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 23 of 33.
Gauri said:
1 decade ago
thanks vaishu, for solving the problem for the last ratio
Vinitha said:
1 decade ago
Any shortcut to solve this type of problem? please share.
Harvinder singh said:
1 decade ago
Two and half mean.. 2+1/2= 2/1+1/2=(2*2+1)/2=(4+1)/2=5/2
Damo said:
8 years ago
For ratio sub 8 in x so we get (4(8)+16)÷(8+16)=2.
Hassan said:
1 decade ago
How can such a question be solved with in one minute?
Sonu said:
9 years ago
How is it 4x? I didn't get it please tell me clearly.
Jagadeesh said:
1 decade ago
Thanks vaishu I understood your explanation clearly.
Nitin said:
10 years ago
Very helpful to practise the problems. Thankyou all.
Preethi said:
1 decade ago
Thank you vaisu & uttam for your explainations.
SHIV said:
8 years ago
How the ratio be 4x+16/x+8?
It should be 4x+8/x+8.
It should be 4x+8/x+8.
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