Question

The sum of two digit and the number formed by interchanging its digit is 110 .if ten is subtracted from the first number the new number is 4 more than 5 times of the sum of the digits on the girst first numbers . find the first number

Answer 1

Let the tens digit be x and the units digit be y.
So the number=10x+y
After interchanging the number
=10y+x
A.q.
(10x+y)+(10y+x)=110
10x+y+10y+x=110
11x+11y=110
11(x+y)=110
x+y=110/11
x+y=10
x=10-y (equation 1)
Also,
(10x+y)-10=4+5(x+y(
10x+y-10=4+5x+5y
10x-5x+y-5y=10+4
5x-4y=14
From equation 1 x=10-y
So,
5(10-y)-4y=14
50-5y-4y=14
9y=50-14
9y=36
y=36/9
y=4
And x=10-y
=10-4
=6
Hence the number is 64.

Answer 2

Here is your solution

Let the digit at ones place = y
Let the digit at tens place = x

Original number = 10x + y 

Number formed by reversing the digits =10y + x

On adding original number and Reversed number.

10x+y+10y+x = 110
11x + 11y = 110
11 (x + y ) = 110
x + y = 10
x = 10 - y ................................................. (1)

on subtracting 10 from the original number , we get = 10x + y -10
and


4 more than 5 times the sum of digits = 5(x + y) + 4

A/q
 
10x + y -10 = 5(x + y) + 4
10x + y -10 = 5x +5y + 4
10x -5x + y -5y = 10 +4
5x - 4y = 14.................................................. (2)


putting the value of x in equation (2)

5x - 4y = 14
5×(10-y) - 4y = 14
50-5y-4y =14
-9y = 14 -50
-9y = -36
y = 4


Now

x = 10-y
x = 10 - 4
x = 6


original number = 10x + y  = 10×6+4 = 64

Hope it helps you