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# If $1,000 is deposited in a certain bank account and remains  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Intern Joined: 19 Oct 2009 Posts: 44 Followers: 0 Kudos [?]: 47 [1] , given: 4 If$1,000 is deposited in a certain bank account and remains [#permalink]  13 Nov 2009, 14:45
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If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? (1) The deposit earns a total of$210 in interest in the first two years
(2) (1 + r/100 )^2 > 1.15

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-1-000-is-deposited-in-a-certain-bank-account-and-remains-104084.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 May 2014, 00:26, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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If $$I = 1000*[(1+\frac{r}{100})^n - 1)]$$, is r > 8%?

Essentially, you have 3 unknowns (I, r, and n), and you need to solve for r.

Statement 1: The deposit earns a total of $210 in interest in the first two years. You are given I ($210) and n (2), and therefore can solve for r. Sufficient. If you want to actually solve for this, you can do so, but there really is no need for it.

Statement 2: $$(1 + \frac{r}{100})^2 > 1.15$$.

This is a bit more tricky. The above formula can be rearranged to the following:

$$1 + \frac{r}{100} > \sqrt{1.15}$$

Now assuming we don't have a calculator handy, you can mentally approximate this (erring on the side of being conservative) as:

$$r > 7.5%$$ (It's actually less than this, but being conservative is fine)

We can make this approximation safely because, simply put, if r > 0, than $$(1+r)^2 = 1 + 2r + r^2$$ will always be greater than $$(1 + 2r)$$.

Since we need to know if r > 8%, this is insufficient.
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Note that this is a YES/NO DS question. So finding the solution regardless of whether it is YES or NO is sufficient.

Statement 1

We are given both the value of I and the value of n. That is two out of the three variables in the formula given above. No further calculation is necessary. Again, because this is a YES/NO question, even if the value is less than 8 percent, you can still answer the question.

SUFFICIENT

Statement 2

Because the statement given is an inequality, we must solve the equation to determine whether it is sufficient. If the r given in the statement is greater than a value less than 8%, then we don't if that specific value of r is greater than or less than 8%.

(1+r/100)>(1.15)^(1/2)
1+r/100>1.07
r>.07

NOT SUFFICIENT

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compound interest [#permalink]  13 Mar 2010, 00:09
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if $1,000 is deposited in a certain bank account and remains in the account along with any accrued interest, the dollar amount of interest, I, earned by deposit in the first n years is given by: I = 1,000 ((1+r/100)^n -1) where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank > 8%? 1) the deposit earns a total of$210 in interest in the first 2 years.

2) (1+r/100)^2 > 1.15
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Re: compound interest [#permalink]  13 Mar 2010, 01:57
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mustdoit wrote:
if $1,000 is deposited in a certain bank account and remains in the account along with any accrued interest, the dollar amount of interest, I, earned by deposit in the first n years is given by: I = 1,000 ((1+r/100)^n -1) where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank > 8%? 1) the deposit earns a total of$210 in interest in the first 2 years.

2) (1+r/100)^2 > 1.15

In other words we're asked whether $$r>8$$.
Stat. 1: the deposit earns a total of $210 in interest in the first 2 years. I replaced the "I" and "n" with the numbers from stat. 1: $$210=(1+\frac{r}{100})^{2-1}$$ - from this we can get "r" - sufficient. Stat. 2: $$(1+\frac{r}{100})^2>1.15$$ $$=>$$ $$1+\frac{r}{100}>\sqrt{1.15}$$ $$=>$$ $$r>(\sqrt{1.15}-1)*100$$ $$=>$$ $$r>(\approx1.07-1)*100$$ $$=>$$ $$r>7$$ - not sufficient So, the answer is A. Senior Manager Joined: 13 Dec 2009 Posts: 263 Followers: 10 Kudos [?]: 156 [0], given: 13 Re: compound interest [#permalink] 16 Mar 2010, 09:31 [quote="Igor010"] Stat. 1: the deposit earns a total of$210 in interest in the first 2 years.
I replaced the "I" and "n" with the numbers from stat. 1:
$$210=(1+\frac{r}{100})^{2-1}$$ - from this we can get "r" - sufficient.

[quote]

Isn't in statement 1 we do not have principle on which this interest was calculated. How can we assume it will be 1 only?
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Re: compound interest [#permalink]  16 Mar 2010, 09:57
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sidhu4u wrote:
Igor010 wrote:
Stat. 1: the deposit earns a total of $210 in interest in the first 2 years. I replaced the "I" and "n" with the numbers from stat. 1: $$210=(1+\frac{r}{100})^{2-1}$$ - from this we can get "r" - sufficient. Quote: Isn't in statement 1 we do not have principle on which this interest was calculated. How can we assume it will be 1 only? Sorry, don't understand your q. We have $$I=1000*(1+\frac{r}{100})^{n-1}$$ as given and statement 1 gives us some figures to use in this formula... Intern Joined: 16 Mar 2010 Posts: 1 Followers: 0 Kudos [?]: 0 [0], given: 4 Re: compound interest [#permalink] 16 Mar 2010, 11:22 Quote: if$1,000 is deposited in a certain bank account and remains in the account along with any accrued interest, the dollar amount of interest, I, earned by deposit in the first n years is given by:

I = 1,000 ((1+r/100)^n -1)

where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank > 8%?

1) the deposit earns a total of $210 in interest in the first 2 years. 2) (1+r/100)^2 > 1.15 A quick look suggests both statements are correct unless you calculate interest rate in each case. Since statement 1 one will give a concrete value of r no need to calculate it, it's sufficient. BUT statement 2 needs calculation of r since r>1 or r>2 etc. these types of answers are not sufficient. r > 7.2%, not sufficient. Manager Joined: 18 Feb 2010 Posts: 174 Schools: ISB Followers: 8 Kudos [?]: 158 [1] , given: 0 Re: compound interest [#permalink] 18 Mar 2010, 21:27 1 This post received KUDOS Igor010 wrote: mustdoit wrote: if$1,000 is deposited in a certain bank account and remains in the account along with any accrued interest, the dollar amount of interest, I, earned by deposit in the first n years is given by:

I = 1,000 ((1+r/100)^n -1)

where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank > 8%?

1) the deposit earns a total of $210 in interest in the first 2 years. 2) (1+r/100)^2 > 1.15 In other words we're asked whether $$r>8$$. Stat. 1: the deposit earns a total of$210 in interest in the first 2 years.
I replaced the "I" and "n" with the numbers from stat. 1:
$$210=(1+\frac{r}{100})^{2-1}$$ - from this we can get "r" - sufficient.

Stat. 2: $$(1+\frac{r}{100})^2>1.15$$ $$=>$$ $$1+\frac{r}{100}>\sqrt{1.15}$$ $$=>$$
$$r>(\sqrt{1.15}-1)*100$$ $$=>$$ $$r>(\approx1.07-1)*100$$ $$=>$$ $$r>7$$ - not sufficient

Statement 1 is pretty easy to evaluate and you can get the answer...

About statement 2 we can do this way as well....

given $$(1+\frac{r}{100})^2>1.15$$

this means we have received a interest greater than 15% ( 1+ .15) for 2 years. That could be 7.5 for each year (approx) or 10% for each year if the 1.15 figure becomes 1.21. Obviously we cannot confirm any value of R.

What do you say...??
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DS [#permalink]  10 May 2010, 16:40
29.If $1000 is deposited in a certain bank account and remains in the account along with any accumulated intrest, the dollar amount of intrest, I, earned by the deposit in the first n years is given by the formula I= 1000((1 + r/100)^n – 1), where r percent is the annual interest rate paid by the bank. Is the annual rate paid by the bank greater than 8 percent? 1.The deposit earns a total of$210 in interest in the first teo years.
2.(1 + r/100)^2 > 1.15

Thanks
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Second option does not give a precise value of r as the one given by option one.
Thus A is the option that holds true.
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Re: If $1,000 is deposited in a certain bank account and remains [#permalink] 18 Jan 2012, 07:52 I calculated A too. But OA is given D. In option 2 r>7. Help please. _________________ Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html Math Expert Joined: 02 Sep 2009 Posts: 31297 Followers: 5357 Kudos [?]: 62390 [9] , given: 9455 Re: If$1,000 is deposited in a certain bank account and remains [#permalink]  18 Jan 2012, 07:59
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Baten80 wrote:
I calculated A too. But OA is given D.
In option 2 r>7.

Answer is A. Discussed here: annual-interest-104084.html

If $1,000 is deposited in a certain bank account and remains in the account along with any accumulated interest, the dollar amount of interest, I, earned by the deposit in the first n years is given by the formula I=1,000((1+r/100)^n-1), where r percent is the annual interest rate paid by the bank. Is the annual interest rate paid by the bank greater than 8 percent? Given: $$I=1,000((1+\frac{r}{100})^n-1)$$. Question: is $$r>8$$. (1) The deposit earns a total of$210 in interest in the first two years --> $$I=210$$ and $$n=2$$ --> $$210=1,000((1+\frac{r}{100})^2-1)$$ --> note that we are left with only one unknown in this equation, $$r$$, and we'll be able to solve for it and say whether it's more than 8, so even withput actual solving we can say that this statement is sufficient.

(2) (1 + r/100 )^2 > 1.15 --> if $$r=8$$ then $$(1+\frac{r}{100})^2=(1+\frac{8}{100})^2=1.08^2\approx{1.16}>1.15$$ so, if $$r$$ is slightly less than 8 (for example 7.99999), $$(1+\frac{r}{100})^2$$ will still be more than 1.15. So, this statement is not sufficient to say whether $$r>8$$.

This topics is locked, so in case of any question please post it here: annual-interest-104084.html

Hope it helps.
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