My Math Forum a word problem involving a system of linear equations

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 June 12th, 2016, 11:57 PM #1 Senior Member   Joined: Apr 2008 Posts: 192 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: 2,737 Thanks: 1387 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
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Quote:
 Originally Posted by davedave 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
Here is your error. Since "A" is the percent of raisins in type A, 0.3A, A percent of 0.3 kg, is the weight of raisins in type A. Similarly 0.5B is the weight of raisins in type B. There sum is the total weight of raisins in the mixture, not the percent. That is why skeeter used (0.3+ 0.5)11= (0.8 )11= 0.88 instead of 11.

Quote:
 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.

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