Mopey & ZegroDate: Thu, 3 Nov 1994 12:17:18 -0700 (MST) From: Laura Adams Subject: Math Boggs can carry two hartsels or three choloes. Mopey can carry three hartsels or five choloes. Zegro can carry 7 hartsels or nine choloes. They have to move 23 hartsels and 23 choloes out to the school playground. You will tell them what to carry you will have to carry whats left over. If each one makes 2 trips. What will be left over for you to carry? Laura Adams (ladams@mtview.d20.co.edu) Date: Thu, 3 Nov 1994 16:52:30 -0500 From: Phil Spector Subject: Re: Math Laura - I think I have a solution. Here's a hint: Every time Mopey and Zegro carry objects, they can clearly carry more choloes than hartsels. Here, it may be a good idea just to get as many objects carried as possible, regardless of what they are, in order to get as close as possible to carrying your total quota. However, Boggs is special, in that he can carry -relatively- more hartsels than choloes; you don't get the usual tradeoff of carrying 2 less hartsels than choloes. Take advantage of Boggs' uniqueness and have him carry those hartsels! See what kind of answers you get then. The above plan should provide an answer, if the goal is for you to have to carry the least amount of objects. In fact, I get two answers. Do you think you can come up with a reason as to why one of them is a better answer for you? Dr. Math a la Phil __________ Date: Thu, 3 Nov 1994 19:56:04 -0500 From: Phil Spector Subject: Laura's problem Anybody (particularly Ethan/Sydney, who I heard were working on it) get a more precise answer to Laura's problem? I think I got an answer (maximum you can get them to carry is 30 objects, with the best solution being having them carry 19 C and 11 H = (Zegro carrying 9 C and 7 H, Mopey carrying 5 C twice, and Boggs carrying 2 H twice, for 19 C and 11 H. This is better than the other solution, with 23 C and 7 H, since it seems as if H is the bulkiest (everyone can carry less of them than the others), so you want your friends to carry it as much as possible. The above solution also assumes we're looking to carry the least -number- of total objects. Note that I don't have a definite equation for the above math... it's a bit of trial and error. Dr. Phil __________ Date: Thu, 3 Nov 1994 20:02:15 -0500 (EST) From: Dr. Ethan Subject: Re: Laura's problem Phil, that is exactly the way I did it and the answer I got. I think we need to know more about the goal of the problem to really give a helpful answer, although given the information given yours was excellent. Maybe the person left can carry 2 h's and 10 c's or maybe they carry them evenly? So given what we had your answer was great. Ethan |
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