I agree with the general consensus that the question is still ambiguous.

"Which of the following inequalities must be true if the values of x are between -1 and 5?"
-> "if" implies that the condition -1 < x < -5 is a given.

However the intention of the question from the explanation seems to be that the condition -1 < x < -5 is something that we can trying to satisfy.

Therefore, the question should read something like:
Which of the following inequalities must be true for the value of x to be between -1 and 5?
OR
Which of the following inequalities satisfies the inequality -1 < x < -5?

I solved this like so -

Given: the values of are between -1 and 5.

Hence, -1 < x < 5
Adding -2 throughout we get;

=> -1 - 2 < x - 2 < 5 - 2
=> -3 < (x - 2) < 3

This means that |x-2| < 3.

Hence we can directly select the last option (E).

Please let me know if you think I assumed something incorrectly. Thanks.

Can someone help me understand why in the answer explanations, it states Option B : where absolute value of x is between 1 & 5, it states that absolute value of x will always be greater than -1, so we are left with absolute value of x is < 5 .

I have a problem solving 2 inequalities simultaneously. I don't know which set to choose from: -1<x<1 or -5<x<5.

Can someone help please?

Thanks..

(B) is most definitely still correct. The question states, "Which of the following inequalities must be true if the values of x are between -1 and 5?" So you're GIVEN that -1 < x < 5. Then we know that 0 <= |x| < 5, and since -1 < 0, (B) is absolutely correct.

The people who are saying (B) is wrong are claiming that it's wrong because (B) expands out to -5 < x < 5, and then x can be in the range -5 < x < -1, which doesn't satisfy the original inequality. But that's completely contradicting the given information - you're TOLD that x can't be less than -1. So x can't be in the range -5 < x < -1.

It sounds like people are doing the problem backwards - using the answers to solve the given information - rather than vice versa. This one needs to be fixed... in the meantime, I'll count my score out of 44 instead of 45 problems.

Guys it's very clear that B is still a correct answer. To ask "which inequality must be true if" means that the inequality in question can be WIDER in range than the statement that is a given fact (that x is between -1 and 5). Thus, |x|< 5 (all values between -5 and 5) includes all solutions of the given equation (-1 < x < 5), and must therefore be true for any value of x between the given range of -1 and 5.

Hope that helps.

other people's confusions are starting to confuse me as well.. i concluded on (E) as well.. not sure if previous posts are still referring to incorrect wording, so hopefully someone could confirm if I'm going about this the right way?

13. Which of the following inequalities must be true for the value of x to be between -1 and 5 ?
i.e. (-1 < x < 5)

A. |3 – x| < -3
if x is (+):
3-x < -3
-x < -6
x > 6
if x is (-):
-3 - (-x) < -3
x < 0
** x<0 or x>6 ; doesn't satisfy



B. -1< |x| < 5
similarly,
-1 < x < 5 (if x is positive)
1 > x > -5 (if x is negative)

** even though first condition (when x is positive) satisfies the desired inequality, the second one (when x is negative), does not... for example, under the restriction (-1 < x < 5), x can be 0,1,2,3,4... but this answer choice allows for x to be 0, -1, -2, -3, -4 (or any fraction in between, since the question doesn't necessarily mention that x is an integer).. anyways, we can't assume x is positive or negative, which means the answer choice must satisfy what the question is asking under both conditions for it to be correct. Thus (B) is a wrong answer choice.

C. |x| - 2 > 2
Same reasoning,
x > 4
x < -4

wrong -- (allows for x to be outside required range)

D. |2 + x| > 3

x > 1 (x is positive)
x < -5 (x is negative)

wrong -- (allows for x to be outside required range)


E. |x – 2| < 3

x < 5
x > -1

** this answer choice gives (-1 < x < 5) --- the question was asking which inequality must be true in order for x to be somewhere between -1 and 5... (E) satisfies this condition.

not sure if I'm on the right track here; perhaps someone could verify?
thanks!