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Math中四麻煩題

發問:

1. when the price of a dozen bottles of wine is raised by$150,12 fewer bottles of wine can be bought at$900.find the original price per dozen bottles of wine. 2. a hawker bought a number of apples at $320 and sold them at$10 for four apples.he gained as he paid for 40 apples.how many apples did he buy?

最佳解答:

1.Let x be the original price per bottle of wine, the original number of bottle of wine can be bought = $900/n since the price of a dozen bottles of wine is raised by $150, then each bottle of wine is raised $12.5.Thereforeo, $900/(n+12.5) is number of bottle of wine can be bought after price is raised, the different is 12 bottles. $900/n﹣$900/(n+12.5) = 12 n2+12.5n﹣937.5 = 0 (n+37.5)(n﹣25) = 0 n = 25 or n = -37.5 (Rejected) ∴the original price per dozen bottles of wine is $25×12 = $300 2.Let x be the number of apples bought by the hawker, the total sales = $10 (x/4) = $2.5x the cost = $320 therefore, unit cost of apple is $320/x profit = ( $320/x )40 = $12800/x for sales﹣profit = cost 2.5n﹣12800/n = 320 2.5n2﹣12800 = 320n 2.5n2﹣320n - 12800 = 0 n2﹣128n﹣5120 = 0 (n﹣160)(n + 32) = 0 n = 160 or n = -32 (rejected) ∴The hawker bought 160 apples

其他解答:

1.Let the original price per dozen bottle of wine be x. 900/x - 900/(x+150)=12/12 900(x+150)-900x / x(x+150) =1 135000=x(x+150) (x+450)(x-300)=0 x=-450)rej.) OR 300 2.Let the original price he bought be x. 10/4x - 320=40(320/x) 2.5x-320-12800=0 2.5x^2 -320-12800=0 x^2-128x-5120=0 (x-160)(x+32)=0 y=160 OR -32(rej.)A9A3995996B435E3