Aptitude - Problems on Ages - Discussion
Discussion Forum : Problems on Ages - General Questions (Q.No. 2)
2.
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Answer: Option
Explanation:
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20
x = 4.
Age of the youngest child = x = 4 years.
Discussion:
102 comments Page 5 of 11.
Priyanka said:
1 decade ago
@Sylivia.
Look the options and add 3 to the 1st option and we can find it out i.e., 4+3=7.
Then 7+3=10 actually dat is de middle age: 10+3=13, 13+3=16.
Good luck.
Look the options and add 3 to the 1st option and we can find it out i.e., 4+3=7.
Then 7+3=10 actually dat is de middle age: 10+3=13, 13+3=16.
Good luck.
Kartikey sharma said:
1 month ago
Now let’s use your formula:
Youngest = (S-d⋅(n(n-1))/2)/n.
Where:
S = 50.
d = 3.
n = 5.
(50-3⋅(5(4))/2)/5.
(50-3⋅10)/5.
(50-30)/5.
20/5 = 4.
So, the answer is 4.
Youngest = (S-d⋅(n(n-1))/2)/n.
Where:
S = 50.
d = 3.
n = 5.
(50-3⋅(5(4))/2)/5.
(50-3⋅10)/5.
(50-30)/5.
20/5 = 4.
So, the answer is 4.
(1)
Durgesh said:
1 decade ago
Avg. of age of 5 children = 50/5 = 10
Now there are 5 no. having sum 50 and avg. as 10 so no. can only be
= (10-6),(10=3),(10),(10+3),(10+6)
=4 , 7 ,10 ,13 ,16
Now there are 5 no. having sum 50 and avg. as 10 so no. can only be
= (10-6),(10=3),(10),(10+3),(10+6)
=4 , 7 ,10 ,13 ,16
Mitali said:
1 decade ago
I have a other way to solve it.
Avg of 50 = 50/5 = 10.
Hence, the middle No. In the series i.e. x+6=10 (middle no is the avg value of the series).
So x = 10-6= 4.
Avg of 50 = 50/5 = 10.
Hence, the middle No. In the series i.e. x+6=10 (middle no is the avg value of the series).
So x = 10-6= 4.
Vineshgujjari said:
1 decade ago
Since given 5 children=50
let one children be x
given interval 3 each
so x+(x+3)+(x+6)+(x+9)+(x+12)=50
5x+30=50
x=20/5
x=4..
let one children be x
given interval 3 each
so x+(x+3)+(x+6)+(x+9)+(x+12)=50
5x+30=50
x=20/5
x=4..
PRATEESH K said:
3 years ago
sum = 50
x,x+3,x+6,x+9,x+12 => eq1.
eq 1 is in arithmetic mean so the average is x+6
average of 5 children is 50/5 = 10,
So, x+6 = 10,
x = 4.
x,x+3,x+6,x+9,x+12 => eq1.
eq 1 is in arithmetic mean so the average is x+6
average of 5 children is 50/5 = 10,
So, x+6 = 10,
x = 4.
(14)
Prasad Baanu said:
2 decades ago
Hai Nandish..., They have already gave in the question that the sum of ages of those boys are 50:-)
so x + (x+3)+ (x+6)+ (x+9)+(x+12) = 50
so x + (x+3)+ (x+6)+ (x+9)+(x+12) = 50
Vishal Joseph leo said:
2 weeks ago
Let 5 children be born 3 years apart.
x + x + 3x + x + 6 + x + 9 + x + 12 = 50yrs.
5x + 30 = 50yrs,
5x = 50 - 30,
5x = 20.
And x = 4yrs.
x + x + 3x + x + 6 + x + 9 + x + 12 = 50yrs.
5x + 30 = 50yrs,
5x = 50 - 30,
5x = 20.
And x = 4yrs.
Jinju said:
2 decades ago
We can easily understand if the avg is 8 then youngestg should be less thgan 8. Then after we calculate the validity of remaining values.
Akhil said:
8 years ago
According to me;
The youngest children age = x+12.
= 4+12.
= 16.
The youngest children age = x+12.
= 4+12.
= 16.
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