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:
99 comments Page 2 of 10.
ISMAIL said:
9 years ago
If we consider the age of younger child x then after 3 years that child will (x + 3) again 3 years 2nd child born and he become (x + 6) like this up to date the child become (x + 12).
Then the age of younger brother should (x + 12) not only x.
Means answer should be (4 + 12). i.e =16.
Then why not getting 4 as the answer. Can anyone explain it?
Then the age of younger brother should (x + 12) not only x.
Means answer should be (4 + 12). i.e =16.
Then why not getting 4 as the answer. Can anyone explain it?
V Seddhartha said:
2 years ago
Let the age of the youngest child is a.
Then the ages of other children are a+3, a+6, a+6, a+9, a+12.
By this first term is a.
The Common difference is 3.
No of the terms is 5.
We know that sum of n terms in AP.
n/2[2a+(n-1)d].
5/2[2a+(4)d] = 50
5/2[2a+12] = 50
5[a+6] = 50
a+6 = 50/5
a+6 = 10
a = 4.
The Age of the youngest child is 4.
Then the ages of other children are a+3, a+6, a+6, a+9, a+12.
By this first term is a.
The Common difference is 3.
No of the terms is 5.
We know that sum of n terms in AP.
n/2[2a+(n-1)d].
5/2[2a+(4)d] = 50
5/2[2a+12] = 50
5[a+6] = 50
a+6 = 50/5
a+6 = 10
a = 4.
The Age of the youngest child is 4.
(1)
Yahaya said:
2 years ago
Let the ages of the 5 children be a, b, c, d, e.
Intervals of 3
i.e if a=3,
b= a+3,
c= a+3+3,
d=a+3+3+3 and e= a+3+3+3+3.
Therefore their sum becomes:
a+a+3+a+3+3+a+3+3+3+a+3+3+3+3 = 50
Then;
5a + 3(10 times) = 50,
5a + 30 = 50.
5a = 50 - 30.
5a =20.
Divide both sides by 5.
5a/5 = 20/5,
a = 4 years.
Intervals of 3
i.e if a=3,
b= a+3,
c= a+3+3,
d=a+3+3+3 and e= a+3+3+3+3.
Therefore their sum becomes:
a+a+3+a+3+3+a+3+3+3+a+3+3+3+3 = 50
Then;
5a + 3(10 times) = 50,
5a + 30 = 50.
5a = 50 - 30.
5a =20.
Divide both sides by 5.
5a/5 = 20/5,
a = 4 years.
(8)
Rags said:
1 decade ago
Guys its very simple.
5 children s sum is 50;
So 50/5=10;
Will get 10 as the middle number, it means the age of 3rd child is 10yrs, so each child born 3 yrs diff, so 10 is middle means,
4 (+3yr gap) 7 (+3yr gap) 10 (middle child(+3yr gap)) 13 (+3yr gap) 16...
So youngest child age is 4.
5 children s sum is 50;
So 50/5=10;
Will get 10 as the middle number, it means the age of 3rd child is 10yrs, so each child born 3 yrs diff, so 10 is middle means,
4 (+3yr gap) 7 (+3yr gap) 10 (middle child(+3yr gap)) 13 (+3yr gap) 16...
So youngest child age is 4.
Gopinath said:
3 years ago
First, you notice every 5 children birth in 3 years cap in 50 years what's is the age in the youngest child.
X, (x+3), (x+6), (x+9), (x+12) years.
x+(x+3)+(x+6)+(x+9)+(x+12) = 50.
Then, here it's using the method in adding x in out 5X +30 = 50.
5X= 50-30.
5X= 20, divisible by 5,
X= 4.
X, (x+3), (x+6), (x+9), (x+12) years.
x+(x+3)+(x+6)+(x+9)+(x+12) = 50.
Then, here it's using the method in adding x in out 5X +30 = 50.
5X= 50-30.
5X= 20, divisible by 5,
X= 4.
(29)
Lucky said:
9 years ago
@Nikil,
The sum of ages of 5 children and their difference is 3 years.
We consider that 1st children age be "x".
Then the second one be "x + 3".
3rd one be"x + 6".
4th one be "x + 9".
5th one be "x + 12" (because they all are born with three years difference).
The sum of ages of 5 children and their difference is 3 years.
We consider that 1st children age be "x".
Then the second one be "x + 3".
3rd one be"x + 6".
4th one be "x + 9".
5th one be "x + 12" (because they all are born with three years difference).
Vikram said:
7 years ago
Divide the sum of the ages by total no of children 50/5=10.
Now as we know that the age varies by 3 years...so take 10+3 &10-3=13&7.
then again take the age and sub and add 3. i.e; 13+3&7-3=16&4,
Now add all the operations 4 + 7 + 10 + 13 + 16 = 50.
Now as we know that the age varies by 3 years...so take 10+3 &10-3=13&7.
then again take the age and sub and add 3. i.e; 13+3&7-3=16&4,
Now add all the operations 4 + 7 + 10 + 13 + 16 = 50.
Abhishek said:
2 years ago
Short trick.
5 children born at interval of 3 years in 50 years.
1st child. 1+3 = 4 year.
2nd child. 2+3 = 5 year.
3rd child. 3+3 = 6 year.
4th child. 4+3 = 7 year.
5th child 5+3 = 8 year.
=>who is youngest child of all above this?
=> Ans= 4 years.
5 children born at interval of 3 years in 50 years.
1st child. 1+3 = 4 year.
2nd child. 2+3 = 5 year.
3rd child. 3+3 = 6 year.
4th child. 4+3 = 7 year.
5th child 5+3 = 8 year.
=>who is youngest child of all above this?
=> Ans= 4 years.
(84)
Hariharan said:
5 years ago
@Sonam.
let youngest son Tashi age be x and father's age be y age of Dema be a, And the age of Sangay be b.
y=x+30.
a=x+2.
b=a+3.
a+b+x=52.
x+2+a+3+x=52.
x+2+x+2+3+x=52.
3x+7=52.
3x=45.
x=15.
y=15+30.
y=45.
The age of the father is 45.
let youngest son Tashi age be x and father's age be y age of Dema be a, And the age of Sangay be b.
y=x+30.
a=x+2.
b=a+3.
a+b+x=52.
x+2+a+3+x=52.
x+2+x+2+3+x=52.
3x+7=52.
3x=45.
x=15.
y=15+30.
y=45.
The age of the father is 45.
(1)
Sonam said:
6 years ago
Can anyone help me to solve this?
A Father's age is 30 years older than the youngest son Tashi. Dema is 2 years older than Tashi. Sangay is 3 years older than Dema. The sum of the ages of the the children are 52. How old is the father?
A Father's age is 30 years older than the youngest son Tashi. Dema is 2 years older than Tashi. Sangay is 3 years older than Dema. The sum of the ages of the the children are 52. How old is the father?
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