Question

In the given question, there are two statements marked I and II. Decide which of the statements are sufficient to answer the question. Choose your answer from the given alternative.

Is $z>0$ ?
(1) $(z+1)\left(z\right)(z-1)<0$
(2) $\left|z\right|<1$

• A. Statement I alone is sufficient to answer the problem.
• B. Statement II alone is sufficient to answer the problem.
• C. Statement I and II both are needed.
• D. Statement I and II both are not sufficient.

Answer 1

Answer: (C)

Statement I and II both are needed.

This is tough one. The question stem asks whether $z$ is positive. Both statements look fairly complicated so start with whichever one looks better to you.

(1) INSUFFICIENT: Many people will try $z=0,1$ or $2$ first. All of these cases are invalid ($0$ and $1$ return a product of $0$, which is not less than $0$ and $2$ returns a positive product). Even $-1$ doesn't work! Think outside of the box: what weirder numbers can you try?

Case $z$ $(z+1)\left(z\right)(z-1)<0$ Is $z>0$

#1 $-2$ $(-1)(-2)(-3)=-6$ No

#2 $\frac{1}{2}$ $\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)(-\frac{1}{2})=-\frac{3}{8}$ Yes

Careful! While the three terms $z+1,z$ and $z-1$ appear to represent consecutive integers, the problem never specifies that $z$ is an integer. When you pick a fraction, $z=\frac{1}{2}$, you find a case that makes the statement true and also answers the question with a Yes.

(2)INSUFFICIENT: If you understand absolute value, then you might recognize that the statement $\left|z\right|<1$ establishes that $z$ is between $-1$ and $1$. If not, test some cases.

In general, start by trying the numbers that worked in the lase statement. Are they valid for this statement as well? If so this could save you time in evaluating these cases and also make it easier if and when you get to the stage of combining the two statements:

Case $z$ $\left|z\right|<1$ Is $z>0$

#1 $-2$ $2<1$

#2 $\frac{1}{2}$ $\frac{1}{2}<1$ Yes

#3 $-\frac{1}{2}$ $-\frac{1}{2}<1$ No

For this statement $-2$ is an invalid case, but $\frac{1}{2}$ is valid. That valid case returns a Yes answer, so try to find a case that will return a No answer instead. Case 3 does just that

(1) and (2) SUFFICIENT: Statement (2) allows any values between $-1$ and $1$. From among these possible values, any numbers between $0$ and $1$ will also make statement (1) true. Therefore, $z$ can be greater than $0$.

The answer could still be (E), though, if any numbers between $-1$ and $0$ make statement (1) true as well.Try plugging $z=-\frac{1}{2}$ into statement (1):

$(-\frac{1}{2}+1)(-\frac{1}{2})(-\frac{1}{2}-1)<0=$?

$\left(\frac{1}{2}\right)(-\frac{1}{2})(-\frac{3}{2})<0$

Don't multiply out the let side of the equation! The two negative terms will multiply to a positive number, leaving:

positive $<0$

This, of course, is never true. It may take a little more work or reasoning to realize that this result will be repeated for any $z$ you pick between $-1$ and $0$. Therefore, only numbers between $0$ and $1$ work for both statements. The value of $z$ has to be positive.

The correct answer is (C)