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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Regards,

PKN

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