Question

A certain two-digit number whose digits when added give a sum of 9. If the digits are reversed, a new number is formed, which when decreased by 9 is equal to 4 times the original number. Find the original number.

• A. 12
• B. 14
• C. 16
• D. 18
• E. 20

Answer 1

The correct option is D

18

Let the digit in tens place be x and digit in unit’s place be y.

Then, the number is 10x + y and sum of digits is x + y.

When the digits are reversed, the new number is 10y + x .

x+y=9.........................(i)

10y+x-9=4(10x+y)

$\Rightarrow$ x-40x+10y-4y=9

$\Rightarrow$ -39x+6y=9

$\Rightarrow$39x-6y=-9

$\Rightarrow$13x-2y=-3 .........................(ii)

From (i), x+y=9

x=9-y

On substituting x=9-y in (i), we get

13(9-x)-2y=-3

$\Rightarrow$117-13y-2y=-3

$\Rightarrow$-15y=-3-117

$\Rightarrow$-15y=-120

$\Rightarrow y=\frac{-120}{-15}=8$

On substituting y=8 in (i), we get

x+8=9

$\Rightarrow$ x=9-8=1

$\therefore$ The required original number is

10x+y=10 $\times$1+8

=10+8=18