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

 It is currently 24 Jan 2019, 04:52

### 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
• ### 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.
• ### Free GMAT Number Properties Webinar

January 27, 2019

January 27, 2019

07:00 AM PST

09:00 AM PST

Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.

# At a family summer party, each of the x members of the family

Author Message
TAGS:

### Hide Tags

Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2190
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

23 Aug 2017, 21:43
19
00:00

Difficulty:

45% (medium)

Question Stats:

66% (01:43) correct 34% (02:40) wrong based on 201 sessions

### HideShow timer Statistics

At a family summer party, each of the x members of the family chose whether or not to have a hamburger and whether or not to have a hotdog. If 1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog, then how many family members chose NOT to have both.

A. x/21
B. x/10
C. 9x/10
D. 10x/21
E. 20x/21

_________________

When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
+1 Kudos if you find this post helpful

Intern
Joined: 22 Aug 2017
Posts: 1
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

23 Aug 2017, 21:50
Skywalker18 wrote:
At a family summer party, each of the x members of the family chose whether or not to have a hamburger and whether or not to have a hotdog. If 1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog, then how many family members chose NOT to have both.

A. x/21
B. x/10
C. 9x/10
D. 10x/21
E. 20x/21

9x/10

Sent from my Redmi Note 3 using GMAT Club Forum mobile app
Manager
Status: Math Tutor
Joined: 12 Aug 2017
Posts: 73
GMAT 1: 750 Q50 V42
WE: Education (Education)
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

23 Aug 2017, 22:12
1
Let x=21
$$\frac{1}{3}$$*21 = 7 chose to have hamburger
$$\frac{1}{7}$$of7 = 1 chose to have both
so remaining 20 (i.e.21-1) will not have both.
Option E, $$\frac{20*21}{21}$$ = 20 gives that answer
So option E
_________________

Abhishek Parikh
Math Tutor
Whatsapp- +919983944321
Mobile- +971568653827
Website: http://www.holamaven.com

Manager
Joined: 08 Oct 2015
Posts: 241
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

23 Aug 2017, 22:50
The questions asks for neither x nor y. I believe that some info is missing from this problem.

We have information for both x and y and x and not y.

Math Expert
Joined: 02 Aug 2009
Posts: 7215
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

24 Aug 2017, 06:30
1
1
rahulkashyap wrote:
The questions asks for neither x nor y. I believe that some info is missing from this problem.

We have information for both x and y and x and not y.

hi..

what is it asking - person who did not have BOTH.
WHAT will it include - who ate any ONE only or NONE : SAME as ALL-BOTH

ALL is x
BOTH is $$\frac{1}{7} of \frac{1}{3}$$ of x=$$\frac{x}{21}$$

ans $$x-\frac{x}{21}=\frac{20x}{21}$$
E
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 20 Apr 2017
Posts: 6
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

21 Sep 2017, 08:40
I get how it's figured out, but what about the people who chose to have a hotdog and not a hamburger? Do we not need to know this?

X is all the people - let's call this 21
Therefore, 7 chose to have a hamburger and 14 chose not to have a hamburger.
Of this 7, 1 chose to also have a hotdog. This leaves 6 people who only had a hamburger.
Of the 14 who chose not to have a hamburger, how many chose to have a hotdog and not a hamburger? - We don't know this piece of information, so surely it's a proportion of the 20x/21?
Intern
Joined: 29 May 2017
Posts: 8
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

06 Oct 2017, 05:42
1
Say, Hamburger = H & Hotdog = D
Given n(H) =X/3 , Total = X
and n(D) = X- X/3 = 2X/3
Both = X/3 * 1/7 = X/21
Only H = X/3-X/21 = 6X/21
Only D = 2X/3-X/21 =13X/21
We know that, Total = n(H) + n(D) - n(H&B) + Neither
X = X/3 + 2X/3 - X/21 + Neither
Neither = X/21
So, who eat none = only H+ only D + Neither
= 6X/21+13X/21+X/21
= 20X/21.
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 273
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

26 Nov 2017, 13:43
1
Skywalker18 wrote:
At a family summer party, each of the x members of the family chose whether or not to have a hamburger and whether or not to have a hotdog. If 1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog, then how many family members chose NOT to have both.

A. x/21
B. x/10
C. 9x/10
D. 10x/21
E. 20x/21

(1/3) * (1/7)
= 1/21 = both

anything other than both
= 1 - (1/21)
20/21

thanks
cheers through the kudos button if this helps
CEO
Joined: 11 Sep 2015
Posts: 3360
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

13 Oct 2018, 07:43
Top Contributor
1
Skywalker18 wrote:
At a family summer party, each of the x members of the family chose whether or not to have a hamburger and whether or not to have a hotdog. If 1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog, then how many family members chose NOT to have both.

A. x/21
B. x/10
C. 9x/10
D. 10x/21
E. 20x/21

These kinds of questions (Variables in the Answer Choices - VIACs) can be answered algebraically or using the INPUT-OUTPUT approach.
Here's the INPUT-OUTPUT approach.

It might be useful to choose a number that works well with the fractions given in the question (1/3 and 1/7).
So, let's say there are 21 family members at the party.
In other words, we're saying that x = 21

1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog.
1/3 of 21 is 7, so 7 people chose to have a hamburger.
1/7 of 7 = 1, so 1 person had BOTH a hamburger AND a hotdog.

How many family members chose NOT to have both?
If 1 person had BOTH a hamburger AND a hotdog, then the remaining 20 people did not have BOTH a hamburger AND a hotdog.

So, when we INPUT x = 21, the answer to the question is "20 people did not have BOTH a hamburger AND a hotdog"

Now we'll INPUT x = 21 into each answer choice and see which one yields the correct OUTPUT of 20

A. 21/21 = 1. We want an output of 20. ELIMINATE A.
B. 21/10 = 2.1. We want an output of 20. ELIMINATE B.
C. (9)(21)/10 = 18.9. We want an output of 20. ELIMINATE C.
D. (10)(21)/21 = 10. We want an output of 20. ELIMINATE D.
E. (20)(21)/21 = = 20. PERFECT!

RELATED VIDEO FROM OUR COURSE

_________________

Test confidently with gmatprepnow.com

Manager
Joined: 07 Apr 2018
Posts: 73
Location: United States
Concentration: General Management, Marketing
GPA: 3.8
At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

13 Oct 2018, 18:53
How to get the missing variable? I think the question is missing some data. HI Bunuel can you please check this?
Attachments

hotdog.PNG [ 2.11 KiB | Viewed 696 times ]

File comment: ven diagram

hotdog.PNG [ 1.97 KiB | Viewed 701 times ]

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4612
Location: United States (CA)
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

04 Nov 2018, 17:08
Skywalker18 wrote:
At a family summer party, each of the x members of the family chose whether or not to have a hamburger and whether or not to have a hotdog. If 1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog, then how many family members chose NOT to have both.

A. x/21
B. x/10
C. 9x/10
D. 10x/21
E. 20x/21

Since x * 1/3 * 1/7 = x/21 of the family members have both a hamburger and a hotdog, 20x/21 of the family members do not have both.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Intern
Joined: 19 Aug 2015
Posts: 36
Re: At a family summer party, each of the x members of the family  [#permalink]

### Show Tags

06 Jan 2019, 11:56
HAPPYatHARVARD wrote:
How to get the missing variable? I think the question is missing some data. HI Bunuel can you please check this?

No data is missing..Que asks how many people did not have both which includes people who had one (either hamburger or hotdog) + people who had neither (No ham no hotdog)

Hope that helps.
Re: At a family summer party, each of the x members of the family &nbs [#permalink] 06 Jan 2019, 11:56
Display posts from previous: Sort by