|
Author |
Message |
|
Manager
Joined: 23 Jun 2008
Posts: 144
Followers: 3
Kudos [?]:
11
[1] , given: 0
|
1
This post received
KUDOS
Question Stats:
59% (01:50) correct
40% (00:18) wrong based on 1 sessions
The box contains 10 tablets of medicine A and 15 tablets of medicine B. N tablets are taken from the box at random. What is the least possible N that ensures that at least two tablets of each kind are among the extracted? (A) 12 (B) 15 (C) 17 (D) 19 (E) 21 Source: GMAT Club Tests - hardest GMAT questions I am trying to decide between a & c. Possible explanations?
|
|
|
|
|
|
|
|
|
Manager
Joined: 20 May 2008
Posts: 57
Followers: 0
Kudos [?]:
1
[1] , given: 1
|
1
This post received
KUDOS
The ans is C. In the worst case scenario, the first 15 tablets that you remove will be of medicine B. The next 2 tablets you remove will ensure that you have atleast 2 tablets of each kind among the extracted. So 15 + 2 = 17
A is not the answer beause you can remove 12 tablets from the box and all of them could be of medicine B.
|
|
|
|
|
|
Manager
Joined: 12 Aug 2008
Posts: 63
Followers: 2
Kudos [?]:
2
[0], given: 2
|
What if I remove four tablets, N=4, and there happens to be two tablets of medicine A and two tablets of medicine B.
So the least number of tablets that ensure that at least two tablets of each kind are among the extracted tablets is four. Is this thinking correct? If so, I think the question should be reworded.
Someone please comment.
|
|
|
|
|
|
Manager
Joined: 12 Aug 2008
Posts: 63
Followers: 2
Kudos [?]:
2
[1] , given: 2
|
1
This post received
KUDOS
sid3699 wrote: What if I remove four tablets, N=4, and there happens to be two tablets of medicine A and two tablets of medicine B.
So the least number of tablets that ensure that at least two tablets of each kind are among the extracted tablets is four. Is this thinking correct? If so, I think the question should be reworded.
Someone please comment. Anyone  Please? 
|
|
|
|
|
|
Senior Manager
Joined: 13 Dec 2009
Posts: 269
Followers: 8
Kudos [?]:
67
[1] , given: 13
|
1
This post received
KUDOS
sid3699 wrote: sid3699 wrote: What if I remove four tablets, N=4, and there happens to be two tablets of medicine A and two tablets of medicine B.
So the least number of tablets that ensure that at least two tablets of each kind are among the extracted tablets is four. Is this thinking correct? If so, I think the question should be reworded.
Someone please comment. Anyone Please? Your approach above takes the best case scenario. However such an event happening (4 tablets = 2A + 2B) is extremely unlikely. The question asks for a guarantee - what is the least number of tablets you pull out so that you DEFINITELY GET 2 of each kind? In this case you have to take the worst case scenario - lets say you keep pulling out tablets and you keep getting tablet B. In this case you can pull out 15 consecutive B's and then you will start getting A's. So in the worst case you can pull out 15 straight B's and then the next 2 will be A. This satisfies the condition that you have 2 of each kind. Hope it is clear now.
_________________
My debrief: done-and-dusted-730-q49-v40
|
|
|
|
|
|
Manager
Joined: 15 Sep 2009
Posts: 148
Followers: 1
Kudos [?]:
8
[0], given: 2
|
I will go with option c)17.
|
|
|
|
|
|
Director
Joined: 21 Dec 2009
Posts: 592
Concentration: Entrepreneurship, Finance
Followers: 13
Kudos [?]:
130
[0], given: 20
|
To ensure two tablets from each of A and B are extracted: 15 of B's must have been exhausted. Then followed by 2A's. Correct response = 17 (C)
_________________
KUDOS me if you feel my contribution has helped you.
|
|
|
|
|
|
Manager
Joined: 12 Jul 2010
Posts: 69
Followers: 1
Kudos [?]:
2
[0], given: 3
|
Its C.
Worst possibility is 15 (Only B is selected) and then obviously 2 for A. So its 17.
|
|
|
|
|
|
Intern
Joined: 02 Nov 2010
Posts: 47
Followers: 0
Kudos [?]:
0
[0], given: 23
|
Go with C, just had to reason it out.
|
|
|
|
|
|
Senior Manager
Joined: 23 Oct 2010
Posts: 335
Location: Azerbaijan
Followers: 6
Kudos [?]:
68
[0], given: 67
|
I also got ans C, but my way of thinking was following- we have 10 A's and 15 B's , so 150 variations and N!/2!(N-2)! a)N!/2!(N-2)! =12!/2!*10!=66 b)105 c)136 d)171 (above 150, so exclude D and E) since C is the greatest , then answ C I agree that my way of thinking is more complicated than yours, but please let me know if it was a right one )) if not, where was I wrong? p.s. I got my answ before looking at the right answ
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
|
|
|
|
|
|
Manager
Joined: 16 May 2011
Posts: 208
Concentration: Finance, Real Estate
GMAT Date: 12-27-2011
WE: Law (Law)
Followers: 0
Kudos [?]:
26
[0], given: 37
|
you look at the highest value:15
you can take 15 and still have them in the same color. so 15 only promises 2 of kind A.
now there are only kind B. taking 2 of kind B will ensuere: 2A and 2B (15+2=17)
|
|
|
|
|
|
Manager
Joined: 11 Feb 2011
Posts: 91
Followers: 1
Kudos [?]:
0
[0], given: 0
|
easier one ...it is c
|
|
|
|
|
|
Manager
Joined: 28 Feb 2011
Posts: 95
Followers: 0
Kudos [?]:
29
[0], given: 2
|
LalaB wrote: I also got ans C, but my way of thinking was following-
we have 10 A's and 15 B's , so 150 variations and N!/2!(N-2)!
a)N!/2!(N-2)! =12!/2!*10!=66
b)105
c)136
d)171 (above 150, so exclude D and E)
since C is the greatest , then answ C
I agree that my way of thinking is more complicated than yours, but please let me know if it was a right one )) if not, where was I wrong?
p.s. I got my answ before looking at the right answ Looks good..should work for more difficult questions..
_________________
Fight till you succeed like a gladiator..or doom your life which can be invincible
|
|
|
|
|
|
Intern
Joined: 24 May 2010
Posts: 47
Followers: 0
Kudos [?]:
5
[0], given: 6
|
LalaB wrote: I also got ans C, but my way of thinking was following-
we have 10 A's and 15 B's , so 150 variations and N!/2!(N-2)!
a)N!/2!(N-2)! =12!/2!*10!=66
b)105
c)136
d)171 (above 150, so exclude D and E)
since C is the greatest , then answ C
I agree that my way of thinking is more complicated than yours, but please let me know if it was a right one )) if not, where was I wrong?
p.s. I got my answ before looking at the right answ How are you ensuring in this approach, that the pick has to be at least 2 each (4) from both A & B ? Shouldn't it be N!/ 4!(N-4)!
|
|
|
|
|
|
Senior Manager
Joined: 15 Sep 2009
Posts: 272
GMAT 1: 750 Q V
Followers: 4
Kudos [?]:
35
[0], given: 6
|
Straight C. Taking 15 ensures that all B are taken (a number greater than 2 of course), then taking two additional tabs from the remaining, which consists of only A. Thus, 15 + 2= 17 Cheers
_________________
+1 Kudos me - I'm half Irish, half Prussian.
|
|
|
|
|
|
Senior Manager
Joined: 29 Mar 2012
Posts: 251
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 10
Kudos [?]:
52
[0], given: 20
|
balboa wrote: The box contains 10 tablets of medicine A and 15 tablets of medicine B. N tablets are taken from the box at random. What is the least possible N that ensures that at least two tablets of each kind are among the extracted? (A) 12 (B) 15 (C) 17 (D) 19 (E) 21 Source: GMAT Club Tests - hardest GMAT questions I am trying to decide between a & c. Possible explanations? Such questions remind me of the limiting reagents!  Regards,
_________________
My posts: Solving Inequalities, Solving Simultaneous equations, Divisibility Rules
My story: 640 What a blunder!
My page:
|
|
|
|
|
|
Intern
Joined: 23 Oct 2011
Posts: 3
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Straight C -nicely explained by OldFritz
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
