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?
2 times
1 times
2
3 times
4
3 times
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:
319 comments Page 2 of 32.

Lokesh said:   2 years ago
As per the question, Father's age is 3 times more than the sons age.
Not 3 times of son's age..
That's why here is 3x+x.
Sons present age is = x
Father present age is = x+3x= 4x
After 8 years father's age would be -
4x+8 = 5/2(x+8).

By solving we get x=8,
now son's present age is x=8.
Father's age is 4x = 4*8= 32.

The required ratio 32+16/8+16 = 48/24 = 2 ans.
(24)

Kamleshkumar said:   2 years ago
Say Sons age is X;

F = 3X
3X + 8 = 5/2(X+8).
6X + 16 = 5X+40.
X = 40 - 16,
X = 24.

So, 3X+16/X+16.
= 3 * 24+16/24+16.
= 88/40.
= 2.2.
Means -> 2 1/5.
(18)

Mani said:   2 years ago
In the question, son's age is 2 (1/2) age.

So, 2x1/2 is 5/2fraction form.
(18)

Barekye Samuel said:   10 months ago
The given answer is Incorrect, below is the right answer with a detailed explanation.

let's denote the son's current age as R and the father's current age as F.

Step 1: Setting up equations
From the problem: The father is three times older than the son.
F=3R
After 8 years, the father will be two and a half times older than Ronit. After 8 years, the father's age will be
F+8 and Ronit's age will be
R+8. According to the problem, after 8 years:
F+8=2.5(R+8)
Step 2: Solve the system of equations
From the first equation, substitute
F=3R into the second equation:
3R+8=2.5(R+8)
Simplifying the equation:
3R+8=2.5R+20
Now, subtract
2.5R from both sides:
0.5R+8=20
Subtract 8 from both sides:
0.5R=12
Now, divide by 0.5:
R=24
So, Ronit's current age is
R=24.

Finding the father's age:
Since
F=3R, we substitute
R=24:
F=3×24=72
So, the father's current age is
F=72.

Step 3: After 16 years
After 8 more years, the father's age will be
F+16 and Ronit's age will be
R+16.
The father's age after 16 years is:
F+16=72+16=88
Ronit's age after 16 years is:
R+16=24+16=40
Now, the question asks how many times the father's age will be compared to Ronit's age after further 8 years (16 years in total). So, the ratio of the father's age to Ronit's age after 16 years is:

88/40 =2.2

Thus, after 16 years, the father will be 2.2 times as old as Ronit.
(17)

Naveen said:   2 years ago
F:S
3:1 *3 = 9p:3p
5:2 *2 = 10p:4p
1p = 8years.
5 * 8 + 8 : 2 * 8 + 8.
48 : 24.

While comparing both father is 2 times more than the son.
(16)

Amey said:   2 years ago
@All.

Here, If x = 8 then 3 times more is 24 which is the father's age and after 8 more years x = 16 and the father's age will be 32 which is not 2 and a half times 16..
So, 3x is correct and x + 3x is wrong.
So, the correct answer will be 2.5.
(15)

Vipul said:   3 years ago
Can someone please tell me how?

=> (4x+8) =5x+40 Became
=> 8x+16?

I really didn't understand this part. Anyone help me to get it.
(14)

Bhavesh said:   3 years ago
A) 2 and a half times old= 2 1/2 = (2*2+1)/2 = 5/2.

B) Initially father is 3 times son's age.
After 8 years father is 2 1/2 times the son's age.
Again after 8 years father is? Times sons age?

Therefore, 0+8+8= 16.

Hence the equation 4x + 16/ x+ 16.
(14)

Madhu said:   2 years ago
Why 5/2 is used? Please explain to me.
(13)

Dhaval said:   2 years ago
Father (F) = 3S (son).

After 8 year;
F + 8 = 5/2(S + 8)
Put F = 3S.

After Solution S = 24.
Now in the present time so before 8 year;
S =24 - 8 = 16.
Now F = 3S.
F = 3×16 = 48.
After further 8 years.
So, Father/son = 48/24.
Ans. 2.
(12)


Post your comments here:

Your comments will be displayed after verification.