Simple Interest Question

[prasadspraveen]

Pat invested a total of $3000. Part of the money was invested at a simple interest account that paid 10% interest annually, and the remainder was invested in another account that paid 8% interest annually. If the interest earned at the end of the first year from these investments was $256, how much did Pat invest at 10% and how much at 8%?

[mkamony]

Pat invested a total of $3000. Part of the money was invested at a simple interest account that paid 10% interest annually, and the remainder was invested in another account that paid 8% interest annually. If the interest earned at the end of the first year from these investments was $256, how much did Pat invest at 10% and how much at 8%?

let amount invested at 10% = x

amount invested at 8% = 3000 - x

10% *x + 8% (3000-x) = 256

0.1x + 240 - 0.08x = 256

0.02x = 16

x = 800 <-- amount invested at 10%

amount invested at 8% = 3000 - 800 = 2200

[tommywallach]

Well said mkamony,

On a more general note, always consider the basic rules of algebra when trying to figure out how to solve a question, namely that you should always have as many equations as variables (there are numerous exceptions to this rule, but it remains a rule). We clearly have two variables here (amount invested at 10% and amount invested at 8%), so we need two equations:

x + y = 3000

AND

.1x + .08y = 256

Then we can combine. Mkamony's explanation kinda skipped the two variable step, so I thought I'd add it back in. Hope that helps!

-t