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# The compound interest on a certain sum of money invested at a certain

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Joined: 02 Sep 2009
Posts: 59095
The compound interest on a certain sum of money invested at a certain  [#permalink]

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02 Jul 2017, 02:42
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Difficulty:

45% (medium)

Question Stats:

67% (01:52) correct 33% (02:14) wrong based on 99 sessions

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The compound interest on a certain sum of money invested at a certain rate of interest in the 2nd year and in the 3rd year was $600 and$720 respectively. What was the rate of interest at which the sum of money was invested?

(A) 12.0%
(B) 12.5%
(C) 15.0%
(D) 20.0%
(E) 25.0%

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Joined: 26 Feb 2016
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The compound interest on a certain sum of money invested at a certain  [#permalink]

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02 Jul 2017, 02:52
1
Given :
On a given sum of money invested at a certain rate of interest,
Compound interest earned(2nd year) = 600$Compound interest earned(3rd year) = 720$

for the third year is $720. The difference in interest is$120, which is the interest earned on $600. We can calculate the rate of interest as follows - $$\frac{120}{600} * 100 = 20%$$ Therefore, the rate of interest on which the amount was invested is 20%(Option D) _________________ You've got what it takes, but it will take everything you've got Manager Joined: 11 Mar 2018 Posts: 150 The compound interest on a certain sum of money invested at a certain [#permalink] ### Show Tags Updated on: 26 Oct 2019, 18:23 Bunuel wrote: The compound interest on a certain sum of money invested at a certain rate of interest in the 2nd year and in the 3rd year was$600 and $720 respectively. What was the rate of interest at which the sum of money was invested? (A) 12.0% (B) 12.5% (C) 15.0% (D) 20.0% (E) 25.0% Let x be the principal amount Let y be the rate of interest. We also know $$Compound Interest = Principal [1 +\frac{R}{100}]^N - Principal$$ Now using the same formula and details as per the question we get, $$600 = x[1 + \frac{y}{100}]^2 - x$$ --- (1) $$720 = x[1 + \frac{y}{100}]^3 - x$$ --- (2) Subtracting (1) from (2), we get $$120 = x[1 + \frac{y}{100}]^3 - x[1 + \frac{y}{100}]^2$$ $$120 = x[1 + \frac{y}{100}]^2 (1 + \frac{y}{100} - 1)$$ $$120 = x[1 + \frac{y}{100}]^2 (\frac{y}{100})$$ --- (3) Using (1) in (3), we get $$120 = 600 (\frac{y}{100})$$ $$\frac{1}{5} = \frac{y}{100}$$ y = 20 Hence D _________________ Regards AD --------------------------------- A Kudos is one more question and its answer understood by somebody !!! Originally posted by adstudy on 26 Oct 2019, 05:17. Last edited by adstudy on 26 Oct 2019, 18:23, edited 1 time in total. Manager Joined: 10 Dec 2017 Posts: 132 Location: India Re: The compound interest on a certain sum of money invested at a certain [#permalink] ### Show Tags 26 Oct 2019, 06:51 Bunuel wrote: The compound interest on a certain sum of money invested at a certain rate of interest in the 2nd year and in the 3rd year was$600 and \$720 respectively. What was the rate of interest at which the sum of money was invested?

(A) 12.0%
(B) 12.5%
(C) 15.0%
(D) 20.0%
(E) 25.0%

Let the interest rate is X
{(1+X/100)^3}/{(1+X/100)^2}=720/600
1+X/100=6/5
X=20
D:)
How to get the equations
Principle amount is P and interest is X, after one year principle amount becomes P1
P1=P+P*X/100
P1=P(1+X/100)
P2=P1+P1*X/100
P2=P1(1+X/100)
P2=P(1+X/100)^2
P3=P(1+X/100)^3
(1+X/100)^3 is nothing but interest after 3 years that is 720 in this problem
same way (1+X/100)^2 interest after 2 years-620
Re: The compound interest on a certain sum of money invested at a certain   [#permalink] 26 Oct 2019, 06:51
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