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01 Nov 2010, 17:32
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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
[Reveal] Spoiler: OA

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02 Nov 2010, 18:53
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butterfly wrote:
It's much easier to multiply 1.08 by 1.08 than to take the square root of 1.15.

Shortcut to multiply numbers of the form (100 + a) or (100 - a)
Write $$a^2$$ on the right hand side. Add a to the original number and write it on left side. The square is ready.

e.g. $$108^2 = (100 + 8)^2$$ Write 64 on right hand side

________ 64

Add 8 to 108 to get 116 and write that on left hand side

11664 - Square of 108

e.g. $$91^2 = (100 - 9)^2$$ => ______81 => 8281
(Here, subtract 9 from 91)

Note: a could be a two digit number as well.
e.g $$112^2 = (100 + 12)^2$$ = ______44 => 12544
(Only last two digit of the square of 12 are written on the right hand side. The 1 of 144 is carried over and added to 112 + 12)

This is Vedic Math though the trick uses basic algebra.
$$(100 + a)^2 = 10000 + 200a + a^2$$
(100 + 8)^2 = 10000 + 200 x 8 + 64 = 10000 + 1600 + 64 = 11664

This is a useful trick that saves time.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 39040 Followers: 7750 Kudos [?]: 106487 [8] , given: 11626 Re: Annual interest [#permalink] ### Show Tags 01 Nov 2010, 20:07 8 This post received KUDOS Expert's post 11 This post was BOOKMARKED 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 $$r>8$$. Answer: A. _________________ Intern Joined: 10 Apr 2012 Posts: 23 Concentration: Finance, Economics GMAT 1: 760 Q50 V44 Followers: 1 Kudos [?]: 38 [4] , given: 0 Re: If$1,000 is deposited in a certain bank account and remains [#permalink]

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15 Mar 2013, 16:47
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I think the gmat is always about insight and not about arithemetic
.
Question: Is r>8%? If r can be 8 percent or greater, the statement will be insufficient.

Stmt 2) Overall interest earned over two years, is greater than 15% (If you read this far down, you know what I am talking about)
Lets say the interest was 8%, then overall compound interest earned over two years will be greater than 16% and so greater than 15%
It goes without saying that if the interest rate was greater than 8%, then the amount of interest earned over two years, will still be greater than 15%.

It took me 1:30 seconds to see this, and another 40 seconds to type this entire post because I was not satisfied with the explanations given
.No messy calculations or nail biting necessary.
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17 Jul 2013, 23:49
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Im2bz2p345 wrote:
greatps24 wrote:
Bunuel wrote:
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.

Bunuel,

As this is a quadratic equation , how did you concluded that we will get one value after solving this equation?

Same question for Bunuel or any of the other experts here.

My calculations:

1) $$210 = 1000 [(1+\frac{r}{100})^2-1)$$

2) $$210 = 1000 [1+\frac{2r}{100}+\frac{r^2}{10000}-1]$$

3) $$210=1000(\frac{200r+r^2}{10,000})$$

4) $$210=\frac{200r+r^2}{10}$$

5) $$2100=r(200+r)$$

How do you solve for the variable r at this point?

Any further explanation would help.

~ Im2bz2p345

Actually you don't need to solve this way:

$$1,000((1+\frac{r}{100})^2-1)=210$$

$$(1+\frac{r}{100})^2-1=\frac{210}{1,000}$$

$$(1+\frac{r}{100})^2=\frac{21}{100}+1$$

$$(1+\frac{r}{100})^2=\frac{121}{100}$$

$$1+\frac{r}{100}=\frac{11}{10}$$ ($$1+\frac{r}{100}$$ cannot equal to $$-\frac{11}{10}$$ because it would men that r is negative.)

$$1+\frac{r}{100}=\frac{11}{10}$$

$$\frac{r}{100}=\frac{1}{10}$$

$$r=10$$
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Re: If $1,000 is deposited in a certain bank [#permalink] ### Show Tags 02 Jan 2013, 21:32 2 This post received KUDOS 1 This post was BOOKMARKED kiyo0610 wrote: 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 [Reveal] Spoiler: A statement 1) I=$210, n=2
Putting this in the equation given in the question, we will be able to find the value of r and thereby be able to answer the question. Suffiicient.

Statement 2) Using Binomial theorem, we can infer $$(1+r/100)^2 > 1.15$$ as $$(1+2r/100) > 1.15$$.
On solving this relation we will get, r>7.5.
But since its not given that r is an integer then r can be 7.51, 7.6,9, 11 etc. Hence insufficient.

+1A
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Re: If $1,000 is deposited in a certain bank account and remains [#permalink] ### Show Tags 14 Mar 2013, 21:16 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED Responding to a pm: Question: (1 + .08)^2 = ? $$1.08 = \frac{108}{100}$$ (it's trickier to deal with decimal so remove it) $$(\frac{108}{100})^2 = \frac{108^2}{10000}$$ We know how to get the square of 108 $$108^2 = 11664$$ (discussed in the post above) So, $$(1 + .08)^2 = 11664/10000 = 1.1664$$ Or you can use (a + b)^2 = a^2 + b^2 + 2ab (the shortcut is anyway based on this formula only) (1 + .08)^2 = 1 + .0064 + 2*1*.08 = 1.1664 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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17 Jul 2013, 21:05
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Im2bz2p345 wrote:
greatps24 wrote:
Bunuel wrote:
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.

Bunuel,

As this is a quadratic equation , how did you concluded that we will get one value after solving this equation?

Same question for Bunuel or any of the other experts here.

My calculations:

1) $$210 = 1000 [(1+\frac{r}{100})^2-1)$$

2) $$210 = 1000 [1+\frac{2r}{100}+\frac{r^2}{10000}-1]$$

3) $$210=1000(\frac{200r+r^2}{10,000})$$

4) $$210=\frac{200r+r^2}{10}$$

5) $$2100=r(200+r)$$

How do you solve for the variable r at this point?

Any further explanation would help.

~ Im2bz2p345

Solving this quadratic is a little time consuming though we will see how to do in a minute. But you don't really need to solve it to figure out that you will have only one solution.

$$2100=r(200+r)$$
$$r^2 + 200r - 2100 = 0$$

In a quadratic, $$ax^2 + bx + c = 0$$, sum of the roots = -b/a and product of the roots = c/a
Notice that the product of the roots (-2100) is negative. This means one root is positive and the other is negative. So we will have only one acceptable solution (the positive one)

Now, if you would like to solve it:
$$r^2 + 200r - 2100 = 0$$

2100 = 2*2*5*5*3*7
Now you need to split 2100 into two factors such that one is a little larger than 200 and the other is a small factor e.g. 5 or 7 or 10 etc. Once you think this way, you easily get 210 and 10

$$r^2 + 210r - 10r - 2100 = 0$$
(r + 210)(r - 10) = 0
r = -210 or 10
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29 Oct 2015, 09:54
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Bunuel wrote:
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 $$r>8$$.

I did silly mistake while calculating 1.08^2 and marked D
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01 Nov 2010, 20:16
ah... for S2, I approached it from the other angle and had to take the square root of 1.15. I got stuck there and time was running out, so I took a guess. It's much easier to multiply 1.08 by 1.08 than to take the square root of 1.15.

Thanks!
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03 Nov 2010, 08:59
Very neat trick Karishma! This should save me a lot of time!! THANKS
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25 Dec 2012, 20:42
Bunuel wrote:
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 $$r>8$$.

Hello Bunuel, your explanation for second DS choice suggests that, if we have only 1 variable in the equation, then we need not solve it. However, I have observed few of the GMAT problems that have similar quadratic equations (with second degree) solve to two different positive roots, hence the DS choice could not be true.

I believe it would be safe to solve the equation until you know if its only going to give you "one" root.
e.g. $$ax^2+bx-c=0$$, this equation will have one positive and one negative root. As rate in this case is supposed to be positive, hence only 1 root.

However, if the equation resolves to $$ax^2-bx+c=0$$ then it can have two positive roots (one of which may be less than 8 and other more than 8), hence the choice may not be true. Only if both positive roots are more than 8, then the choice can be taken as true.

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Re: If $1,000 is deposited in a certain bank [#permalink] ### Show Tags 15 Mar 2013, 00:40 Marcab wrote: kiyo0610 wrote: 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 $210 in interest in the first two years. (2) (1+r/100)^2 >1.15 The total interest is given as I=1,000[(1+r/100)^n - 1]. From F.S 1 we have that I = 210. Thus, we have a quadratic equation and we know that it can be solved leading to a fixed value for r. Sufficient.Also, one can notice that an interest of 210$ is obtained when r=10% and this is greater than 8%. Sufficient.

From F.S 2, we know that n=2. And the Interest earned would be greater than 150.Thus, I=1,000[(1+r/100)^2 - 1] = $$1000[r/100*(2+\frac{r}{100})]$$. We know for r=8% we have this equal to 2.08*80 = 166.4 which is anyways greater than 150. Now, for r=7%, the expression equals 2.07*70 = 144.9. Thus, for a value between 7 and 8 , this value will change and become more than 150. Thus we wouldn't know for sure if r>8 or not. Insufficient.

A.
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19 Mar 2013, 11:37
hi all , why not D?
From first statement, one can answer that interest rate is greater than 8%
From second second statement, one can answer that interest rate is less than 8%

So, either statement can be used to answer the question. Am I missing any thing? Please reply.
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Re: If $1,000 is deposited in a certain bank account and remains [#permalink] ### Show Tags 19 Mar 2013, 20:39 chandrak wrote: hi all , why not D? From first statement, one can answer that interest rate is greater than 8% From second second statement, one can answer that interest rate is less than 8% So, either statement can be used to answer the question. Am I missing any thing? Please reply. How can you say that the interest rate is less than 8% from the second statement? If r were 8%, we would have (1 + r/100 )^2 = 1.08^2 = 1.1664 Now all that statement 2 tells us is that (1 + r/100 )^2 > 1.15 We don't know whether it is less than 1.1664 or greater. Hence statement 2 alone is not sufficient. Besides, it is not possible that statement 1 tells you that r is greater than 8% and statement 2 tells you that it is less than 8%. This is a conflict. If both statements independently give you the answer, the answer you will get will be the same i.e. either both will tell that r is greater than 8% or both will tell that r is less than 8%. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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29 Mar 2013, 08:54
Bunuel wrote:
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 $$r>8$$.

Bunuel,

As this is a quadratic equation , how did you concluded that we will get one value after solving this equation?
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17 Jul 2013, 11:15
greatps24 wrote:
Bunuel wrote:
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.

Bunuel,

As this is a quadratic equation , how did you concluded that we will get one value after solving this equation?

Same question for Bunuel or any of the other experts here.

My calculations:

1) $$210 = 1000 [(1+\frac{r}{100})^2-1)$$

2) $$210 = 1000 [1+\frac{2r}{100}+\frac{r^2}{10000}-1]$$

3) $$210=1000(\frac{200r+r^2}{10,000})$$

4) $$210=\frac{200r+r^2}{10}$$

5) $$2100=r(200+r)$$

How do you solve for the variable r at this point?

Any further explanation would help.

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17 Jul 2013, 21:17
VeritasPrepKarishma wrote:
Solving this quadratic is a little time consuming though we will see how to do in a minute. But you don't really need to solve it to figure out that you will have only one solution.

$$2100=r(200+r)$$
$$r^2 + 200r - 2100 = 0$$

In a quadratic, $$ax^2 + bx + c = 0$$, sum of the roots = -b/a and product of the roots = c/a
Notice that the product of the roots (-2100) is negative. This means one root is positive and the other is negative. So we will have only one acceptable solution (the positive one)

Now, if you would like to solve it:
$$r^2 + 200r - 2100 = 0$$

2100 = 2*2*5*5*3*7
Now you need to split 2100 into two factors such that one is a little larger than 200 and the other is a small factor e.g. 5 or 7 or 10 etc. Once you think this way, you easily get 210 and 10

$$r^2 + 210r - 10r - 2100 = 0$$
(r + 210)(r - 10) = 0
r = -210 or 10

Perfect! Thank you Karishma for filling in these last steps for me. You're always a big help!

As PraPon pointed out earlier in the thread, I believe it's important to bring the first statement down to the $$r^2 + 200r - 2100 = 0$$ level, otherwise there is no "guarantee" that r will only have one root.

Once you realize there is will be a positive & negative solution and that you can simply "ignore" the negative solution since we checking to see if r > 8 (leaving you with truly one solution), then you can make the judgement that the first statement is sufficient.

I just don't know how you can assume the solution of $$210=1,000((1+\frac{r}{100})^2-1)$$ will have only ONE positive solution (as you pointed out Karishma, if the simplified form gave you "+ 2100" instead of "- 2100," the statement would be insufficient due to having TWO positive solutions).

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18 Jul 2013, 08:47
Bunuel wrote:
Actually you don't need to solve this way:

$$1,000((1+\frac{r}{100})^2-1)=210$$

$$(1+\frac{r}{100})^2-1=\frac{210}{1,000}$$

$$(1+\frac{r}{100})^2=\frac{21}{100}+1$$

$$(1+\frac{r}{100})^2=\frac{121}{100}$$

$$1+\frac{r}{100}=\frac{11}{10}$$ ($$1+\frac{r}{100}=\frac{11}{10}$$ cannot equal to $$-\frac{11}{10}$$ because it would men that r is negative.)

$$1+\frac{r}{100}=\frac{11}{10}$$

$$\frac{r}{100}=\frac{1}{10}$$

$$r=10$$

Thank you Bunuel for this alternate calculation method!

It's much easier this way you showed. Greatly appreciate your follow-up.

~ Im2bz2p345
Re: Annual interest   [#permalink] 18 Jul 2013, 08:47

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