Ages is another Quantitative Aptitude topic that needs you to expertise and views a clear picture of understanding. Following some important or shortcut points to remember:
Suppose
If a person is in present Age = X
Two years before Age= X2 (Present
Age  Two years before)
Three years later / Three years hence
/ After Three Years/ Age= X+3 (Present Age + 3 years later)
 In before/ Ago cases, years will be subtracted.
 In Later/hence and After Cases, years will be added.
Let’s Discuss with Problems:
1. The Sum of
the ages of a son and father is 56 years. After four years, the age of the
father will be three times of the son . Their ages respectively are:
1 (a) 2 years,
44 years
1 (b) 6 years,
42 years
1 (c) 6 years,
48 years
(d) 8 years,
36 years
Solution:
Age of
Father + Age of Son= 56
Take Age of son = X
Then, the Age of father = 56X
Father

Son


Present Age

56X

X

After 4 Years

56X+4= 60X

X+4

Father is three times of son

60X

3

It becomes
60X = 3( X+ 4)
X= 48%4=12 years (Son’s Age)
Father’s age =56 –X = 5612 = 44 years
Right Answer is A.

2. The sum of
the ages of the mother and daughter is 50 Years. Also, 5 years ago, the mother
age was 7 times the age of the daughter. The present ages of the mother and
daughter respectively are:
(a) 35 years,
15 years
(b) 38 years,
12 years
(c) 40 years, 10
years
(d) 42 years, 8 years
Solution:
Mother Age
+ Daughter Age = 50
Let’s take Daughter Age= X
Mother Age= 50X
Mother

Daughter


Present Age

50X

X

5 Years Ago

50 – X 5= 45X

X5

Mother is 7 times the age of the daughter

45X

7

It becomes:
45X = 7 (X5)
X = 10
Daughter Age= 10 Years
Mother Age= 50X = 40 Years
Right Answer is C.

3. The
difference between the ages of two persons is 10 years. 15 years ago, the elder
one was twice as old as the younger one. The present age of elder person is:
a) 25 years
b) 35 years
c) 45 years
d) 55 years
Solution:
Difference
in Younger and Elder = 10 years
Elder – Younger = 10
Take younger age = X
Elder age = 10 + X
Younger

Elder


Present Age

X

10+X

15 Years Ago

X 15

10 + X 15=
X5

The elder one was twice as old as the younger one

2

X5

It becomes:
2 (X15) = X5
X=25 = Younger’s Age
Elder’s Age = 25 + 10= 35 years
Right Answer is B

4. The sum of
the ages of a father and son is 45 years. Five years ago the product of their
ages was 4 times the father’s age at that time. The present Ages of the father
and son respectively are:
a) 25 years, 10 years
b) 36 years, 9
years
c) 39 years, 6
years
d) None of
these
Solution:
Sum of Age
of Father and Son = 45 years
Take Age of
Son = X
Then,
Father’s age = 45X
Father

Son


Present Age

45X

X

5 Years Ago

45 X 5= 40X

X 5

The product of their ages was 4 times the father’s age
at that time

(40X) x (X5) = 4 ( 40 –X)

It becomes:
(40X) * (X5) = 4 ( 40 –X)
X= 9 (Son’s Age)
Father’s Age =45 9= 36 Years
Right Answer is B

5. Pushpa is
twice as old as Rita was two years ago. If the difference between their ages be
2 years, how old is Pushpa today?
a) 6 years
b) 8 years
c) 10 years
d) 12 years
Solution:
The Difference
between their age = 2 years
Pushpa

Rita


2 years ago



X

Present Age

2X

X+ 2

The difference between their ages be 2 years

2X (X+2

It becomes:
2X (X+2) = 2
X=4
Rita’s Age = X+ 2= 4+2= 6 years
Right
Answer is B.
