GMAT Changed on April 16th - Read about the latest changes here

 It is currently 22 Apr 2018, 12:54

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M17-27

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44599

### Show Tags

16 Sep 2014, 01:02
Expert's post
15
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

31% (01:23) correct 69% (01:19) wrong based on 199 sessions

### HideShow timer Statistics

Is the product of four consecutive even integers positive?

(1) The sum of these integers is positive but smaller than 20

(2) The product of the middle two of these integers is positive
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 44599

### Show Tags

16 Sep 2014, 01:02
Expert's post
3
This post was
BOOKMARKED
Official Solution:

First of all: the product of four consecutive even integers can be either 0 (when one of the eve integers is zero) or positive (when all integers are positive or when all integers are negative).

(1) The sum of these integers is positive but smaller than 20:

If the greatest term is $$\ge 8$$, e.g. {2, 4, 6, 8 }, then the sum will be more than or equal to 20.

If the smallest term $$\le -4$$, e.g. {-4, -2, 0, 2 }, then the sum won't be positive.

Hence this statement gives only TWO possible sets {0, 2, 4, 6 } and {-2, 0, 2, 4 } ($$sum \lt 20$$ and $$product =0$$). The product of the terms of either of 2 sets is zero, so the answer to the question is NO. Sufficient.

(2) The product of the middle two of these integers is positive. The product could be zero as well as positive. For example consider {0, 2, 4, 6 } ($$product = 0$$) and {2, 4, 6, 8 } ($$product=positive$$). Not sufficient.

_________________
Intern
Joined: 14 Mar 2015
Posts: 34
Schools: ISB '18

### Show Tags

23 Oct 2015, 02:18
Bunuel wrote:
Official Solution:

First of all: the product of four consecutive even integers can be either 0 (when one of the eve integers is zero) or positive (when all integers are positive or when all integers are negative).

(1) The sum of these integers is positive but smaller than 20:

If the greatest term is $$\ge 8$$, e.g. {2, 4, 6, 8 }, then the sum will be more than or equal to 20.

If the smallest term $$\le -4$$, e.g. {-4, -2, 0, 2 }, then the sum won't be positive.

Hence this statement gives only TWO possible sets {0, 2, 4, 6 } and {-2, 0, 2, 4 } ($$sum \lt 20$$ and $$product =0$$). The product of the terms of either of 2 sets is zero, so the answer to the question is NO. Sufficient.

(2) The product of the middle two of these integers is positive. The product could be zero as well as positive. For example consider {0, 2, 4, 6 } ($$product = 0$$) and {2, 4, 6, 8 } ($$product=positive$$). Not sufficient.

For (1), i would just assume, just to simplify, the integers to be x, x+2, x+4, x+6.
So, their sum will be (4x + 12).
Now (4x + 12) < 20.
This now follows the only two possible sets for x = 0 and x = -2, as pointed out by you.
Bunuel wrote:
TWO possible sets {0, 2, 4, 6 } and {-2, 0, 2, 4 } ($$sum \lt 20$$ and $$product =0$$)

I just feel better playing with the equations than the numbers directly.
_________________

It ain’t about how hard you hit. It’s about how hard you can get hit and keep moving forward;
how much you can take and keep moving forward.

That’s how winning is done!

Intern
Joined: 07 May 2012
Posts: 8

### Show Tags

30 Apr 2016, 08:01
vivekgautam1 wrote:
Bunuel wrote:
Official Solution:

First of all: the product of four consecutive even integers can be either 0 (when one of the eve integers is zero) or positive (when all integers are positive or when all integers are negative).

(1) The sum of these integers is positive but smaller than 20:

If the greatest term is $$\ge 8$$, e.g. {2, 4, 6, 8 }, then the sum will be more than or equal to 20.

If the smallest term $$\le -4$$, e.g. {-4, -2, 0, 2 }, then the sum won't be positive.

Hence this statement gives only TWO possible sets {0, 2, 4, 6 } and {-2, 0, 2, 4 } ($$sum \lt 20$$ and $$product =0$$). The product of the terms of either of 2 sets is zero, so the answer to the question is NO. Sufficient.

(2) The product of the middle two of these integers is positive. The product could be zero as well as positive. For example consider {0, 2, 4, 6 } ($$product = 0$$) and {2, 4, 6, 8 } ($$product=positive$$). Not sufficient.

I went the equation route too. Felt comfortable with that.

Let the numbers be 2n , 2n+2 , 2n+4 , 2n+6 ( 4 consecutive even int)

1 says that sum is +ve but less than 20.

Hence 0 < 8n + 12 < 20
It gives -1.5 < n < 1

Hence n = 0,-1.
So we have 0 , 2 , 4 , 6 --> Prod = 0
or
-2,0,2,4 ---> Prod = 0

Hence st 1 is suff

St 2:
We only know middle numbers prod > 0.
So prod could be > or = 0.
Not suff
Retired Moderator
Joined: 26 Jul 2016
Posts: 298
GMAT 1: 750 Q50 V42
WE: Engineering (Energy and Utilities)

### Show Tags

06 Sep 2016, 01:03
I think this is a high-quality question and I don't agree with the explanation. In many of your explanations, Zero is considered "neither positive nor negatice". Infact the answer hinges on this point.

Here you have considered Zero as postive, why?
Math Expert
Joined: 02 Sep 2009
Posts: 44599

### Show Tags

06 Sep 2016, 01:31
1
KUDOS
Expert's post
bitun10 wrote:
I think this is a high-quality question and I don't agree with the explanation. In many of your explanations, Zero is considered "neither positive nor negatice". Infact the answer hinges on this point.

Here you have considered Zero as postive, why?

Where in the solution is 0 considered positive? That's not so.
_________________
Intern
Joined: 22 Jul 2017
Posts: 1

### Show Tags

27 Aug 2017, 12:43
It says in the second stem that the product is positive. Zero is neither positive nor negative?
Correct me if i am wrong
Intern
Joined: 17 Feb 2017
Posts: 5
Location: India
GMAT 1: 670 Q49 V32
GPA: 3.3

### Show Tags

22 Nov 2017, 23:02
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Shouldn't the set be {-2, 2, 4, 8} instead of considering the 0 in the set as 0 is NOT an even integer?

By this logic also, statement 1 seems is correct. I marked 1 by this logic.

Please explain. Should we consider 0 or not?
Math Expert
Joined: 02 Sep 2009
Posts: 44599

### Show Tags

23 Nov 2017, 00:11
Expert's post
2
This post was
BOOKMARKED
jasanisanket24 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Shouldn't the set be {-2, 2, 4, 8} instead of considering the 0 in the set as 0 is NOT an even integer?

By this logic also, statement 1 seems is correct. I marked 1 by this logic.

Please explain. Should we consider 0 or not?

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check below for more:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
_________________
Re: M17-27   [#permalink] 23 Nov 2017, 00:11
Display posts from previous: Sort by

# M17-27

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.