St1:- 20 < lx+10l < 30

You know, |x+10| is always positive. Also, (a) |x+10|=x+10 when \(x+10\geq{0}\) or, \(x\geq{-10}\) (b) |x+10|=-(x+10) when \((x+10)<{0}\) or, \(x<{-10}\)

Case-I:- When x< -10 then |x+10|=-(x+10)

We can write 20 < lx+10l < 30 =\(20<\left(-\left(x+10\right)\right)<30\)
On solving, we have, -40 < x < -30---------------(1)

Case-II:- When \(x+10\geq{0}\) then |x+10|=x+10
We can write 20 < lx+10l < 30 =20 < x+10 < 30
On solving, we have, 10<x<20------------------------(2)
From (1) & (2) , we have range of x: -40 < x < -30 or 10<x<20
Answer to question stem is NO.
Sufficient.

St2:- 10 < lx+5l < 30

Applying the same procedures as we have applied in st1, we have

-35 < x < -15 or 5 < x < 25
Question stem is inconsistent.
Insufficient.

Ans. (A)

PS:- We can solve this question using graphical method too.
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I have one query regarding this question.though I had solved this question by absolute method but can't we just try to put the value of X =20 and confirm
In the first case it can not be equal to 20 as it will make X+10 as 30 and it is not smaller than 30 so the answer is No

For the second scenario X can be 20 or something else also eg 20 or 21.

Is this method correct or should we stick to absolute method

Posted from my mobile device

Statement 1
\(20 < |x + 10| < 30\)

Case 1
\(20 < x + 10 < 30\)
\(20 - 10 < x + 10 - 10 < 30 - 10\)
\(10 < x < 20\)

Case 2
\(20 + 10 < - x - 10 + 10 < 30 + 10\)
\(30< - x < 40\)
\(- 40< x < - 30\)

From the two cases we can conclude x < 20. This means x ≠ 20
Sufficient

Statement 1
\(10 < |x + 5| < 30\)

Case 1
\(10 < x + 5 < 30\)
\(10 - 5< x + 5 - 5< 30 - 5\)
\(5< x < 25\)

Case 2
\(10 < - x - 5 < 30\)
\(10 + 5 < - x - 5 + 5 < 30 + 5\)
\(15 < - x < 35\)
\(- 35 < x < - 15\)

From the two cases we can conclude x < 25. This means x ≠ 20 or x = 20
Insufficient

Hence, A
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