BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

If there are more than two numbers in a certain list, is

Expert replies
by BTGmoderatorDC » Thu Oct 18, 2018 6:08 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0?

(1) The product of any two numbers in the list is equal to 0.
(2) The sum of any two numbers in the list is equal to 0.

OA B

Source: GMAT Prep
Join the discussion
Source: — Data Sufficiency |

by Jay@ManhattanReview » Thu Oct 18, 2018 8:48 pm
BTGmoderatorDC wrote:If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0?

(1) The product of any two numbers in the list is equal to 0.
(2) The sum of any two numbers in the list is equal to 0.

OA B

Source: GMAT Prep
Given: There are more than two numbers in a certain list.

Question: Is each of the numbers in the list equal to 0?

Let's take each statement one by one.

(1) The product of any two numbers in the list is equal to 0.

Case 1: Say the set is: { 0, 0, 0, ...}. The answer is yes.
Case 2: Say the set is: { 0, 1, 2, 3, ...}. The answer is No.

No unique answer. Insufficient.

(2) The sum of any two numbers in the list is equal to 0.

Case 1: Say the set is: { 0, 0, 0, ...}. The answer is yes.
Case 2: Say the set is: { -1, 1, -2, 2, -3, 3}. Though the sum of the set of numbers is 0, this is a not a valid case.

The key word in the statement "The sum of any two numbers in the list is equal to 0." is 'any'.

From the above set, if we take -1 and -2 (any two numbers), the sum is not 0.

Thus, only one set is possible, i.e., when each number of the set is 0. Sufficient.

The correct answer: B

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: Pasadena | Warsaw | Cairo | Kuala Lumpur | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
Join the discussion

by Brent@GMATPrepNow » Fri Oct 19, 2018 4:51 am
Jay@ManhattanReview wrote:
BTGmoderatorDC wrote:If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0?

(1) The product of any two numbers in the list is equal to 0.
(2) The sum of any two numbers in the list is equal to 0.

OA B



(1) The product of any two numbers in the list is equal to 0.

Case 1: Say the set is: { 0, 0, 0, ...}. The answer is yes.
Case 2: Say the set is: { 0, 1, 2, 3, ...}. The answer is No.

No unique answer. Insufficient.
Hey Jay,

I thought I should point out that case 2 doesn't satisfy the statement.
If we choose 1 and 2 from the set, then the product isn't 0.
That said, {0, 0, 0, 0, 1} meets the given condition, as does {0, 0, 0, 0, 0, 0, 0, 3} etc

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

BTGmoderatorDC wrote:If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0?

(1) The product of any two numbers in the list is equal to 0.
(2) The sum of any two numbers in the list is equal to 0.
Source: GMAT Prep
\[L = \left\{ {\,{x_1}\,,\,{x_2}\,,\, \ldots \,\,,\,\,{x_n}} \right\}\,\,\,\,,\,\,\,n \geqslant 3\]
\[?\,\,\,:\,\,\,{\text{all}}\,\,{\text{zero}}\]
\[\left( 1 \right)\,\,\,{x_j} \cdot {x_k} = 0\,\,\,\,\,\left( {j \ne k} \right)\,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,L = \left\{ {0,0, \ldots ,0,0} \right\}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,L = \left\{ {0,0, \ldots ,0,1} \right\}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\, \hfill \\
\end{gathered} \right.\]

What about statement (2)? Do you "feel" this statement is sufficient... but you cannot be 100% sure?

EMBRACE MATHEMATICS and develop your quantitative maturity to EXCEL IN YOUR EXAM (and in the MBA that goes right after it)!

\[\left( 2 \right)\,\,\left\{ \begin{gathered}
\,{x_j} + {x_k} = 0 \hfill \\
{x_k} + {x_m} = 0 \hfill \\
\end{gathered} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,\,\,{x_j} - {x_m} = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x_j} = {x_m}\,\,\,\,{\text{for}}\,\,\,\underline {{\text{ANY}}} \,\,\,\,{x_j}\,,\,\,{x_k}\,,\,\,{x_m}\,\,\,{\text{in}}\,\,L\]
\[\,\left\{ \begin{gathered}
\,{x_j} = {x_m} \hfill \\
\,0 = {x_j} + {x_m} = 2\,\, \cdot {x_j} \hfill \\
\end{gathered} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{x_j} = 0\,\,\,{\text{for}}\,\,{\text{all}}\,\,j\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Join the discussion

by fskilnik@GMATH » Sun Oct 21, 2018 7:10 am
I was asked if there is another formal proof of the sufficiency of the statement (2) but in a more "down-to-earth" arguments.

Certainly!

Let´s imagine (at first) that there is a negative number among the elements in the given list, say A.
In this case, there is another number in the list (say B) such that A+B= 0, hence B must be positive (B=-A).
Let´s consider any third number (say C) of the list. (We know the list has at least three elements.)
It is impossible to have A+C = 0 (C would be positive) and B+C = 0 (C would be negative) simultaneously,
therefore there is NO negative number among the elements of the given list.

Let´s now imagine that there is a positive number among the elements in the given list, say B.
In this case, there is a negative number (say A) so that B+A = 0 (A=-B), but we have already proven
(in the previous paragraph) that there are NO negative elements in the given list.

From both paragraphs above, we are sure all numbers (elements) in the given list must be non-negative
and also non-positive, hence all of them are equal to zero.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Join the discussion