Five years ago father's age was 2 times the sum of ages of his 3 children . after 5 years fathers age can be obtain by substracting 8 from the sum of the ages of the 3 children then father's present age is
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PRESENT AGE
Father=x
sum of the ages of his children= y
5 YEARS AGO
father= x-5
sum....=y-15 (coz there are three children)
it is given that 5 years ago fathers age was 2 times the sum
∵x-5=2(y-15)
x-5=2y-30
2y-x=25......................(eq 1)
AFTER 5 YEARS
FATHER= X+5
sum= y+15
it is given that fathers age can be obtained..........
∵x+5=y+15-8
x+5=y+7
x-y=2.........................(eq 2)
from eq 1 and 2
by elimination method
x=29
hence father is 29 years
Father=x
sum of the ages of his children= y
5 YEARS AGO
father= x-5
sum....=y-15 (coz there are three children)
it is given that 5 years ago fathers age was 2 times the sum
∵x-5=2(y-15)
x-5=2y-30
2y-x=25......................(eq 1)
AFTER 5 YEARS
FATHER= X+5
sum= y+15
it is given that fathers age can be obtained..........
∵x+5=y+15-8
x+5=y+7
x-y=2.........................(eq 2)
from eq 1 and 2
by elimination method
x=29
hence father is 29 years
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