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Pat invested a total of $3,000. Part of the money was invest
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24 May 2019, 01:52
24 May 2019, 01:52
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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?
Show: :: OA
$800 at 10% and $2,200 at 8%
Math Review
Question: 13
Page: 244
Difficulty: medium/hard
Question: 13
Page: 244
Difficulty: medium/hard
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Re: Pat invested a total of $3,000. Part of the money was invest
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25 May 2019, 04:08
25 May 2019, 04:08
4
1
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?
Show: :: OA
$800 at 10% and $2,200 at 8%
Math Review
Question: 13
Page: 244
Difficulty: medium/hard
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 = 0.08x
1 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 = 2200
So, $2200 was invested at 8%, and $800 was invested at 10%
Cheers,
Brent
Re: Pat invested a total of $3,000. Part of the money was invest
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08 May 2020, 12:11
08 May 2020, 12:11
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SI = Principal * Rate * Time
Let x = investment at 8%
then, 3000-x is remaining investment at 10%
Apply formula and add both investments
SI=P*R*T
=x*.08*1
=.08x
SI=P*R*T
=(3000-x)*.10*1
=300-.1x
Add both and equate to given SI
.08x-.10x+300=256
.02x=44
x=2200
Pat invested at 8%
2200*8%
=176
Remainder: 3000-2200=800
800*10%=80
so,
176+80=256
Let x = investment at 8%
then, 3000-x is remaining investment at 10%
Apply formula and add both investments
SI=P*R*T
=x*.08*1
=.08x
SI=P*R*T
=(3000-x)*.10*1
=300-.1x
Add both and equate to given SI
.08x-.10x+300=256
.02x=44
x=2200
Pat invested at 8%
2200*8%
=176
Remainder: 3000-2200=800
800*10%=80
so,
176+80=256
