Backward Through the StepsDate: 5/20/96 at 21:1:40 From: Anonymous Subject: Solving strategies How do you do equivalent fractions, and mixed numbers - such as solving strategies? Here is one of the problems: Fred gave one half of his baseball cards to Sally. Sally gave Jeff half of the cards that she got from fred. Jeff gave Allen half of the cards that he got from Sally. Allen got 6 cards. How many cards did Fred originally have? Thank you for helping! Joshua Chen Date: 7/13/96 at 14:38:32 From: Doctor Jodi Subject: Re: Solving strategies Hi there! There are quite a few ways to answer a question like this one, but in this case, one of the easiest ways is to go backward through the steps: Allen got 6 cards, which is half of Jeff's from Sally, or a quarter of Sally's from Fred, or an eighth of Fred's cards. If 6 cards is 1/8 of Fred's total, Fred had 8 * 6 cards, or 48. A similar technique is to look at the ratios of each person's cards. 2/1 Fred 2/1 Sally 2/1 Jeff to to to Sally Jeff Allen Since we're looking for the ratio of Fred's cards to Allen, we find: For every 2 cards that Fred has, Sally has 1. For every two cards that Sally has, Jeff has 1. To compare Fred's cards to Jeff's, we can rewrite the ratios: 4 Fred 2 Sally ------ * ------- 2 Sally 1 Jeff Does that make sense? That says that for every 4 cards Fred has, Jeff has one. We can check this by saying: If Fred gave Sally 2 cards (1/2 of 4), then Sally gave Jeff 1 card (1/2 of 2). So that works. But to find out how many cards Fred has we need the ratio of Fred's cards to Allen's. We start off with the ratio from above, 4 Fred ------ 1 Jeff Now, we know that Jeff gives half of his cards to Allen. Again, we'll rewrite the ratio, this time to: 8 Fred ------ 2 Jeff and multiply by the ratio of Jeff's cards to Allen's. 8 Fred 2 Jeff ------ * ------- 2 Jeff 1 Allen So, for every 1 card that Allen has, Fred had 8. But Allen has 6 cards, so Fred had 6 * 8 or 48 cards. I hope this makes sense. Let us know if you need more help. -Doctor Jodi, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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