Aptitude - Problems on Ages - Discussion
Discussion Forum : Problems on Ages - General Questions (Q.No. 15)
15.
The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is:
Answer: Option
Explanation:
Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10]
3x + 20 = 2x + 40
x = 20.
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
Discussion:
77 comments Page 1 of 8.
Bibek paudel said:
3 years ago
@All.
Here's my explanation.
Let's take the present age of
Son= s
Father =f
10 year before,
son= s-10.
father= f-10.
Equation i
3(s-10)=f-10
3s-f=20 --> (i)
After 10 year;
son=s+10
father= f+ 10
equation (ii)
2(s+10) = f+10
2s-f = -10 --> i)
By equation;
S = 30
F = 70.
So, the ratio = 7:3.
Here's my explanation.
Let's take the present age of
Son= s
Father =f
10 year before,
son= s-10.
father= f-10.
Equation i
3(s-10)=f-10
3s-f=20 --> (i)
After 10 year;
son=s+10
father= f+ 10
equation (ii)
2(s+10) = f+10
2s-f = -10 --> i)
By equation;
S = 30
F = 70.
So, the ratio = 7:3.
(76)
Ravi Teja said:
2 years ago
I think the answer is incorrect because the year is in ago so, we would subtract the 3x-10 but there is 3x+10.
How it is possible?
How it is possible?
(51)
Ayushi gujarati said:
6 years ago
Here is my solution;
Son's present age = x.
Father's present age=y.
10 years ago
(y-10) = 3(x-10)
y-10=3x-30
y-3x=-20 --------> eq 1
10 years after
(y+10)=2(x+10)
y+10=2x+20
y-2x=10 --------> eq2.
now
y-3x=-20,
y-2x=10.
10(y-3x)=-20(y-2x).
10y-30x=-20y+40x.
10y+20y=40x+30x.
30y=70x.
So now y:x = 3:7.
Son's present age = x.
Father's present age=y.
10 years ago
(y-10) = 3(x-10)
y-10=3x-30
y-3x=-20 --------> eq 1
10 years after
(y+10)=2(x+10)
y+10=2x+20
y-2x=10 --------> eq2.
now
y-3x=-20,
y-2x=10.
10(y-3x)=-20(y-2x).
10y-30x=-20y+40x.
10y+20y=40x+30x.
30y=70x.
So now y:x = 3:7.
(14)
Deepika and kavipriya said:
2 months ago
The son age is x.
10 years ago, father is age was 3x-10.
After 10 years, father's age is 2x+10.
then,
3x-10 = 2x + 10.
x = 20.
The required ratio is = (3x+10):(x+10).
= 70:30,
= 7:3.
10 years ago, father is age was 3x-10.
After 10 years, father's age is 2x+10.
then,
3x-10 = 2x + 10.
x = 20.
The required ratio is = (3x+10):(x+10).
= 70:30,
= 7:3.
(9)
Destello said:
4 years ago
Son age is "x".
Father age is "3x".
10 year hence (after).
son age will become "x+10".
father age will become "3x+10" but father age is twice of son's age So it will become 2(x+10).
So, the equation will be.
(3x+10) = 2(x+10),
3x+10=2x+20,
3x-2x=20-10.
x=10.
For 10 years after age we need to add 10 in x (the most important).
x=10+10.
x=20.
Now, substitute the value of x in the x+10 equation.
Son age will become "x+10" => 20+10 = 30.
Father age will become "3x+10" => 3*20+10 = 70.
70/30 = 7/3 = 7:3.
Father age is "3x".
10 year hence (after).
son age will become "x+10".
father age will become "3x+10" but father age is twice of son's age So it will become 2(x+10).
So, the equation will be.
(3x+10) = 2(x+10),
3x+10=2x+20,
3x-2x=20-10.
x=10.
For 10 years after age we need to add 10 in x (the most important).
x=10+10.
x=20.
Now, substitute the value of x in the x+10 equation.
Son age will become "x+10" => 20+10 = 30.
Father age will become "3x+10" => 3*20+10 = 70.
70/30 = 7/3 = 7:3.
(9)
Pravallika said:
5 years ago
10 years ago : father age = thrice son ---> 3 x Son's age - 10(10 years ago) = father's age.
After 3 years : father age = twice son ---> 2 x Son's age + 10(after 10 years) = father's age.
So, ,
3S-10 : F = 2S+10 : F
3S-10=2S+10
S=20.
Substituting S value in 3S -10: 2S + 10.
We get 7:3.
After 3 years : father age = twice son ---> 2 x Son's age + 10(after 10 years) = father's age.
So, ,
3S-10 : F = 2S+10 : F
3S-10=2S+10
S=20.
Substituting S value in 3S -10: 2S + 10.
We get 7:3.
(5)
Akshat said:
2 years ago
Father's age ten years ago: (3x + 10),
Ten years hence: (2x + 10),
Let son be x years:
(x+10) = (2x + 10),
x = 20.
3x+10 : x+10 = 70 : 30 = 7 : 3.
Ten years hence: (2x + 10),
Let son be x years:
(x+10) = (2x + 10),
x = 20.
3x+10 : x+10 = 70 : 30 = 7 : 3.
(4)
Shayan said:
2 years ago
Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10],
3x + 20 = 2x + 40.
x = 20..
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
Then, (3x + 10) + 10 = 2[(x + 10) + 10],
3x + 20 = 2x + 40.
x = 20..
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
(3)
Nagu said:
2 decades ago
Hi all,
There is no like we should subtract
We need to finish with simple way so
They are taking 10 years ago they boy age is x
Then the father's age is 3x
So now the age of boy and father is x+10 and 3x+10
So 10 years hence it will be x+10+10 and 3x+10+10
But as per problem fathers age is double than his son
So (3x + 10) + 10 = 2[(x + 10) + 10]
Hope this will help you. Have a nice day.!
There is no like we should subtract
We need to finish with simple way so
They are taking 10 years ago they boy age is x
Then the father's age is 3x
So now the age of boy and father is x+10 and 3x+10
So 10 years hence it will be x+10+10 and 3x+10+10
But as per problem fathers age is double than his son
So (3x + 10) + 10 = 2[(x + 10) + 10]
Hope this will help you. Have a nice day.!
(2)
Gimmeda Pussi said:
5 years ago
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
How are we supposed to arrive at this step? Please explain me.
How are we supposed to arrive at this step? Please explain me.
(2)
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