Since exactly one of the three printed messages is true and the other two are false, it is often quicker to test each message for being the one true statement, rather than testing each carton for holding the cash.
Suppose the third carton's message is the true one, 'cash is in the second carton'. Then the cash is in the second carton, which would also make the first carton's message, 'cash is not in this carton', true, since the cash is not in the first carton either. That gives two true messages at once, which is not allowed, so the third message cannot be the true one.
Suppose instead the first carton's message is the true one, 'cash is not in this carton'. Then the cash is in either the second or the third carton. If it is in the second carton, the third carton's message would also be true, again giving two true messages. If it is in the third carton, the second carton's message, 'no cash in this carton', would also be true, once again giving two true messages. Either way this breaks the one-true-statement condition, so the first message cannot be the true one either.
That leaves the second carton's message as the only one that can be true, 'no cash in this carton', meaning the cash is not in the second carton. Since the third message is then false, the cash is not in the second carton, already established, and since the first message must also be false, the cash must be in the first carton.
Only the scenario with the second message true is consistent, and in that scenario the cash sits in the first carton.
the correct answer is Option A: First.
