Most of the problems on ages could be solved by
- forming linear equations from the relationships or conditions given in the questions
- solving those linear equations to get the required answer.
Here are some solved problems on the ages:
Q1. A got married 8 years ago. A’s 1 present age is 1(1/4) times his age at the time of marriage.A’s son’s age is 1/10 times his present age. His son’s age in years, is
(a) 2 (b) 3
(c) 4 (d) 5
Answer:(c) 4
Solution:
Let A's present age be x age, then according to the conditions given in the questions, we have
x = 1(1/4) (x-8)
=> x = 1.25(x-8)
=> x = 1.25x - 10
=> 1.25x -x = 10
0.25 x =10
and x = 40 years
Thus, A's son age = (1/10)40 = 4 years
Q2.Three years ago the average age of a family of 5 members was 17 years . A baby having been born , the average age of the family remains the same today. The age of the baby is
(a) 3 years (b) 2 years
(c) 1 year (d) 1.5 years
Answer: (b) 2 years
Solution:
Let the age of the baby today is y years.
Then average of the family member today = (20x5+y) /6
But this average is equal to 17.
Thus, (100 + y) = 17x6
=> 100 + y = 102
=> y =2 years
Q3. The sum of ages of two brothers , having a difference of 8 years between them, will double after 10 years. What is the ratio of the younger brother to that of the elder brother?
(a) 3:7 (b) 8:9
(c) 10:11 (d) 7:11
Answer: (a) 3:7
Solution:
Let the age of elder brother be x years and smaller brother be y years. Then given that
x - y = 8 ----- (i)
and sum is x+y
Again, it is given that the sum of their ages will double after 10 years. The equations thus formed is
2(x+y) = (x+10) + (y+10)
Thus, 2x + 2y = x+y+20
=> x+y=20 ------ (ii)
adding (i) and (ii), we get
2x = 28
=> x = 14 years
and y is 14- 8 = 6 years
Therefore, ratio of younger to elder brother is 6/14 = 3/7
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