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 3 of 10.
Shawon said:
9 years ago
Can anybody solve the problem? Please.
Three times the present age of a father is equal to eight times the present of his son. Eight years hence the father will twice as old as his son at that time. What are their present age?
Three times the present age of a father is equal to eight times the present of his son. Eight years hence the father will twice as old as his son at that time. What are their present age?
SIDDHARTHA ROY NANDI said:
1 decade ago
The Problem can be solved using Arithmetic Progression:
S = n/2{2a+(n-1)d}
a = youngest age(We have to calculate this)
n = Total no. of people(5)
d = Difference of Ages in the progression(3)
S = Summation of their ages(50)
S = n/2{2a+(n-1)d}
a = youngest age(We have to calculate this)
n = Total no. of people(5)
d = Difference of Ages in the progression(3)
S = Summation of their ages(50)
Mkt said:
1 year ago
it is in arithmetic progression with a common difference of 3.
So,
Sn = n/2 [2a + (n-1) d],
50 = 5/2 [2a + (5-1)3],
10 = 1/2[2a + 12],
20 = 2a + 12,
8 = 2a.
a = 4.
Where a is the first term or the age of the youngest child.
So,
Sn = n/2 [2a + (n-1) d],
50 = 5/2 [2a + (5-1)3],
10 = 1/2[2a + 12],
20 = 2a + 12,
8 = 2a.
a = 4.
Where a is the first term or the age of the youngest child.
(8)
Pramod ( nepal) said:
8 years ago
@Pooja,
According to your Q 1st eq x+ (x+5) + (x+10) but if after 5 years.
The sum of there age 60 the new eq become (x-5) +x+ (x+5) = 60.
After solving x=20.
Youngest bro age is 15 but 5 years ago his age is 10.
According to your Q 1st eq x+ (x+5) + (x+10) but if after 5 years.
The sum of there age 60 the new eq become (x-5) +x+ (x+5) = 60.
After solving x=20.
Youngest bro age is 15 but 5 years ago his age is 10.
Rgauri said:
1 decade ago
Let us assume that:
a, b, c, d, e are five childs to him.
Let,
a = x;
b = x+3;
c = x+6;
d = x+9;
e = x+12;
Then,
x+(x+3)+(x+6)+(x+9)+(x+12) = 50;
Therefore,
5x = 20;
x = 4.
So,
a = 4.
Youngest one.
a, b, c, d, e are five childs to him.
Let,
a = x;
b = x+3;
c = x+6;
d = x+9;
e = x+12;
Then,
x+(x+3)+(x+6)+(x+9)+(x+12) = 50;
Therefore,
5x = 20;
x = 4.
So,
a = 4.
Youngest one.
Malak said:
7 years ago
The sum of age of 3 sisters now is 25 years. If the oldest was born before the middle by 3 years and the middle was born before the youngest by two years. Find the age of each of them.
Please, anyone, answer it.
Please, anyone, answer it.
Suriya khumar said:
8 years ago
HI guys.
First see the option.
In the question, they said that difference between the children is 3 years, and the total age is 50 years.
Hence,
Option A.
4+7+10+13+16 = 50 thats all.
Have a nice day!
First see the option.
In the question, they said that difference between the children is 3 years, and the total age is 50 years.
Hence,
Option A.
4+7+10+13+16 = 50 thats all.
Have a nice day!
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...
Pooja said:
8 years ago
A man had three sons and they were born at an interval of five years. The sum of their ages after 5years will be 60 years. What was the age of youngest son five years ago?
Can anyone solve this?
Can anyone solve this?
Manish said:
10 years ago
Take the ages as x-6, x-3, x, x+3, x+6.
Now add all, (x-6)+(x-3)+(x)+(x+3)+(x+6) = 50.
All numbers will be cancelled and you will get:
5x = 50.
x = 10.
Youngest one will be x-6 = 10-6 = 4.
Now add all, (x-6)+(x-3)+(x)+(x+3)+(x+6) = 50.
All numbers will be cancelled and you will get:
5x = 50.
x = 10.
Youngest one will be x-6 = 10-6 = 4.
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