Question 698137: A store keeper sells ping pong balls in two different boxes. One size holds 5 balls and the other 12 balls. How many boxes of each kind did the store keeper sell if he sold 99 balls and used more than 10 boxes?
Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
This is really a trial and error problem (or "guess and check", if you prefer). But we can do some analysis at the beginning so that the set of things you can try (or guess) is limited.
So what do we know? Some boxes have 5, some have 12. The total number of boxes is MORE than 10. While it is not specifically stated, I think it is reasonable to assume that the guy only sold whole boxes -- hence the number of 5 ball boxes and the number of 12 ball boxes have to both be integers.
Divide 99 by 5. You get 19 with a remainder of 4. Divide 99 by 12. You get 8 with a remainder of 3. This tells us that if there is a combination of integers that fits the parameters of the problem, then neither number can be zero. That gives us a lower bound on our trail of 1. We can also see that if we want to limit our set of trials we should try numbers of 12 ball boxes in the range 1 through 8 as opposed to trying numbers of 5 ball boxes in the range 1 through 19. Less than 1/2 the calculations that way.
Start with 8 12-ball boxes. 8 times 12 is 96, leaving 3 balls which don't make a full box of 5. Reject 8.
Next try 7 12-ball boxes. 7 times 12 is 84, leaving 99 - 84 equals 15. Ah ha! 15 is evenly divisible by 5, so that would mean 3 5-ball boxes. But this still doesn't work. The problem says he sold MORE than 10 boxes, but 7 plus 3 is exactly 10, not more than 10. Reject 7.
Continue using the same analysis for 6, 5, 4, 3, 2, and 1 12-ball boxes. Let me know if you find one that gives you integers for both sizes of box and where the sum is greater than 10.
There is nothing that says you have to do the trials in the order I gave. But even if you choose at random and happen to find one that works, you aren't done until you have tried all of them. There could be more than one answer to the problem. Not saying that there is more than one answer, but you have an obligation demonstrate whether the answer is unique or not.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Answer by josmiceli(19441)  (Show Source):
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