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 10 of 33.
Janu said:
1 decade ago
After Means We have to add plus symbol. So after 8 years means 4x+8 then another after 8 years means add another plus 8 so it become 4x+8+8. So equation become 4x+16. Like that For Rohit x+16.
Payal said:
7 years ago
Ronit age = X.
Father age = 3x.
After 8 years.
(X+8) = (3x+8).
Half of of Ronit age.
(X+8) =1/2(3x+8).
X=8.
After further 8 years.
Both age ratio, where X = 8.
(3x+8)/X+8 = 32/16.
=> 2.
Father age = 3x.
After 8 years.
(X+8) = (3x+8).
Half of of Ronit age.
(X+8) =1/2(3x+8).
X=8.
After further 8 years.
Both age ratio, where X = 8.
(3x+8)/X+8 = 32/16.
=> 2.
(1)
Santhiya said:
7 years ago
@All.
Simply, the solution is;
Given:
Ist father age= 3 times more than son age.
2nd after 8yrs father age= 2 1/2 times more than son age.
3rd after 8 yrs age =2 time more than son age.
Simply, the solution is;
Given:
Ist father age= 3 times more than son age.
2nd after 8yrs father age= 2 1/2 times more than son age.
3rd after 8 yrs age =2 time more than son age.
TUHIN PALI said:
1 decade ago
I solved in this way.
Per 8 years age decrease: (3x-5x/2) = x/2.
"1" " " x/16.
After 16 years = 16*x/16 = x.
3x-x = 2x.
After 16 years. So 2 times of Ronit's age. As father's age x.
Per 8 years age decrease: (3x-5x/2) = x/2.
"1" " " x/16.
After 16 years = 16*x/16 = x.
3x-x = 2x.
After 16 years. So 2 times of Ronit's age. As father's age x.
Aviskar chapagain said:
1 decade ago
Let,
Father = X.
Ronit = Y.
1st case:
X = 3Y.
2nd case:
X+8 = Y/2.
Third case:
X+16 = Y.
2X+16 = 3Y. From 1st and second case.
Then subtract third case in it. You will get,
X=2Y.
Father = X.
Ronit = Y.
1st case:
X = 3Y.
2nd case:
X+8 = Y/2.
Third case:
X+16 = Y.
2X+16 = 3Y. From 1st and second case.
Then subtract third case in it. You will get,
X=2Y.
Jyothi sajjan said:
1 decade ago
Hi @Rashmi (4x+8)is cross multiply with 2(denominator)then it becomes 8x+16. and, 5(numerator) is multiplied with (x+8)then it becomes 5x+40 this is the correct manner.
Have a nice day.
Have a nice day.
Rohan said:
1 decade ago
Answer to 5/2(x+8) ...in example its is written that After 8 years, he would be two and a half times of Ronit's age
So two and a half is 2 1/2, now 2x2 is 4 and 4+1 is 5 so its 5/2..
So two and a half is 2 1/2, now 2x2 is 4 and 4+1 is 5 so its 5/2..
Anil said:
1 decade ago
@Poonam And @Thara in question this given that two and half.
Means 2and half which is 2.5 and 5/2 is same equal to 2.5.
Don't get confused in 2/5 and 2.5. Because 2/5 is .4 not 2.5.
Means 2and half which is 2.5 and 5/2 is same equal to 2.5.
Don't get confused in 2/5 and 2.5. Because 2/5 is .4 not 2.5.
Bharathviswa said:
2 years ago
It's 16 because we assume the son's age is x and the father is x + 3x = 4x.
We equate it and find the son's present age is 8.
So, we need to find 8+8 years, that's how it became 16.
We equate it and find the son's present age is 8.
So, we need to find 8+8 years, that's how it became 16.
(8)
Nivetha said:
1 decade ago
Fathers age is 3 times more than his son 4x.
But after getting full stop considering and comparing other sums I can understand like this (x+8). How we put (4x+8) in fathers age?
But after getting full stop considering and comparing other sums I can understand like this (x+8). How we put (4x+8) in fathers age?
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