# Important Questions Class 8 Maths Chapter 16

## Important Questions Class 8 Mathematics Chapter 16 – Playing With Numbers

Mathematics requires analytical thinking and problem-solving skills. One should do lots of practice and develop a deep understanding of the concepts of the subject in order to excel in Mathematics.

Chapter 16 of Class 8 Mathematics is about Playing with Numbers. The important topics covered in this chapter are:

• Introduction
• Numbers in general form
• Games with numbers
• Letters for digits
• Tests of divisibility

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As we all have come across the phrase “Practice makes a man perfect” and hence to gain mastery in Mathematics, one needs to practise solving mathematical questions on a regular basis. Extramarks is a promising online learning platform that understands the significance of solving problems when it comes to Mathematics.

Our experienced Mathematics faculty experts have prepared question-bank Class 8 Mathematics Chapter 16 Important Questions, after doing proper research and the past years’ question paper analysis. Important Questions Class 8 Mathematics Chapter 16 will provide the students with substantial questions relevant to the chapter Play with Numbers so that they can learn better and practise them for their examinations. Our teachers have given step-by-step instructions for each solution making it easy for students to comprehend and remember the concepts used in these solutions.

Along with Important Questions Class 8 Mathematics Chapter 16, the website of Extramarks also provides the students with the CBSE revision notes, CBSE sample papers, and CBSE past year question papers, strictly abiding by the CBSE syllabus.

## Get Access to CBSE Class 8 Maths Important Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

 CBSE Class 8 Maths Important Questions Sr No. Chapters Chapters Name 1 Chapter 1 Rational Numbers 2 Chapter 2 Linear Equations in One Variable 3 Chapter 3 Understanding Quadrilaterals 4 Chapter 4 Practical Geometry 5 Chapter 5 Data Handling 6 Chapter 6 Squares and Square Roots 7 Chapter 7 Cubes and Cube Roots 8 Chapter 8 Comparing Quantities 9 Chapter 9 Algebraic Expressions and Identities 10 Chapter 10 Visualising Solid Shapes 11 Chapter 11 Mensuration 12 Chapter 12 Exponents and Powers 13 Chapter 13 Direct and Inverse Proportions 14 Chapter 14 Factorisation 15 Chapter 15 Introduction to Graphs 16 Chapter 16 Playing with Numbers

### Important Questions Class 8 Mathematics Chapter 16 – With Solutions

Given below are a few of the important questions and their solutions those are included in our Mathematics Class 8 Chapter 16 important questions:

Question 1: Write in the normal form 10 × 6 + 5.

• 65
• 56
• 25
• 54

Explanation 1: 10 × 6 + 5 = 60 + 5 = 65

Question 2: If the division of N ÷ 5 leaves a remainder of 1, what might be the one’s digit of N?

• 1
• Either 7 or 2
• 6
• 5

Explanation 2: 5 + 1= 6

Therefore, one’s digit of N is 6.

Question 3: The difference between a two-digit number and the number obtained by reversing its digits is always divisible by ____________

Explanation 3: Taking a and b as the two digits, we get X = ab

On reversing the digits, we get Y = ba

Hence, the sum of digits in X is 10a + b and the sum of digits of Y is 10b + a

On subtracting X and Y, we get

9a – 9b = 9 (ab)

Therefore, the number is divisible by 9

Question 4: A four-digit number abcd is divisible by 11, if d + b = ___________ or ___________.

Answer 4: a + c or b + d

Explanation 4: Implying the divisibility rule of 11, if abcd is divisible by 11,

then a – b + c – d = 0

Therefore, we get a + b = b + d or each should be zero.

Question 5:     A       B

+  3       7

.                 —————-

6        A

—————-

Answer 5: Here, we see that B = 5 so that 7 + 5 = 12

Putting 2 at one’s place and carrying over 1 and A = 2, we get

2 + 3 + 1 = 6

This A = 2 and B = 5

Question 6: Find the values of the letters in the following and give reasons for each step involved

•          2          A

+         6             A           B

—————————————

A             0           9

—————————————

Answer 6: Taking A = 8 and B = 1, we find that 8 + 1 = 9

On again adding 2 + 8 = 10

Thus, 10’s place will have a 0 and carry over 1.

Now, 1 + 6 + 1 = 8 = A

Therefore, A = 8 and B = 1

Question 7: Express the given number in a normal form:

(2×1000) + (2×10)

Answer 7: e can write (2×1000) + (2×10) in a normal form in the following way

(2×1000) + (2×10) = 2000 + 20 = 2020

Question 8:  What is the smallest number you have to add to 100000 to get a multiple of 1234?

Answer 8: 00000 can also be written as 1234 × 81 + 46

Thus, 1234 + 46 = 1188

Hence, the required number is 1188.

Question 9: Check the divisibility of 2146587 by 3.

Answer 9: The sum of the digits of  2146587 is 2 + 1 + 4 + 6 + 5 + 8 + 7 = 33

33 is divisible by 3 since 3 × 11 = 33

Hence, 2146587 is divisible by 3

Question 10: If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Answer 10: We are given that 31z5 is a multiple of 3.

Following the divisibility rule of 3,

We understand that the sum of the digits 31z5 need to be divisible by 3 for the number to be divisible by 3.

Which is, 3 + 1 + z + 5 = 9 + z

Therefore, 9 + z will be a multiple of 3

And it will be possible only when 9 + z is any of the values – 0, 3, 6, 9, 12, 15, 18 and further.

If z = 0, then 9 + z = 9 + 0 = 9

if z = 3, then 9 + z = 9 + 3 = 12

if z = 6, then 9 + z = 9 + 6 = 15

if z = 9, then 9 + z = 9 + 9 = 18

So we see that the value of 9 + z can be 9 or 12 or 15 or 18.

Hence we can conclude that 0, 3, 6 or 9 are four possible answers for z.

Question 11: If 24x is a multiple of 3, where x is a digit, what is the value of x?

Answer 11: Given that 24x is a multiple of 3.

Then we know that following the visibility rule off 3, the sum of all the digits of 24x should be a multiple of 3.

2 + 4 + x = 6 + x

Therefore, 6 + x is a multiple of 3, which is possible only when x is one the following numbers – 0, 3, 6, 9, 12, 15, 18 and so on.

X being a digit, the value of x shall either be 0 or 3 or 6 or 9, and the sum shall be either 6 or 9 or 12 or 15, respectively.

Hence, x shall be any of the four values – 0, 3, 6 or 9.

Question 12: A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.

Answer 12: Let the unit place digit be x and tens place digit be y.

Therefore, Equation (1) is 10y + x

From the question, a two-digit number is 3 more than 4 times the sum of its digits

Now, the above condition suggests that the Equation (2) is 4(y + x) + 3

Equating equation 1 and 2

4(y + x) + 3 = 10y + x

• 4x + 4y + 3 = 10y + x
• 3x – 6y = -3
• X – 2y = -1                  …equation (3)

Now, the second condition suggests that if 18 is added to the number, it’s digit is reversed

Therefore, Equation 4 is 10x + y

By the given condition,

( 10y + x ) + 18 = 10x + y

• 10y + x = 10x + y -18
• 9y -9x = -18
• y – x = -2.                   …equation (5)

Solving equations 3 and 5, we get x = 5 and y = 3.

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5. CBSE extra questions

Q.1 Find A, B, C in the addition.

$\begin{array}{l}\text{}3\text{}4\text{A}\\ \text{+}3\text{A}\mathrm{B}\\ \overline{\text{}\mathrm{C}\text{}2\text{}9\text{}}\end{array}$

Marks:3
Ans

The addition of A and B is giving 9 i.e., a number whose ones digits is 9. The sum can be 9 only as the sum of two single digit numbers cannot be 19. Therefore, there will not be any carry in this step.

In the next step, 4 + A = 2

It is possible only when A = 8

4 + 8 = 12 and 1 will be the carry for the next step.

1 + 3 + 3 = C

Therefore, C is 7.

We know that the addition of A and B is giving 9. As A is 8, therefore, B is 1.

Q.2 Find a such that the five digit number 91a92 is divisible by 9.

Marks:2
Ans

A number is divisible by 9, if its sum is divisible by 9 i.e. 9 + 1 + a + 9 + 2 is divisible by 9 .
21 + a is divisible by 9
Therefore, the number can be 27, 36, …
But since a is a one-digit number
So, 21 + a = 27
a = 6

Q.3 Find the values of A, B and C in the multiplication.

$\begin{array}{l}\underset{¯}{\begin{array}{l}\text{A B}\\ \text{× 5}\end{array}}\\ \text{C A B}\end{array}$

Marks:4
Ans

The multiplication of B and 5 is giving a number whose ones digit is B again. This is possible when B = 5 or B = 0 only.

In case of B = 5, the product, B — 5 = 5 — 5 = 25

2 will be a carry for the next step.

We have, 5 — A + 2 = CA, which is possible for A = 2 or 7

The multiplication is as follows.

If B = 0,

B — 5 = B 0 — 5 = 0

There will not be any carry in this step.

In the next step, 5 — A = CA

It can happen only when A = 5 or A = 0

However, A cannot be 0 as AB is a two-digit number.

Hence, A can be 5 only. The multiplication is as follows.

$\begin{array}{l}\text{}50\\ \frac{—5}{250}\end{array}$

Hence, there are 3 possible values of A, B, and C.

(i) 5, 0 and 2 respectively

(ii) 2, 5 and 1 respectively

(iii) 7, 5 and 3 respectively

Q.4 The ones digit of a two-digit number is 3 and the sum of digits is 1/7 of the number itself. What is the number

Marks:2
Ans

Let the two digit number having 3 as its units digit be 10a + 3.
Sum of the digits = a + 3
Now, a + 3 = (1/7)(10 a + 3)
7a + 21 = 10a + 3
3a = 18
a = 6
The number is 63.

Q.5 Check the divisibility of 2147681 by 3.

Marks:1
Ans

2 + 1 + 4 + 7 + 6 + 8 + 1 = 29. Here 29 is not a multiple of 3, so 2147681 is not divisible by 3.