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 9 of 10.
Neeraj dixit said:
1 decade ago
Just go through answer select 4 first will be 4 than 7 than 10, 13, 16 add all five you will get 50.
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
Sangeetha said:
1 decade ago
Thank you ishu subham, very simple and good logic.
Ishu Subham said:
1 decade ago
Let the ages be (x-6), (x-3), (x), (x+3), (x+6).
then sum of ages = 5x
=> 5x = 50
x = 10
youngest = x-6
=4
then sum of ages = 5x
=> 5x = 50
x = 10
youngest = x-6
=4
Gkgokulan said:
1 decade ago
Its simple logic frnds 4+7+10+13+16=50.
Then the answer is 4.
Then the answer is 4.
Satheesh said:
1 decade ago
It seems these calculation are merely mathematical wise not logic wise. Whats the relation between X and younger child?
Stella said:
1 decade ago
Hi cherry, you just see your calculations: x,3x,6x,9x,12x.(wrong)
u have to add x,x+3,x+6,x+9,x+12.(this is the correct format)
now u have to add and get the result ya.
u have to add x,x+3,x+6,x+9,x+12.(this is the correct format)
now u have to add and get the result ya.
Jinju said:
1 decade 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.
Jitu said:
1 decade ago
How can I submitt a question?
Raj said:
1 decade ago
Use arithmetic prog... formula
s=n/2[2a+(n-1)d]
where s=total sum of numbers,n-count the numbers ,d-diff&a-intial no. so they ask intial no i.e a?
s=50,n=5,d=3,a?
50=5/2[2a+12]=>a=4...
s=n/2[2a+(n-1)d]
where s=total sum of numbers,n-count the numbers ,d-diff&a-intial no. so they ask intial no i.e a?
s=50,n=5,d=3,a?
50=5/2[2a+12]=>a=4...
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