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A sum of money invested under simple interest, amounts to $1200 in
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02 Jul 2017, 02:23
Question Stats:
72% (02:42) correct 28% (02:32) wrong based on 78 sessions
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A sum of money invested under simple interest, amounts to $1200 in three years and $1500 in five years. What is the rate at which the sum of money was invested? (A) 10% (B) 15% (C) 20% (D) 25% (E) 45%
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A sum of money invested under simple interest, amounts to $1200 in
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02 Jul 2017, 03:15
Bunuel wrote: A sum of money invested under simple interest, amounts to $1200 in three years and $1500 in five years. What is the rate at which the sum of money was invested?
(A) 10% (B) 15% (C) 20% (D) 25% (E) 45% Principal, Rate is same for \(3\) years and \(5\) years.
\(Amount = P + SI\)  (\(SI =\) Simple interest. \(SI = \frac{Principle * Rate * Time}{100}\))
\(1200 = P + SI_3 => P = 1200  SI_3\)  (i)  (\(SI_3 = SI\) for \(3\) years)
\(1500 = P + SI_5 => P = 1500  SI_5\)  (ii)  (\(SI_5 = SI\) for \(5\) years) Equating (i) and (ii), we get;
\(1200  SI_3 = 1500  SI_5\)
\(1200  \frac{PR*3}{100} = 1500  \frac{PR*5}{100}\)
\(120000  PR*3 = 150000  PR*5\)
\(PR*5  PR*3 = 150000  120000\)
\(PR(5  3) = 30000\)
\(PR = \frac{30000}{2} = 15000\)
Substituting value of \(PR\) in \(SI_3\), we get;
\(1200 = P + \frac{PR*3}{100}\)
\(1200 = P + \frac{15000*3}{100} = P + 450\)
\(P = 1200  45 = $750\)
\(SI\) for amount \(1200\) for \(3\) years \(= $450\)
\(450 = \frac{PR*3}{100}\)
\(R = \frac{450 * 100}{3*750} = 20\)%.
Answer (C)...



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Re: A sum of money invested under simple interest, amounts to $1200 in
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02 Jul 2017, 03:44
The amount at the end of 5 years is 1500 and at the end of 3 years is 1200 Total interest earned in these 2 years = 15001200 = 300 Interest earned per year = \(\frac{300}{2}\) = 150
The original amount invested at the beginning of year 1 = 12003*(150) = 1200450 = 750 Annual rate of interest = \(\frac{150}{750}\) = \(\frac{1}{5}\) = 20%
Answer C



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Re: A sum of money invested under simple interest, amounts to $1200 in
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02 Jul 2017, 04:34
Rate of SI =p*r*t according to the question we have 1200=p(1+r*3/100).......(1) 1500=P(1+r*5/100).......(2) Divide 1 by 2 we have 4/5=(100+3r)/(100+5r) solving we get r=20% Thus C
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Re: A sum of money invested under simple interest, amounts to $1200 in
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03 Jul 2017, 11:42
It is given that: 3PR/100 + P = 1200  (1)
PR = 100 (1200  P)/3  (2)
Also given: 5PR/100 + P = 1500 (3)
substituting equation 2 into 3, 500/300*(1200P) + P = 1500
or P = 750 (4)
further substituting (4) into (2)
R is 20%



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Re: A sum of money invested under simple interest, amounts to $1200 in
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03 Jul 2017, 12:29
Bunuel wrote: A sum of money invested under simple interest, amounts to $1200 in three years and $1500 in five years. What is the rate at which the sum of money was invested?
(A) 10% (B) 15% (C) 20% (D) 25% (E) 45% I got the answer of C  20% however my method was a bit longer one. Some really good solutions out there.
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Re: A sum of money invested under simple interest, amounts to $1200 in
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17 May 2019, 13:23
3Pr=1200P 5Pr=1500P 2Pr= 15001200=300 so, Pr=150 So, P=1200450=750 ; r=150/750=0,2 or 20%
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Re: A sum of money invested under simple interest, amounts to $1200 in
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17 Aug 2019, 07:28
(100+3R)/(100+5R)=1200/1500 \implies R=20
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Re: A sum of money invested under simple interest, amounts to $1200 in
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