CBSE Class 8 Mathematics Number Play Practice Paper Worksheet Set 02 with Answers

This is CBSE Class 8 Mathematics Ganita Prakash Chapter 5 Number Play Practice Paper 16: 20 questions, 40 marks, five sections (A to E) and a clear step-by-step answer for each question. It tests odd numbers written as 4n + 1, remainder forms like 9k + 4 and 8k + 5, the divisibility tests for 4, 6 and 8 (last two and last three digits), digital roots, the sum of three consecutive integers, and why changing plus signs to minus never changes the parity of a result.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Write the algebraic form of a number leaving remainder 5 when divided by 8.
Answer:
Such a number is 5 more than a multiple of 8.
Algebraic form: 8k + 5, where k is a whole number.
Examples: k = 0 → 5, k = 1 → 13, k = 2 → 21. Check: 13 = 8 × 1 + 5.
In simple words: Take a multiple of 8 and add 5: 8k + 5.

Teacher's Note:
1. Form: divisor × k + remainder, with the remainder less than the divisor.
2. Giving one or two examples makes the answer complete.

Question 12. State whether the number 7356 is divisible by 4 and justify.
Answer:
Test for 4: a number is divisible by 4 if the number formed by its last two digits is divisible by 4 (because 100 is a multiple of 4).
Last two digits of 7356 = 56 = 4 × 14.
So yes, 7356 is divisible by 4 (7356 = 4 × 1839).
In simple words: 56 is a multiple of 4, so 7356 is also a multiple of 4.

Teacher's Note:
1. 7356 = 7300 + 56, and 7300 is a multiple of 100, hence of 4.
2. Justify by stating the test and applying it.

Question 13. Find the remainder when 946 is divided by 9.
Answer:
Digit sum = 9 + 4 + 6 = 19; then 1 + 9 = 10; then 1 + 0 = 1.
So 946 leaves remainder 1 when divided by 9.
Check: 9 × 105 = 945, and 946 − 945 = 1.
In simple words: The digits reduce to 1, so 946 is 1 more than a multiple of 9.

Teacher's Note:
1. The digital root (when it is not 9) is the remainder on division by 9.
2. You can ignore the digit 9 straight away: 4 + 6 = 10 → 1.

Question 14. Write two consecutive multiples of 7.
Answer:
Consecutive multiples of 7 are 7 × k and 7 × (k + 1); they differ by 7.
Taking k = 3: 7 × 3 = 21 and 7 × 4 = 28.
(Two consecutive numbers such as 21 and 22 cannot both be multiples of 7, since they differ by only 1.)
In simple words: Multiples of 7 next to each other in the 7 times table, such as 21 and 28.

Teacher's Note:
1. 'Consecutive multiples' means multiples next to each other in the times table, not numbers 1 apart.
2. Consecutive multiples of any number m differ by m.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find the smallest number that leaves remainder 2 when divided by 3 and remainder 4 when divided by 5.
Answer:
Numbers leaving remainder 4 when divided by 5 (form 5k + 4): 4, 9, 14, 19, …
Test for remainder 2 when divided by 3:
4 → remainder 1; 9 → remainder 0; 14 = 3 × 4 + 2 → remainder 2 (yes).
Verification: 14 ÷ 3 = 4 remainder 2, and 14 ÷ 5 = 2 remainder 4.
Smallest number = 14.
In simple words: Of the numbers 4, 9, 14, … (4 more than multiples of 5), the first that is 2 more than a multiple of 3 is 14.

Teacher's Note:
1. Next answers: 29, 44, … (adding LCM(3, 5) = 15).
2. Notice 14 is 1 less than 15 in both cases: 14 = 15 − 1, and remainders 2 and 4 are each 1 less than the divisor.

Question 16. Explain why every multiple of 6 is also a multiple of 3.
Answer:
A multiple of 6 can be written as 6n, where n is a whole number.
Since 6 = 3 × 2, we get 6n = 3 × (2n).
2n is a whole number, so 6n is 3 times a whole number, i.e. a multiple of 3.
Example: 42 = 6 × 7 = 3 × 14. (The reverse is not true: 9 is a multiple of 3 but not of 6.)
In simple words: 6 is 3 × 2, so every 6 in a multiple of 6 is two 3s; the total is still a multiple of 3.

Teacher's Note:
1. If m is a factor of a number, m is a factor of all its multiples.
2. Give an algebraic reason, then an example.

Question 17. Prove that the sum of three consecutive integers is always divisible by 3.
Answer:
Let the three consecutive integers be n − 1, n, n + 1.
Sum = (n − 1) + n + (n + 1) = 3n.
3n is a multiple of 3, so the sum is always divisible by 3.
Example: 14 + 15 + 16 = 45 = 3 × 15 (and the sum is 3 times the middle number).
In simple words: The −1 and +1 cancel, leaving three times the middle number.

Teacher's Note:
1. Choosing the middle number as n makes the algebra shortest.
2. With n, n + 1, n + 2 the sum is 3n + 3 = 3(n + 1), the same result.

Section D: Long Answer Question (5 Marks)

Question 18. Explain the divisibility test for 8 with a suitable example.
Answer:
Rule: A number is divisible by 8 if the number formed by its last three digits (hundreds, tens and units) is divisible by 8.
Reason: 1000 = 8 × 125 is a multiple of 8, so 10000, 100000, … are also multiples of 8.
So any number = 1000 × (number formed by the remaining digits) + (number formed by the last three digits).
The first part is always a multiple of 8. Hence the whole number is divisible by 8 exactly when the last three digits form a multiple of 8.
Example: 1456 = 1000 × 1 + 456. Now 456 ÷ 8 = 57 with remainder 0.
So 1456 is divisible by 8 (1456 = 8 × 182).
Non-example: 3812: 812 ÷ 8 = 101 remainder 4, so 3812 is not divisible by 8.
In simple words: Thousands are always multiples of 8, so you only need to check whether the last three digits make a multiple of 8.

Teacher's Note:
1. Compare: 10 → test for 2 and 5, 100 → test for 4, 1000 → test for 8.
2. Showing both an example and a non-example earns full marks.
3. Last two digits are not enough: 116 ends in 16 (a multiple of 8) but 116 is not divisible by 8.

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

Question 19. Case Study 1: Algebraic Form and Remainders
During a class discussion, students explored numbers formed using algebraic expressions. Aarav wrote a number in the form 4k + 3 and wondered why such numbers always leave the same remainder when divided by 4. His teacher explained that algebraic forms help represent infinite sets of numbers with the same remainder property. By substituting different whole numbers for k, Aarav generated several numbers and verified the division results. This activity helped students understand how algebra connects with real number behaviour.
(a) What remainder does a number of the form 4k + 3 leave when divided by 4?
(b) Write two numbers of this form.
(c) Check your answer by actual division.
(d) Write the general form of numbers leaving remainder 1 when divided by 5.
Answer:
(a) 4k is a multiple of 4, and 3 < 4, so 4k + 3 leaves remainder 3.
(b) k = 1: 4 × 1 + 3 = 7; k = 2: 4 × 2 + 3 = 11.
(c) 7 ÷ 4 = 1 remainder 3 (4 × 1 = 4, 7 − 4 = 3); 11 ÷ 4 = 2 remainder 3 (4 × 2 = 8, 11 − 8 = 3). ✓
(d) 5k + 1 (for example 1, 6, 11, 16, …).
In simple words: 4k + 3 is always 3 more than a multiple of 4, like 7 and 11; numbers 1 more than a multiple of 5 are 5k + 1.

Teacher's Note:
1. One letter stands for infinitely many numbers with the same remainder.
2. Numbers of the form 4k + 3 are always odd.

Question 20. Case Study 2: Parity and Sign Change
In a number reasoning contest, students were given four consecutive numbers and asked to apply different arithmetic operations using plus and minus signs. Surprisingly, they observed that no matter how the signs were changed, the final result always remained even. The teacher explained that changing a sign affects the value by an even number and therefore does not change the parity. This observation helped students understand the deeper concept of parity beyond simple addition.
(a) Find the sum of four consecutive numbers starting from 12.
(b) Change one plus sign to minus and find the new result.
(c) State whether the result is even or odd.
(d) Explain why the parity does not change.
Answer:
(a) 12 + 13 + 14 + 15 = 54.
(b) Changing the last plus to minus: 12 + 13 + 14 − 15 = 24. (Any choice works, e.g. 12 − 13 + 14 + 15 = 28.)
(c) 24 is even, just like 54.
(d) Changing +a to −a reduces the total by 2a (here 54 − 24 = 30 = 2 × 15). 2a is always even, and subtracting an even number does not change whether a number is odd or even. So the parity stays the same (even).
In simple words: Turning a plus into a minus takes away double that number, which is even, so an even total stays even.

Teacher's Note:
1. Four consecutive numbers contain two odd numbers, so their sum is even.
2. Parity of a + b and a − b is always the same.