Determine whether z 0?

Question

Determine whether z > 0?

(1) (z + 1)(z)(z – 1) < 0

(2) |z| < 1

gmatophobia · Answer

Question: Is z positive Statement 1 (1) (z + 1)(z)(z – 1) < 0 Let's plot the points 0, -1, and + 1 on a number line. ----------- (-1) ----------- (0) ----------- (+1) ----------- (z + 1)(z)(z – 1) < 0 indicates that z can either be between 0 and 1 or z can be less than -1. If 0 < z < 1 → Is z > 0-- Yes If z < -1 → Is z > 0 -- No Statement 1 alone is not sufficient. We can eliminate A and D. Statement 2 (2) |z| > 1 The distance of 'z' from 0 is greater than 1. This implies that either z > 1 or z < -1 ----------- (-1) ----------- (0)----------- (+1) ----------- If z > 1 → Is z > 0 -- Yes If z < -1 → Is z > 0 -- No Statement 2 alone is not sufficient. We can eliminate B. Combined The region common to statement 1 and statement 2 is z < -1. If z < -1 → Is z > 0 -- No We now have a definite answer. Hence, the statements combined are sufficient. Option C Similar Question: https://gmatclub.com/forum/find-out-whe ... 14306.html

Nabneet · Answer

St 1:
(z + 1) (z) (z – 1) < 0
The condition satisfies when,
0<z<1 & z<-1
Not Sufficient

St2:
z>1 & z<-1
Not sufficient

Combining:
z<-1

Answer C

Laila12618 · Answer

I don’t understand… for statement 1 the inequality sign does not have the equal sign so none of z-1, z, z+1 can be zero.
Consequently, all three numbers, which are consecutive, must be negative in order for their product to be negative >>> z must be negative and A would look like the correct answer.

Anyone can explain why statement 1 is not sufficient?

Posted from my mobile device

Nabneet · Answer

A is not sufficient because

z can be between 0 and 1 also along with z<-1

Bunuel · Answer

(z + 1)(z)(z – 1) < 0 Here is how to solve the above inequality easily. The "roots", in ascending order, are -1, 0, and 1, which gives us 4 ranges: z < -1 ; -1 < z < 0 ; 0 < z < 1 ; z > 1 . Next, test an extreme value for z : if z is some large enough number, say 10, then all three multiples will be positive, giving a positive result for the whole expression. So when z > 1 , the expression is positive. Now the trick: as in the 4th range, the expression is positive, then in the 3rd it'll be negative, in the 2nd it'll be positive, and finally in the 1st, it'll be negative: ext{(- + - +)} . So, the ranges when the expression is negative are: z < -1 and 0 < z < 1 . P.S. You can apply this technique to any inequality of the form (ax - b)(cx - d)(ex - f)... > 0 or < 0 . If one of the factors is in the form (b - ax) instead of (ax - b) , simply rewrite it as -(ax - b) and adjust the inequality sign accordingly. For example, consider the inequality (4 - x)(2 + x) > 0 . Rewrite it as -(x - 4)(2 + x) > 0 . Multiply by -1 and flip the sign: (x - 4)(2 + x) < 0 . The "roots", in ascending order, are -2 and 4, giving us three ranges: x < -2 , -2 < x < 4 , and x > 4 . If x is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when x > 4 , the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: ext{(+ - +)} . Therefore, the ranges where the expression is negative are: -2 < x < 4 .

Determine whether z > 0?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

Determine whether z > 0?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards