Question

Armys, Please Help me.
In a two digit number, digit at the tens
place is twice the digit at units place. If the
number obtained by interchanging the digits is
added to the original number the sum is 66.
Find
the
number. (it's equation in One variable). (please give answer in step by step)​

Answer 1

Let the number at tens digit be x and ones digit ne y.

Number = 10x + y { As the number is of two digit }

A.t.q,

x = 2y ---------( i )

The number obtained by interchanging the digits ( 10y + x ) is added to the original number ( 10x + y ) the sum is 66.

This can be represented in equation as :

=> 10y + x + 10x + y = 66

=> 11y + 11x = 66

=> 11( x + y) = 66

=> x + y = 66/11

=> x + y = 6 ------( ii )

Substituting value of x from ( i ) in (ii) , we get

=> x + y = 6

=> 2y + y = 6 { from (i) }

=> 3y = 6

=> y = 6/3 => y = 2

Substituting value of y in (ii)

=> x + y = 6

=> x + 2 = 6

=> x = 6 - 2 => x = 4

Therefore the number is 10×4 + 2 or 42

Answer : 42