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A college student earned $7200 during summer vacation working as a waiter in a popular restaurant. The students invested part of the money at 10% and the rest a 9%. If the student received a total of $680 in interest at the end of the year, how much was invested at 10%?
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You could use systems of equations: X = amount invested in 10% Y = amount invested in 9% X + Y = 7200 0.1X + 0.09Y = 680 Solving for Y @ 1st equation: Y = 7200 - X Substitute to 2nd equation: 0.1X + 0.09(7200 - X) = 680 0.01X = 32 X = 3200 Therefore, the amount invested @ 10% interest is $3200
$3200 at 10% and $4000 at 9% 7200 = a + b 680 = 0.10a +0.09b then solve simultaneously
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