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Intern
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A person invested $20000 for one year, part at 8.5 percent [#permalink]
 25 Feb 2005, 14:09
A person invested $20000 for one year, part at 8.5 percent simple annual interest and part at 12.5 percent simple annual interest. For a $2500 interest, at least how much of the money is to be deposited at 12.5 percent?
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Manager
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Well the answer is deposit the entire amount at 12.5% .
since 12.5% of 20,000 = 2500
Equation is
2500 = (8.5/100)*x + (12.5/100)*(20,000-x)
2500 = 2500 - 4x/100
=> x =0
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Intern
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You are missing something
The question asking for AT LEAST how much of the money is to be deposited at 12.5 percen
The minimum money needed
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Manager
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That's it 20,000
Suppose we deposit $1000 at 8.5% and $19,000 at 12.5%
Then the total interest is 85 + 2375 = 2460 < 2500
You can not deposit anything less than 20,000 for 12.5%
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SVP
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let me reword the problem as under:
A has $20,000.00. He/she has two options to make his/her investment: Bond X offers 10% annual coupon intrate rate and Bond Y offers 15% annual coupon interest rate. A plans to make an income of $4500, how much should he invest in X and Y?
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Manager
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Is that a new question? Or are you trying to rephrase the problem?
Anyway the answer is 20,000.
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SVP
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anirban16 wrote: Is that a new question? Or are you trying to rephrase the problem?
Anyway the answer is 20,000.
It was a new question from MA, anirban. You can see the numbers are different. And you answer may not answer his question. 
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Intern
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.125x + .085 (20000-x) = 2500
Solve for x, x = 20000
_________________
The biggest risk in life is not taking one!!!
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Director
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MA wrote: let me reword the problem as under:
A has $20,000.00. He/she has two options to make his/her investment: Bond X offers 10% annual coupon intrate rate and Bond Y offers 15% annual coupon interest rate. A plans to make an income of $4500, how much should he invest in X and Y?
What am I doing wrong here?
Let's say Mr A. allocated $X and $Y for bond X and Y resp.
0.1x + 0.15y = 4500 (sum of interests for 1 year)
x + y = 20000 (sum of money invested should be 20,000)
I solve for x and y and i get y = 50K and x = -30K 
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Director
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Arsene_Wenger
you are right in your approach, your answer just means that this made up problem is not solvable. I am sure MA was just trying to convey basic concept behind this problem. If you change 20000 to some number heigher than 50000 problem is solvable
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Praveen
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anirban16 wrote: Is that a new question? Or are you trying to rephrase the problem? Anyway the answer is 20,000.
yes, it is rephrased-new question.
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close
