NCERT Solutions Class 8 Maths Chapter 16 Exercise 16.1

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

$\begin{array}{l}\text{1.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}A}\\ \underset{¯}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}5}}\\ \underset{¯}{\mathrm{B}2}\\ \\ \text{2.}\\ 4\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}A}\\ \underset{¯}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}98}\\ \underset{¯}{\mathrm{C}\mathrm{B}3}\\ \\ \text{3.}\\ 1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}A}\\ \underset{¯}{\times \mathrm{A}}\\ \underset{¯}{9\mathrm{A}}\\ \\ \text{4.}\\ \mathrm{A}\mathrm{B}\\ \underset{¯}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}37}\\ \underset{¯}{6\mathrm{A}}\\ \\ \text{5.}\\ \mathrm{A}\mathrm{B}\\ \underset{¯}{\times 3}\\ \underset{¯}{\mathrm{C}\mathrm{A}\mathrm{B}}\\ \\ \text{6.}\\ \mathrm{A}\mathrm{B}\\ \underset{¯}{\times 5}\\ \underset{¯}{\mathrm{C}\mathrm{A}\mathrm{B}}\\ \\ \text{7.}\\ \mathrm{A}\mathrm{B}\\ \underset{¯}{\times 6}\\ \underset{¯}{\mathrm{B}\mathrm{B}\mathrm{B}}\\ \\ \text{8.}\\ \mathrm{A}1\\ \underset{¯}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}1\mathrm{B}}\\ \underset{¯}{\mathrm{B}0}\\ \\ \text{9.}\\ 2\mathrm{A}\mathrm{B}\\ \underset{¯}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\mathrm{B}1}\\ \underset{¯}{\mathrm{B}18}\\ \\ \text{10.}\\ 12\mathrm{A}\\ \underset{¯}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}6\mathrm{A}\mathrm{B}}\\ \underset{¯}{\mathrm{A}09}\end{array}$

Ans

\begin{array}{l}\text{1}\text{. 3 A}\\ \underset{¯}{\text{+ 2 5}}\\ \underset{¯}{\text{B 2}}\\ \\ \text{The addition of A and 5 is 2 i}.\text{e}.,\text{a number whose}\\ \text{one’s digit is 2}.\text{This is possible only when A = 7}.\\ \\ \text{Hence},\text{the addition of A and 5 will give 12 and}\\ \text{then 1 will be the carryover for the next step}.\\ \text{In the next step},\text{1}+\text{3}+\text{2}=\text{6}\text{.}\\ \text{So,B = 6}\text{.Therefore},\text{addition will be:}\\ \\ \text{3 7}\\ \underset{¯}{\text{+ 2 5}}\\ \underset{¯}{\text{6 2}}\\ \\ \text{Hence,}\text{A=7 and B=6}\text{.}\end{array}

\begin{array}{l}\text{2}\text{. 4 A}\\ \underset{¯}{\text{+ 9 8}}\\ \underset{¯}{\text{C B 3}}\\ \\ \\ \text{The addition of A and 8 gives us 3}\text{.}\\ \text{This is possible only when digit A is 5 since the addition of 5}\\ \text{and 8 will gives us 13 and thus},\text{1 will be the carry for the}\\ \text{next step}.\\ \\ \text{Now,}1+4+9=14.\\ \text{So B will be equal to 4 and 1 will be carry over to the next step}\text{.}\\ \text{Since there are no numbers in the next step which are}\\ \text{to be added,therefore C = 1}\\ \text{Hence, the addition will be as follows:}\\ \text{4 5}\\ \underset{¯}{\text{+ 9 8}}\\ \underset{¯}{\text{1 4 3}}\\ \\ \therefore \text{A}=5,\text{B= 4 and C=1}\end{array} \begin{array}{l}\end{array}

$\begin{array}{l}\begin{array}{l}\text{3. 1 A}\\ \underset{¯}{\text{× A}}\\ \underset{¯}{\text{9 A}}\end{array}\end{array}$

The multiplication of A with A itself gives a number whose ones digit is A again. This happens only when A = 1, 5, or 6.
Case 1: If A = 1
The multiplication will be 11 × 1 = 11 but the tens digit given here is 9. Therefore, A = 1 is not possible.
Case 2: If A = 5
The multiplication will be 15 × 5 = 75. Thus, A = 5 is also not possible since the tens digit given is 9.
Case 3: If A = 6
If we take A = 6, then 16 × 6 = 96. Since the tens and one’s digit are same as the given question, therefore, A should be 6.
The multiplication will be as follows:

$\begin{array}{l}\text{1 6}\\ \underset{¯}{\text{× 6}}\\ \underset{¯}{\text{9 6}}\end{array}$

\begin{array}{l}\text{4}\text{. A B}\\ \underset{¯}{\text{+ 3 7}}\\ \underset{¯}{\text{6 A}}\end{array}

We can observe that the addition of A and 3 is giving 6. There can be two cases.
Case 1: First step is not producing a carryover.
If first step is not producing a carryover then A comes to be 3 as 3 + 3 = 6.
Consider the first step in which the addition of B and 7 is giving A (i.e., 3), then B should be a number such that the unit’s digit of this addition is 3.This is only possible when B = 6.
Then A = 6 + 7 = 13. However, A is a single digit number. Hence, it is not possible.
Case 2: First step is producing a carry
If we are getting a carryover on adding B and 7, then A comes out to be 2 as 1 + 2 + 3 = 6.
Consider the first step in which the addition of B and 7 is giving A (i.e., 2).That means B should be a number such that the units digit of this addition is 2. It is only possible when B = 5 since 5 + 7 = 12.
The addition will be as follows:

\begin{array}{l}\text{2 5}\\ \underset{¯}{\text{+ 3 7}}\\ \underset{¯}{\text{6 2}}\end{array}

Hence, the values of A and B are 2 and 5 respectively.

\begin{array}{l}\text{5}\text{. A B}\\ \underset{¯}{\times \text{3}}\\ \underset{¯}{\text{C A B}}\end{array}

The multiplication of 3 and B gives a number whose ones digit is B again. This is possible only when B is 0 or 5.
Case 1: If B = 5
Multiplication of first step = 3 × 5 = 15
1 will be a carry for the next step.
We have, 3 × A + 1 = CA
This is not possible for any value of A.
Case 2: If B = 0
If B = 0, then there will be no carry for the next step.
We should obtain, 3 × A = CA
That is, the one’s digit of 3 × A should be A.
This is possible when A = 5 or 0.
Here, A cannot be 0 as AB is a two-digit number.
Therefore, A must be 5 only. The multiplication is as follows.

\begin{array}{l}\text{5 0}\\ \underset{¯}{\times \text{3}}\\ \underset{¯}{\text{1 5 0}}\end{array}

Hence, the values of A, B, and C are 5, 0, and 1 respectively.

\begin{array}{l}\text{6}\text{. A B}\\ \underset{¯}{\times \text{5}}\\ \underset{¯}{\text{C A B}}\end{array}

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.
Case 1: If B = 5
Then, B × 5 = 5 × 5 = 25
Thus, 2 will be a carry for the next step.
We have, 5 × A + 2 = CA, which is possible for A = 2 or 7
Again we have two cases.
The multiplication is as follows.

\begin{array}{l}\text{2 5}\\ \underset{¯}{\times \text{5}}\\ \underset{¯}{\text{1 2 5}}\end{array}

\begin{array}{l}\text{7 5}\\ \underset{¯}{\times \text{5}}\\ \underset{¯}{\text{3 7 5}}\end{array}

Case 2: 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{5 0}\\ \underset{¯}{\times \text{5}}\\ \underset{¯}{\text{2 5 0}}\end{array}

Hence, there are 3 possible values of A, B, and C.
(i) 5, 0, and 2 (ii) 2, 5, and 1 (iii) 7, 5, and 3

\begin{array}{l}\text{7}\text{. A B}\\ \underset{¯}{\times \text{6}}\\ \underset{¯}{\text{B B B}}\end{array}

The multiplication of 6 and B gives a number whose one’s digit is B again.
It is possible only when B = 0, 2, 4, 6, or 8
Case 1: If B = 0
Then the product will be 0. Therefore, this value of B is not possible.
Case 2: If B = 2
If B = 2, then B × 6 = 12 and 1 will be a carry for the next step.
6A + 1 = BB = 22 ⇒ 6A = 21 which is not possible.
Case 3: If B = 6
If B = 6, then B × 6 = 36 and 3 will be a carry for the next step.
6A + 3 = BB = 66 ⇒ 6A = 63 .Therefore the value of A is not possible.
Case 4: If B = 8
Then B × 6 = 48 and 4 will be a carry for the next step.
6A + 4 = BB = 88 ⇒ 6A = 84 and hence, A = 14, but A is a single digit number. Therefore, this value of A is not possible.
Case 5: If B = 4
If B = 4, then B × 6 = 24 and 2 will be a carry for the next step.
6A + 2 = BB = 44 ⇒ 6A = 42 and hence, A = 7
The multiplication is as follows.

\begin{array}{l}\text{7 4}\\ \underset{¯}{\times \text{6}}\\ \underset{¯}{\text{4 4 4}}\end{array}

Hence, the values of A and B are 7 and 4 respectively.

\begin{array}{l}\text{8}\text{. A 1}\\ \underset{¯}{\text{+ 1 B}}\\ \underset{¯}{\text{B 0}}\end{array}

The addition of 1 and B is giving 0 i.e., a number whose ones digits is 0. This is possible only when digit B is 9. In that case, the addition of 1 and B will give 10 and thus, 1 will be the carry for the next step. In the next step,
1 + A + 1 = B
Clearly, A is 7 as 1 + 7 + 1 = 9 = B
Therefore, the addition is as follows.

\begin{array}{l}\text{7 1}\\ \underset{¯}{\text{+ 1 9}}\\ \underset{¯}{\text{9 0}}\end{array}

Hence, the values of A and B are 7 and 9 respectively.

\begin{array}{l}\text{9}\text{. 2 A B}\\ \underset{¯}{\text{+ A B 1}}\\ \underset{¯}{\text{B 1 8}}\end{array}

The addition of B and 1 is 8. This is possible only when digit B is 7. In the next step,
A + B = 1
Clearly, A is 4 since, 4 + 7 = 11 and 1 will be a carry for the next step.
In the next step,
1 + 2 + A = B
1 + 2 + 4 = 7
Therefore, the addition is as follows.

\begin{array}{l}\text{2 4 7}\\ \underset{¯}{\text{+ 4 7 1}}\\ \underset{¯}{\text{7 1 8}}\end{array}

Hence, the values of A and B are 4 and 7 respectively.

\begin{array}{l}\text{10}\text{. 1 2 A}\\ \underset{¯}{\text{+ 6 A B}}\\ \underset{¯}{\text{A 0 9}}\end{array}

Since both A and B are unknown to us in the first step, so let’s find the value of A from the next step.
In the next step, 2 + A = 0
It is possible only when A = 8
If we put the value of A = 8 in the first step, we get the value of B as 1.
In the second step, 2 + 8 = 10 and 1 will be the carry for the next step.
1 + 1 + 6 = A
Therefore, the addition is as follows.

\begin{array}{l}\text{1 2 8}\\ \underset{¯}{\text{+ 6 8 1}}\\ \underset{¯}{\text{8 0 9}}\end{array}

Hence, the values of A and B are 8 and 1 respectively.