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 7 of 8.
Iba Krish said:
7 years ago
Very useful @Nagu.
Rohini Priya said:
7 years ago
Here they considered ages 10 years ago 3x and x, these are not present ages.
Since we solve all problems by having present ages we convert them to the present age.
So the present ages will be 3x+10 and x+10.
Since we solve all problems by having present ages we convert them to the present age.
So the present ages will be 3x+10 and x+10.
(1)
SHUBHAM said:
7 years ago
Ago means subtract so, X+10-10 this statement may be right.
Rachana said:
6 years ago
@Shubham.
Actually, it is not right.
Because initially the age of the boy and his dad are 'x' & '3x' respectively.
10yrs ago it was the case.
Presently, their ages are (x+10) & (3x+10) respectively.
Actually, it is not right.
Because initially the age of the boy and his dad are 'x' & '3x' respectively.
10yrs ago it was the case.
Presently, their ages are (x+10) & (3x+10) respectively.
Pammy said:
6 years ago
@Archi simple why they have taken minus means 3x is 10 year ago.
3x+10 present age.
3x+10 +10 means 10 years after present age.
3x+10 present age.
3x+10 +10 means 10 years after present age.
(1)
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)
Gimid-pussi said:
5 years ago
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
Anyone explain this step in detail.
Anyone explain this step in detail.
(1)
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)
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)
Prakash said:
4 years ago
Thank you @Nagu.
(1)
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