# A man has 2 daughters and 1 son. the sum of ages of children is equal to the age of father . in 15 years the sum of ages of his children will be one and a half times their father's age then. what is the father's age now?

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let son's age = A

1st daughter's age= B

2nd. ,, ,, = C

A.T.Q.

A + B + C = D...... (i)

ages after 15 years...

father's age = 3/2D

son's age = A+ 15 daughter's age= B+15

2nd daughter ,,= C+ 15

acco. to condition.

A+15+B+15+C+15=3/2 D

A+B+C+45 = 3/2D

D + 45= 3/2 D

(....A+B+C=D)

2 (D+ 45) = 3D

2D + 90 = 3D

3D - 2D = 90

D = 90.

so age of father is 90..

Let the daughter be b years

Let the 2nd daughter be c years

And father be x years

a+b+c=d

>a+b+c+45=1½*d+15

45-15=30 30 means half times father's age .

30*3=90

Father's age is 90-15=75years i.e 15years back