Mathematics
tyanovski6
12

Money is invested at two rates of interest. One rate is 8% and the other is 2%. If there is $700 more invested at 8% than at 2%, find the amount invested at each rate if the total annual interest received is $380. Let x = amount invested at 8% and y = amount invested at 2%.

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(1) Answers
rekeyia21

In the given problem, two assumptions are already provided. Let the amount of money invested at 8% = x Let the amount of money invested in 2% = y Then from the above question we can find that x = y + 700 Also (8x/100) + (2y/100) = 380 8x + 2y = 380 * 100 8x + 2y = 38000 Now replacing the value of x from the first equation we get 8(y + 700) + 2y = 38000 8y + 5600 + 2y = 38000 10y = 38000 - 5600 10 y = 32400 y = 32400/10    = 3240 Putting the value of y in the first equation, we get x = y + 700    = 3240 + 700    = 3940 So the amount of money invested at 8% is $3940 and the amount of money invested at 2% is $3240.

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