If a man takes a portion of 8000 dollars and invests it at 3 percent and invests the rest of it at 4 percent and earns 275 dollars in interest in one year how much did he invest at each rate?

Answer 1

Many times when solving word problems, it's harder to figure out the question and the formulas to use than it is to do the algebra itself. So let's try to figure this out.

We'll let x be the amount of money the guy invested at 3 percent.

We'll let 8000 - x be the rest of it, which he invested at 4 percent.

That means that 0.03 * x is the amount of interest he earned on the money invested at 3 percent, and 0.04 * (8000 - x) is the amount he earned at 4 percent.

We also know that those two values add up to 275 dollars, so we can write the following equation: 0.03x + 0.04(8000 - x) = 275. Solving for x, we get

0.03x + 0.04(8000 - x) = 275

0.03x + 320 - 0.04x = 275

-0.01x = -45

x = 4500

So 4500 is the amount invested at 3 percent. The rest, 8000 - 4500 = 3500, is invested at 4 percent. How do you get -0.01x=-45