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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?

Let father's age = D 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 son be of a years
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