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June 12th, 2016, 11:57 PM  #1 
Senior Member Joined: Apr 2008 Posts: 176 Thanks: 3  a word problem involving a system of linear equations
Suppose that you want to buy 0.3 kg of Type A raisin bran and 0.5 kg of Type B raisin bran. You want a mixture to have 11% of raisins. In the mixture, there is twice as much Type A as Type B. What percent of raisins do you need in each type? This is how I set up the question. Let A and B be the percents of Type A raisins and Type B raisins respectively 0.3A + 0.5B = 11 A = 2B Solving for A and B gives me the wrong answer. The answer key says A = 16% and B=8%. Can someone explain it? Thanks a lot. 
June 13th, 2016, 03:19 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,640 Thanks: 1319 
Let p = percentage of raisins in type B ... 2p = percentage in type A $2p(0.3)+p(0.5)=11(0.8 )$ $0.6p+0.5p=8.8$ $1.1p=8.8$ $p=8$% in type B $2p=16$% in type A 
June 14th, 2016, 06:30 PM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,740 Thanks: 709  Quote:
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