CBSE Class 8 Mathematics Number Play Practice Paper Worksheet Set 01 with Answers

CBSE Class 8 Maths Ganita Prakash Chapter 5 Number Play Practice Paper 14 is a 40-mark test with 20 questions in Sections A to E, given here with step-by-step answers. It checks how well you can write numbers in algebraic form (9n, 2n + 1, 4k + 3), use the divisibility tests for 3, 9 and 11, find remainders and digital roots, and prove simple facts such as why the product of three consecutive numbers is divisible by 6.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Write the general form of a number divisible by 8 and give two examples.
Answer:
A number divisible by 8 is 8 times some whole number.
General form: 8n, where n is a whole number.
Examples: n = 3 gives 8 × 3 = 24; n = 5 gives 8 × 5 = 40.
In simple words: Every multiple of 8 looks like 8 × something, such as 24 and 40.

Teacher's Note:
1. Write the general form with a letter, and say what values the letter can take.
2. Any value of n gives a valid example.

Question 12. Prove that the sum of two multiples of 5 is also a multiple of 5.
Answer:
Let the two multiples of 5 be 5a and 5b, where a and b are whole numbers.
Sum = 5a + 5b = 5(a + b).
Since a + b is a whole number, the sum is 5 times a whole number.
Hence the sum is a multiple of 5. (Example: 15 + 35 = 50 = 5 × 10.)
In simple words: Take 5 common out of both numbers; what is left is still a whole number, so the sum is a multiple of 5.

Teacher's Note:
1. Use two different letters, since the two multiples need not be equal.
2. Taking out a common factor (distributive property) is the key step.

Question 13. Find the remainder when 735 is divided by 9.
Answer:
A number leaves the same remainder on division by 9 as its digit sum.
Digit sum of 735 = 7 + 3 + 5 = 15.
15 = 9 + 6, so 15 leaves remainder 6.
Hence 735 leaves remainder 6. (Check: 9 × 81 = 729 and 735 − 729 = 6.)
In simple words: Add the digits to get 15; 15 is 6 more than 9, so the remainder is 6.

Teacher's Note:
1. Keep reducing the digit sum until it is less than 9 (or use the digital root).
2. A quick multiplication check confirms the answer.

Question 14. Write two numbers that leave remainder 3 when divided by 4.
Answer:
Such numbers are 3 more than a multiple of 4, i.e. of the form 4k + 3.
k = 1: 4 × 1 + 3 = 7; k = 2: 4 × 2 + 3 = 11.
Check: 7 = 4 × 1 + 3 and 11 = 4 × 2 + 3, so both leave remainder 3.
In simple words: Add 3 to any multiple of 4, for example 4 + 3 = 7 and 8 + 3 = 11.

Teacher's Note:
1. k = 0 gives 3, which also leaves remainder 3 (3 = 4 × 0 + 3).
2. Always verify by writing the division.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find the smallest number that leaves remainder 1 when divided by 4 and remainder 3 when divided by 6.
Answer:
Numbers leaving remainder 3 when divided by 6: 3, 9, 15, 21, 27, …
Now test each for remainder 1 when divided by 4:
3 = 4 × 0 + 3 → remainder 3 (no); 9 = 4 × 2 + 1 → remainder 1 (yes).
Verification: 9 ÷ 4 = 2 remainder 1, and 9 ÷ 6 = 1 remainder 3.
Smallest such number = 9.
In simple words: List numbers that are 3 more than a multiple of 6, and the first one that is 1 more than a multiple of 4 is 9.

Teacher's Note:
1. Listing the numbers for the larger divisor first is quicker.
2. The next such numbers are 21, 33, … (adding LCM(4, 6) = 12 each time).

Question 16. Explain why divisibility by 9 implies divisibility by 3, but not vice versa.
Answer:
If a number is divisible by 9, it is 9n = 3 × (3n) for some whole number n.
So it is also a multiple of 3: divisible by 9 ⟹ divisible by 3.
The converse is false: a multiple of 3 need not have 9 as a factor.
Counter-example: 15 = 3 × 5 is divisible by 3, but 15 ÷ 9 leaves remainder 6. Similarly 12, 21, 24 …
Hence divisibility by 3 does not imply divisibility by 9.
In simple words: 9 is 3 × 3, so multiples of 9 are multiples of 3, but numbers like 15 are multiples of 3 and not of 9.

Teacher's Note:
1. One counter-example is enough to show that a statement is not always true.
2. Digit sums: 15 has digit sum 6, divisible by 3 but not by 9.

Question 17. Show that the product of three consecutive numbers is divisible by 6.
Answer:
Let the numbers be n, n + 1, n + 2.
Divisible by 2: among any two consecutive numbers one is even, so at least one of the three is even.
Divisible by 3: when divided by 3, three consecutive numbers leave remainders 0, 1 and 2 in some order, so exactly one of them is a multiple of 3.
So the product has a factor 2 and a factor 3. Since 2 and 3 are co-prime, it is divisible by 2 × 3 = 6.
Example: 4 × 5 × 6 = 120 = 6 × 20.
In simple words: In any three numbers in a row, one is even and one is a multiple of 3, so the product is a multiple of 6.

Teacher's Note:
1. Divisible by 2 and by 3 gives divisible by 6 because 2 and 3 are co-prime.
2. Similarly the product of four consecutive numbers is divisible by 24.

Section D: Long Answer Question (5 Marks)

Question 18. Explain the divisibility test for 9 using place-value expansion.
Answer:
Take a 4-digit number with digits a, b, c, d, i.e. the number abcd.
Place-value expansion: number = 1000a + 100b + 10c + d.
Write each power of 10 as a multiple of 9 plus 1: 1000 = 999 + 1, 100 = 99 + 1, 10 = 9 + 1.
So number = (999a + 99b + 9c) + (a + b + c + d).
= 9(111a + 11b + c) + (sum of digits).
The first part is always a multiple of 9. So the number and its digit sum differ by a multiple of 9, and they leave the same remainder on division by 9.
Hence: a number is divisible by 9 if and only if the sum of its digits is divisible by 9.
Example: 5832 → 5 + 8 + 3 + 2 = 18, divisible by 9, so 5832 is divisible by 9 (5832 = 9 × 648). But 5833 → 19, not divisible by 9.
In simple words: Every 10, 100, 1000 … is a multiple of 9 plus 1, so only the digits' sum decides whether the number is a multiple of 9.

Teacher's Note:
1. The same argument works for any number of digits.
2. Since 9 is a multiple of 3, the same proof also gives the test for 3.
3. Write the steps in order: expand, split, take 9 common, conclude.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: Digital Root Game
In a digital root game, the teacher wrote several large numbers on the board and asked students to find their digital roots. Ramesh noticed that whenever the digital root was 9, the number was divisible by 9. This observation helped students verify divisibility quickly without long division. The activity strengthened their understanding of place value and digit behaviour.
(a) Find the digital root of 99972.
(b) Is the number divisible by 9?
(c) Explain the reason behind this rule.
(d) Give another example of a number with digital root 9.
Answer:
(a) 9 + 9 + 9 + 7 + 2 = 36, and 3 + 6 = 9. Digital root = 9.
(b) Yes. The digital root is 9, so 99972 is divisible by 9 (99972 = 9 × 11108).
(c) Each place value is 1 more than a multiple of 9 (10 = 9 + 1, 100 = 99 + 1, …). So a number minus its digit sum is always a multiple of 9. Hence the number is divisible by 9 exactly when its digit sum (and so its digital root) is 9 or a multiple of 9.
(d) For example 45693: 4 + 5 + 6 + 9 + 3 = 27 and 2 + 7 = 9. (Any multiple of 9 works, e.g. 18, 729.)
In simple words: Digits of 99972 add to 36 and then to 9, so it is a multiple of 9; this works because 10, 100, … are all one more than a multiple of 9.

Teacher's Note:
1. A digital root of 9 means remainder 0 on division by 9.
2. To make your own example, choose digits that add up to 9, 18, 27, …

Question 20. Case Study 2: Classroom Challenge
During a classroom challenge, students were asked to form numbers that leave specific remainders when divided by given numbers. Sita formed numbers that left remainder 3 when divided by 4 and checked her answers by division. She learned how algebraic expressions like 4k + 3 help represent such numbers and allow easy verification.
(a) Write the algebraic form of numbers leaving remainder 3 when divided by 4.
(b) Write two such numbers.
(c) Verify one of them by division.
(d) Write a similar form for numbers leaving remainder 2 when divided by 5.
Answer:
(a) 4k + 3, where k is a whole number.
(b) k = 1 gives 7; k = 2 gives 11.
(c) 7 ÷ 4: 4 × 1 = 4 and 7 − 4 = 3, so quotient 1, remainder 3. ✓
(d) 5k + 2, e.g. 7, 12, 17, …
In simple words: Remainder 3 on division by 4 means 4k + 3 (like 7 and 11), and remainder 2 on division by 5 means 5k + 2.

Teacher's Note:
1. General form: (divisor) × k + (remainder), with remainder less than the divisor.
2. Notice that 7 appears in both lists: 7 = 4 × 1 + 3 = 5 × 1 + 2.