The average age of three boys is 15 years. If their ages are in ratio 3 : 5 : 7, the age of the youngest boy is

3*15=45
3+5+7= 15
15=45
1u = 45/15
1u= 3
Youngest boy's 3*3=9 years

Total 3x15=45
Ratio 3+5+7=15
Find 1 part => 45/15 =3
3x3=9, 3x5=15,3x7=21
9 smallest

First we find sum?
A=15,N=3
A=s/n
S= 15*3
S=45
3x+5x+7x=45
15x=45
X=3
So, age of youngest boy = 3x
= 3 × 3
= 9 years

ratio 3:5:7 and 3+5+7=15 average =15 so 15(3)=45 so 3/15 *45/1= 3*3=9 for 5= 5/15*45/1= 5*3=15 similarly for 7 =7*3=21 (9+15+21)=45

Sum of ages of 3 boys =45 how 45??

3x+5x+7x=45
15x=45
X=3
3*3=9

3*15=45
(45/(3+5+7))*3=9

(x+y+z)/3=15
(3x+5x+7x)/3=15
15x/3=15
5x=15
x=15/5
x=3
age of youngest boy is 3 so.......
3*3=9
age of second boy is 5 so.....
3*5=15
age of third boy is 7 so......
3*7=21

Tanq

the age of the youngest boy = 3*15 of 3/(3+5+7)
= 9 years

the age of three boy are in this ration 3:4:5, if the difference between the ages of the oldest and the youngest is 18years find the sum of the age of the three boys

3x+2x+7x/3 = 15
x=15×3/15
x=3
Age of youngest boy= 3x=>3×3
=9

good