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Potwe 13 Ae Np 15 s

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  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the WeekProblem E and SolutionCoffee ... Problem A coffee merchant mixes two types of beans, bean   1  and bean   2 , to make different blends of coffee. All of these blends are sold at the same price. If he mixes bean   1  and bean   2  in theratio (by weight) 2 : 1, then he makes a profit of 26%. If he mixes bean   1  and bean   2  in theratio (by weight) 3 : 2, then he makes a profit of 20%. In what ratio should he blend bean   1 and bean   2  for his profit to be 40%? Solution Let  x  be the price the merchant pays for bean   1 , in $ per kg.Let  y  be the price the merchant pays for bean   2 , in $ per kg.Let  p  be the price the merchant sells the different blends, in $ per kg.If he mixes bean   1  and bean   2  in the ratio (by weight) 2 : 1, then he makes a profit of 26%.This means that in 1 kg of the blend,  23  kg is bean   1  and  13  kg is bean   2 .1 kg of this blend will cost the merchant  23 x  +  13 y . Since he makes a profit of 26%, we have1 . 26  23 x  + 13 y   =  p  (1)If he mixes bean   1  and bean   2  in the ratio (by weight) 3 : 2, then he makes a profit of 20%.This means that in 1 kg of the blend,  35  kg is bean   1  and  25  kg is bean   2 .1 kg of this blend will cost the merchant  35 x  +  25 y . Since he makes a profit of 20%, we have1 . 20  35 x  + 25 y   =  p  (2)Since  p  =  p  in (1) and (2), we have 1 . 26  23 x  +  13 y   =  p  = 1 . 20  35 x  +  25 y  , and so1 . 26  23 x  + 13 y   = 1 . 20  35 x  + 25 y  Dividing both sides by 1.20, 1 . 05  23 x  + 13 y   = 35 x  + 25 y Expanding the left side, 0 . 7 x  + 0 . 35 y  = 0 . 6 x  + 0 . 4 y 0 . 1 x  = 0 . 05 yy  = 2 x Substituting  y  = 2 x  back into (1), we have  p  = 1 . 26  23 x  + 13 y   = 1 . 26  23 x  + 13(2 x )   = 1 . 26  23 x  + 23 x   = 1 . 26  43 x   = 1 . 68 x  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING We want the profit to be 40% for our new blend.If we make 1 kg of the new blend, let  b  be the amount of bean   1  in the blend. Therefore(1 − b ) will be the amount of bean   2  in the blend.It will cost the merchant  bx  + (1 − b ) y  to make the blend. The merchant will sell the blend for  p  $/kg and we want a profit of 40%.In other words, we want 1 . 4[ bx  + (1 − b ) y ] =  p . (3)Substituting  y  = 2 x  and  p  = 1 . 68 x  in (3), we have1 . 4[ bx  + (1 − b )2 x ] = 1 . 68 x 1 . 4[ bx  + 2 x − 2 bx ] = 1 . 68 x 1 . 4 bx  + 2 . 8 x − 2 . 8 bx  = 1 . 68 x 1 . 12 x − 1 . 4 bx  = 01 . 12 x (1 − 1 . 25 b ) = 0Since  x  = 0 (in other words, the beans are not free), we have 1 − 1 . 25 b  = 0, and so  b  = 0 . 8.So in 1 kg of the new blend, 0.8 kg is bean   1  and (1 − 0 . 8) = 0 . 2 kg is bean   2 .Therefore, the merchant should blend bean   1  and bean   2  in the ratio 0 . 8 : 0 . 2 = 4 : 1.
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