Question

The sum of a two digit number and the number obtained by reversing its digits is 121. Find the number if its units place digit is greater than the tens place digit by 7.

Answer 1

Tip: Recall the concept of solving algebraic equations

Explanation:

Let the digit in the tens place be $x$ and that in the ones place be $y$

So the two-digit number can be written as $10x+y$

If the digits are reversed, it can be written as $10y+x$

Given that,

$(10x+y)+(10y+x)=121$

$\implies 11x+11y=121$

Dividing by $11$

$\implies x+y=11$    .......(1)

Also,  the units place digit is greater than the tens place digit by $7$.

$\implies y=x+7$     .....(2)

Substituting (2) in (1)

$\implies x+x+7=11$

$\implies 2x+7=11$

$\implies 2x=11-7$

$\implies 2x=4$

$\implies x=\frac{4}{2}$

$\implies x=2$

Substituting the value in (2)

$\implies y=2+7$

$\implies 9$

So the two-digit number is $10(2)+9=29$