Problem leading equations

Question

The question is : "A and B begin to play with 60$ each. If they play till A's money is double B's, what does A win?"

Now i tried to solve it like they both have 60\$ each, then A got his money doubled than B so he must have got 120\$ so i think that A wins 60\$ . But the real answer is 20\$.

How please help , and sorry if i sound like a dumb but i could'nt understand it.

Thanks

Answer 1

The game in question is a simple $2$-participant zero-sum game. In other words, whatever $A$ wins, $B$ has lost, and vice versa. The fact that the question didn't explicitly state this makes this a poorly set question, since there can easily be other sorts of games, e.g. $A$ and $B$ playing against the "house" in a casino.

Regardless, assuming a simple zero-sum game, let the amount won by $A$ be $x$. This is also the amount lost by $B$.

Hence, at the end of the game, $A$ has $60 + x$ and $B$ has $60-x$.

You're given that $A$ has twice what $B$ has at the end of the game.

That means $60 + x = 2(60 - x)$

Expanding,

$60 + x = 120 - 2x$

Rearranging,

$3x = 60$

$x = 20$.

Hence $A$ wins $\$20$.

Answer 2

Both drop $120\$$ into the pool.

At the end of the game, B is left with x, and A is left with 2x. Since no other addition of money from anywhere.

$x+2x=120$

$\implies x=40$.

So, B is left with $40\$$, so A is left with $80\$$.

But A started with 60. So He won, $80-60$=$20\$$