Fathers age is three times the sum of ages of two children. After five years his age will be twice the sum of age of two children. Find the age of the father
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let the ages of two sons be 1x and 2x
the father's age be y
3(1x + 2x) = y
1x + 2x = y÷3
2(1x+2x + 10)=y+5
2÷3y +20=y-(2÷3)y
20-5=y-(2÷3)y
15=1÷3y
y=45
the father's age be y
3(1x + 2x) = y
1x + 2x = y÷3
2(1x+2x + 10)=y+5
2÷3y +20=y-(2÷3)y
20-5=y-(2÷3)y
15=1÷3y
y=45
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Let sum of age of two children be x
father age =3×X =3X
after 5 years,
children age =x+10
father age =2(x+10)/ 3x+5
2(x+10)=3x+5
2x+20=3x+5
2x-3x=5-20
-x= -15
x=15
father age =3×X =3X
after 5 years,
children age =x+10
father age =2(x+10)/ 3x+5
2(x+10)=3x+5
2x+20=3x+5
2x-3x=5-20
-x= -15
x=15
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