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AGES - Important Quantitative Aptitude Topic


Ages - Important Quantitative Analysis




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 =

Two years before Age= X-2  (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 = 56-X


Father
Son
Present Age
56-X
X
After 4 Years
56-X+4= 60-X
X+4
Father is three times of son
60-X
3(X+4)


It becomes

60-X = 3(X+ 4)
X= 48%4=12 years (Son’s Age)
Father’s age =56 –X = 56-12 = 44 years

Right Answer is A.
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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= 50-X


Mother
Daughter
Present Age
50-X
X
5 Years Ago
50 – X- 5= 45-X
X-5
Mother is 7 times the age of the daughter
45-X
7(X-5)

It becomes:

45-X = 7 (X-5)
X = 10
Daughter Age= 10 Years
Mother Age= 50-X = 40 Years

Right Answer is C.
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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= X-5
The elder one was twice as old as the younger one
2(X-15)
X-5

It becomes:
2 (X-15) = X-5
X=25 = Younger’s Age
Elder’s Age = 25 + 10= 35 years

Right Answer is B

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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 = 45-X


Father
Son
Present Age
45-X
X
5 Years Ago
45- X -5= 40-X
X- 5
The product of their ages was 4 times the father’s age at that time
(40-X) x (X-5) = 4 ( 40 –X)

It becomes:

(40-X) * (X-5) = 4 ( 40 –X)
X= 9 (Son’s Age)
Father’s Age =45- 9= 36 Years

Right Answer is B

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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)= 2

It becomes:
2X- (X+2) = 2
X=4

Pushpa’s Age= 2X= 2*4= 8 years
Rita’s Age = X+ 2= 4+2= 6 years

Right Answer is B.                          

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For more Ages based Problem, Click on Practice Papers