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A merchant invested $10,000 at 5% annual interest, compounded semiann [#permalink]
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25 Jan 2017, 02:48
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A merchant invested $10,000 at 5% annual interest, compounded semiannually, and an amount of $X at 5% simple annual interest. At the end of the first year, the total interest earned on each investment was the same. What is the value of X? a. $10,000 b. $10,125 c. $10,250 d. $10,500 e. $10,825
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Last edited by Bunuel on 25 Jan 2017, 06:03, edited 1 time in total.
Renamed the topic, edited the question, moved to PS forum and added the OA.

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Re: A merchant invested $10,000 at 5% annual interest, compounded semiann [#permalink]
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25 Jan 2017, 03:56
Here we go,
I don't think that's a difficult question to someone who knows the formula to calculate Compound Interes and Simple Interest.
Compound Interest is calculated using 
A = P (1 + r/n)^(nt)  (1)
A = the future value of the investment/loan, including interest P = the principal investment amount (the initial deposit or loan amount) r = the annual interest rate (decimal) n = the number of times that interest is compounded per year t = the number of years the money is invested or borrowed for
and
Simple Interest is calculated using 
SI = (P x R x T)/100  (2)
Principal = P, Rate = R% per annum (p.a.) and Time = T years
Back to the question now.
A merchant invested $10,000 at 5% annual interest, compounded semiannually
Using equation 1
10000 * (1 + 0.05/2)^2 = 10506.25
Total interest earned in this case = 10506.25  10000 = 506.25  (a)
The question mentioned that tt the end of the first year, the total interest earned on each investment was the same..
Using equation (2)
506.25 = X * 0.05 * 1
so X = 10125
Option B should be the answer....
Now to answer to "im looking for a fast way to solve this."
I would have suggested to use the approximation technique by removing the decimal .25 in (a), but considering that the options are pretty close to each other, I don't think of any fast way to solve it as of now....

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Re: A merchant invested $10,000 at 5% annual interest, compounded semiann [#permalink]
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25 Jan 2017, 06:02
magniv123 wrote: A merchant invested $10,000 at 5% annual interest, compounded semiannually, and an amount of $X at 5% simple annual interest. At the end of the first year, the total interest earned on each investment was the same. What is the value of X?
a. $10,000 b. $10,125 c. $10,250 d. $10,500 e. $10,825
im looking for a fast way to solve this.
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Re: hi can anyone help me with this difficult question? [#permalink]
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25 Jan 2017, 10:03
magniv123 wrote: A merchant invested $10,000 at 5% annual interest, compounded semiannually, and an amount of $X at 5% simple annual interest. At the end of the first year, the total interest earned on each investment was the same. What is the value of X?
a. $10,000 b. $10,125 c. $10,250 d. $10,500 e. $10,825
im looking for a fast way to solve this.
thank you:) Hi, I can think of 2 ways.. 1) substitution.. Start from the middle value so that you can eliminate 2 choices either one on top or below.. Equation is \(10000(1+\frac{2.5}{100})^210000=\frac{10250*5}{100}\) 10000*1.025*1.02510000~500.. But 10250*.05~ 612.. So it has to be less than 10250.. AND ofcourse it has to be more than 10000 Only value left is 10125.. 2) method.. Leave 10000 and take it as 1.. So \((1+\frac{2.5}{100})^21=\frac{x*5}{100}\) 1.025*1.0251=x/20.......0.00506=x/20.....X=0.506*20=1.012.. Ans= 1.012*10000=10120 B
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A merchant invested $10,000 at 5% annual interest, compounded semiann [#permalink]
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16 Apr 2017, 00:20
chetan2u wrote: magniv123 wrote: A merchant invested $10,000 at 5% annual interest, compounded semiannually, and an amount of $X at 5% simple annual interest. At the end of the first year, the total interest earned on each investment was the same. What is the value of X?
a. $10,000 b. $10,125 c. $10,250 d. $10,500 e. $10,825
im looking for a fast way to solve this.
thank you:) Hi, I can think of 2 ways.. 1) substitution.. Start from the middle value so that you can eliminate 2 choices either one on top or below.. Equation is \(10000(1+\frac{2.5}{100})^210000=\frac{10250*5}{100}\) 10000*1.025*1.02510000~500.. But 10250*.05~ 612.. So it has to be less than 10250.. AND ofcourse it has to be more than 10000 Only value left is 10125.. 2) method.. Leave 10000 and take it as 1.. So \((1+\frac{2.5}{100})^21=\frac{x*5}{100}\) 1.025*1.0251=x/20.......0.00506=x/20.....X=0.506*20=1.012.. Ans= 1.012*10000=10120 B I came up with another fast way.. Because the amount in the first case has been invested semiannually, we can divide it into two periods of 6 months each. In the first 6 months the interest earned will be = 250 In the next 6 months the interest will be = 250 + interest on 250 which is 250 + 25/4 Now this total interest earned is equal to 5% of X.. \(250 + 250 + \frac{25}{4} = \frac{5}{100}*X\) \(X = 10125\) (B) This method probably minimizes the calculations.
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Re: A merchant invested $10,000 at 5% annual interest, compounded semiann [#permalink]
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21 Apr 2017, 21:21
Bunuel wrote: magniv123 wrote: A merchant invested $10,000 at 5% annual interest, compounded semiannually, and an amount of $X at 5% simple annual interest. At the end of the first year, the total interest earned on each investment was the same. What is the value of X?
a. $10,000 b. $10,125 c. $10,250 d. $10,500 e. $10,825
im looking for a fast way to solve this.
thank you:) Hi Bunuel, Can we solve the prblm in the below way ? Total amount= 10000+5/100 * 10000 =10500 CI amount should be little greater than > 10500 So in order to equal to an amount little greater than 10500, the nearest amount on which SI is calculated will be little greater than 10000, which is 10125. Please let me know if this method can be used.

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Re: A merchant invested $10,000 at 5% annual interest, compounded semiann
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