Score Higher by Avoiding Algebra
To maximize your GMAT score, you need to be familiar with a few unconventional methods. Guessing and checking and following your intuition are methods that are neither as systematic nor as elegant as algebra, but they are your most valuable assets on the GMAT. During a timed test, doing algebra is often counter-productive because it is slow and error-prone. Instead, quick and decisive methods are your preferred, time-saving tactics.
This results from the structure of the GMAT itself. On the Quant, you’re only given two minutes per question, which elevates time management skills over math skills. This is precisely what the test-makers intended: as a business executive, the ability to recall math formulas is not as important as the ability to reason quickly. On the GMAT, then, half the battle is learning how to use valuable shortcuts effectively.
As an example, let’s take a look at the following data sufficiency problem:
Is b <
?
(1) b < a
(2) b = -2
A) Statement 1) alone is sufficient, but statement 2) alone is not sufficient
B) Statement 2) alone is sufficient, but statement 1) alone is not sufficient
C) BOTH statements 1) and 2) TOGETHER are sufficient, but NEITHER statement ALONE is sufficient)
D) EACH statement ALONE is sufficient
E) Statements 1) and 2) TOGETHER are NOT sufficient
To prove whether statement 1) or 2) is sufficient, you might be tempted to logically reason through each statement. You could set up inequalities, perhaps perform some substitution or logical reasoning, and then manipulate the inequalities to fit your proof. Such an approach, however, would likely exceed five minutes, and might possibly be riddled with careless errors. For these reasons, you should ignore traditional methodology and instead use some unconventional methods. First up, let’s follow our intuition.
Just by looking at statement 1) alone, we can probably suspect that it provides insufficient data. We’re told that b < a, but that seems to tell us very little about whether b <
. After all, we don’t know whether the variable “a” is positive or negative, and the exponent in
can change “a” from a negative number to a positive number. We can’t prove this suspicion for certain, since it was merely an educated guess, but we now have a starting point. If statement 1) alone is insufficient, we can prove this without doing any algebra simply by demonstrating a contradiction.
Demonstrating a contradiction is a powerful technique for data sufficiency questions on the Quant. For a statement to provide sufficient data for a Yes/No-type DS question, it must guarantee that the answer to the question stem is either always “Yes” or always “No”. In this problem, statement 1) would provide sufficient data if we could prove that b <
is always true, or always false. If we can find even one single set of contradictory examples, so that the answer is at times “Yes” but at times “No”, we have proven that the statement is insufficient. Demonstrating a contradiction obviates the need for algebra altogether — all we need is a single set of examples that fit the statement, yet give contradictory answers to the question stem.
We’ll use the technique of plugging-in to find two contradictory examples. For the first example, let’s choose a few small prime numbers, such as a = 3 and b = 2. If we plug these numbers in, we see that b < a, so these numbers satisfy statement 1). Now, let’s answer the question stem: is b <
? We see that 2 is indeed less than
= 9, so the answer is “Yes”.
We now need to find a single example of a contradictory “No” answer to the question stem. To do so, let’s switch the type of numbers used for the plug-in. We used positive, prime integers at first, but there are no restrictions which state that the numbers must be positive values — or even integers. We should therefore consider using fractions or negative numbers.
Let’s try one set of fractional plug-ins: b=
and a=
. We see that it satisfies statement 1), but because
is not smaller than
, we have a “No” answer to the question stem. With two answers that lead to contradictory results (sometimes “yes”, other times “No”), we have insufficient data. We can eliminate choices A/D, all without touching even the slightest bit of algebra.
In general, if your hunch suggests that you have insufficient data, avoid algebra. Prove you have insufficiency by demonstrating a contradiction by plugging-in a few numbers. Only use algebra if you suspect you have sufficient data.
For the following questions, I’d like you to try to solve the problem by avoiding algebra whenever possible. Instead, try to demonstrate a contradiction to prove that you have insufficient data:
Is |
| > |
|?
(1) x > y
(2) x > 0

| > |
|?
18 comments
Deeps on June 7th, 2012 at 10:10 pm
D
Vicky on June 13th, 2012 at 4:37 am
Its D! Look at the modulus sign the difference of any 2 nos within a modulus will always remain smaller than the sum of the nos within the modulus. even if x=5 and y=-10 then x2 will be 25 and y2 will be 100 so modulus (25+100) = 125 and modulus(25-100)= -75 but within modulus so we only consider positive. hence 125 as a no greater than 75!
Vikram on June 7th, 2012 at 10:12 pm
B
AbhiJ on June 7th, 2012 at 11:44 pm
The inequality will always be true except when X=Y = 0 when LHS = RHS.
So anything that makes impossible the above case will be sufficient.
Hence D.
Ankita on June 8th, 2012 at 1:56 am
I think it would be E because the inequality does not hold when either x= 0 or y = 0 and even by combining the 2 equations we can't know whether y = 0 or not
Vicky on June 13th, 2012 at 8:25 pm
I agree. Had an after thought of this just when I posted D!
E on June 8th, 2012 at 7:16 am
The answer should be E
statement 1
x = 5, y=1 works
X=5, y = -10 does not work
statement 1 insufficient
statement 2
we can use the same examples as above and prove the statement is insufficient
same thing for statements 1 and 2 combined. Therefore E
somsubhra on June 8th, 2012 at 12:25 pm
The ans is D
Cory McAboy on June 9th, 2012 at 6:40 am
Its E.
Think of the case where X=1 and Y=0. This satisfies both statement 1 and 2 and it does not hold true:
| 1^2 + 0^2 | > | 1^2 - 0^2 |
1 > 1
Not true
Now take X=2 Y=1:
| 2^2 + 1^2 | > | 2^2 - 1^2 |
|4 + 1 | > | 4 - 1 |
5 > 3
This is True
Therefore you have a contradiction for both statements together and thus the answer is E.
Tara on June 10th, 2012 at 9:37 pm
E.
A single plug in of y=0 leads to contradictions in all the cases with LHS=RHS.
Correct Answer is E on June 12th, 2012 at 9:37 am
X^2 and Y^2 are each always positive, so you can drop the absolute value signs from the first part of the stem equation. It simplifies to "Is X^2 + (positive number or zero) > |X^2 - (positive number or zero)|". This is false when Y=0, true otherwise, and isn't addressed in either statement (a) or (b). Answer is E.
Srikanth on June 12th, 2012 at 10:12 am
Is b < a^2 ?(1) b < a(2) b = -2
Regarding above question, it looks obvious that answer is B. Since square of a number is always greater than a negative number. However we don't have any information about a. It could also be a complex number right?
Mark G on June 12th, 2012 at 12:02 pm
I don't believe the gmat takes complex numbers into consideration for these types of problems. The answer is B due to the fact that a square of a number is always greater than a negative number.
Srikanth on June 12th, 2012 at 3:07 pm
Complex numbers are out of scope or out of syllabus
Siying on June 13th, 2012 at 12:13 am
E.
The inequality should work in most cases. But here's a specific example where the inequality does NOT work:
let y=0, then the relation should be "=", not ">"
Since even if we combine (1) and (2), we still have possibility of ">" (in most cases) and "=" (if y=0), the answer is E.
Sheila on June 13th, 2012 at 11:13 am
Answer is E
Denysse on December 10th, 2012 at 9:23 am
I am confused from all of the answers. Is the answer D or E. From what I got from above it is E because for D, you can substitute x=1 and y=1 and the statement wouldn't hold true anymore? I did understand statement A. These articles would be helpful if the person who wrote it answered at the bottom.
Maximiliano Isa on February 13th, 2013 at 4:38 am
I think it´s B
I) b not sufficient.
II) b=-2 --> sufficient, because a^2 will be always positive, and then b<a^2 (even if a=0).