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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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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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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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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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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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Babineaux
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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Bunuel
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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Simplest way to solve a must be true question is to take the critical values of ur final equation and check if it does satisfy the option, if it does then thats ur answer, otherwise not. In this case values lie between 3 and 15 and hence all values MUST be greater than -5. But when you start taking intervals if you even have one value on the other side of equality then it does not satisfy must be true questions.
Bunuel
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
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Bunuel
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

For must be true questions, write down "given" and get the kind of values that x will take.

Given:
\(|-\frac{x}{3} + 3| < 2\)

\(|-\frac{x}{3} + 3| = |\frac{x}{3} - 3|= 3|x - 9| \)

\(3|x - 9| < 2\)
\(|x - 9| < 2/3\)

So x is at a distance of 2/3 from 9. Visualize it on the number line - all values of x lie somewhere between 8 and 10.
So x is definitely more than -5 for every possible value of x.

Answer (D)

Theory discussed here: https://youtu.be/3mmzzEJSLSQ
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