## NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers (EX 16.1) Exercise 16.1

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## Download PDF of NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1

1) $6.4,6.46,0.646,4.66$

2) $9.21,19.2,2.91,1.29$

3) $8.62,86.2,0.82,862.0$

$3 \times 1000000 + 4 \times 100000 + 6 \times 1000 + 5 \times 100 + 7 \times 1$

\[9,25,46,73,107{\text{ }}........\]

$\left( a \right){\text{ 133}}$

$\left( b \right){\text{ 149}}$

$\left( c \right){\text{ 148}}$

$\left( d \right){\text{ 145}}$

$\left( e \right){\text{ 151}}$

A.28

B.36

C.64

D.224

6 : 35 :: 11: ?

(a) 120

(b) 121

(c) 115

(d) 122

_{}

## NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers (Ex 16.1) Exercise 16.1

### Exercise 16.1

1. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{ }}3{\text{ A}} \\ \underline {{\text{ + 2 5}}} \\ \underline {{\quad\text{ B 2}}} \\ \]

Ans: The sum of $A$ and $5$ is $2$ i.e., a number whose ones digit is $2$. This is possible only when $A = 7$. In that case, the sum of $A\left( 7 \right)$ and $5$ will give $12$ and thus, $1$ is carried to the next step. In the next step,

$1 + 3 + 2 = 6$

Therefore, the addition is as follows.

\[{\quad\text{ }}3{\text{ 7}} \\ \underline {{\text{ + 2 5}}} \\ \underline {{\quad\text{ 6 2}}} \\ \]

Clearly, $B$ is $6$.

The values of $A$ and $B$ are $7$ and $6$ respectively.

2. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{ 4 A}} \\ \underline {{\text{ + 9 8}}} \\ \underline {{\quad\text{CB 3}}} \\ \]

Ans: The sum of $A$ and $8$ is $3$ i.e., a number whose ones digit is $3$. This is possible only when $A$ is $5$. In that case, the sum of $A$ and $8$ will give $13$ and thus, $1$ is carried for the next step. In the next step,

$1 + 4 + 9 = 14$

Therefore, the addition is as follows.

\[{\quad\text{ 4 5}} \\ \underline {{\text{ + 9 8}}} \\ \underline {{\quad\text{14 3}}} \\ \]

Clearly, $B = 4$ and $C = 1$.

The values are $A = 5$, $B = 4$, and $C = 1$.

3. Find the values of the letters in the following and give reasons for the steps involved.

\[\begin{align} &\quad \text{ 1 A } \\ & \underline{\quad\times \text{ A}} \\ & \underline{\quad\text{ 9 A }} \\ \end{align}\]

Ans: Only numbers that give the same number when multiplied by themselves are $0,1,5,6$.

If $A=0$, not possible because $0\times 1\ne 9$.

If $A=1$, then also not possible because $1\times 1\ne 9$.

If $A=5$, then $5\times 5=25$, $2$ is carried and $5\times 1+2=7$. So, $5$ is also not possible.

Only possibility is of $6$, $6\times 6=36$ and $3$ is carried and $6\times 1+3=9$. Hence, the value of $A$ is $6$.

4. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{A B}} \\ \underline {{\text{ + 3 7}}} \\ \underline {{\quad\text{6 A}}} \\ \]

Ans: The addition of $A$ and $3$ is giving $6$. There can be two cases.

(1) First case is not producing a carry

In that case, $A$ is $3$ as $3 + 3 = 6$. Consider the first step in which the

sum of $B$ and $7$ is $A$ (i.e., $3$), $B$ should be a number whose unit digit of this addition comes to be $3$. This is possible only when $B = 6$. In that situation, $A = 6 + 7 = 13$. But, $A$ is a single digit number so it is not possible.

(2) Second case is producing a carry

In that case, $A$ comes to be $2$ as $1 + 2 + 3 = 6$. Consider the first step in which the addition of $B$ and $7$ gives $A$ (i.e., 2), then $B$ is a number whose unit digit of this addition is $2$. It is possible only when $B = 5$ and $5 + 7 = 12$. So, the values are $A = 2$ and $B = 5$.

5. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{ A B}} \\ \underline {\quad\times {\text{3 }}} \\ \underline {{\text{ C A B}}} \\ \]

Ans: The multiplication of $3$ and $B$ gives a number whose ones digit is $B$ again.

Hence, $B$ can be $0$ or $5$.

Let $B = 5$.

${\text{Multiplication of first step}} = 3 \times 5 = 15$

And $1$ is a carry for the next step.

We have, $3 \times A + 1 = CA$

This is not true for any value of $A$.

Hence, $B = 0$ and if this happens, then there will be no carry for the next step.

We have to obtain, $3 \times A = CA$

That is, the one’s digit of $3 \times A$ is $A$. This is possible only when $A$ is $5$ or $0$.

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

Therefore, $A$ can only be $5$ and the multiplication is as follows.

\[{\text{ 50}} \\ \underline {\times {\text{3}}} \\ \underline {{\text{150 }}} \\ \]

Hence, the values are $A = 5$, $B = 0$, and $C = 1$

6. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{ A B}} \\ \underline { \quad\times {\text{ 5 }}} \\ \underline {{\text{ C A B}}} \\ \]

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

If $B = 5$, then the product, $B \times 5 = 5 \times 5 = 25$

$2$ is the carry for the next step.

We have, $5 \times A + 2 = CA$, which is possible for $A = 2$ or $7$.

The multiplication is as follows.

\[{\text{ 25}} \\ \underline { \times {\text{ 5 }}} \\ \underline {{\text{125 }}} \\ \]

\[{\text{ 75}} \\ \underline { \times {\text{ 5 }}} \\ \underline {{\text{375 }}} \\ \]

If $B = 0$,

$B \times 5 = B \Rightarrow 0 \times 5 = 0$

So, no carry in this step.

In the next step, $5 \times A = CA$.

This can happen only when $A = 5$ or $A = 0$.

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

Hence, $A = 5$ and the multiplication is as follows.

\[{\text{ 50}} \\ \underline { \times {\text{ 5 }}} \\ \underline {{\text{250 }}} \\ \]

Hence, the possible values of $A$, $B$, and $C$ are:

(i) $5$, $0$, and $2$ respectively

(ii) $2$, $5$, and $1$ respectively

(iii) $7$, $5$, and $3$ respectively

7. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{A B}} \\ \underline { \quad\times {\text{6}}} \\ \underline {{\text{B B B}}} \\ \]

Ans: The multiplication of $6$ and $B$ is a number whose one’s digit is $B$ again.

It is possible only if $B = 0,2,4,6,8$

If $B = 0$, thai value is not possible as then the product is $0$.

If $B = 2$, then $B \times 6 = 12$ and $1$ will be a carry for the next step.

$6A + 1 = BB = 22 \Rightarrow 6A = 21$ and hence, any integer value of $A$ is not possible.

If $B = 6$, then $B \times 6 = 36$ and $3$ will be a carry for the next step.

$6A + 3 = BB = 66 \Rightarrow 6A = 63$ and hence, any integer value of $A$ is not possible.

If $B = 8$, then $B \times 6 = 48$ and $4$ will be a carry for the next step.

$6A + 4 = BB = 88 \Rightarrow 6A = 84$ and hence, $A = 14$.

Since, $A$ is a single digit number so this value is also not possible.

If $B = 4$, then $B \times 6 = 24$ and $2$ will be a carry for the next step.

$6A + 2 = BB = 44 \Rightarrow 6A = 42$ and hence, $A = 7$.

The multiplication is as follows.

\[{\text{ 7 4}} \\ \underline { \times {\text{ 6 }}} \\ \underline {444} \\ \]

The values are $A = 7$ and $B = 4$.

8. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{ A 1}} \\ \underline {{\text{ + 1 B}}} \\ \underline {{\quad\text{ B 0}}} \\ \]

Ans: The addition of $1$ and $B$ is $0$ i.e., a number whose ones digits is $0$. This is possible only if $B = 9$. In that case, the sum of $1$ and $B$ is $10$ and

Thus, $1$ will be the carry for the next step. Then,

$1 + A + 1 = B$.

Now it’s clear that $A$ is $7$ as \[1 + 7 + 1 = 9 = B\].

The addition is as follows.

\[{\quad\text{ 7 1}} \\ \underline {{\text{ + 1 9}}} \\ \underline {{\quad\text{ 9 0}}} \\ \]

The values of $A$ and $B$ are $7$ and $9$ respectively.

9. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{2 A B}} \\ \underline {{\text{ + A B 1}}} \\ \underline {{\quad\text{ B 1 8}}} \\ \]

Ans: The sum of $1$ and $B$ is $8$ i.e., a number whose ones digits is $8$. This is

possible only when $B = 7$. In that case, the sum of $1$ and $B$ will give $8$. In the next step,

$A + B = 1$

Clearly, $A$ is $4$.

$4 + 7 = 11$ and $1$ is the carry for the next step.

$1 + 2 + A = B$

$1 + 2 + 4 = 7$

Therefore, the addition is as follows.

\[{\quad\text{ 2 4 7}} \\ \underline {{\text{ + 4 7 1}}} \\ \underline {{\quad\text{ 7 1 8}}} \\ \]

The values of $A$ and $B$ are $4$ and $7$ respectively.

10. Find the values of the letters in the following and give reasons for the steps involved.

\[{\quad\text{1 2 A}} \\ \underline {{\text{ + 6 A B }}} \\ \underline {{\quad\text{ A 0 9}}} \\ \]

Ans: The sum of $1$ and $B$ is $9$ i.e., a number whose ones digits is $9$. The sum can only be $9$ as the sum of two single digit numbers cannot be $19$ so, no carry in this step.

In the next step, $2 + A = 0$

This is possible only when $A = 8$ as $2 + 8 = 10$ and $1$ is the carry for the next step.

$1 + 1 + 6 = A$

Now it is clear that $A = 8$. We know that the sum of $A$ and $B$ is $9$. As $A$ is $8$, therefore, $B = 1$.

The addition is as follows.

\[{\quad\text{ 1 2 8}} \\ \underline {{\text{ + 6 8 1 }}} \\ \underline {{\quad\text{ 8 0 9}}} \\ \]

The values of $A$ and $B$ are $8$ and $1$ respectively.

### NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers (Ex 16.1) Exercise 16.1

Opting for the NCERT solutions for Ex 16.1 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 16.1 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 16 Exercise 16.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 16 Exercise 16.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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