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Re: If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
Explanation:

|−x/3 + 3| < 2
−2 < −x/3 + 3 < 2
−5 < −x/3 <−1
1 < x/3 < 5
3 < x < 15

IMO-D
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Re: If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
Given, |-\(\frac{x}{3}\) + 3| < 2

=> -2 < -\(\frac{x}{3}\) + 3 < 2

=> -5 < -\(\frac{x}{3}\) < -1

=> -15 < - x < -3

=> 3 < x < 15

Now, x > -5 gives the range from 3, 3 - (- 5) = 8 and within -5 < x < 3 none is satisfying

And, x < 7 gives the range from 3, 7 - 3 = 4 and withing 3 < x < 7 several points satisfy the inequality.

To me E is satisfying.
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Re: If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
IMO-D

Given|-|-x/3+3|<2

-2 <-x/3 +3 <2
-5< -x/3< -1
-15<-x<-3
3<x<15

Only D statisfy the condition

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Re: If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
Please clarify how come D is the answer??

We are seeing that X>3 and X<15.
In between 3 and 15 the possible value of X would be?? How come X>-5??

Please clarify with logic.
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Re: If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
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PriyankarPaul wrote:
Please clarify how come D is the answer??

We are seeing that X>3 and X<15.
In between 3 and 15 the possible value of X would be?? How come X>-5??

Please clarify with logic.

Let's see the constraint.
We have 3<X<15.
Let's sat X = 14
Now check which constraint among option have 14 in between them.

A. -3 < x < 0 (Incorrect, as 14 is greater than 0)

B. 0 < x < 9 (Incorrect, as 14 is greater than 9)

C. -3 < x < 8 (Incorrect, as 14 is greater than 8)

D. x > -5 (X is greater than -5. So it can surely take value of +14)

E. x < 7 (Incorrect, as 14 is greater than 7)

Hence D is the answer.
When we are dealing with inequality question we have to see which option satisfies both extreme values. In 3>X>15
Extreme values are 4&14
The options must satisfy both

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If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
Babineaux wrote:
Given, |-\(\frac{x}{3}\) + 3| < 2

=> -2 < -\(\frac{x}{3}\) + 3 < 2

=> -5 < -\(\frac{x}{3}\) < -1

=> -15 < - x < -3

=> 3 < x < 15

Now, x > -5 gives the range from 3, 3 - (- 5) = 8 and within -5 < x < 3 none is satisfying

And, x < 7 gives the range from 3, 7 - 3 = 4 and withing 3 < x < 7 several points satisfy the inequality.

To me E is satisfying.



Previously I posted the above one,

But opined with previous explanation,

To incorporate both the min max solutions for x, it is definite that, both points are > - 5;

Thus x > - 5

Option D is correct.

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Re: If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
Expert Reply
Bunuel wrote:
If \(|-\frac{x}{3} + 3| < 2\), which of the following must be true ?

A. -3 < x < 0
B. 0 < x < 9
C. -3 < x < 8
D. x > -5
E. x < 7


Solution:

Rewriting the inequality without the absolute value sign, we have:

-2 < -x/3 + 3 < 2

-5 < -x/3 < -1

5 > x/3 > 1

15 > x > 3

Since 3 < x < 15, then x > -5.

Note that the answer x > -5 doesn’t necessarily mean that x can take on any value greater than -5; rather , it means that the entire interval of 3 < x < 15 (the true solution set) is indeed greater than -5. There is no other answer choice that contains all values in the true solution set.

Answer: D
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Re: If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
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Re: If |-x/3 + 3| < 2, which of the following must be true ? [#permalink]
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