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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 $$I=1,000((1+\frac{r}{100})^n-1)$$, 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) $$(1 + \frac{r}{100})^2 > 1.15$$

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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? 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 $$r>8$$.

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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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Re: If $1,000 is deposited in a certain bank account and remains in the [#permalink] Show Tags 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 $$r>8$$. Answer: A. 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. Please advice. _________________ Thanks, Prashant Ponde Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7 Reading Comprehension notes: Click here VOTE GMAT Practice Tests: Vote Here PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here Finance your Student loan through SoFi and get$100 referral bonus : Click here
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Re: If $1,000 is deposited in a certain bank account and remains in the [#permalink] Show Tags 2 1 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 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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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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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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Re: If $1,000 is deposited in a certain bank account and remains in the [#permalink] Show Tags 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 Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 9706 Location: Pune, India Re: If$1,000 is deposited in a certain bank account and remains in the  [#permalink]

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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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Re: If $1,000 is deposited in a certain bank account and remains in the [#permalink] Show Tags 5 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.) $$1+\frac{r}{100}=\frac{11}{10}$$ $$\frac{r}{100}=\frac{1}{10}$$ $$r=10$$ _________________ Intern  Joined: 23 Mar 2011 Posts: 26 Re: If$1,000 is deposited in a certain bank account and remains in the  [#permalink]

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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.

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Re: If $1,000 is deposited in a certain bank account and remains in the [#permalink] Show Tags Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. 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 There are 3 variables (r,I,n) and 1 equation (I=1,000((1+r/100)^n-1) in the original condition, 2 equations from the 2 conditions; there is high chance (C) will be our answer. In condition 1, there are 2 equations, and if the interest is$210, the interest rate is either greater (yes) or smaller (no) than 8%, therefore sufficient.
In condition 2, (1.08)^2=1.1664, the interest rate is not greater than 8%, so this is insufficient.
Therefore the answer becomes (A).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Th3Ap3xPr3dator wrote:
Dear Experts Bunuel, VeritasPrepKarishma,

In case of inequalities, can we take square roots on both sides?

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Check for more here: Manipulating Inequalities (adding, subtracting, squaring etc.).

Hope it helps.
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Re: If $1,000 is deposited in a certain bank account and remains in the [#permalink] Show Tags Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: If$1,000 is deposited in a certain bank account and remains in the   [#permalink] 12 Oct 2018, 07:05
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