NCERT Solutions | Class 8 Maths Chapter 16 | Playing with Numbers

CBSE Solutions | Maths Class 8

Check the below NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Pdf free download. NCERT Solutions Class 8 Maths were prepared based on the latest exam pattern. We have Provided Playing with Numbers Class 8 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 8 Maths

NCERT Solutions Class 8 Maths
Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	8th
Subject:	Maths
Chapter:	16
Chapters Name:	Playing with Numbers
Medium:	English

Playing with Numbers | Class 8 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 8 Maths Chapter 16 Playing with Numbers to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.1

Ex 16.1 Class 8 Maths Question 1.

Find the values of the letters in each of the following and give reasons for the steps involved.
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 1

Solution:

1. There are two letters whose values are to be found out.
Study the addition in the ones column, i. e., from A + 5 we get 2, i. e., the number whose ones digit is 2.
For this to happen, A must be 7 (∵ A + 5 = 7 + 5 = 12).
So for the addition in tens column, we have
1 + 3 + 2 = B ⇒ B = 6
So, the puzzle has been decoded as
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 2

∴ The possible values of A and B are 7 and 6 respectively.

2. There are three letters whose values are to be found out.
Study the addition in the ones column, i. e., from A + 8 we get 3, i. e., the number whose ones digit is 3.
For this to happen, A must be 5 (∵ A + 8 = 5 + 8 = 13).
So for the addition in tens column, we have 1 + 4 + 9 = B ⇒ 14 = B
∴Clearly, B is 4 and C is 1.
So the puzzle has been decoded as
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 3
∴ The possible values of A, B and C are 5, 4 and 1 respectively.

3. Since the ones digit of A x A is A, it must be that A = 1 or A = 5 or A = 6.
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 4

4. There are two letters whose values are to be found : A and B.
Study the addition in the given puzzle. .
i. e., from B + 7 we get A and from A + 3 we get 6
Possible values can be
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 5

5. This has three letters whose values are to be found : A, B and C.
Since the ones digit of 3 x B = B, it must be B = 0. So, the puzzle becomes
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 6

6. This has three letters whose values are to be found: A, B and C. Since the ones digit of 5 x B is B, it must be that B = 0 or B = 5 If B = 0, then the puzzle becomes
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 7

7. This has two letters whose values are to be found : A and B.
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 8
Possible values of BBB are 111, 222, 333, etc.
Let us divide these numbers by 6.
111 ÷ 6 =18, remainder 3. So, 111 is rejected.
222 ÷ 6 = 37, remainder 0. So, the quotient 37 is not of the form A2. Thus,
222 is rejected.
333 ÷ 6 = 55, remainder 3. So, 333 is rejected.
444 ÷ 6 = 74, remainder 0. The quotient 74 is of the form A4.
So, the puzzle has been decoded as
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 9
This one clearly works out correctly.
So, the answer is A = 7 and B = 4.

8. This has two letters whose values are to be found : A and B.
Study the addition in the ones column : from 1 + B we get 0, that is, a number whose ones digit is 0.
For this to happen, the ones digit of B should be 9. And since B itself is a digit, we get B =9. So, the puzzle becomes
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 10
But 90 – 19 = 71 so, A1 = 71 ⇒ A = 7
Thus, A = 7 and B = 9 is the answer.

9. There are two letters whose values are to be found out.
Study the addition in the ones column, i. e., from B + 1 we get 8, that is, a number whose ones digit is 8.
For this to happen, the ones digit of B should be 7. As B itself is a digit, we get B = 7. So the puzzle becomes
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 11
Now study the addition in the tens column, i. e., from A+ 7 we get 1, that is, a number whose ones digit is 1.
For this to happen, the ones digit of A should be 4. As A itself is a digit, we get A = 4. So, the puzzle has been decoded as
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 12
Therefore, possible values of A and B are 4 and 7 respectively.

10. This has two letters whose values are to be found : A and B.
Study the addition in the tens column : from 2 + A we get 0, that is, a number whose ones digit is 0.
For this to happen, the ones digit of A should be 8. And since A is a digit, we get A = 8. So, the puzzle becomes
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 13
Now, study the addition in the ones column: from 8 + B we get 9. For this to happen, we must have B = 1. Therefore, the puzzle has been decoded as

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.2

Ex 16.2 Class 8 Maths Question 1.

If 21 y 5 is a multiple of 9, where y is a digit, what is the value of y?

Solution:

Since the number 21y5 is a multiple of 9.
So, the sum of its digits 2 + 1 + y + 5 = 8 + y is a multiple of 9.
∴ (8 + y) is either 0 or 9 or 18 or 27 …
But since y is a digit, so (8 + y) must be equal to 9.
i.e.,8 + y = 9 ⇒ y = 9- 8= l

Ex 16.2 Class 8 Maths Question 2.

If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

Solution:

Since the number 31z5 is a multiple of 9.
So, the sum of its digits 3 + 1 + z + 5 = 9+ z is a multiple of 9.
∴ (9 + z) is either 0 or 9 or 18 or 27 …
But since z is a digit, so (9 + z) must be equal to 9 or 18…
i.e., 9 + z = 9 ⇒ z = 9 + z = 18 ⇒ z = 9

Ex 16.2 Class 8 Maths Question 3.

If 24x is a multiple of 3, where x is a digit, what is the value of x?

Solution:

Since 24 x is a multiple of 3. So, the sum of its digits 2 + 4 + x = (6 + x) is
a multiple of 3.
∴ (6 + x) is one of the numbers 0, 3, 6, 9, 12, 15, 18…
But x is a digit. Therefore, (6 + x) must be equal to 6 or 9 or 12 or 15.
i.e., 6 + x = 6 or 9 or 12 or 15
⇒ x = 0 or 3 or 6 or 9
Thus, x can have any of the four different values, namely, 0,3,6 or 9.

Ex 16.2 Class 8 Maths Question 4.

31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Solution:

Since 31z5 is a multiple of 3. So, the sum of its digits 3 + 1 + z + 5 = (9 + z) is a multiple of 3.
∴ (9 + z) is one of the numbers 0, 3, 6, 9, 12, 15, 18, …
But z is a digit. Therefore (9 + z) must be equal to 9 or 12 or 15 or 18.
i.e., 9 + z = 9 or 12 or 15 or 18
⇒ z = 0 or 3 or 6 or 9
Thus, z can have any of the four different values, namely, 0, 3, 6 or 9.