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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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.
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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.

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!

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.
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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.

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,

As this is a quadratic equation , how did you concluded that we will get one value after solving this equation?
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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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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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Thank you Bunuel for this alternate calculation method!

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

~ Im2bz2p345

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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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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