CLASS 8 MATHEMATICS

Interactive CBSE Worksheet

Algebra • Distributive Property • Number Patterns • Divisibility • Cryptarithms

Click Show Answer after each question to reveal the answer.

Section A — Distributive Property & Algebraic Products

1. Expand and simplify:

(i) (3 + x)(y − 2)

(ii) (5 − a)(a + 4)

(iii) (2p + 3)(p − 5)

(iv) (4m − n)(m + 2n)

(v) (7x + 2y)(x − y)

Show Answer
(i) 3y − 6 + xy − 2x
(ii) 5a + 20 − a² − 4a = −a² + a + 20
(iii) 2p² − 10p + 3p − 15 = 2p² − 7p − 15
(iv) 4m² + 8mn − mn − 2n² = 4m² + 7mn − 2n²
(v) 7x² − 7xy + 2xy − 2y² = 7x² − 5xy − 2y²
2. Expand and simplify:

(i) (a + ab − 2b²)(3 + b)

(ii) (3x + 5)(x + 2y − 1)

(iii) (2p − q)(p + 3q)

Show Answer
(i) 3a + 3ab − 6b² + ab + ab² − 2b³ = 3a + 4ab + ab² − 6b² − 2b³
(ii) 3x² + 6xy − 3x + 5x + 10y − 5 = 3x² + 6xy + 2x + 10y − 5
(iii) 2p² + 6pq − pq − 3q² = 2p² + 5pq − 3q²
3. Use the distributive property to calculate mentally:

(i) 387 × 11    (ii) 495 × 11    (iii) 728 × 101    (iv) 2635 × 101

Show Answer
(i) 4257    (ii) 5445    (iii) 73528    (iv) 266135
4. Write a one-line general rule for multiplying a number N by 11 using the distributive property.
Show Answer
N × 11 = N(10 + 1) = 10N + N.

Section B — Special Algebraic Patterns

5. Expand and identify the pattern:

(i) (a − b)(a + b)

(ii) (a − b)(a² + ab + b²)

(iii) (a − b)(a³ + a²b + ab² + b³)

Show Answer
(i) a² − b²
(ii) a³ − b³
(iii) a⁴ − b⁴
Pattern: (a − b)(an−1 + an−2b + … + bn−1) = an − bn.
6. Without fully multiplying, find:

(i) 49 × 51    (ii) 103 × 97    (iii) 998 × 1002

Show Answer
(i) (50 − 1)(50 + 1) = 2499
(ii) (100 + 3)(100 − 3) = 9991
(iii) (1000 − 2)(1000 + 2) = 999996
7. Complete the pattern and verify by expansion.
Show Answer
(a − b)(a + b) = a² − b²
(a − b)(a² + ab + b²) = a³ − b³
(a − b)(a³ + a²b + ab² + b³) = a⁴ − b⁴
Next: (a − b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ − b⁵.

Section C — Algebraic Expressions & Parity

8. Decide whether each expression is ALWAYS even, ALWAYS odd, or SOMETIMES even/odd.

(i) 4m + 2n    (ii) 3p + 5q    (iii) x² + 2    (iv) 6a − 3b    (v) 4x × 3y

Show Answer
(i) Always even.
(ii) Sometimes even/odd.
(iii) Sometimes even/odd.
(iv) Sometimes even/odd.
(v) Always even.
9. Give one example and one non-example for any two expressions in Question 8 that are not always even.
Show Answer
Example for (ii): p = 1, q = 1 → 8 (even). Non-example: p = 1, q = 2 → 13 (odd).
Example for (iii): x = 2 → 6 (even). Non-example: x = 3 → 11 (odd).

Section D — Number Patterns & Consecutive Numbers

10. The sum of four consecutive integers is 86. Find the integers.
Show Answer
Let the integers be n, n+1, n+2, n+3. Then 4n + 6 = 86, so n = 20. The integers are 20, 21, 22, 23.
11. Let p be the greatest of five consecutive even numbers. Express the other four numbers in terms of p.
Show Answer
p − 8, p − 6, p − 4, p − 2.
12. Find the missing terms:

(i) 4, 9, 14, 19, □, □

(ii) 50, 44, 38, 32, □, □

(iii) 3, 6, 12, 24, □, □

(iv) 2, 6, 12, 20, 30, □

Show Answer
(i) 24, 29
(ii) 26, 20
(iii) 48, 96
(iv) 42
13. The nth term of a sequence is 4n − 1. Write the first six terms.
Show Answer
3, 7, 11, 15, 19, 23.
14. A sequence begins 7, 12, 17, 22, … . Write an algebraic rule for its nth term.
Show Answer
an = 5n + 2.

Section E — Divisibility & Remainders

15. Without dividing, determine whether each number is divisible by 3 and by 9:

12357, 40509, 8888, 93546, 358095

Show Answer
12357: 3 Yes, 9 No
40509: 3 Yes, 9 Yes
8888: 3 No, 9 No
93546: 3 Yes, 9 Yes
358095: 3 Yes, 9 Yes
16. A number leaves remainder 2 when divided by 3 and remainder 2 when divided by 4. Write the first five positive numbers with this property.
Show Answer
2, 14, 26, 38, 50.
17. When divided by 7, 661 leaves remainder 3 and 4779 leaves remainder 5. Find the remainder when:

(i) 4779 + 661 is divided by 7
(ii) 4779 − 661 is divided by 7

Show Answer
(i) 5 + 3 = 8 ≡ 1 (mod 7), so remainder 1.
(ii) 5 − 3 = 2, so remainder 2.
18. If a number is divisible by 18, what divisibility conditions can you use to check it quickly?
Show Answer
Since 18 = 2 × 9, the number must be divisible by both 2 and 9. Thus its last digit must be even and the sum of its digits must be divisible by 9.
19. If 48a23b is divisible by 18, find all possible pairs (a, b).
Show Answer
For divisibility by 2, b is even. Digit sum = 4+8+a+2+3+b = 17+a+b, which must be divisible by 9. Hence a+b ≡ 1 (mod 9). With a,b digits and b even, possible pairs are: (1,0), (8,2), (6,4), (4,6), (2,8), (9,0), (7,2), (5,4), (3,6), (1,8), (9,2), (7,4), (5,6), (3,8).

Section F — Cryptarithms & Reasoning

20. Solve the cryptarithms. Different letters represent different digits.

(i) AB × 5 = BC
(ii) UT × 3 = PUT
(iii) GH × H = 9K
(iv) BYE × 6 = RAY

Show Answer
(i) AB = 18 gives 18 × 5 = 90, so B=8 and C=0, A=1. Thus 18 × 5 = 90.
(ii) UT = 42 gives 42 × 3 = 126, so U=4, T=2, P=1. Thus 42 × 3 = 126.
(iii) One valid solution is 23 × 3 = 69, so G=2, H=3, K=9.
(iv) One valid solution is 158 × 6 = 948, so B=1, Y=5, E=8, R=9, A=4.
21. A two-digit number multiplied by 8 gives another two-digit number in the 90s. List all possible original numbers.
Show Answer
12 × 8 = 96, so the only possible original number is 12.

Section G — HOTS / Case-Based Questions

22. A student claims: “The sum of any three multiples of 3 is always a multiple of 9.”

(a) Test the claim with two examples. (b) Give a counterexample if false. (c) State the correct general conclusion.

Show Answer
False. Example: 3 + 6 + 9 = 18, which is divisible by 9. But 3 + 3 + 6 = 12, which is not divisible by 9. Correct conclusion: the sum of any three multiples of 3 is always a multiple of 3, but not necessarily a multiple of 9.
23. A number is 4 more than a multiple of 12. Another number is 8 less than the same multiple of 12.

(a) Represent both numbers algebraically. (b) Find their sum. (c) Is the sum always divisible by 6?

Show Answer
Let the multiple be 12n.
(a) Numbers: 12n + 4 and 12n − 8.
(b) Sum = 24n − 4 = 4(6n − 1).
(c) No. For n=1, sum = 20, which is not divisible by 6.
24. A six-digit number is divisible by 15. When its digits are reversed, the new number is divisible by 6. Construct one such number and explain why it works.
Show Answer
One example is 120030. It is divisible by 15 because it is divisible by 3 (digit sum = 6) and by 5 (last digit 0). Its reverse is 030021 = 30021, which is divisible by 3 but not by 2, so this example does NOT satisfy the second condition.
A valid example is 150060: it is divisible by 15, and its reverse is 060051 = 60051, which is not even. Therefore, constructing such a number requires the original number to end in 0 or 5 for divisibility by 15, while the reversed number must end in an even digit. Hence it must begin with an even digit and end in 0. One valid example is 210030; reverse = 030012 = 30012, divisible by 6.
25. ⭐ Challenge: The product of two consecutive integers is always even. What can you say about the product of three consecutive integers? What about four consecutive integers? Explain using algebra.
Show Answer
For two consecutive integers, one is even, so the product is always even.
For three consecutive integers n(n+1)(n+2), at least one factor is even, so the product is always even; in fact it is always divisible by 6 because among three consecutive integers one is divisible by 3.
For four consecutive integers n(n+1)(n+2)(n+3), there are at least two even factors, and one is divisible by 4. Hence the product is always divisible by 24.