We could do it algebraically. We'll make the amount with which she starts each play blue, and the amount she loses on that play red.

After 0 plays: x chips

After 1 play: x - (x/2 + 1) chips, or (x/2 - 1) chips

After 2 plays: (x/2 - 1) - ((x/2 - 1)/2 + 1) chips, or (x - 6)/4 chips

We know that (x - 6)/4 = 5, so x = 26.

We could also try answers. Start with answer C, so you know whether to go up (if C is too small) or down (if C is too big).

Let's say x = 24. After the first play she has 11 chips. After the next play she has 4.5 chips. 24 was too small, so we'll move up.

We also learn (from that last example) that x must be even, or we'll get fractional chips on the first play. (There are other conditions too to ensure that x is an integer after both plays, but we don't need to bother sorting them out: we'll just try the next few even numbers up until we find the answer.)

Let's say x = 26. After the first play she has 12 chips. After the next play she has 5 chips. Success! With clever use of the answer choices, we only had to try two numbers, and we're done much faster than we would have been with the algebraic approach I gave earlier.

A final idea would be to work backwards.

If she ends with 5 chips, she must have had (5 + 1) * 2, or 12 chips before that.

If she had 12 chips after the first step, she must had had (12 + 1) * 2, or 26 chips before that. Aha!

This might be the fastest of all, though it also might not be the easiest to think of.