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:
322 comments Page 33 of 33.
Kolimi said:
4 weeks ago
Present ages F : S = 4:1(because father is 3 times more than son which is 4-1 =3)
Father = 4R.
Son = R.
After 8 years
F: S = 2.5 : 1.
(4R + 8)(R+8) = 2.5
4R + 8 = 2.5(R + 8),
4R + 8 = 2.5R + 20,
4R - 2.5R = 20 - 8.
1.5R = 12.
R=12/1.5 = 8.
If R is 8 then 4R is 4(8)= 32, that is, father's age after 8 years.
So, after 16(that is, after 8 years + after 8 years).
F = 32 + 16 = 48.
S = 8 + 16 = 24.
F:S = 48:24 = 2:1.
That is Father's age 2 times more than son after another 8 years.
Father = 4R.
Son = R.
After 8 years
F: S = 2.5 : 1.
(4R + 8)(R+8) = 2.5
4R + 8 = 2.5(R + 8),
4R + 8 = 2.5R + 20,
4R - 2.5R = 20 - 8.
1.5R = 12.
R=12/1.5 = 8.
If R is 8 then 4R is 4(8)= 32, that is, father's age after 8 years.
So, after 16(that is, after 8 years + after 8 years).
F = 32 + 16 = 48.
S = 8 + 16 = 24.
F:S = 48:24 = 2:1.
That is Father's age 2 times more than son after another 8 years.
(1)
Ayush Pandey said:
1 week ago
Very clear, thanks all for explaining the answer.
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