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 1 of 10.
Vincent said:
2 decades ago
How did we get 5x = 20 ?
Martin said:
2 decades ago
x+3+x+6+x+9+x+12=30
5x+30=50
5x=50-30
5x=20
x=4.
5x+30=50
5x=50-30
5x=20
x=4.
(1)
Priya said:
2 decades ago
How to form this step : x+3+x+6+x+9+x+12 ?
Gyan said:
2 decades ago
Jus Add x+0
X+3
X+6
X+9
X+12
X+3
X+6
X+9
X+12
Cherry said:
2 decades ago
x+3x+6x+9x+12x=50
x(1+3+6+9+12)=50
x=50/31
x=1.64
If we do the same like this what is that x mean?
x(1+3+6+9+12)=50
x=50/31
x=1.64
If we do the same like this what is that x mean?
Mani Ramani said:
1 decade ago
Hi cherry..
you just add all the X first to get 5X then add numbers to get 30.
Then add 5x+30=50
5X=50-30(shift 30 to right side)
5x=20
X=20/5
X=4
you just add all the X first to get 5X then add numbers to get 30.
Then add 5x+30=50
5X=50-30(shift 30 to right side)
5x=20
X=20/5
X=4
Nandish said:
1 decade ago
x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
how 50 come
how 50 come
Prasad Baanu said:
1 decade 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
NANDAN said:
1 decade ago
If 1st child age x then 3year gap for next child so x+3
5 children = 5x
5x+30=50
5x=50-30
5x=20
x=20/5
x=4
so 4 years
5 children = 5x
5x+30=50
5x=50-30
5x=20
x=20/5
x=4
so 4 years
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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