GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Jan 2019, 17:09

### 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 January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winners of the GMAT game show

January 22, 2019

January 22, 2019

10:00 PM PST

11:00 PM PST

In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
• ### Key Strategies to Master GMAT SC

January 26, 2019

January 26, 2019

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

# M22-22

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52390

### Show Tags

16 Sep 2014, 00:17
12
00:00

Difficulty:

95% (hard)

Question Stats:

27% (01:20) correct 73% (01:34) wrong based on 210 sessions

### HideShow timer Statistics

Is $$a+b+c$$ even?

(1) $$a-c-b$$ is even.

(2) $$\frac{a-c}{b}$$ is odd

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52390

### Show Tags

16 Sep 2014, 00:17
3
1
Official Solution:

Note that we are not told that $$a$$, $$b$$ and $$c$$ are integers.

(1) $$a-c-b$$ is even. If the variables are integers then $$a+b+c$$ will be even but if they are not, for example if $$a=3.5$$, $$b=1$$ and $$c=0.5$$ ($$a-c-b=2=even$$), then $$a+b+c=5=odd$$. Not sufficient.

(2) $$\frac{a-c}{b}$$ is odd. The same here: if the variables are integers then $$a+b+c$$ will be even but if they are not, for example if $$a=3.5$$, $$b=1$$ and $$c=0.5$$ ($$\frac{a-c}{b}=3=odd$$), then $$a+b+c=5=odd$$. Not sufficient

(1)+(2) $$a+b+c$$ may or may not be even (again if variables are integers: YES but if $$a=3.5$$, $$b=1$$, $$c=0.5$$ answer is NO). Not sufficient.

_________________
Intern
Joined: 11 Jan 2013
Posts: 4

### Show Tags

23 Dec 2014, 15:27
Would the answer be D if it was given that a, b, and c are integers?
Math Expert
Joined: 02 Sep 2009
Posts: 52390

### Show Tags

24 Dec 2014, 06:23
amishra1 wrote:
Would the answer be D if it was given that a, b, and c are integers?

Yes, in that case the answer would be D.
_________________
Intern
Joined: 06 Mar 2013
Posts: 1

### Show Tags

26 Jun 2015, 04:03
Here's how I would look at Statement A,

a - c - b = even
a + c + b - 2c - 2b = even
a + c + b - 2(c+b) = even

For this statement to be true,
a + c + b = even, since only "even - even = even"

Am I missing something?
Math Expert
Joined: 02 Sep 2009
Posts: 52390

### Show Tags

26 Jun 2015, 05:06
mukulc wrote:
Here's how I would look at Statement A,

a - c - b = even
a + c + b - 2c - 2b = even
a + c + b - 2(c+b) = even

For this statement to be true,
a + c + b = even, since only "even - even = even"

Am I missing something?

Yes.

I guess you haven't read the solution. Very first sentence there says: "note that we are not told that a, b and c are integers".
_________________
Intern
Joined: 05 Aug 2015
Posts: 42

### Show Tags

12 Mar 2016, 22:50
For statement 1, is it correct to say that if A+B+C=Even and A,B,C are integers, then A-B-C is also even under any order of A,B,C (i.e., also even C-A-B or B-A-C)?
_________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

Math Expert
Joined: 02 Sep 2009
Posts: 52390

### Show Tags

13 Mar 2016, 07:22
happyface101 wrote:
For statement 1, is it correct to say that if A+B+C=Even and A,B,C are integers, then A-B-C is also even under any order of A,B,C (i.e., also even C-A-B or B-A-C)?

Yes, this would be correct.
_________________
Intern
Joined: 06 May 2016
Posts: 36

### Show Tags

01 Aug 2016, 11:02
I think this is a high-quality question and I agree with explanation. Good one!
Senior Manager
Joined: 31 Mar 2016
Posts: 384
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

### Show Tags

22 Aug 2016, 04:36
I think this is a high-quality question and I agree with explanation. Bombastic Question Lol great trap! You people never fail to surprise me!
Manager
Joined: 17 Mar 2015
Posts: 116

### Show Tags

01 Dec 2016, 06:33
Sigh, how many more times am I going to forget to check if I'm dealing with integers. Quickest and proudest answer D, thinking I'm all that smart, just to be blown by the "might not be integers" detail.
Intern
Joined: 15 Aug 2017
Posts: 14

### Show Tags

30 Sep 2017, 11:30
1
1
Here is a method I used to solve this question in the most efficient manner that I haven't seen discussed here.

Stem:$$A+B+C= Even$$
(1)$$A-B-C=E$$
This becomes $$A=E+B+C$$
Important property: In a term of the form (Even number) + X, the (Even number) plays no role in the Even-Odd nature of the term
In turn, this becomes A=B+C,
Plugging into the equation we get:$$B+C+B+C=E?$$this becomes $$2B+2C=E$$
If A & B are integers, this is sufficient, but A & B can both be .1

(2)$$\frac{A-C}{B}=O$$
Or $$O*B=A-C$$
Important property: In a term of the form (Odd number)*(X), the (Odd number) plays no role in the Even-Odd nature of the term
$$B=A-C$$
Plugging into the equation we are left with $$2A=E?$$
If A is an integer, this is sufficient, but A can be .1

Therefore E
Intern
Joined: 20 Sep 2016
Posts: 12
Location: Indonesia
Schools: Cambridge"20 (S)

### Show Tags

24 Jul 2018, 06:02
wow bamboozled.
Intern
Status: All our dreams can come true, if we have the courage to pursue them
Joined: 03 Jul 2015
Posts: 25
Location: India
Concentration: Technology, Finance
WE: Information Technology (Computer Software)

### Show Tags

19 Nov 2018, 05:26
I think this is a high-quality question and I agree with explanation. A very good question.

Completely stumped ignoring the fact that they are not necessarily integers. Chose D after solving the question for about 4 mts.
Re M22-22 &nbs [#permalink] 19 Nov 2018, 05:26
Display posts from previous: Sort by

# M22-22

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®.