Question

A number consists of two digits. When it is divided by the sum of the digits, the quotient is 7. If 27 is subtracted from the number, the digits are reversed. Find the number.

Answer 1

Let the number on the ten's place be 'x' nad unit place be 'y'

So when it is divided by the sum of the digit, the quotient is  7 :-

10x + y / x + y = 7 so,
3x - 6y = 0 ----------( equation 1)

Then, If 27 is sub from the no the digits are reversed:

10 x + y - 27 = 10 y + x
x - y = 27 ------------( equation 2)

Multiply Eq (1) and Eq (2), we get,

3x - 3y = 9-------------( 3)

Sub Eq (1) and (), we get

3x - 6y  = 0 - ( 3x - 3y = 9)
= 3y = 9
y = 9/3
= 3
( y = 3)-------( 4)

Put (4) in (2)
 
x - 3 = 27
x = 6

( x = 6)

So the number is 63





 

Answer 2

Let x is the tenth digit and y is unit digit.
then the number is 10x+y.

When it is divided by sum of the digits, quotient is 7

thus $\frac{10x+y}{x+y} =7$

⇒$10x+y=7x+7y$
⇒$10x-7x=7y-y$
⇒$3x=6y$
⇒$x=2y$

When 27 is subtracted from it, digits are reversed(y comes to tenth place and x goes to unit place), thus

$(10x+y)-27=10y+x$
⇒$10x-x=10y-y+27$
⇒$9x=9y+27$
⇒$x=y+3$     (dividing by 9)
⇒$2y=y+3$
⇒$2y-y=3$
⇒$y=3$
⇒$x=2y=2*3=6$

The number is 10x+y = 10*6+3 = 63