Aptitude - Problems on Ages - Discussion
Discussion Forum : Problems on Ages - General Questions (Q.No. 1)
1.
Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?
Answer: Option
Explanation:
Let Ronit's present age be x years. Then, father's present age =(x + 3x) years = 4x years.
 |
(4x + 8) = | 5 | (x + 8) |
| 2 |
8x + 16 = 5x + 40
3x = 24
x = 8.
| Hence, required ratio = | (4x + 16) | = | 48 | = 2. |
| (x + 16) | 24 |
Discussion:
326 comments Page 5 of 33.
Diksha said:
4 years ago
If we will take Rohit's age I think it should be 3x (x+3).
If we will take Rohit's age as 1 it can be 4 only.
Correct me, If I am wrong.
If we will take Rohit's age as 1 it can be 4 only.
Correct me, If I am wrong.
(7)
Hamza said:
4 years ago
@All.
Please explain, father is two and a half mean 5/2. Then why we write 5/2 With son's age?
Because when any question ask father is 5 times of son then we put father is 5x and son is x.
Anyone, please explain this.
Please explain, father is two and a half mean 5/2. Then why we write 5/2 With son's age?
Because when any question ask father is 5 times of son then we put father is 5x and son is x.
Anyone, please explain this.
Chinnu said:
4 years ago
Anyone, Can you explain this sum by taking 2.5?
Chinnu said:
4 years ago
If we take 2.5 or 5/2 will we get the same answer? Anyone explain.
Bibek said:
5 years ago
It should be 3x + 8 = 5/2 (x+8).
(7)
Bibek said:
5 years ago
It should be 3x + 8 = 5/2 (x+8).
(4)
Sunil said:
5 years ago
The right answer is 3.
At first it is;
1 : 4.
then;
2 : 5.
Here difference after 8 years is 1.
1 part = 8 years.
Again we increase 8 years;
3 : 6.
At first it is;
1 : 4.
then;
2 : 5.
Here difference after 8 years is 1.
1 part = 8 years.
Again we increase 8 years;
3 : 6.
(3)
Manoj said:
5 years ago
How (4x+16) / (x+16) is = 48/24? Explain please.
(2)
Gora4663 said:
5 years ago
Let son age = x
Father age = y.
X + 3x =y( acc to 1st statemnt) ---> eq1
5/2(X+8) = y+8
5x+40 = 2y+16
5x-2y=-24 ---> eq2.
Solve eq 1 and 2 we get;
X=8 then y = 24,
After 16 year,
Son age 24 & father age 48.
Therefore 2 times.
Father age = y.
X + 3x =y( acc to 1st statemnt) ---> eq1
5/2(X+8) = y+8
5x+40 = 2y+16
5x-2y=-24 ---> eq2.
Solve eq 1 and 2 we get;
X=8 then y = 24,
After 16 year,
Son age 24 & father age 48.
Therefore 2 times.
Sree said:
5 years ago
Let Rohit age=X.
By condition:Fathers age=X + 3X,
After 8 yrs,
Fathers age=5/2(X+8).
So, 4x+8=5/2(x+8).
8x+16=5x+40.
X=8.
Sub x=8 in 5/2(x+8).
Ans:40.
After 8 yrs,
40+8=48.
Sons age present=8,
After 16 yrs, age=24,
24*2=48.
So, the answer is 3 times.
Thank you
By condition:Fathers age=X + 3X,
After 8 yrs,
Fathers age=5/2(X+8).
So, 4x+8=5/2(x+8).
8x+16=5x+40.
X=8.
Sub x=8 in 5/2(x+8).
Ans:40.
After 8 yrs,
40+8=48.
Sons age present=8,
After 16 yrs, age=24,
24*2=48.
So, the answer is 3 times.
Thank you
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