Question

The sum of a two digit number and the number obtained by reversing the order of its digits is $121$, and the two digits differ by $3$. Find the number.

• A. $47$
• B. $74$
• C. $44$
• D. $34$

Answer 1

Answer: (A), (B)

The correct options are
A $47$
B $74$

Let the digit in the unit's place be $x$ and the digit at the ten's place be $y$. Then,

Number $=10y+x$

The number obtained by reversing the order of the digits is $10x+y$.

According to the given conditions, we have
$(10y+x)+(10x+y)=121$

$\Rightarrow 11(x+y)=121$

$\Rightarrow x+y=11$

and, $x-y=\pm 3$ [ $\because$ Difference of digits is $3$]

Thus, we have the following sets of simultaneous equations and,

$\begin{array}{cc}x+y=11& \left(i\right)\\ x-y=3& \left(ii\right)\end{array}$ or $\{\begin{array}{cc}x+y=11& .\left(iii\right)\\ x-y=-3& \left(iv\right)\end{array}$

On adding equation (i) and (ii), we get,

$2x=14,x=7$, Putting this in (ii)

$y=7-3=4$

$x=7,y=4$

On adding equations (iii) and (iv), we get

$2x=8,x=4$,

Putting this in (iv)

$y=4+3=7$

$x=4,y=7$

When $x=7,y=4$, we have

Number $=10y+x=10\times 4+7=47$

When $x=4,y=7$, we have

Number $=10y+x=10\times 7+4=74$

Hence, the required number is either $47$ or $74$.