Come out of the Dilemma: Switch To Government Jobs # AGES - Important Quantitative Aptitude Topic

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

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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

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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

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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

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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