Question

the sum of the digit of a two digit number is12 if the digits are reversed the new number becomes 4/7 times the original .find the original number

Answer 1

x+y = 12

10y+x = 4/7 (10x+y)

70y+7x = 40x + 4y

66y = 33x

2y = x

x+y = 12 replacing value of x from above

2y+y = 12

3y = 12

y = 4

x+y = 12 placing the value of y from above

x+4 = 12

x = 8

the number is 84

Answer 2

Let,
The original number be 10x+y
When,the digits are reserved the new number be 10x+y

Given,
x+y = 12

A.T.Q,
10y+x = 4/7 (10x+y)
=> 7(10y+x) = 4(10x+y)
=> 70y+ 7x = 40x+4y
=> 70y-4y = 40x-7x
=> 66y = 33x
=> y = 33x/66
=> y = 1/2x
=> 2y = x
=> x= 2y

Now,
(1) => x+ y = 12
=> 2y+y = 12
=> 3y = 12
=> y = 12/3
=> y = 4

So,(2) => x = 2y
=> x = 2×4
=> x = 8

Hence,the original number is = 10x+y
= 10×8+4
= 80+4
= 84