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

Practice Paper 15 for CBSE Class 8 Maths Ganita Prakash Chapter 5 Number Play has 20 questions worth 40 marks, arranged in Sections A to E, and every question comes with a worked solution. This paper focuses on even and odd numbers in algebraic form, the remainder form 7k + 6 and 5k + 1, the divisibility tests for 4, 6, 9 and 11 explained through place value, and using prime factors and the LCM to see why a number divisible by 4 and 6 need not be divisible by 24.

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 9 and give two examples.
Answer:
General form: 9n, where n is a whole number.
Examples: n = 3 gives 27; n = 7 gives 63.
Check: 2 + 7 = 9 and 6 + 3 = 9, so both have digit sum 9.
In simple words: A multiple of 9 is 9 × something, like 27 or 63.

Teacher's Note:
1. Mention what the letter stands for (a whole number).
2. You can confirm examples with the digit-sum test.

Question 12. Show that the sum of two odd numbers is always even.
Answer:
Any odd number is 1 more than an even number, so let the two odd numbers be 2m + 1 and 2n + 1 (m, n whole numbers).
Sum = (2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1).
This is 2 times the whole number m + n + 1, so the sum is always even.
Example: 7 + 9 = 16 = 2 × 8.
In simple words: Each odd number has one extra 1 left over; the two extra 1s make a pair, so the total is even.

Teacher's Note:
1. Use two different letters so that the two odd numbers can be different.
2. Examples illustrate, but only the algebra proves it for all cases.

Question 13. Find the remainder when 456 is divided by 9.
Answer:
Digit sum of 456 = 4 + 5 + 6 = 15.
15 ÷ 9 = 1 remainder 6.
So 456 leaves remainder 6. (Check: 9 × 50 = 450, 456 − 450 = 6.)
In simple words: The digits add up to 15, which is 6 more than 9, so the remainder is 6.

Teacher's Note:
1. A number and its digit sum leave the same remainder on division by 9.
2. Digital root 6 gives remainder 6.

Question 14. Write two numbers that leave remainder 1 when divided by 5.
Answer:
Such numbers have the form 5k + 1.
k = 1: 6; k = 2: 11.
Check: 6 = 5 × 1 + 1 and 11 = 5 × 2 + 1.
In simple words: Add 1 to a multiple of 5, for example 6 and 11.

Teacher's Note:
1. All such numbers end in 1 or 6.
2. 1 itself also works (1 = 5 × 0 + 1).

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find the smallest number that leaves remainder 2 when divided by 3 and remainder 3 when divided by 5.
Answer:
Numbers leaving remainder 3 when divided by 5 (form 5k + 3): 3, 8, 13, 18, 23, …
Test each for remainder 2 when divided by 3:
3 → remainder 0 (no); 8 = 3 × 2 + 2 → remainder 2 (yes).
Verification: 8 ÷ 3 = 2 remainder 2, and 8 ÷ 5 = 1 remainder 3.
Smallest number = 8.
In simple words: Of the numbers 3, 8, 13, … (3 more than multiples of 5), the first that is 2 more than a multiple of 3 is 8.

Teacher's Note:
1. The next answers are 23, 38, … (add LCM(3, 5) = 15).
2. Always check both conditions at the end.

Question 16. Explain why divisibility by 6 requires divisibility by both 2 and 3.
Answer:
6 = 2 × 3, so a multiple of 6 is 6n = 2 × (3n) = 3 × (2n).
So every multiple of 6 is a multiple of 2 and a multiple of 3: both conditions are necessary.
One condition alone is not enough: 8 is divisible by 2 but not by 6; 9 is divisible by 3 but not by 6.
When both hold, the number has the factors 2 and 3, and since 2 and 3 are co-prime it is divisible by 2 × 3 = 6. So both together are sufficient.
Example: 642 is even and its digit sum 12 is divisible by 3, so 642 is divisible by 6 (642 = 6 × 107).
In simple words: 6 is 2 × 3, so a number must be even and have a digit sum divisible by 3 to be a multiple of 6.

Teacher's Note:
1. The two tests are combined because 2 and 3 are co-prime.
2. Do not wrongly extend this: divisible by 2 and 4 does not mean divisible by 8.

Question 17. Prove that the difference of two multiples of 7 is also a multiple of 7.
Answer:
Let the two multiples be 7a and 7b, with a ≥ b (whole numbers).
Difference = 7a − 7b = 7(a − b).
a − b is a whole number, so the difference is 7 times a whole number.
Hence the difference is a multiple of 7. Example: 63 − 28 = 35 = 7 × 5.
In simple words: Take 7 common from both numbers; what remains is a whole number, so the difference is a multiple of 7.

Teacher's Note:
1. This is the distributive property used in reverse (taking out a common factor).
2. The same proof works for any divisor, not just 7.

Section D: Long Answer Question (5 Marks)

Question 18. Explain the divisibility test for 11 using place value with an example.
Answer:
Write the place values as multiples of 11 plus or minus 1:
10 = 11 − 1, 100 = 99 + 1, 1000 = 1001 − 1, 10000 = 9999 + 1 (and 99, 1001, 9999 are multiples of 11).
For a 5-digit number with digits a, b, c, d, e (from left to right):
number = 10000a + 1000b + 100c + 10d + e
= (9999a + 1001b + 99c + 11d) + (a − b + c − d + e).
The first bracket is a multiple of 11. So the number is divisible by 11 exactly when (e + c + a) − (d + b) is divisible by 11.
Test: subtract the sum of the digits in the even places from the sum of the digits in the odd places (counting from the right). If the result is 0 or a multiple of 11, the number is divisible by 11.
Example: 52844. Odd places: 4 + 8 + 5 = 17. Even places: 4 + 2 = 6. Difference = 17 − 6 = 11, a multiple of 11.
So 52844 is divisible by 11 (52844 = 11 × 4804).
In simple words: Because the place values are alternately just below and just above multiples of 11, the alternating digit sum decides divisibility by 11.

Teacher's Note:
1. Powers of 10 alternate between remainders 1 and 10 (that is, −1) when divided by 11.
2. The difference can be 0, 11, 22, … (or negative multiples); all mean divisible.
3. Always state the rule clearly and show it on an example.

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

Question 19. Case Study 1: Divisibility Booths
During a school math fair, students created “Divisibility Booths”. One booth focused on testing divisibility by 3. Meera demonstrated that instead of dividing large numbers, adding their digits was enough to check divisibility. She showed how place values differ from their digits by multiples of 3, making digit sum a reliable method. This booth attracted many students as they could test large numbers quickly and accurately.
(a) Find the digit sum of 73956.
(b) Determine whether the number is divisible by 3.
(c) Explain why digit sum works for divisibility by 3.
(d) Give another example to support your answer.
Answer:
(a) 7 + 3 + 9 + 5 + 6 = 30.
(b) 30 = 3 × 10 is divisible by 3, so 73956 is divisible by 3 (73956 = 3 × 24652).
(c) 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, … So each digit's place value differs from the digit itself by a multiple of 9, and hence of 3. Therefore number − digit sum is a multiple of 3, and the number is divisible by 3 exactly when its digit sum is.
(d) 4521: 4 + 5 + 2 + 1 = 12, divisible by 3, so 4521 is divisible by 3 (4521 = 3 × 1507). On the other hand 4522 has digit sum 13, so it is not.
In simple words: Add the digits of 73956 to get 30; since 30 is a multiple of 3, so is 73956.

Teacher's Note:
1. Digit sum test works for 3 and 9 because 10 leaves remainder 1 when divided by either.
2. A good example shows both a number that passes and one that fails.

Question 20. Case Study 2: Classroom Challenge
A teacher asked students to identify numbers divisible by both 4 and 6. Initially, some students assumed that such numbers would always be divisible by 24. However, through examples and prime factor discussion, the teacher showed that this is not always true. Students learned the importance of least common multiples and prime factorisation in understanding divisibility relationships.
(a) Write the prime factors of 4 and 6.
(b) Find their least common multiple.
(c) Check whether 12 is divisible by 4 and 6.
(d) Explain why 12 is not divisible by 24.
Answer:
(a) 4 = 2 × 2; 6 = 2 × 3.
(b) Take each prime the greatest number of times it appears: 2 × 2 × 3 = 12. LCM(4, 6) = 12.
(c) 12 ÷ 4 = 3 and 12 ÷ 6 = 2, both with remainder 0. So 12 is divisible by both 4 and 6.
(d) 24 = 2 × 2 × 2 × 3 needs three factors of 2, but 12 = 2 × 2 × 3 has only two. Also 12 is smaller than 24. So 12 is not divisible by 24; numbers divisible by 4 and 6 need only be multiples of 12.
In simple words: 4 and 6 share a factor 2, so their LCM is 12, not 24; 12 works for both but is not a multiple of 24.

Teacher's Note:
1. Product of two numbers = LCM only when they are co-prime.
2. Here 4 × 6 = 24 = 2 × LCM because HCF(4, 6) = 2.