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Last edited by Bunuel on 04 Jan 2013, 01:54, edited 1 time in total.

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]
04 Jan 2013, 01:59

Expert's post

If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is x^2 - x - 6\geq{x^2 - 2x - 8}? --> is x\geq{-2}? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is x\geq{-2}? and (2) answers this question with a NO. Sufficient.

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]
12 May 2013, 07:02

Bunuel wrote:

If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is x^2 - x - 6\geq{x^2 - 2x - 8}? --> is x\geq{-2}? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is x\geq{-2}? and (2) answers this question with a NO. Sufficient.

Answer: C.

Hope it's clear.

hi banuel,

From statement 1 , how you conclude we hace solve for whether x<= -2 ?
_________________

Kabilan.K Kudos is a boost to participate actively and contribute more to the forum

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]
12 May 2013, 09:55

Expert's post

kabilank87 wrote:

Bunuel wrote:

If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is x^2 - x - 6\geq{x^2 - 2x - 8}? --> is x\geq{-2}? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is x\geq{-2}? and (2) answers this question with a NO. Sufficient.

Answer: C.

Hope it's clear.

hi banuel,

From statement 1 , how you conclude we hace solve for whether x<= -2 ?

x^2 - x - 6\geq{x^2 - 2x - 8} Cancel x^2 on both sides and re-arrange - x +2x\geq{ - 8+6} -->x\geq{-2}.

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]
12 May 2013, 10:33

Question: f(x) = x^2 - x - 6. Is f(x) >= g(x)?

1) g(x) = x^2 - 2x - 8 2) x < - 2

Consider statement (1):

For f(x) to be greater than or equal to g(x), x^2-x-6 should be greater than or equal to x^2-2x-8. This implies that: x^2-x-6>=x^2-2x-8 (or) x>=-2.

We don't know for sure that this is true. So this is not sufficient.

Consider statement (2):

x < -2. However, we know nothing about the form of the function g(x). So this is not sufficient, either.

Consider both together:

If g(x) is given by the function in option (1), then we require x>=-2 in order for f(x) to be greater than or equal to g(x). From (2) we know that x < -2. Therefore, both statements together are sufficient to prove that f(x) is NOT greater than or equal to g(x).

The correct answer, therefore, is C.
_________________

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]
13 May 2013, 02:14

Expert's post

shankg06 wrote:

mridulparashar1 wrote:

If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8

(2) x < -2

I have tried this Question and got it wrong first time (Wrong approach). But I don't agree with OA.

How to answer it?? where are the options??

This is a data sufficiency question.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked. C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked. D. EACH statement ALONE is sufficient to answer the question asked. E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]
14 May 2013, 01:13

Bunuel wrote:

If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is x^2 - x - 6\geq{x^2 - 2x - 8}? --> is x\geq{-2}? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is x\geq{-2}? and (2) answers this question with a NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel.. I understood Your solution but I have a doubt.

x^2 -x -6 = (x-3)*(x+2) x^2-2x - 8 = (x-4)*(x+2)

now the question becomes is (x-3)*(x+2) >= (x-4)*(x+2)

if x is not equal to -2 then it will become is x-3 > x-4 so YES if x is equal to -2 then both the sides are equal so YES

So isn't A sufficient ??
_________________

"Kudos" will help me a lot!!!!!!Please donate some!!!

Completed Official Quant Review OG - Quant

In Progress Official Verbal Review OG 13th ed MGMAT IR AWA Structure

Yet to do 100 700+ SC questions MR Verbal MR Quant

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]
15 May 2013, 01:44

Expert's post

SrinathVangala wrote:

Bunuel wrote:

If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is x^2 - x - 6\geq{x^2 - 2x - 8}? --> is x\geq{-2}? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is x\geq{-2}? and (2) answers this question with a NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel.. I understood Your solution but I have a doubt.

x^2 -x -6 = (x-3)*(x+2) x^2-2x - 8 = (x-4)*(x+2)

now the question becomes is (x-3)*(x+2) >= (x-4)*(x+2)

if x is not equal to -2 then it will become is x-3 > x-4 so YES if x is equal to -2 then both the sides are equal so YES

So isn't A sufficient ??

The red part is not correct.

If x is not equal to -2, then x+2 could be more (for example, if x=0) as well as less than zero (for example if x=-3).

If x+2 is less than zero, then when reducing (x-3)*(x+2)\geq{ (x-4)*(x+2)} by negative x+2 we should flip the sign and we'll get x-3\leq{{x-4} --> so the answer is NO.

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]
15 May 2013, 02:06

Bunuel wrote:

SrinathVangala wrote:

Bunuel wrote:

If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is x^2 - x - 6\geq{x^2 - 2x - 8}? --> is x\geq{-2}? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is x\geq{-2}? and (2) answers this question with a NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel.. I understood Your solution but I have a doubt.

x^2 -x -6 = (x-3)*(x+2) x^2-2x - 8 = (x-4)*(x+2)

now the question becomes is (x-3)*(x+2) >= (x-4)*(x+2)

if x is not equal to -2 then it will become is x-3 > x-4 so YES if x is equal to -2 then both the sides are equal so YES

So isn't A sufficient ??

The red part is not correct.

If x is not equal to -2, then x+2 could be more (for example, if x=0) as well as less than zero (for example if x=-3).

If x+2 is less than zero, then when reducing (x-3)*(x+2)\geq{ (x-4)*(x+2)} by negative x+2 we should flip the sign and we'll get x-3\leq{{x-4} --> so the answer is NO.

Hope it's clear.

Yes!!!!!!! Awesome!!!! I didn't take into account that we cannot cancel negative numbers on the both sides of a inequality without changing the sign.

All I was thinking about was to avoid the condition where x = -2 so that we cancel the terms.

Thanks a lot!!!! Will remember this!!!!
_________________

"Kudos" will help me a lot!!!!!!Please donate some!!!

Completed Official Quant Review OG - Quant

In Progress Official Verbal Review OG 13th ed MGMAT IR AWA Structure

Yet to do 100 700+ SC questions MR Verbal MR Quant

Verbal is a ghost. Cant find head and tail of it.

gmatclubot

Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?
[#permalink]
15 May 2013, 02:06