Author 
Message 
TAGS:

Hide Tags

VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1029
Location: India
GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)

An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
27 Oct 2012, 00:54
3
This post received KUDOS
16
This post was BOOKMARKED
Question Stats:
67% (02:29) correct 33% (02:10) wrong based on 403 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1029
Location: India
GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)

Re: An amount is deposited into an account [#permalink]
Show Tags
27 Oct 2012, 00:56
1
This post was BOOKMARKED
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



Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 617
Location: India
Concentration: Strategy, General Management
Schools: Olin  Wash U  Class of 2015
WE: Information Technology (Computer Software)

Re: An amount is deposited into an account [#permalink]
Show Tags
27 Oct 2012, 01:03
4
This post received KUDOS
4
This post was BOOKMARKED
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.
_________________
Lets Kudos!!! Black Friday Debrief



VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1029
Location: India
GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)

Re: An amount is deposited into an account [#permalink]
Show Tags
27 Oct 2012, 01:27
$900 in the third year
t= 3 year .....why have you taken t= 2 years



Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 617
Location: India
Concentration: Strategy, General Management
Schools: Olin  Wash U  Class of 2015
WE: Information Technology (Computer Software)

Re: An amount is deposited into an account [#permalink]
Show Tags
27 Oct 2012, 01:44
1
This post received KUDOS
1
This post was BOOKMARKED
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.
_________________
Lets Kudos!!! Black Friday Debrief



Senior Manager
Joined: 27 May 2012
Posts: 473

Re: An amount is deposited into an account [#permalink]
Show Tags
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?
_________________
 Stne



Manager
Joined: 21 Oct 2013
Posts: 189
Location: Germany
GPA: 3.51

Re: An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
21 Jan 2014, 05:47
1
This post received KUDOS
2
This post was BOOKMARKED
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!



Senior Manager
Joined: 27 May 2012
Posts: 473

Re: An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
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.
_________________
 Stne



Math Expert
Joined: 02 Sep 2009
Posts: 45498

Re: An amount is deposited into an account [#permalink]
Show Tags
22 Jan 2014, 03:13
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 27 May 2012
Posts: 473

Re: An amount is deposited into an account [#permalink]
Show Tags
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.
_________________
 Stne



Manager
Joined: 21 Oct 2013
Posts: 189
Location: Germany
GPA: 3.51

Re: An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
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



Current Student
Joined: 06 Sep 2013
Posts: 1891
Concentration: Finance

Re: An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
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



Intern
Joined: 14 May 2014
Posts: 41

Re: An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
21 May 2014, 02:25
1
This post was BOOKMARKED
Attachment: File comment: Amount at different times
gmatintrest.jpg [ 11.83 KiB  Viewed 3287 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
_________________
Help me with Kudos if it helped you "
Mathematics is a thought process.



Intern
Joined: 12 Sep 2012
Posts: 25

Re: An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
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.



Intern
Joined: 16 Apr 2015
Posts: 42
Concentration: Operations, Strategy

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



Board of Directors
Joined: 17 Jul 2014
Posts: 2734
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
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



Intern
Joined: 11 Oct 2017
Posts: 22

Re: An amount is deposited into an account accruing interest ann [#permalink]
Show Tags
17 Oct 2017, 14:50
1
This post was BOOKMARKED
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 At the end of the second year/ beginning of third year, the value was 900. At the end of the third year/beginning of the fourth year, the value was 1080. The difference between the amounts: 1080900 = 180 Beginning of year three divided into the difference 180/900 = 0.2 = 20% rate plug into the formula x(1+r)^2 = 900 x(1+0.2)^2 = 900 x(1.44) = 900 x = 625 or into the other formula: x(1+0.2)^3 = 1080 x(1.728) = 1080 x = 625 The easier approach would be the x(1+0.2)^2 = 900 since there is less multiplication.




Re: An amount is deposited into an account accruing interest ann
[#permalink]
17 Oct 2017, 14:50






