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Only A is always true, as ANY \(x\) from the TRUE range \(x\geq{7}\) will be more than 6.

Answer: A.

But with A) x can be 6.1, which will not satisfy the given equation..... shouldn't option C) be valid choice ....

It should be other way around.

We are given that \(x\geq{7}\). The question is: which of the following is true about \(x\)?

\(x>6\) is true about \(x\), because as \(x\) is more than (or equal to) 7 then it's definitely more than 6.

To elaborate more. Question uses the same logic as in the examples below:

If \(x=5\), then which of the following must be true about \(x\): A. x=3 B. x^2=10 C. x<4 D. |x|=1 E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or: If \(-1<x<10\), then which of the following must be true about \(x\): A. x=3 B. x^2=10 C. x<4 D. |x|=1 E. x<120

Again answer is E, because ANY \(x\) from \(-1<x<10\) will be less than 120 so it's always true about the number from this range to say that it's less than 120.

Or: If \(-1<x<0\) or \(x>1\), then which of the following must be true about \(x\): A. x>1 B. x>-1 C. |x|<1 D. |x|=1 E. |x|^2>1

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand for example A is not always true as it says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Re: MGMAT Inequalities [#permalink]
25 May 2012, 02:47

Bunuel wrote:

If \(-1<x<0\) or \(x>1\), then which of the following must be true about \(x\): A. x>1 B. x>-1 C. |x|<1 D. |x|=1 E. |x|^2>1

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand for example A is not always true as it says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Hope it's clear.

just for the sake of learning please tell me how to evaluate option E. |X|^2>1

is it |x|>1 and |x|>-1

if |x|>1 then -1 >x>1

if |x|>-1 then how do we evaluate this part, confused here.

Re: MGMAT Inequalities [#permalink]
25 May 2012, 04:27

Expert's post

Joy111 wrote:

Bunuel wrote:

If \(-1<x<0\) or \(x>1\), then which of the following must be true about \(x\): A. x>1 B. x>-1 C. |x|<1 D. |x|=1 E. |x|^2>1

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand for example A is not always true as it says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Hope it's clear.

just for the sake of learning please tell me how to evaluate option E. |X|^2>1

is it |x|>1 and |x|>-1

if |x|>1 then -1 >x>1

if |x|>-1 then how do we evaluate this part, confused here.

|x|^2>1 is the same as x^2>1, so it holds true for x<-1 and x>1.

As for |x|>-1: it holds true for ANY value of x, because absolute value of a number is always non-negative.

Re: MGMAT Inequalities [#permalink]
25 May 2012, 05:57

Bunuel wrote:

Joy111 wrote:

Bunuel wrote:

If \(-1<x<0\) or \(x>1\), then which of the following must be true about \(x\): A. x>1 B. x>-1 C. |x|<1 D. |x|=1 E. |x|^2>1

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand for example A is not always true as it says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Hope it's clear.

just for the sake of learning please tell me how to evaluate option E. |X|^2>1

is it |x|>1 and |x|>-1

if |x|>1 then -1 >x>1

if |x|>-1 then how do we evaluate this part, confused here.

|x|^2>1 is the same as x^2>1, so it holds true for x<-1 and x>1.

As for |x|>-1: it holds true for ANY value of x, because absolute value of a number is always non-negative.

Hope it helps.

ok , thank you for that .

Now suppose |x|>-1 was one of the options of the question , then couldn't this have been one of the solutions .

can you please show how |x|>-1 fails to satisfy all the values of x in the equation

If -1< x <0 or x >1 then which of the following must be true about x : A. x>1 B. x>-1 C. |x|<1 D. |x|=1 E. |x|>-1

now for all values of x i.e x=-1/2, x=3 ,x=4, option E also is also satisfied .

seems to me that both A and E ,could be the solutions? please correct me . Thank you.

Re: If 4x-12 >= x + 9, which of the following must be true [#permalink]
10 Aug 2012, 07:07

4x-12>=x+9 3x>=21 x>=7

Only option A must be true. The rest of the options need not be true _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Only A is always true, as ANY \(x\) from the TRUE range \(x\geq{7}\) will be more than 6.

Answer: A.

Bunuel, I must be reading the question incorrectly because I feel that after you get x>=7 you think 7=7 and 8>7 so the answer of x could 7 or 8 but since it is asking which is always greater than the answer would be D. 8. I don't see why we would look at it as x>6 when we solved the equation and got x>=7. How would the question need to be worded for that to be the case because I am really confused. I don't feel that in other OG questions we we replaced the x with a number smaller when it is asking what number is greater than or equal to a number larger. If that's the case, that means C, D, and E were the trap answers? I'd appreciate your help. Thanks!

If , then which of the following must be true about : A. x=3 B. x^2=10 C. x<4 D. |x|=1 E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

And for this one I know it can't be A-D, but for E, x could be -9 which wouldn't necessarily mean that it is greater than 5? Would you choose it anyways?

Only A is always true, as ANY \(x\) from the TRUE range \(x\geq{7}\) will be more than 6.

Answer: A.

Bunuel, I must be reading the question incorrectly because I feel that after you get x>=7 you think 7=7 and 8>7 so the answer of x could 7 or 8 but since it is asking which is always greater than the answer would be D. 8. I don't see why we would look at it as x>6 when we solved the equation and got x>=7. How would the question need to be worded for that to be the case because I am really confused. I don't feel that in other OG questions we we replaced the x with a number smaller when it is asking what number is greater than or equal to a number larger. If that's the case, that means C, D, and E were the trap answers? I'd appreciate your help. Thanks!

If , then which of the following must be true about : A. x=3 B. x^2=10 C. x<4 D. |x|=1 E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

And for this one I know it can't be A-D, but for E, x could be -9 which wouldn't necessarily mean that it is greater than 5? Would you choose it anyways?

I think you are confused about what is given and what is asked.

Given that \(x\geq{7}\), so \(x\) is some number which is more than or equal to 7. Now, the question asks, what must be true about \(x\) (which we know is more than or equal to 7).

If \(x\) is more than or equal to 7, then it must be true that \(x\) is greater than 6, thus A must be true.

The same with another example in your post: given that \(x=5\). The question asks, what must be true about \(x\). Since, \(x=5\), thus it's true to say that it's greater than -10.

Re: If 4x-12 >= x + 9, which of the following must be true [#permalink]
13 Dec 2012, 15:02

Hi Bunuel, I am not quite satisfied with the OA. Since the question is not clear if x is an integer, for example x = 6.5 <7 => then it is not true. So the correct choice answer should be C

Re: If 4x-12 >= x + 9, which of the following must be true [#permalink]
14 Dec 2012, 01:27

Expert's post

Moralhazard wrote:

Hi Bunuel, I am not quite satisfied with the OA. Since the question is not clear if x is an integer, for example x = 6.5 <7 => then it is not true. So the correct choice answer should be C

What do you think?

We are not told that x is an integer but it has nothing to do with the question.

x cannot be 6.5 because we are told that \(x\geq{7}\).

Option C (x>7) is not always true, since x can be 7 and in this case x>7 won't hold true.

Re: 4x - 12 ≥ x + 9, which of the following must be true? [#permalink]
27 Dec 2013, 07:26

uwengdori wrote:

A) x>6 B) x<7 C) x>7 D) x>8 E) x<8

This is MGMAT's problem set question (algebra 5th edition pg 104) The answer's reasoning is that for C), one of the most possible answers, can include 7, 7.3, 8 , 9.2 itself, so it's wrong which makes sense.

However, what I don't get is the reasoning behind A being the correct answer is that x could be 7, 7.3, 8, 9.2. Then, what about 6.3, 6.4 and so on?

Re: If 4x-12 >= x + 9, which of the following must be true [#permalink]
14 Jan 2014, 04:55

Bunuel, I don't understand why OA is x>6 we already know for a fact that x>=7 so why not choose x>7 like the answer suggest .You have said that x>6 will always be true but so will be x>7.

Why cant we select the answer that is directly within the x>=7 range?

Re: If 4x-12 >= x + 9, which of the following must be true [#permalink]
14 Jan 2014, 06:01

Expert's post

mumbijoh wrote:

Bunuel, I don't understand why OA is x>6 we already know for a fact that x>=7 so why not choose x>7 like the answer suggest .You have said that x>6 will always be true but so will be x>7.

Why cant we select the answer that is directly within the x>=7 range?

aryabhatta wrote:

I fell for D. It says which of the following must be true. x>8 holds good for all values.

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