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An amount is deposited into an account accruing interest ann [#permalink]
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27 Oct 2012, 00:54
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An amount is deposited into an account accruing interest annually at a fixed percentage rate. It is valued at $900 in the third year (after interest has compounded twice), and $1080 in the fourth year (after interest has compounded three times). What is the original amount? A. 520 B. 540 C. 600 D. 625 E. 650
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Re: An amount is deposited into an account [#permalink]
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27 Oct 2012, 00:56
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correct me if i am wrong
R= 6/5
Now , 900 = P (1+x/100)^3
900 = P {6/5 *6/5 *6/5}
P = 521



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27 Oct 2012, 01:03
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Archit143 wrote: correct me if i am wrong
R= 6/5
Now , 900 = P (1+x/100)^3
900 = P {6/5 *6/5 *6/5}
P = 521 Yup wrong approach. if R is interest rate and P is original amount then from question : \(900 = P (1+R/100)^2\) and \(1080 = P (1+R/100)^3\) => \(1080/900 = 1+R/100\) => \(R = 20%\) Substituting R in any one of the equations, you can obtain P. eg \(900 = P (1+0.2)^2\) => \(P =625\) Ans D it is. Hope it helps.
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Re: An amount is deposited into an account [#permalink]
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27 Oct 2012, 01:27
$900 in the third year
t= 3 year .....why have you taken t= 2 years



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Re: An amount is deposited into an account [#permalink]
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27 Oct 2012, 01:44
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Archit143 wrote: $900 in the third year
t= 3 year .....why have you taken t= 2 years Question says in 3rd year, not after 3rd year. In 3rd year interest would have been compounded twice. Hope it clarifies ur doubt.
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Re: An amount is deposited into an account [#permalink]
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19 Jan 2014, 01:42
Vips0000 wrote: Archit143 wrote: correct me if i am wrong
R= 6/5
Now , 900 = P (1+x/100)^3
900 = P {6/5 *6/5 *6/5}
P = 521 Yup wrong approach. if R is interest rate and P is original amount then from question : \(900 = P (1+R/100)^2\) and \(1080 = P (1+R/100)^3\) => \(1080/900 = 1+R/100\) => \(R = 20%\) Substituting R in any one of the equations, you can obtain P. eg \(900 = P (1+0.2)^2\) => \(P =625\) Ans D it is. Hope it helps. Am I missing anything ? Compound interest formula is > p(1+r/n)^(nt) where t is time in years , r is the rate in decimal and n is the number of times interest is compounded in a year based on this formula > equation should be >900 = p(1+r/2)^2 and >1080 = p(1+r/3)^3 if these are the correct equations, how do we solve these 2 to get the correct value of P?
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Re: An amount is deposited into an account accruing interest ann [#permalink]
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21 Jan 2014, 05:47
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Interest rate: 1080/900 = 6/5 = 1.2
so you would have x * 1.2 * 1.2 = 900
900 / 1.2 = 750 (easy approach: 90/12 = 7.5) 2nd year 750 / 1.2 = 625 (same as above, make it easy: 75/12 = 6.25) 1st year.
Hence D!



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Re: An amount is deposited into an account accruing interest ann [#permalink]
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22 Jan 2014, 02:51
unceldolan wrote: Interest rate: 1080/900 = 6/5 = 1.2
so you would have x * 1.2 * 1.2 = 900
900 / 1.2 = 750 (easy approach: 90/12 = 7.5) 2nd year 750 / 1.2 = 625 (same as above, make it easy: 75/12 = 6.25) 1st year.
Hence D! Sorry couldn't get you, can you elaborate? Bunuel can you have a look at this? I believe vips0000 has some things missing in the equation,Thank you.
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Re: An amount is deposited into an account [#permalink]
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22 Jan 2014, 03:13
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stne wrote: Vips0000 wrote: Archit143 wrote: An amount is deposited into an account accruing interest annually at a fixed percentage rate. It is valued at $900 in the third year (after interest has compounded twice), and $1080 in the fourth year (after interest has compounded three times). What is the original amount?
A. 520 B. 540 C. 600 D. 625 E. 650
correct me if i am wrong
R= 6/5
Now , 900 = P (1+x/100)^3
900 = P {6/5 *6/5 *6/5}
P = 521 Yup wrong approach. if R is interest rate and P is original amount then from question : \(900 = P (1+R/100)^2\) and \(1080 = P (1+R/100)^3\) => \(1080/900 = 1+R/100\) => \(R = 20%\) Substituting R in any one of the equations, you can obtain P. eg \(900 = P (1+0.2)^2\) => \(P =625\) Ans D it is. Hope it helps. Am I missing anything ? Compound interest formula is > p(1+r/n)^(nt) where t is time in years , r is the rate in decimal and n is the number of times interest is compounded in a year based on this formula > equation should be >900 = p(1+r/2)^2 and >1080 = p(1+r/3)^3 if these are the correct equations, how do we solve these 2 to get the correct value of P? The interest is compounded once a year. Why are you divide r by 2 and 3? Vips0000 solution is 100% correct. After interest has compounded twice the amount is 900: \(p(1+\frac{r}{100})^2=900\). After interest has compounded thrice the amount is 1,080: \(p(1+\frac{r}{100})^3=1,080\). Divide the second equation by the first one: \(1+\frac{r}{100}=\frac{1,080}{900}\) > \(1+\frac{r}{100}=\frac{6}{5}\) > \(r=20\). Substitute the value of r in either equations above: \(p(1+\frac{20}{100})^2=900\) > \(p=625\). Answer: D. Hope it's clear.
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Re: An amount is deposited into an account [#permalink]
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22 Jan 2014, 03:41
Bunuel wrote: stne wrote: Am I missing anything ? Compound interest formula is > p(1+r/n)^(nt) where t is time in years , r is the rate in decimal and n is the number of times interest is compounded in a year
based on this formula > equation should be >900 = p(1+r/2)^2 and >1080 = p(1+r/3)^3
if these are the correct equations, how do we solve these 2 to get the correct value of P? The interest is compounded once a year. Why are you divide r by 2 and 3? Vips0000 solution is 100% correct. After interest has compounded twice the amount is 900: \(p(1+\frac{r}{100})^2=900\). After interest has compounded thrice the amount is 1,080: \(p(1+\frac{r}{100})^3=1,080\). Divide the second equation by the first one: \(1+\frac{r}{100}=\frac{1,080}{900}\) > \(1+\frac{r}{100}=\frac{6}{5}\) > \(r=20\). Substitute the value of r in either equations above: \(p(1+\frac{20}{100})^2=900\) > \(p=625\). Answer: D. Hope it's clear. oh I see ! When it said in the third year after interest was compounded twice , I thought in the third year interest was compounded twice , but that is not so.In this question interest is compounded annually. Yes its clear now, thank you.
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Re: An amount is deposited into an account accruing interest ann [#permalink]
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22 Jan 2014, 04:36
stne wrote: unceldolan wrote: Interest rate: 1080/900 = 6/5 = 1.2
so you would have x * 1.2 * 1.2 = 900
900 / 1.2 = 750 (easy approach: 90/12 = 7.5) 2nd year 750 / 1.2 = 625 (same as above, make it easy: 75/12 = 6.25) 1st year.
Hence D! Sorry couldn't get you, can you elaborate? Bunuel can you have a look at this? I believe vips0000 has some things missing in the equation,Thank you. For interest problems you can either use the compound interest approach, as vips did, or my approach, repeating percentages. First, you want to know the annual interest rate. As you yield 1080 in the 4th year and 900 in the third year, you have 900 * x = 1080. Hence 1080/900 = x = 1.2 This is our interest rate, our repeating percentage. Thus, since we know that after the interest has compunded twice, the amount of the deposit is 900, we can say that X (original amount) * 1.2 * 1.2 = 900. You could compute and write 1.44x = 900 but in my opinion it is easier to divide 900 by 1.2 and the result by 1.2 again. For my computing approach: 900/1.2 looks difficult at first. BUT you can simplify this by shifting the decimal point. E.g. 900/12 = something. still too difficult. Better: 90/12 = 7.5 But since you have shifted the decimal point in the numerator one to the left and the decimal point in the denominator one to the right you have to shift the decimal point in the result two to the right. hence 7.5 > 75 > 750. Then you get x * 1.2 = 750 > x = 750/1.2. Repeat the steps above and get X! Hope it's clearer now! Greets



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Re: An amount is deposited into an account accruing interest ann [#permalink]
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26 Mar 2014, 10:04
Please refer to the explanation by Vips0000 or Bunuel.
For the last part one may want to do the following. 1.44x =900. x  900/1.44. Now take square root from both sides hence 30/1.2=25. Nos sqrt (x) = 25, so x must equal 625. Therefore D is the correct answer choice
Hope this adds Cheers J



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Re: An amount is deposited into an account accruing interest ann [#permalink]
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21 May 2014, 02:25
Attachment: File comment: Amount at different times
gmatintrest.jpg [ 11.83 KiB  Viewed 2400 times ]
Above image explains the total amount at different times.. Amount in third year = Amount at the end of second year = P(1+r)^2 Amount in the fourth year = Amount at the end of third year = P(1+r)^3 Given, P(1+r)^2 = 900 P(1+r)^3 = 1080 dividing second eq by first eq will give (1+r) = 1080/900 = 1.2 From first eq, P = 900/(1+r)^2 = 900/1.44 = 625 Answer is D
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Re: An amount is deposited into an account accruing interest ann [#permalink]
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12 Aug 2014, 11:31
Hi, It is hard to understand that why the rate is not divided by 2 in the third year and is not divided by 3 in the fourth year, while in the question it is said that the amount is compounded twice in the third year and compounded three times in the fourth year. So, according to the formula for third year it should be p(1+r/2)^2 and for fourth year it should be p(1+r/3)^3. Kindly tell me where I went wrong.



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Re: An amount is deposited into an account accruing interest ann [#permalink]
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15 Sep 2015, 05:24
took me more than 2.5 minutes to solve this.is there a shorter way ?



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Re: An amount is deposited into an account accruing interest ann [#permalink]
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20 Sep 2016, 06:30
interest compounded in the fourth year is 180. 180 is 20% of 900. if interest is the same then at the beginning of third year, we had 900*5/6. then ad the beginning of second year, we had 900*5/6 * 5/6. and at the beginning of first year, 900*5*5*5/6*6*6. let's split 900 into 3*3*10*10. now..do some simplifications. we get to 625




Re: An amount is deposited into an account accruing interest ann
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