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lana [24]
3 years ago
7

a father is now 38 years older than his son. Ten years ago he was twice as old as his son. let x represent the age of the son no

w.
Mathematics
1 answer:
Helga [31]3 years ago
7 0

The present age of father is 86 years old and present age of son is 48 years old

<em><u>Solution:</u></em>

Given that, a father is now 38 years older than his son

Ten years ago he was twice as old as his son

Let "x" be the age of son now

Therefore, from given,

Father age now = 38 + age of son now

Father age now = 38 + x

<em><u>Ten years ago he was twice as old as his son</u></em>

Age of son ten years ago = age of son now - 10

Age of son ten years ago = x - 10

Age of father ten years ago = 38 + x - 10

Then we get,

Age of father ten years ago = twice the age of son ten years ago

38 + x - 10 = 2(x - 10)

28 + x = 2x - 20

2x - x = 28 + 20

x = 48

Thus son age now is 48 years old

Father age now = x + 38 = 48 + 38 = 86

Thus present age of father is 86 years old and present age of son is 48 years old

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