Question

If the digits of my present age are reversed then i get the age of my son.If 1 year ago my age was twice as that of my son.Find my present age. Ans. Father-73, son-37

Answer 1

Hi there!
Here's the answer:

Let Units digit in present age of father be x.
and tens digit in present age of father be y.

•°• The father's age= 10y+x

If the digits are interchanged, we get son's age
=> Son's age= 10x+y


One year ago,
Father's age= (10y+x)-1
Son's age = (10x+y)-1

Given,
Father's age= 2(Son's age)

(10y+x)-1= 2(10x+y-1)
=> 10y+x-1 = 20x+2y-2
=> 8y= 19x-1
=> y=(19x-1)/8

Now, Check for which value of x, 19x-1 is divisible by 8

1- Not divisible -19(1)-1= 18
2- Not divisible - 19(2)-1= 37
3- Not divisible - 19(3)-1= 56. (¶¶¶)

•°• When x= 3, 19x-1 is divisible by 8

°°° x= 3 & y= (19(3)-1)/8 = 56/8 = 7

•°• Father's present age = 73
and Son's present age = 37

:)
Hope it helps

Answer 2

Answer:

The age of the son is 37 and the age of the father is 73.

Step-by-step explanation:

Given

• If the digits of my present age are reversed then I get the age of my son.
• If 1 year ago my age was twice that of my son.

To find

The present age of son and father.

Step 1

Let Units digit in the current age of the father be x.

and tens digit in the present age of the father is y.

My age is $10 x+y$

Son age is  $10 y+x$

so $10 x+y=10 y+x$

$10 x+y-1=2(10 y+x-1)$

$x=(19 y-1) / 8$

for$y=3$

$(19 y-1)$ is completely divisible by 8 $50 x=7$

So my present age :$10 \div 7+3$

$=73$

son's age : $10* 3+7$

$=37$

#SPJ3