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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Answered by
11
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 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..
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Answered by
6
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
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
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