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Pat invested a total of $3,000. Part of the money was invest [#permalink]
24 May 2019, 01:52
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
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Re: Pat invested a total of $3,000. Part of the money was invest [#permalink]
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? $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, 3000x = 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
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Re: Pat invested a total of $3,000. Part of the money was invest [#permalink]
08 May 2020, 12:11
SI = Principal * Rate * Time Let x = investment at 8% then, 3000x is remaining investment at 10% Apply formula and add both investments SI=P*R*T =x*.08*1 =.08x SI=P*R*T =(3000x)*.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: 30002200=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]
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? $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, 3000x = 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 Hi Brent, If we assume x at 10% and 3000x 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]
08 May 2020, 12:27
Farina wrote: Hi Brent,
If we assume x at 10% and 3000x 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, 3000x = 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 = 0.1x1 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 = 800So, $800 was invested at 10%, and $2200 was invested at 0.08%Cheers, Brent
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Re: Pat invested a total of $3,000. Part of the money was invest
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08 May 2020, 12:27





