Question

In a two digit number, the digit at the ten's place is thrice the digit at the unit's place. If the number obtained by Interchanging the digits is added to the original number, the sum is 44. Find the number.

• A. 31
• B. 22
• C. 62
• D. 35

Answer 1

The correct option is A

31

Let the digit at the ten's place be x.

Let the digit at the unit's place by y.

Then, the number would be $10x+y.$ ...... (1)

The number obtained by interchanging x and y $=10y+x$ ............ (2)

The digit at the ten's place is thrice the digit at the units place.

$\therefore$ x = 3y.

put x = 3y in expression (1)

$10x+y$

put x = 3y

$30y+y=31y$ .... (3)

Put x = 3y in expression (2)

$10y+x$

Put $x=3y$

$10y+3y=13y$ .... (4)

Given that,

(Number) + (Reverse of the Number)

= 44

$(10x+y)+(10y+x)=44$

$10x+y+x+10y=44$

$11x+11y=44$

$x=3y$

$11x+11y=44$

$11\times \left(3y\right)+11y=44$

$33y+11y=44$

$44y=44$

Dividing by 44 on both sides,

$\frac{44y}{44}=\frac{44}{44}$

$y=1$

as, $x=3y$

$x=3\times 1,x=3$

Put x = 3 and y = 1 in (1)

The original number was 10x+y

$10\left(3\right)+1$

=31 $\to$ Number we require.

The number obtained by Interchanging x and y

$y=10y+x$

$=10\left(1\right)+3$

$=13$

Cross check:

Given that,

(Number) + (Number obtained by reversing the digits) = 44

(31) + (13) = 44

44 = 44

LHS = RHS.