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A sum of money invested under simple interest, amounts to $1200 in

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Bunuel
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A sum of money invested under simple interest, amounts to $1200 in [#permalink] New post   02 Jul 2017, 02:23
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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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danlew
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Re: A sum of money invested under simple interest, amounts to $1200 in [#permalink] New post   02 Jul 2017, 03:44
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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 = 1500-1200 = 300
Interest earned per year = \(\frac{300}{2}\) = 150

The original amount invested at the beginning of year 1 = 1200-3*(150) = 1200-450 = 750
Annual rate of interest = \(\frac{150}{750}\) = \(\frac{1}{5}\) = 20%

Answer C
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A sum of money invested under simple interest, amounts to $1200 in [#permalink] New post   02 Jul 2017, 03:15
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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 [#permalink] New post   02 Jul 2017, 04:34
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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 [#permalink] New post   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*(1200-P) + 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 [#permalink] New post   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 [#permalink] New post   17 May 2019, 13:23
3Pr=1200-P 5Pr=1500-P
2Pr= 1500-1200=300 so, Pr=150
So,
P=1200-450=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 [#permalink] New post   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 [#permalink] New post   15 Apr 2024, 02:45
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Re: A sum of money invested under simple interest, amounts to $1200 in [#permalink]
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