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June 11th, 2019, 08:35 AM  #1 
Newbie Joined: Jun 2019 From: Ralaigh Posts: 1 Thanks: 0  Shopping Word Problem
How would I solve the following? 1 ) a b c d e went shopping 2 ) each had a whole dollar amount to spend 3 ) together they had 56 dollars 4 ) The absolute difference between what a and b had to spend was 19 5 ) The absolute difference between what b and c had to spend was 7 6 ) The absolute difference between what c and d had to spend was 5 7 ) The absolute difference between what d and e had to spend was 4 8 ) The absolute difference between what e and a had to spend was 11 9 ) How much did each have to spend? I used a spreadsheet to find the solution, but wanted to know the algebra way: a 21 b 2 c 9 d 14 e 10 Last edited by skipjack; June 11th, 2019 at 11:58 AM. 
June 11th, 2019, 06:37 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  
June 16th, 2019, 09:09 PM  #3 
Member Joined: Jun 2019 From: AZ, Seattle, San Diego Posts: 30 Thanks: 21 
Hi dbmathis. Denis' link didn't work for me, but I don't think Wolfram Alpha will answer your question, anyway (unless you pay to see their fancy steps). When we consider absolute value statements like ab=19, we need to consider both possibilities: Either a is the larger number or b is. That is, ab=19 OR ba=19. If we knew the increasing order were b,c,e,d,a, then we could write a system of five equations and solve it. In this exercise, we don't know the order. We're given only the distance between five pairs of values and that sum. I started by noting (of the pairs listed) that a and b are farthest apart. Next, I noted that e is about half as far away from a than b is. So, it seemed natural to sketch e about halfway between a and b, as a first guess. Next, I started taking cases, using facts that b is 7 units away from c AND d is 5 units away from c. My sketches quickly eliminated a number of possibilities because those results don't jive with what's known about the relative distances of a,b,e. Eventually, I arrived at the following order and interstitial distances: b∙∙∙7∙∙∙c∙∙∙1∙∙∙e∙∙∙4∙∙∙d∙∙∙7∙∙∙a I solved the following for b, and the rest followed: 5(b) + 4(7) + 3(1) + 2(4) + 1(7) = 56 ~ Cheers PS: I think this is what Denis was trying to do. 

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