (ii) and (ix) contain two terms. So, both of them are binomial.
The expressions given in (iv) and (v) contain  three terms. So, both of them are trinomial.
The expression given in (vii) contains four terms. So, it does not represents any of the given types.

Page No 133:

Question 8:

Write all the terms of the algebraic expressions:
(i) 4x⁵ − 6y⁴ + 7x²y − 9
(ii) 9x³ − 5z⁴ + 7z³ y − xyz

ANSWER:

(i) Expression 4x⁵ − 6y⁴ + 7x²y − 9 has four terms, namely 4x⁵ ,-6y⁴ , 7x²y and -9.
(ii) Expression 9x³ − 5z⁴ + 7z³ y − xyz has four terms, namely 9x³ , -5z⁴ , 7z³ y and -xyz.

Page No 133:

Question 9:

Identify the like terms in the following:
(i) a² , b², −2a², c², 4a
(ii) 3x, 4xy, −yz,12zy3x, 4xy, -yz,12zy
(iii) −2xy², x²y, 5y²x, x²z
(iv) abc, ab²c, acb², c²ab, b²ac, a²bc, cab²

ANSWER:

The terms that have same literals are called like terms.
(i)  a² and 2a² are like terms.
(ii) −yz and 12zy-yz and 12zy are like terms.
(iii) −2xy² and 5y²x are like terms.
(iv) ab²c , acb² , b²ac and cab² are like terms.

Page No 134:

Exercise 8B

Question 1:

Add:
(i) 3x, 7x
(ii) 7y, −9y
(iii) 2xy, 5xy, −xy
(iv) 3x, 2y
(v) 2x², − 3x², 7x²
(vi) 7xyz, − 5xyz, 9xyz, −8xyz
(vii) 6a³, − 4a³, 10a³, −8a³
(viii) x² − a², −5x² + 2a², −4x² + 4a²

ANSWER:

  (i) Required sum = 3x + 7x
     = (3+7)x = 10x

(ii) Required sum = 7y +(−9y)
     = (7-9)y = -2y

(iii) Required sum = 2xy +5xy + (−xy)
    = (2+5-1)xy = 6xy

(iv) Required sum = 3x+2y

(v) Required sum = 2x² + (− 3x²) + 7x²
    =(2-3+7)x² = 6x²

(vi)Required sum =  7xyz + (− 5xyz) + 9xyz + (−8xyz)
       = (7-5+9-8)xyz = 3xyz

(vii) Required sum = 6a³ +(− 4a³) + 10a³ +( −8a³)
     =(6-4+10-8)a³ = 4a³
 
(viii) Required sum = x² − a² + (−5x² + 2a²) +( −4x² + 4a² )
       Rearranging and collecting the like terms =  x² -5x² -4x² -a² + 2a² +4a²
      = (1-5-4)x² +(-1+2+4)a²
     = -8x² + 5a²

Page No 134:

Question 2:

Add the following:
(i)    x − 3y − 2z  5x + 7y − z− 7x − 2y + 4z                            x – 3y – 2z  5x + 7y – z- 7x – 2y + 4z                         

(ii)    m2 − 4m + 5− 2m2 + 6m − 6  − m2 − 2m − 7                               m2 – 4m + 5- 2m2 + 6m – 6  – m2 – 2m – 7                            

(iii)   2×2 − 3xy + y2− 7×2 − 5xy − 2y2  4×2 + xy − 6y2                                     2×2 – 3xy + y2- 7×2 – 5xy – 2y2  4×2 + xy – 6y2                                   

(iv)   4xy − 5yz − 7zx− 5xy + 2yz + zx− 2xy − 3yz + 3zx                               4xy – 5yz – 7zx- 5xy + 2yz + zx- 2xy – 3yz + 3zx                             

ANSWER:

(i)
     x −  3y    −2z  5 x + 7y    −  z−7x − 2y + 4z−x   +2y   +  z     x –  3y    -2z  5 x + 7y    –  z-7x – 2y + 4z-x   +2y   +  z

(ii)
          m2 − 4m + 5   − 2m2 + 6m − 6     − m2 − 2m − 7       −2m2  +0×m−8= −2m2  + 0 −8 = −2m2−8          m2 – 4m + 5   – 2m2 + 6m – 6     – m2 – 2m – 7       -2m2  +0×m-8= -2m2  + 0 -8 = -2m2-8

(iii)
     2×2 − 3xy   + y2− 7×2 − 5xy − 2y2     4×2 + xy  − 6y2                                      −x2−7xy  −7y2     2×2 – 3xy   + y2- 7×2 – 5xy – 2y2     4×2 + xy  – 6y2                                      -x2-7xy  -7y2

(iv)
     4xy − 5yz − 7zx− 5xy + 2yz + zx− 2xy − 3yz + 3zx−3xy   −6yz −3zx     4xy – 5yz – 7zx- 5xy + 2yz + zx- 2xy – 3yz + 3zx-3xy   -6yz -3zx

Page No 134:

Question 3:

Add:
(i) 3a − 2b + 5c, 2a + 5b − 7c, − a − b + c
(ii) 8a − 6ab + 5b, −6a − ab − 8b, −4a + 2ab + 3b
(iii) 2x³ − 3x² + 7x − 8, −5x³ + 2x² − 4x + 1, 3 − 6x + 5x² − x³
(iv) 2x² − 8xy + 7y² − 8xy², 2xy² + 6xy − y² + 3x², 4y² − xy − x² + xy²
(v) x³ + y³ − z³ + 3xyz, − x³ + y³ + z³ − 6xyz, x³ − y³ − z³ − 8xyz
(vi) 2 + x − x² + 6x³, −6 −2x + 4x² −3x³, 2 + x², 3 − x³ + 4x − 2x²

ANSWER:

(i) Sum of the given expressions
 = (3a − 2b + 5c)+(2a + 5b − 7c)+ (− a − b + c)
Rearranging and collecting the like terms
= 3a+2a-a-2b+5b-b+5c-7c+c
= (3+2-1)a + (-2+5-1)b + (5-7+1)c
= 4a+2b-c


(ii) Sum of the given expressions
 = (8a − 6ab + 5b) + (−6a − ab − 8b) + (−4a + 2ab + 3b)
Rearranging and collecting the like terms
  =(8−6−4)a + (- 6 −1+2)ab + (5− 8+ 3)b
  = -2a-5ab+0 = -2a – 5ab

(iii) Sum of the given expressions
    = (2x³ − 3x² + 7x − 8) + (−5x³ + 2x² − 4x + 1) + ( 3 − 6x + 5x² − x³ )
      Rearranging and collecting the like terms
     =2x³−5x³ − x³ − 3x² + 2x² + 5x² +7x-4x-6x-8+1+3
     = (2-5-1)x³ +(-3+2+5)x²+(7-4-6)x-4
     = -4x³ +4x²-3x-4


(iv) Sum of the given expressions
= (2x² − 8xy + 7y² − 8xy²)+( 2xy² + 6xy − y² + 3x²)+( 4y² − xy − x² + xy² )
     Rearranging and collecting the like terms
  = 2x² +3x² − x²   + 7y² − y² +4y² − 8xy + 6xy − xy− 8xy² +2xy² + xy²
  = (2 +3− 1)x²   + (7 − 1 +4)y² + (-8 + 6 −1)xy + (− 8 +2 +1)xy²
 = 4x²   + 10y² − 3xy -5xy²

(v) Sum of the given expressions
 = (x³ + y³ − z³ + 3xyz)+(− x³ + y³ + z³ − 6xyz)+(x³ − y³ − z³ − 8xyz)
      Rearranging and collecting the like terms
   = x³ -x³ + x³ + y³ + y³ − y³ -z³ + z³ − z³ + 3xyz-6xyz-8xyz
  = (1-1+1)x³ + (1+1-1)y³ + (-1+1-1)z³ +(3-6-8)xyz
  = x³ + y³ − z³ -11xyz

(vi) Sum of the given expressions
  = (2 + x − x² + 6x³)+(−6 −2x + 4x² −3x³)+( 2 + x²)+( 3 − x³ + 4x − 2x² )
   Rearranging and collecting the like terms
  = 6x³ −3x³− x³− x² +4x²+ x²− 2x²+ x −2x+ 4x+2-6+2+3
=  (6-3-1)x³+(-1+4+1-2)x²+(1-2+4)x+1
= 2x³+2x²+3x+1

Page No 135:

Question 4:

Subtract:
(i) 5x from 2x
(ii) −xy from 6xy
(iii) 3a from 5b
(iv) −7x from 9y
(v) 10x² from −7x²
(vi) a² − b² from b² − a²

ANSWER:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5x
 Changing the sign of each term of the expression gives -5x.
 On adding:

   2x+(-5x) = 2x-5x
   = (2-5)x
   = -3x

(ii)  Term to be subtracted = -xy
 Changing the sign of each term of the expression gives xy.
 On adding:

   6xy+xy
   = (6+1)xy
   = 7xy

(iii) Term to be subtracted = 3a
 Changing the sign of each term of the expression gives -3a.
 On adding:

   5b+(-3a)
   = 5b-3a

(iv) Term to be subtracted = -7x
 Changing the sign of each term of the expression gives 7x.
On adding:
9y+7x


(v) Term to be subtracted = 10x²
 Changing the sign of each term of the expression gives -10x².
On adding:
−7x² + (-10x²) = −7x² −10x²
   = (−7−10)x²
   = −17x²

(vi) Term to be subtracted = a² − b²
 Changing the sign of each term of the expression gives -a² + b².
On adding:
   b² − a² + (-a² + b²) = b² − a² -a² + b²
   = (1+1)b²  +(−1-1) a²
   = 2b² − 2a²

Page No 135:

Question 5:

Subtract:
(i) 5a + 7b − 2c from 3a − 7b + 4c
(ii) a − 2b − 3c from −2a + 5b − 4c
(iii) 5x² − 3xy + y² from 7x² − 2xy − 4y²
(iv) 6x³ − 7x² + 5x − 3 from 4 − 5x + 6x² − 8x³
(v) x³ + 2x²y + 6xy² − y³ from y³ − 3xy² − 4x²y
(vi) −11x²y² + 7xy −6 from 9x²y² −6xy + 9
(vii) −2a + b + 6d from 5a − 2b − 3c

ANSWER:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5a + 7b − 2c
 Changing the sign of each term of the expression gives -5a -7b + 2c.
 On adding:
  (3a − 7b + 4c)+(-5a -7b + 2c ) = 3a − 7b + 4c-5a -7b + 2c
   = (3-5)a+( − 7-7)b + (4+2)c
   = -2a − 14b + 6c

(ii) Term to be subtracted = a − 2b − 3c
 Changing the sign of each term of the expression gives -a +2b + 3c.
  On adding:
  (−2a + 5b − 4c)+(-a +2b + 3c ) = −2a + 5b − 4c-a +2b + 3c
   = (−2-1)a + (5+2)b +(−4+3)c
   = −3a + 7b − c

(iii) Term to be subtracted = 5x² − 3xy + y²
 Changing the sign of each term of the expression gives  -5x² + 3xy – y².

   On adding:
(7x² − 2xy − 4y²)+(-5x² + 3xy – y²) = 7x² − 2xy − 4y²-5x² + 3xy – y²
   = (7-5)x² +(−2+3)xy +(−4-1)y²
   = 2x² +xy − 5y²


(iv) Term to be subtracted = 6x³ − 7x² + 5x − 3
 Changing the sign of each term of the expression gives  -6x³ + 7x² – 5x + 3.
  On adding:
  (4 − 5x + 6x² − 8x³)+(-6x³ + 7x² – 5x + 3) = 4 − 5x + 6x² − 8x³-6x³ + 7x² – 5x + 3
   = (-8-6)x³ +(6+7)x² +(-5- 5)x + 7
   = -14x³ + 13x² – 10x + 7

(v) Term to be subtracted = x³ + 2x²y + 6xy² − y³
 Changing the sign of each term of the expression gives  -x³ – 2x²y – 6xy² + y³.
  On adding:
(y³ − 3xy² − 4x²y)+(-x³ – 2x²y – 6xy² + y³) = y³ − 3xy² − 4x²y-x³ – 2x²y – 6xy² + y³
   = -x³ +(- 2-4)x²y +(-6-3)xy² + (1+1)y³
   = -x³ – 6x²y – 9xy² + 2y³

(vi) Term to be subtracted = −11x²y² + 7xy −6
 Changing the sign of each term of the expression gives  11x²y² -7xy +6.
  On adding:
(9x²y² −6xy + 9)+(11x²y² -7xy +6) = 9x²y² −6xy + 9+11x²y² -7xy +6
   = (9+11)x²y² (-7−6)xy + 15
   = 20x²y² −13xy +15

(vii) Term to be subtracted = −2a + b + 6d
 Changing the sign of each term of the expression gives 2a-b-6d.
 On adding:
(5a − 2b -3c)+(2a-b-6d ) = 5a − 2b -3c +2a-b-6d
   = (5+2)a+(− 2-1)b -3c -6d
   = 7a − 3b-3c -6d

Page No 135:

Question 6:

Simplify:
(i) 2p³ − 3p² + 4p − 5 − 6p³ + 2p² − 8p − 2 + 6p + 8
(ii) 2x² − xy + 6x − 4y + 5xy − 4x + 6x² + 3y
(iii) x⁴ − 6x³ + 2x − 7 + 7x³ − x + 5x² + 2 − x⁴

ANSWER:

(i) 2p³ − 3p² + 4p − 5 − 6p³ + 2p² − 8p − 2 + 6p + 8
   Rearranging and collecting the like terms
 = (2-6)p³ +(−3+2)p² + (4-8+6)p − 5-2+8
 = -4p³ −p² +2p +1

(ii) 2x² − xy + 6x − 4y + 5xy − 4x + 6x² + 3y
     Rearranging and collecting the like terms
  = (2+6)x² +(−1+5) xy + (6-4)x +(− 4+3)y
 = 8x² + 4xy + 2x − y

(iii) x⁴ − 6x³ + 2x − 7 + 7x³ − x + 5x² + 2 − x⁴
       Rearranging and collectingthe like terms
    = (1-1)x⁴ +(− 6+7)x³ + 5x² +(2-1)x-7+ 2
    = 0 +  x³ + 5x² +x-5
    = x³ + 5x² +x-5

Page No 135:

Question 7:

From the sum of 3x² − 5x + 2 and −5x² − 8x + 6, subtract 4x² − 9x + 7.

ANSWER:

Adding:
(3x² − 5x + 2) + (−5x² − 8x + 6)
Rearranging and collecting the like terms:
(3-5)x² +(− 5-8)x + 2 +6
= -2x² − 13x + 8

Subtract 4x² − 9x + 7 from -2x² − 13x + 8.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 4x² − 9x + 7
Changing the sign of each term of the expression gives -4x² + 9x – 7.
On adding:
   ( -2x² − 13x + 8 )+(-4x² + 9x – 7 )    = -2x² − 13x + 8 -4x² + 9x – 7
   = ( -2-4)x² +(−13+9)x + 8 -7
      = -6x² − 4x + 1

Page No 135:

Question 8:

If A = 7x² + 5xy − 9y², B = −4x² + xy + 5y² and C = 4y² − 3x² − 6xy then show that A + B + C = 0.

ANSWER:

A = 7x² + 5xy − 9y²
B = −4x² + xy + 5y² 
C = 4y² − 3x² − 6xy

Substituting the values of A, B and C in A+B+C:
= (7x² + 5xy − 9y²)+(−4x² + xy + 5y²)+(4y² − 3x² − 6xy)
= 7x² + 5xy − 9y²−4x² + xy + 5y²+4y² − 3x² − 6xy

Rearranging and collecting the like terms:
(7-4-3)x² + (5+1-6)xy +(−9+5+4)y²
= (0)x² + (0)xy + (0)y²
= 0
⇒A+B+C=0⇒A+B+C=0

Page No 135:

Question 9:

What must be added to 5x³ − 2x² + 6x + 7 to make the sum x³ + 3x² − x + 1?

ANSWER:

Let the expression to be added be X.
(5x³ − 2x² + 6x + 7)+X = (x³ + 3x² − x + 1)
X = (x³ + 3x² − x + 1) – (5x³ − 2x² + 6x + 7)
Changing the sign of each term of the expression that is to be subtracted and then adding:
X = (x³ + 3x² − x + 1) + (-5x³ + 2x² – 6x – 7)
X = x³ + 3x² − x + 1-5x³ + 2x² – 6x – 7

Rearranging and collecting the like terms:
X = (1-5)x³ + (3+2)x² +(−1-6) x + 1-7
X = -4x³ + 5x² − 7x -6

So, -4x³ + 5x² − 7x -6 must be added to 5x³ − 2x² + 6x + 7 to get the sum as x³ + 3x² − x + 1.

Page No 135:

Question 10:

Let      P = a² − b² + 2ab,      Q = a² + 4b² − 6ab,      R = b² + 6,      S = a² − 4ab      and      T = −2a² + b² − ab + a. Find P + Q + R + S − T.

ANSWER:

P = a² − b² + 2ab
Q = a² + 4b² − 6ab
R = b² + 6
S = a² − 4ab 
T = −2a² + b² − ab + a

Adding P, Q, R and S:
P+Q+R+S
= (a² − b² + 2ab)+(a² + 4b² − 6ab)+(b² + 6)+(a² − 4ab )
= a² − b² + 2ab+a² + 4b² − 6ab+b² + 6+a² − 4ab

Rearranging and collecting the like terms:
= (1+1+1)a² +(−1+4+1) b² + (2-6-4)ab+6
P+Q+R+S = 3a² +4b² – 8ab+6

To find P + Q + R + S − T, subtract T = (−2a² + b² − ab + a) from P+Q+R+S = (3a² +4b² – 8ab+6).

On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a² + b² − ab + a
Changing the sign of each term of the expression gives 2a² – b² + ab – a.
 Now add:
(3a² +4b² – 8ab+6)+(2a² – b² + ab – a) = 3a² +4b² – 8ab+6+2a² – b² + ab – a
= (3+2)a² +(4-1) b² +(-8+1) ab – a+6
 
P + Q + R + S − T = 5a² +3b² -7 ab – a+6

Page No 135:

Question 11:

What must be subtracted from a³ − 4a² + 5a − 6 to obtain a² − 2a + 1?

ANSWER:

Let the expression to be subtracted be X.
(a³ − 4a² + 5a − 6)-X = (a² − 2a + 1)
X = (a³ − 4a² + 5a − 6)- (a² − 2a + 1)
Since ‘-‘ sign precedes the parenthesis, we remove it and change the sign of each term within the parenthesis.
X = a³ − 4a² + 5a − 6- a² + 2a – 1
Rearranging and collecting the like terms:
X = a³ +(− 4-1)a² + (5+2)a − 6 – 1
X = a³ −5a² + 7a − 7
So, a³ −5a² + 7a − 7 must be subtracted from a³ − 4a² + 5a − 6 to obtain a² − 2a + 1.

Page No 135:

Question 12:

How much is a + 2a − 3c greater than 2a − 3b + c?

ANSWER:

To calculate how much is a + 2b − 3c greater than 2a − 3b + c, we have to subtract 2a − 3b + c from a + 2b − 3c.

Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = 2a − 3b + c
Changing the sign of each term of the expression gives -2a + 3b – c.
On adding:
   (a + 2b − 3c )+(-2a + 3b – c )   
   = a + 2b − 3c -2a + 3b – c
   = (1-2)a + (2+3)b +(− 3-1)c
   = -a + 5b − 4c

Page No 135:

Question 13:

How much less than x − 2y + 3z is 2x − 4y − z?

ANSWER:

To calculate how much less than x − 2y + 3z is 2x − 4y − z, we have to subtract 2x − 4y − z from x − 2y + 3z.

Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = 2x − 4y − z
Changing the sign of each term of the expression gives -2x + 4y + z.
On adding:
    (x − 2y + 3z)+(-2x + 4y + z )   
   = x − 2y + 3z-2x + 4y + z 
   = (1-2)x +(−2+4)y + (3+1)z
   = -x + 2y + 4z

Page No 135:

Question 14:

By how much does 3x² − 5x + 6 exceed x³ − x² + 4x − 1?

ANSWER:

To calculate how much does 3x² − 5x + 6 exceed x³ − x² + 4x − 1, we have to subtract x³ − x² + 4x − 1 from 3x² − 5x + 6.
Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = x³ − x² + 4x − 1
Changing the sign of each term of the expression gives -x³ + x² – 4x + 1.
On adding:
(3x² − 5x + 6)+(-x³ + x² – 4x + 1 )   
   = 3x² − 5x + 6-x³ + x² – 4x + 1
   = -x³ + (3+1)x² +(-5-4)x+6 + 1
   = -x³ +4 x² – 9x + 7

Page No 135:

Question 15:

Subtract the sum of 5x − 4y + 6z and −8x + y − 2z from the sum of 12x − y + 3z and −3x + 5y − 8z.

ANSWER:

Add 5x − 4y + 6z and −8x + y − 2z.

(5x − 4y + 6z )+(−8x + y − 2z)
= 5x − 4y + 6z −8x + y − 2z
= (5-8)x +(−4+1)y + (6-2)z
= -3x − 3y + 4z

Adding 12x − y + 3z and −3x + 5y − 8z:
(12x − y + 3z )+(−3x + 5y − 8z)
= 12x − y + 3z −3x + 5y − 8z
= (12-3)x +(−1+5)y + (3-8)z
= 9x +4y -5z

Subtract -3x − 3y + 4z from 9x +4y -5z.
Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = -3x − 3y + 4z
Changing the sign of each term of the expression gives 3x + 3y – 4z.
On adding:
  (9x +4y -5z)+(3x + 3y – 4z )   
   = 9x +4y -5z+3x + 3y – 4z   
   = (9+3)x +(4+3)y + (-5-4)z
  = 12x +7y -9z

Page No 135:

Question 16:

By how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2?

ANSWER:

To calculate how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2, we have to subtract 2x + 5y − 6z + 2 from 2x − 3y + 4z.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2x + 5y − 6z + 2
Changing the sign of each term of the expression gives -2x – 5y + 6z – 2.
On adding:
    (2x − 3y + 4z )+(-2x – 5y + 6z – 2 )   
   = 2x − 3y + 4z-2x – 5y + 6z – 2   
   = (2-2)x + (-3-5)y +(4+6)z-2
   = 0-8y+10z-2
   = -8y+10z-2

Page No 135:

Question 17:

By how much does 1 exceed 2x − 3y − 4?

ANSWER:

To calculate how much does 1 exceed 2x-3y-4, we have to subtract 2x-3y-4 from 1.
Change the sign of each term of the expression to be subtracted and then add.

 Term to be subtracted = 2x-3y-4
Changing the sign of each term of the expression gives -2x+3y+4.
On adding:
   (1)+(-2x+3y+4 )   
   = 1-2x+3y+4
   = 5-2x+3y

Page No 136:

Exercise 8C

Question 1:

Simplify:
a − (b − 2a)

ANSWER:

a – (b – 2a)
Here,  ‘-‘ sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
=a – b + 2a
=3a – b

Page No 136:

Question 2:

Simplify:
4x − (3y − x + 2z)

ANSWER:

4x − (3y − x + 2z)
Here, ‘−’ sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
= 4x − 3y + x − 2z
= 5x − 3y − 2z

Page No 136:

Question 3:

Simplify:
(a² + b² + 2ab) − (a² + b² −2ab)

ANSWER:

(a² + b² + 2ab) − (a² + b² − 2ab)
Here, ‘−’ sign precedes the second parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
 a² + b² + 2ab − a² − b² +2ab

Rearranging and collecting the like terms:
a² − a² +b² − b² + 2ab + 2ab
=(1 − 1)a² + (1− 1)b² + (2 + 2)ab
=0 + 0 + 4ab
= 4ab

Page No 136:

Question 4:

Simplify:
−3(a + b) + 4(2a − 3b) − (2a − b)

ANSWER:

−3(a + b) + 4(2a − 3b) − (2a − b)
Here, ‘−’ sign precedes the first and the third parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −3a − 3b + (4××2a )−(4××3b) − 2a + b
= − 3a − 3b + 8a − 12b − 2a + b

Rearranging and collecting the like terms:
 −3a + 8a − 2a − 3b − 12b + b
= (−3 + 8 − 2)a + (−3 − 12 + 1)b
= 3a −14b

Page No 136:

Question 5:

Simplify:
−4x² + {(2x² − 3) − (4 − 3x²)}

ANSWER:

−4x² + {(2x² − 3) − (4 − 3x²)}

We will first remove the innermost grouping symbol (  ) and then {  }.

∴ −4x² + {(2x² − 3) − (4 − 3x²)}
= −4x² + {2x² − 3 − 4 + 3x²}
= −4x² + {5x² − 7}
= −4x² + 5x² − 7
= x² − 7

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Question 6:

Simplify:
−2(x² − y² + xy) −3(x² + y² − xy)

ANSWER:

−2(x² − y² + xy) −3(x² + y² − xy)
Here a ‘−’ sign precedes both the parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −2x² +2y² − 2xy −3x² − 3y² + 3xy
= (−2 − 3)x² +(2 − 3)y² + (− 2 + 3)xy
= −5x² − y² + xy

Page No 136:

Question 7:

Simplify:
a − [2b − {3a − (2b − 3c)}]

ANSWER:

a − [2b − {3a − (2b − 3c)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ a − [2b − {3a − (2b − 3c)}]
= a − [2b − {3a − 2b + 3c}]
= a − [2b − 3a + 2b − 3c]
= a − [4b − 3a − 3c]
= a − 4b + 3a + 3c
= 4a − 4b + 3c

Page No 136:

Question 8:

Simplify:
−x + [5y − {x − (5y − 2x)}]

ANSWER:

−x + [5y − {x − (5y − 2x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ −x + [5y − {x − (5y − 2x)}]
= −x + [5y − {x − 5y + 2x}]
= −x + [5y − {3x − 5y}]
= −x + [5y − 3x + 5y]
= −x + [10y − 3x]
= −x + 10y − 3x
=  − 4x + 10y

Page No 137:

Question 9:

Simplify:
86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]

ANSWER:

86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
= 86 − [15x − 42x + 63 −2{10x − 10 + 15x}]
= 86 − [15x − 42x + 63 −2{25x − 10}]
= 86 − [15x − 42x + 63 −50x + 20]
= 86 − [− 77x + 83 ]
= 86 + 77x − 83
= 77x + 3

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Question 10:

Simplify:
12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]

ANSWER:

12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]
= 12x − [3x³ + 5x² − {7x² − 4 + 3x + x³+ 6x³} − 3x]
= 12x − [3x³ + 5x² − {7x² − 4 + 3x + 7x³} − 3x]
= 12x − [3x³ + 5x² − 7x² + 4 − 3x − 7x³ − 3x]
= 12x − [ − 2x² + 4 − 4x³ − 6x]
= 12x + 2x² − 4 + 4x³ + 6x
= 4x³ + 2x² +18x-4

Page No 137:

Question 11:

Simplify:
5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a

ANSWER:

5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a
= 5a − [a² − {2a − 2a² + 8a³ − 3a³ + 15a² + 9a}] −8a
= 5a − [a² − {5a³ + 13a² + 11a}] − 8a
= 5a − [a² − 5a³ − 13a² −11a] − 8a
= 5a − [ − 5a³ − 12a² − 11a] − 8a
= 5a + 5a³ + 12a² + 11a − 8a
= 5a³ + 12a² + 8a

Page No 137:

Question 12:

Simplify:
3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]

ANSWER:

3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]
= 3 − [x − {2y − 5x – y + 3 + 2x²} − x² + 3y]
= 3 − [x − {y − 5x + 3 + 2x²} − x² + 3y]
= 3 − [x − y + 5x − 3 − 2x² − x² + 3y]
= 3 − [ 6x − 3 − 3x² + 2y]
= 3 − 6x + 3 + 3x² − 2y
= 3x² − 2y − 6x+6

Page No 137:

Question 13:

Simplify:
xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]

ANSWER:

xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]
= xy − [yz − zx − {yx − 3y + xz − xy + zy}]
= xy − [yz − zx − {− 3y + xz + zy}]  (∵xy=yx)(∵xy=yx)
= xy − [yz − zx + 3y – xz – zy] 
= xy − [− 2zx + 3y ] (∵ yz=zy, zx=xz)(∵ yz=zy, zx=xz)
= xy + 2zx − 3y 

Page No 137:

Question 14:

Simplify:
2a − 3b − [3a − 2b − {a − c − (a − 2b)}]

ANSWER:

2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
= 2a − 3b − [3a − 2b − {a − c − a + 2b}]
= 2a − 3b − [3a − 2b − {− c  + 2b}]
= 2a − 3b − [3a − 2b + c  − 2b]
= 2a − 3b − [3a − 4b + c ]
= 2a − 3b − 3a + 4b − c
= − a + b − c

Page No 137:

Question 15:

Simplify:
−a − [a + {a + b − 2a − (a − 2b)} − b]

ANSWER:

-a − [a + {a + b − 2a − (a − 2b)} − b]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].


∴ −a − [a + {a + b − 2a − (a − 2b)} − b]
= −a − [a + {a + b − 2a − a + 2b} − b]
= −a − [a + {3b − 2a } − b]
= −a − [a + 3b − 2a  − b]
= −a − [2b − a ]
= −a − 2b + a
= −2b

Page No 137:

Question 16:

Simplify:
2a − [4b − {4a − (3b − 2a + 2b)}]2a – [4b – {4a – (3b – 2a + 2b)}]

ANSWER:

2a-[4b-{4a-(3b-2a+2b¯¯¯¯¯¯¯¯¯)}]2a-[4b-{4a-(3b-2a+2b¯)}]
We will first remove the innermost grouping symbol bar bracket. Next, we will remove (  ), followed by {  } and then [   ].

∴ 2a-[4b-{4a-(3b-2a+2b¯¯¯¯¯¯¯¯¯)}]2a-[4b-{4a-(3b-2a+2b¯)}]
= 2a-[4b-{4a-(3b-2a-2b)}]
= 2a-[4b-{4a-(b-2a)}]
= 2a-[4b-{4a-b+2a}]
=2a-[4b-{6a-b}]
= 2a-[4b-6a+b]
= 2a-[5b-6a]
= 2a-5b+6a
= 8a-5b

Page No 137:

Question 17:

Simplify:
5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]

ANSWER:

5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
= 5x − [4y − {7x − 3z + 2y + 4z − 3x − 9y + 6z}]
= 5x − [4y − {4x + 7z − 7y}]
= 5x − [4y − 4x − 7z + 7y]
= 5x − [11y − 4x − 7z ]
= 5x − 11y + 4x + 7z
= 9x − 11y + 7z