CBSE Solved Sample Papers for Class 8 – Maths

CBSE Solved Sample Papers for Class 8 for Maths – Second Term

CBSE Solved Sample Papers for Class 8 for Maths Subject (Term 2)  is given below. The solved sample papers are created as per the latest CBSE syllabus and curriculum keeping in mind the latest marking scheme.

SECTION – A

 

Q 1 The value of (5˚ + 7^-1) ^* 7 are:

(a)    35
(b)   36
(c)    8
(d)   84

Ans (c)

Q 2 Factories x^{2 +} 5x + 6:

(a)    (x+3) (x+2)
(b)   (x-3) (x+2)
(c)    (x+3) (x- 2)
(d)   All of above

Ans (a)

Q 3 A train is moving at a uniform speed of 80 km/hr. How for it will travel in 15 minutes.

(a)    85 km
(b)   120km
(c)    20km
(d)   30km

Ans (c)

Q 4 (1.02)² – (0.98)² are equal to:

(a)    0.04
(b)   0.08
(c)    0.8
(d)   2

Ans (b)

Q 5 what is the coordinate of point P on the graph:

(a)    7

(b)   (7,0)
(c)    (1,5)
(d)   (0,7)

Ans (d)

Q 6 the cost of 5 pens is Rs. 25.50. What is the cost of 8 pens?

(a)    40.80
(b)   40
(c)    50
(d)   45.20

Ans (a)

Q 7 Which one of the following pair forms like terms:

(a)    7x – 5y
(b)   2x – 3x²
(c)    2/3x – 5/7x
(d)   2/5x – 3xy

Ans (c)

Q 8 The relation between F,V and E are represented by Euler’s formula as follows:

(a)    F – V + E = 0
(b)   F + E + V = 1
(c)    F – V + E = 2
(d)   F + V – E = 2

Ans (d)

Q 9 which one is binomial:

(a)    4l + 5m
(b)   3x²– 5x + 2
(c)    4 + 3/x
(d)   X + 1/y

Ans (a)

Q 10 if the longer side of a trapezium is 10 cm and the distance between the parallel sides is 8 cm. The area is 50 cm^2. What is the smaller parallel side?

(a)    10 cm
(b)   2 cm
(c)    2.5 cm
(d)   3.5 cm

Ans (c)

SECTION – B

Q 11 How many sides of the following have?

(a)    Nonagon
(b)   Octagon

Ans

(a)    Nonagon: Nonagon has 9 sides.
(b)   Octagon: octagon has 8 sides.

Q 12 What is the volume of a cube with side 6 cm?

Ans side of cube = 6 cm

Volume of cube= side * side * side

= (6 * 6 * 6) cm

= 216 cm³                             

Q 13 Simplify (-5)^-2

Ans (-5)^-2 = 1 / (-5)^-2 = 1/-5 * -5

=1 / 25

Q 14 Find the value of: (2^-1 + 4^-1) / 2)^-2

Ans (2^-1 + 4^-1) / 2)^-2

= [1/2+ ¼] / 1/2²

= (2 + 1 / 4) / (1/4)

= (¾) / (1/4)

= ¾ * 4 = 3

Q 15 Write 2 – dimensional or 3 – dimensional for the following shapes:

Square, cone, circle, cylinder

AnsSquare, circle = 2 – dimensional

Cone, cylinder = 3 – dimensional

Q 16 What formula would you use to calculate the area of a rectangle?

AnsArea of a rectangle = length * breadth

Q 17 what is range of the data?

Ans The difference between the highest and the lowest values of the observation in a data is called the range of the data.

Q 18 Evaluate (2/3)^-3

Ans(2/3)^{-3 =} (3/2)³

= 3/2 * 3/2 * 3/2

= 27/8

SECTION – C

Q 19 The lateral surface area of a hollow cylinder is 4224 cm^2. It is cut along its height and formed a rectangle sheet of width 33 cm. Find the perimeter of the rectangle sheet.

Ans lateral surface area of cylinder = 4224 cm²

Width of sheet (b) = 33 cm

Length of sheet (l) = area/width

L = 4224/33 = 128 cm

Perimeter of sheet = 2(l + b)

= 2 (128 + 33)

= 2(161)

= 322 cm

Q 20 Factories 16x⁵ – 121x³

Ans        16x⁵ – 121x³

= x³(16x²– 121)

= x³[(4x) ²– (11)²]

= [(a ²– b ²⁾ = (a +b) (a – b)]

= x³[(4x + 11)– (4x – 11)]

Q 21 Find the value of 9^-1 * 5^-3 / 3^-4

Ans 9^-1 * 5^-3 / 3^-4= 3⁴ * 5³ / 9

= 3⁴ * 5³ / 3²

= 3^{4 – 2} * 5³

                        = 3² * 5³

                        = (3 * 3) (5 * 5 * 5)

= 9 * 125

= 1125

Q 22 A loaded truck travels 14 km in 25 minute. If the speed remains the same. How far can it ravel in 5 hours?

AnsDistance covered in 25 minute = 14 km

Distance covered in 1 minute = 14 / 25 km

Distance covered in 5 hours = (5 hours = 5 * 60 = 300 minute)

= 14 / 25 * 300

= 14 * 12 = 168 km

He can travel 168 km in 5 hours.

Q 23 In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m. How long is the mast on the ship?

Ans The height of the mast and the length of the model ship are indirect proportion.

Let x =length of model

Y= length of ship

X = 9 cm               ,               y =12 cm

X =?                       ,               y = 28 cm

X₁/y₁         =            x₂/y₂

9 / 12     =             x₂/28

_X2 = 9 * 28 /12

                        _X2= 3 * 7s

_X2 = 21 cm

Thus Model ship is 21 cm long.

Q 24 Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³.

Ans Base area of a cuboid = 180 cm²

Volume of a cuboid = 900 cm³

Height of a cuboid = volume/base area

= 900 / 180

Thus height of a cuboid = 5 cm.

Q 25 Using suitable identity, evaluate:

(a)    1.05 * 0.95

(b)   9.7 * 9.8

Ans

(a)    1.05 * 0.95

= (1 + 0.05)(1-0.05)

= 1 – 0.0025 = .9975

(b)   9.7 * 9.8

= (9 + .7) (9 + .8)

= (9)² + (.7 + .8) * 9 + (.7 * .8)

=81 + (1.5) * 9 + (.56)

= 95.06

Q 26 multiply (2 / 3 xy) * (-9 / 10 x²y²)

Ans (2 / 3 xy) * (-9 / 10 x²y²)

= (2 / 3 * -9 / 10) xy * x²y²

= -3 / 5 x³y³

Q 27 Using identity solve 99²

Ans        99²          = (100 – 1)²

(a – b) ²  = a² +b² -2ab

= (100) ² + (1) ² – 2(100) (1)

= 10000 + 1 -200

= 10001 – 200

= 9801

Q 28 Subtract 3xy – 5yz- 7zx from 5xy – 2yz -2zx + 10xyz

Ans (5xy – 2yz -2zx + 10xyz) – (3xy – 5yz- 7zx)

= 5xy – 2yz -2zx + 10xyz – 3xy + 5yz + 7zx

= 5xy – 3xy – 2yz + 5yz -2zx + 7zx + 10xyz

= 2xy +3yz + 5zx +10xyz

Q 29 A field is 60 m long and 24 m broad. In one corner of the filed, a pit which is 8 m long 6.5m broad and 4 m deep has been dug out. The earth taken out of it is evenly spread over the remaining part of the field. Find the rise in the level of the filed.

Ans area of the filed = (60 * 24) m

= 1440 m²

Area of th pit = (8 * 6.5) m

= 52m²

Area over the remaining part of the filed = (1440 – 52) m²

= 1388 m²

Volume of earth dug out = 8 * 6.5 * 4

= 208 m²

Rise in level = Volume of earth dug out/area over the remaining part of filed

=208/1388 m

= 208 * 100/1388 cm

= 10.4 cm

Q 30 If x – 1/x = 6 find (x² + 1 / x²) and x⁴ + 1 / x⁴

Ans x – 1 / x = 6 = (x – 1 / x) ² = (6)2

= x² + 1 / x²– 2 = 36

= x² + 1 / x²= 38

=(x2 + 1 / x²⁾²= (38)²

= x⁴ + 1 / x⁴+ 2 = 1444

= x⁴ + 1 / x⁴= 1444

Q 31 A square park has been each side of 100 m. At each corner there is a flower bed in the form of a quadrant of radius 14m as shown in fig. Find the area of the remaining part of the park. (Takeπ =22 / 7)

AnsArea of each quadrant of radius = 14 cm

= 1 / 4 (πr²)

=1 / 4 π * 14²

= 1 / 4 * 22 / 7 * 14 *14

= 154 m²

Area of 4 quadrants = (4 * 154) m² = 616 m²

Side of square = 100 m

Area of square = (100 * 100) m²

= 10000 m²

Area of the remaining part of the park

= (1000 – 616) m²

= 9384 m²

Q 32 Factories:

(a)    8x³ – 18 xy²

(b)   3x² + 12x – 36

(c)    X² + 4xy +4y² – 9z²

Ans

(a)    8x³ – 18 xy²

= 2x (4x² – 9y²)

= 2x (2x + 3y) (2x – 3y)                   [By using a² – b² = (a + b) (a – b)]

= 2x (2x + 3y) (2x – 3y)

(b)   3x² + 12x – 36

= 3 (x² + 4x – 12)

= 3 (x² + 6x -2x – 12)

= 3 [x (x + 6) – 2 (x + 6)]

= 3 [(x – 2) (x +6)]

(c)     X² + 4xy +4y² – 9z²

= (x² + 4xy + 4y²) – 9z²

= (x + 2y) ² – (9z²)

=(x +2y + 3z) (x + 2y -3z)

= (x + 2y + 3z) (x + 2y – 3z)

Q 33 Subtract: 3a (a + b + c) -2b (a – b + c) from 4c (- a + b + c)

Ans 3a (a + b + c) -2b (a – b + c) from 4c (- a + b + c)

= 4c (- a + b + c) – [3a (a + b + c) -2b (a – b + c)]

= – 4ca + 4bc +4c² – [(3a² + 3ab + 3ac) – 2ba +2b² – 2bc)]

= – 4ca + 4bc +4c² – [3a² + 3ab + 3ac – 2ba +2b² – 2bc]

= – 4ca +4bc +4c² – [+ 3ac -2bc + 3a² + 2b² + 3ab]

= – 4ca +4bc +4c² – 3ac +2bc – 3a² – 2b² – 3ab

= – 7ac + 6bc – 3a² – 2b² +4c² – 3ab

= – 7ac + 6bc – 3ab – 3a² – 2b² +4c²

Q 34 Plot the graph from the following table. Is the graph linear?

Side of square (in cm)	2	3	3.5	5	6
Perimeter (in cm)	8	12	14	20	24

 

Ans

 

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