Playing with Numbers Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 16 Playing with Numbers will help students prepare well for the exams.

Class 8 Maths Chapter 16 Extra Questions Playing with Numbers

Playing with Numbers Class 8 Extra Questions

Question 1.
Find Q in the addition.
Playing with Numbers Class 8 Extra Questions with Answers 1
Solution:
There is just one letter Q whose value we have to find.
Study the addition in the ones column: from Q + 3, we get ‘1’ that is, a number whose ones digit is 1.
For this to happen, the digit Q should be 8. So the puzzle can be solve as shown below.
Playing with Numbers Class 8 Extra Questions with Answers 2
That is, Q = 8

Playing with Numbers Class 8 Extra Questions with Answers

Question 2.
Find A and B in the addition.
Playing with Numbers Class 8 Extra Questions with Answers 3
Solution:
This has two letters A and B whose values are to be found.

Study the addition in the ones column: the sum of three A’s is a number whose ones digit is A. Therefore, the sum of two A’s must be a number whose ones digit is 0.

This happens only for A = 0 and A = 5. If A = 0, then the sum is 0 + 0 + 0 = 0, which makes B = 0 too. We do not want this (as it makes A = B, and then the tens digit of BA too becomes 0), so we reject this possibility. So, A = 5.
Therefore, the puzzle is solved as shown below.
Playing with Numbers Class 8 Extra Questions with Answers 4
That is A = 5 and B = 1.

Question 3.
Find the digits A and B.
Playing with Numbers Class 8 Extra Questions with Answers 5
Solution:
This also has two letters A and B whose values are to be found.
Since the ones digit of 3 × A is A, it must,, be that A = 0 or A = 5.
Now look at B. If B = 1, then BA × B3 would at most be equal to 19 × 19; that is, it would at most be equal to 361. But the product here is 57A, which is mote than 500. So we cannot have B = 1.
If B = 3, then BA × B3 would be more than 30 × 30; that is, more than 900. But 57A is less than 600. So, B can not be equal to 3.
Putting these two facts together, we see that B = 2 only. So the multiplication is either 20 × 23, or 25 × 23.
The first possibility fails, since 20 × 23 = 460. But, the second one works out correctly, since 25 × 23 = 575.
So the answer is A = 5, B – 2.
Playing with Numbers Class 8 Extra Questions with Answers 6

Question 4.
Check the divisibility of 21436587 by 9.
Solution:
The sum of the digits of 21436587 is 2 + 1 + 4 + 3 + 6 + 5 + 8 + 7 = 36.
This number is divisible by 9
(for 36 ÷ 9 = 4). We conclude that 21436587 is divisible by 9.
We can double-check :
\(\frac{21436587}{9}\) = 2381843 (the division is exact).

Question 5.
Check the divisibility of 152875 by 9.
Solution:
The sum of the digits of 152875 is 1 + 5 + 2 + 8 + 7 + 5 = 28.
This number is not divisible by 9.
We conclude that 152875 is not divisible by 9.

Playing with Numbers Class 8 Extra Questions with Answers

Question 6.
If the three digit number 24x is divisible by 9, what is the value of x ?
Solution:
Since 24x is divisible by 9, sum of it’s digits, i.e., 2 + 4 + x should be divisible by 9, i.e., 6 + x should be divisible by 9. This is possible when 6 + x = 9 or 18, ….
But, since x is a digit, therefore, 6 + x = 9, i.e., x = 3.

Question 7.
Check the divisibility of 2146587 by 3.
Solution:
The sum of the digits of 2146587 is 2 + 1 + 4 + 6 + 5 + 8 + 7 = 33.
This number is divisible by 3 (for 33 ÷ 3 = 11).
We conclude that 2146587 is divisible by 3.

Question 8.
Check the divisibility of 15287 by 3.
Solution:
The sum of the digits of 15287 is 1 + 5 + 2 + 8 + 7 = 23.
This number is not divisible by 3. We conclude that 15287 too Is not divisible by 3.

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