Name: Duane
Who is asking: Other
Level of the question: All

Question: Hey i am pretty lost with this problem.

Pat invested a total of $3000 dollars. part of the money yields 10 percent interest per year and the rest yields 8 percent interest per year if the total yearly interest is $256 how much did pat invest at 10 per cent and how much at 8 percent.

I am not sure of the formula to solve this equation i though it was to divide out 10 and 8 percent out of 3000 then subtract.


Hi Duane,

You have the correct idea. You need to divide the $3000 into 2 parts and apply 10% to one part and 8% to the other. If you only knew one part you could find the other since the sum is $3000. This is a place where the notation of algebra really helps.

Suppose that $x was invested at 10% them $(3000 - x) was invested at 8%.

In one year, $x invested at 10% would yield an interest of 0.10 x.
In one year, $(3000 - x) invested at 8% would yield an interest of 0.08 (3000 - x).

But you know that the total interest is $256 and hence

0.10 x + 0.08 (3000 - x) = 256

Solve for x.