Q6 of 38 Page 190

Raju wrote a number having 2 digit where the digit in the ten’s position is thrice the digit in the unit position and the new number obtained by reversing the positions of the digits is 36 less than the initial number. Let’s find the number writer by Raju.

Let the digit in the unit’s place be x


Then given the digit in the ten’s place is thrice the digit in the unit position = 3x


So the 2 digit number is 10(3x)+x………(i)


By reversing the position of the digits we get the new number as


10(x)+(3x)……..(ii)


Now according to the given criteria,


36 less than the initial number = 10(3x)+x-36=31x-36…..(iii)


So from question equation (ii) and equation (iii) are equal,


Hence


10(x)+(3x)= 31x-36


⇒ 13x=31x-36


⇒ 36=31x-13x


⇒ 18x=36



From equation (i), the 2 digit number is


10(3x)+x=10(3×2)+2=10(6)+2=62


Hence the 2 digit number written by Raju is 62


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