# SOLUTION: Kelly has a coffee shop. Tea is \$1.05, Coffee is \$1.35, and Hot Chocolate is \$2.25. One day she sold 20 more cups of coffee than hot chocolate, 30 more cups of coffee than tea fo

Algebra ->  -> SOLUTION: Kelly has a coffee shop. Tea is \$1.05, Coffee is \$1.35, and Hot Chocolate is \$2.25. One day she sold 20 more cups of coffee than hot chocolate, 30 more cups of coffee than tea fo      Log On

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 Click here to see ALL problems on Expressions-with-variables Question 189966: Kelly has a coffee shop. Tea is \$1.05, Coffee is \$1.35, and Hot Chocolate is \$2.25. One day she sold 20 more cups of coffee than hot chocolate, 30 more cups of coffee than tea for a total of \$202.50. How many cups of each drink did she sell?Found 2 solutions by nerdybill, stanbon:Answer by nerdybill(7090)   (Show Source): You can put this solution on YOUR website!Let x = number of cups of hot chocolate sold then 20+x = number of cups of coffee sold (20+x)-30 = number of cups of tea sold or x-10 = number of cups of tea sold . 1.05(x-10) + 1.35(20+x) + 2.25x = 202.50 1.05x - 10.5 + 27 + 1.35x + 2.25x = 202.50 4.65x + 16.5 = 202.50 4.65x = 186 x = 40 (cups of hot chocolate) . Cups of coffee: 20+x = 20+40 = 60 (cups of coffee) . Cups of tea: x-10 = 40-10 = 30 (cups of tea) Answer by stanbon(60772)   (Show Source): You can put this solution on YOUR website!Kelly has a coffee shop. Tea is \$1.05, Coffee is \$1.35, and Hot Chocolate is \$2.25. One day she sold 20 more cups of coffee than hot chocolate, 30 more cups of coffee than tea for a total of \$202.50. How many cups of each drink did she sell? ------------------------ Let # of hot chocolate be "h". Then # of coffee sold was "h+20" And # of tea sold was h+20-30 = "h-10" ------------------------ Value Equation: 2.25h + 1.35(h+20) + 1.05(h-10) = 202.50 Multiply thru by 100 to get: 225h + 135h + 2700 + 105h - 1050 = 20250 --- h = 40 (# of cups of hot chocolate sold) h+20 = 60 (# of cups of coffee sold) h-10 = 30 (# of cups of tea sold) ====================================== Cheers, Stan H.