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Pat invested a total of $3,000. Part of the money was invest

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Pat invested a total of $3,000. Part of the money was invest [#permalink] New post 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?


[Reveal] Spoiler: OA
$800 at 10% and $2,200 at 8%


Math Review
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 [#permalink] New post 25 May 2019, 04:08
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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?


[Reveal] Spoiler: OA
$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 = 0.08x
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
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Re: Pat invested a total of $3,000. Part of the money was invest [#permalink] New post 08 May 2020, 12:11
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
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Re: Pat invested a total of $3,000. Part of the money was invest [#permalink] New post 08 May 2020, 12:14
GreenlightTestPrep wrote:
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?


[Reveal] Spoiler: OA
$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 = 0.08x
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


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
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Re: Pat invested a total of $3,000. Part of the money was invest [#permalink] New post 08 May 2020, 12:27
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Expert's post
Farina wrote:
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
_________________

Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course.
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Re: Pat invested a total of $3,000. Part of the money was invest   [#permalink] 08 May 2020, 12:27
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