CBSE Sample Papers For Class 8 Maths Mock Paper 1

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CBSE Sample Paper for Class 8 Maths with Solutions – Mock Paper-1

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Download Sample Paper Class 8 Math 2021

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Q1. A retailer buys a radio for ₹215. He expends ₹25 on its repairing. If he sells the radio for ₹300, his profit percentage will be

Opt:

43.6%

39.5%

25%

20%

Ans: 25%

Q2. Number of edges in a pyramid with square base is

Opt:

10

8

6

4

Ans: 8

Q3. The value of ‘a’ in the given mathematical statement x2 – 10x + 25 = (x + a)(x + a) is

Opt:

–10

–5

5

10

Ans: –5

Q4. If the time (t) taken to travel from one point to another is inversely proportional to the speed (s) of the travel, then the relation can be represented as (where k is a constant of variation)

Opt:

s/t = k

s/t = k2

s × t = k

s = t + k

Ans: s × t = k

Q5. If the division X ÷ 5 leaves a remainder of 2, then which of the following can be one’s digit of X?

Opt:

3

4

7

8

Ans: 7

Q6.

Opt:

1

2

3

4

Ans: 2

Q7. The value of 192 – 172 is

Opt:

650

361

289

72

Ans: 72

Q8. A box of cereals is 11.2 inch high, 7.5 inch long and 5.2 inch wide. The volume of box is

Opt:

873.6 inch3

436.8 inch3

362.48 inch2

181.24 inch2

Ans: 436.8 inch3

Q9. On subtracting the reciprocal of 5/7 from the additive inverse of (–3/4) we get

Opt:

11/20

13/20

–11/20

–13/20

Ans: –13/20

Q10. A number is tripled, and then increased by five. The result is 47. The number is

Opt:

11

12

13

14

Ans: 14

Q11. The number of sides of a regular polygon whose each exterior angle has a measure of 40° is

Opt:

4

6

8

9

Ans: 9

Q12. Digit at the one’s place of the square root of 4624 is:

Opt:

2

5

7

8

Ans: 8

Q13.

Evalutate:700×2×49×53

Ans:

700 × 2× 49× 53    =7×10×10 ×2× 7×7×5 3  =  7×7×7×10×10×103  =73 × 1033   =733  ×1033    =7 × 10 = 70

Q14. Find the side of a square whose area is 1024 m2.

Ans:

Let s be the side of a square,
s2 = area of the square = 1024   (Given)

∴ s =

1024

Therefore, the side of the square is 32 m.

Q15.

In parallelogram PQRS, given that OQ = 4 cm, and PR is 5 more than SQ. Find OP.

Ans:

OQ = OS = 4 cm

OQ + OS = SQ = 8 cm (diagonals of parallelogram bisect each other)

∴  PR = 8 + 5 = 13 cm

Now,

OP =

132

= 6.5 cm

Q16.

By what number should 3-14 be multiplied so as to get 512.

Ans:

Product of two numbers is 512.One of these numbers is 3–14.Other number is 512÷3–14                          =512×–143=–3518

Q17. A car can finish a journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?

Ans: Let the speed be x km/hr for covering the same distance in 8 hours.

Time Taken in hours 10 8
Speed in km/hr 48 x

48 × 10 = 8 × x x= 48 × 108 = 60 km/hours

Therefore, the speed needs to be increased by 12 km/hr.

Q18. Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.

Ans:

(2, 3) and (3, 2) meets the x-axis at (5, 0) and y-axis at (0, 5) .

Q19. A bag has 4 red balls and 6 yellow balls. A ball is drawn from the bag without looking into the bag. Find the probability of getting a red ball.

Ans: Number of outcomes of getting a red ball = 4
Total number of outcomes = 4 + 6 = 10
Probability of getting a red ball = 4/10 = 2/5

Q20. The ratio of two adjacent sides of a parallelogram is 4:5. If its perimeter is 72 cm, find its adjacent sides.

Ans:

Let two adjacent sides of parallelogram be 4x and 5x.Then,Perimeter of parallelogram = 2 × sum of adjecent sides72 cm = 24x + 5x72 cm = 18x     x = 7218        = 4So, the sides of  parallelogram are     4x = 44         =  16 cm     5x = 54         = 20 cm

Q21. Divide x2 + 7x + 10 by x + 5.

Ans:

x2+7x+10=x2+5x+2x+10       {Splitting the middle term}                   = xx+5+2x+5                 = x+2x+5We have to divide x2+7x+10 by (x+5)x+2x+5(x+5)=x+2        

Q22. Simplify: (xy + yz)2 – (xy – yz)2

Ans: Simplify the given expression as:

(xy + yz)2 – (xy – yz)2  = (xy)2 + 2xy2z + (yz)2 – [x2y2 – 2xy2z + y2z2]

= x2y2 + 2xy2z + y2z2 –  x2y2 + 2xy2z – y2z2

= 4xy2z

Q23. Find the volume of 64 cubes whose one side is 4 cm.

Ans:

Given, volume of cube = side3
Length of one side of cube = 4 cm
∴ Volume of one cube = 43
= 64 cm3
∴ Volume of 64 cubes = 64 × 64
= 4096 cm3

Q24. Find x such that (-2)x+1 × (-2)5 = (-2)7.

Ans: (-2)x + 1 + 5 = (-2)7 ⇒  (-2)x + 6 = (-2)7

On both the sides powers have the same base different from 1 and -1, their exponents must be equal.

∴  x + 6 = 7

⇒  x = 7 – 6 = 1

Q25. By using Euler’s formula find the unknown.

a) Vertices = 12, Faces = 4, Edges = ?

b) Faces = 5, Edges = 8, Vertices = ?

c) Edges = 2, Vertices = 3, Faces = ?

Ans: By Euler’s formula
F + V = E + 2

Where F = Number of faces of polyhedron

V = Number of vertices of polyhedron and

E = Number of edges of polyhedron

(a) F = 4; V = 12; E = ?
 ∵  F + V = E + 2
4 + 12 = E + 2
 ⇒          E = 16 – 2 = 14

(b) F = 5; E = 8; V = ?
∵ F + V = E + 2
5 + V = 8 + 2
⇒         V = 10 – 5 = 5

(c) E = 2; V = 3; F = ?
∵   F + V = E + 2
F + 3 = 2 + 2
 ⇒       F = 4 – 3 = 1

Q26. Each side of a triangle is increased by 10 cm; if the ratio of the perimeter of the new triangle and the given triangle is 5 : 4, find the perimeter of the given triangle.

Ans: Let the perimeter of the triangle be x cm.

New perimeter after increasing the sides = (x + 30) cm

x+30:x=5:4⇒ x+30x=54⇒ 4x+120=5x⇒ x=120cm

Therefore, the perimeter of the given triangle = 120 cm

Q27. Construct a rhombus with side 4.5 cm and one diagonal 6 cm.

Ans: The given rhombus can be drawn as follows:

Step 1: Draw AC of 6 cm.

Step 2: From A with radius 4.5 cm, draw arcs above and below AC.

Step 3: From C with radius 4.5 cm, draw arcs above and below AC intersecting the arcs drawn in step 2. Let the points of intersection be D and B.

Step 4 : Join AD, CD, AB and BC.

ABCD is the required rhombus.

Q28. Find the smallest number by which 1600 must be divided so that the quotient is a perfect cube, further find its cube root.

Ans:

1600=2×2×2¯×2×2×2¯×5×5The prime factor 5 does not appear in a group of three. So 1600 is not a perfect cube.If we divide the number by 25, then the prime factorisation of the quotient will not contain 5×5.So, 1600÷25=2×2×2×2×2×2                  =64             643=2×2                   =4

Q29. Find the smallest square number which is divisible by each of the numbers 4, 9 and 10.

Ans: First find the smallest common multiple of 4, 9 and 10.

LCM of 4, 9 and 10 = 180

Prime factors of 180 = 2 × 2 × 3 × 3 × 5

Prime factor 5 is not in pair. Thus, 180 is not a perfect square number. It has to be multiplied by 5 so that it’s a perfect square number.

Hence, 180 × 5 = 900 is the perfect square number divisible by 4, 9 and 10.

Q30.

Simplify 25× a-45-3 × 10 × a -8 ,Where a ≠0

Ans:

25×a−45−3×10×a−8=525−3×a−4+85×2                 =525−3+1×a−4+82                 =525−2×a42                 =54×a42                 =625×a42.

Q31.

Find A, B, C in the addition.     3     4     A+   3     A     B     C     2     9 ¯

Ans: The addition of A and B is giving 9 i.e., a number whose ones digits is 9. The sum can be 9 only as the sum of two single digit numbers cannot be 19. Therefore, there will not be any carry in this step.

In the next step, 4 + A = 2

It is possible only when A = 8

4 + 8 = 12 and 1 will be the carry for the next step.

1 + 3 + 3 = C

Therefore, C is 7.

We know that the addition of A and B is giving 9. As A is 8, therefore, B is 1.

Q32.

Using appropriate properties, find –23×35–35×16+52

Ans:

–23×35–35×16+52= 23×−35+−35×16+52= −3523+16+52= −354+16+52= −35×56+52=−12+52=2

Q33. Srishti has a total of ₹ 780 as currency notes in the denominations of ₹ 10, ₹ 20 and ₹ 50. The ratio of the number of 50 notes and Rs 10 notes is 3 : 2. If she has a total of 32 notes, how many notes of each denomination she has?

Ans: Let Srishti has 3x notes of ₹ 50 and 2x notes of ₹ 10 in numbers.

So, number of notes of ₹ 20 = (32 – 3x – 2x) = (32 – 5x)

Amount of ₹ 50 notes = 50(3x) = ₹ (150x)

Amount of ₹ 20 notes = 20(32 – 5x) = ₹ (640 – 100x)

Amount of ₹ 10 notes = 10(2x) = ₹ (20x)

Total money she has = ₹ 780

150x + 640 – 100x + 20x = 780

or  70x = 140

or  x = 2

The number of ₹ 50 notes Srishti has = 6

The number of ₹ 20 notes Srishti has = 22

The number of ₹ 10 notes Srishti has = 4

Q34. The area of a trapezium is 384 cm2. Its parallel sides are in the ratio 3:5 and the distance between them is 12 cm. Find the length of each parallel side.

Ans:

Let parallel sides of trapezium be 3x and 5x.The distance between parallel sides is 12 cm.Area of trapezium =12(3x + 5x)×12               384=12(8x)×12                       x = 3846×8                        x = 8cm∴3x = 3 × 8 = 24cm   5x = 5 × 8 = 40cmThus, the parallel sides of trapezium are 24 cm and 40 cm. \begin{array}{l}{Let parallel sides of trapezium be 3x and 5x.}\\ {The distance between parallel sides is 12 cm.}\\ {Area of trapezium =}\frac{{1}}{{2}}\left({3x + 5x}\right){×12}\\ {               \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}384=}\frac{{1}}{{2}}\left({8x}\right){×12}\\ {\hspace{0.17em}                       \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x = }\frac{{384}}{{6×8}}\\ {                        \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x = 8\hspace{0.17em}cm}\\ \therefore {\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3x = 3 × 8 = 24\hspace{0.17em}cm}\\ {\hspace{0.17em}   \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5x = 5 × 8 = 40\hspace{0.17em}cm}\\ {Thus, the parallel sides of trapezium }\\ {are 24 cm and 40 cm.}\end{array}

Q35.

Simplify 53x + 34y2 – 53x – 34y2and also evaluate it when x = 2 and y = -1.

Ans:

53x + 34y2 − 53x - 34y2= 53x2 + 2 × 53x × 34y + 34y2 - 53x2 - 2 × 53x × 34y + 34y2= 259x2 + 52xy + 916y2 -259x2 + 52xy -916y2= 2 × 52xy= 5xy= 5 2 -1=-10

Q36. (i) In the figure given below, ABCD is a parallelogram. Find the value of x, y and z.

(ii) Figure HELP shown below is a parallelogram. It is given that OE = 3 cm and HL is 7 more than PE, find OH.

Ans:

(i)
 

Since opposite angles of parallelogram are equal.So, ∠A=∠C⇒      x=110°∠ABE=∠DCBCorresponding​ angles are equal⇒     y=x=110°⇒     y=110°  ∠ABC+∠ABE=180°   ∵ Sum of linear pair of angles is 180°z+y=180°  z+110°=180°       z=70°Thus,  x=110°,y=110° and z=70°.

(ii) 

If OE = 3 cm then OP = 3 cm ∵ OE = OPPE = OE + OPPE = 3 + 3 = 6 cm ∵ Diagonals of parallelogram bisecteach other.Now,HL = PE + 7 GivenHL = 6 + 7   = 13 cm∵  OH = HL2  ∵ Diagonals of parallelogram bisecteach other.OH = 132     = 6.5 cm.

 

Q37. A washing machine was sold for ₹ 5760 after giving successive discounts of 15% and 10% respectively. What was the marked price?

Ans:

Selling price of washing machine=₹ 5760Two successive discounts are 15% and 10%.Let marked price of washing machine=₹ xS.P. of washing machine after first discount =x100−15100                                                                  =85x100                                                                  =17x20S.P. of washing machine after second discount=17x20100−10100                                                                       =17x20×90100                                                                       =153x200Then, according to condition                                                                153x200=Rs.5760                                                                           x=₹5760×200153                                                                            x=₹7529.40 ​approxThus, the marked price of washing machine is ₹7529.40. 

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