Question

If $AB+7C=102$, where $B\ne 0$,$C\ne 0$, then$A+B+C=14$. State, whether the statement given, is true or false .

• A. True
• B. False

Answer 1

The correct option is A True

Find the value of $A,B,C$.

Given: $AB+7C=102$, (Here $\mathrm{AB}$ is a two digit number) and $A+B+C=14$

$AB+7C102$

Since $A,B,C$ are singal digit numbers, hence $A$ could be $3$ (no carry) or $2$ (when there is a carry after adding $B$ and $C$)

Since, $B+C$ has units digit $2$. and $A+B+C=14$

It shows that the value of $A$ can't be $3$, thus the value of $A$ is $2$.

From, $A+B+C=14$

$\Rightarrow$ $B+C=14-2$

$\Rightarrow$ $B+C=12$

We have more than one possible value for $B$ and $C$, i.e. $B=5,C=7$ or $B=7,C=5$ or $B=8,C=4$ etc.

Ex. $B+C=14-2$

Verifying the values of $A,B,C$:

$A=2,B=5andC=7$

$$\begin{gathered} AB+7C=102 \\ \Rightarrow 25+77=102 \end{gathered}$$

Hence, the given statement is true.