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Ten years before Alice's age is 4 times John's age. Ten years later, Alice's age is 2 times John's age. How old is John?

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John is

20. Ten years earlier, John was 10 and Alice was 40. Ten years from now, Alice will be 60 and John will be 30.

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10
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John is

20

Equations

Let age of Alice now be a and age of John now be j. Therefore 10 years ago,
$$a-10=4(j-10)$$

and

After 10 years
$$a+10=2(j+10)$$

Solving two equations we get

j = 20 and a = 50. So today age of John is 20 years.

This proves that this is a $6^{th}$ grade math problem.

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