Aptitude::Problems on Ages

Let Jack's present age be x
Jack's age before 5 years = (x - 5)
Jack's age after 22 years = (x + 22)
We are given that, Jack's age after 22 years (x + 22) is 4 times his age 5 years back (x – 5)
Therefore,
(x + 22) = 4(x – 5)
Solving the equation, we get
x + 22 = 4x – 20
3x = 42
x = 14 years Ans.

Let age of Rohan be y
Rahul is 16 years elder than Rohan = (y + 16).
So Rahul's age 4 years ago = (y + 16 – 4)
Rohan's age before 4 years = (y – 4)
4 years ago, Rahul is 5 times as old as Rohan
Thus, we can have the equation as,
(y + 16 – 4) = 5 (y – 4)
(y + 12) = (5y – 20)
4y = 32
y = 8
Rohan's age = 8 years
Rahul's age = (y + 16) = (8+16) = 24 years Ans.

We are given that age ratio of Paul: Peter = 5:6
Let,
Paul's age = 5x and Peter's age = 6x
1 year ago, their age was 5x and 6x.
Hence at present,
Paul's age = 5x +1
Peter's age = 6x +1
Now after 4 years,
Paul's age = (5x +1) + 4 = (5x + 5)
Peter's age = (6x +1) + 4 = (6x + 5)
Hence,
After 4 years, this ratio becomes 6 : 7. Therefore,

\(\frac{Paul's age}{6}\)= \(\frac{Peter's age}{7}\)
\(\frac{(5x+5}{6x+5}\)= \(\frac{6}{7}\)
7 (5x + 5) = 6 (6x + 5)
X = 5
Peter's present age = (6x + 1) = (6 x 5 + 1) = 31 years
Paul's present age = (5x + 1) = (5 x 5 + 1) = 26 years Ans.

Age of y= \(\frac{ratio of y}{sum of ratios}\)* sum of ages
Age of y= \(\frac{q}{p+q}\)* A Ans.

Let 10 years ago, age of son be x and so mother's age= 6x.
At present,
Mother's age will be (6x + 10) and son's age will be (x + 10)
After 10 years,
Mother's age will be (6x + 10) +10 and son's age will be (x + 10) + 10
Mother's age is twice that of son
(6x + 10) +10 = 2 [(x + 10) + 10]
(6x + 20) = 2[x + 20]
Solving the equation, we get x = 5
Therefore, the present ratio.
(6x + 10) : (x + 10) = 70 : 15 = 14:3 Ans.

Let us assume x years ago.
At present,
Raj's age= 40 years
Surya's age= 60 years
x years ago;
Raj's age = (40 – x)
Surya's age = (60 – x)
Also,
Ratio of their ages x years ago was 4 : 8

\(\frac{(40-x)}{(60-x)}\)= \(\frac{4}{8}\)
8(40 – x) = 4(60 – x)
320 – 6x = 240 – 4x
x = 40
Therefore, 40 years ago, the ratio of their ages was 4 : 8.

At present;
Ratio of their ages = 5 : 3.
Therefore, let it be 5x and 3x.
Rohan's age 4 years ago = 5x – 4
Rahul's age after 4 years = 3x + 4
Now;
Ratio of Rohan's age 4 years ago and Rahul's age after 4 years is 1 : 1
Therefore,
Solving, we get;;
\(\frac{(5x-4)}{(3x+4)}\)= \(\frac{1}{1}\)
So, x=4.
Now;
the ratio between Rohan's age 4 years hence and Rahul's age 4 years ago will be=
Rohan's age : (5x + 4)
Rahul's age: (3x – 4)
So, Rahul's age: Rohan's age
\(\frac{(5x+4)}{(3x-4)}\)
= \(\frac{24}{8}\)
= \(\frac{3}{1}\)= 3:1

We are given, 5 years ago sister's age was 5 times the age of her brother.
Therefore,
(34 – x) – 5 = 5 (x – 5)
34 – x – 5 = 5x – 25
5x + x = 34 – 5 +25
6x = 54
x = 9
So, after 6 yrs
(x + 6)
= (9 + 6) = 15 years

Let daughter's age be x and father's age be 2x.
Father's age is 3 times more aged than his daughter, therefore father's present age = x + 2x = 3x
After 5 years, father's age is 3 times more than his daughter age.
(3x + 6) = 2(x + 6)
x = 6

After 6 years it is (3x + 5), then after further 6 years, father's age = (3x +12) and daughter's age = (x + 12)
\(\frac{(3x+12)}{(x+12)}\)= ?

Substitute the value of x, we get
\(\frac{30}{18}\)=1.6
After further 6 years, father will be 1.6 times of daughter's age.

Let the son's present age be x years.
Then, (40 - x) = x
=> x= 20.
Son's age 5 years back = (20 - 5) = 15 years.

Let the present ages of son and father be x and (60 -x) years respectively.
Then, (60 - x) - 5= 4(x - 5)
55 - x = 4x - 20
5x = 75
Thus,
=> x = 15

let their present ages be 3X and 4X.
5 years ago, the ratio of their ages was 5 : 7,
Thus, (3x-5) : (4x-5) = 5 : 7.
Solving, we get X = 10.
Hence, their present ages are 3X = 30 and 4X = 40

Let the present ages of Dhoni= 4x and Ganguly = 3x
So, 4x + 6 = 26
x = 5
Now,
Present age of Ganguly = 3x
So,
=> 3*5 = 15 years.

Let Riya's present age be x years.
Riya's age after 20 years = (x + 20) years.
Riya's age 4 years back = (x - 4) years
Then,
x + 20 = 5 (x - 4)
x + 20 = 5x - 20
=> x = 10 Ans.

Let A's age = X
Now,
C is 12 yrs older than A
So, C's age = x+12
B is 4 times as old as A
so B's age = 4x
The ages of A,B and C together are 57 years
x + (x+12) + 4x = 60
=> 6x = 48
x = 8
Thus,
A's age = 8
B's age = 4x= 32
C's age = x+12=20 Ans.

Average age of 80 boys = 15 x 80 = 1200 yrs.-----(i)
average age of 15 boys = 15 *16 = 240 yrs-----(ii)
average age of 25 boys = 14 * 25 = 350 yrs.-----(iii)
Remaining boys= 80-(15+25)= 40 boys
Average age of remaining 40 boys =
\(\frac{1200 - (240+350)}{80 - (25+15)}\)
= \(\frac{610}{41}\)
= 15.25 years. Ans.

Ram age when he was born = 0 years
=> His brother's age = 6 year
=> His father's age = brother age + 32years
= 6+32 = 38
=> His mother's age = father's age – 3
=38-3 = 35
So sister's age = 35-25 = 10 years. Ans.

Let Rishi's age now be A
Rahul's age= B
(A - 6) = X(B - 6)
But Rahul is 17 and therefore (17-6)=11 = X(B - 6)
\(\frac{11}{X}\) = B-6
\(\frac{11}{X}\) + 6 = B
B = \(\frac{11}{X}\) + 6 Ans.

Let Peter's age = x.
Then Paul's age = x + 2
After 6 years the total of their ages will be 7 times of their current age.
Not clear.
So the given data are inadequate. Ans.

Let Ravi's age at the time of his marriage = X years
Then,
=> X + 10 = \(\frac{6}{5}\) X
X = 50.
Since got married 10 years ago,
So, his present age = 50+10= 60
His sister is 5 years younger => 60-5 = 55 years. Ans.

Let the Present age of Ram = x
Rahul's age = 5-2 = 3
So, as per the question:
\(\frac{x-6}{18}\) = 3
x-6 = 54
x = 60
Ram's present age = 60 years Ans.

As per given in question:
(A+B)=12+(B+C)-----(i)
So,
(A+B) - (B+C)= 12
A-C= 12
Thus;
C is younger than A by 12 years.
Ans.