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 The sum of two digits 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 in the first number. Find the first number.       
  • a)
      46       
  • b)
      48       
  • c)
      64       
  • d)
      84
Correct answer is 'C'. Can you explain this answer?
Verified Answer
The sum of two digits and the number formed by interchanging its digit...
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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Most Upvoted Answer
The sum of two digits and the number formed by interchanging its digit...
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 x 6+4= 60+4= 64

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