CBSE Class 8 Mathematics Chapters 1 to 4 Revision Practice Paper 01 with Answers (Old NCERT Syllabus)

This Class 8 Maths Chapters 1 to 4 Revision Practice Paper 01 is a 40-mark paper of 17 questions in Sections A to E, with a worked step-by-step answer for each question. It follows the earlier (old) NCERT Class 8 Maths syllabus, so state-board students and anyone revising the first four chapters can use it. It mixes rational number operations and the distributive property, linear equations in one variable, exterior angles and parallelogram angle problems, simple probability, and drawing and reading pie charts.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Simplify: \( -\frac{8}{25} \) × \( \frac{15}{7} \) × \( \frac{15}{32} \) × \( \left( -\frac{14}{9} \right) \)
Answer:
Two of the factors are negative, so the product is positive.
Cancel common factors between numerators and denominators: 8 and 32 (by 8), 14 and 7 (by 7), 15 and 25 (by 5), 15 and 9 (by 3).
What is left: numerators 1, 3, 5, 2 and denominators 5, 1, 4, 3. Product = (1 × 3 × 5 × 2) ÷ (5 × 1 × 4 × 3) = 30 ÷ 60.
\( \frac{30}{60} \) = \( \frac{1}{2} \).
So the value is \( \frac{1}{2} \).
In simple words: The two minus signs cancel, and after cancelling common factors the product reduces to one-half.

Teacher's Note:
1. Cancel across any numerator and any denominator in a product.
2. Check by direct multiplication: (8 × 15 × 15 × 14) ÷ (25 × 7 × 32 × 9) = 25200 ÷ 50400 = \( \frac{1}{2} \).

Question 8. In the below figure. Find x.

Answer:
The marked angles 50°, 80°, x and 120° are the exterior angles of the quadrilateral, one at each vertex (each side is extended in the same turning direction).
The exterior angles of any polygon add up to 360°.
x + 50° + 80° + 120° = 360°, so x + 250° = 360°.
x = 360° − 250°, so x = 110°.
In simple words: Going once round the shape you turn through 360° in total, so x = 360° − 250° = 110°.

Teacher's Note:
1. Check that each marked angle is between one side and the extension of the next side; only then is the 360° rule valid.
2. The interior angles are 130°, 100°, 70° and 60°, which also add to 360°.
3. The figure is drawn to the stated angles, so the two upright sides are not parallel as they appear in the paper's sketch.

Question 9. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of
(i) getting a number 6?
(ii) getting a number less than 6?
Answer:
Total number of equally likely outcomes = 10.
(i) Only one slip has the number 6, so P(6) = \( \frac{1}{10} \).
(ii) Numbers less than 6 are 1, 2, 3, 4, 5: 5 outcomes.
P(less than 6) = \( \frac{5}{10} \) = \( \frac{1}{2} \).
In simple words: One slip out of ten shows 6, and five slips out of ten show a number below 6.

Teacher's Note:
1. Probability = favourable outcomes ÷ total outcomes, when all outcomes are equally likely.
2. 'Less than 6' does not include 6 itself.

Section C: Short Answer Questions (3 Marks Each)

Question 10. In the below figure both RISK and CLUE are parallelograms. Find the value of x.

Answer:
Let the dotted lines IS and CE cross at X; x is the angle at X, equal to its vertically opposite angle ∠IXC.
In parallelogram RISK, ∠RIS = ∠K = 120° (opposite angles). R, I and C lie on one line, so ∠SIC = 180° − 120° = 60°.
In parallelogram CLUE, adjacent angles are supplementary: ∠ECL = 180° − ∠L = 180° − 70° = 110°. So ∠ECI = 180° − 110° = 70°.
In triangle XIC: x + 60° + 70° = 180°.
x = 180° − 130°, so x = 50°.
In simple words: The parallelograms give angles of 60° at I and 70° at C, so the third angle of the small triangle is 50°.

Teacher's Note:
1. Find the two base angles of the small triangle first; the angle sum then gives x.
2. ∠ECI = ∠L can also be seen as corresponding angles, because EC ∥ UL.

Question 11. Find the value of \( -\frac{3}{4} \) × \( \left\{ \frac{2}{3} + \left( -\frac{5}{6} \right) \right\} \) using distributive property
Answer:
Distributive property: a × (b + c) = a × b + a × c.
= \( \left( -\frac{3}{4} \times \frac{2}{3} \right) + \left( -\frac{3}{4} \times -\frac{5}{6} \right) \)
= \( -\frac{1}{2} \) + \( \frac{5}{8} \)
= \( -\frac{4}{8} \) + \( \frac{5}{8} \) = \( \frac{1}{8} \).
So the value is \( \frac{1}{8} \).
In simple words: Multiply −3/4 by each term inside the bracket separately and add: −1/2 + 5/8 = 1/8.

Teacher's Note:
1. A negative times a negative is positive, so the second product is +\( \frac{5}{8} \).
2. Check without the property: \( \frac{2}{3} \) − \( \frac{5}{6} \) = \( -\frac{1}{6} \), and \( -\frac{3}{4} \) × \( -\frac{1}{6} \) = \( \frac{1}{8} \).

Question 12. Find \( \frac{3}{7} \) + \( \left( -\frac{6}{11} \right) \) + \( \left( -\frac{8}{21} \right) \) + \( \frac{5}{22} \)
Answer:
LCM of 7, 11, 21 and 22 = 2 × 3 × 7 × 11 = 462.
\( \frac{3}{7} \) = \( \frac{198}{462} \), \( -\frac{6}{11} \) = \( -\frac{252}{462} \), \( -\frac{8}{21} \) = \( -\frac{176}{462} \), \( \frac{5}{22} \) = \( \frac{105}{462} \).
Sum = \( \frac{198 - 252 - 176 + 105}{462} \) = \( \frac{303 - 428}{462} \).
= \( -\frac{125}{462} \).
In simple words: Change all four fractions to the common denominator 462, then add the numerators: 198 − 252 − 176 + 105 = −125.

Teacher's Note:
1. Grouping the positive numerators (198 + 105 = 303) and the negative ones (252 + 176 = 428) reduces mistakes.
2. 125 = 5 × 5 × 5 shares no factor with 462, so the answer is already in lowest terms.

Question 13. Solve: \( \frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21 \)
Answer:
LCM of 2, 4 and 6 is 12. Multiply every term by 12: 6n − 9n + 10n = 252.
7n = 252.
n = 252 ÷ 7 = 36.
So n = 36.
Check: 18 − 27 + 30 = 21.
In simple words: Clearing the fractions with 12 gives 7n = 252, so n = 36.

Teacher's Note:
1. Multiply both sides by the LCM of the denominators to remove all fractions at once.
2. Remember to multiply the right-hand side (21) by 12 as well.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Consider the following parallelograms. Find the values of the unknowns x, y, z.

Answer:
(i) The right side is extended upward. z lies between this extension and the top side, and the 80° angle lies between the same line and the bottom side. Top ∥ bottom, so these are corresponding angles: z = 80°.
y is the angle opposite the 80° angle, so y = 80°.
x and the 80° angle are adjacent angles, so x = 180° − 80° = 100°.
(ii) y is opposite the 112° angle, so y = 112°.
The angle at the bottom-left vertex is adjacent to 112°, so 40° + z = 180° − 112° = 68°, which gives z = 28°.
The top and bottom sides are parallel and the dotted diagonal is a transversal, so x and z are alternate angles: x = 28°.
So (i) x = 100°, y = 80°, z = 80°; (ii) x = 28°, y = 112°, z = 28°.
In simple words: Use opposite angles equal, adjacent angles adding to 180°, and the parallel-line angle facts.

Teacher's Note:
1. In (ii) the diagonal splits the 68° angle into 40° and z.
2. Check (ii): in the triangle below the diagonal, 28° + 112° + 40° = 180°, where 40° is the angle between the diagonal and the right side.

Question 15. Simplify and solve the following linear equations:
15(y − 4) − 2(y − 9) + 5(y + 6) = 0
Answer:
Expand the brackets: 15y − 60 − 2y + 18 + 5y + 30 = 0.
Collect like terms: (15 − 2 + 5)y + (−60 + 18 + 30) = 0, so 18y − 12 = 0.
18y = 12, so y = \( \frac{12}{18} \) = \( \frac{2}{3} \).
So y = \( \frac{2}{3} \).
Check: 15(\( \frac{2}{3} \) − 4) − 2(\( \frac{2}{3} \) − 9) + 5(\( \frac{2}{3} \) + 6) = −50 + \( \frac{50}{3} \) + \( \frac{100}{3} \) = −50 + 50 = 0.
In simple words: Open the brackets, combine the y terms and the numbers, and you get 18y = 12, so y = 2/3.

Teacher's Note:
1. Watch the sign in −2(y − 9): it becomes −2y + 18.
2. A fractional answer is fine; always reduce it to lowest terms.

Teacher's Note:
1. Always check that the central angles add up to exactly 360° before drawing.
2. Ordinary Bread brings in nearly half the sales, so its sector is the largest.

Section E: Case Study Based Question (4 Marks)

Question 17. Family income is the income from all sources like salary of family members, rents, and interest received from banks and savings from using skill of family members.
Whatever money we spend from the income for buying various things to fulfil our needs is called expenditure. Income brings money into a family while expenditure takes money out so that it is not available for anything else. Saving is the money set aside by the family for use in times of need.
Below pie chart gives the expenditure (in percentage) on various items and savings of a family during a month.
(a) If the monthly savings of the family is Rs 3000, what is the total expenditure?
(b) If the monthly savings of the family is Rs 3000, what is the monthly expenditure on clothes?
(c) If the monthly savings of the family is Rs 3000, what is the monthly expenditure on Savings?
(d) If the monthly savings of the family is Rs 3000, what is the monthly expenditure on House rent?
Answer:
The whole pie (100%) is the family's monthly income. Savings are 15% of it.
15% of income = Rs 3000, so income = 3000 × \( \frac{100}{15} \) = Rs 20,000.
(a) Expenditure is everything except savings: 100% − 15% = 85%. Total expenditure = \( \frac{85}{100} \) × 20,000 = Rs 17,000 (out of a total monthly income of Rs 20,000).
(b) Clothes = 10% of 20,000 = \( \frac{10}{100} \) × 20,000 = Rs 2000.
(c) Amount set aside as savings = 15% of 20,000 = Rs 3000 (as given).
(d) House rent = 10% of 20,000 = Rs 2000.
In simple words: Rs 3000 is 15% of the income, so the income is Rs 20,000; each item is its percentage of Rs 20,000.

Teacher's Note:
1. First find the whole (100%) from the part you know, then find the other parts.
2. Savings are not spending: the total spent is 85% of the income, while the total income is Rs 20,000.
3. Check: the percentages 15 + 25 + 10 + 15 + 20 + 5 + 10 add up to 100.