A father is 7 times as old as his son. Two years ago, the father was 13 times as old as his son. How old are they now?
Let the present age of son be x years
Then, age of his father will be 7x years
Two years ago, age of son = (x – 2) years
Two years ago, age of his father = (7x – 2) years
According to the given problem,
7x – 2 = 13 (x – 2)
7x – 2 = 13x – 26
7x – 13x = - 26 + 2
-6x = - 24
x = - 24 /- 6
We get,
x = 4
Therefore, age of son = 4 years and
Age of his father = 7x = 7 × 4 = 28 years
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