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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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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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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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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
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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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chandrak
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%.
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Bunuel
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.

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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Bunuel
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 :)
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Im2bz2p345
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Bunuel
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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Im2bz2p345
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Bunuel
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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Bunuel

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 :)
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Dear Experts Bunuel, VeritasPrepKarishma,

In case of inequalities, can we take square roots on both sides?
Thanks in advance.
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Th3Ap3xPr3dator
Dear Experts Bunuel, VeritasPrepKarishma,

In case of inequalities, can we take square roots on both sides?
Thanks in advance.

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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This is a good question that tests your understanding of Compound interest as an application of percentage increase. Solving this question in less than 2 min is about how well you apply the conceptual understanding and logic rather than calculation-intensive approaches.

Let's take an example: 1000$ is invested at a rate of r% for n years at C. I annually.

The amount, A=\(1,000((1+r/100)^n\)

The first step is to understand what this formula is doing. It is very simple. Here, 1000 $ is increased at a rate of r% every year. Since it was deposited for n years, 1000$ will be increased by r %, n times successively. So it is a pure application of successive percentage increase.

Let me make it more clear for you., if r = 8% and n = 2 years.

A = \(1,000((1+8/100)^2\) = \(1000 * \frac{108}{100} * \frac{108}{100}\)

Multiplying 1000 by 108/100 means you are increasing 1000 by 8%.

So here, 1000$ is successively increased by 8%,2 times. I hope you all got a hang of how compound interest works.

Interest amount = Final amount after successive increase - Initial amount deposited.

I=\(1,000(1+r/100)^n- 1000\) = \(1,000((1+r/100)^n-1).\)

Now it's time to analyze the question stem.
Question stem: Is the annual interest rate paid by the bank greater than 8 percent? so is r > 8%?

There is no info given in the question regarding the no of years you have deposited. Also, it will not affect the interest rate charged by the bank.

But, if you analyze both the statements given we will know that the data is given for 2 years. So let's try to re-frame the question stem w.rt 2 years.

If 1000$ is deposited for 2 years at 8 % CI annually, then the interest would = 80 + 80 + 6.4 = 166.4

Explaining my calculation in detail below:

Total interest = First-year interest + Second-year interest.

First year interest = 8%of initial amount = 8 % of 1000 = 80$

Second year interest = 8%of initial amount + 8% of ( first year interest) = 8 % of 1000 + 8 % 80 = 80 + 6.4

So total interest in 2 years = 80 + 80 + 6.4 = 166.4

Instead of using intensive calculation formula to find Compound interest, I prefer to stick to this logical approach. Now, it's your choice. ;)

If r =8%, we found that interest for 2 years is 166.4$

If r is greater than 8 %, the interest amount for 2 years also should be greater than 166.4$

Hence, we can re-frame the question stem is r > 8%? as is Interest for 2 years > 166.4$?

So here we need to check if the interest for 2 years is greater than 166.4 $ or not.

St1: The deposit earns a total of $210 in interest in the first two years.

You will get a definite Yes as it is given that the deposit earned a total of $210 in interest in the first two years. Therefore, Statement 1 alone is sufficient to answer the question.

St2: \((1+r/100)^2 >1.15\)

\((1+r/100)^2 \) will give you an idea about the overall percentage increase for 2 years.

Analyzing in detail : \(A=1,000(1+r/100)^2\)

If \((1+r/100)^2\) = 1.15

A = 1000 * 1.15 = 1000 * 115/100 = 1150

Multiplying 1000 by 1.15 means 1000 is increased by 15 % in 2 years i.e 1000 + 150 = 1150.


So here, \((1+r/100)^2 > 1.15\) means the amount after 2 years > 1150 Or we can say that the overall percentage change happened in 2 years is greater than 15 %

Since Interest = Final amount - initial deposit, We can conclude that total interest for 2 years is greater than 150.

If you could recollect how we re-framed the question stem in terms of the total interest in St 1, Can we use it here as well?

is r > 8%? as is Interest for 2 years > 166.4$?


Greater than 150 means it could be between 150 and 166.4 or greater than 166.4. So you will get a NO as well as YES to the question stem.
Therefore, statement 2 alone is not sufficient.

:idea: There is also an alternative approach you can think about in st 2, in terms of the overall percentage change.

Let's try to re-frame the question stem in terms of overall percentage increase for 2 years.
If you are increasing an amount by 8 % successively 2 times, the overall percentage increase would be 8 + 8 + 8*8/100 = 16.64 %

I hope all of you are aware of this formula for overall percentage change.

If not, please have a look here.
If a value is increased by a % then again increased by b %, the overall percentage change = a + b + ab/100.
Note: This formula is applicable only for 2 changes. Since the value is increased by a%, we are putting + a. So if it's a decrease, use the '-' sign instead of '+'.

For eg. If a value is decreased by a % then increased by b %, then overall percentage change = -a + b + (-a)(b)/100 = -a +b -ab/100

Coming back to the question stem, can we re-frame is r > 8% ? as is the overall percentage increase for 2 years > 16.64 %?

In the st 2, (1+r/100)^2 >1.15 means the overall percentage increase for 2 years is greater than 15 %. This will not give you a definite answer to the question stem.

Greater than 15 % means it can be greater than 16.64 % or it can also be between 15 and 16.64 %. So, it can be a yes or a no.
Hence, St 2 alone is not sufficient.

Option A is the correct answer.

I hope this explanation helps to have a better understanding of compound interest and how to apply concepts and logic instead of calculation-intensive approaches.

Thanks,
Clifin J Francis,
GMAT QUANT SME
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Hi, could anyone help me with this question?

I recently came across this question in one of my official mocks, however, with a slight change in statement 1.


Question
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 equals 1 comma 000 times open paren open paren 1 plus r over 100 close paren to the nth power negative 1 close paren​, 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 few years.
(2) ​open paren 1 plus r over 100 close paren squared is greater than 1 period 15

Classic statement 1, when n=2, is sufficient, I understand, though when it says first few years, I am not sure. Could someone pls help?
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Bunuel , this seems like a weird question, but is it really given to us that the principle amount deposited is 1000 dollars?

The opening sentence is posed in the form of a conditional stating " If 1000 dollars are deposited, then the interest earned is given by ...." The purpose of this sentence seems to be to introduce the formula for compound interest.Then the actual question is posed.
I know this seems very obvious that the question means for us to take 1000 usd as the principle, but I have seen some new-age DS questions which really punish you for assuming obvious stuff.
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KarishmaB

butterfly
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.
­Hi,

Does this also work with other 3 digit numbers than ones starting with 1 such as (3.05)^2?

305= 300+5

-----25

Now adding 5 to 305 gives me 310


So the square now becomes 31025 right? But instead 305^2 is 93025. How do I work with numbers like 3.05^2 or 7.17^2 ?
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