Question

The digit at the tens place of a two digit number is four times the the digit at ones place if the sum of this number and the number formed by reversing the digits is 55 find the number

Answer 1

$\mathfrak{AnSwEr}$

Let the digits at tens and ones be x and y respectively

Equations formed :

x = 4y ...(1)

10x + y + 10y + x = 55 ...(2)

By further solving, the values you'll get are :

x = 4

y = 1

Number = 41

Answer 2

Solution:-

Let the digit at once place be "y" and digit at Tens place be "x".

=) Original Number = 10x + y.

Condition I,

x = 4y

Condition II,

( 10x + y) + ( 10y + x) = 55

=) 11x + 11y = 55

=) 11( x + y) = 5 × 11

=) x + y = 5

Substituting [ x = 4y ] here. we get,

=) 4y + y = 5

=) y = 1

Now, Substituting [ y = 1 ] in Condition I. we get,

=) x = 4×1

=) x = 4.

Hence,

The Number is ;

=) 10x + y

=) 10×4 + 1

=) 41.