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:
289 comments Page 1 of 29.
Bhavesh said:
2 months 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.
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.
(6)
Vipul said:
4 months 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.
=> (4x+8) =5x+40 Became
=> 8x+16?
I really didn't understand this part. Anyone help me to get it.
(8)
L.Haritha said:
8 months ago
At present, F and R.
R=R,
Father's age would 3 times more than Ronit's age
F=3R+R -> F=4R -----> (1)
After 8 years, F+8 and R+8.
Father would be 2 and half a times of Ronit's age
F+8=2 1/2(R+8).
Mixed fraction 2 1/2 = ((2*2)+1)/2 = 5/2.
F + 8=5/2(R+8),
2F+16 = 5R + 40,
2F-5R = 40 - 16,
2F-5R = 24 -------> (2)
Sub F=4R in (2).
2(4R)-5R = 24,
8R-5R = 24,
3R= 24.
R = 8 years, F = 32 years.
After further 8 years means;
F + 16 = 32 + 16= 48,
R + 16 = 8 + 16= 24,
Required ratio= 48/24 = 2.
Hence, After further 8 years, 2 times he would be of Ronit's age.
R=R,
Father's age would 3 times more than Ronit's age
F=3R+R -> F=4R -----> (1)
After 8 years, F+8 and R+8.
Father would be 2 and half a times of Ronit's age
F+8=2 1/2(R+8).
Mixed fraction 2 1/2 = ((2*2)+1)/2 = 5/2.
F + 8=5/2(R+8),
2F+16 = 5R + 40,
2F-5R = 40 - 16,
2F-5R = 24 -------> (2)
Sub F=4R in (2).
2(4R)-5R = 24,
8R-5R = 24,
3R= 24.
R = 8 years, F = 32 years.
After further 8 years means;
F + 16 = 32 + 16= 48,
R + 16 = 8 + 16= 24,
Required ratio= 48/24 = 2.
Hence, After further 8 years, 2 times he would be of Ronit's age.
(10)
Diksha said:
9 months 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.
(4)
Hamza said:
9 months 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:
9 months ago
Anyone, Can you explain this sum by taking 2.5?
Chinnu said:
9 months ago
If we take 2.5 or 5/2 will we get the same answer? Anyone explain.
Bibek said:
2 years ago
It should be 3x + 8 = 5/2 (x+8).
(4)
Bibek said:
2 years ago
It should be 3x + 8 = 5/2 (x+8).
(2)
Sunil said:
2 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.
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