CBSE Class 8 Mathematics Chapters 1 to 6 Revision Practice Paper 01 with Answers (Old NCERT Syllabus)

This Class 8 Maths revision practice paper covers Chapters 1 to 6 (Rational Numbers, Linear Equations in One Variable, Understanding Quadrilaterals, Data Handling, Squares and Square Roots, Cubes and Cube Roots) in a 40-mark paper of 17 questions across Sections A to E, with step-by-step answers to every question. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it suits state-board students and anyone revising these topics. The questions test square and cube roots, multiplicative inverses, the distributive property, probability, polygon angle sums, reading a pie chart and turning a word problem into an equation.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Find the sum: \( \frac{2}{3} + \frac{-3}{5} + \frac{1}{6} + \frac{-8}{15} \)
Answer:
The LCM of 3, 5, 6 and 15 is 30.
\( \frac{2}{3} = \frac{20}{30}, \; \frac{-3}{5} = \frac{-18}{30}, \; \frac{1}{6} = \frac{5}{30}, \; \frac{-8}{15} = \frac{-16}{30} \)
Sum = \( \frac{20 - 18 + 5 - 16}{30} = \frac{-9}{30} \)
= \( \frac{-3}{10} \)
In simple words: Write every fraction with denominator 30, add the tops, and simplify: −9/30 = −3/10.

Teacher's Note:
1. Group the positives (20 + 5 = 25) and the negatives (−18 − 16 = −34) to add quickly.
2. Always reduce the answer: −9/30 divides by 3 to give −3/10.

Question 8. In a survey of 200 ladies, it was found that 82 like coffee while 118 dislike it. From these ladies, one is chosen at random. What is the probability that the chosen lady dislikes coffee?
Answer:
Total number of outcomes = 200.
Number of ladies who dislike coffee = 118 (favourable outcomes).
P(dislikes coffee) = \( \frac{118}{200} = \frac{59}{100} \)
So the probability is \( \frac{59}{100} \) (that is, 0.59).
In simple words: 118 of the 200 ladies dislike coffee, so the chance is 118/200 = 59/100.

Teacher's Note:
1. Probability = favourable outcomes ÷ total outcomes.
2. Check: 82/200 + 118/200 = 1, as the two events cover every lady.

Question 9. Find the measure x in the following figure:

Answer:
The figure is a pentagon whose base lies on a straight line.
Interior angle at the bottom left = 180° − 70° = 110° (linear pair).
Interior angle at the bottom right = 180° − 60° = 120°.
Sum of the interior angles of a pentagon = (5 − 2) × 180° = 540°.
So 110° + x + 30° + x + 120° = 540°
260° + 2x = 540°, so 2x = 280°
x = 140°
In simple words: Turn the outside angles into inside angles, then make all five inside angles add up to 540°.

Teacher's Note:
1. An exterior angle and its interior angle form a linear pair and add to 180°.
2. Angle sum of an n-sided polygon = (n − 2) × 180°.

Section C: Short Answer Questions (3 Marks Each)

Question 10. Verify the property x × (y + z) = (x × y) + (x × z) by taking x = \( \frac{-3}{4} \), y = \( \frac{5}{2} \) and z = \( \frac{-1}{6} \).
Answer:
LHS: y + z = \( \frac{5}{2} + \frac{-1}{6} = \frac{15 - 1}{6} = \frac{14}{6} = \frac{7}{3} \)
x × (y + z) = \( \frac{-3}{4} \times \frac{7}{3} = \frac{-7}{4} \)
RHS: x × y = \( \frac{-3}{4} \times \frac{5}{2} = \frac{-15}{8} \)
x × z = \( \frac{-3}{4} \times \frac{-1}{6} = \frac{3}{24} = \frac{1}{8} \)
(x × y) + (x × z) = \( \frac{-15}{8} + \frac{1}{8} = \frac{-14}{8} = \frac{-7}{4} \)
LHS = RHS = \( \frac{-7}{4} \), so x × (y + z) = (x × y) + (x × z) is verified.
In simple words: Work out both sides separately with the given numbers; both come to −7/4.

Teacher's Note:
1. This is the distributive property of multiplication over addition.
2. Simplify each side fully before comparing, so equal values are easy to see.

Question 11. Given below is a pie chart showing the time spent by a group of 350 children in different games. Observe it and answer the questions that follow.
(a) How many children spend at least one hour in playing games?
(b) How many children spend more than 2 hours in playing games?
(c) How many children spend 3 or fewer hours in playing games?
Answer:
From the chart: less than 1 hr = 6%, 1 hr = 16%, 2 hrs = 30%, 3 hrs = 34%, 4 hrs = 10%, 5 hrs = 4%.
(a) Less than 1 hour = 6% of 350 = \( \frac{6 \times 350}{100} \) = 21 children.
At least one hour = 350 − 21 = 329 children.
(b) More than 2 hours = 3 hrs + 4 hrs + 5 hrs = 34% + 10% + 4% = 48%.
48% of 350 = \( \frac{48 \times 350}{100} \) = 168 children.
(c) 3 or fewer hours = 6% + 16% + 30% + 34% = 86%.
86% of 350 = \( \frac{86 \times 350}{100} \) = 301 children.
In simple words: Add the percentages of the groups asked for, then take that percentage of 350.

Teacher's Note:
1. Read each sector's label against the legend before adding percentages.
2. Shortcut: 1% of 350 = 3.5, so multiply the total percentage by 3.5.
3. Check that all the sector percentages add to 100%.

Question 12. Find the value of \( \frac{\sqrt[3]{729} - \sqrt[3]{27}}{\sqrt[3]{512} + \sqrt[3]{343}} \).
OR
Simplify: \( \sqrt[3]{(1.1)^3} \times \sqrt[3]{1.331} \)
Answer:
∛729 = 9, ∛27 = 3, ∛512 = 8, ∛343 = 7.
Value = \( \frac{9 - 3}{8 + 7} = \frac{6}{15} \) = \( \frac{2}{5} \)
OR
\( \sqrt[3]{(1.1)^3} = 1.1 \)
\( \sqrt[3]{1.331} = \sqrt[3]{\frac{1331}{1000}} = \frac{11}{10} = 1.1 \)
Product = 1.1 × 1.1 = 1.21
In simple words: Find each cube root first (9, 3, 8, 7), then do the subtraction, addition and division.

Teacher's Note:
1. Cube root of a decimal: write it as a fraction with denominator 1000, then take cube roots of top and bottom.
2. ∛(a³) = a, so ∛(1.1)³ is simply 1.1.

Question 13. Find the number of plants in each row if 1024 plants are arranged so that the number of plants in a row is the same as the number of rows.
Answer:
Let the number of rows = x. Then the number of plants in each row = x.
Total plants = x × x = x² = 1024.
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2 × 2 × 2)²
x = √1024 = 2 × 2 × 2 × 2 × 2 = 32
So there are 32 plants in each row.
In simple words: Rows × plants per row = 1024 with both equal, so each is √1024 = 32.

Teacher's Note:
1. "Rows equal to plants in a row" means the arrangement is a square, so take the square root.
2. Pair the prime factors; one factor from each pair gives the square root.

Section D: Long Answer Questions (4 Marks Each)

Question 14. (a) The four angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. Find the angles.
(b) Two adjacent angles of a parallelogram are in the ratio 4 : 5. Find their measures.
Answer:
(a) Let the angles be 3x, 4x, 5x and 6x.
Sum of the angles of a quadrilateral = 360°, so 18x = 360° and x = 20°.
The angles are 60°, 80°, 100° and 120°.
(b) Adjacent angles of a parallelogram are supplementary. Let them be 4x and 5x.
4x + 5x = 180°, so 9x = 180° and x = 20°.
The angles are 80° and 100°.
In simple words: Share 360° in the ratio 3 : 4 : 5 : 6, and 180° in the ratio 4 : 5.

Teacher's Note:
1. Quadrilateral angles total 360°; adjacent angles of a parallelogram total 180°.
2. Check: 60 + 80 + 100 + 120 = 360 and 80 + 100 = 180.

Question 15. Find the smallest square number which is exactly divisible by 4, 9 and 10.
Answer:
The smallest number divisible by 4, 9 and 10 is their LCM = 180.
180 = 2 × 2 × 3 × 3 × 5
The factors 2 and 3 are in pairs, but 5 is not paired.
Multiply 180 by 5 to complete the pair: 180 × 5 = 900.
900 = 30², and it is divisible by 4, 9 and 10.
The required number is 900.
In simple words: Take the LCM 180 and multiply by the unpaired factor 5 to make it a perfect square: 900.

Teacher's Note:
1. Any number divisible by 4, 9 and 10 is a multiple of their LCM.
2. To turn a number into a perfect square, multiply by the prime factors that are left without a pair.

Question 16. Difference of two perfect cubes is 189. If the cube root of the smaller of the numbers is 3, find the cube root of the larger number.
Answer:
Smaller number = 3³ = 27.
Let the cube root of the larger number be x, so the larger number is x³.
x³ − 27 = 189, so x³ = 216.
216 = 6 × 6 × 6, so x = 6.
The cube root of the larger number is 6.
In simple words: The smaller cube is 27; add 189 to get 216, whose cube root is 6.

Teacher's Note:
1. Translate "difference of cubes is 189" into an equation: larger − smaller = 189.
2. Check: 6³ − 3³ = 216 − 27 = 189.

Section E: Case Study Based Question (4 Marks)

Question 17. The teacher tells the class that the lowest marks obtained by a student in his class is half the highest marks plus 5. The lowest score is 45. To just pass the examination, the borderline marks are 12 less than the lowest marks obtained by a student in the class.
(a) Find the highest marks. (2)
(b) Find the required passing marks. (1)
(c) Aditya, a student of another class, scored 9 marks more than double the lowest marks of this class. Find Aditya's marks. (1)
Answer:
(a) Let the highest marks be x. Lowest marks = \( \frac{1}{2} \)x + 5.
45 = \( \frac{1}{2} \)x + 5, so \( \frac{1}{2} \)x = 40 and x = 80.
Highest marks = 80.
(b) Passing marks = lowest marks − 12 = 45 − 12 = 33.
(c) Aditya's marks = 2 × 45 + 9 = 90 + 9 = 99.
In simple words: Half of the top score plus 5 is 45, so the top score is 80; pass mark is 33 and Aditya got 99.

Teacher's Note:
1. Turn each sentence into an equation before solving.
2. Check (a): half of 80 is 40, and 40 + 5 = 45.