Max went to the store and bought 22 apples. He bought twice as many red apples as he did green, and he bought one-third as many yellow as he did red. How many of each type of apple did he buy?
@Mashy HELP
Solve the following for g, y and r:\[{g + y + r = 22, r = 2 g, 3 y = r}\]
{g = 6, y = 4, r = 12} where g, y and r is the number of green, yellow and red apples respectively.
yummy, for God's sake SPIT IT OUT.
\[r=red\]\[g=green\]\[y=yellow\] \[r=2g\]\[y={r\over3}~~\implies~~y={2g\over3}\]so lets write the equation:\[r+y+g=22\]substitute all the values:\[2g+{2g\over3}+g=22\]find common denominators:\[{6g\over3}+{2g\over3}+{3g\over3}=22\]add:\[{11g\over3}=22\]multiply by the reciprocal to isolate \(g\):\[{\cancel{3}^1\over\cancel{11}^1}\times{\cancel{11}^1g\over\cancel{3}^1}=\cancel{22}^2\times{3\over\cancel{11}^1}\]lets write that a little cleaner:\[g=2\times3\]multiply:\[g=6\] now plug that into \(r=2g\) and \(y={2g\over3}\):\[r=2(6)~~~\implies~~~r=12\] \[y={2(6)\over3}~~~\implies~~~y={12\over3}~~~\implies~~~y=4\]now lets test it out:\[r=12\]\[g=6\]\[y=4\]\[12+4+6=22~?\]\[22=22~\checkmark\] and thus:\[\text{Max bought 12 red, 6 green, and 4 yellow apples.}\]
@breja hope that helps! :)
@yummydum THANKS
youre welcome!
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