Class 8th

MCQ Questions for Class 8 Maths: Ch 12 Exponents and Powers

1. Multiplicative  inverse of 7^-2 is _________ 

(a) 49

(b) 5

(c) 7

(d) -14

► (a) 49

2. (a^m)ⁿ is equal to

(a) a^m+n

(b) a^m-n

(c) a^mn

(d) a^n-m

► (c) a^mn

3. Very small numbers can be expressed in standard form using __________ exponents.

(a) equal

(b) negative

(c) positive

(d) none of these

► (b) negative

4. The Base in the expression 10²⁴ is __________.

(a) 1

(b) 10

(c) 0

(d) 24

► (b) 10

5. Fill in the blank  a^m × aⁿ = a …….. where m and n are natural numbers:-

(a) mn                        

(b) m + n              

(c) m – n

(d) m/n

► (b) m + n

6. The value of 1/3^-2 is equal to

(a) 9

(b) 1

(c) -6

(d) 1/3

► (a) 9

7. When we have to add numbers in standard form, we convert them into numbers with the ________ exponents.
(a) same
(b) different
(c) not equal
(d) None of these
► (a) same

8. The value of 3⁰ is ________.
(a) 0
(b) 3
(c) 1
(d) None of these
► (c) 1

9. Value of (3⁰ + 2⁰) × 5⁰ is
(a) 1
(b) 25
(c) 2
(d) 0
► (c) 2

10. In exponential form 149,600,000,000 m is given by :

(a) 1.496 × 10¹¹ m      

(b) 1.496 × 10⁸ m  

(c) 14.96 × 10⁸ m              

(d) 14.96 × 10¹¹ m

► (a) 1.496 × 10¹¹ m   

11. Evaluate the exponential expression (−n)⁴× (−n)², for n = 5.

(a) 25

(b) 15625

(c) 3125

(d) 625
► (b) 15625

12. Evaluate exponential expression − 2⁵.

(a) 15

(b) -32

(c) 16

(d) none of these
► (b) -32

13. Charge of an electron is 0.000,000,000,000,000,000,16 coulomb and in exponential form it can be written as
(a) 16 × 10^-18 coulomb
(b) 1.6 × 10^-21 coulomb
(c) 1.6 × 10^-19 coulomb           
(d) 16 × 10^-21  coulomb
► (c) 1.6 × 10^-19 coulomb         

14. Find the value of the expression a² for a = 10.

(a) 100

(b) 1

(c) 10

(d) None of these
► (a) 100

15. In 10² the base is

(a) 1

(b) 0

(c) 10

(d) 100
► (c) 10

16. The multiplicative inverse of 2-3 is

(a) 2

(b) 3

(c) 3

(d) 23

► (d) 23

17. 16 is the multiplicative inverse of  

(a) 2^-4

(b) 2⁸

(c) 8²

(d) 2⁴

► (a) 2^-4

18. The standard form of 9030000000 is given by

(a) 9.03 × 10⁹

(b) 90.3 × 10⁷

(c) 903 × 10⁶

(d) 9.03 × 10^-9

► (a) 9.03 × 10⁹

19. The value of 7² is

(a) 7

(b) 49

(c) 2

(d) 14

► (b) 49

20. The value of 1000⁰ is

(a) 0

(b) 1000

(c) 1

(d) None of these

► (c) 1

Exponents and Powers Class 8 Extra Questions Very Short Answer Type

Question 1.
Find the multiplicative inverse of:
(i) 3^-3
(ii) 10^-10
Solution:

Question 2.
Expand the following using exponents.
(i) 0.0523
(ii) 32.005
Solution:

Question 3.
Simplify and write in exponential form.

Solution:

Question 4.
Simplify the following and write in exponential form.

Solution:

Question 5.
Express 8^-4 as a power with the base 2.
Solution:
We have 8 = 2 × 2 × 2 = 2³
8^-4 = (2³)^-4 = 2^3×(-4) = 2^-12

Question 6.
Simplify the following and write in exponential form.
(i) (3⁶ ÷ 3⁸)⁴ × 3^-4
(ii) 127 × 3^-3
Solution:

Question 7.
Find the value of k if (-2)^k+1 × (-2)³ = (-2)⁷
Solution:
(-2)^k+1 × (-2)³ = (-2)⁷
⇒ (-2)^k+1+3 = (-2)⁷
⇒ (-2)^k+4 = (-2)⁷
⇒ k + 4 = 7
⇒ k = 3
Hence, k = 3.

Question 8.
Simplify the following:

Solution:

Question 9.
Find the value of [(−34)−2]2
Solution:

Question 10.
Write the following in standard form
(i) 0 0035
(ii) 365.05
Solution:

Exponents and Powers Class 8 Extra Questions Short Answer Type

Question 11.
Find the value of P if

Solution:

Question 12.

Solution:

Question 13.
Find the value of x if

Solution:

⇒ 3 + x = 18 [Equating the powers of same base]
x = 18 – 3 = 15

Question 14.
Solve the following: (81)^-4 ÷ (729)^2-x = 9^4x
Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.
Find x so that (-5)^x+1 × (-5)⁵ = (-5)⁷ (NCERT Exemplar)
Solution:
(-5)^x+1 × (-5)⁵ = (-5)⁷
(-5)^x+1+5 = (-5)⁷ {a^m × aⁿ = a^m+n}
(-5)^x+6 = (-5)⁷
On both sides, powers have the same base, so their exponents must be equal.
Therefore, x + 6 = 7
x = 7 – 6 = 1
x = 1.

MCQ Questions for Class 8 Maths: Ch 11 Mensuration

1. In a quadrilateral, half of the product of the sum of the lengths of parallel sides  and the parallel distance between them gives the area of

(a) rectangle

(b) parallelogram

(c) triangle

(d) trapezium

► (d) trapezium

2. Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm. 

(a) 215 cm³

(b) 172 cm³

(c) 150 cm³

(d) 168 cm³

►(d) 168 cm³

3. Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm.
(a) 168 cm²
(b) 168 cm³
(c) 215 cm³
(d) 150 cm³
► (b) 168 cm³

4. The formula for finding total surface area of cuboid is  

(a) 2 (lb x bh x hl)

(b) 2 (lb + bh + hl)

(c) 2h (l + b)

(d) 2 lb (bh + hl)

► (b) 2 (lb + bh + hl)

5. Which of the following is an example of two dimensions  

(a) cuboid

(b) cone

(c) sphere

(d) circle

► (d) circle

6. The area of a trapezium is  

(a) 1/2 (sum of parallel sides) × h

(b) 2 (sum of parallel sides) × h

(c) (sum of parallel sides) × h

(d) 1/2 (sum of parallel sides) + h

► (a) 1/2 (sum of parallel sides) × h

7. The formula for finding lateral surface area  of cylinder is   

(a) 2πrh

(b) πr²

(c) 2πr(r+h)

(d) 2πr

► (a) 2πrh

8. Solid figures are
(a) 2 D
(b) 3 D
(c) 1 D
(d) 4 D
► (b) 3 D

9. A rectangular paper of width 7 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder.

(a) 8800 cm³
(b) 8800 cm
(c) 8800 cm²
(d) none of these
► (a) 8800 cm³

10. Find the total surface area of a cube whose volume is 343 cm3.
(a) 350 cm²
(b) 294 cm²
(c) 494 cm²
(d) 200 cm²
► (b) 294 cm2

11. Two dimensional figure is a  

(a) solid figure

(b) plane figure

(c) cylinder figure

(d) None of these

► (b) plane figure

12. A cylindrical tank has a capacity of 5632 m³. If the diameter of its base is 16 m, find its depth.

(a) 66m

(b) 30 m

(c) 26 m

(d) 28 m

► (d) 28 m

13. The length of parallel sides of trapezium is 14 cm and 6 cm and its height is 5 cm. Its area will be

(a) 50 cm²

(b) 100 cm²

(c) 210 cmsup>2

(d) 10 cm²

► (a) 50 cm²

14. The amount of space occupied by a three dimensional objects is called its
(a) area
(b) surface area
(c) volume
(d) lateral surface area
► (c) volume

15. Surface area of a cuboid = __________
(a) 2 h (l + b)
(b) 2lbh
(c) 2(lb + bh + hl)
(d) None of these
► (c) 2(lb + bh + hl)

16. Find the height of cuboid whose volume is 490 cmsup>3

and base area is 35 cmsup>3.
(a) 12 cm
(b) 14 cm
(c) 10 cm
(d) 16 cm
► (b) 14 cm

17. The cost of papering the wall of a room, 12 m long, at the rate of Rs. 1.35 per square meter is Rs. 340.20. The cost of matting the floor at Re. 0.85 per square metre is Rs. 91.80. Find the height of the room.
(a) 12 m
(b) 8 m
(c) 6 m
(d) 10 m
► (c) 6 m

18. The area of four walls of the room is  

(a) 2 (lb + bh + hl) 

(b) 2l (h + b) 

(c) 2 (lb x bh x hl) 

(d) 2h (l + b)

► (d) 2h (l + b)

19. The formula for lateral surface area of cuboid is  

(a) 2h (l + b)

(b) 2l (h + b)

(c) 2b (l + h)

(d) 2 (lb + bh + hl)

► (a) 2h (l + b)

20. Diagonals of rhombus are

(a) equal

(b) half of one diagonal

(c) of different length

(d) none of above

► (c) of different length

Short Answer Type Questions:

1. The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm.

Solution:

Area of trapezium = ½ × perpendicular distance between parallel sides × sum of parallel sides

= ½ × 15 × (12 + 20)

= 1/2 × 15 × 32

= 15 × 16

= 240  cm²

2. Calculate the height of a cuboid which has a base area of 180 cm² and volume is 900 cm³.

Solution:

Volume of cuboid = base area × height

900 = 180 × height

So, height = 900/180 = 5 cm

3. A square and a rectangle have the same perimeter. Calculate the area of the rectangle if the side of the square is 60 cm and the length of the rectangle is 80 cm.

Solution:

Perimeter of square formula = 4 × side of the square

Hence, P (square) = 4 × 60 = 240 cm

Perimeter of rectangle formula = 2 × (Length + Breadth)

Hence, P (rectangle) = 2 (80 + Breadth)

= 160 + 2 × Breadth

According to the given question,
160 + 2 × Breadth = 240 cm
2 × Breadth = 240 – 160
Breadth = 80/2
The breadth of the rectangle = 40 cm

Now, the area of rectangle = Length × Breadth = 80 × 40 = 3200 cm²

4. A lawnmower takes 750 complete revolutions to cut grass on a field. Calculate the area of the field if the diameter of the lawnmower is 84 cm and length is 1 m.

Solution:

Given, length of lawnmower = 1m = 100cm

Its circumference = π × D = 22/7 × 84 = 264 cm

Length of field will be = 264 × 750 = 198000 cm

Here, the width of field = length of the lawnmower i.e. 100 cm

So, area of field = 198000 × 100 = 19,800,000 cm²

Or, 1980 m²

5. The area of a rhombus is 16 cm² and the length of one of its diagonal is 4 cm. Calculate the length of other the diagonal.

Solution:

Area of rhombus = ½ × d₁ × d₂

⇒ 16 = ½ × 4 × d₂

So, d₂ = 32/4 = 8 cm

Long Answer Type Questions:

6. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed.

Solution:

The area of the remaining sheet after the smaller circle is removed will be = Area of the entire circle with radius 4 cm – Area of the circle with radius 3 cm

We know,

Area of circle = πr²

So,

Area of the entire circle = π(4)² = 16π cm²

And,

Area of the circle with radius 3 cm which is cut out = π(3)² = 9π cm²

Thus, the remaining area = 16π – 9π = 7π cm²

7. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted.

Solution:

Given,

Length of the box, l = 2 m,

Breadth of box, b = 1 m

Height of box, h = 1.5 m

We know that the surface area of a cuboid = 2(lb + lh + bh)

But here the bottom part is not to be painted.

So,

Surface area of box to be painted = lb + 2(bh + hl)

= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)

= 2 + 2 (1.5 + 3.0)

= 2 + 9.0

= 11

Hence, the required surface area of the cuboidal box = 11 m²

8. In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.

Solution:

From the question statement draw the diagram.

Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.

Now, CM will be the distance between the two parallel sides or the height of the trapezium.

We know,

Area of trapezium = ½ × sum of parallel sides × height.

So, height has to be found.

In the diagram, draw CL || AD

Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm

As AD = CB,

CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height.

Here, BL = AB – AL = (40 – 20) = 20 cm. So,

LM = MB = ½ BL = ½ × 20 = 10 cm

Now, in ΔCLM,

CL² = CM² + LM² (Pythagoras Theorem)

26² = CM² + 10²

CM² = 26² – 10²

Using algebraic identities, we get; 26² – 10² = (26 – 10) (26 + 10)

hence,

CM² = (26 – 10) (26 + 10) = 16 × 36 = 576

CM = √576 = 24 cm

Now, the area of trapezium can be calculated.

Area of trapezium, ABCD = ½ × (AB + CD) × CM

= ½ × (20 + 40) × 24

Or, Area of trapezium ABCD = 720 cm²

MCQ Questions for Class 8 Maths: Ch 10 Visualising Solid Shapes

1. Which is the three-dimensional figure formed by rotating a triangle?

(a) Cone

(b) Quadrilateral

(c) Prism

(d) Square

► (a) Cone

2. How many congruent edges does a cube have?

(a) 12

(b) 6

(c) 8

(d) none of these

► (a) 12

3. A square pyramid always has ___.

(a) Four lateral faces, which are parallel to each other.

(b) Four lateral faces, which are congruent equilateral triangles and a rectangular base.

(c) Two bases which are congruent and parallel.

(d) Four lateral faces, which are congruent isosceles triangles and a square base.

► (d) Four lateral faces, which are congruent isosceles triangles and a square base.

4. How many circular bases does a cylinder have?

(a) 1

(b) 2

(c) 3

(d) 4

► (b) 2

5. How many vertices does a triangular pyramid have ?

(a) 1

(b) 2

(c) 3

(d) 4

► (d) 4

6. The lateral faces of a pyramid are____.

(a) triangles

(b) pentagons

(c) rectangles

(d) None of these

► (a) triangles

7. How many triangles would you find in a net that folds into a square pyramid?  

(a) 3

(b) 4

(c) 5

(d) 6

► (b) 4

8. Identify the statement that is false for a prism. 

(a) A right prism has rectangular lateral faces.

(b) A right prism has two bases.

(c) A right prism has triangular lateral faces.

(d) A right prism has identical parallel faces.

9. What is the difference between a rectangle and a cube?

(a) A rectangle is 3-dimensional and a cube is 2-dimensional.

(b) A rectangle and a cube are both 2-dimensional.

(c) A rectangle is 2-dimensional and a cube is 3-dimensional.

(d) A rectangle and a cube are both 3-dimensional.

► (c) A rectangle is 2-dimensional and a cube is 3-dimensional.

10. Which of the following statements is true? 

(a) The lateral faces of a square prism are triangles.

(b) The lateral faces of a triangular prism can be squares or rectangles.

(c) The lateral faces of a square pyramid can be squares.

(d) The lateral faces of a triangular pyramid can be squares or rectangles.

► (b) The lateral faces of a triangular prism can be squares or rectangles.

11. How many vertices does a cone have?

(a) 2

(b) 1

(c) 3

(d) 4

► (b) 1

12. For a polyhedron, if ‘F’ stands for number of faces, V stands for number of vertices and E stands for number of edges, then which of the following relationships is named as Euler’s formula ?

(a) F + V = E + 2

(b) F + E = V + 2

(c) V + E = F + 2

(d) F+ V = E – 2

► (a) F + V = E + 2

13. The top-view of a cube looks like a:

(a) Circle

(b) Square

(c) Rectangle

(d) Triangle

► (b) Square

14. The top-view of a cuboid looks like a:

(a) Circle

(b) Square

(c) Rectangle

(d) Triangle

► (c) Rectangle

15. How many circular bases does a cylinder have?

(a) 2

(b) 3

(c) 4

(d) 1

► (a) 2

16. The faces of a triangular pyramid consist of

(a) 1 square and 3 triangles

(b) 2 triangles and 3 rectangles

(c) 4 triangles

(d) 2 rectangles and 3 triangles

► (c) 4 triangles

17. A three dimensional shape is _________ object.

(a) solid

(b) 2d

(c) plane

(d) None of these

► (a) solid

18. What do you call solid figures with line segments as their edges?

(a) Polygons                            

(b) Squares              

(c) Cylinders           

(d) Polyhedrons

► (d) Polyhedrons

19. Which of the following can be other name of a cylinder?

(a) A triangular prism

(b) A rectangular prism

(c) A vertical prism

(d) A circular prism

► (d) A circular prism

20. Which is/are three dimensional shapes?

(a) Sphere               

(b) Cylinder

(c) Cone       

(d) All of the above

► (d) All of the above

21. Which of the following statements is false?

(a) A sphere has one flat surface.

(b) A cone has one flat face.

(c) A cylinder has two circular faces.

(d) A sphere has one curved face.

► (a) A sphere has one flat surface.

22. The faces of a cube consist of

(a) 8 squares

(b) 4 squares and 2 rectangles

(c) 2 squares and 4 rectangle

(d) 6 squares

► (d) 6 squares

MCQ Questions for Class 8 Maths: Ch 9 Algebraic Expressions and Identities

1. Expressions consists of _____________ and _______________.

(a) variables, constants

(b) identities

(c) expressions

(d) none of these

► (a) variables, constants

2. Which of the following is an expression?

(a) 1/2

(b) 3

(c) 3x-2

(d) 2

► (c) 3x-2

3. The coefficient of x in the expression -7x +5 is

(a) 5        

(b) -7

(c) 7

(d) 0

► (b) -7

4. The number of terms in the expression 2x²+3x+5 is

(a) 1      

(b) 2

(c) 3

(d) 5
► (c) 3

5. Like terms in the expression 7x,5x²,7y, -5yx, -9x², are
(a) 7x, -5yx     
(b) 5x², -5yx
(c) 5x², -9x²
(d) 7x, 7y
► (c) 5x², -9x²

6. Terms are added to form ___________.
(a) expressions
(b) terms
(c) identities
(d) none of these
► (a) expressions

7. Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y, –3xz + 5x – 2xy.
(a) 5xy + 3zx + 5x – 4y
(b) 5xy + 9yz +2zx + 5x – 4y
(c) 5xy + 9yz +3zx + 5x – 4y
(d) 5xy + 9yz +3zx + 4y
► (c) 5xy + 9yz +3zx + 5x – 4y

8. The expression x + y + z is in
(a) one variable
(b) no variable
(c) three variables
(d) two variables
► (c) three variables

9. Which of the following is a monomial ?
(a) 4x²
(b) a + 6
(c) a + 6 + c
(d) a + b + c + d

► (a) 4x²

10. How many terms are there in the expression 5 – 3xy ?

(a) 1

(b) 2

(c) 3

(d) 5

► (b) 2

11. How many terms are there in the expression 5xy + 9yz + 3zx + 5x – 4y ?

(a) 1

(b) 3

(c) 4

(d) 5

► (d) 5

12. If x = 3 is solution of x² + kx + 15, value of k is

(a) k = -8, x = 3

(b) k = 8, x = 5

(c) k = 6, x = 5

(d) None of above

► (a) k = -8, x = 3

13. The coefficient in the term -5x is

(a) 5

(b) -5

(c) 1

(d) 2

► (b) -5

14. n (4 + m) = 4n + ___ 

(a) 4m

(b) 4n

(c) 4mn

(d) nm

► (d) nm

15. The like terms of the following are

(a) x, 3x

(b) x, 2y

(c) 2y, 6xy

(d) 3x, 2y

► (a) x, 3x

16. Value of expression ‘a(a²+a +1)+5’ for ‘ a’ = 0 is

(a) a+5      

(b) 1

(c) 6

(d) 5
► (d) 5

17. Which of the following is like term as 7xy?
(a) 9
(b) 9x
(c) 9y
(d) 9xy
► (d) 9xy

18. The area of triangle is’ xy’ where’ x’ is length and ‘y’ is breadth. If the length of rectangle is increased by 5 units and breadth is decreased by 3 units, the new area of rectangle will be
(a) (x-y)(x+3) 
(b) (xy+15)
(c) (x+5)(y-3)
(d) (xy+5-3)
► (c) (x+5)(y-3)

19. The expression 7xy has the factors
(a) 7,x,y         
(b) x,y
(c) 7,x
(d) 7,y
► (a) 7,x,y         

20. Which of the following is a binomial?
(a) 4x + y + 2
(b) 2x + 7
(c) 3x + 4y – 6
(d) 3x
► (b) 2x + 7

21. When numbers/literals are added or subtracted, they are called _________.
(a) identities
(b) expressions
(c) variables
(d) terms
► (d) terms

22. The volume of rectangular box whose length, breadth and height is 2p,4q.8r respectively is
(a) 14pqr   
(b) 2p+4q+8r
(c) 64pqr
(d) 64
► (c) 64pqr

Algebraic Expressions and Identities Class 8 Extra important Questions Short Answer Type

Question 11.
Simplify the following:
(i) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)
(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)
Solution:
(i) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)
= a²b² – a²c²) + b²c² – b²a²) + c²a² – c²b²)
= 0
(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)
= x³ – 3x²y² – xy³ + 2x²y² – xy³ + 5x³
= x³ + 5x³ – 3x²y² + 2x²y² – xy³ – xy³
= 6x³ – x²y² – 2xy³

Question 12.
Multiply (3x² + 5y²) by (5x² – 3y²)
Solution:
(3x² + 5y²) × (5x² – 3y²)
= 3x²(5x² – 3y²) + 5y²(5x² – 3y²)
= 15x⁴ – 9x²y² + 25x²y² – 15y⁴
= 15x⁴ + 16x²y² – 15y⁴

Question 13.
Multiply (6x² – 5x + 3) by (3x² + 7x – 3)
Solution:
(6x² – 5x + 3) × (3x² + 7x – 3)
= 6x²(3x² + 7x – 3) – 5x(3x² + 7x – 3) + 3(3x² + 7x – 3)
= 18x⁴ + 42x³ – 18x² – 15x³ – 35x² + 15x + 9x² + 21x – 9
= 18x⁴ + 42x³ – 15x³ – 18x² – 35x² + 9x² + 15x + 21x – 9
= 18x⁴ + 27x³ – 44x² + 36x – 9

Question 14.
Simplify:
2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)
Solution:
2x²(x + 2) – 3x(x² – 3) – 5x(x + 5)
= 2x³ + 4x² – 3x³ + 9x – 5x² – 25x
= 2x³ – 3x³ – 5x² + 4x² + 9x – 25x
= -x³ – x² – 16x

Question 15.
Multiply x² + 2y by x³ – 2xy + y³ and find the value of the product for x = 1 and y = -1.
Solution:
(x² + 2y) × (x³ – 2xy + y³)
= x²(x³ – 2xy + y³) + 2y(x³ – 2xy + y³)
= x⁵ – 2x³y + x²y³ + 2x³y – 4xy² + 2y⁴
= x⁵ + x²y³ – 4xy² + 2y⁴
Put x = 1 and y = -1
= (1)⁵ + (1)² (-1)³ – 4(1)(-1)² + 2(-1)⁴
= 1 + (1) (-1) – 4(1)(1) + 2(1)
= 1 – 1 – 4 + 2
= -2

Question 16.
Using suitable identity find:
(i) 48² (NCERT Exemplar)
(ii) 96²
(iii) 231² – 131²
(iv) 97 × 103
(v) 181² – 19² = 162 × 200 (NCERT Exemplar)
Solution:

Question 17.

Solution:

Question 18.
Verify that (11pq + 4q)² – (11pq – 4q)² = 176pq² (NCERT Exemplar)
Solution:
LHS = (11pq + 4q)² – (11pq – 4q)² = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)
[using a² -b² = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]
= (22pq) (8q)
= 176 pq²
= RHS.
Hence Verified.

Question 19.
Find the value of \(\frac { { 38 }^{ 2 }-{ 22 }^{ 2 } }{ 16 }\), using a suitable identity. (NCERT Exemplar)
Solution:

Question 20.
Find the value of x, if 10000x = (9982)² – (18)² (NCERT Exemplar)
Solution:
RHS = (9982)² – (18)² = (9982 + 18)(9982 – 18)
[Since a² -b² = (a + b) (a – b)]
= (10000) × (9964)
LHS = (10000) × x
Comparing L.H.S. and RHS, we get
10000x = 10000 × 9964
x = 9964

MCQ Questions for Class 8 Maths: Ch 8 Comparing Quantities

1. Find the ratio of speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

(a) It is 1:2

(b) It is 1:3

(c) It is 2:1

(d) It is 3:1

► (a) It is 1:2

2. The ratio of 10m to 10 km is:

(a) 1/10

(b) 1/100

(c) 1/1000

(d) 1000

► (c) 1/1000

3. The price of a house was Rs 34,00,000 last year. It has increased by 20% this year. What is the price now?

(a) Rs 40,40,000

(b) Rs 40,80,000

(c) Rs 30,40,000

(d) Rs 30,80,000

► (b) Rs 40,80,000

4. Calculate compound interest on Rs 10,800 for 3 years at 12.5% per annum compounded annually.

(a) 13,377.34

(b) 4577.34

(c) 14,377.34

(d) None of these

► (b) 4577.34

5. An item marked at Rs 840 is sold for Rs 714. What is the discount %?

(a) 20%

(b) 10%

(c) 15%

(d) none of these

► (c) 15%

6. A sum of money, at compound interest, yields Rs. 200 and Rs. 220 at the end of first and second years respectively. What is the rate percent?

(a) 20%

(b) 15%

(c) 10%

(d) 5%

► (c) 10%

7. A picnic is being planned in a school. Girls are 60% of the total number of students and are 300 in number. Find the ratio of the number of girls to the number of boys in the class.

(a) It is 3:2

(b) It is 3:1

(c) It is 2:3

(d) It is 2:1

► (a) It is 3:2

8. The price of a scooter was Rs 34,000 last year. It has increased by 20% this year. What is the price now?

(a) Rs 30,800

(b) Rs 30,400

(c) Rs 40,800

(d) Rs 40,400

► (c) Rs 40,800

9. A shopkeeper purchased 500 pieces for Rs 20 each. However 50 pieces were spoiled in the way and had to be thrown away. The remaining were sold at Rs 25 each. Find the gain or loss %.

(a) 18%

(b) 15%

(c) 12.5%

(d) none of these

► (c) 12.5%

10. The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find the population after 3 years.

(a) 29484

(b) 28696

(c) 24576

(d) 30184

► (a) 29484

11. Find the ratio of speed of a cycle 20 km per hour to the speed of scooter 30 km per hour.

(a) It is 3:1

(b) It is 2:1

(c) It is 2:3

(d) It is 1:3

► (c) It is 2:3

12. The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.

(a) 25153

(b) 24153

(c) 23153

(d) None of these

► (c) 23153

13. Sohan bought a washing machine for Rs 40,000, then spent Rs 5,000 on its repairs and sold it for Rs 50,000. Find his loss or gain per cent.

(a) Loss 10%

(b) Loss 20%

(c) Profit 11%

(d) none of these

► (c) Profit 11%

14. ________ means comparing two quantities.

(a) Ratio

(b) Proportion

(c) Percent

(d) None of these

► (a) Ratio

15. Find selling price (SP) if a profit of 5% is made on a cycle of Rs 700 with Rs 50 as overhead charges.

(a) Rs 600

(b) Rs 787.50

(c) Rs 780

(d) None of these

► (b) Rs 787.50

16. Find the ratio of speed of a car 50 km per hour to the speed of scooter 40 km per hour.

(a) It is 4:5

(b) It is 4:1

(c) It is 5:4

(d) It is 1:5

► (c) It is 5:4

17. The sale price of a shirt is Rs.176. If a discount of 20% is allowed on its marked price, what is the marked price of the shirt?

(a) Rs.160

(b) Rs.180

(c) Rs. 200

(d) Rs. 220

► (d) Rs. 220

18. The difference in S.I. and C.I. on a certain sum of money in 2 years at 15% p.a. is Rs.144. Find the sum.

(a) Rs. 6000

(b) Rs. 6200

(c) Rs. 6300

(d) Rs. 6400

► (d) Rs. 6400

19. Rohan bought a second hand refrigerator for Rs 2,500, then spent Rs 500 on its repairs and sold it for Rs 3,300. Find his loss or gain per cent.

(a) Loss 15% 2a

(b) Loss 10%

(c) Profit 10%

(d) None of these

► (c) Profit 10%

20. The value of an article which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs.9680, what is the price at which it was purchased?

(a) Rs.10000

(b) Rs.12500

(c) Rs.14575

(d) Rs.16250

► (b) Rs.12500

21. Find the ratio of 5 m to 10 km.

(a) It is 1:3

(b) It is 3000:1

(c) It is 2000:1

(d) It is 1:2000

► (d) It is 1:2000

22. A sum of money, at compound interest, yields Rs. 200 and Rs. 220 at the end of first and second years respectively. What is the rate percent?

(a) 20%

(b) 15%

(c) 10%

(d) 5%

► (c) 10%

23. A shopkeeper purchased 300 bulbs for Rs 10 each. However 10 bulbs were fused and had to be thrown away. The remaining were sold at Rs 12 each. Find the gain or loss %.

(a) 15%

(b) 13%

(c) 16%

(d) none of these

► (c) 16%

24. In what time will Rs.1000 amount to Rs.1331 at 10% p.a. compounded annually?

(a) 4 years

(b) 3 years

(c) 2 years

(d) 1 year

► (b) 3 years

Comparing Quantities Class 8 Extra Questions Very Short Answer Type

Question 1.
Express the following in decimal form:
(a) 12%
(b) 25%
Solution:
(a) 12% = 12100 = 0.12
(b) 25% = 25100 = 0.25

Question 2.
Evaluate the following:
(a) 20% of 400
(b) 1212% of 625
Solution:

Question 3.
If 20% of x is 25, then find x.
Solution:
20% of x = 25

Hence x = 125

Question 4.
Express the following as a fraction
(a) 35%
(b) 64%
Solution:

Question 5.
Express the following into per cent
(а) 135
(b) 2 : 5
Solution:

Question 6.
There are 24% of boys in a school. If the number of girls is 456, find the total number of students in the school.
Solution:
Let the total number of students be 100.
Number of boys = 24% of 100 = 24100 × 100 = 24
Number of girls = 100 – 24 = 76
⇒ If number of girls is 76, then total number of students = 100
⇒ If Number of girls is 1, then total number of students = 10076
If Number of girls is 456, then total number of students = 100×45676 = 600
Hence, the total number of students in the school = 600

Question 7.
The cost of 15 articles is equal to the selling price of 12 articles. Find the profit per cent.
Solution:
Let CP of 15 articles be ₹ 100
CP of 1 article = ₹ 10015
SP of 12 articles = ₹ 100
SP fo 1 article = ₹ 10012
SP > CP


Hence, profit = 25%

Question 8.
An article is marked at ₹ 940. If it is sold for ₹ 799, then find the discount per cent.
Solution:
MP = ₹ 940
SP = ₹ 799
Discount = MP – SP = 940 – 799 = ₹ 141

Hence, discount = 15%

Question 9.
A watch was bought for ₹ 2,700 including 8% VAT. Find its price before the VAT was added.
Solution:
Cost of watch including VAT = ₹ 2,700
Let the initial cost of the watch be ₹ 100
VAT = 8% of ₹ 100 = ₹ 8
Cost of watch including VAT = ₹ 100 + ₹ 8 = ₹ 108
If cost including VAT is ₹ 108, then its initial cost = ₹ 100
If cost including VAT is ₹ 1, then its initial cost = ₹ 100108
If cost including VAT is ₹ 2,700, then its initial cost = ₹ 100108 × 2700 = ₹ 2500
Hence, the required cost = ₹ 2,500

Question 10.
Find the amount if ₹ 2,000 is invested for 2 years at 4% p.a. compounded annually.
Solution:

Comparing Quantities Class 8 Extra Questions Short Answer Tpye

Question 11.
A number is increased by 20% and then it is decreased by 20%. Find the net increase or decrease per cent. (NCERT Exemplar)
Solution:
Let the number be 100
20% increase = 20100 × 100 = 20
Increased value = 100 + 20 = 120
Now it is decreased by 20%
Decreased value = 120 – 120100 × 20 = 120 – 24 = 96
Net decrease = 100 – 96 = 4
Decrease per cent = 4100 × 100 = 4%
Hence, the net decrease per cent = 4%

Question 12.
Two candidates Raman and Rajan contested an election. Raman gets 46% of the valid votes and is#defeated by 1600 votes. Find the total number of valid votes cast in the election.
Solution:
Let the total number of valid votes be 100
Number of votes got by Raman = 46% of 100 = 46100 × 100 = 46
Number of votes got by Rajan = 100 – 46 = 54
Difference between the votes = 54 – 46 = 8
8% of Valid votes = 1,600
⇒ 8100 × Valid votes = 1,600
⇒ Valid votes = 1600×1008 = 20,000
Hence, the total number of valid votes = 20,000

Question 13.
A man whose income is ₹ 57,600 a year spends ₹ 43,200 a year. What percentage of his income does he save?
Solution:
Annual income of a man = ₹ 57,600
Amount spent by him in the year = ₹ 43,200
Net amount saved by him = ₹ 57,600 – ₹ 43,200 = ₹ 14,400
Percentage of his annual saving Saving = SavingIncome × 100
= 1440057600 × 100
= 25%
Hence, the saving percentage = 25%

Question 14.
A CD player was purchased for ₹ 3,200 and ₹ 560 were spent on its repairs. It was then sold at a gain of 1212 %. How much did the seller receive?
Solution:
Cost price of the CD player = ₹ 3,200
Amount spent on its repairing = ₹ 560
Net cost price = ₹ 3,200 + ₹ 560 = ₹ 3,760

Hence, the required amount = ₹ 4,230

Question 15.
A car is marked at ₹ 3,00,000. The dealer allows successive discounts of 6%, 4% and 212 % on it. What is the net selling price of it?
Solution:
Marked price of the car = ₹ 3,00,000
Net selling price after the successive discounts

Hence, the net selling price = ₹ 2,63,952

Question 16.
Ramesh bought a shirt for ₹ 336, including 12% ST and a tie for ₹ 110 including 10% ST. Find the list price (without sales tax) of the shirt and the tie together.
Solution:
List price of the shirt = 110112 × 336 = ₹ 300
List price of the tie = 100110 × 110 = ₹ 100
List price of both together = ₹ 300 + ₹ 100 = ₹ 400

Question 17.
Find the amount of ₹ 6,250 at 8% pa compounded annually for 2 years. Also, find the compound interest.
Solution:

Question 18.
Find the compound interest on ₹ 31,250 at 12% pa for 1212 years.
Solution:

Question 19.
Vishakha offers a discount of 20% on all the items at her shop and still makes a profit of 12%. What is the cost price of an article marked at ₹ 280? (NCERT Exemplar)
Solution:
Marked Price = ₹ 280
Discount = 20% of ₹ 280
= 12 × 280 = ₹ 56
So selling price = ₹ (280 – 56) = ₹ 224
Let the cost price be ₹ 100
Profit = 12% of ₹ 100 = ₹ 12
So selling price = ₹ (100 + 12) = ₹ 112
If the selling price is ₹ 112, cost price = ₹ 100
If the selling price is ₹ 224, cost price = ₹ (100112 × 224) = ₹ 200

Question 20.
Find the compound interest on ₹ 48,000 for one year at 8% per annum when compounded half yearly. (NCERT Exemplar)
Solution:
Principal (P) = ₹ 48,000
Rate (R) = 8% p.a.
Time (n) = 1 year
Interest is compounded half yearly

Therefore Compound Interest = A – P = ₹ (519,16.80 – 48,000) = ₹ 3,916.80.

MCQ Questions for Class 8 Maths: Ch 7 Cubes and Cube Roots

1. Ones place digit in the cube of 5832 is ______.
(a) 5
(b) 7
(c) 2
(d) 8
► (d) 8

2. What is the cube of double of ‘a’?
(a) 16a³
(b) 2a
(c) 8a³
(d) 4a²
► (c) 8a³

3. If (2744)^1/3 = 2p+2, then the value of P is
(a) 3
(b) 6
(c) 2
(d) 8
► (b) 6

4. Each prime factor appears _________ times in its cube?
(a) 2
(b) 3
(c) 1                             
(d) 4
► (b) 3

5. The value of 5³ is __________.
(a) 125
(b) 15
(c) 10                         
(d) 75
► (a) 125

6. A natural number is said to be a perfect cube, if it is the cube of some _________.
(a) natural number
(b) square number
(c) cube number
(d) cuboid number
► (a) natural number

7. The square of a natural number subtracts from its cube comes 100. The number is __________.
(a) 2
(b) 3
(c) 5
(d) 1
► (c) 5

8. The expansion of a³ is ___________.
(a) 3 × a
(b) a + a + a
(c) 3 × 3 × 3
(d) a × a × a
► (d) a × a × a

9. If volume of cube is 4913cm³ then length of side of cube is
(a) 16 cm
(b) 17 cm
(c) 18 cm
(d) 19 cm
► (b) 17 cm

10. What will be the unit digit of the cube of a number ending with 6 ?
(a) 4             
(b) 6

(c) 2                           
(d) 8
► (b) 6

11. The numbers 1, 8, 27… are ______.
(a) negative numbers
(b) cube numbers
(c) square numbers
(d) none of these
► (b) cube numbers

12. The cube of 23 is ___________
(a) 2304
(b) 23
(c) 12167
(d) 529
► (c) 12167

13.The smallest natural number by which 704 must be divided to obtain a perfect cube is
(a) 22
(b) 12
(c) 11                           
(d) 13
► (c) 11                           

14. Express 6³ as the sum of odd numbers.
(a) 31 + 33 + 35 + 37 + 39 + 41 + 43
(b) 31 + 33 + 35 + 37 + 39 + 41
(c) 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45
(d) none of these
► (b) 31 + 33 + 35 + 37 + 39 + 41

15. The symbol for cube root is __________.
(a) √3
(b) ∛
(c) 2√3
(d) 3√3
► (b) ∛

16. The cube of an odd number is always __________.
(a) odd number       
(b) even number
(c) prime number     
(d) none of these
► (a) odd number       

17. What will be the unit digit of the cube root of a number ends with 8?
(a) 2     
(b) 8
(c) 4
(d) 6
► (a) 2     

18. The cube root of the 216 x (−32) x 54 is ________
(a) -36
(b) -72
(c) -48
(d) none of these
► (b) -72

19. Ones digit of cube of a number depends on the ________ of the number.
(a) tens digit
(b) ones digit
(c) hundred digit
(d) none of these
► (b) ones digit

20. If (504 + p) is a perfect cube number, whose cube root is p, then p = ______.
(a) 6
(b) 4
(c) 2
(d) 8
► (d) 8

21. How many digits will be there in the cube root of 46656 ?
(a) 2
(b) 1
(c) 3 
(d) 4
► (a) 2

22. Find the cube root of 0.001331.
(a) 0.111
(b) 0.101
(c) 0.11
(d) none of these
► (c) 0.11

IMPORTANT QUESTIONS

Question 1:

Which of the following numbers are notperfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

ANSWER:

(i) The prime factorisation of 216 is as follows.

2	216
2	108
2	54
3	27
3	9
3	3
	1

216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 216 is a perfect cube.

(ii)The prime factorisation of 128 is as follows.

2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, each prime factor is not appearing as many times as a perfect multiple of 3. One 2 is remaining after grouping the triplets of 2. Therefore, 128 is not a perfect cube.

(iii) The prime factorisation of 1000 is as follows.

2	1000
2	500
2	250
5	125
5	25
5	5
	1

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube.

(iv)The prime factorisation of 100 is as follows.

2	100
2	50
5	25
5	5
	1

100 = 2 × 2 × 5 × 5

Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grouping the triplets. Therefore, 100 is not a perfect cube.

(v)The prime factorisation of 46656 is as follows.

2	46656
2	23328
2	11664
2	5832
2	2916
2	1458
3	729
3	243
3	81
3	27
3	9
3	3
	1

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 46656 is a perfect cube.

NCERT Solution for Class 8 math – cubes and cube roots 114 , Question 1

Page No 114:

Question 2:

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

ANSWER:

(i) 243 = 3 × 3 × 3 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.

In that case, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect cube.

Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required.

Then, we obtain

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii) 72 = 2 × 2 × 2 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.

Then, we obtain

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv) 675 = 3 × 3 × 3 × 5 × 5

Here, two 5s are left which are not in a triplet. To make 675 a cube, one more 5 is required.

Then, we obtain

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v) 100 = 2 × 2 × 5 × 5

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.

Then, we obtain

100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.

NCERT Solution for Class 8 math – cubes and cube roots 114 , Question 2

Page No 114:

Question 3:

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81

(ii) 128

(iii) 135

(iv) 192

(v) 704

ANSWER:

(i) 81 = 3 × 3 × 3 × 3

Here, one 3 is left which is not in a triplet.

If we divide 81 by 3, then it will become a perfect cube.

Thus, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, one 2 is left which is not in a triplet.

If we divide 128 by 2, then it will become a perfect cube.

Thus, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

(iii) 135 = 3 × 3 × 3 × 5

Here, one 5 is left which is not in a triplet.

If we divide 135 by 5, then it will become a perfect cube.

Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.

(iv) 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, one 3 is left which is not in a triplet.

If we divide 192 by 3, then it will become a perfect cube.

Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.

(v) 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Here, one 11 is left which is not in a triplet.

If we divide 704 by 11, then it will become a perfect cube.

Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.

Page No 114:

Question 4:

Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

ANSWER:

Here, some cuboids of size 5 × 2 × 5 are given.

When these cuboids are arranged to form a cube, the side of this cube so formed will be a common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid.

LCM of 5, 2, and 5 = 10

Let us try to make a cube of 10 cm side.

For this arrangement, we have to put 2 cuboids along with its length, 5 along with its width, and 2 along with its height.

Total cuboids required according to this arrangement = 2 × 5 × 2 = 20

With the help of 20 cuboids of such measures, a cube is formed as follows.

Alternatively

Volume of the cube of sides 5 cm, 2 cm, 5 cm

= 5 cm × 2 cm × 5 cm = (5 × 5 × 2) cm³

Here, two 5s and one 2 are left which are not in a triplet.

If we multiply this expression by 2 × 2 × 5 = 20, then it will become a perfect cube.

Thus, (5 × 5 × 2 × 2 × 2 × 5) = (5 × 5 × 5 × 2 × 2 × 2) = 1000 is a perfect cube. Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.

)

NCERT Solution for Class 8 math – cubes and cube roots 114 , Question 4

Page No 116:

Question 1:

Find the cube root of each of the following numbers by prime factorisation method.

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

ANSWER:

(i) Prime factorisation of 

∴ 

(ii) Prime factorisation of 

∴ 

(iii) Prime factorisation of 

∴ 

(iv) Prime factorisation of 

∴ 

(v) Prime factorisation of 

∴ 

(vi) Prime factorisation of 

∴ 

(vii) Prime factorisation of 

∴ 

(viii) Prime factorisation of 

∴ 

(ix) Prime factorisation of 

∴ 

(x)Prime factorisation of 

∴ 

NCERT Solution for Class 8 math – cubes and cube roots 116 , Question 1

Page No 116:

Question 2:

State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeroes.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

ANSWER:

For finding the cube of any number, the number is first multiplied with itself and this product is again multiplied with this number.

(i) False. When we find out the cube of an odd number, we will find an odd number as the result because the unit place digit of an odd number is odd and we are multiplying three odd numbers. Therefore, the product will be again an odd number.

For example, the cube of 3 (i.e., an odd number) is 27, which is again an odd number.

(ii) True. Perfect cube will end with a certain number of zeroes that are always a perfect multiple of 3.

Foe example, the cube of 10 is 1000 and there are 3 zeroes at the end of it.

The cube of 100 is 1000000 and there are 6 zeroes at the end of it.

(iii) False. It is not always necessary that if the square of a number ends with 5, then its cube will end with 25.

For example, the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.

(iv) False. There are many cubes which will end with 8. The cubes of all the numbers having their unit place digit as 2 will end with 8.

The cube of 12 is 1728 and the cube of 22 is 10648.

(v) False. The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.

(vi) False. The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.

(vii)True, as the cube of 1 and 2 are 1 and 8 respectively.

Page No 116:

Question 3:

You are told that 1331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768

ANSWER:

Firstly, we will make groups of three digits starting from the rightmost digit of the number as.

There are 2 groups, 1 and 331, in it.

Considering 331,

The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1.

Taking the other group i.e., 1,

The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the unit place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.

Hence, 

The cube root of 4913 has to be calculated.

We will make groups of three digits starting from the rightmost digit of 4913, as. The groups are 4 and 913.

Considering the group 913,

The number 913 ends with 3. We know that if the digit 3 is at the end of a perfect cube number, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7.

Taking the other group i.e., 4,

We know that, 1³ = 1 and 2³ = 8

Also, 1 < 4 < 8

Therefore, 1 will be taken at the tens place of the required cube root.

Thus, 

The cube root of 12167 has to be calculated.

We will make groups of three digits starting from the rightmost digit of the number 12167, as. The groups are 12 and 167.

Considering the group 167,

167 ends with 7. We know that if the digit 7 is at the end of a perfect cube number, then its cube root will have its unit place digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3.

Taking the other group i.e., 12,

We know that, 2³ = 8 and 3³ = 27

Also, 8 < 12 < 27

2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root.

Thus, 

The cube root of 32768 has to be calculated.

We will make groups of three digits starting from the rightmost digit of the number 32768, as .

Considering the group 768,

768 ends with 8. We know that if the digit 8 is at the end of a perfect cube number, then its cube root will have its unit place digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2.

Taking the other group i.e., 32,

We know that, 3³ = 27 and 4³ = 64

Also, 27 < 32 < 64

3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root.

Thus, 

MCQ Questions for Class 8 Maths: Ch 6 Squares and Square Roots

1. Find the least number that must be subtracted from 5607 so as to get a perfect square.

(a) 130

(b) 135

(c) 131

(d) none of these

► (c) 131

2. Which of the following would end with digit 1?

(a) 123²

(b) 161²

(c) 77²

(d) 82²
► (b) 161²

3. Find the square of 39.
(a) 1500
(b) 78
(c) 1521
(d) none of these
► (c) 1521

4. The square of 23 is :
(a) 529
(b) 526
(c) 461
(d) 429
► (a) 529

5. Sum of squares of two numbers is 145. If square root of one number is 3, find the other number.
(a) 136
(b) 8
(c) 9
(d) 64
► (b) 8

6. If a number has 1 or 9 in the unit’s place, then it’s square ends in ________.
(a) 3
(b) 9
(c) 1
(d) none of these
► (c) 1

7. The largest perfect square between 4 and 50 is
(a) 25
(b) 36
(c) 49
(d) 45
► (c) 49

8. What will be the number of digits in the square root of 1296?
(a) 2 
(b) 3
(c) 1
(d) 4
► (a) 2 

9. How many natural numbers lie between 9² and 102?
(a) 15
(b) 19
(c) 18
(d) 17
► (c) 18

10. Without adding,find the sum of 1+3+5+7+9+11+13+15+17+19
(a) 100
(b) 64
(c) 49
(d) 81
► (a) 100

11. Which is the greatest three-digit perfect square?
(a) 999
(b) 961
(c) 962
(d) 970
► (b) 961

12. Find the perfect square numbers between 30 and 40.
(a) 936
(b) 49
(c) 25
(d) none of these
► (a) 936

13. How many numbers lie between square of 12 and 13
(a) 22
(b) 23
(c) 24
(d) 25
► (c) 24

14. Without doing any calculation, find the numbers which are surely perfect squares.

(a) 441

(b) 408

(c) 153

(d) 257

► (a) 441

15. The square root of 1.21 is

(a) 1.1

(b) 11

(c) 21

(d) 2.1

► (a) 1.1

16. Find the greatest 4-digit number which is a perfect square.

(a) 9990

(b) 9801

(c) 9999

(d) none of these

► (b) 9801

17. Sum of squares of two numbers is 145. If square root of one number is 3, find the other number. 

(a) 136

(b) 9

(c) 64

(d) 8

► (d) 8

18. Which of the following are the factors of ac+ ab + bc + ca  

(a) (a – c)(a – b)

(b) (a + c)(a + b)

(c) (a – c)(a + b)

(d) (a + c)(a – b)

► (b) (a + c)(a + b)

19. What could be the possible “one’s digit” of the square root of 625?

(a) 5        

(b) 0

(c) 4

(d) 8

► (a) 5        

20. What is the length of the side of a square whose area is 441 cm² ?

(a) 21

(b) 22

(c) 20

(d) 12

► (a) 21

21. The square root of 169 is
(a) 13
(b) 1.3
(c) -1.3 
(d) 13/10
► (a) 13

22. What is smallest number with which 5400 may be multiplied so that the product is perfect cube?
(a) 5
(b) 3
(c) 4
(d) 6
► (a) 5

Squares and Square Roots Class 8 Extra Questions Very Short Answer Type

Question 1.
Find the perfect square numbers between 40 and 50.
Solution:
Perfect square numbers between 40 and 50 = 49.

Question 2.
Which of the following 24², 49², 77², 131² or 189² end with digit 1?
Solution:
Only 49², 131² and 189² end with digit 1.

Question 3.
Find the value of each of the following without calculating squares.
(i) 27² – 26²
(ii) 118² – 117²
Solution:
(i) 27² – 26² = 27 + 26 = 53
(ii) 118² – 117² = 118 + 117 = 235

Question 4.
Write each of the following numbers as difference of the square of two consecutive natural numbers.
(i) 49
(ii) 75
(iii) 125
Solution:
(i) 49 = 2 × 24 + 1
49 = 25² – 24²
(ii) 75 = 2 × 37 + 1
75 = 38² – 37²
(iii) 125 = 2 × 62 + 1
125 = 63² – 62²

Question 5.
Write down the following as sum of odd numbers.
(i) 7²
(ii) 9²
Solution:
(i) 7² = Sum of first 7 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 9² = Sum of first 9 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

Question 6.
Express the following as the sum of two consecutive integers.
(i) 15²
(ii) 19²
Solution:

Question 7.
Find the product of the following:
(i) 23 × 25
(ii) 41 × 43
Solution:
(i) 23 × 25 = (24 – 1) (24 + 1) = 24² – 1 = 576 – 1 = 575
(ii) 41 × 43 = (42 – 1) (42 + 1) = 42² – 1 = 1764 – 1 = 1763

Question 8.
Find the squares of:
(i) −37
(ii) −917
Solution:

Question 9.
Check whether (6, 8, 10) is a Pythagorean triplet.
Solution:
2m, m² – 1 and m² + 1 represent the Pythagorean triplet.
Let 2m = 6 ⇒ m = 3
m² – 1 = (3)² – 1 = 9 – 1 = 8
and m² + 1 = (3)² + 1 = 9 + 1 = 10
Hence (6, 8, 10) is a Pythagorean triplet.
Alternative Method:
(6)² + (8)² = 36 + 64 = 100 = (10)²
⇒ (6, 8, 10) is a Pythagorean triplet.

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Question 10.
Using property, find the value of the following:
(i) 19² – 18²
(ii) 23² – 22²
Solution:
(i) 19² – 18² = 19 + 18 = 37
(ii) 23² – 22² = 23 + 22 = 45

Squares and Square Roots Class 8 Extra Questions Short Answer Type

Question 11.
Using the prime factorisation method, find which of the following numbers are not perfect squares.
(i) 768
(ii) 1296
Solution:

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, 3 is not in pair.
768 is not a perfect square.

1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
Here, there is no number left to make a pair.
1296 is a perfect square.

Question 12.
Which of the following triplets are Pythagorean?
(i) (14, 48, 50)
(ii) (18, 79, 82)
Solution:
We know that 2m, m² – 1 and m² + 1 make Pythagorean triplets.
(i) For (14, 48, 50),
Put 2m =14 ⇒ m = 7
m² – 1 = (7)² – 1 = 49 – 1 = 48
m² + 1 = (7)² + 1 = 49 + 1 = 50
Hence (14, 48, 50) is a Pythagorean triplet.
(ii) For (18, 79, 82)
Put 2m = 18 ⇒ m = 9
m² – 1 = (9)² – 1 = 81 – 1 = 80
m² + 1 = (9)² + 1 = 81 + 1 = 82
Hence (18, 79, 82) is not a Pythagorean triplet.

Question 13.
Find the square root of the following using successive subtraction of odd numbers starting from 1.
(i) 169
(ii) 81
(iii) 225
Solution:
(i) 169 – 1 = 168, 168 – 3 = 165, 165 – 5 = 160, 160 – 7 = 153, 153 – 9 = 144, 144 – 11 = 133, 133 – 13 = 120, 120 – 15 = 105, 105 – 17 = 88, 88 – 19 = 69,
69 – 21 = 48, 48 – 23 = 25, 25 – 25 = 0
We have subtracted odd numbers 13 times to get 0.
√169 = 13
(ii) 81 – 1 = 80, 80 – 3 = 77, 77 – 5 = 72, 72 – 7 = 65, 65 – 9 = 56, 56 – 11 = 45, 45 – 13 = 32, 32 – 15 = 17, 17 – 17 = 0
We have subtracted 9 times to get 0.
√81 = 9
(iii) 225 – 1 = 224, 224 – 3 = 221, 221 – 5 = 216, 216 – 7 = 209, 209 – 9 = 200, 200 – 11 = 189, 189 – 13 = 176, 176 – 15 = 161, 161 – 17 = 144, 144 – 19 = 125,
125 – 21 = 104, 104 – 23 = 81, 81 – 25 = 56, 56 – 27 = 29, 29 – 29 = 0
We have subtracted 15 times to get 0.
√225 = 15

Question 14.
Find the square rootofthe following using prime factorisation
(i) 441
(ii) 2025
(iii) 7056
(iv) 4096
Solution:
(i) 441 = 3 × 3 × 7 × 7
√441 = 3 × 7 = 21

(ii) 2025 = 3 × 3 × 3 × 3 × 5 × 5
√2025 = 3 × 3 × 5 = 45

(iii) 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
√7056 = 2 × 2 × 3 × 7 = 84

(iv) 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

Question 15.
Find the least square number which is divisible by each of the number 4, 8 and 12.
Solution:
LCM of 4, 8, 12 is the least number divisible by each of them.
LCM of 4, 8 and 12 = 24
24 = 2 × 2 × 2 × 3
To make it perfect square multiply 24 by the product of unpaired numbers, i.e., 2 × 3 = 6
Required number = 24 × 6 = 144

Question 16.
Find the square roots of the following decimal numbers
(i) 1056.25
(ii) 10020.01
Solution:

Question 17.
What is the least number that must be subtracted from 3793 so as to get a perfect square? Also, find the square root of the number so obtained.
Solution:
First, we find the square root of 3793 by division method.

Here, we get a remainder 72
61² < 3793
Required perfect square number = 3793 – 72 = 3721 and √3721 = 61

Question 18.
Fill in the blanks:
(а) The perfect square number between 60 and 70 is …………
(b) The square root of 361 ends with digit …………..
(c) The sum of first n odd numbers is …………
(d) The number of digits in the square root of 4096 is ………..
(e) If (-3)² = 9, then the square root of 9 is ……….
(f) Number of digits in the square root of 1002001 is …………
(g) Square root of 36625 is ………..
(h) The value of √(63 × 28) = …………
Solution:
(a) 64
(b) 9
(c) n²
(d) 2
(e) ±3
(f) 4
(g) 625
(h) 42

Question 19.
Simplify: √900 + √0.09 + √0.000009
Solution:
We know that √(ab) = √a × √b
√900 = √(9 × 100) = √9 × √100 = 3 × 10 = 30
√0.09 = √(0.3 × 0.3) = 0.3
√0.000009 = √(0.003 × 0.003) = 0.003
√900 + √0.09 + √0.000009 = 30 + 0.3 + 0.003 = 30.303

Squares and Square Roots Class 8 Extra Questions Higher Order Thinking Skills (HOTS)

Question 20.
Find the value of x if
1369−−−−√+0.0615+x−−−−−−−−−√=37.25
Solution:

Question 21.
Simplify:

Solution:

Question 22.
A ladder 10 m long rests against a vertical wall. If the foot of the ladder is 6 m away from the wall and the ladder just reaches the top of the wall, how high is the wall? (NCERT Exemplar)

Solution:
Let AC be the ladder.
Therefore, AC = 10 m
Let BC be the distance between the foot of the ladder and the wall.
Therefore, BC = 6 m
∆ABC forms a right-angled triangle, right angled at B.
By Pythagoras theorem,
AC² = AB² + BC²
10² = AB² + 6²
or AB² = 10² – 6² = 100 – 36 = 64
or AB = √64 = 8m
Hence, the wall is 8 m high.

Question 23.
Find the length of a diagonal of a rectangle with dimensions 20 m by 15 m. (NCERT Exemplar)
Solution:
Using Pythagoras theorem, we have Length of diagonal of the rectangle = l2+b2−−−−−√ units

Hence, the length of the diagonal is 25 m.

Question 24.
The area of a rectangular field whose length is twice its breadth is 2450 m2. Find the perimeter of the field.
Solution:
Let the breadth of the field be x metres. The length of the field 2x metres.
Therefore, area of the rectangular field = length × breadth = (2x)(x) = (2x²) m²
Given that area is 2450 m².
Therefore, 2x² = 2450
⇒ x² = 1225
⇒ x = √1225 or x = 35 m
Hence, breadth = 35 m
and length = 35 × 2 = 70 m
Perimeter of the field = 2 (l + b ) = 2(70 + 35) m = 2 × 105 m = 210 m.

MCQ Questions for Class 8 Maths: Ch 5 Data Handling

1. In the class- interval 70-80, 80 is the

(a) upper limit

(b) frequency

(c) range

(d) lower limit

► (a) upper limit

2. If a coin is flipped in the air, what is the probability of getting a tail?

(a) 0

(b) ½

(c) 1

(d) 2

► b) ½

3. The class mark of 95-100 is

(a) 95.5

(b) 97.5

(c) 95

(d) 100

► (b) 97.5

4. The number of times an observation occurs in a data is called its

(a) Range

(b) Interval

(c) Frequency

(d) Raw data

► (c) Frequency

5. The pie-chart is divided into

(a) circles

(b) squares

(c) sectors

(d) segments

► (c) sectors

6. Numbers 1 to 10 are written on ten separates slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking in to it. What is the probability of getting a number less than 6?

(a) 1

(b) 0

(c) 1/10

(d) 1/2

► (d) 1/2

7. When a die is thrown, total number of possible outcomes is ______.

(a) 6

(b) 36

(c) 2

(d) None of these

► (a) 6

8. There are 2 Red, 3 Blue and 5 Black balls in a bag. A ball is drawn from the bag without looking in to the bag. What is the probability of getting a red ball?

(a) 2/5

(b) 3/5

(c) 1/5

(d) None of these

► (c) 1/5

9. 18 out of 36 people love reading, so reading in the pie chart will be represented by

(a) 36 degree sector

(b) quarter sector

(c) semi circular sector

(d) None of these

► (c) semi circular sector

10. Which of the following is the probability of an impossible event?

(a) 0

(b) 1

(c) 2

(d) None of these

► (a) 0

11. There are 2 Red, 3 Blue and 5 Black balls in a bag. A ball is drawn from the bag without looking in to the bag. What is the probability of getting a blue ball?

(a) 3/5

(b) 2/5

(c) 3/10

(d) None of these

► (c) 3/10

12. Two dice are thrown, find and number of outcomes.

(a) 36

(b) 6

(c) 12

(d) None of these

► (a) 36

13. The central total angle in a pie chart is

(a) 180°

(b) 210°

(c) 360°

(d) None of these

► (c) 360°

14. There are 2 Red, 3 Blue and 5 Black balls in a bag. A ball is drawn from the bag without looking in to the bag. What is the probability of getting a non-black ball?

(a) 3/5

(b) 2/5

(c) 1/2

(d) None of these

► (c) 1/2

15. There are 2 Red, 3 Blue and 5 Black balls in a bag. A ball is drawn from the bag without looking in to the bag. What is the probability of getting a non-red ball?

(a) 3/5

(b) 4/5

(c) 2/5

(d) None of these

► (b) 4/5

16. Numbers 1 to 10 are written on ten separates slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking in to it .What is the probability of getting a number 6?

(a) 1

(b) 0

(c) 1/10

(d) 1/2

► (c) 1/10

17. A coin is tossed. Which of the following is the probability of getting a head or tail?

(a) 0

(b) 1

(c) 1/2

(d) None of these

► (b) 1

18. There are 2 Red, 3 Blue and 5 Black balls in a bag. A ball is drawn from the bag without looking in to the bag. What is the probability of getting a black ball?

(a) 3/5

(b) 2/5

(c) 1/2

(d) None of these

► (c) 1/2

19. There are 2 Red, 3 Blue and 5 Black balls in a bag. A ball is drawn from the bag without looking in to the bag. What is the probability of getting a non-blue ball?

(a) 7/10

(b) 3/5

(c) 2/5

(d) None of these

► (a) 7/10

20. Numbers 1 to 10 are written on ten separates slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking in to it. What is the probability of getting a 1-digit number ?

(a) 1

(b) 0

(c) 1/10

(d) 9/10

► (d) 9/10

21. Numbers 1 to 10 are written on ten separates slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking in to it. What is the probability of getting a number greater than 6?

(a) 1

(b) 0

(c) 1/2

(d) 1/10

► (d) 1/10

22. When a coin is thrown, total number of possible outcomes is ______.

(a) 2

(b) 5

(c) 6

(d) None of these

► (a) 2

MCQ Questions for Class 8 Maths: Ch 4 Practical Geometry

1. A parallelogram whose all sides are equal is called ________.

(a) triangle

(b) trapezium

(c) square

(d) rectangle

► (c) square

2. What do we require to construct a quadrilateral if lengths of four sides are given?

(a) One of the angle

(b) Length of a diagonal

(c) Length of two diagonals

(d) None of these

► (b) Length of a diagonal

3. The quadrilateral whose diagonals are equal and bisect each other at right angle is ________. 

(a) Triangle

(b) Square

(c) Rhombus

(d) None of these

► (b) Square

4. A polygon with minimum number of sides is  

(a) Pentagon

(b) Square

(c) triangle 

(d) angle

► (c) triangle 

5. The diagonals of a square bisect each other at  _________  angle.

(a) acute

(b) right

(c) obtuse

(d) reflex

► (b) right

6. What do we require to construct a square?

(a) Length of one side

(b) Lengths of three sides

(c) Lengths of two sides

(d) None of these

► (a) Length of one side

7. All the angles of a regular polygon are of ________________.

(a) 90°

(b) 60°

(c) equal length

(d) equal measure

► (d) equal measure

8. What do we require to construct a quadrilateral if measures of two adjacent angles are given?

(a) Lengths of three sides

(b) Length of one side

(c) Lengths of two sides

(d) None of these

► (a) Lengths of three sides

9. To construct a quadrilateral uniquely, it is necessary to know at least_________ of its parts.

(a) 5

(b) 4

(c) 3

(d) 2

► (a) 5

10. A quadrilateral can be constructed uniquely if the lengths of its four sides and ____ diagonal are given.

(a) 3

(b) 2

(c) 1

(d) none of these

► (c) 1

11.A simple closed curve made up of only _____________ is called  a polygon .

(a) lines

(b) curves

(c) closed curves

(d) line segments

► (d) line segments

12. A quadrilateral can be constructed uniquely if the lengths of its ______ sides and a diagonal are given.

(a) 3

(b) 1

(c) 2

(d) 4

► (d) 4

13. Diagonals of a rectangle:

(a) equal to each other

(b) not equal

(c) one is double of the other

(d) none of these

► (a) equal to each other

14. The measure of each interior angle of a regular polygon is 140o, then number of sides that regular polygon has ___

(a) 15

(b) 12

(c) 9

(d) 10

► (c) 9

15. A parallelogram must be a rectangle if its diagonals 

(a) bisect the angles to which they are drawn

(b) are perpendicular to each other

(c) bisect each other

(d) are congruent

► (d) are congruent

16. A parallelogram each of whose angles measures 90o is _____________.

(a) rectangle

(b) rhombus

(c) kite

(d) trapezium

► (a) rectangle

17. What is the number of sides in Hexagon ?

(a) 4      

(b) 7    

(c) 6    

(d) 5

► (c) 6    

18. What do we require to construct a quadrilateral if measures of three angles are given?

(a) Length of one side

(b) Two adjacent sides

(c) Length of one diagonal

(d) None of these

► (b) Two adjacent sides

19. Polygons that have no portions of their diagonals in their exteriors are called 

(a) triangles

(b) convex

(c) concave

(d) squares

► (b) convex

20. The ratio of two adjacent sides of a parallelogram is 4:5. If its perimeter is 72 cm, find its adjacent sides. 

(a) 18 cm and 25 cm

(b) 16 cm and 25 cm

(c) 18 cm and 20 cm

(d) 16 cm and 20 cm

►(d)16 cm and 20 cm

21. Sum of all interior angles of a polygon with (n) sides is given by

(a) (n – 2) x 180°

(b) n – 2 x 180°

(c) (n + 2) x 180°

(d) n + 2 x 180°

► (a) (n – 2) x 180°

22. A quadrilateral can be constructed uniquely if its _____ sides and two included angles are given.

(a) 1

(b) 2

(c) 3

(d) none of these

► (c) 3

Practical Geometry Class 8 Extra Questions Maths Chapter 4

Extra Questions for Class 8 Maths Chapter 4 Practical Geometry

Question 1.
Construct a quadrilateral PQRS, given that QR = 4.5 cm, PS = 5.5 cm, RS = 5 cm and the diagonal PR = 5.5 cm and diagonal SQ = 7 cm.

Solution:
Construction:
Step I: Draw QR = 4.5 cm.
Step II: Draw an arc with centre R and radius 5 cm.
Step III: Draw another arc with centre Q and radius 7 cm to meet the previous arc at S.
Step IV: Join RS and QS.
Step V: Draw two arcs with centre S and R and radius 5.5 cm each to meet each other at P.
Step VI: Join RP, SP and PQ.
Thus PQRS is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD in which AB = 4 cm, BC = 3.5 cm, CD = 5 cm, AD = 5.5 cm and ∠B = 75°.

Solution:
Construction:
Step I: Draw AB = 4 cm.
Step II: Draw an angle of 75° at B and cut BC = 3.5 cm.
Step III: Draw an arc with centre C and radius 5 cm.
Step IV: Draw another arc with centre A and radius 5.5 cm to meet the previous arc at D.
Step V: Join CD and AD.
Thus ABCD is the required quadrilateral.

Question 3.
Construct a square whose side is 5 cm.

Solution:
Construction:
Step I: Draw AB = 5 cm.
Step II: Draw an angle of 90° at B and cut BC = 5 cm.
Step III: Draw two arcs with centre A and C and same radii of 5 cm which meet each other at D.
Step IV: Join AD and CD.
Thus, ABCD is the required square.

Question 4.
Construct a rhombus ABCD in which AB = 5.8 cm and AC = 7.5 cm.

Solution:
Construction:
Step I: Draw AB = 5.8 cm.
Step II: Draw an arc with centre B and radius 5.8 cm.
Step III: Draw another arc with centre A and radius 7.5 cm to meet the previous arc at C.
Step IV: Draw two arcs with centres A and C and of the same radius 5.8 cm to meet each other at D.
Step V: Join BC, AC, CD and AD.
Thus ABCD is the required rhombus.

Question 5.
Construct a rhombus whose diagonals are 6 cm and 8 cm.

Solution:
Construction:
Step I: Draw SQ = 8 cm.
Step II: Draw a right bisector of SQ at O.
Step III: Draw two arcs with centre O and radius 3 cm each to cut the right bisector at P and R.
Step TV: Join PQ, QR, RS and SP.
Thus PQRS is the required rhombus.

Question 6.
Construct a rectangle whose diagonal is 5 cm and the angle between the diagonal is 50°.


Solution:
Construction:
Step I: Draw AC = 5 cm.
Step II: Draw the right bisector of AC at O.
Step III: Draw an angle of 50° at O and product both sides.
Step IV: Draw two arcs with centre O and of the same radius 2.5 cm to cut at B and D.
Step V: Join AB, BC, CD and DA.
Thus, ABCD is the required rectangle.

Question 7.
Construct a quadrilateral ABCD in which BC = 4 cm, ∠B = 60°, ∠C = 135°, AB = 5 cm and ∠A = 90°.

Solution:
Construction:
Step I: Draw AB = 5 cm.
Step II: Draw the angle of 60° at B and cut BC = 4 cm.
Step III: Draw an angle of 135° at C and angle of 90° at A which meet each other at D.
Thus, ABCD is the required quadrilateral.

Question 8.
Construct a parallelogram ABCD in which AB = 5.5 cm, AC = 7 cm and BD = 8 cm.

Solution:
Construction:
Step I: Draw AB = 5.5 cm.
Step II: Draw an arc with centre B and radius 82 cm = 4 cm.
Step III: Draw another arc with centre A and radius 72 cm = 3.5 cm which cuts the previous arc at O.
Step IV: Join AO and produce to C such that AO = OC.
Step V: Join BO and produce to D such that BO = OD.
Step VI: Join BC, CD and AD.
Thus ABCD is the required parallelogram.

Question 9.
Construct a rhombus PAIR, given that PA = 6 cm and angle ∠A = 110°.
Solution:
Since in a rhombus, all sides are equal, so PA = AI = IR = RP = 6 cm
Also, rhombus is a parallelogram
so, adjacent angle, ∠I = 180° – 110° = 70°

Steps of construction
Step I. Draw AI = 6 cm
Step II. Draw ray AX¯ such that ∠IAX = 110° and draw IY¯ such that ∠AIY = 70°.
Step III. With A and I as centres and radius 6 cm draw arcs intersecting AX and IY at P and R respectively.
Step IV. Join PR.
Thus, PAIR is the required rhombus.

MCQ Questions for Class 8th Maths: Ch 3 Quadrilaterals

1. The sum of all the angles of a quadrilateral is equal to:

(a) 180°

(b) 270°

(c) 360°

(d) 90°

► (c) 360°

2. A diagonal of a parallelogram divides it into two congruent:

(a) Square

(b) Parallelogram

(c) Triangles

(d) Rectangle

► (c) Triangles

3. The diagonals of a parallelogram:

(a) Equal

(b) Unequal

(c) Bisect each other

(d) Have no relation

► (c) Bisect each other

4. Each angle of rectangle is:

(a) More than 90°

(b) Less than 90°

(c) Equal to 90°

(d) Equal to 45°

► (c) Equal to 90°

5. If ABCD is a trapezium in which AB || CD and AD = BC, then:

(a) ∠A = ∠B

(b) ∠A > ∠B

(c) ∠A < ∠B

(d) None of the above

► (a) ∠A = ∠B

6. The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32°, ∠AOB = 70°, then ∠DBC is equal to:

(a) 32°

(b) 88°

(c) 24°

(d) 38°

► (d) 38°

7. In parallelogram ABCD, if ∠A = 2x + 15°, ∠B = 3x – 25°, then value of x is:

(a) 91°

(b) 89°

(c) 34°

(d) 38°

► (d) 38°

8. The opposite angles of a parallelogram are (3x – 2)° and (50 – x)° the measure of these angles is ______.​

(a) 140°, 140°

(b) 20°, 160°

(c) 37°, 143°

(d) 37°, 37°

► (d) 37°, 37°

9. In quadrilateral PQRS, if ∠P = 60° and ∠Q : ∠R : ∠S = 2 : 3 : 7, then ∠S =

(a) 175°

(b) 210°

(c) 150°

(d) 135°

► (a) 175°

10. Two angles of a quadrilateral are 50° and 80° and other two angles are in the ratio 8 : 15. Find the measure of the remaining two angles.​

(a) 100°, 130°

(b) 140°, 90°

(c) 80°, 150°

(d) 70°, 160°

► (c) 80°, 150°

11. The diagonals of rhombus are 12 cm and 16 cm. The length of the side of rhombus is:

(a) 12 cm

(b) 16 cm

(c) 8 cm

(d) 10 cm

► (d) 10 cm

12. In a parallelogram the sum of two consecutive angles is

(a) 360°

(b) 100°

(c) 180°

(d) 90°

► (c) 180°

13. The angles of a quadrilateral are (5x)°, (3x + 10)°, (6x – 20)° and (x + 25)°. Now, the measure of each angle of the quadrilateral will be

(a) 115°, 79°, 118°, 48°

(b) 100° 79°, 118°, 63°

(c) 110°, 84°, 106°, 60°

(d) 75°, 89°, 128°, 68°

► (a) 115°, 79°, 118°, 48°

14. Which of the following is not a parallelogram?

(a) Rectangle

(b) Rhombus

(c) Square

(d) Trapezium

► (d) Trapezium

15. In a Quadrilateral ABCD, AB = BC and CD = DA, then the quadrilateral is a

(a) Triangle

(b) Kite

(c) Rhombus

(d) Rectangle

► (b) Kite

16. If ABCD is a Parallelogram with 2 Adjacent angles ∠A =∠B, then the parallelogram is a

(a) Rhombus

(b) Triangle

(c) Rectangle

(d) Square

► (c) Rectangle

17. All the angles of a convex quadrilateral are congruent. However, not all its sides are congruent. What type of quadrilateral is it?

(a) Parallelogram

(b) Square

(c) Rectangle

(d) Trapezium

► (c) Rectangle

18. Perimeter of a parallelogram is 22 cm. If the longer side, measures 6.5 cm, the measure of the shorter side will be

(a) 4.5 cm

(b) 6.5 cm

(c) 2.5 cm

(d) 3.0 cm

► (a) 4.5 cm

19. If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then ABCD is 

(a) rhombus

(b) parallelogram

(c) trapezium

(d) kite

► (c) trapezium

20. Angles of a quadrilateral are in the ratio 3 : 6 : 8: 13. The largest angle is :

(a) 178°

(b) 156°

(c) 90°​

(d) 36°​

► (b) 156°

21. The diagonals of a rectangle PQRS intersects at O. If ∠QOR = 44°, ∠OPS =?

(a) 82°

(b) 52°

(c) 68°

(d) 75°

► (c) 68°

22. In a parallelogram ABCD, if ∠A = 75°, then ∠B = ?

(a) 95°

(b) 80°

(c) 105°

(d) 15°

► (c) 105°

23. A diagonal of a Rectangle is inclines to one side of the rectangle at an angle of 25∘. The Acute Angle between the diagonals is :

(a) 115°

(b) 50°

(c) 40°

(d) 25°

► (b) 50°

24. If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is:

(a) 81°

(b) 54°

(c) 108°

(d) 72°

► (d) 72°

Understanding Quadrilaterals Class 8 Extra Questions Very Short Answer Type

Question 1.
In the given figure, ABCD is a parallelogram. Find x.

Solution:
AB = DC [Opposite sides of a parallelogram]
3x + 5 = 5x – 1
⇒ 3x – 5x = -1 – 5
⇒ -2x = -6
⇒ x = 3

Question 2.
In the given figure find x + y + z.

Solution:
We know that the sum of all the exterior angles of a polygon = 360°
x + y + z = 360°

Question 3.
In the given figure, find x.

Solution:
∠A + ∠B + ∠C = 180° [Angle sum property]
(x + 10)° + (3x + 5)° + (2x + 15)° = 180°
⇒ x + 10 + 3x + 5 + 2x + 15 = 180
⇒ 6x + 30 = 180
⇒ 6x = 180 – 30
⇒ 6x = 150
⇒ x = 25

Question 4.
The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle.
Solution:
Sum of all interior angles of a quadrilateral = 360°
Let the angles of the quadrilateral be 2x°, 3x°, 5x° and 8x°.
2x + 3x + 5x + 8x = 360°
⇒ 18x = 360°
⇒ x = 20°
Hence the angles are
2 × 20 = 40°,
3 × 20 = 60°,
5 × 20 = 100°
and 8 × 20 = 160°.

Question 5.
Find the measure of an interior angle of a regular polygon of 9 sides.
Solution:
Measure of an interior angle of a regular polygon

Question 6.
Length and breadth of a rectangular wire are 9 cm and 7 cm respectively. If the wire is bent into a square, find the length of its side.
Solution:
Perimeter of the rectangle = 2 [length + breadth]
= 2[9 + 7] = 2 × 16 = 32 cm.

Now perimeter of the square = Perimeter of rectangle = 32 cm.
Side of the square = 324 = 8 cm.
Hence, the length of the side of square = 8 cm.

Question 7.
In the given figure ABCD, find the value of x.

Solution:
Sum of all the exterior angles of a polygon = 360°
x + 70° + 80° + 70° = 360°
⇒ x + 220° = 360°
⇒ x = 360° – 220° = 140°

Question 8.
In the parallelogram given alongside if m∠Q = 110°, find all the other angles.

Solution:
Given m∠Q = 110°
Then m∠S = 110° (Opposite angles are equal)
Since ∠P and ∠Q are supplementary.
Then m∠P + m∠Q = 180°
⇒ m∠P + 110° = 180°
⇒ m∠P = 180° – 110° = 70°
⇒ m∠P = m∠R = 70° (Opposite angles)
Hence m∠P = 70, m∠R = 70°
and m∠S = 110°

Question 9.
In the given figure, ABCD is a rhombus. Find the values of x, y and z.

Solution:
AB = BC (Sides of a rhombus)
x = 13 cm.
Since the diagonals of a rhombus bisect each other
z = 5 and y = 12
Hence, x = 13 cm, y = 12 cm and z = 5 cm.

Question 10.
In the given figure, ABCD is a parallelogram. Find x, y and z.

Solution:
∠A + ∠D = 180° (Adjacent angles)
⇒ 125° + ∠D = 180°
⇒ ∠D = 180° – 125°
x = 55°
∠A = ∠C [Opposite angles of a parallelogram]
⇒ 125° = y + 56°
⇒ y = 125° – 56°
⇒ y = 69°
∠z + ∠y = 180° (Adjacent angles)
⇒ ∠z + 69° = 180°
⇒ ∠z = 180° – 69° = 111°
Hence the angles x = 55°, y = 69° and z = 111°

Question 11.
Find x in the following figure. (NCERT Exemplar)

Solution:
In the given figure ∠1 + 90° = 180° (linear pair)
∠1 = 90°
Now, sum of exterior angles of a polygon is 360°, therefore,
x + 60° + 90° + 90° + 40° = 360°
⇒ x + 280° = 360°
⇒ x = 80°

Understanding Quadrilaterals Class 8 Extra Questions Short Answer Type

Question 12.
In the given parallelogram ABCD, find the value of x andy.

Solution:
∠A + ∠B = 180°
3y + 2y – 5 = 180°
⇒ 5y – 5 = 180°
⇒ 5y = 180 + 5°
⇒ 5y = 185°
⇒ y = 37°
Now ∠A = ∠C [Opposite angles of a parallelogram]
3y = 3x + 3
⇒ 3 × 37 = 3x + 3
⇒ 111 = 3x + 3
⇒ 111 – 3 = 3x
⇒ 108 = 3x
⇒ x = 36°
Hence, x = 36° and y – 37°.

Question 13.
ABCD is a rhombus with ∠ABC = 126°, find the measure of ∠ACD.

Solution:
∠ABC = ∠ADC (Opposite angles of a rhombus)
∠ADC = 126°
∠ODC = 12 × ∠ADC (Diagonal of rhombus bisects the respective angles)
⇒ ∠ODC = 12 × 126° = 63°
⇒ ∠DOC = 90° (Diagonals of a rhombus bisect each other at 90°)
In ΔOCD,
∠OCD + ∠ODC + ∠DOC = 180° (Angle sum property)
⇒ ∠OCD + 63° + 90° = 180°
⇒ ∠OCD + 153° = 180°
⇒ ∠OCD = 180° – 153° = 27°
Hence ∠OCD or ∠ACD = 27°

Question 14.
Find the values of x and y in the following parallelogram.

Solution:
Since, the diagonals of a parallelogram bisect each other.
OA = OC
x + 8 = 16 – x
⇒ x + x = 16 – 8
⇒ 2x = 8
x = 4
Similarly, OB = OD
5y + 4 = 2y + 13
⇒ 3y = 9
⇒ y = 3
Hence, x = 4 and y = 3

Question 15.
Write true and false against each of the given statements.
(a) Diagonals of a rhombus are equal.
(b) Diagonals of rectangles are equal.
(c) Kite is a parallelogram.
(d) Sum of the interior angles of a triangle is 180°.
(e) A trapezium is a parallelogram.
(f) Sum of all the exterior angles of a polygon is 360°.
(g) Diagonals of a rectangle are perpendicular to each other.
(h) Triangle is possible with angles 60°, 80° and 100°.
(i) In a parallelogram, the opposite sides are equal.
Solution:
(a) False
(b) True
(c) False
(d) True
(e) False
(f) True
(g) False
(h) False
(i) True

Question 16.
The sides AB and CD of a quadrilateral ABCD are extended to points P and Q respectively. Is ∠ADQ + ∠CBP = ∠A + ∠C? Give reason.
(NCERT Exemplar)
Solution:
Join AC, then
∠CBP = ∠BCA + ∠BAC and ∠ADQ = ∠ACD + ∠DAC (Exterior angles of triangles)

Therefore,
∠CBP + ∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC
= (∠BCA + ∠ACD) + (∠BAC + ∠DAC)
= ∠C + ∠A

Understanding Quadrilaterals Class 8 Extra Questions Higher Order Thinking Skills (HOTS)

Question 17.
The diagonal of a rectangle is thrice its smaller side. Find the ratio of its sides.

Solution:
Let AD = x cm
diagonal BD = 3x cm
In right-angled triangle DAB,
AD² + AB² = BD² (Using Pythagoras Theorem)
x² + AB² = (3x)²
⇒ x² + AB² = 9x²
⇒ AB² = 9x² – x²
⇒ AB² = 8x²
⇒ AB = √8x = 2√2x
Required ratio of AB : AD = 2√2x : x = 2√2 : 1

Question 18.
If AM and CN are perpendiculars on the diagonal BD of a parallelogram ABCD, Is ∆AMD = ∆CNB? Give reason. (NCERT Exemplar)
Solution:

In triangles AMD and CNB,
AD = BC (opposite sides of parallelogram)
∠AMB = ∠CNB = 90°
∠ADM = ∠NBC (AD || BC and BD is transversal.)
So, ∆AMD = ∆CNB (AAS)