Question

The sum of a two-digit number and ths number obtained by reversing the digit is 110. If ones' digit is one-fourth of the tens' digit find the number.

Answer 1

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

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Hope this will help you...