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# Last year Luis invested x dollars for one year, half at 8

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Manager
Joined: 23 Mar 2008
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Last year Luis invested x dollars for one year, half at 8 [#permalink]

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18 Apr 2008, 07:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Last year Luis invested x dollars for one year, half at 8 percent simple annual interest and the other half at 12 percent simple annual interest. Now he wants to reinvest the x dollars for one year in the same two types of investments, but the lower rate has decreased. If the higher rate is unchanged , what fraction of the x dollars must be reinvested at the 12 percent rate so that the total interest earned from the x dollars will be the same for both years?

1) the lower rate is now 6%
2) the total amount of interest earned from the two investments last eyar was $3000 Manager Joined: 02 Mar 2008 Posts: 210 Concentration: Finance, Strategy Followers: 1 Kudos [?]: 45 [0], given: 1 Re: investment [#permalink] ### Show Tags 18 Apr 2008, 08:49 A last year: 8%*0.5x + 12%*0.5x 1) this year: 6%*kx + (1-k)x*12% equal 2 sides, we can solve for k : sufficient 2) the total amount does not help. we have 2 variables the ratio k and the new interest VP Joined: 10 Jun 2007 Posts: 1459 Followers: 7 Kudos [?]: 255 [0], given: 0 Re: investment [#permalink] ### Show Tags 18 Apr 2008, 12:54 puma wrote: Last year Luis invested x dollars for one year, half at 8 percent simple annual interest and the other half at 12 percent simple annual interest. Now he wants to reinvest the x dollars for one year in the same two types of investments, but the lower rate has decreased. If the higher rate is unchanged , what fraction of the x dollars must be reinvested at the 12 percent rate so that the total interest earned from the x dollars will be the same for both years? 1) the lower rate is now 6% 2) the total amount of interest earned from the two investments last eyar was$3000

A.

(1) Set n = fraction of x needed
Original interest earned = 0.12*(x/2) + 0.08*(x/2) = 0.1x
New interest = 0.12*(x/2 + nx) + 0.06*(x/2)
We want to same interest, so
0.1x = 0.12*(x/2 + nx) + 0.06*(x/2) = 0.06x + 0.12nx + 0.03x
0.1 = 0.06 + 0.12n + 0.03
We can find n, SUFFICIENT

(2) Without knowing the decreased rate, you cannot find how much more money you need. INSUFFICIENT
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Re: investment [#permalink]

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18 Apr 2008, 18:37
puma wrote:
Last year Luis invested x dollars for one year, half at 8 percent simple annual interest and the other half at 12 percent simple annual interest. Now he wants to reinvest the x dollars for one year in the same two types of investments, but the lower rate has decreased. If the higher rate is unchanged , what fraction of the x dollars must be reinvested at the 12 percent rate so that the total interest earned from the x dollars will be the same for both years?

1) the lower rate is now 6%
2) the total amount of interest earned from the two investments last eyar was $3000 A little more information on why #2 is wrong: Year 1 - 8%(x/2) + 12%(x/2) =$3,000
4%x + 6%x = $3,000 10%x =$3,000
x = $30,000 Year 2 - I%(x) + 12%(1-x) =$3,000

You have two variables I and x to solve for and thus #2 is unsolvable.
Re: investment   [#permalink] 18 Apr 2008, 18:37
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# Last year Luis invested x dollars for one year, half at 8

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