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CBSE Class 8  >  Class 8 Test  >  Mathematics (Maths)   >  Test: Number Play - Class 8 MCQ

Number Play - Class 8 Maths Free MCQ Test with solutions


MCQ Practice Test & Solutions: Test: Number Play (15 Questions)

You can prepare effectively for Class 8 Mathematics (Maths) Class 8 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Number Play". These 15 questions have been designed by the experts with the latest curriculum of Class 8 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 15

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Test: Number Play - Question 1
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Which of the following numbers cannot be expressed as a sum of consecutive natural numbers?

Detailed Solution: Question 1

A number can be expressed as a sum of consecutive natural numbers if and only if it is not a power of 2. Powers of 2 (2, 4, 8, 16, 32, …) cannot be written in this form.

  • 21 = 10 + 11
  • 29 = 14 + 15
  • 45 = 22 + 23
  • 32 = 25 (a power of 2), so it cannot be expressed as a sum of consecutive natural numbers.

Correct answer: C) 32

Test: Number Play - Question 2
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How many expressions can be formed using + or – signs between the three consecutive numbers n, n+1, n+2?

Detailed Solution: Question 2

There are two places where an operation can be placed between the three numbers

Each place has two choices: plus or minus

Therefore the total number of expressions = 2 × 2 = 22 = 4 

Hence, the correct answer is 4

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Test: Number Play - Question 3
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According to parity rules, what is the result of Even × Odd?

Detailed Solution: Question 3

Multiplying any number with an even number always gives an even product.

Test: Number Play - Question 4
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Which of the following is always odd?

Detailed Solution: Question 4

The form 2n + 1 always gives an odd number for any integer n.

Test: Number Play - Question 5
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The least common multiple (LCM) of 6 and 8 is:

Detailed Solution: Question 5

6 = 2 × 3, 8 = 23 → LCM = 23 × 3 = 24.

Test: Number Play - Question 6
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Numbers that leave a remainder of 2 when divided by 7 can be written in the form:

Detailed Solution: Question 6

1) Division algorithm (the rule behind remainders)

For any whole number n and any positive divisor 7, there exist integers k (the quotient) and r (the remainder) such that
n = 7k + r, with 0 ≤ r ≤ 6.

So the possible remainders on division by 7 are only 0, 1, 2, 3, 4, 5, 6.

2) Apply it to “remainder = 2”

If a number leaves remainder 2 when divided by 7, then r = 2 in the formula:
n = 7k + 2.
This is exactly option (b) 7k + 2.

3) Quick checks with examples

Take k = 0 → n = 7·0 + 2 = 2 → 2 ÷ 7 = 0 remainder 2
k = 1 → n = 9 → 9 ÷ 7 = 1 remainder 2
k = 2 → n = 16 → 16 ÷ 7 = 2 remainder 2
k = 5 → n = 37 → 37 ÷ 7 = 5 remainder 2
So all numbers of the form 7k + 2 give remainder 2.

4) Why the other options are wrong

  • (a) 7k + 1 → remainder is 1 (example: 8 ÷ 7 leaves 1).

  • (c) 7k + 3 → remainder is 3 (example: 10 ÷ 7 leaves 3).

  • (d) 7k – 2 → this looks tempting, but note:
    7k – 2 = 7(k – 1) + 5, so the standard (non-negative) remainder is 5, not 2.
    Example: k = 2 → 7k – 2 = 12, and 12 ÷ 7 leaves remainder 5.

5) One-line “test”

To see which form matches remainder r, just reduce it mod 7:

  • 7k + r ≡ r (mod 7).
    Only 7k + 2 is ≡ 2 (mod 7).

Therefore, numbers that leave remainder 2 when divided by 7 are exactly those of the form 7k + 2 (option b).

Test: Number Play - Question 7
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For divisibility by 8, which digits of a number are checked?

Detailed Solution: Question 7

A number is divisible by 8 if the number formed by its last 3 digits is divisible by 8.

Test: Number Play - Question 8
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A number is divisible by 6 if:

Detailed Solution: Question 8

Rule: Divisible by 6 ⇔ divisible by 2 and 3.

Test: Number Play - Question 9
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The digital root of 4725 is:

Detailed Solution: Question 9

4+7+2+5=18 → 1+8=9.

Test: Number Play - Question 10
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Which of the following numbers is divisible by 11?

Detailed Solution: Question 10

Divisiblity Rule: A number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is divisible by 11 (including 0).
For Option A:
(1+0) – (0+1) = 0, divisible by 11 → so 1001 is divisible by 11.

So, Correct option: A

Test: Number Play - Question 11
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If n is an integer, then which expression is always divisible by 2?

Detailed Solution: Question 11

n2 + n = n(n+1)
Here, n and n+1 are two consecutive integers → one of them is always even.
So their product is always even → divisible by 2.

Test: Number Play - Question 12
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The remainder when 105 is divided by 9 is:

Detailed Solution: Question 12

To find the remainder when 105 is divided by 9, we can use the divisibility rule for 9:

Step 1: Add the digits of 105: 1 + 0 + 5 = 6

Step 2: Find remainder when 6 is divided by 9. Since 6 is less than 9, the remainder is 6.

Test: Number Play - Question 13
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If XY × 9 = ZW, where XY and ZW are both two-digit numbers, which of the following equations is correct

Detailed Solution: Question 13

Check each option.
12 × 9 = 108 is three-digit, not acceptable.
13 × 9 = 117 is three-digit, not acceptable.
11 × 9 = 99 is two-digit, fits the requirement.
14 × 9 = 126 is three-digit, not acceptable.
Therefore, the valid equation is 11 × 9 = 99.

Test: Number Play - Question 14
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Which of the following pairs are co-prime?

Detailed Solution: Question 14

To find which pair is co-prime, we check which pair has HCF (GCD) = 1.
Co-prime numbers have no common factor other than 1.

Let’s check each option:

a) 6 and 8
Factors of 6 → 1, 2, 3, 6
Factors of 8 → 1, 2, 4, 8
Common factors → 1, 2
HCF = 2 → Not co-prime

b) 9 and 15
Factors of 9 → 1, 3, 9
Factors of 15 → 1, 3, 5, 15
Common factors → 1, 3
HCF = 3 → Not co-prime

c) 8 and 25
Factors of 8 → 1, 2, 4, 8
Factors of 25 → 1, 5, 25
Common factor → Only 1
HCF = 1 → Co-prime 

d) 12 and 18
Common factors include 1, 2, 3, 6
HCF = 6 → Not co-prime

Correct answer: c) 8 and 25
 

Test: Number Play - Question 15
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If the sum of digits of a number is 27, then the number is definitely divisible by:

Detailed Solution: Question 15

Divisibility rule: If sum of digits is divisible by 9, then number is divisible by 9.

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About this Class 8 Practice Test

This test contains 15 MCQs on Number Play with detailed solutions for each question. It is part of the Mathematics (Maths) Class 8 course on EduRev, designed to align with the current Class 8 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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