1. The age of a man is
three times the sum of the ages of his
two sons. After five years, his age will be double of the sum of the ages of
his sons. What’s the present age of the father?
Solution:
Let the sum of the ages of 2 sons be x.
Then the father’s age will be 3x.
After 5 years,
3x+5 = 2(x+10) [Note: 10 is due to the sum of the 2 sons
will rise to 10 after 5 years]
3x+5 = 2x+20
x = 15
The
present age of the father is 3x = 45 years.
2. The sum of the
ages of a mother and her son is 45 years. Five years ago, the product of their
ages was 34. The ages of the son and the mother are respectively:
Solution:
Let the son’s age be x and the mother’s age is (45-x). [Given: The sum of their ages is 45]
Their
product before 5 years was,
(x-5)(45-x-5) = 34
(x-5)(40-x) = 34
40x-x2-200+5x=34
-x2-200+45x-34 = 0
x2-45x+234
= 0
x2-39x-6x+234=0
x(x-39)-6(x-39)=0
(x-6)(x-39)=0
x = 6 or 39(not
possible)
Therefore, x = 6.
Their ages are x and (45-x)
= 6 and 39 years respectively.
3. A got married 8
years ago. His present age is 6/5 times his age at the time of his marriage. A’s
brother was 10 years younger to him at the time of his marriage. The age of A’s
brother is:
Solution:
Let the present age of A be x.
His present age is 6/5 times his age at the
time of his marriage,
i.e., x =
6/5(x-8)
5x=6x-48
x =48
The age of A’s brother at the time of A’s marriage is x-8-10 = 48-18 =30
years.
The present age of brother is 30+8 = 38 years.
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