Author 
Message 
TAGS:

Hide Tags

Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 1950
Location: India
Concentration: General Management, Strategy
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, 22:43
11
This post was BOOKMARKED
Question Stats:
61% (01:35) correct 39% (02:47) wrong based on 98 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
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, 22:50
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: 74
WE: Education (Education)

Re: At a family summer party, each of the x members of the family [#permalink]
Show Tags
23 Aug 2017, 23:12
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.211) 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: 09 Oct 2015
Posts: 95

Re: At a family summer party, each of the x members of the family [#permalink]
Show Tags
23 Aug 2017, 23:50
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. chetan2u , could you please help with this



Math Expert
Joined: 02 Aug 2009
Posts: 5732

Re: At a family summer party, each of the x members of the family [#permalink]
Show Tags
24 Aug 2017, 07:30
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. chetan2u , could you please help with this hi.. what is it asking  person who did not have BOTH.WHAT will it include  who ate any ONE only or NONE : SAME as ALLBOTHALL 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
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.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, 09:40
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: 9

Re: At a family summer party, each of the x members of the family [#permalink]
Show Tags
06 Oct 2017, 06:42
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/3X/21 = 6X/21 Only D = 2X/3X/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: 285

Re: At a family summer party, each of the x members of the family [#permalink]
Show Tags
26 Nov 2017, 14:43
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




Re: At a family summer party, each of the x members of the family
[#permalink]
26 Nov 2017, 14:43






