Question

if the unit digit is thrice the ten's digit of a two digit number and the new number formed by reversing the digit is greater than the original number by 54 , find the original number​

Answer 1

SOLUTION :-

Let,

Digit in ten's place = x

Digit in one's place = y

Two digit number = 10x + y

According to the first condition,

$\longrightarrow\sf y = 3x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(1)$

According to the second condition,

$\longrightarrow\sf 10x+y+54 = 10y+x\\\\\longrightarrow 10x-x+y-10y = -54 \\\\\longrightarrow 9x-9y = -54 \\\\\longrightarrow x-y = - 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(2)$

Substitute the value of y in equation (2),

$\longrightarrow\sf x-3x= -6 \\\\\longrightarrow -2x= -6 \\\\\longrightarrow \bf x = 3$

Digit in ten's place = 3

Digit in one's place = 3 × 3

                                =  9

Two digit number = 39