CBSE Sample Papers For Class 8 Maths Mock Paper 1 with Answer Key and Solutions

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

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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– 17²is

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.

$\begin{matrix} Evalutate: \\ \sqrt[3]{700 \times 2 \times 49 \times 5} \end{matrix}$

Ans:

$\begin{matrix} \sqrt[3]{700 \times 2 \times 49 \times 5} \\ = \sqrt[3]{7 \times 10 \times 10 \times 2 \times 7 \times 7 \times 5} \\ = \sqrt[3]{7 \times 7 \times 7 \times 10 \times 10 \times 10} \\ = \sqrt[3]{7^{3} \times 10^{3}} \\ = \sqrt[3]{7^{3}} \times \sqrt[3]{10^{3}} \\ = 7 \times 10 \\ = 70 \end{matrix}$

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 =

$\sqrt{1024}$

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 =

$\frac{13}{2}$

= 6.5 cm

Q16.

$By what number should \frac{3}{-14} be multiplied so as to get \frac{5}{12} .$

Ans:

$\begin{matrix} Product of two numbers is \frac{5}{12} . \\ One of these numbers is \frac{3}{–14} . \\ Other number is \frac{5}{12} ÷ \frac{3}{–14} \\ = \frac{5}{12} × \frac{–14}{3} = \frac{–35}{18} \end{matrix}$

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

$\begin{matrix} 48 \times 10 = 8 \times x \\ x= \frac{48 \times 10}{8} = 60 km/hours \end{matrix}$

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:

$\begin{matrix} Let two adjacent sides of parallelogram be 4x and 5x. \\ Then, \\ Perimeter of parallelogram = 2 × sum of adjecent sides \\ 72 cm = 2 (4x + 5x) \\ 72 cm = 18x \\ x = \frac{72}{18} \\ = 4 \\ So, the sides of parallelogram are \\ 4x = 4 (4) \\ = 16 cm \\ 5x = 5 (4) \\ = 20 cm \end{matrix}$

Q21. Divide x² + 7x + 10 by x + 5.

Ans:

$\begin{matrix} x^{2} + 7 x + 10 \\ = x^{2} + 5 x + 2 x + 10 {Splitting the middle term} \\ = x (x + 5) + 2 (x + 5) \\ = (x + 2) (x + 5) \\ We have to divide x^{2} + 7 x + 10 by (x + 5) \\ \frac{(x + 2) (x + 5)}{(x + 5)} = (x + 2) \end{matrix}$

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

$\begin{matrix} x + 30 : x = 5 : 4 \\ \Rightarrow \frac{x + 30}{x} = \frac{5}{4} \\ \Rightarrow 4 x + 120 = 5 x \\ \Rightarrow x = 120 cm \end{matrix}$

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:

$\begin{matrix} 1600 = \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times 5 \times 5 \\ The prime factor 5 does not appear in a group of three . \\ So 1600 is not a perfect cube . \\ If we divide the number by 25, then \\ the prime factorisation of the quotient will not contain 5 \times 5 . \\ So, 1600 \div 25 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \\ = 64 \\ \sqrt[3]{64} = 2 \times 2 \\ = 4 \end{matrix}$

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.

$Simplify \frac{{25× a}^{-4}}{5^{-3} {× 10 × a}^{-8}},Where a \neq 0$

Ans:

$\begin{matrix} \frac{25 \times a^{- 4}}{5^{- 3} \times 10 \times a^{- 8}} = \frac{5^{2}}{5^{- 3}} \times \frac{a^{- 4 + 8}}{5 \times 2} \\ = \frac{5^{2}}{5^{- 3 + 1}} \times \frac{a^{- 4 + 8}}{2} \\ = \frac{5^{2}}{5^{- 2}} \times \frac{a^{4}}{2} \\ = \frac{5^{4} \times a^{4}}{2} \\ = \frac{625 \times a^{4}}{2} . \end{matrix}$

Q31.

$\begin{matrix} Find A, B, C in the addition. \\ 3 4 A \\ + 3 A B \\ \bar{C 2 9} \end{matrix}$

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.

$Using appropriate properties, find \frac{–2}{3} × \frac{3}{5} – \frac{3}{5} × \frac{1}{6} + \frac{5}{2}$

Ans:

$\begin{matrix} \frac{–2}{3} × \frac{3}{5} - \frac{3}{5} \times \frac{1}{6} + \frac{5}{2} \\ = \frac{2}{3} \times (\frac{- 3}{5}) + \frac{- 3}{5} \times \frac{1}{6} + \frac{5}{2} \\ = - \frac{3}{5} \frac{2}{3} + \frac{1}{6} + \frac{5}{2} \\ = \frac{- 3}{5} \frac{4 + 1}{6} + \frac{5}{2} \\ = \frac{- 3}{5} \times \frac{5}{6} + \frac{5}{2} = \frac{- 1}{2} + \frac{5}{2} = 2 \end{matrix}$

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:

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

$L e t p a r a ll e l s i d es o f t r a p ez i u m b e 3 x an d 5 x . T h e d i s t an ce b e tw ee n p a r a ll e l s i d es i s 12 c m . A re a o f t r a p ez i u m = 2 1 (3 x + 5 x) \times 12 384 = 2 1 (8 x) \times 12 x = 6 \times 8 384 x = 8 c m ∴ 3 x = 3 \times 8 = 24 c m 5 x = 5 \times 8 = 40 c m T h u s, t h e p a r a ll e l s i d es o f t r a p ez i u m a re 24 c m an d 40 c m .$

Q35.

$\begin{matrix} Simplify {(\frac{5}{3} x + \frac{3}{4} y)}^{2} – {(\frac{5}{3} x – \frac{3}{4} y)}^{2} and also evaluate \\ it when x = 2 and y = -1. \end{matrix}$

Ans:

$\begin{matrix} {(\frac{5}{3} x + \frac{3}{4} y)}^{2} - {(\frac{5}{3} x - \frac{3}{4} y)}^{2} \\ = {(\frac{5}{3} x)}^{2} + 2 × \frac{5}{3} x × \frac{3}{4} y + {(\frac{3}{4} y)}^{2} - {(\frac{5}{3} x)}^{2} - 2 × \frac{5}{3} x × \frac{3}{4} y + {(\frac{3}{4} y)}^{2} \\ = \frac{25}{9} x^{2} + \frac{5}{2} xy + \frac{9}{16} y^{2} - \frac{25}{9} x^{2} + \frac{5}{2} xy - \frac{9}{16} y^{2} \\ = 2 × \frac{5}{2} xy \\ = 5 xy \\ = 5 (2) (-1) \\ =-10 \end{matrix}$

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)

$\begin{matrix} Since oppos i te angles of parallelogram are equal. \\ So, ∠ A = ∠ C \\ \Rightarrow x = 110 ° \\ ∠ ABE = ∠ DCB Corresponding angles are equal \\ \Rightarrow y = x = 110 ° \\ \Rightarrow y = 110 ° \\ ∠ ABC + ∠ ABE = 180 ° ∵ Sum of l i near pa i r of angles i s 180° \\ z + y = 180 ° \\ z + 110 ° = 180 ° \\ z = 70 ° \\ Thus, x = 110 °, y = 110 ° and z=70 ° . \end{matrix}$

(ii)

$\begin{matrix} If OE = 3 cm then \\ OP = 3 cm ∵ OE = OP \\ PE = OE + OP \\ PE = 3 + 3 = 6 cm \begin{matrix} ∵ Diagonals of parallelogram bisect \\ each other. \end{matrix} \\ Now, \\ HL = PE + 7 Given \\ HL = 6 + 7 \\ = 13 cm \\ ∵ OH = \frac{HL}{2} \begin{matrix} ∵ Diagonals of parallelogram bisect \\ each other. \end{matrix} \\ OH = \frac{13}{2} \\ = 6 .5 cm . \end{matrix}$

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

Ans:

$\begin{matrix} Selling price of washing machine = ₹ 5760 \\ Two successive discounts are 15% and 10%. \\ Let marked price of washing machine = ₹ x \\ S . P . of washing machine after first discount = x (\frac{100 - 15}{100}) \\ = \frac{85 x}{100} \\ = \frac{17 x}{20} \\ S . P . of washing machine after second discount = \frac{17 x}{20} (\frac{100 - 10}{100}) \\ = \frac{17 x}{20} \times \frac{90}{100} \\ = \frac{153 x}{200} \\ Then, according to condition \\ \frac{153 x}{200} = Rs . 5760 \\ x = ₹ 5760 \times \frac{200}{153} \\ x = ₹ 7529 .40 (approx) \\ Thus, the marked price of washing machine is ₹ 7529 .40. \end{matrix}$

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FAQs (Frequently Asked Questions)

It is strongly advised that students prepare for the CBSE Class 8 Examination of Mathematics using the CBSE Sample Papers For Class 8 Maths Mock Paper 1 because they have been prepared by experts in the field. The CBSE Sample Papers For Class 8 Maths Mock Paper 1 have been compiled by professionals in an easy-to-understand format. They have been extensively elaborated and systematically broken down into smaller steps for students to understand easily. The CBSE Sample Papers For Class 8 Maths Mock Paper 1 may be acquired from the learning application or website of Extramarks.

It is highly recommended that students of Class 8 go through the syllabus of CBSE Class 8 Mathematics in depth. Students are suggested to practice sample papers and review past years’ papers in order to be well-prepared for the Mathematics Examination. Past years’ papers may help give students a better understanding of the examination question paper pattern. It is also suggested that students of Class 8 make it a point to memorise all the Formulas covered in the Mathematics syllabus of Class 8.

Extramarks provides a number of learning tools for students of all classes. Students can acquire a variety of study materials as per their requirements from the Extramarks website or learning application. The NCERT Books exercise problems and solutions, CBSE Important Question, CBSE Extra Questions, solved examples, practice papers, sample papers, revision notes, and many more can be acquired from the Extramarks website. Students appearing for competitive examinations can access preparatory learning resources designed with the examination point of view in mind from the Learning application or website of Extramarks.

CBSE Sample Papers For Class 8 Maths Mock Paper 1

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

Download Sample Paper Class 8 Math 2021

Chapter-Wise Weightage of Class 8 Math Sample Paper

Benefits of solving Math Sample Papers

More Benefits of solving Sample Paper of Class 8 Math

Conclusion

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CBSE Sample Papers For Class 8 Maths Mock Paper 1

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

Download Sample Paper Class 8 Math 2021

Chapter-Wise Weightage of Class 8 Math Sample Paper

Benefits of solving Math Sample Papers

More Benefits of solving Sample Paper of Class 8 Math

Conclusion

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Please register to view this section

FAQs (Frequently Asked Questions)

1. Why are students of Class 8 advised to use the CBSE Sample Papers For Class 8 Maths Mock Paper 1 for their preparation?

2. What are the strategies that students may employ to perform well in the CBSE Mathematics Examination?

3. What are the different types of pedagogical resources provided by Extramarks?

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