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A says to B

My age is twice of what your age was when I was of your age.

B says to A

If you add your current age to twice of your age when I was born, the answer is $63$.

Assuming they both told the truth, how old are they?

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3 Answers 3

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At first there seems to be a contradiction. The first sentence makes A's age even, since it's double something. The second makes it odd, since when added to double something, you get an odd number. But if you write out the equations and solve them, you do get a consistent pair of numbers.

I think it's rather a cheat, but:

A is 42 and B is 31.5

to check it:

When A was 31.5, B, was 21 (the difference is 10.5 years.) Double 21 is 42.

and

When B was born, A was 10.5. Double that is 21. Add to 42 you get 63.

Nobody said it had to be integers I guess.

As to how to do it, for me it was easier to solve for A and d (the difference between their ages) than A and B. Spoilers below because the MathJax was hard enough but getting the spoilers to interact properly with it appears impossible.

$$A = 2 (A -2d)$$ $$A = 2A - 4d$$ $$-A = 4d$$ $$ A = 4d$$

$$A + 2d = 63$$ $$substitute A = 4d $$ $$2d + 4d = 63$$ $$6d = 63$$ $$d = 10.5$$

$$A = 4d = 42$$ $$B = A - d = 31.5$$

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  • $\begingroup$ I think you got it! $\endgroup$
    – Alex
    Commented May 11, 2015 at 17:22
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Kate Gregory got the right solution first, but this answer shows how one could derive the solution from the puzzle as stated.


Let's look at A's statement first:

My age is twice of what your age was when I was of your age.

If $a$ is A's age and $b$ is B's, this becomes:

$$a=2*(b-(a-b))$$

(since A was B's age $a-b$ years ago). Simplifying:

$$a=2*(2b-a)=4b-2a,$$

so

$$3a=4b.$$


Now let's look at B's statement:

If you add your current age to twice of your age when I was born, the answer is $63$.

This becomes:

$$a+2*(a-b)=63$$

(since A's age when B was born was $a-b$). Simplifying:

$$3a-2b=63.$$


Substitute the first equation into the second to get $4b-2b=63$, so $b=31.5$. Substitute back into the first equation to get $a=\frac{4}{3}(31.5)=42$.

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A: When I was your age [A-[A-B]], my age is twice [2*] of what your age was: [B-[A-B]]

[A-[A-B]] = 2*[B-[A-B]] --> B = 4*B - 2*A --> A = 3/2*B

B: If you add your current age [+A] to twice [*2] your age when I was born [A-B] you get 63 [= 63]

A + 2*[A-B] = 63 --> 3*A - 2*B = 63 --> 3*(3/2*B) - 2B = 63
B = 25.2
A = 37.8

I'm pretty sure this isn't correct...but this is what my equations get me. Maybe if these people didn't talk such nonsense it would be easier :)

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  • $\begingroup$ It looks like your first equation is backwards; "twice of what your age was" would be 2[b-(a-b)], not 2[a-(a-b)]. $\endgroup$
    – goldPseudo
    Commented May 12, 2015 at 2:59

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