Question

The sum of digits of a two digit number is 11. If the digit at ten’s place is increased by 5 and the digit at unit place is decreased by 5, the digits of the number are found to be reversed. Find the original number.
A. 40
B. 33
C. 16
D. 38

Answer 1

Hint: Let the digit at ones place be $y$ and the digit at tens place be $x$, then $x + y = 11$. Again, use the given condition to find another equation. Then, solve both the equations simultaneously to find the value of $x$ and $y$.

Complete step-by-step answer:
We are given that the sum of digits is 11.
Let the digit at ones place be $y$ and the digit at tens place be $x$, then
 $x + y = 11$ eqn.(1)
And, the value of the number will be $10x + y$
Also, we are given that digit at ten’s place is increased by 5 and the digit at unit place is decreased by 5, the digits of the number are found to be reversed.
If we increase the tens place by 5, we will get, $x + 5$ and if we decrease unit place by 5, we will get, $y - 5$
The value of the number becomes,
$
  10\left( {x + 5} \right) + \left( {y - 5} \right) = 10x + 50 + y - 5 \\
   = 10x + y + 45 \\
$
This the reverse of the original number.
We can write the reverse of he original number by interchanging the values at tens and ones place.
The reverse of the original number is $10y + x$
Therefore, according to the given condition,
$10x + y + 45 = 10y + x$
On rearranging the above equation and bringing like terms together, we will get,
$9x - 9y = - 45$
Divide the equation throughout by 9
$x - y = - 5$ eqn. (2)
Solve equation (1) and equation (2) .
Add equation (1) and equation (2) to eliminate the value of $y$.
$
  x + y + x - y = 11 - 5 \\
   \Rightarrow 2x = 6 \\
   \Rightarrow x = 3 \\
$
Now, substitute the value of \[x = 3\] in equation (1) to find the value of $y$
$
  3 + y = 11 \\
   \Rightarrow y = 8 \\
$
Then, the original number is 38.
Hence, option D is correct.

Note: In any number, the rightmost digit is always the unit place and the digit left to it is tens place. That is if the number is 45, then 4 is at tens place and 5 is at one place. Also, we have solved two equations using elimination and substitution, but one can also use cross-multiplication and graphical methods.