Question

a woman's son is 2years older than her daughter. her age is three times the sum of the ages of her children after 5 years she will be 47. find the present age of her children

Answer 1

Let the age of her son be x
Let the age of her daughter be y
Let her age be z
x=y+2········(1)
z=3(x+y)·····(2)
z+5=47·······(3)
⇒z=47-5
     =42
∴42=3(x+y)
substituting equation (1) in the above equation, we get 
42=3[(y+2)+y]
42=3(2y+2)
42/3=2y+2
14=2y+2
14-2=2y
12=2y
⇒y=6
from this we get that 
x=y+2
x=6+2
x=8
∴the present age of the son is 8 and the present age of the daughter is 6.


hope this helps:)

Answer 2

Solution:-

• The present age of the woman is 47 - 5 = 42.

• Since the woman's age is 3 times the sum of the ages of the children,

• This sum is 42/3 = 14.

• So for children ages you have two equations:

=> s + d = 14,

=> s - d = 2.

• Every time when you see two equations like these, add them.

• You will get

=> 2s = 14 + 2,

=> 2s = 16,

=> s = 16/2 = 8.

Thus son is 8 years now.

The daughter is 8 - 2 = 6 years.

i hope it helps you.