Investment: You earned $462 after 1 year by investing $11,400 split into amount paying 2% and 3% annual interest. How much money was invested in each rate?

Answer 1

I believe this problem is either a trick question or has an error.

Let's say we put x dollars in the 2% interest investment, and therefore 11400-x in the 3% investment. If the total amount we accrued in interest after one year was $462, then

0.02x+0.03(11400-x)=462

This is a simple enough algebra expression which we can solve for x. The value we get is

x=$-12,000

Which should pretty clearly not be a sensible answer. We have to invest -12000 dollars at 2% and 23400 dollars at 3% when we only have 11400 to work with?

So let's test that there's no sensible answer by investing ALL of our money in the 3% account (which, to be fair, why wouldn't we just do that anyway?). The amount we'd get after one year is just 0.03(11400), which is $342. But this is the highest possible interest we could hope to get; we invested all of our money in the highest possible account, and it's still not as high at $462! So getting to $462 is simply impossible.

Answer 2

(11400-x).02 + x(.03) = 462

228 - .02x + .03x = 462

,01x = 462-228 = 234

x = 234/.01 = $23,400 invested at 3%

11400-23400 = negative $12,00 invested at 2%

which doesn't seem to make sense

possibly the problem was supposed to read 20 and 30%

then

.1x = 234

x = $2,340 invested at 30%, $9,060 invested at 20%