Question

The sum of two digits and the number formed by interchanging its digit is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the digits in the first number. Find the number

Answer 1

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 or first numbe = 10x + y  = 10×6+4 = 64

Hope it helps you

Answer 2

Answer:

Step-by-step explanation:

Let the unit digit be y & tens digit be x.

Original number = (10x+y)

After interchanging the digits

New number = (10y+x)

(10x+y) + (10y+x) = 110

11x +11y = 110

11(x+y)= 110

x+y = 110/11

x+y= 10…………...(1)

x= 10-y……………(2)

(10x+y) - 10 = 4+ 5(x+y)

(10x+y) - 10 = 4+ 5(10)

(10x+y) = 4+ 50+10

(10x+y) = 64

10(10-y) +y = 64

100-10y +y= 64

100 -9y = 64

-9y = 64-100

-9y = -36

y= 36/9= 4

y= 4

putting the value of y in eqn 2

x= 10-y

x= 10-4

x= 6

Hence , the first number is 6 & second number is 4.

Original Number is 10x+y = 10× 6+4= 60+4= 64