Carcass wrote:
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. 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?
$800 at 10% and $2,200 at 8%
Math Review
Question: 13
Page: 244
Difficulty: medium/hard
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
8%So,
3000-x = the amount of money invested at
10%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
8% account =
0.08x1 year interest on
10% account =
0.1(3000 - x)We can write:
0.08x +
0.1(3000 - x) = 256
Expand: 0.08x + 300 - 0.1x = 256
Simplify: 300 - 0.02x = 256
Subtract 300 from both sides to get: -0.02x = -44
Solve:
x = -44/-0.02 = 44/0.02 = 4400/2 =
2200So,
$2200 was invested at
8%, and
$800 was invested at
10%Cheers,
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
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