Question

The sum of the digits of a two-digit number is 15. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.

Answer 1

Hint: We are given the relation between two digits of a number in the given question so we have two unknown quantities. That’s why we assume the digits of the number to be any two different variables such that one of the digits comes first and the other comes later. Then using the given conditions, we can construct two different equations in terms of the two variables and can obtain their values by solving the equations.

Complete step-by-step answer:
Let, the digit at the left side of the number be X and the digit at the right side be Y. so the number can be written as $XY$ . The sum of the two digits is 15, so –
$X + Y = 15...(1)$
We know that the variable X is in tens place and Y is in ones place, then the number is –
$10X + Y$
After interchanging the digits of the number, the number can be written as $YX$ . So, the number after reversing the digits is $10Y + X$
The new number exceeds the original number by 9. So,
$
  10X + Y + 9 = 10Y + X \\
   \Rightarrow 9X - 9Y = - 9 \\
   \Rightarrow X - Y = - 1...(2) \;
 $
Solving the equations (1) and (2) by elimination method –
$
  X + Y = 15 \\
  \underline {X - Y = - 1} \\
  \underline {\,\,\,\,2X\,\,\, = 14} \\
   \Rightarrow X = \dfrac{{14}}{2} \\
   \Rightarrow X = 7 \;
 $
Put the obtained value of X in (1),
$
  7 + Y = 15 \\
   \Rightarrow Y = 8 \;
 $
Hence, the original number is 78.
So, the correct answer is “78”.

Note: The first digit of a number is called the ones place, the second digit is called the tens place, the third digit is called the hundreds place and so on. For example, a 2-digit number is 50, where 0 is in one place and 5 is in the tens place, this is how the number becomes $5 \times 10 + 0 = 50$ . The equations obtained involve two unknown quantities and we have two equations to find the value, so we solve the equations by the elimination method.