Author Archives: Swati

Exercise 15.1

Question: 1

Take three non-collinear points A. B and C on a page of your notebook. Join AB, BC and CA. What figure do you get? Name the triangle. Also, name

(i) The side opposite to ∠B

(ii) The angle opposite to side AB

(iii) The vertex opposite to side BC

(iv) The side opposite to vertex B.

Solution:

(i) AC

(ii) ∠B

(iii) A

(iv) AC

Question: 2

Take three collinear points A, B and C on a page of your note book. Join AB. BC and CA. Is the figure a triangle? If not, why?

Solution:

No, the figure is not a triangle. By definition a triangle is a plane figure formed by three non-parallel line segments

Question: 3

Distinguish between a triangle and its triangular region.

Solution:

A triangle is a plane figure formed by three non-parallel line segments, whereas, its triangular region includes the interior of the triangle along with the triangle itself.

Question: 4

D is a point on side BC of a ∆CAD is joined. Name all the triangles that you can observe in the figure. How many are they?

Solution:

We can observe the following three triangles in the given figure

(i) ∆ABC

(ii) ∆ACD

(iii) ∆ADB

Question: 5

A, B. C and D are four points, and no three points are collinear. AC and BD intersed at O. There are eight triangles that you can observe. Name all the triangles

Solution:

(i) ∆ABC

(ii) ∆ABD

(iii) ∆ABO

(iv) ∆BCD

(v) ∆DCO

(vi) ∆AOD

(vii) ∆ACD

(viii) ∆BCD

Question: 6

What is the difference between a triangle and triangular region?

Solution:

Plane of figure formed by three non-parallel line segments is called a triangle where as triangular region is the interior of triangle ABC together with the triangle ABC itself is called the triangular region ABC

Question: 7

Explain the following terms:

(i) Triangle

(a) Parts or elements of a triangle

(iii) Scalene triangle

(iv) Isosceles triangle

(v) Equilateral triangle

(vi) Acute triangle

(vii) Right triangle

(viii) Obtuse triangle

(ix) Interior of a triangle

(x) Exterior of a triangle

Solution:

(i) A triangle is a plane figure formed by three non-parallel line segments.

(ii) The three sides and the three angles of a triangle are together known as the parts or elements of that triangle.

(iii) A scalene triangle is a triangle in which no two sides are equal.

(iv) An isosceles triangle is a triangle in which two sides are equal. Isosceles triangle

(v) An equilateral triangle is a triangle in which all three sides are equal. Equilateral triangle

(vi) An acute triangle is a triangle in which all the angles are acute (less than 90°).

(vii) A right angled triangle is a triangle in which one angle is right angled, i.e. 90°.

(viii) An obtuse triangle is a triangle in which one angle is obtuse (more than 90°).

(ix) The interior of a triangle is made up of all such points that are enclosed within the triangle.

(x) The exterior of a triangle is made up of all such points that are not enclosed within the triangle.

Question: 8

In Figure, the length (in cm) of each side has been indicated along the side. State for each triangle angle whether it is scalene, isosceles or equilateral:

Solution:

(i) This triangle is a scalene triangle because no two sides are equal.

(ii) This triangle is an isosceles triangle because two of its sides, viz. PQ and PR, are equal.

(iii) This triangle is an equilateral triangle because all its three sides are equal.

(iv) This triangle is a scalene triangle because no two sides are equal.

(v) This triangle is an isosceles triangle because two of its sides are equal.

Question: 9

There are five triangles. The measures of some of their angles have been indicated. State for each triangle whether it is acute, right or obtuse.

Solution:

(i) This is a right triangle because one of its angles is 90°.

(ii) This is an obtuse triangle because one of its angles is 120°, which is greater than 90°.

(iii) This is an acute triangle because all its angles are acute angles (less than 90°).

(iv) This is a right triangle because one of its angles is 90°.

(v) This is an obtuse triangle because one of its angles is 110°, which is greater than 90°.

Question: 10

Fill in the blanks with the correct word/symbol to make it a true statement:

(i) A triangle has _____ sides.

(ii) A triangle has ______ vertices.

(iii) A triangle has _____ angles.

(iv) A triangle has ________ parts.

(v) A triangle whose no two sides are equal is known as ________ 

(v0 A triangle whose two sides are equal is known as ________ 

(vii) A triangle whose all the sides are equal is known as ________

(viii) A triangle whose one angle is a right angle is known as _________

(ix) A triangle whose all the angles are of measure less than 90′ is known as _________

(x) A triangle whose one angle is more than 90′ is known as _________

Solution:

(i) three

(ii) three

(iii) three

(iv) six (three sides + three angles)

(v) a scalene triangle

(vi) an isosceles triangle

(vii) an equilateral triangle

(viii) a right triangle

(ix) an acute triangle

(x) an obtuse triangle

Question: 11

In each of the following, state if the statement is true (T) or false (F):

(i) A triangle has three sides.

(ii) A triangle may have four vertices.

(iii) Any three line-segments make up a triangle.

(iv) The interior of a triangle includes its vertices.

(v) The triangular region includes the vertices of the corresponding triangle.

(vi) The vertices of a triangle are three collinear points.

(vii) An equilateral triangle is isosceles also.

(viii) Every right triangle is scalene.

(ix) Each acute triangle is equilateral.

(x) No isosceles triangle is obtuse.

Solution:

(i) True.

(ii) False. A triangle has three vertices.

(iii) False. Any three non-parallel line segments can make up a triangle.

(iv) False. The interior of a triangle is the region enclosed by the triangle and the vertices are not enclosed by the triangle.

(v) True. The triangular region includes the interior region and the triangle itself.

(vi) False. The vertices of a triangle are three non-collinear points.

(vii) True. In an equilateral triangle, any two sides are equal.

(viii) False. A right triangle can also be an isosceles triangle.

(ix) False. Each acute triangle is not an equilateral triangle, but each equilateral triangle is an acute triangle.

(x) False. An isosceles triangle can be an obtuse triangle, a right triangle or an acute triangle

Exercise 15.2

Question: 1

Two angles of a triangle are of measures 150° and 30°. Find the measure of the third angle.

Solution:

Let the third angle be x

Sum of all the angles of a triangle = 180°

105° + 30° + x = 180°

135° + x = 180°

x = 180° – 135°

x = 45°

Therefore the third angle is 45°

Question: 2

One of the angles of a triangle is 130°, and the other two angles are equal. What is the measure of each of these equal angles?

Solution:

Let the second and third angle be x

Sum of all the angles of a triangle = 180°

130° + x + x = 180°

130° + 2x = 180°

2x = 180°– 130°

2x = 50°

x = 50/2

x = 25°

Therefore the two other angles are 25° each

Question: 3

The three angles of a triangle are equal to one another. What is the measure of each of the angles?

Solution:

Let the each angle be x

Sum of all the angles of a triangle =180°

x + x + x = 180°

3x = 180°

x = 180/3

x = 60°

Therefore angle is 60° each

Question: 4

If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.

Solution:

If angles of the triangle are in the ratio 1: 2: 3 then take first angle as ‘x’, second angle as ‘2x’ and third angle as ‘3x’

Sum of all the angles of a triangle=180°

x + 2x + 3x = 180°

6x = 180°

x = 180/6

x = 30°

2x = 30° × 2 = 60°

3x = 30° × 3 = 90°

Therefore the first angle is 30°, second angle is 60° and third angle is 90°

Question: 5

The angles of a triangle are (x − 40) °, (x − 20) ° and (1/2 − 10) °. Find the value of x.

Solution:

Question: 6

The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°. Find the three angles.

Solution:

Let the first angle be x

Second angle be x + 10°

Third angle be x + 10° + 10°

Sum of all the angles of a triangle = 180°

x + x + 10° + x + 10° +10° = 180°

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3

x = 50°

First angle is 50°

Second angle x + 10° = 50 + 10 = 60°

Third angle x + 10° +10° = 50 + 10 + 10 = 70°

Question: 7

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle

Solution:

Let the first and second angle be x

The third angle is greater than the first and second by 30° = x + 30°

The first and the second angles are equal

Sum of all the angles of a triangle = 180°

x + x + x + 30° = 180°

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3

x = 50°

Third angle = x + 30° = 50° + 30° = 80°

The first and the second angle is 50° and the third angle is 80°

Question: 8

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

Solution:

One angle of a triangle is equal to the sum of the other two

x = y + z

Let the measure of angles be x, y, z

x + y + z = 180°

x + x = 180°

2x = 180°

x = 180/2

x = 90°

If one angle is 90° then the given triangle is a right angled triangle

Question: 9

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Solution:

Each angle of a triangle is less than the sum of the other two

Measure of angles be x, y and z

x > y + z

y < x + z

z < x + y

Therefore triangle is an acute triangle

Question: 10

In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:

(i) 63°, 37°, 80°

(ii) 45°, 61°, 73°

(iii) 59°, 72°, 61°

(iv) 45°, 45°, 90°

(v) 30°, 20°, 125°

Solution:

(i) 63°, 37°, 80° = 180°

Angles form a triangle

(ii) 45°, 61°, 73° is not equal to 180°

Therefore not a triangle

(iii) 59°, 72°, 61° is not equal to 180°

Therefore not a triangle

(iv) 45°, 45°, 90° = 180°

Angles form a triangle

(v) 30°, 20°, 125° is not equal to 180°

Therefore not a triangle

Question: 11

The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle

Solution:

Given that

Angles of a triangle are in the ratio: 3: 4: 5

Measure of the angles be 3x, 4x, 5x

Sum of the angles of a triangle =180°

3x + 4x + 5x = 180°

12x = 180°

x = 180/12

x = 15°

Smallest angle = 3x

=3 × 15°

= 45°

Question: 12

Two acute angles of a right triangle are equal. Find the two angles.

Solution:

Given acute angles of a right angled triangle are equal

Right triangle: whose one of the angle is a right angle

Measured angle be x, x, 90°

x + x + 180°= 180°

2x = 90°

x = 90/2

x = 45°

The two angles are 45° and 45°

Question: 13

One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?

Solution:

Angle of a triangle is greater than the sum of the other two

Measure of the angles be x, y, z

x > y + z  or

y > x + z   or

z > x + y

x or y or z > 90° which is obtuse

Therefore triangle is an obtuse angle

Question: 14

AC, AD and AE are joined. Find

∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA

∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA

Solution:

We know that sum of the angles of a triangle is 180°

Therefore in ∆ABC, we have

∠CAB + ∠ABC + ∠BCA = 180° — (i)

In ∆ACD, we have

∠DAC + ∠ACD + ∠CDA = 180° — (ii)

In ∆ADE, we have

∠EAD + ∠ADE + ∠DEA =180° — (iii)

In ∆AEF, we have

∠FAE + ∠AEF + ∠EFA = 180° — (iv)

Adding (i), (ii), (iii), (iv) we get

∠CAB + ∠ABC + ∠BCA + ∠DAC + ∠ACD + ∠CDA + ∠EAD + ∠ADE + ∠DEA + ∠FAE + ∠AEF +∠EFA = 720°

Therefore ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720°

Question: 15

Find x, y, z (whichever is required) from the figures given below:

Solution:

(i) In ∆ABC and ∆ADE we have:

∠ADE = ∠ABC (corresponding angles)

x = 40°

∠AED = ∠ACB (corresponding angles)

y = 30°

We know that the sum of all the three angles of a triangle is equal to 180°

x + y + z = 180° (Angles of ∆ADE)

Which means: 40° + 30° + z = 180°

z = 180° – 70°

z = 110°

Therefore, we can conclude that the three angles of the given triangle are 40°, 30° and 110°.

(ii) We can see that in ∆ADC, ∠ADC is equal to 90°.

(∆ADC is a right triangle)

We also know that the sum of all the angles of a triangle is equal to 180°.

Which means: 45° + 90° + y = 180° (Sum of the angles of ∆ADC)

135° + y = 180°

y = 180° – 135°.

y = 45°.

We can also say that in ∆ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180°.

(Sum of the angles of ∆ABC)

40° + y + (x + 45°) = 180°

40° + 45° + x + 45° = 180°  (y = 45°)

x = 180° –130°

x = 50°

Therefore, we can say that the required angles are 45° and 50°.

(iii) We know that the sum of all the angles of a triangle is equal to 180°.

Therefore, for △ABD:

∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of ∆ABD)

50° + x + 50° = 180°

100° + x = 180°

x = 180° – 100°

x = 80°

For ∆ABC:

∠ABC + ∠ACB + ∠BAC = 180° (Sum of the angles of ∆ABC)

50° + z + (50° + 30°) = 180°

50° + z + 50° + 30° = 180°

z = 180° – 130°

z = 50°

Using the same argument for ∆ADC:

∠ADC + ∠ACD + ∠DAC = 180° (Sum of the angles of ∆ADC)

y +z + 30° =180°

y + 50° + 30° = 180° (z = 50°)

y = 180° – 80°

y = 100°

Therefore, we can conclude that the required angles are 80°, 50° and 100°.

(iv) In ∆ABC and ∆ADE we have:

∠ADE = ∠ABC (Corresponding angles)

y = 50°

Also, ∠AED = ∠ACB  (Corresponding angles)

z = 40°

We know that the sum of all the three angles of a triangle is equal to 180°.

Which means: x + 50° + 40° = 180° (Angles of ∆ADE)

x = 180° – 90°

x = 90°

Therefore, we can conclude that the required angles are 50°, 40° and 90°.

Question: 16

If one angle of a triangle is 60° and the other two angles are in the ratio 1: 2, find the angles

Solution:

We know that one of the angles of the given triangle is 60°. (Given)

We also know that the other two angles of the triangle are in the ratio 1: 2.

Let one of the other two angles be x.

Therefore, the second one will be 2x.

We know that the sum of all the three angles of a triangle is equal to 180°.

60° + x + 2x = 180°

3x = 180° – 60°

3x = 120°

x = 120/3

x = 40°

2x = 2 × 40

2x = 80°

Hence, we can conclude that the required angles are 40° and 80°.

Question: 17

It one angle of a triangle is 100° and the other two angles are in the ratio 2: 3. find the angles.

Solution:

We know that one of the angles of the given triangle is 100°.

We also know that the other two angles are in the ratio 2: 3.

Let one of the other two angles be 2x.

Therefore, the second angle will be 3x.

We know that the sum of all three angles of a triangle is 180°.

100° + 2x + 3x = 180°

5x = 180° – 100°

5x = 80°

x = 80/5

2x = 2 ×16

2x = 32°

3x = 3×16

3x = 48°

Thus, the required angles are 32° and 48°.

Question: 18

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

Solution:

We know that for the given triangle, 3∠A = 6∠C

∠A = 2∠C — (i)

We also know that for the same triangle, 4∠B = 6∠C

∠B = (6/4)∠C  — (ii)

We know that the sum of all three angles of a triangle is 180°.

Therefore, we can say that:

∠A + ∠B + ∠C = 180° (Angles of ∆ABC) — (iii)

On putting the values of ∠A and ∠B in equation (iii), we get:

2∠C + (6/4)∠C +∠C = 180°

(18/4) ∠C = 180°

∠C = 40°

From equation (i), we have:

∠A = 2∠C = 2 × 40

∠A = 80°

From equation (ii), we have:

∠B = (6/4)∠C = (6/4) × 40°

∠B = 60°

∠A = 80°, ∠B = 60°, ∠C = 40°

Therefore, the three angles of the given triangle are 80°, 60°, and 40°.

Question: 19

Is it possible to have a triangle, in which

(i) Two of the angles are right?

(ii) Two of the angles are obtuse?

(iii) Two of the angles are acute?

(iv) Each angle is less than 60°?

(v) Each angle is greater than 60°?

(vi) Each angle is equal to 60°

Solution:

Give reasons in support of your answer in each case.

(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.

(ii) No, because as we know that the sum of all three angles of a triangle is always 180°. If there are two obtuse angles, then their sum will be more than 180°, which is not possible in case of a triangle.

(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.

(iv) No, because if each angle is less than 60°, then the sum of all three angles will be less than 180°, which is not possible in case of a triangle.

Proof:

Let the three angles of the triangle be ∠A, ∠B and ∠C.

As per the given information,

∠A < 60°  … (i)

∠B< 60°   … (ii)

∠C < 60°   … (iii)

On adding (i), (ii) and (iii), we get:

∠A + ∠B + ∠C < 60°+ 60°+ 60°

∠A + ∠B + ∠C < 180°

We can see that the sum of all three angles is less than 180°, which is not possible for a triangle.

Hence, we can say that it is not possible for each angle of a triangle to be less than 60°.

(v) No, because if each angle is greater than 60°, then the sum of all three angles will be greater than 180°, which is not possible.

Proof:

Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,

∠A > 60°  … (i)

∠B > 60°   … (ii)

∠C >  60°   … (iii)

On adding (i), (ii) and (iii), we get:

∠A + ∠B + ∠C > 60°+ 60°+ 60°

∠A + ∠B + ∠C > 180°

We can see that the sum of all three angles of the given triangle are greater than 180°, which is not possible for a triangle.

Hence, we can say that it is not possible for each angle of a triangle to be greater than 60°.

(vi) Yes, if each angle of the triangle is equal to 60°, then the sum of all three angles will be 180° , which is possible in case of a triangle.

Proof:

Let the three angles of the triangle be ∠A, ∠B and ∠C. As per the given information,

∠A = 60°  … (i)

∠B = 60°   …(ii)

∠C = 60°   … (iii)

On adding (i), (ii) and (iii), we get:

∠A + ∠B + ∠C = 60°+ 60°+ 60°

∠A + ∠B + ∠C =180°

We can see that the sum of all three angles of the given triangle is equal to 180°, which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to 60°.

Question: 20

In ∆ABC, ∠A = 100°, AD bisects ∠A and AD perpendicular BC. Find ∠B

Solution:

Consider ∆ABD

∠BAD = 100/2   (AD bisects ∠A)

∠BAD = 50°

∠ADB = 90°   (AD perpendicular to BC)

We know that the sum of all three angles of a triangle is 180°.

Thus,

∠ABD + ∠BAD + ∠ADB = 180° (Sum of angles of ∆ABD)

Or,

∠ABD + 50° + 90° = 180°

∠ABD =180° – 140°

∠ABD = 40°

Question: 21

In ∆ABC, ∠A = 50°, ∠B = 100° and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC

Solution:

We know that the sum of all three angles of a triangle is equal to 180°.

Therefore, for the given △ABC, we can say that:

∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)

50° + 70° + ∠C = 180°

∠C= 180° –120°

∠C = 60°

∠ACD = ∠BCD =∠C2 (CD bisects ∠C and meets AB in D. )

∠ACD = ∠BCD = 60/2= 30°

Using the same logic for the given ∆ACD, we can say that:

∠DAC + ∠ACD + ∠ADC = 180°

50° + 30° + ∠ADC = 180°

∠ADC = 180°– 80°

∠ADC = 100°

If we use the same logic for the given ∆BCD, we can say that

∠DBC + ∠BCD + ∠BDC = 180°

70° + 30° + ∠BDC = 180°

∠BDC = 180° – 100°

∠BDC = 80°

Thus,

For ∆ADC: ∠A = 50°, ∠D = 100° ∠C = 30°

∆BDC: ∠B = 70°, ∠D = 80° ∠C = 30°

Question: 22

In ∆ABC, ∠A = 60°, ∠B = 80°, and the bisectors of ∠B and ∠C,  meet at O. Find

(i) ∠C

(ii) ∠BOC

Solution:

We know that the sum of all three angles of a triangle is 180°.

Hence, for △ABC, we can say that:

∠A + ∠B + ∠C = 180° (Sum of angles of ∆ABC)

60° + 80° + ∠C= 180°.

∠C = 180° – 140°

∠C = 140°.

For △OBC,

∠OBC = ∠B2 = 80/2 (OB bisects ∠B)

∠OBC = 40°

∠OCB =∠C2 = 40/2 (OC bisects ∠C)

∠OCB = 20°

If we apply the above logic to this triangle, we can say that:

∠OCB + ∠OBC + ∠BOC = 180° (Sum of angles of ∆OBC)

20° + 40° + ∠BOC = 180°

∠BOC = 180° – 60°

∠BOC = 120°

Question: 23

The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.

Solution:

We know that the sum of all three angles of a triangle is 180°.

Hence, for ∆ABC, we can say that:

∠A + ∠B + ∠C = 180°

∠A + 90° + ∠C = 180°

∠A + ∠C = 180° – 90°

∠A + ∠C = 90°

For ∆OAC:

∠OAC = ∠A2           (OA bisects LA)

∠OCA   = ∠C2         (OC bisects LC)

On applying the above logic to △OAC, we get:

∠AOC + ∠OAC + ∠OCA = 180°    (Sum of angles of ∆AOC)

∠AOC + ∠A2 + ∠C2 = 180°

∠AOC + ∠A + ∠C2 = 180°

∠AOC + 90/2 = 180°

∠AOC = 180° – 45°

∠AOC = 135°

Question: 24

In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.

Solution:

In the given triangle,

∠ACD = ∠A + ∠B. (Exterior angle is equal to the sum of two opposite interior angles.)

We know that the sum of all three angles of a triangle is 180°.

Therefore, for the given triangle, we can say that:

Question: 25

In ∆ABC, ∠B = 60°, ∠C = 40°, AL perpendicular BC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD

Solution:

We know that the sum of all angles of a triangle is 180°

Therefore, for ∆ABC, we can say that:

Question: 26

Line segments AB and CD intersect at O such that AC perpendicular DB. It ∠CAB = 35° and ∠CDB = 55°. Find ∠BOD.

Solution:

We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.

∠CAB = ∠DBA (Alternate interior angles)

∠DBA = 35°

We also know that the sum of all three angles of a triangle is 180°.

Hence, for △OBD, we can say that:

∠DBO + ∠ODB + ∠BOD = 180°

35° + 55° + ∠BOD = 180° (∠DBO = ∠DBA and ∠ODB = ∠CDB)

∠BOD = 180° – 90°

∠BOD = 90°

Question: 27

In Figure, ∆ABC is right angled at A, Q and R are points on line BC and P is a point such that QP perpendicular to AC and RP perpendicular to AB. Find ∠P

Solution:

In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.

∠QCA = ∠CQP (Alternate interior angles)

Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,

∠ABC = ∠PRQ (alternate interior angles).

We know that the sum of all three angles of a triangle is 180°.

Hence, for ∆ABC, we can say that:

∠ABC + ∠ACB + ∠BAC = 180°

∠ABC + ∠ACB + 90° = 180° (Right angled at A)

∠ABC + ∠ACB = 90°

Using the same logic for △PQR, we can say that:

∠PQR + ∠PRQ + ∠QPR = 180°

∠ABC + ∠ACB + ∠QPR = 180° (∠ABC = ∠PRQ and ∠QCA = ∠CQP)

Or,

90°+ ∠QPR =180° (∠ABC+ ∠ACB = 90°)

∠QPR = 90°

Exercise 15.3

Question: 1

∠CBX is an exterior angle of ∆ABC at B. Name

(i) the interior adjacent angle

(ii) the interior opposite angles to exterior ∠CBX

Also, name the interior opposite angles to an exterior angle at A.

Solution:

(i) ∠ABC

(ii) ∠BAC and ∠ACB

Also the interior angles opposite to exterior are ∠ABC and ∠ACB

Question: 2

In the fig, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?

Solution:

In ∆ABC, ∠A = 50° and ∠B = 55°

Because of the angle sum property of the triangle, we can say that

∠A + ∠B + ∠C = 180°

50°+ 55°+ ∠C = 180°

Or

∠C = 75°

∠ACB = 75°

∠ACX = 180°− ∠ACB = 180°−75° = 105°

Question: 3

In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angles is 55°. Find all the angles of the triangle.

Solution:

We know that the sum of interior opposite angles is equal to the exterior angle.

Hence, for the given triangle, we can say that:

∠ABC+ ∠BAC = ∠BCO

55° + ∠BAC = 95°

Or,

∠BAC= 95°– 95°

= ∠BAC = 40°

We also know that the sum of all angles of a triangle is 180°.

Hence, for the given △ABC, we can say that:

∠ABC + ∠BAC + ∠BCA = 180°

55° + 40° + ∠BCA = 180°

Or,

∠BCA = 180° –95°

= ∠BCA = 85°

Question: 4

One of the exterior angles of a triangle is 80°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?

Solution:

Let us assume that A and B are the two interior opposite angles.

We know that ∠A is equal to ∠B.

We also know that the sum of interior opposite angles is equal to the exterior angle.

Hence, we can say that:

∠A + ∠B = 80°

Or,

∠A +∠A = 80° (∠A= ∠B)

2∠A = 80°

∠A = 40/2 =40°

∠A= ∠B = 40°

Thus, each of the required angles is of 40°.

Question: 5

The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.

Solution:

In the given figure, ∠ABE and ∠ABC form a linear pair.

∠ABE + ∠ABC =180°

∠ABC = 180°– 136°

∠ABC = 44°

We can also see that ∠ACD and ∠ACB form a linear pair.

∠ACD + ∠ACB = 180°

∠AUB = 180°– 104°

∠ACB = 76°

We know that the sum of interior opposite angles is equal to the exterior angle.

Therefore, we can say that:

∠BAC + ∠ABC = 104°

∠BAC = 104°– 44° = 60°

Thus,

∠ACE = 76° and ∠BAC = 60°

Question: 6

In Fig, the sides BC, CA and BA of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°; find all the angles of the ∆ABC

Solution:

In a ∆ABC, ∠BAC and ∠EAF are vertically opposite angles.

Hence, we can say that:

∠BAC = ∠EAF = 45°

Considering the exterior angle property, we can say that:

∠BAC + ∠ABC = ∠ACD = 105°

∠ABC = 105°– 45° = 60°

Because of the angle sum property of the triangle, we can say that:

∠ABC + ∠ACS +∠BAC = 180°

∠ACB = 75°

Therefore, the angles are 45°, 65° and 75°.

Question: 7

In Figure, AC perpendicular to CE and C ∠A: ∠B: ∠C= 3: 2: 1. Find the value of ∠ECD.

Solution:

In the given triangle, the angles are in the ratio 3: 2: 1.

Let the angles of the triangle be 3x, 2x and x.

Because of the angle sum property of the triangle, we can say that:

3x + 2x + x = 180°

6x = 180°

Or,

x = 30° … (i)

Also, ∠ACB + ∠ACE + ∠ECD = 180°

x + 90° + ∠ECD = 180° (∠ACE = 90°)

∠ECD = 60° [From (i)]

Question: 8

A student when asked to measure two exterior angles of ∆ABC observed that the exterior angles at A and B are of 103° and 74° respectively. Is this possible? Why or why not?

Solution:

Here,

Internal angle at A + External angle at A = 180°

Internal angle at A + 103° =180°

Internal angle at A = 77°

Internal angle at B + External angle at B = 180°

Internal angle at B + 74° = 180°

Internal angle at B = 106°

Sum of internal angles at A and B = 77° + 106° =183°

It means that the sum of internal angles at A and B is greater than 180°, which cannot be possible.

Question: 9

In Figure, AD and CF are respectively perpendiculars to sides BC and AB of ∆ABC. If ∠FCD = 50°, find ∠BAD

Solution:

We know that the sum of all angles of a triangle is 180°

Therefore, for the given ∆FCB, we can say that:

∠FCB + ∠CBF + ∠BFC = 180°

50° + ∠CBF + 90°= 180°

Or,

∠CBF = 180° – 50°– 90° = 40° … (i)

Using the above rule for ∆ABD, we can say that:

∠ABD + ∠BDA + ∠BAD = 180°

∠BAD = 180° – 90°– 40° = 50° [from (i)]

Question: 10

In Figure, measures of some angles are indicated. Find the value of x.

Solution:

Here,

∠AED + 120° = 180° (Linear pair)

∠AED = 180°– 120° = 60°

We know that the sum of all angles of a triangle is 180°.

Therefore, for △ADE, we can say that:

∠ADE + ∠AED + ∠DAE = 180°

60°+ ∠ADE + 30° =180°

Or,

∠ADE = 180°– 60°– 30° = 90°

From the given figure, we can also say that:

∠FDC + 90° = 180° (Linear pair)

∠FDC = 180°– 90° = 90°

Using the above rule for △CDF, we can say that:

∠CDF + ∠DCF + ∠DFC = 180°

90° + ∠DCF + 60° =180°

∠DCF = 180°−60°− 90°= 30°

Also,

∠DCF + x = 180° (Linear pair)

30° + x = 180°

Or,

x = 180°– 30° = 150°

Question: 11

In Figure, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If ∠AFE = 130°, find

(i) ∠BDE

(ii) ∠BCA

(iii) ∠ABC

Solution:

(i) Here,

∠BAF + ∠FAD = 180° (Linear pair)

∠FAD = 180°- ∠BAF = 180°– 90° = 90°

Also,

∠AFE = ∠ADF + ∠FAD (Exterior angle property)

∠ADF + 90° = 130°

∠ADF = 130°− 90° = 40°

(ii) We know that the sum of all the angles of a triangle is 180°.

Therefore, for ∆BDE, we can say that:

∠BDE + ∠BED + ∠DBE = 180°.

∠DBE = 180°– ∠BDE ∠BED = 180°− 90°− 40°= 50° — (i)

Also,

∠FAD = ∠ABC + ∠ACB (Exterior angle property)

90° = 50° + ∠ACB

Or,

∠ACB = 90°– 50° = 40°

(iii) ∠ABC = ∠DBE = 50° [From (i)]

Question: 12

ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70°. Find ∠ACB.

Solution:

Here,

∠CAX = ∠DAX (AX bisects ∠CAD)

∠CAX =70°

∠CAX +∠DAX + ∠CAB =180°

70°+ 70° + ∠CAB =180°

∠CAB =180° –140°

∠CAB =40°

∠ACB + ∠CBA + ∠CAB = 180° (Sum of the angles of ∆ABC)

∠ACB + ∠ACB+ 40° = 180° (∠C = ∠B)

2∠ACB = 180°– 40°

∠ACB = 140/2

∠ACB = 70°

Question: 13

The side BC of ∆ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC= 30° and ∠ACD = 115°, find ∠ALC

Solution:

∠ACD and ∠ACL make a linear pair.

∠ACD+ ∠ACB = 180°

115° + ∠ACB =180°

∠ACB = 180°– 115°

∠ACB = 65°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ABC, we can say that:

∠ABC + ∠BAC + ∠ACB = 180°

30° + ∠BAC + 65° = 180°

Or,

∠BAC = 85°

∠LAC = ∠BAC/2 = 85/2

Using the above rule for ∆ALC, we can say that:

∠ALC + ∠LAC + ∠ACL = 180°

Question: 14

D is a point on the side BC of ∆ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If ∠A = 80°, ∠ABC = 60° and ∠PDC = 15°, find

(i) ∠AQD

(ii) ∠APD

Solution:

∠ABD and ∠QBD form a linear pair.

∠ABC + ∠QBC =180°

60° + ∠QBC = 180°

∠QBC = 120°

∠PDC = ∠BDQ (Vertically opposite angles)

∠BDQ = 75°

In ∆QBD:

∠QBD + ∠QDB + ∠BDQ = 180° (Sum of angles of ∆QBD)

120°+ 15° + ∠BQD = 180°

∠BQD = 180°– 135°

∠BQD = 45°

∠AQD = ∠BQD = 45°

In ∆AQP:

∠QAP + ∠AQP + ∠APQ = 180° (Sum of angles of ∆AQP)

80° + 45° + ∠APQ = 180°

∠APQ = 55°

∠APD = ∠APQ

Question: 15

Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y

Solution:

The interior angles of a triangle are the three angle elements inside the triangle.

The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Using these definitions, we will obtain the values of x and y.

(I) From the given figure, we can see that:

∠ACB + x = 180° (Linear pair)

75°+ x = 180°

Or,

x = 105°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ABC, we can say that:

∠BAC+ ∠ABC +∠ACB = 180°

40°+ y +75° = 180°

Or,

y = 65°

(ii) x + 80°= 180° (Linear pair)

= x = 100°

In ∆ABC:

x+ y+ 30° = 180° (Angle sum property)

100° + 30° + y = 180°

= y = 50°

(iii) We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ACD, we can say that:

30° + 100° + y = 180°

Or,

y = 50°

∠ACB + 100° = 180°

∠ACB = 80° … (i)

Using the above rule for ∆ACD, we can say that:

x + 45° + 80° = 180°

= x = 55°

(iv) We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆DBC, we can say that:

30° + 50° + ∠DBC = 180°

∠DBC = 100°

x + ∠DBC = 180° (Linear pair)

x = 80°

And,

y = 30° + 80° = 110° (Exterior angle property)

Question: 16

Compute the value of x in each of the following figures

Solution:

(i) From the given figure, we can say that:

∠ACD + ∠ACB = 180° (Linear pair)

Or,

∠ACB = 180°– 112° = 68°

We can also say that:

∠BAE + ∠BAC = 180° (Linear pair)

Or,

∠BAC = 180°– 120° = 60°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ABC:

x + ∠BAC + ∠ACB = 180°

x = 180°– 60°– 68° = 52°

= x = 52°

(ii) From the given figure, we can say that:

∠ABC + 120° = 180° (Linear pair)

∠ABC = 60°

We can also say that:

∠ACB+ 110° = 180° (Linear pair)

∠ACB = 70°

We know that the sum of all angles of a triangle is 180°.

Therefore, for ∆ABC:

x + ∠ABC + ∠ACB = 180°

x = 50°

(iii) From the given figure, we can see that:

∠BAD = ∠ADC = 52° (Alternate angles)

We know that the sum of all the angles of a triangle is 180°.

Therefore, for ∆DEC:

x + 40°+ 52° = 180°

= x = 88°

(iv) In the given figure, we have a quadrilateral whose sum of all angles is 360°.

Thus,

35° + 45° + 50° + reflex ∠ADC = 360°

Or,

reflex ∠ADC = 230°

230° + x = 360° (A complete angle)

= x = 130°

Exercise 15.4

Question: 1

In each of the following, there are three positive numbers. State if these numbers could possibly be the lengths of the sides of a triangle:

(i) 5, 7, 9

(ii) 2, 10.15

(iii) 3, 4, 5

(iv) 2, 5, 7

(v) 5, 8, 20

Solution:

(i) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side. Here, 5 + 7 > 9, 5 + 9 > 7, 9 + 7 > 5

(ii) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.

(iii) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of triangle is always greater than the third side. Here, 3 + 4 > 5, 3 + 5 > 4, 4 + 5 > 3

(iv) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 2 + 5 = 7

(v) No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case. Here, 5 + 8 < 20

Question: 2

In Fig, P is the point on the side BC. Complete each of the following statements using symbol ‘=’,’ > ‘or ‘< ‘so as to make it true:

(i) AP… AB+ BP

(ii) AP… AC + PC

(iii) AP…. 1/2(AB + AC + BC)

Solution:

(i) In triangle APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side.

(ii) In triangle APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side.

(iii) AP < 12(AB + AC + BC) In triangles ABP and ACP, we can see that:

AP < AB + BP…(i) (Because the sum of any two sides of a triangle is greater than the third side)

AP < AC + PC…(ii) (Because the sum of any two sides of a triangle is greater than the third side)

On adding (i) and (ii), we have:

AP + AP < AB + BP + AC + PC

2AP < AB + AC + BC (BC = BP + PC)

AP < (AB – FAC + BC)

Question: 3

P is a point in the interior of △ABC as shown in Fig. State which of the following statements are true (T) or false (F):

(i) AP + PB < AB

(ii) AP + PC > AC

(iii) BP + PC = BC

Solution:

(i) False

We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.

(ii) True

We know that the sum of any two sides of a triangle is greater than the third side: it is true for the given triangle.

(iii) False

We know that the sum of any two sides of a triangle is greater than the third side: it is not true for the given triangle.

Question: 4

O is a point in the exterior of △ABC. What symbol ‘>’,’<’ or ‘=’ will you see to complete the statement OA+OB….AB? Write two other similar statements and show that

OA + OB + OC > 1/2(AB + BC +CA)

Solution:

Because the sum of any two sides of a triangle is always greater than the third side, in triangle OAB, we have:

OA + OB > AB — (i)

OB + OC > BC — (ii)

OA + OC > CA — (iii)

On adding equations (i), (ii) and (iii) we get:

OA + OB + OB + OC + OA + OC > AB + BC + CA

2(OA + OB + OC) > AB + BC + CA

OA + OB + OC > (AB + BC + CA)/2

Question: 5

In ∆ABC, ∠B = 30°, ∠C = 50°. Name the smallest and the largest sides of the triangle.

Solution:

Because the smallest side is always opposite to the smallest angle, which in this case is 30°, it is AC. Also, because the largest side is always opposite to the largest angle, which in this case is 100°, it is BC.

Exercise 15.5

Question: 1

State Pythagoras theorem and its converse.

Solution:

The Pythagoras Theorem: In a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides.

Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle.

Question: 2

In right ∆ABC, the lengths of the legs are given. Find the length of the hypotenuse

(i) a = 6 cm, b = 8 cm

(ii) a = 8 cm, b = 15 cm

(iii) a = 3 cm, b = 4 cm

(iv) a = 2 cm, b =1.5 cm

Solution:

According to the Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Height)²

(i) c² = a² + b²

c² = 6² + 8²

c² = 36 + 64 = 100

c = 10 cm

(ii) c² = a² + b²

c² = 8² + 15²

c² = 64 + 225 = 289

c = 17cm

(iii) c² = a² + b²

c² = 3² + 4²

c² = 9 + 16 = 25

c = 5 cm

(iv) c² = a² + b²

c² = 2² + 1.5²

c² = 4 + 2.25 = 6.25

c = 2.5 cm

Question: 3

The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm. find the length of the other side.

Solution:

Let the hypotenuse be ” c ” and the other two sides be ” b ” and ” c”.

Using the Pythagoras theorem, we can say that:

c² = a² + b²

2.52 = 1.52 + b²

b² = 6.25 −2.25 = 4

c = 2 cm

Hence, the length of the other side is 2 cm.

Question: 4

A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches.

Solution:

Let the hypotenuse be h.

Using the Pythagoras theorem, we get:

3.72 = 1.22 + h²

h² = 13.69 – 1.44 = 12.25

h = 3.5 m

Hence, the height of the wall is 3.5 m.

Question: 5

If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle.

Solution:

In the given triangle, the largest side is 6 cm.

We know that in a right angled triangle, the sum of the squares of the smaller sides should be equal to the square of the largest side.

Therefore,

3² + 42 = 9 + 16 = 25

But,

6² = 36

3² + 4² not equal to 6²

Hence, the given triangle is not a right angled triangle.

Question: 6

The sides of certain triangles are given below. Determine which of them are right triangles.

(i) a = 7 cm, b = 24 cm and c= 25 cm

(ii) a = 9 cm, b = 16 cm and c = 18 cm

Solution:

(i) We know that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides.

Here, the larger side is c, which is 25 cm.

c² = 625

We have:

a²+ b² = 7² + 24² = 49 + 576 = 625 = c²

Thus, the given triangle is a right triangle.

(ii) We know that in a right angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides.

Here, the larger side is c, which is 18 cm.

c² = 324

We have:

a²+ b² = 9²+16² = 81 + 256 = 337 not equal to c²

Thus, the given triangle is not a right triangle.

Question: 7

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m.  Find the distance between their tops.

(Hint: Find the hypotenuse of a right triangle having the sides (11 – 6) m = 5 m and 12 m)

Solution:

The distance between the tops of the poles is the distance between points A and B.

We can see from the given figure that points A, B and C form a right triangle, with AB as the hypotenuse.

On using the Pythagoras Theorem in ∆ABC, we get:

(11−6)² + 12² = AB²

AB² = 25 + 144

AB² = 169

AB = 13

Hence, the distance between the tops of the poles is 13 m.

Question: 8

A man goes 15 m due west and then 8 m due north. How far is he from the starting point?

Solution:

Let O be the starting point and P be the final point.

By using the Pythagoras theorem, we can find the distance OP.

OP² = 15² + 8²

OP² = 225 + 64

OP² = 289

OP = 17

Hence, the required distance is 17 m.

Question: 9

The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?

Solution:

Given Let the length of the ladder be L m.

By using the Pythagoras theorem, we can find the length of the ladder.

6² + 8² = L²

L² = 36 + 64 =100

L = 10

Thus, the length of the ladder is 10 m.

When the ladder is shifted:

Let the height of the ladder after it is shifted be H m.

By using the Pythagoras theorem, we can find the height of the ladder after it is shifted.

8² + H² = 10²

H² = 100 – 64 = 36

H = 6

Thus, the height of the ladder is 6 m.

Question: 10

A ladder 50 dm long when set against the wall of a house just reaches a window at a height of 48 dm. How far is the lower end of the ladder from the base of the wall?

Solution:

Let the distance of the lower end of the ladder from the wall be x m.

On using the Pythagoras theorem, we get:

x² + 48² = 50²

x² = 50² − 48² = 2500 – 2304 = 196

H = 14 dm

Hence, the distance of the lower end of the ladder from the wall is 14 dm.

Question: 11

The two legs of a right triangle are equal and the square of the hypotenuse is 50. Find the length of each leg. 

Solution:

Let the length of each leg of the given triangle be x units.

Using the Pythagoras theorem, we get:

x² + x² = (Hypotenuse)²

x² + x² = 50

2x² =  50

x² = 25

x = 5

Hence, we can say that the length of each leg is 5 units.

Question: 12

Verity that the following numbers represent Pythagorean triplet:

(i) 12, 35, 37

(ii) 7, 24, 25

(iii) 27, 36, 45

(iv) 15, 36, 39

Solution:

We will check for a Pythagorean triplet by checking if the square of the largest side is equal to the sum of the squares of the other two sides.

(i) 372 =1369

122 + 352 = 144 + 1225 = 1369

122 + 352 = 372

Yes, they represent a Pythagorean triplet.

(ii) 25² = 625

7² + 24² = 49 + 576 = 625

7² + 24² = 25²

Yes, they represent a Pythagorean triplet.

(iii) 45² = 2025

27² + 36² = 729 + 1296 = 2025

27² + 36² = 45²

Yes, they represent a Pythagorean triplet.

(iv) 39² = 1521

15² + 36² = 225 + 1296 = 1521

15² + 36² = 39²

Yes, they represent a Pythagorean triplet.

Question: 13

In ∆ABC, ∠ABC = 100°, ∠BAC = 35° and BD perpendicular to AC meets side AC in D. If BD = 2 cm, find ∠C, and length DC.

Solution:

We know that the sum of all angles of a triangle is 180°

Therefore, for the given ∆ABC, we can say that:

∠ABC + ∠BAC + ∠ACB = 180°

100° + 35° + ∠ACB = 180°

∠ACB = 180° –135°

∠ACB = 45°

∠C = 45°

If we apply the above rule on ∆BCD, we can say that:

∠BCD + ∠BDC + ∠CBD = 180°

45° + 90° + ∠CBD = 180° (∠ACB = ∠BCD and BD parallel to AC)

∠CBD = 180°– 135°

∠CBD = 45°

We know that the sides opposite to equal angles have equal length.

Thus, BD = DC

DC = 2 cm

Question: 14

In a ∆ABC, AD is the altitude from A such that AD = 12 cm. BD = 9 cm and DC = 16 cm. Examine if ∆ABC is right angled at A.

Solution:

In ∆ADC,

∠ADC = 90° (AD is an altitude on BC)

Using the Pythagoras theorem, we get:

12² + 16² = AC²

AC² = 144 + 256 = 400

AC = 20 cm

In ∆ADB,

∠ADB = 90° (AD is an altitude on BC)

Using the Pythagoras theorem, we get:

12² + 9² = AB²

AB² = 144 + 81 = 225

AB = 15 cm

In ∆ABC,

BC² = 25² = 625

AB² + AC² = 15² + 20² = 625

AB² + AC² = BC²

Because it satisfies the Pythagoras theorem, we can say that ∆ABC is right angled at A.

Question: 15

Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 105°. Measure AB. Is (AC)² + (BC)²? If not which one of the following is true: (AB)² > (AC)² + (BC)² or (AB)² < (AC)² + (BC)²

Solution:

Draw ∆ABC.

Draw a line BC = 3 cm.

At point C, draw a line at 105° angle with BC.

Take an arc of 4 cm from point C, which will cut the line at point A.

Now, join AB, which will be approximately 5.5 cm.

AC² + BC² = 4² + 3² = 9 +16 = 25

AB² = 5.52 = 30.25

AB² not equal to AC²+ BC²

Here,

AB² > AC² + BC²

Question: 16

Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 80°. Measure AB. Is (AC)² + (BC)²? If not which one of the following is true: (AB)² > (AC)² + (BC)² or (AB)² < (AC)² + (BC)²

Solution:

Draw ∆ABC.

Draw a line BC = 3 cm.

At point C, draw a line at 80° angle with BC.

Take an arc of 4 cm from point C, which will cut the line at point A.

Now, join AB; it will be approximately 4.5 cm.

AC² + BC² = 4² + 3² = 9 +16 = 25

AB² = 4.5² = 20.25

AB² not equal to AC² + BC²

Here,

AB² < AC² + BC²

Exercise14.1

Question: 1

Write down each pair of adjacent angles shown in Figure

Solution:

The angles that have common vertex and a common arm are known as adjacent angles

The adjacent angles are:

∠DOC and ∠BOC

∠COB and ∠BOA

Question: 2

In figure, name all the pairs of adjacent angles.

Solution:

In fig (i), the adjacent angles are

∠EBA and ∠ABC

∠ACB and ∠BCF

∠BAC and ∠CAD

In fig (ii), the adjacent angles are

∠BAD and ∠DAC

∠BDA and ∠CDA

Question: 3

In fig , write down

(i) each linear pair

(ii) each pair of vertically opposite angles.

Solution:

(i) The two adjacent angles are said to form a linear pair of angles if their non – common arms are two opposite rays.

∠1 and ∠3

∠1 and ∠2

∠4 and ∠3

∠4 and ∠2

∠5 and ∠6

∠5 and ∠7

∠6 and ∠8

∠7 and ∠8

(ii) The two angles formed by two intersecting lines and have no common arms are called vertically opposite angles.

∠1 and ∠4

∠2 and ∠3

∠5 and ∠8

∠6 and ∠7

Question: 4

Are the angles 1 and 2 in figure adjacent angles?

Solution:

No, because they do not have common vertex.

Question: 5

Find the complement of each of the following angles:

(i) 35°

(ii) 72°

(iii) 45°

(iv) 85°

Solution:

The two angles are said to be complementary angles if the sum of those angles is 90°

Complementary angles for the following angles are:

(i) 90° – 35° = 55°

(ii) 90° – 72° = 18°

(iii) 90° – 45° = 45°

(iv) 90° – 85° = 5°

Question: 6

Find the supplement of each of the following angles:

(i) 70°

(ii) 120°

(iii) 135°

(iv) 90°

Solution:

The two angles are said to be supplementary angles if the sum of those angles is 180°

(i) 180° – 70° = 110°

(ii) 180° – 120° = 60°

(iii) 180° – 135° = 45°

(iv) 180° – 90° = 90°

Question: 7

Identify the complementary and supplementary pairs of angles from the following pairs

(i) 25°, 65°

(ii) 120°, 60°

(iii) 63°, 27°

(iv) 100°, 80°

Solution:

(i) 25° + 65° = 90° so, this is a complementary pair of angle.

(ii) 120° + 60° = 180° so, this is a supplementary pair of angle.

(iii) 63° + 27° = 90° so, this is a complementary pair of angle.

(iv) 100° + 80° = 180° so, this is a supplementary pair of angle.

Here, (i) and (iii) are complementary pair of angles and (ii) and (iv) are supplementary pair of angles.

Question: 8

Can two obtuse angles be supplementary, if both of them be

(i) obtuse?

(ii) right?

(iii) acute?

Solution:

(i) No, two obtuse angles cannot be supplementary

Because, the sum of two angles is greater than 90 degrees so their sum will be greater than 180degrees.

(ii) Yes, two right angles can be supplementary

Because, 90° + 90° = 180°

(iii) No, two acute angle cannot be supplementary

Because, the sum of two angles is less than 90 degrees so their sum will also be less tha 90 degrees.

Question: 9

Name the four pairs of supplementary angles shown in Fig.

Solution:

The supplementary angles are

∠AOC and ∠COB

∠BOC and ∠DOB

∠BOD and ∠DOA

∠AOC and ∠DOA

Question: 10

In Figure, A, B, C are collinear points and ∠DBA = ∠EBA.

(i) Name two linear pairs.

(ii) Name two pairs of supplementary angles.

Solution:

(i) Linear pairs

∠ABD and ∠DBC

∠ABE and ∠EBC

Because every linear pair forms supplementary angles, these angles are

∠ABD and ∠DBC

∠ABE and ∠EBC

Question: 11

If two supplementary angles have equal measure, what is the measure of each angle?

Solution:

Let p and q be the two supplementary angles that are equal

∠p = ∠q

So,

∠p + ∠q = 180°

=> ∠p + ∠p = 180°

=> 2∠p = 180°

=> ∠p = 180°/2

=> ∠p = 90°

Therefore, ∠p = ∠q = 90°

Question: 12

If the complement of an angle is 28°, then find the supplement of the angle.

Solution:

Here, let p be the complement of the given angle 28°

Therefore, ∠p + 28° = 90°

=> ∠p = 90° – 28°

= 62°

So, the supplement of the angle = 180° – 62°

= 118°

Question: 13

In Figure, name each linear pair and each pair of vertically opposite angles.

Solution:

Two adjacent angles are said to be linear pair of angles, if their non-common arms are two opposite rays.

∠1 and ∠2

∠2 and ∠3

∠3 and ∠4

∠1 and ∠4

∠5 and ∠6

∠6 and ∠7

∠7 and ∠8

∠8 and ∠5

∠9 and ∠10

∠10 and ∠11

∠11 and ∠12

∠12 and ∠9

The two angles are said to be vertically opposite angles if the two intersecting lines have no common arms.

∠1 and ∠3

∠4 and ∠2

∠5 and ∠7

∠6 and ∠8

∠9 and ∠11

∠10 and ∠12

Question: 14

In Figure, OE is the bisector of ∠BOD. If ∠1 = 70°, Find the magnitude of ∠2, ∠3, ∠4

Solution:

Given,

∠1 = 70°

∠3 = 2(∠1)

= 2(70°)

= 140°

∠3 = ∠4

As, OE is the angle bisector,

∠DOB = 2(∠1)

= 2(70°)

= 140°

∠DOB + ∠AOC + ∠COB +∠DOB = 360°

=> 140° + 140° + 2(∠COB) = 360°

Since, ∠COB = ∠AOD

=> 2(∠COB) = 360° – 280°

=> 2(∠COB) = 80°

=> ∠COB = 80°/2

=> ∠COB = 40°

Therefore, ∠COB = ∠AOB = 40°

The angles are,

∠1 = 70°,

∠2 = 40°,

∠3 = 140°,

∠4 = 40°

Question: 15

One of the angles forming a linear pair is a right angle. What can you say about its other angle?

Solution:

One of the Angle of a linear pair is the right angle (90°)

Therefore, the other angle is

=> 180° – 90° = 90°

Question: 16

One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?

Solution:

One of the Angles of a linear pair is obtuse, then the other angle should be acute, only then their sum will be 180°.

Question: 17

One of the angles forming a linear pair is an acute angle. What kind of angle is the other?

Solution:

One of the Angles of a linear pair is acute, then the other angle should be obtuse, only then their sum will be 180°.

Question: 18

Can two acute angles form a linear pair?

Solution:

No, two acute angles cannot form a linear pair because their sum is always less than 180°.

Question: 19

If the supplement of an angle is 65°, then find its complement.

Solution:

Let x be the required angle

So,

=> x + 65° = 180°

=> x = 180° – 65°

= 115°

But the complement of the angle cannot be determined.

Question:20

In Fig. 22, it being given that ∠1 = 65°, find all the other angles.

Solution:

Given,

∠1 = ∠3 are the vertically opposite angles

Therefore, ∠3 = 65°

Here, ∠1 + ∠2 = 180° are the linear pair

Therefore, ∠2 = 180° – 65°

= 115°

∠2 = ∠4 are the vertically opposite angles

Therefore, ∠2 = ∠4 = 115°

And ∠3 = 65°

Question: 21

In Fig. 23 OA and OB are the opposite rays:

(i) If x = 25°, what is the value of y?

(ii) If y = 35°, what is the value of x?

Solution:

∠AOC + ∠BOC = 180° – Linear pair

=> 2y + 5 + 3x = 180°

=> 3x + 2y = 175°

(i) If x = 25°, then

=> 3(25°) + 2y = 175°

=> 75° + 2y = 175°

=> 2y = 175° – 75°

=> 2y = 100°

=> y = 100°/2

=> y = 50°

(ii) If y = 35°, then

3x + 2(35°) = 175°

=> 3x + 70° = 175°

=> 3x = 175° – 70°

=> 3x = 105°

=> x = 105°/3

=> x = 35°

Question: 22

In Figure, write all pairs of adjacent angles and all the linear pairs.

Solution:

Pairs of adjacent angles are:

∠DOA and ∠DOC

∠BOC and ∠COD

∠AOD and ∠BOD

∠AOC and ∠BOC

Linear pairs:

∠AOD and ∠BOD

∠AOC and ∠BOC

Question: 23

In Figure, find ∠x. Further find ∠BOC, ∠COD, ∠AOD

Solution:

(x + 10)° + x° + (x + 20) ° = 180°

=> 3x° + 30° = 180°

=> 3x° = 180° – 30°

=> 3x° = 150°

=> x° = 150°/3

=> x° = 50°

Here,

∠BOC = (x + 20)°

= (50 + 20)°

= 70°

∠COD = 50°

∠AOD = (x + 10)°

= (50 + 10)°

= 60°

Question: 24

How many pairs of adjacent angles are formed when two lines intersect in a point?

Solution:

If the two lines intersect at a point, then four adjacent pairs are formed and those are linear.

Question: 26

How many pairs of adjacent angles, in all, can you name in Figure?

Solution:

There are 10 adjacent pairs

∠EOD and ∠DOC

∠COD and ∠BOC

∠COB and ∠BOA

∠AOB and ∠BOD

∠BOC and ∠COE

∠COD and ∠COA

∠DOE and ∠DOB

∠EOD and ∠DOA

∠EOC and ∠AOC

∠AOB and ∠BOE

Question: 25

In Figure, determine the value of x.

Solution:

Linear pair:

∠COB + ∠AOB = 180°

=> 3x° + 3x° = 180°

=> 6x° = 180°

=> x° = 180°/6

=> x° = 30°

Question: 26

In Figure, AOC is a line, find x.

Solution:

∠AOB + ∠BOC = 180°

Linear pair

=> 2x + 70° = 180°

=> 2x = 180° – 70°

=> 2x = 110°

=> x = 110°/2

=> x = 55°

Question: 27

In Figure, POS is a line, find x.

Solution:

Angles of a straight line,

∠QOP + ∠QOR + ∠ROS = 108°

=> 60° + 4x + 40° = 180°

=> 100° + 4x = 180°

=> 4x = 180° – 100°

=> 4x = 80°

=> x = 80°/4

=> x = 20°

Question: 28

In Figure, lines l₁ and l₂ intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.

Solution:

Given that,

∠x = 45°

∠x = ∠z = 45°

∠y = ∠u

∠x + ∠y + ∠z + ∠u = 360°

=> 45° + 45° + ∠y + ∠u = 360°

=> 90° + ∠y + ∠u = 360°

=> ∠y + ∠u = 360° – 90°

=> ∠y + ∠u = 270°

=> ∠y + ∠z = 270°

=> 2∠z = 270°

=> ∠z = 135°

Therefore, ∠y = ∠u = 135°

So, ∠x = 45°,

∠y = 135°,

∠z = 45°,

∠u = 135°

Question: 29

In Figure, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u

Solution:

Given that,

∠x + ∠y + ∠z + ∠u + 50° + 90° = 360°

Linear pair,

∠x + 50° + 90° = 180°

=> ∠x + 140° = 180°

=> ∠x = 180° – 140°

=> ∠x = 40°

∠x = ∠u = 40° are vertically opposite angles

=> ∠z = 90° is a vertically opposite angle

=> ∠y = 50° is a vertically opposite angle

Therefore, ∠x = 40°,

∠y = 50°,

∠z = 90°,

∠u = 40°

Question: 32

In Figure, find the values of x, y and z

Solution:

∠y = 25° vertically opposite angle

∠x = ∠y are vertically opposite angles

∠x + ∠y + ∠z + 25° = 360°

=> ∠x + ∠z + 25° + 25° = 360°

=> ∠x + ∠z + 50° = 360°

=> ∠x + ∠z = 360° – 50°

=> 2∠x = 310°

=> ∠x = 155°

And, ∠x = ∠z = 155°

Therefore, ∠x = 155°,

∠y = 25°,

∠z = 155°

Exercise 14.2

Question: 1

In Figure, line n is a transversal to line l and m. Identify the following:

(i) Alternate and corresponding angles in Figure.(i)

(ii) Angles alternate to ∠d and ∠g and angles corresponding to ∠RQF and angle alternate to ∠PQE in Figure. (ii)

(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Figure. (iii)

(iv) Pairs of interior and exterior angles on the same side of the transversal in Figure. (iii)

Solution:

(i) Figure (i)

Corresponding angles:

∠EGB and ∠GHD

∠HGB and ∠FHD

∠EGA and ∠GHC

∠AGH and ∠CHF

The alternate angles are:

∠EGB and ∠CHF

∠HGB and ∠CHG

∠EGA and ∠FHD

∠AGH and ∠GHD

(ii) Figure (ii)

The alternate angle to ∠d is ∠e.

The alternate angle to ∠g is ∠b.

The corresponding angle to ∠f is ∠c.

The corresponding angle to ∠h is ∠a.

(iii) Figure (iii)

Angle alternate to ∠PQR is ∠QRA.

Angle corresponding to ∠RQF is ∠ARB.

Angle alternate to ∠POE is ∠ARB.

(iv) Figure (ii)

Pair of interior angles are

∠a is ∠e.

∠d is ∠f.

Pair of exterior angles are

∠b is ∠h.

∠c is ∠g.

Question: 2

In Figure, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively, If ∠CMQ = 60°, find all other angles in the figure.

Solution:

Corresponding angles:

∠ALM = ∠CMQ = 60°

Vertically opposite angles:

∠LMD = ∠CMQ = 60°

Vertically opposite angles:

∠ALM = ∠PLB = 60°

Here,

∠CMQ + ∠QMD = 180° are the linear pair

= ∠QMD = 180° – 60°

= 120°

Corresponding angles:

∠QMD = ∠MLB = 120°

Vertically opposite angles

∠QMD = ∠CML = 120°

Vertically opposite angles

∠MLB = ∠ALP = 120°

Question: 3

In Figure, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.

Solution:

Given that,

∠LMD = 35°

∠LMD and ∠LMC is a linear pair

∠LMD + ∠LMC = 180°

= ∠LMC = 180° – 35°

= 145°

So, ∠LMC = ∠PLA = 145°

And, ∠LMC = ∠MLB = 145°

∠MLB and ∠ALM is a linear pair

∠MLB + ∠ALM = 180°

= ∠ALM = 180° – 145°

= ∠ALM = 35°

Therefore, ∠ALM = 35°, ∠PLA = 145°.

Question: 4

The line n is transversal to line l and m in Figure. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.

Solution:

Given that, l ∥ m

So,

The angle alternate to ∠13 is ∠7

The angle corresponding to ∠15 is ∠7

The angle alternate to ∠15 is ∠5

Question: 5

In Figure, line l ∥ m and n is transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.

Solution:

Given that,

∠1 = 40°

∠1 and ∠2 is a linear pair

= ∠1 + ∠2 = 180°

= ∠2 = 180° – 40°

= ∠2 = 140°

∠2 and ∠6 is a corresponding angle pair

So, ∠6 = 140°

∠6 and ∠5 is a linear pair

= ∠6 + ∠5 = 180°

= ∠5 = 180° – 140°

= ∠5 = 40°

∠3 and ∠5 are alternative interior angles

So, ∠5 = ∠3 = 40°

∠3 and ∠4 is a linear pair

= ∠3 + ∠4 = 180°

= ∠4 = 180° – 40°

= ∠4 = 140°

∠4 and ∠6 are a pair interior angles

So, ∠4 = ∠6 = 140°

∠3 and ∠7 are pair of corresponding angles

So, ∠3 = ∠7 = 40°

Therefore, ∠7 = 40°

∠4 and ∠8 are a pair corresponding angles

So, ∠4 = ∠8 = 140°

Therefore, ∠8 = 140°

So, ∠1 = 40°, ∠2 = 140°, ∠3 = 40°, ∠4 = 140°, ∠5 = 40°, ∠6 = 140°, ∠7 = 40°, ∠8 = 140°

Question: 6

In Figure, line l ∥ m and a transversal n cuts them P and Q respectively. If ∠1 = 75°, find all other angles.

Solution:

Given that, l ∥ m and ∠1 = 75∘

We know that,

∠1 + ∠2 = 180° → (linear pair)

= ∠2 = 180° – 75°

= ∠2 = 105°

here, ∠1 = ∠5 = 75° are corresponding angles

∠5 = ∠7 = 75° are vertically opposite angles.

∠2 = ∠6 = 105° are corresponding angles

∠6 = ∠8 = 105° are vertically opposite angles

∠2 = ∠4 = 105° are vertically opposite angles

So, ∠1 = 75°, ∠2 = 105°, ∠3 = 75°, ∠4 = 105°, ∠5 = 75°, ∠6 = 105°, ∠7 = 75°, ∠8 = 105°

Question: 7

In Figure, AB ∥ CD and a transversal PQ cuts at L and M respectively. If ∠QMD = 100°, find all the other angles.

Solution:

Given that, AB ∥ CD and ∠QMD = 100°

We know that,

Linear pair,

∠QMD + ∠QMC = 180°

= ∠QMC = 180° – ∠QMD

= ∠QMC = 180° – 100°

= ∠QMC = 80°

Corresponding angles,

∠DMQ = ∠BLM = 100°

∠CMQ = ∠ALM = 80°

Vertically Opposite angles,

∠DMQ = ∠CML = 100°

∠BLM = ∠PLA = 100°

∠CMQ = ∠DML = 80°

∠ALM = ∠PLB = 80°

Question: 8

In Figure, l ∥ m and p || q. Find the values of x, y, z, t.

Solution:

Give that, angle is 80°

∠z and 80° are vertically opposite angles

= ∠z = 80°

∠z and ∠t are corresponding angles

= ∠z = ∠t

Therefore, ∠t = 80°

∠z and ∠y are corresponding angles

= ∠z = ∠y

Therefore, ∠y = 80°

∠x and ∠y are corresponding angles

= ∠y = ∠x

Therefore, ∠x = 80°

Question: 9

In Figure, line l ∥ m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.

Solution:

Given that, ∠1 = 120° and ∠2 = 100°

∠1 and ∠5 a linear pair

= ∠1 + ∠5 = 180°

= ∠5 = 180° – 120°

= ∠5 = 60°

Therefore, ∠5 = 60°

∠2 and ∠6 are corresponding angles

= ∠2 = ∠6 = 100°

Therefore, ∠6 = 100°

∠6 and ∠3 a linear pair

= ∠6 + ∠3 = 180°

= ∠3 = 180° – 100°

= ∠3 = 80°

Therefore, ∠3 = 80°

By, angles of sum property

= ∠3 + ∠5 + ∠4 = 180°

= ∠4 = 180° – 80° – 60°

= ∠4 = 40°

Therefore, ∠4 = 40°

Question: 10

In Figure, l ∥ m. Find the values of a, b, c, d. Give reasons.

Solution:

Given that, l ∥ m

Vertically opposite angles,

∠a = 110°

Corresponding angles,

∠a = ∠b

Therefore, ∠b = 110°

Vertically opposite angle,

∠d = 85°

Corresponding angles,

∠d = ∠c

Therefore, ∠c = 85°

Hence, ∠a = 110°, ∠b = 110°, ∠c = 85°, ∠d = 85°

Question: 11

In Figure, AB ∥ CD and ∠1 and ∠2 are in the ratio of 3: 2. Determine all angles from 1 to 8.

Solution:

Given that,

∠1 and ∠2 are 3: 2

Let us take the angles as 3x, 2x

∠1 and ∠2 are linear pair

= 3x + 2x = 180°

= 5x = 180°

= x = 180°/5

= x = 36°

Therefore, ∠1 = 3x = 3(36) = 108°

∠2 = 2x = 2(36) = 72°

∠1 and ∠5 are corresponding angles

= ∠1 = ∠5

Therefore, ∠5 = 108°

∠2 and ∠6 are corresponding angles

= ∠2 = ∠6

Therefore, ∠6 = 72°

∠4 and ∠6 are alternate pair of angles

= ∠4 = ∠6 = 72°

Therefore, ∠4 = 72°

∠3 and ∠5 are alternate pair of angles

= ∠3 = ∠5 = 108°

Therefore, ∠5 = 108°

∠2 and ∠8 are alternate exterior of angles

= ∠2 = ∠8 = 72°

Therefore, ∠8 = 72°

∠1 and ∠7 are alternate exterior of angles

= ∠1 = ∠7 = 108°

Therefore, ∠7 = 108°

Hence, ∠1 = 108°, ∠2 = 72°, ∠3 = 108°, ∠4 = 72°, ∠5 = 108°, ∠6 = 72°, ∠7 = 108°, ∠8 = 72°

Question: 12

In Figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.

Solution:

Linear pair,

= ∠4 + 60° = 180°

= ∠4 = 180° – 60∘

= ∠4 = 120°

∠4 and ∠1 are corresponding angles

= ∠4 = ∠1

Therefore, ∠1 = 120°

∠1 and ∠2 are corresponding angles

= ∠2 = ∠1

Therefore, ∠2 = 120°

∠2 and ∠3 are vertically opposite angles

= ∠2 = ∠3

Therefore, ∠3 = 120°

Question: 13

In Figure, if l ∥ m ∥ n and ∠1 = 60°, find ∠2

Solution:

Given that,

Corresponding angles:

∠1 = ∠3

= ∠1 = 60°

Therefore, ∠3 = 60°

∠3 and ∠4 are linear pair

= ∠3 + ∠4 = 180°

= ∠4 = 180° – 60°

= ∠4 = 120°

∠3 and ∠4 are alternative interior angles

= ∠4 = ∠2

Therefore, ∠2 = 120°

Question: 14

In Figure, if AB ∥ CD and CD ∥ EF, find ∠ACE

Solution:

Given that,

Sum of the interior angles,

= ∠CEF + ∠ECD = 180°

= 130° + ∠ECD = 180°

= ∠ECD = 180° – 130°

= ∠ECD = 50°

We know that alternate angles are equal

= ∠BAC = ∠ACD

= ∠BAC = ∠ECD + ∠ACE

= ∠ACE = 70° – 50°

= ∠ACE = 20°

Therefore, ∠ACE = 20°

Question: 15

In Figure, if l ∥ m, n ∥ p and ∠1 = 85°, find ∠2.

Solution:

Given that, ∠1 = 85°

∠1 and ∠3 are corresponding angles

So, ∠1 = ∠3

= ∠3 = 85°

Sum of the interior angles

= ∠3 + ∠2 = 180°

= ∠2 = 180° – 85°

= ∠2 = 95°

Question: 16

In Figure, a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l ∥ m?

Solution:

We know that if the alternate exterior angles of the two lines are equal, then the lines are parallel.

Here, ∠1 and ∠7 are alternate exterior angles, but they are not equal

= ∠1 ≠ ∠7 ≠ 80°

Question: 17

In Figure, a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?

Solution:

vertically opposite angels,

∠2 = ∠3 = 65°

∠8 = ∠6 = 65°

Therefore, ∠3 = ∠6

Hence, l ∥ m

Question: 18

In Figure, Show that AB ∥ EF.

Solution:

We know that,

∠ACD = ∠ACE + ∠ECD

= ∠ACD = 35° + 22°

= ∠ACD = 57° = ∠BAC

Thus, lines BA and CD are intersected by the line AC such that, ∠ACD = ∠BAC

So, the alternate angles are equal

Therefore, AB ∥ CD  — 1

Now,

∠ECD + ∠CEF = 35° + 45° = 180°

This, shows that sum of the angles of the interior angles on the same side of the transversal CE is 180 degrees

So, they are supplementary angles

Therefore, EF ∥ CD    ——- 2

From eq 1 and 2

We can say that, AB ∥ EF

Question: 19

In Figure, AB ∥ CD. Find the values of x, y, z.

Solution:

Linear pair,

= ∠x + 125° = 180°

= ∠x = 180° – 125°

= ∠x = 55°

Corresponding angles

= ∠z = 125°

Adjacent interior angles

= ∠x + ∠z = 180°

= ∠x + 125° = 180°

= ∠x = 180° – 125°

= ∠x = 55°

Adjacent interior angles

= ∠x + ∠y = 180°

= ∠y + 55° = 180°

= ∠y = 180° – 55°

= ∠y = 125°

Question: 20

In Figure, find out ∠PXR, if PQ ∥ RS.

Solution:

We need to find ∠PXR

∠XRS = 50°

∠XPR = 70°

Given, that PQ ∥ RS

∠PXR = ∠XRS + ∠XPR

∠PXR = 50° + 70°

∠PXR = 120°

Therefore, ∠PXR = 120°

Question: 21

In Figure, we have

(i) ∠MLY = 2∠LMQ

(ii) ∠XLM = (2x – 10)° and ∠LMQ = (x + 30) °, find x.

(iii) ∠XLM = ∠PML, find ∠ALY

(iv) ∠ALY = (2x – 15) °, ∠LMQ = (x + 40) °, find x.

Solution:

(i) ∠MLY and ∠LMQ are interior angles

= ∠MLY + ∠LMQ = 180°

= 2∠LMQ + ∠LMQ = 180°

= 3∠LMQ = 180°

= ∠LMQ = 180°/3

= ∠LMQ = 60°

(ii) ∠XLM = (2x – 10)° and ∠LMQ = (x + 30) °, find x.

∠XLM = (2x – 10) ° and ∠LMQ = (x + 30) °

∠XLM and ∠LMQ are alternate interior angles

= ∠XLM = ∠LMQ

= (2x – 10) ° = (x + 30) °

= 2x – x = 30° + 10°

=      x = 40°

Therefore, x = 40°

(iii) ∠XLM = ∠PML, find ∠ALY

∠XLM = ∠PML

Sum of interior angles is 180 degrees

= ∠XLM + ∠PML = 180°

= ∠XLM + ∠XLM = 180°

= 2∠XLM = 180°

= ∠XLM = 180°/2

= ∠XLM = 90°

∠XLM and ∠ALY are vertically opposite angles

Therefore, ∠ALY = 90°

(iv) ∠ALY = (2x – 15) °, ∠LMQ = (x + 40) °, find x.

∠ALY and ∠LMQ are corresponding angles

= ∠ALY = ∠LMQ

= (2x – 15) °= (x + 40) °

= 2x – x = 40° + 15°

=x = 55° 

Therefore, x = 55°

Question: 22

In Figure, DE ∥ BC. Find the values of x and y.

Solution:

We know that,

ABC, DAB are alternate interior angles

∠ABC = ∠DAB

So, x = 40°

And ACB, EAC are alternate interior angles

∠ACB = ∠EAC

So, y = 40°

Question: 23

In Figure, line AC ∥ line DE and ∠ABD = 32°, Find out the angles x and y if ∠E = 122°.

Solution:

∠BDE = ∠ABD = 32° – Alternate interior angles

= ∠BDE + y = 180°– linear pair

= 32° + y = 180°

= y = 180° – 32°

= y = 148°

∠ABE = ∠E = 32° – Alternate interior angles

= ∠ABD + ∠DBE = 122°

= 32° + x = 122°

= x = 122° – 32°

= x = 90°

Question: 24

In Figure, side BC of ΔABC has been produced to D and CE ∥ BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD, ∠ACD.

Solution:

Corresponding angles,

∠ABC = ∠ECD = 55°

Alternate interior angles,

∠BAC = ∠ACE = 65°

Now, ∠ACD = ∠ACE + ∠ECD

= ∠ACD = 55° + 65°

= 120°

Question: 25

In Figure, line CA ⊥ AB ∥ line CR and line PR ∥ line BD. Find ∠x, ∠y, ∠z.

Solution:

Given that, CA ⊥ AB

= ∠CAB = 90°

= ∠AQP = 20°

By, angle of sum property

In ΔAPD

= ∠CAB + ∠AQP + ∠APQ = 180∘

= ∠APQ = 180° – 90° – 20°

= ∠APQ = 70°

y and ∠APQ are corresponding angles

= y = ∠APQ = 70°

∠APQ and ∠z are interior angles

= ∠APQ + ∠z = 180°

= ∠z = 180° – 70°

= ∠z = 110°

Question: 26

In Figure, PQ ∥ RS. Find the value of x.

Solution:

Given,

Linear pair,

∠RCD + ∠RCB = 180°

= ∠RCB = 180° – 130°

= 50°

In ΔABC,

∠BAC + ∠ABC + ∠BCA = 180°

By, angle sum property

= ∠BAC = 180° – 55° – 50°

= ∠BAC = 75°

Question: 27

In Figure, AB ∥ CD and AE ∥ CF, ∠FCG = 90° and ∠BAC = 120°. Find the value of x, y and z.

Solution:

Alternate interior angle

∠BAC = ∠ACG = 120°

=   ∠ACF + ∠FCG = 120°

So, ∠ACF = 120° – 90°

= 30°

Linear pair,

∠DCA + ∠ACG = 180°

= ∠x = 180° – 120°

= 60°

∠BAC + ∠BAE + ∠EAC = 360°

∠CAE = 360° – 120° – (60° + 30°)

= 150°

Exercise 13.1

Question: 1

Find the simple interest , when :

Solution:

(i) . Principal = Rs . 2000 , Rate of interest = 5% per annum , and Time = 5 years

(ii) . Principal = Rs . 500 , Rate of interest = 12.5% per annum , and Time = 4 years

(iii) . Principal = Rs . 4500 , Rate of interest = 4% per annum , and Time = 6 months

(iv) . Principal =Rs . 12000 , Rate of interest = 18% per annum , and Time = 4 months

(v) . Principal = Rs . 1000 , Rate of interest = 10% per annum , and Time = 73 days

Question: 2

Find the interest on Rs . 500 for a period of 4 years at the rate of 8% per annum . Also , find the amount to be paid at the end of the period .

Solution:

Question: 3

A sum of Rs . 400 is lent at the rate of 5% per annum . Find the interest at the end of 2 years .

Solution:

Question: 4

A sum of Rs . 400 is lent for 3 years at the rate of 6% per annum . Find the interest .

Solution:

Question: 5

A person deposits Rs . 25000 in a firm who pays an interest at the rate of 20 % per annum . Calculate the income he gets from it annually .

Solution:

Question: 6

A man borrowed Rs . 8000 from a bank at 8% per annum . Find the amount he has to pay after 412 years .

Solution:

Question: 7

Rakesh lent out Rs . 8000 for 5 years at 15 % per annum and borrowed Rs . 6000 for 3 years at 12% per annum . How much did he gain or lose ?

Solution:

Question: 8

Anita deposits Rs . 1000 in a savings bank account . The bank pays interest at the rate of 5 % per annum . What amount can Anita get after 1 year ?

Solution:

Question: 9

Nalini borrowed Rs . 550 from her friend at 8% per annum . She returned the amount after six months . How much did she pay ?

Solution:

Question: 10

Rohit borrowed Rs . 60000 from a bank at 9% per annum for 2 years . He lent this sum of money to Rohan at 10% per annum for 2 years . How much did Rohit earn from this transaction ?

Solution:

Question: 11

Romesh borrowed Rs . 2000 at 2% per annum and Rs . 1000 at 5% per annum . He cleared his debt after 2 years by giving Rs . 2800 and a watch . What is the cost of watch ?

Solution:

Question: 12

Mr.Garg lent Rs . 15000 to his friend . He charged 15% per annum on Rs. 12500 and 18% on the rest . How much interest does he earn in 3 years ?

Solution:

Question: 13

Shikha deposited Rs . 2000 in a bank which pays 6% simple interest . She withdrew Rs . 700 at the end of first year .What will be her balance after 3 years ? 

Solution:

Question: 14

Reema took a loan of Rs . 8000 from a money lender , who charged interest at the rate of 18% per annum . After 2 years , Reema paid him Rs . 10400 and wrist watch to clear the debt . What is the price of the watch ?

Solution:

Question: 15

Mr. Sharma deposited Rs. 20000 as a fixed deposit in a bank at 10% per annum . If 30% is deducted as income tax on the interest earned , find his annual income .

Solution:

Exercise 12.1

Question: 1

Given the following values, find the unknown values:

(i) C.P. = Rs 1200, S.P. = Rs 1350 Profit/Loss , ?

(ii) C.P. = Rs 980, S.P. = Rs 940 Profit/Loss = ?

(iii) C.P. = Rs 720, S.P. ?, Profit = Rs 55.50

(iv) C.P. ? S.P. = Rs 1254, Loss = Rs 32

Solution:

(i) CP = Rs. 1200, SP = Rs. 1350

CP < SP. So, profit.

Profit = Rs. (1350 – 1200) = Rs. 150

(ii) CP = Rs. 980, SP = Rs. 940

CP > SP. So, loss.

Loss = Rs. (980 – 940) = Rs. 40

(iii) CP = Rs. 720, SP = ?, profit = Rs. 55.50

Profit = SP – CP

Rs. 55.50 = SP – Rs. 720

SP = Rs. (55.50 + 720) = Rs. 775.50

(iv) CP = ?, SP = Rs. 1254, loss = Rs. 32

Loss = CP – SP

Rs. 32 = CP – Rs. 1254

CP = Rs. (1254 + 32) = Rs. 1286

Question: 2

Fill in the blanks in each of the following:

(I )C.P. = Rs 1265, S.P. = Rs 1253, Loss = Rs ______ 

(ii) C.P. = Rs______ , S.P. = Rs 450, Profit = Rs 150

(iii) C.P. = Rs 3355, S.P. = Rs 7355, profit/loss = Rs______

(iv) C.P. = Rs _______, S.P. = Rs 2390, Loss = Rs 5.50

Solution:

(i) CP = Rs. 1265, SP = Rs. 1253

Loss = CP – SP = Rs. (1265 – 1253) = Rs. 12

(ii) CP = 7, SP = Rs. 450, profit = Rs. 150

Profit = SP – CP

Rs. 150 = Rs. 450 – CP = CP = Rs. (450 – 150) = Rs. 300

(iii) CP = Rs. 3355, SP = Rs. 7355,

Here SP > CP, so profit.

Profit = SP – CP Profit = Rs. (7355 – 3355) = Rs. 4000

(iv) CP = 7, SP = Rs. 2390, loss = Rs. 5.50

Loss = CP – SP

Rs. 5.50 = CP – Rs. 2390 CP = Rs. (5.50 + 2390) = Rs. 2395.50

Question: 3

Calculate the profit or loss and profit or loss per cent in each of the following cases:

(i) C.P. = Rs 4560, S.P. = Rs 5000

(ii) C.P. = Rs 2600, S.P. = Rs 2470

(iii) C.P. = Rs 332, S.P. = Rs 350

(iv) C.P. = Rs 1500, S.P. = Rs 1500

Solution:

(i) CP = Rs. 4560, SP = Rs. 5000

Here, SP > CP. So, profit.

Profit = SP – CP = Rs. (5000 – 4560)= Rs. 440

Profit % = {(Profit/CP) x 100}% = {(440/4560) x 100}% = {0.0965 x 100}% = 9.65%

(ii) CP = Rs. 2600, SP = Rs. 2470.

Here, CP > SR. So, loss.

Loss = CP – SP = Rs. (2600 – 2470) = Rs. 130 Profit% = {(Profit/CP) x 100}%= {(130/2600) x 100}%

= {0.05 x 100}% = 5%

(iii) CP = Rs. 332, SP= Rs. 350.

Here, SP > CP. So, profit.

Profit = SP – CP = Rs. (350 – 332) = Rs. 18 Profit% = {(Profit/CP) x 100}% = {(18/332) x 100}%

= {0.054 x 100}% = 5.4%

(iv) CP = Rs. 1500, SP = Rs. 1500

SP = CP.

So, neither profit nor loss.

Question: 4

Find the gain or loss per cent, when:

 (i) C.P. = Rs 4000 and gain = Rs 40.

(ii) S.P. = Rs 1272 and loss = Rs 328

(iii) S.P. = Rs 1820 and gain = Rs 420.

Solution:

(i) CP = Rs. 4000, gain = Rs. 40

Gain % = {(Gain/CP) x 100)% = {(40/4000) x 100}%= (0.01 x 100)%= 1%

(ii) SP = Rs. 1272, loss = Rs. 328

Loss = CP – SP

Hence, CP = Loss+ SP = Rs. 328 + Rs. 1272 = Rs. 1600

Loss % = {(Loss/CP) x 100}% = {(328/1600) x 100% = 20.5%

(iii) SP = Rs. 1820, gain = Rs. 420

Gain = SP – CP CP = 1820 – 420 = Rs. 1400

Gain c/o= {(Gain/CP) x 100}% = {(420/1400) x 100% = 30%

Question: 5

Find the gain or loss per cent, when:

(i) C.P. = Rs 2300, Overhead expenses = Rs 300 and gain = Rs 260.

(ii) C.P. = Rs 3500, Overhead expenses = Rs 150 and loss = Rs 146

Solution:

(i) CP = Rs. 2300, overhead expenses = Rs. 300, gain = Rs. 260

Gain % = {(Gain/(CP + overhead expenses)} x 100 = {260/(2300 + 300} x 100

= {260/2600} x 100 = 10%

(ii) CP = Rs. 3500, overhead expenses = Rs. 150, loss = Rs. 146

Loss % = {(Loss/(CP + overhead expenses)} x 100 = {146/(3500+ 150)} x 100

= {146/3650} x 100 = 14600/3650 = 4%

Question: 6

A grain merchant sold 600 quintals of rice at a profit of 7%. If a quintal of rice cost him Rs 250 and his total overhead charges for transportation, etc. were Rs 1000 find his total profit and the selling price of 600 quintals of rice.

Solution:

Cost of 1 quintal of rice = Rs. 250

Cost of 600 quintals of rice = 600 x 250 = Rs. 150000

Overhead expenses = Rs. 1000

Total CP = Rs. (150000 + 1000) = Rs. 151000

Profit % = (Profit/CP) x 100

7 = (P/151000) x 100

P = 1510 x 7 = Rs. 10570

Profit = Rs. 10570

SP = CP + profit = Rs. (151000 + 10570) = Rs. 161570

Question: 7

Naresh bought 4 dozen pencils at Rs 10.80 a dozen and sold them for 80 paise each. Find his gain or loss percent.

Solution:

Cost of 1 dozen pencils = Rs. 10.80

Cost of 4 dozen pencils = 4 x 10.80 = Rs. 43.2

Selling price of each pencil = 80 paise

Total number of pencils = 12 x 4 = 48

SP of 48 pencils = 48 x 80 paise = 3840 paise = Rs. 38.40

Here, SP < CP.

Loss = CP – SP = Rs. (43.2 – 38.4) = Rs. 4.8

Loss % = (Loss/CP) x 100 = (4.8/43.2) x 100 = 480/43.2 = 11.11%

Question: 8

A vendor buys oranges at Rs 26 per dozen and sells them at 5 for Rs 13. Find his gain percent.

Solution:

CP of 1 dozen oranges = Rs. 26

CP of 1 orange = 26/12 = Rs. 2.16

CP of 5 oranges = 2.16 x 5 = Rs. 10.8

Now, SP of 5 oranges = Rs. 13

Gain = SP – CP = Rs. (13- 10.8) = Rs. 2.2

Gain %= (Gain/CP) x 100 = (2.2/10.8) x 100 = 20.3%

Question: 9

Mr Virmani purchased a house for Rs 365000 and spent Rs 135000 on its repairs. If he sold it for Rs 550000, find his gain percent.

Solution:

Amount Mr. Virmani paid to purchase the house = Rs. 365000

Amount he spent on repair = Rs. 135000

Total amount he spent on the house (CP) = Rs. (365000 + 135000) = Rs. 500000

SP of the house = Rs. 550000

Gain = SP – CP = Rs. (550000 – 500000) = Rs. 50000

Gain % = (Gain/CP) x 100= (50000/500000) x 100

= 5000000/500000 = 10%

Question: 10

Shikha purchased a wrist watch for Rs 840 and sold it to her friend Vidhi for Rs 910. Find her gain percent.

Solution:

The cost price of the wristwatch that Shikha purchased, CP = Rs. 840

The price at which she sold it, SP = Rs. 910

Gain = SP – CP = (910 – 840) = Rs. 70

Gain % = (Gain/CP) x 100 = (70/840) x 100 = 7000/840 = 8.3%

Question: 11

A business man makes a 10% profit by selling a toy costing him Rs 120. What is the selling price?

Solution:

CP = Rs. 12

Profit % = 10

We now that

SP = {(100 + profit %)/100} x CP = {(100+ 10)/100} x 120

= {(110/100)} x 120 = 1.1 x 120

= Rs. 132

Question: 12

Harish purchased 50 dozen bananas for Rs 135. Five dozen bananas could not be sold because they were rotten. At what price per dozen should Harish sell the remaining bananas so that he makes a profit of 20%?

Solution:

Cost price of 50 dozens bananas that Harish purchased, CP = Rs. 135

Bananas left after removing 5 dozen rotten bananas = 45 dozens

Effective CP of one dozen bananas = Rs. 135/45 = Rs. 3

Calculating the price at which Harish should sell each dozen bananas to make a profit of 20% (or 1/5), we get

Profit % = (Gain/CP) x100

To get a gain of 20% we give profit % = 20 and substitute

20 = (gain/135) x100

Gain = 270/10 = 27

We know ; SP = CP + Gain

SP =  27 + 135

SP = 162

Now that SP is for 45 Dozens of bananas

Calculating for one dozen

→ 162/45

=Rs. 3.6

Harish should sell the bananas at Rs. 3.60 a dozen in order to make a profit of 20%.

Question: 13

A woman bought 50 dozen eggs at Rs 6.40 a dozen. Out of these 20 eggs were found to be broken. She sold the remaining eggs at 55 paise per egg. Find her gain or loss percent.

Solution:

Cost of one dozen eggs = Rs. 6.40

Cost of 50 dozen eggs = 50 x 6.40 = Rs. 320

Total number of eggs = 50 x 12 = 600

Number of eggs left after removing the broken ones = 600 – 20 = 580

SP of 1 egg = 55 paise

So, SP of 580 eggs = 580 x 55 = 31900 paise = Rs. 31900/100 = Rs. 319

Loss = CP – SP = Rs. (320-319) = Re. 1

Loss % = (Loss/CP) x 100 = (1/320) x 100 = 0.31%

Question: 14

Jyotsana bought 400 eggs at Rs 8.40 a dozen. At what price per hundred must she sell them so as to earn a profit of 15%?

Solution:

Cost of eggs per dozen = Rs. 8.40

Cost of 1 egg = 8.40/12 = Rs. 0.7

Cost of 400 eggs = 400 x 0.7 = Rs. 280

Calculating the price at which Jyotsana should sell the eggs to earn a profit of 15%,

we get 15% of 280 + 280

= {(15/100) x 280} + 280 = {4200/100} + 280 = 42 + 280 = Rs. 322

So, Jyotsana must sell the 400 eggs for Rs. 322 in order to earn a profit of 15%. Therefore, the SP per one hundred eggs = Rs. 322/4 = Rs. 80.50.

Question: 15

A shopkeeper makes a profit of 15% by selling a book for Rs 230. What is the C.P. and the actual profit ?

Solution:

Given that the SP of a book = Rs. 230

Profit % = 15

Since

CP = (SP x 100) + (100 + profit %)

CP = (230x 100) + (100 + 15)

CP = 23000 + 115 = Rs. 200

Also, Profit = SP – CP = Rs. (230 – 200) = Rs. 30

Actual profit = Rs. 30

Question: 16

A bookseller sells all his books at a profit of 10%. If he buys a book from the distributor at Rs 200, how much does he sell it for ?

Solution:

Given

Profit % = 10% CP = Rs. 200

Since

SP = {(100 + profit %)/100} x CP = {(100 + 10)/100} x 200

= {110/100} x 200 = Rs. 220

The bookseller sells the book for Rs. 220.

Question: 17

A floweriest buys 100 dozen roses at Rs 2 a dozen. By the time the flowers are delivered, 20 dozen roses are mutilated and are thrown away. At what price should he sell the rest if he needs to make a 20% profit on his purchase ?

Solution:

Cost of 1 dozen roses = Rs. 2

Number of roses bought by the florist = 100 dozens

Thus, cost price of 100 dozen roses = 2 x 100 = Rs. 200

Roses left after discarding the mutilated ones = 80 dozens

Calculating the price at which the florist should sell the 80 dozen roses in order to make a profit of 20%, we have

Profit % = ((SP-CP)/CP) x100 = ((SP-200)/200)x100

40 = SP – 200

SP = Rs. 240

Therefore, the SP of the roses should be Rs. 240/80 = Rs. 3 per dozen.

Question: 18

By selling an article for Rs 240, a man makes a profit of 20%.What is his C.P. ? What would his profit percent be if he sold the article for Rs 275 ?

Solution:

Let CP = Rs. x SP = Rs. 240

Let profit be Rs. P.

Now, profit % = 20%

Since Profit % = (Profit/CP) x 100

→ 20 = (P/x) x 100

→ P = 20x/100 = x/5

Profit = SP – CP = 240 – x

 → P = 240 – x

→ x/5 = 240 – x

→ 240 = x + x/5

→ 240 = 6×15

→ x = 1200/6

→ 200

So, CP = Rs. 200

New SP = Rs. 275 and CP = Rs. 200

Profit % = {(SP – CP)/CP} x 100

→ {(275 – 200)1200} x 100 = (75/200) x 100

= 7500/200

= 37.5%

Exercise 11.1

Question: 1

Express each of the following per cents as fractions in the simplest forms:

(i) 45%

(ii) 0.25%

(iii) 150%

(iv) 6(1/4) %

Solution:

Question: 2

Express each of the following fractions as a per cent:

(i) (3/4) %

(ii) (53/100) %

(iii) 1(3/5) %

(iv) (7/20) %

Solution:

Exercise 11.2

Question: 1

Express each of the following ratios as per cents

(i) 4:5

(ii) 1:5

(iii) 11: 125

Solution:

Question: 2

Express each of the following per cents as ratios in simplest forms

(i) 2.5%

(ii) 0.4%

(iii) 13(3/4)%

Solution:

Exercise 11.3

Question: 1

Express each of the following per cents as decimals:

(i) 12.5%

(ii) 75%

(iii) 128.8%

Solution:

Question: 2

Express each of the following decimals as per cent:

(i) 0.004

(ii) 0.24

(iii) 0.02

(iv) 0.275

Solution:

Question: 3

Write each of the following as whole numbers or mixed numbers:

(i) 136%

(ii) 250%

(iii) 300%

Solution:

Exercise 11.4

Question: 1

Find each of the following:

(i) 7% of Rs.7150

(ii) 40% of 400Kg

(iii) 20% of 15.125litres

(iv) 3(1/3)% of 90 km

(v) 2.5% of 600metres

Solution:

Question: 2

Find the number whose 12(1/2) % is 64.

Solution:

Question: 3

What is the number, 6(1/4)% of which is 2?

Solution:

Question: 4

If 6 is 50% of a number, what is the number?

Solution:

Exercise 11.5

Question: 1

What per cent of

(i) 24 is 6?

(ii) Rs.125 is Rs.10?

(iii) 4km is 160 metres?

(iv) Rs. 8 is 25 paise?

(v) 2 days is 8 hours?

(vi) 1 lire is 175 ml

Solution:

Question: 2

What per cent is equivalent to 3/8?

Solution:

Question: 3

Find the following:

(i) 8 is 4% of which number

(ii) 6 is 60% of which number

(iii) 6 is 30% of which number

(iv) 12 is 25% of which number

Solution:

Question: 4

Convert each of the following pairs into percentages and find out which is more?

(i) 25 marks out of 30, 35 marks out of 40

(ii) 100 runs scored off 110 balls, 50 runs scored off 55 balls

Solution:

Question: 5

Find 20% more than Rs.200.

Solution:

Question: 6

Find 10% less than Rs.150

Solution:

Exercise 11.6

Question: 1

Ashu had 24 pages to write. By the evening, he had completed 25% of his work. How many pages were left?

Solution:

Question: 2

A box contains 60 eggs. Out of which 16(2/3)% are rotten ones. How many eggs are rotten?

Solution:

Question: 3

Rohit obtained 45 marks out of 80. What per cent marks did he get?

Solution:

Question: 4

Mr Virmani saves 12% of his salary. If he receives Rs 15900 per month as salary, find his monthly expenditure.

Solution:

Question: 5

A lawyer willed his 3 sons Rs 250000 to be divided into portions 30%, 45% and 25%. How much did each of them inherit?

Solution:

Question: 6

Rajdhani College has 2400 students, 40% of whom are girls. How many boys are there in the college?

Solution:

Question: 7

Aman obtained 410 marks out of 500 in CBSE XII examination while his brother Anish gets 536 marks out of 600 in IX class examination. Find whose performance is better?

Solution:

Question: 8

Rahim obtained 60 marks out of 75 in Mathematics. Find the percentage of marks obtained by Rahim in Mathematics.

Solution:

Question: 9

In an orchard, 16(2/3) % of the trees are apple trees. If the number of trees in the orchard is 240, 3 find the number of other type of trees in the orchard.

Solution:

Question: 10

Ram scored 553 marks out of 700 and Gita scored 486 marks out of 600 in science. Whose performance is better?

Solution:

Question: 11

Out of an income of Rs 15000, Nazima spends Rs 10200. What per cent of her income does she save?

Solution:

Question: 12

45% of the students in a school are boys. If the total number of students in the school is 880, find the number of girls in the school.

Solution:

Question: 13

Mr. Sidhana saves 28% of his income. If he saves As 840 per month, find his monthly income.

Solution:

Question: 14

In an examination, 8% of the students fail. What percentage of the students pass? If 1650 students appeared in the examination, how many passed?

Solution:

Question: 15

In an examination, 92% of the candidates passed and 46 failed. How many candidates appeared?

Solution:

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Get your questions answered by the expert for freeRelated ResourcesPercentage Exercise 11.5

Chapter 11: Percentage Exercise – 11.5…Percentage Exercise 11.1

Chapter 11: Percentage Exercise – 11.1…Percentage Exercise 11.2

Exercise 10.1

Question: 1

20 chocolates cost Rs 320. Find the cost of 35 such chocolates.

Solution:

Cost of 20 chocolates = Rs 320

Cost of 1 chocolate = Rs (320/20)

Therefore, the cost of 35 chocolates = Rs ((320/20) × 35) = Rs 560

Question: 2

The cost of 40 meters of cloth is Rs 200. Find the cost of 50 meters of cloth.

Solution:

Cost of 40 meters of cloth = Rs 200

Cost of 1 meter of cloth = Rs (200/40)

Therefore, the cost of 50 chocolates = Rs ((200/40) × 50) = Rs 250

Question: 3

A car can cover a distance of 522 km on 36 liters of petrol. How far can it travel on 14 litres of petrol?

Solution:

Number of kilometers a car can cover by using 36 liters = 522 km

Number of kilometers a car can cover by using 1 liter = 522/36 km

Hence, the number of kilometers a car can cover by using 14 liters = ((522/36) × 14) km = 203 km

Question: 4

Travelling 900 km by rail costs Rs 280. What would be the fare for a journey of 360 km when a person travels by the same class?

Solution:

Cost of travelling 900 km by rail = Rs 280

Cost of travelling 1 km by rail = Rs (280/900)

Hence, Cost of travelling 360 km by rail = Rs (280/900) × 360 = Rs 112

Question: 5

If 6 oil tankers can be filled by a pipe in 4(1/2) hours, how long does the pipe take to fill 4 such oil tankers?

Solution:

Question: 6

3/4 of the salary per month is Rs 600. What is the salary per month?

Solution:

Question: 7

The cost of 32 tables is Rs 23520. Find the number of such tables that can be purchased for Rs 51450.

Solution:

Number of tables bought for Rs 23520 = 32

Number of tables bought for Re 1 = 32/23520

Number of tables bought for Rs 51450 = (32/23520) × 51450 = 70

Question: 8

The yield of wheat from 6 hectares is 280 quintals. Find the number of hectares required for a yield of 225 quintals.

Solution:

Number of hectares required for a yield of 280 quintals = 6 hectares

Number of hectares required for a yield of 1 quintal = 6/280 hectares

Hence, the number of hectares required for a yield of 225 quintals = 6/280 × 225 hectares = 4(23/28) hectares

Question: 9

Fifteen post cards cost Rs 2.25. What will be the cost of 36 post cards? How many postcards can we buy in Rs 45?

Solution:

Question: 10

A rail journey of 75 km costs Rs 215. How much will a journey of 120 km cost?

Solution:

Cost of a rail journey of 75 km = Rs 215

Cost of a rail journey of 1 km = Rs 215/75

Cost of a rail journey of 120 km = Rs 215/75 × 120 = Rs 344

Question: 11

If the sales tax on a purchase worth Rs 60 is Rs 4.20. What will be the sales tax on the purchase worth of Rs 150?

Solution:

Sales tax on the purchase worth of Rs 60 = Rs 4.20

Sales tax on the purchase worth of Re 1 = Rs 4.2060

Sales tax on the purchase worth of Rs 150 = Rs (4.20/60) × 150 = Rs 10.50

Question: 12

52 packets of 12 pencils each, cost Rs 499.20. Find the cost of 65 packets of 10 pencils each.

Solution:

Total number of pencils in 52 packets of 12 pencils each = 52 × 12 = 624

Cost of 624 pencils = Rs 499.20

Cost of 1 pencil = Rs 499.20/624

Now,

Number of pencils in 65 packets of 10 pencils each = 65 x 10 = 650 pencils

Therefore, cost of 650 pencils = Rs (499.20/624) × 650 = Rs 520.

Exercise 9.1

Question: 1

If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y.

Solution:

Question: 2

If x: y = 8: 9, find the ratio (7x – 4y): 3x + 2y.

Solution:

Question: 3

If two numbers are in the ratio 6:13 and their L.C.M is 312, find the numbers.

Solution:

Question: 4

Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2: 3. Find the numbers

Solution:

Question: 5

What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3

Solution:

Question: 6

Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers

Solution:

Question: 7

The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8: 13. Find their present ages.

Solution:

Question: 8

Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, the ratio becomes 2: 3. Find the numbers.

Solution:

Question: 9

Two numbers are in the ratio 2: 7. 11 the sum of the numbers is 810. Find the numbers.

Solution:

Question: 10

Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3.

Solution:

Question: 11

Divide Rs 2000 among P, Q, R in the ratio 2: 3: 5.

Solution:

Question: 12

The boys and the girls in a school are in the ratio 7:4. If total strength of the school be 550, find the number of boys and girls.

Solution:

Question: 13

The ratio of monthly income to the savings of a family is 7: 2. If the savings be of Rs. 500, find the income and expenditure.

Solution:

It is given that

The ratio of income and savings is 7: 2

Savings

2x = 500

So, x = 250

Therefore,

Income = 7 × 250 = 1750

Expenditure = Income – savings

= 1750 – 500

= Rs.1250

Question: 14

The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides.

Solution:

Question: 15

A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get?

Solution:

Question: 16

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

Solution:

Question: 17

In the ratio 7: 8. If the consequent is 40, what a the antecedent

Solution:

Given ratio = 7: 8

Consequent

8x = 40

x = 40/8

x = 5

antecedent = 7x = 7 × 5 = 35

Question: 18

Divide Rs 351 into two parts such that one may be to the other as 2: 7.

Solution:

Question: 19

Find the ratio of the price of pencil to that of ball pen, if pencil cost Rs.16 per score and ball pen cost Rs.8.40 per dozen.

Solution:

Question: 20

In a class, one out of every six students fails. If there are 42 students in the class, how many pass?

Solution:

Given,

One out of 6 student fails

x out of 42 students

16 = x/42

x = 42/6

x = 7

Number of students who fail = 7 students

No of students who pass =Total students – Number of students who fail = 42 – 7 = 35 students.

Exercise 9.2

Question: 1

Which ratio is larger in the following pairs?

(i) 3: 4 or 9: 16

(ii) 15: 16 or 24: 25

(iii) 4: 7 or 5: 8

(iv) 9: 20 or 8: 13

(v) 1: 2 or 13: 27

Solution:

Question: 2

Give the equivalent ratios of 6: 8.

Solution:

Question: 3

Fill in the following blanks:

Solution:

x = 3 and y =15

Exercise 9.3

Question: 1

Which of the following are in proportion?

(i) 33, 44, 66, 88

(ii) 46, 69, 69, 46

(iii) 72, 84, 186, 217

Solution:

Question: 2

Find x in the following proportions

(i) 16:18 = x: 96

(ii) x: 92 = 87: 116

Solution:

Question: 3

The ratio of income to the expenditure of a family is 7: 6. Find the savings if the income is Rs.1400.

Solution:

The ratio of income and expenditure = 7: 6

7x = 1400

x = 200

Expenditure = 6x = 6 × 200 = Rs.1200

Savings = Income – Expenditure

= 1400 -1200

= Rs.200

Question: 4

The scale of a map is 1: 4000000. What is the actual distance between the two towns if they are 5cm apart on the map?

Solution:

The scale of map = 1: 4000000

Let us assume the actual distance between towns is x cm

1: 4000000 =5: x

x = 5 × 4000000 = 20000000 cm

1km = 1000 m

1m = 100 cm

Therefore

x = 200 km

Question: 5

The ratio of income of a person to his savings is 10: 1. If his savings for one year is Rs.6000, what is his income per month?

Solution:

The ratio of income of a person to his savings is 10: 1

Savings per month = 6000/12

= Rs.500

Then let income per month be x

x: 500 = 10:1

x = 500 × 10

x = 5000

Question: 6

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

Solution:

Height of the tree: Length of the shadow of tree

Height of the pole: Length of the shadow of pole

x: 20 = 6: 8

x = 120/8

x = 15

Exercise 7.1

Question: 1

Identify the monomials, binomials, trinomials and quadrinomials from the following expressions:

(i) a²

(ii) a² − b²

(iii) x³ + y³ + z³

(iv) x³ + y³ + z³ + 3xyz

(v) 7 + 5

(vi) abc + 1

(vii) 3x – 2 + 5

(viii) 2x – 3y + 4

(ix) xy + yz + zx

(x) ax³ + bx² + cx + d

Solution:

The monomials, binomials, trinomials and quadrinomials are as follows.

(i) a² is a monomial expression as it contains one term only.

(ii) a² − b² is a binomial expression as it contains two terms.

(iii) x³ + y³ + z³ is a trinomial expression as it contains three terms.

(iv) x³ + y³ + z³ + 3xyz is a quadrinomial expression as it contains four terms.

(v) 7 + 5 = 12 is a monomial expression as it contains one term only.

(vi) abc + 1 is a binomial expression as it contains two terms.

(vii) 3x – 2 + 5 = 3x + 3 is a binomial expression as it contains two terms.

(viii) 2x – 3y + 4 is a trinomial expression as it contains three terms.

(ix) xy + yz + zx is a trinomial expression as it contains three terms.

(x) ax³ + bx² + cx + d is a quadrinomial expression as it contains four terms.

Question: 2

Write all the terms of each of the following algebraic expressions:

(i) 3x

(ii) 2x – 3

(iii) 2x² − 7

(iv) 2x² + y² − 3xy + 4

Solution:

The terms of each of the given algebraic expressions are as follows.

(i) 3x is the only term of the given algebraic expression.

(ii) 2x and -3 are the terms of the given algebraic expression.

(iii) 2x² and −7 are the terms of the given algebraic expression.

(iv) 2x², y², −3xy and 4 are the terms of the given algebraic expression.

Question: 3

Identify the terms and also mention the numerical coefficients of those terms:

(i) 4xy, -5x²y, -3yx, 2xy²

(ii) 7a²bc,-3ca²b,-(5/2) abc², 3/2abc²,-4/3cba²

Solution:

(i) Like terms – 4xy, -3yx and Numerical coefficients – 4, -3

(i) Like terms – {7a²bc, −3ca²b} and Numerical coefficients – 7, -3

{−5/2abc²}   {−5/2}

{3/2 abc²}     {3/2}

{−4/3cba²}    {−4/3}

Question: 4

Identify the like terms in the following algebraic expressions:

(i) a² + b² -2a² + c² + 4a

(ii) 3x + 4xy − 2yz + 52zy

(iii) abc + ab²c + 2acb² + 3c²ab + b²ac − 2a²bc + 3cab²

Solution:

The like terms in the given algebraic expressions are as follows.

(i) The like terms in the given algebraic expressions are a² and −2a².

(ii) The like terms in the given algebraic expressions are -2yz and 5/2zy.

(iii) The like terms in the given algebraic expressions are ab²c, 2acb², b²ac and 3cab².

Question: 5

Write the coefficient of x in the following:

(i) –12x

(ii) –7xy

(iii) xyz

(iv) –7ax

Solution:

The coefficients of x are as follows.

(i) The numerical coefficient of x is -12.

(ii) The numerical coefficient of x is -7y.

(iii) The numerical coefficient of x is yz.

(iv) The numerical coefficient of x is -7a.

Question: 6

Write the coefficient of 2 in the following:

(i) −3x²

(ii) 5x²yz

(iii) 5/7x²z

(iv) –(3/2) ax² + yx

Solution:

The coefficient of x² are as follows.

(i) The numerical coefficient of x² is -3.

(ii) The numerical coefficient of x² is 5yz.

(iii) The numerical coefficient of x² is 57z.

(iv) The numerical coefficient of x² is – (3/2) a.

Question: 7

Write the coefficient of:

(i) y in –3y

(ii) a in 2ab

(iii) z in –7xyz

(iv) p in –3pqr

(v) y² in 9xy²z

(vi) x³ in x³ +1

(vii) x² in − x²

Solution:

The coefficients are as follows.

(i) The coefficient of y is -3.

(ii) The coefficient of a is 2b.

(iii) The coefficient of z is -7xy.

(iv) The coefficient of p is -3qr.

(v) The coefficient of y² is 9xz.

(vi) The coefficient of x³ is 1.

(vii) The coefficient of −x² is -1.

Question: 8

Write the numerical coefficient of each in the following

(i) xy

(ii) -6yz

(iii) 7abc

(iv) -2x3y2z

Solution:

The numerical coefficient of each of the given terms is as follows.

(i) The numerical coefficient in the term xy is 1.

(ii) The numerical coefficient in the term – 6yz is – 6.

(iii) The numerical coefficient in the term 7abc is 7.

(iv) The numerical coefficient in the term −2x³y²z is -2.

Question: 9

Write the numerical coefficient of each term in the following algebraic expressions:

(i) 4x²y – (3/2)xy + 5/2 xy²

(ii) –(5/3)x²y + (7/4)xyz + 3

Solution:

The numerical coefficient of each term in the given algebraic expression is as follows.

Question: 10

Write the constant term of each of the following algebraic expressions:

(i) x²y − xy² + 7xy − 3

(ii) a³ − 3a² + 7a + 5

Solution:

The constant term of each of the given algebraic expressions is as follows.

(i) The constant term in the given algebraic expressions is -3.

(ii) The constant term in the given algebraic expressions is 5.

Question: 11

Evaluate each of the following expressions for x = -2, y = -1, z = 3:

Solution:

Question: 12

Evaluate each of the following algebraic expressions for x = 1, y = -1, z = 2, a = -2, b = 1, c = -2:

(i) ax + by + cz

(ii) ax² + by² – cz

(iii) axy + byz + cxy

Solution:

We have x = 1, y = -1, z = 2, a = -2, b = 1 and c = -2.

Thus,

(i) ax + by + cz

= (-2)(1) + (1)(-1) + (-2)(2)

= –2 – 1 – 4

= –7

(ii) ax² + by² – cz

= (-2) × 1² + 1 × (-1)² – (-2) × 2

= 4 + 1 – (-4)

= 5 + 4

= 9

(iii) axy + byz + cxy

= (-2) × 1 × -1 + 1 × -1 × 2 + (-2) × 1 × (-1)

= 2 + (-2) + 2

= 4 – 2

= 2

Exercise 7.2

Question: 1

Add the following:

(i) 3x and 7x

(ii) -5xy and 9xy

Solution:

We have

(i) 3x + 7x = (3 + 7) x = 10x

(ii) -5xy + 9xy = (-5 + 9)xy = 4xy

Question: 2

Simplify each of the following:

(i) 7x³y +9yx³

(ii) 12a²b + 3ba²

Solution:

Simplifying the given expressions, we have

(i) 7x³y + 9yx³ = (7 + 9)x³y = 16x³y

(ii) 12a²b + 3ba² = (12 + 3)a²b =15a²b

Question: 3

Add the following:

(i) 7abc, -5abc, 9abc, -8abc

(ii) 2x²y, – 4x²y, 6x²y, -5x²y

Solution:

Adding the given terms, we have

(i) 7abc + (-5abc) + (9abc) + (-8abc)

= 7abc – 5abc + 9abc – 8abc

= (7 – 5 + 9 – 8)abc

= (16 – 13)abc

= 3abc

(ii) 2x²y +(-4x²y) + (6x²y) + (-5x²y)

= 2x²y – 4x²y + 6x²y – 5x2y

= (2- 4 + 6 – 5) x 2y

= (8 – 9) x 2y

= -x²y

Question: 4

Add the following expressions:

(i) x³ -2x²y + 3xy²– y³, 2x³– 5xy² + 3x²y – 4y³

(ii) a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴, – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

Solution:

Adding the given expressions, we have

(i) x³ -2x²y + 3xy²– y³, 2x³– 5xy² + 3x²y – 4y³

Collecting positive and negative like terms together, we get

x³ +2x³ – 2x²y + 3x²y + 3xy² – 5xy² – y³– 4y³

= 3x³ + x²y – 2xy² – 5y³

(ii) a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴, – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴ – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

Collecting positive and negative like terms together, we get

a⁴ – 2a⁴– 2a³b + 7a³b + 3ab³ – 5ab³ + 4a²b² – 6a²b² + 3b⁴ + b⁴

= – a⁴ + 5a³b – 2ab³ – 2a²b² + 4b⁴

Question: 5

Add the following expressions:

(i) 8a – 6ab + 5b, –6a – ab – 8b and –4a + 2ab + 3b

(ii) 5x³ + 7 + 6x – 5x², 2x² – 8 – 9x, 4x – 2x² + 3 x 3, 3 x 3 – 9x – x² and x – x² – x³ – 4

Solution:

(i) Required expression = (8a – 6ab + 5b) + (–6a – ab – 8b) + (–4a + 2ab + 3b)

Collecting positive and negative like terms together, we get

8a – 6a – 4a – 6ab – ab + 2ab + 5b – 8b + 3b

= 8a – 10a – 7ab + 2ab + 8b – 8b

= –2a – 5ab

(ii) Required expression = (5 x 3 + 7+ 6x – 5x²) + (2 x 2 – 8 – 9x) + (4x – 2x² + 3 x 3) + (3 x 3 – 9x-x²) + (x – x² – x³ – 4)

Collecting positive and negative like terms together, we get

5x³ + 3x³ + 3x³ – x³ – 5x² + 2x² – 2x²– x² – x² + 6x – 9x + 4x – 9x + x + 7 – 8 – 4

= 10x³ – 7x² – 7x – 5

Question: 6

Add the following:

(i) x – 3y – 2z

5x + 7y – 8z

3x – 2y + 5z

(ii) 4ab – 5bc + 7ca

–3ab + 2bc – 3ca

5ab – 3bc + 4ca

Solution:

(i) Required expression = (x – 3y – 2z) + (5x + 7y – 8z) + (3x – 2y + 5z)

Collecting positive and negative like terms together, we get

x + 5x + 3x – 3y + 7y – 2y – 2z – 8z + 5z

= 9x – 5y + 7y – 10z + 5z

= 9x + 2y – 5z

(ii) Required expression = (4ab – 5bc + 7ca) + (–3ab + 2bc – 3ca) + (5ab – 3bc + 4ca)

Collecting positive and negative like terms together, we get

4ab – 3ab + 5ab – 5bc + 2bc – 3bc + 7ca – 3ca + 4ca

= 9ab – 3ab – 8bc + 2bc + 11ca – 3ca

= 6ab – 6bc + 8ca

Question: 7

Add 2x² – 3x + 1 to the sum of 3x² – 2x and 3x + 7.

Solution:

Sum of 3x² – 2x and 3x + 7

= (3x² – 2x) + (3x +7)

=3x² – 2x + 3x + 7

= (3x² + x + 7)

Now, required expression = 2x² – 3x + 1+ (3x² + x + 7)

= 2x² + 3x² – 3x + x + 1 + 7

= 5x² – 2x + 8

Question: 8

Add x² + 2xy + y² to the sum of x² – 3y²and 2x² – y² + 9.

Solution:

Question: 9

Add a³+ b³ – 3 to the sum of 2a³ – 3b³ – 3ab + 7 and -a³ + b³ + 3ab – 9.

Solution:

Question: 10

Subtract:

(i) 7a²b from 3a²b

(ii) 4xy from -3xy

Solution:

(i) Required expression = 3a²b -7a²b

= (3 -7)a²b

= – 4a²b

(ii) Required expression = –3xy – 4xy

= –7xy

Question: 11

Subtract:

(i) – 4x from 3y

(ii) – 2x from – 5y

Solution:

(i) Required expression = (3y) – (–4x)

= 3y + 4x

(ii) Required expression = (-5y) – (–2x)

= –5y + 2x

Question: 12

Subtract:

(i) 6x³ −7x² + 5x − 3 from 4 − 5x + 6x² − 8x³

(ii) − x² −3z from 5x² – y + z + 7

(iii) x³ + 2x²y + 6xy² − y³ from y³−3xy²−4x²y

Solution:

Question: 13

From

(i) p3 – 4 + 3p², take away 5p² − 3p³ + p − 6

(ii) 7 + x − x², take away 9 + x + 3x² + 7x³

(iii) 1− 5y², take away y³ + 7y² + y + 1

(iv) x³ − 5x² + 3x + 1, take away 6x² − 4x³ + 5 + 3x

Solution:

Question: 14

From the sum of 3x² − 5x + 2 and − 5x² − 8x + 9 subtract 4x² − 7x + 9.

Solution:

Question: 15

Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and   2x + 4y – 7.

Solution:

Sum of (13x – 4y + 7z) and (–6z + 6x + 3y)

= (13x – 4y + 7z) + (–6z + 6x + 3y)

= (13x – 4y + 7z – 6z + 6x + 3y)

= (13x + 6x – 4y + 3y + 7z – 6z)

= (19x – y + z)

Sum of (6x – 4y – 4z) and (2x + 4y – 7)

= (6x – 4y – 4z) + (2x + 4y – 7)

= (6x – 4y – 4z + 2x + 4y – 7)

= (6x + 2x – 4z – 7)

= (8x – 4z – 7)

Now, required expression = (8x – 4z – 7) – (19x – y + z)

= 8x – 4z – 7 – 19x + y – z

= 8x – 19x + y – 4z – z – 7

= –11x + y – 5z – 7

Question: 16

From the sum of x² + 3y² − 6xy, 2x² − y² + 8xy, y² + 8 and x² − 3xy subtract −3x² + 4y² – xy + x – y + 3.

Solution:

Question: 17

What should be added to xy – 3yz + 4zx to get 4xy – 3zx + 4yz + 7?

Solution:

The required expression can be got by subtracting xy – 3yz + 4zx from 4xy – 3zx + 4yz + 7.

Therefore, required expression = (4xy – 3zx + 4yz + 7) – (xy – 3yz + 4zx)

= 4xy – 3zx + 4yz + 7 – xy + 3yz – 4zx

= 4xy – xy – 3zx – 4zx + 4yz + 3yz + 7

= 3xy – 7zx + 7yz + 7

Question: 18

What should be subtracted from x² – xy + y² – x + y + 3 to obtain −x² + 3y² − 4xy + 1?

Solution:

Question: 19

How much is x – 2y + 3z greater than 3x + 5y – 7?

Solution:

Required expression = (x – 2y + 3z) – (3x + 5y – 7)

= x – 2y + 3z – 3x – 5y + 7

Collecting positive and negative like terms together, we get

x – 3x – 2y + 5y + 3z + 7

= –2x – 7y + 3z + 7

Question: 20

How much is x² − 2xy + 3y² less than 2x² − 3y² + xy?

Solution:

Question: 21

How much does a² − 3ab + 2b² exceed 2a² − 7ab + 9b²?

Solution:

Question: 22

What must be added to 12x³ − 4x² + 3x − 7 to make the sum x³ + 2x² − 3x + 2?

Solution:

Question: 23

If P = 7x² + 5xy − 9y², Q = 4y² − 3x² − 6xy and R = −4x² + xy + 5y², show that P + Q + R = 0.

Solution:

Question: 24

If P = a² − b² + 2ab, Q = a² + 4b² − 6ab, R = b² + b, S = a² − 4ab and T = −2a² + b² – ab + a. Find P + Q + R + S – T.

Solution:

Exercise 7.3

Question: 1

Place the last two terms of the following expressions in parentheses preceded by a minus sign:

(i) x + y – 3z + y    

(ii) 3x – 2y – 5z – 4

(iii) 3a – 2b + 4c – 5

(iv) 7a + 3b + 2c + 4

(v) 2a² – b² – 3ab + 6

(vi) a² + b² – c² + ab – 3ac

Solution:

We have

(i) x + y – 3z + y = x + y – (3z – y)

(ii) 3x – 2y – 5z – 4 = 3x – 2y – (5z + 4)

(iii) 3a – 2b + 4c – 5 = 3a – 2b – (–4c + 5)

(iv) 7a + 3b + 2c + 4 = 7a + 3b – (–2c – 4)

(v) 2a² – b² – 3ab + 6 = 2a² – b² – (3ab – 6)

(vi) a² + b² – c² + ab – 3ac = a² + b² – c² – (- ab + 3ac)

Question: 2

Write each of the following statements by using appropriate grouping symbols:

(i) The sum of a – b and 3a – 2b + 5 is subtracted from 4a + 2b – 7.

(ii) Three times the sum of 2x + y – [5 – (x – 3y)] and 7x – 4y + 3 is subtracted from 3x – 4y + 7

(iii) The subtraction of x² – y² + 4xy from 2x² + y² – 3xy is added to 9x² – 3y²– xy.

Solution:

(i) The sum of a – b and 3a – 2b + 5 = [(a – b) + (3a – 2b + 5)].

This is subtracted from 4a + 2b – 7.

Thus, the required expression is (4a + 2b – 7) – [(a – b) + (3a – 2b + 5)]

(ii) Three times the sum of 2x + y – {5 – (x – 3y)} and 7x – 4y + 3 = 3[(2x + y – {5 – (x – 3y)}) + (7x – 4y + 3)]

This is subtracted from 3x – 4y + 7.

Thus, the required expression is (3x – 4y + 7) – 3[(2x + y – {5 – (x – 3y)}) + (7x – 4y + 3)]

(iii) The product of subtraction of x²– y² + 4xy from 2x² + y² – 3xy is given by {(2x² + y² – 3xy) – (x²-y² + 4xy)}

When the above equation is added to 9x² – 3y² – xy, we get

{(2x² + y² – 3xy) – (x² – y² + 4xy)} + (9x² – 3y²– xy))

Exercise 7.4

Question: 1

Simplify, the algebraic expressions by removing grouping symbols.

2x + (5x – 3y)

Solution:

We have

2x + (5x – 3y)

Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in the parentheses when we remove them.

= 2x + 5x – 3y

= 7x – 3y

Question: 2

Simplify, the algebraic expressions by removing grouping symbols.

3x – (y – 2x)

Solution:

We have

3x – (y – 2x)

Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

3x – y + 2x

= 5x – y

Question: 3

Simplify, the algebraic expressions by removing grouping symbols.

5a – (3b – 2a + 4c)

Solution:

We have

5a – (3b – 2a + 4c)

Since the ‘-‘ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them.

= 5a – 3b + 2a – 4c

= 7a – 3b – 4c

Question: 4

Simplify, the algebraic expressions by removing grouping symbols.

-2(x² – y² + xy) – 3(x² +y² – xy)

Solution:

We have

– 2(x² – y² + xy) – 3(x² +y² – xy)

Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

= -2x² + 2y² – 2xy – 3x² – 3y² + 3xy

= -2x² – 3x² + 2y² – 3y² – 2xy + 3xy

= -5x² – y² + xy

Question: 5

Simplify, the algebraic expressions by removing grouping symbols.

3x + 2y – {x – (2y – 3)}

Solution:

We have

3x + 2y – {x – (2y – 3)}

First, we have to remove the small brackets (or parentheses): ( ). Then, we have to remove the curly brackets (or braces): { }.

Therefore,

= 3x + 2y – {x – 2y + 3}

= 3x + 2y – x + 2y – 3

= 2x + 4y – 3

Question: 6

Simplify, the algebraic expressions by removing grouping symbols.

5a – {3a – (2 – a) + 4}

Solution:

We have

5a – {3a – (2 – a) + 4}

First, we have to remove the small brackets (or parentheses): ( ). Then, we have to remove the curly brackets (or braces): { }.

Therefore,

= 5a – {3a – 2 + a + 4}

= 5a – 3a + 2 – a – 4

= 5a – 4a – 2

= a – 2

Question: 7

Simplify, the algebraic expressions by removing grouping symbols.

a – [b – {a – (b – 1) + 3a}]

Solution:

First we have to remove the parentheses, or small brackets, ( ), then the curly brackets, { }, and then the square brackets [ ].

Therefore, we have

a – [b – {a – (b – 1) + 3a}]

= a – [b – {a – b + 1 + 3a}]

= a – [b – {4a – b + 1}]

= a – [b – 4a + b – 1]

= a – [2b – 4a – 1]

= a – 2b + 4a + 1

= 5a – 2b + 1

Question: 8

Simplify, the algebraic expressions by removing grouping symbols.

 a – [2b – {3a – (2b – 3c)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

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

Question: 9

Simplify, the algebraic expressions by removing grouping symbols.

 -x + [5y – {2x – (3y – 5x)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets { }, and then the square brackets, [ ].

Therefore, we have

– x + [5y – {2x – (3y – 5x)}]

= – x + [5y – {2x – 3y + 5x)]

= – x + [5y – {7x – 3y}]

= – x + [5y – 7x + 3y]

= – x + [8y – 7x]

= – x + 8y – 7x

= – 8x + 8y

Question: 10

Simplify, the algebraic expressions by removing grouping symbols.

 2a – [4b – {4a – 3(2a – b)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

2a – [4b – {4a – 3(2a – b)}]

= 2a – [4b – {4a – 6a + 3b}]

= 2a – [4b – {- 2a + 3b}]

= 2a – [4b + 2a – 3b]

= 2a – [b + 2a]

= 2a – b – 2a

= – b

Question: 11

Simplify, the algebraic expressions by removing grouping symbols.

-a – [a + {a + b – 2a – (a – 2b)} – b]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

– a – [a + {a + b – 2a – (a – 2b)} – b]

= – a – [a + {a + b – 2a – a + 2b} – b]

= – a – [a + {- 2a + 3b} – b]

= – a – [a – 2a + 3b – b]

= – a – [- a + 2b]

= – a + a – 2b

= – 2b

Question: 12

Simplify, the algebraic expressions by removing grouping symbols.

2x – 3y – [3x – 2y -{x – z – (x – 2y)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

2x – 3y – [3x – 2y – {x – z – (x – 2y)})

= 2x – 3y – [3x – 2y – {x – z – x + 2y}]

= 2x – 3y – [3x – 2y – {- z + 2y}]

= 2x – 3y – [3x – 2y + z – 2y]

= 2x – 3y – [3x – 4y + z]

= 2x – 3y – 3x + 4y – z

= – x + y – z

Question: 13

Simplify, the algebraic expressions by removing grouping symbols.

 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]

= 5 + [x – {2y – 6x – y + 4 + 2x} – {x – y + 2}]

= 5 + [x – {y – 4x + 4} – {x – y + 2}]

= 5 + [x – y + 4x – 4 – x + y – 2]

= 5 + [4x – 6]

= 5 + 4x – 6

= 4x – 1

Question: 14

Simplify, the algebraic expressions by removing grouping symbols.

x² – [3x + [2x – (x² – 1)] + 2]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

x² – [3x + [2x – (x² – 1)] + 2]

= x² – [3x + [2x – x² + 1] + 2]

= x² – [3x + 2x – x² + 1 + 2]

= x² – [5x – x² + 3]

= x² – 5x + x² – 3

= 2x² – 5x – 3

Question: 15

Simplify, the algebraic expressions by removing grouping symbols.

20 – [5xy + 3[x² – (xy – y) – (x – y)]]

Solution:

20 – [5xy + 3[x² – (xy – y) – (x – y)]]

= 20 – [5xy + 3[x² – xy + y – x + y]]

= 20 – [5xy + 3[x² – xy + 2y – x]]

= 20 – [5xy + 3x² – 3xy + 6y – 3x]

= 20 – [2xy + 3x² + 6y – 3x]

= 20 – 2xy – 3x² – 6y + 3x

= – 3x² – 2xy – 6y + 3x + 20

Question: 16

Simplify, the algebraic expressions by removing grouping symbols.

85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]

= 85 – [12x – 56x + 21 – 2{10x – 10 + 20x}]

= 85 – [12x – 56x + 21 – 2{30x – 10}]

= 85 – [12x – 56x + 21 – 60x + 20]

= 85 – [12x – 116x + 41]

= 85 – [- 104x + 41]

= 85 + 104x – 41

= 44 + 104x

Question: 17

Simplify, the algebraic expressions by removing grouping symbols.

 xy[yz – zx – {yx – (3y – xz) – (xy – zy)}]

Solution:

First we have to remove the small brackets, or parentheses, ( ), then the curly brackets, { }, and then the square brackets, [ ].

Therefore, we have

xy – [yz – zx – {yx – (3y – xz) – (xy – zy)}]

= xy – [yz – zx – {yx – 3y + xz – xy + zy}]

= xy – [yz – zx – {- 3y + xz + zy}]

= xy – [yz – zx + 3y – xz – zy]

= xy – [- zx + 3y – xz]

= xy – [- 2zx + 3y]

= xy + 2xz – 3y

Exercise 6.1

Question: 1

Find the values of each of the following:

(i) 132

 (ii) 73

 (iii) 34

Solution:

(i) 13² =  13 × 13

= 169

(ii) 7³ = 7 × 7 × 7

= 343

(iii) 3⁴ = 3 × 3 × 3 × 3

= 81

Question: 2

Find the value of each of the following:

(i) (-7)²

(ii) (-3)⁴

(iii)  (-5)⁵

Solution:

We know that if ‘a’ is a natural number, then

We have,

(i)  (-7)² = (-7) × (-7)

= 49

(ii) (-3)⁴ = (-3) × (-3) × (-3) × (-3)

= 81

(iii) (-5)⁵ = (-5) × (-5) × (-5) × (-5) × (-5)

= -3125

Question: 3

Simply:

(i) 3 × 10²

(ii) 2² × 5³

(iii) 3³ × 5²

Solution:

(i) 3 × 10² = 3 × 10 × 10

= 3 × 100

= 300

(ii) 2² × 5³ = 2 × 2 × 5 × 5 × 5

= 4 × 125

= 500

(iii) 3³ × 5² = 3 × 3 × 3 × 5 × 5

= 27 × 25

= 675

Question: 4

Simply:

(i)  3² × 10⁴

(ii)  2⁴ × 3²

(iii) 5² × 3⁴

Solution:

(i)  3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000

= 90000

(ii) 2⁴ × 3² = 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144

(iii) 5² × 3⁴ = 5 × 5 × 3 × 3 × 3 × 3

= 25 × 81

= 2025

Question: 5

Simply:

(i) (-2) × (-3)³

(ii) (-3)² × (-5)³

(iii) (-2)⁵ × (-10)²

Solution:

(i) (-2) × (-3)³ = (-2) × (-3) × (-3) × (-3)

= (-2) × (-27)

= 54

(ii) (-3)² × (-5)³ = (-3) × (-3) × (-5) × (-5) × (-5)

= 9 × (-125)

= -1125

(iii) (-2)⁵ × (-10)² = (-2) × (-2) × (-2) × (-2) × (-2) × (-10) × (-10)

= (-32) × 100

= -3200

Question: 6

Simply:

(i) (3/4)²

(ii) (-2/3)⁴

(iii) (- 4/5)⁵

Solution:

Question: 7

Identify the greater number in each of the following

(i) 2⁵ or 5²

(ii) 3⁴ or 4³

(iii) 3⁵ or 5³

Solution:

(i) 2⁵ or 5²

2⁵ = 2 × 2 × 2 × 2 × 2

= 32

5² = 5 × 5

= 25

Therefore, 2⁵ 5²

(ii) 3⁴ or 4³

= 3⁴ = 3 × 3 × 3 × 3

= 81

= 4³ = 4 × 4 × 4

= 64

Therefore, 3⁴ 4³

(iii) 3⁵ or 5³

= 3⁵ = 3 × 3 × 3 × 3 × 3

= 243

= 5³ = 5 × 5 × 5

= 125

Therefore, 3⁵ 5³

Question: 8

Express each of the following in exponential form

(i) (-5) × (-5) × (-5)

Solution:

(i) (-5) × (-5) × (-5) = (-5)³

Question: 9

Express each of the following in exponential form

(i) x × x × x × x × a × a × b × b × b

(ii) (-2) × (-2) × (-2) × (-2) × a × a × a

(iii) (-2/3) × (-2/3) × x × x × x

Solution:

(i) x × x × x × x × a × a × b × b × b = x⁴a²b³

(ii) (-2) × (-2) × (-2) × (-2) × a × a × a = (-2)⁴a³

(iii) (-2/3) × (-2/3) × x × x × x = (-2/3)² x³

Question: 10

Express each of the following numbers in exponential form

(i) 512

(ii) 625

(iii) 729

Solution:

(i) 512 = 2⁹

(iii) 625 = 5⁴

(iii) 729 = 3⁶

Question: 11

Express each of the following numbers as a product of powers of their prime factors

(i) 36

(ii) 675

(iii) 392

Solution:

(i) 36 = 2 × 2 × 3 × 3

= 2² × 3²

(ii) 675 = 3 × 3 × 3 × 5 × 5

= 3³ × 5²

(iii) 392 = 2 × 2 × 2 × 7 × 7

= 2³ × 7²

Question: 12

Express each of the following numbers as a product of powers of their prime factors

(i) 450

(ii) 2800

(iii) 24000

Solution:

(i) 450 = 2 × 3 × 3 × 5 × 5

= 2 × 3² × 5²

(ii) 2800 = 2 × 2 × 2 × 2 × 5 × 5 ×7

= 2⁴ × 5² × 7

(iii) 24000 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5

= 2⁵ × 3 × 5³

Question: 13

Express each of the following as a rational number of the form p/q

(i) (3/7)²

(ii) (7/9)³

(iii) (-2/3)⁴

Solution:

Question: 14

Express each of the following rational numbers in power notation

(i) 49/64

(ii) – 64/125

(iii) -12/16

Solution:

(i) 49/64 = (7/8)²

Because 7² = 49 and 8² = 64

(ii) – 64/125 = (- 4/5)³

Because 4³ = 64 and 5³ = 125

(iii) – (1/216) = – (1/6)³

Because 1³ = 1 and 6³ = 216

Question: 15

Find the value of the following

(i) (-1/2)² × 2³ × (3/4)²

(ii) (-3/5)⁴ × (4/9)⁴ × (-15/18)²

Solution:

(i) (-1/2)² × 2³ × (3/4)² = 1/4 × 8 × 9/16

= 9/8

(ii) (-3/5)⁴ × (4/9)⁴ × (-15/18)² = 81/625 × 256/6561 × 225/324 = 64/18225

Question: 16

If a = 2 and b= 3, the find the values of each of the followimg

(i) (a + b)^a

(ii) (ab)^b

(iii) (b/a)^b

(iv) (a/b + b/a)^a

Solution:

(i) (a + b)^a = (2 + 3)²

= (5)²

= 25

(ii) (ab)^b = (2 × 3)³

= (6)³

= 216

(iii) (b/a)^b = (3/2)³

= 27/8

(iv) (a/b + b/a)^a = (2/3 + 3/2)²

= 169/36

Exercise 6.2

Question: 1

Using laws of exponents, simplify and write the answer in exponential form

(i) 2³ × 2⁴ × 2⁵

(ii) 5¹² ÷ 5³

(iii) (7²)³

(iv) (3²)⁵ ÷ 3⁴

(v) 3⁷ × 2⁷

(vi) (5²¹ ÷ 5¹³) × 5⁷

Solution:

Question: 2

Simplify and express each of the following in exponential form

(i) ((2³)⁴ × 28) ÷ 212

(ii) (8² × 8⁴) ÷ 8³

(iii) (5⁷/5²) × 5³

Solution:

Question: 3

Simplify and express each of the following in exponential form

Solution:

Question: 4

Write 9 × 9 × 9 × 9 × 9 in exponential form with base 3

Solution:

9 × 9 × 9 × 9 × 9 = (9)⁵ = (3²)⁵

= 3¹⁰

Question: 5

Simplify and write each of the following in exponential form

Solution:

Question: 6

Simplify

Solution:

Question: 7

Find the values of n in each of the following

Solution:

Question: 8

Solution:

Exercise 6.3

Question: 1

Express the following numbers in the standard form

(i) 3908.78

(ii) 5, 00, 00, 000

(iii) 3, 18, 65, 00, 000

(iv) 846 × 10⁷

(v) 723 × 10⁹

Solution:

(i) 3908.78 = 3.90878 × 10³

Since, the decimal point is moved three places to the left

(ii) 5, 00, 00, 000 = 5, 00, 00, 000.00

= 5 × 10⁷

Since, the decimal point is moved seven places to the left

(iii) 3, 18, 65, 00, 000 = 3, 18, 65, 00, 000.00

= 3.1865 × 10⁹

Since, the decimal point is moved nine places to the left

(iv) 846 × 10⁷ = 8.46 × 10² × 10⁷

= 8.46 × 10⁹

Since, the decimal point is moved two places to the left

(v) 723 × 10⁹ = 7.23 × 10² × 10⁹

= 7.23 × 10¹¹

Since, the decimal point is moved two places to the left

Question: 2

Write the following numbers in the usual form

(i) 4.83 × 10⁷

(ii) 3.21 × 10⁵

(iii) 3.5 × 10³

Solution:

(i) 4.83 × 10⁷ = 483 × 10^(7–2)

= 483 × 10⁵

= 4, 83, 00, 000

Since, the decimal point is moved two places to the right

(ii) 3.21 × 10⁵ = 321 × 10^(5–2)

= 321 × 103

= 3, 21, 000

Since, the decimal point is moved two places to the right

(ii) 3.5 × 10³ = 35 × 10^(3–1)

= 35 × 10²

= 3,500

Since, the decimal point is moved one place to the right

Question: 4

Express the numbers appearing in the following statements in the standard form

(i) The distance between the earth and the moon is 384,000,000 metres.

(ii) Diameter of the earth is 1, 27, 56,000 metres.

(iii) Diameter of the sun is 1, 400, 000, 000 metres.

(iv) The universe is estimated to be about 12,000,000,000 years old.

Solution:

(i) The distance between the earth and the moon is 3.84 × 10⁸ metres.

Since, the decimal point is moved eight places to the left

(ii) Diameter of the earth is 1.2756 × 10⁷ metres.

Since, the decimal point is moved seven places to the left

(ii) Diameter of the sun is 1.4 × 10⁹ metres.

Since, the decimal point is moved nine places to the left

(iv) The universe is estimated to be about 1.2 × 10¹⁰ years old.

Since, the decimal point is moved ten places to the left

Exercise 5.1

Question: 1

Add the following rational numbers:

Solution:

We have,

We have,

We have,

We have,

Question: 2

Add the following rational numbers:

Solution:

If p/q and r/s are two rational numbers such that q and s do not have a common factor

If p/q and r/s are two rational numbers such that q and s do not have a common factor

LCM of 27 and 18 is 54

LCM of 4 and 8 is 4

Question: 3

Simplify

Solution:

LCM of 5 and 10 is 10

LCM of 16 and 24 is 48

LCM of 12 and 15 is 60

LCM of 19 and 57 is 57

Question: 4

Add and express the sum as a mixed fraction:

Solution:

LCM of 5 and 10 is 10

LCM of 7 and 4 is 28

Exercise 5.2

Question: 1

Subtract the first rational number from the second in each of the following:

Solution:

Question: 2

Evaluate each of the following:

Solution:

LCM of 3 and 5 is 15

LCM of 3 and 7 is 21

Question: 3

The sum of the two numbers is 5/9. If one of the numbers is 1/3, find the other.

Solution:

LCM of 3 and 9 is 9

Question: 4

The sum of two numbers is -1/3. If one of the numbers is -12/3, find the other.

Solution:

Let the required number be x

The required number is 11/3

Question: 5

The sum of two numbers is – 4/3. If one of the numbers is -5, find the other.

Solution:

Let the required number be x

The required number is 11/3

Question: 6

 The sum of two rational numbers is – 8. If one of the numbers is – (15/7), find the other.

Solution:

Let the required number be x

The required number is – (41/7)

Question: 7

What should be added to – (7/8) so as to get 5/9?

Solution:

Let the required number be x

The required number is 103/72

Question: 8

What number should be added to (-5)/11 so as to get 26/33?

Solution:

Let the required number be x

The required number is 41/33

Question: 9

What number should be added to (-5)/7 to get (-2)/3?

Solution:

Let the required number be x

The required number is 1/21

Question: 10

What number should be subtracted from -5/3 to get 5/6?

Solution:

Let the required number be x

The required number is 15/6

Question: 11

What number should be subtracted from 3/7 to get 5/4?

Solution:

Let the required number be x

The required number is 23/28

Question: 12

Solution:

Let the required number be x

The required number is (-7)/5

Question: 13

Solution:

Let the required number be x

The required number is 59/30

Question: 14

Solution:

Let the required number be x

x = 1/4

The required number is ¼

Question: 15

Simplify:

Solution:

Question: 16

Fill in the blanks:

Solution:

Exercise 5.3

Question: 1

Multiply:

Solution:

Question: 2

Multiply:

Solution:

Question: 3

Simplify each of the following and express the result as a rational number in standard form:

Solution:

Question: 4

Simplify:

Solution:

Question: 5

Simplify:

Solution:

Exercise 5.4

Question: 1

Divide:

Solution:

Question: 2

Find the value and express as a rational number in standard form:

Solution:

Question: 3

The product of two rational numbers is 15. If one of the numbers is -10, find the other.

Solution:

Let the number to be found be x

x ×- 10 = 15

x = 15/(-10)

x = 3/(-2)

x = (-3)/2

Hence the number is x = (-3)/2

Question: 4

The product of two rational numbers is – 8/9. If one of the numbers is – 4/15, find the other.

Solution:

Let the number to be found be x

Hence the number is x = 10/3

Question: 5

By what number should we multiply -1/6 so that the product may be -23/9?

Solution:

Let the number to be found be x

Hence the number is x = 46/3

Question: 6

By what number should we multiply -15/28 so that the product may be -5/7?

Solution:

Let the number to be found be x

Hence the number is x = 4/3

Question: 7

By what number should we multiply -8/13 so that the product may be 24?

Solution:

Let the number to be found be x

x = – 39

Hence the number is x = – 39

Question: 8

By what number should -3/4 be multiplied in order to produce -2/3?

Solution:

Let the number to be found be x

x = 8/9

Hence the number is x = 8/9

Question: 9

 Find (x + y) ÷ (x —y), if

Solution:

Question: 10

The cost of 7(2/3) metres of rope is Rs. 12(3/4). Find its cost per metre. 7(2/3) metres of rope cost = Rs. 12(3/4).

Solution:

Question: 11

The cost of 2(1/3) metres of cloth is Rs.75 1/4. Find the cost of cloth per metre. 2(1/3) metres of rope cost = Rs. 75(1/4)

Solution:

Question: 12

By what number should (-33)/16 be divided to get (-11)/4?

Solution:

x = 3/4

The number is x = 3/4

Question: 13

Divide the sum of (-13)/5 and 12/7 by the product of (-31)/7 and (-1)/2

Solution:

Question: 14

Divide the sum of 65/12 and 8/3 by their difference. 

Solution:

Question: 15

If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser?

Solution:

= 54/24

= 9/4 metres

9/4 metres of cloth is required to make each trouser

Exercise 5.5

Question: 1

Find six rational numbers between (-4)/8 and 3/8

Solution:

We know that

– 4, -3, -2, -1, 0, 1, 2, 3

Question: 2

Find 10 rational numbers between 7/13 and (- 4)/13

Solution:

We know that

76543210 -1 -2 -3 –4

Question: 3

State true or false:

(i) Between any two distinct integers there is always an integer.

(ii) Between any two distinct rational numbers there is always a rational number.

(iii) Between any two distinct rational numbers there are infinitely many rational numbers.

Solution:

(i) False
(ii) True
(iii) True