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A. \(|3-x| < -3\) --> absolute value is always non-negative, hence this options is wrong for any value of \(x\); B. \(|x| < 4\) --> not necessarily true: consider \(x=4.5\); C. \(|x| - 2 > 2\) --> \(|x|>4\) --> not necessarily true: consider \(x=3\); D. \(|2 + x| > 3\) --> not necessarily true: consider \(x=0\);

So, we are left with option E only.

Just to check: E. \(|x-2| < 3\) --> \(-3<x-2<3\) --> add 2 to each part: \(-1<x<5\). So, this option is basically the same as the condition given in the stem.

In order to solve the modulus inequality you have to consider two possible values of \(x\): a positive one and a negative one. First, you just strip the modulus signs and solve the inequality: \(x-2<3\) \(x<5\) This is the range of \(x\) for positive values of \(x\). Now you need to solve a different inequality for negative \(x\): \(-x+2<3\) \(x>-1\) What we did here is flip the signs of the expression inside the modulus and solve the new inequality. This is range of \(x\) for negative \(x\). Combining both \(x<5\) and \(x>-1\) you get the solution \(x \in (-1,5)\).

harjaskooner wrote:

pls explain how |x-2|<3 is x(-1,5) ...really need help !

Sorry guys, I had limited access to Internet for 10 days. I agree the question is ambiguous. It has to be reworded somehow. Here are my two suggestions:

Which of the following inequalities satisfies the condition that the values of \(x\) are between -1 and 5? Which of the following inequalities has solution for \(x\) in the range \((-1,5)\)?

What do you all think? Is either of the rewordings any better than the original?

I would go for the following:

Quote:

Which of the following inequalities must be true if the values of x are between -1 and 5?

A. |3 – x| < -3 B. -1< |x| < 5 C. |x| - 2 > 2 D. |2 + x| > 3 E. |x – 2| < 3

Can someone please explain in more detail why B is wrong? I'm plugging in different values and I can't find one between -1 and 5 that doesn't satisfy the inequality.

Can someone please explain in more detail why B is wrong? I'm plugging in different values and I can't find one between -1 and 5 that doesn't satisfy the inequality.

You should approach it the "opposite" way - instead try values outside of the (-1; 5) range and see if they work - for example (-4). The question is asking you, which equation below represents the best a condition where -1 < x < 5, not which of the following equations will work with values of x. _________________

Agree with GMATTiger, the question clearly states that x lies between -1 and 5. so we should just ignore the 'other' values i.e. -5<x<=-1 for B. B should also satisify the equation.

well, I'm so new to gmat but both b and e may seem like a correct answer, in the actual test I will still prefer e because b probably should seem a trick to me, the absolute value of any number can't be below zero, so I think it's a false expression from the start, right? or am I mixing things?

Sorry guys, I had limited access to Internet for 10 days. I agree the question is ambiguous. It has to be reworded somehow. Here are my two suggestions:

Which of the following inequalities satisfies the condition that the values of \(x\) are between -1 and 5? Which of the following inequalities has solution for \(x\) in the range \((-1,5)\)?

What do you all think? Is either of the rewordings any better than the original? _________________

In order to solve the modulus inequality you have to consider two possible values of \(x\): a positive one and a negative one. First, you just strip the modulus signs and solve the inequality: \(x-2<3\) \(x<5\) This is the range of \(x\) for positive values of \(x\). Now you need to solve a different inequality for negative \(x\): \(-x+2<3\) \(x>-1\) What we did here is flip the signs of the expression inside the modulus and solve the new inequality. This is range of \(x\) for negative \(x\). Combining both \(x<5\) and \(x>-1\) you get the solution \(x \in (-1,5)\).

harjaskooner wrote:

pls explain how |x-2|<3 is x(-1,5) ...really need help !

In order to solve the modulus inequality you have to consider two possible values of \(x\): a positive one and a negative one. First, you just strip the modulus signs and solve the inequality: \(x-2<3\) \(x<5\) This is the range of \(x\) for positive values of \(x\). Now you need to solve a different inequality for negative \(x\): \(-x+2<3\) \(x>-1\) What we did here is flip the signs of the expression inside the modulus and solve the new inequality. This is range of \(x\) for negative \(x\). Combining both \(x<5\) and \(x>-1\) you get the solution \(x \in (-1,5)\).

harjaskooner wrote:

pls explain how |x-2|<3 is x(-1,5) ...really need help !

thanks ! now i understand it !

Guys, my understanding from the GMAT is that between means a closed set i.e. and based on your question does not include the values -1 nor 5 but rather 0, 1, 2, 3 & 4 - of course I am assuming x is an integer. This suggest to me that questions of this sort will not appear in the real test as we seem to have two correct answers to the same question, namely B and D

Guys, my understanding from the GMAT is that between means a closed set i.e. and based on your question does not include the values -1 nor 5 but rather 0, 1, 2, 3 & 4 - of course I am assuming x is an integer. This suggest to me that questions of this sort will not appear in the real test as we seem to have two correct answers to the same question, namely B and D

The highlighted part is not correct. X could be a fraction or -ve as well.

-1 < x < 5 means values excluding -1 and 5. _________________