SOLUTION: Susan invests 2 times as much money at 8% as she does at 6%. If her total imterest after 1 year is $1,100. How much does she have invested at each rate?
I made an equation 2x+
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I made an equation 2x+
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Question 1053825: Susan invests 2 times as much money at 8% as she does at 6%. If her total imterest after 1 year is $1,100. How much does she have invested at each rate?
I made an equation 2x+ y =1100 and 0.08x + 0.06y= $1100 but when i solve i get a negative number so i mnow my formulas are nit correct Found 2 solutions by josgarithmetic, MathTherapy:Answer by josgarithmetic(39616) (Show Source):
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Susan invests 2 times as much money at 8% as she does at 6%. If her total imterest after 1 year is $1,100. How much does she have invested at each rate?
I made an equation 2x+ y =1100 and 0.08x + 0.06y= $1100 but when i solve i get a negative number so i mnow my formulas are nit correct
The correct approach is:
Let the amount invested at 6% be S
Then amount invested at 8% = 2S
Do you understand this?
Interest earned from 6% investment: .06S
Interest earned from 8% investment: .08(2S), or .16S
We then get the following TOTAL INTEREST AMOUNT equation: .06S + .16S = 1,100
Solve this for S, the amount invested at 6%
Double or multiply value of S by 2 to get amount invested at 8%.
It's that easy....nothing complex!
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