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
A. 21 years
B. 18 years
C. 15 years
D. 9 years
E. 12 years
Answer: Option D
Solution(By Examveda Team)
Sum of ages of three boys = 45 yearsNow, (3x + 5x + 7x) = 45
⇒ 15x = 45
⇒ x = 3
So, age of youngest boy = 3x
= 3 × 3
= 9 years
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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