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Is a < 0 ? (1) a^3 < a^2 + 2a (2) a^2 > a^3

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by DavidG@VeritasPrep » Thu Feb 23, 2017 6:22 pm
Anaira Mitch wrote:Is a < 0 ?
(1) a^3 < a^2 + 2a
(2) a^2 > a^3
You could pick some simple numbers.

1) Case 1: a = 1/2. (1/8 < 1/4 + 1, so this satisfies the statement.) 1/2 is not less than 0, so the answer is NO
Case 2: a = -10. (-1000 < 100 + (-20), so this satisfies the statement.) -10 is less than 0, so the answer is YES. Statement 1 does not yield a definitive answer to the question, and is not sufficient.

2) Let's see if we can reuse the same values.
Case 1: a = 1/2. (1/4 > 1/8, so this satisfies the statement.) We already know this gives us a NO, as 1/2 is not less than 0.
Case 2: a = -10. (100 > -1000, so this satisfies the statement.) We already know this gives us a YES, as -10 is less than 0. Not sufficient.

Taken together, a could still be either 1/2 or -10, so the answer to the question could be a YES or a NO, and thus together the statements are not sufficient. The answer is E
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by Jay@ManhattanReview » Fri Feb 24, 2017 1:15 am
Anaira Mitch wrote:Is a < 0 ?

(1) a^3 < a^2 + 2a
(2) a^2 > a^3
Hi Anaira,

We need to see whether a < 0 or whether a is negative.

Between the two statements, statement 2 seems to be relatively easier as it has only two terms to deal with compared to statement 1 that has three terms.

S2: a^2 > a^3

Whether a is negative or positive, a^2 is positive, so dividing the inequality by a^2, we get

a^2/a^2 > a^3/a^2

=> 1 > a

1 > a implies that for the range 0 > a, a is negative and for the range 1 > a > 0, a is positive. No unique answer. Insufficient.

Let us see statement 1 now.

S1: a^3 < a^2 + 2a

We cannot divide the inequality by 'a' to simplify it since we do not know the nature of 'a,' it can be negative or positive. Depending upon its nature, the sign of inequality would change.

The optimum way to deal with this is by testing values.

1. Say a is negative. At a = -2,

a^3 < a^2 + 2a => (-2)^3 < (-2)^2 + 2(-2) => -8 < 4 -4 => -8 < 0. The answer is YES.

2. Say a is positive. At a = 1/2 (I deliberative chose a = 1/2 since it qualifies in the range 1 > a > 0 (one of the qualified range for 'a' from statement 2),

a^3 < a^2 + 2a => (1/2)^3 < (1/2)^2 + 2(1/2) => 1/8 < 1/4 + 1 => 1/8 < 5/4. The answer is No. Insufficient.

S1 and S2:

Even after combining both the statements, we cannot conclude since both the cases taken while analyzing statement 1 also qualifies for statement 2. So no help!

The correct answer: E

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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by Brent@GMATPrepNow » Fri Feb 24, 2017 6:44 am
Anaira Mitch wrote:Is a < 0 ?

(1) a³ < a² + 2a
(2) a^2 > a^3
Target question: Is a < 0 ?

Statement 1: a³ < a² + 2a
Subtract a² and 2a from both sides to get: a³ - a² - 2a < 0
Factor: a(a² - a - 2) < 0
Factor more: a(a - 2)(a + 1) < 0
There are several values of a that satisfy this inequality. Here are two:
Case a: a = 0.5. In this case, a > 0
Case b: a = -10. In this case, a < 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: a² > a³
Since a² must be POSITIVE here, we can divide both sides by a² to get: 1 < a
There are several values of a that satisfy this inequality. Here are two:
Case a: a = 0.5. In this case, a > 0
Case b: a = -10. In this case, a < 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
There are several values of a that satisfy BOTH statements Here are two:
Case a: a = 0.5. In this case, a > 0
Case b: a = -10. In this case, a < 0
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

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Brent
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by [email protected] » Fri Feb 24, 2017 10:19 am
Hi Anaira Mitch,

A number of the explanations provided show how TESTing VALUES is perfect for this type of prompt, so I won't rehash any of that here. Instead, it's worth noting that there are various rules/patterns regarding exponents that you should look for (especially when exponents are presented in the way that they are here).

For example:
A positive fraction, when raised to an exponent greater than 1, will get SMALLER
Example: (1/2)^3 = 1/8

A negative fraction, when raised to an exponent greater than 1, will get BIGGER
Examples: (-1/2)^2 = +1/4 (-1/2)^3 = -1/8 BOTH are greater than -1/2

A negative integer (other than -1), when raised to an exponent greater than 1, will get BIGGER or SMALLER based on whether the exponent is even or odd.
Examples (-2)^2 = +4 (-2)^3 = -8

As such, you should consider how positive fractions and/or negative numbers (both fractions and integers) behave in these situations. While it's not something that you'll be tested on often, these patterns are often the key to correctly answering these types of questions.

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by Matt@VeritasPrep » Fri Mar 03, 2017 3:37 am
I think we're missing what might be the simplest explanation, so I'll add it!

S1:

a³ < a² + 2a

a³ - a² - 2a < 0

a * (a² - a - 2) < 0

a * (a - 2) * (a + 1) < 0

For this to be possible, either exactly one of these is negative or all three are negative.

Since (a - 2) is the smallest, it MUST be negative. That means a - 2 < 0, or a < 2. Not quite good enough!

S2:

a² > a³

0 > a³ - a²

0 > a² * (a - 1)

Since a² can't be negative, (a - 1) must be. From this, we know 0 > a - 1, or 1 > a. Ugh!

Taking the statements together is redundant, since S1 adds nothing to S2, so our answer is E.
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by Matt@VeritasPrep » Fri Mar 03, 2017 3:41 am
Fun follow up question:

You may have noticed that my intervals above are not comprehensive. (For instance, in S1, we can't have 0 ≥ a ≥ -1.) Did my algebra go awry, or is my solution still valid? (I promise this has some GMAT implications! :))
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by Matt@VeritasPrep » Fri Mar 03, 2017 3:44 am
One more fun follow up question! (No more in this thread, I promise.)

Suppose I take S1 and S2, and place them like so:

2a + a² > a³

a² > a³

Then I have the brilliant idea to subtract the second from the first!

2a > 0

Wow, so a > 0? All these "experts" are clueless!

But of course they aren't ... so what did I do wrong?
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