Hi Brent,

If we assume x at 10% and 3000-x at 8% we dont get the same answer which is obvious but I still dont know why we cant assume like this. Although I have solved it but still there is some confusion

We should get the same answer no matter what. Let's try:

Pat invested a total of $3,000. Part of the money was invested in a money market account that paid 10 percent simple annual interest, and the remainder of the money was invested in a fund that paid 8 percent simple annual interest.
Let x = the amount of money invested at 10%
So, 3000-x = the amount of money invested at 8%

If the interest earned at the end of the first year from these investments was $256, how much did Pat invest at 10 percent and how much at 8 percent?
1 year interest on 10% account = 0.1x
1 year interest on 8% account = 0.08(3000 - x)

We can write: 0.1x + 0.08(3000 - x) = 256
Expand: 0.1x + 240 - 0.08x = 256
Simplify: 240 + 0.02x = 256
Subtract 240 from both sides to get: 0.02x = 16
Solve: x = 16/0.02 = 1600/2 = 800

So, $800 was invested at 10%, and $2200 was invested at 0.08%

Cheers,
Brent

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