Question

The sum of two $2$-digit numbers is $132$. If their $HCF$ is $11$, the numbers are

• A. $55,77$
• B. $44,88$
• C. $33,99$
• D. $22,110$

Answer 1

Answer: (A)

$55,77$

Let the two numbers be $x$ and $y$ respectively.

It is given that the sum of the two numbers is $132$, therefore,

$x+y=132$

Also $11$ is their HCF, thus both numbers must be divisible by $11$.

So let $x=11a$ and $y=11b$, then

$11a+11b=132\Rightarrow 11(a+b)=132\Rightarrow a+b=\frac{132}{11}\Rightarrow a+b=12$

Therefore, required possible pair of values of $x$ and $y$ which are prime to each other are $(1,11)$ and $(5,7)$.

Hence, the required numbers are $(11,121)$ and $(55,77)$.