SOLUTION: Two years ago, Jim was 1/2 of his fathers age. At that time, adding all the digits (no zeros) in their ages would give you a number that is 1/7 of the fathers age now. Hint Jim’s

Question 1201870: Two years ago, Jim was 1/2 of his fathers age. At that time, adding all the digits (no zeros) in their ages would give you a number that is 1/7 of the fathers age now. Hint Jim’s father is not yet 50 what are their ages
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Because the statement of the problem talks about the sum of the digits of their ages two years ago, this is a problem that is not easily solved using formal algebra.

The answer can be found quickly using logical reasoning and simple arithmetic.

Adding all the digits of their ages two years ago gives a number that is 1/7 of the father's age now, so the father's age now is a multiple of 7. So

let 7x be the father's age now

Two years ago, when the father's age was 7x-2, Jim was half his father's age; that means 7x-2 is an even number, and that means 7x is an even number.

Since the father is less than 50 years old, the only possible ages for the father now are 14, 28, and 42.

From there, you can try each of those possible answers to see if the sum of the digits of their ages two years ago was 1/7 of the father's current age.

Or you could use logical reasoning to see that the possible answers of 14 and 28 are impossible due to the fact that the father's age two years ago of 12 or 26 would make it impossible for him to have a son half his age then.

So...

the father's age now is 7x = 42
the father's age two years ago was 42-2 = 40
the son's age two years ago was 40/2 = 20
the son's age now is 20+2 = 22

ANSWERS: father 42, son 22

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