The digit at the tens place of a two-digit number is three times the digit at the ones place. If the sum of this number and the number found by reversing the digit is 88. Find the number.

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Solution

let the digit at one's place =x
then the digit at ten's place = 3x
then original no. will be= 10(3x)+x
=31x
by reversing the no.
one's digit will be =3x
ten's digit will be= x
then no. will be =10x+3x
=13x
then according to the problem
10x+3x+x+30x=88
x=2
then original no will be 312=62

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The digit at the tens place of a two-digit number is three times the digit at the ones place. If the sum of this number and the number found by reversing the digit is 88. Find the number.

let the digit at one's place =x then the digit at ten's place = 3*x then original no. will be= 10*(3x)+x =31x by reversing the no. one's digit will be =3x ten's digit will be= x then no. will be =10*x+3x =13x then according to the problem 10x+3x+x+30x=88 x=2 then original no will be 31*2=62

let the digit at one's place =x
then the digit at ten's place = 3x
then original no. will be= 10(3x)+x
=31x
by reversing the no.
one's digit will be =3x
ten's digit will be= x
then no. will be =10x+3x
=13x
then according to the problem
10x+3x+x+30x=88
x=2
then original no will be 312=62