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

It is currently 21 Oct 2018, 01:24

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

A box of doughnuts contains glazed, cream-filled, and jelly

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Senior Manager
Senior Manager
User avatar
Joined: 31 Jul 2006
Posts: 436
A box of doughnuts contains glazed, cream-filled, and jelly  [#permalink]

Show Tags

New post 02 Sep 2006, 17:00
2
7
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (01:41) correct 43% (01:34) wrong based on 383 sessions

HideShow timer Statistics

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

(2) There were originally 5 glazed doughnuts in the box.

Spoiler: :: Doubt
Why is 1) Not sufficient? If P of second donut selected = 2/5 isn't P of first donut 2+1 / 5+1 = 3/6? (i.e. we put donut back in the box).

Here is what Kaplan said, but I don't get why 1) is not suff.

1)+2):
If there were originally 5 glazed doughnuts in the box, and one was
selected, then there are 4 glazed doughnuts in the box. Since the
probability of picking another glazed doughnut is 2/5, you know that
2/5 = 4 / num of d left, so
there must be a total of 10 doughnuts left in the box. So there were originally 5 glazed doughnuts out of 11 total.

Thank you kindly.
Kaplan GMAT Prep Discount CodesOptimus Prep Discount CodesMath Revolution Discount Codes
Senior Manager
Senior Manager
avatar
Joined: 14 Jul 2005
Posts: 391
  [#permalink]

Show Tags

New post 02 Sep 2006, 18:18
2
1
Is the answer C i.e. both statements are needed ?

Statement 1 just says that after removing 1 glazed doughnut the probability is 2/5. which is essentially 40%. Which means there could 4/10 glazed doughnuts making orig. prob = 5/10 = 1/2 or say 8/20 in which case the orig. prob = 9/21 or 3/7

Hence INSUFF

Statement 2 says there were 5 glazed doughnuts originally
NOT SUFF

Combining, let x be the total no. of doughnuts, then

2/5 x = 4 (since we know that there were 5 originally)

i.e. x = 20/2 = 10

Hence required prob. = 5/10 = 0.5
Senior Manager
Senior Manager
User avatar
Joined: 31 Jul 2006
Posts: 436
  [#permalink]

Show Tags

New post 02 Sep 2006, 18:58
gmatornot wrote:
Is the answer C i.e. both statements are needed ?

Statement 1 just says that after removing 1 glazed doughnut the probability is 2/5. which is essentially 40%. Which means there could 4/10 glazed doughnuts making orig. prob = 5/10 = 1/2 or say 8/20 in which case the orig. prob = 9/21 or 3/7

Hence INSUFF

Statement 2 says there were 5 glazed doughnuts originally
NOT SUFF

Combining, let x be the total no. of doughnuts, then

2/5 x = 4 (since we know that there were 5 originally)

i.e. x = 20/2 = 10

Hence required prob. = 5/10 = 0.5


Correct the answer is C and your explanation matches except you forgot to add 1 to 10 making it P= 5/11.

But I still don't understand. If after removing 1st donut P=2/5, if we want to go back 1 step can't we just add 1 to both num and denom making it 3/6 =1/2?
Manager
Manager
avatar
Joined: 30 Jan 2006
Posts: 62
  [#permalink]

Show Tags

New post 02 Sep 2006, 19:29
1
The answer is C.

Nsentra,

For the 1 statement we cannot add 1 to the den and num coz we don't know what the fraction would be it can be 2/5 or 4/10 or 6/15 which will give different answers every time.
Senior Manager
Senior Manager
User avatar
Joined: 31 Jul 2006
Posts: 436
  [#permalink]

Show Tags

New post 02 Sep 2006, 19:44
smily_buddy wrote:
The answer is C.

Nsentra,

For the 1 statement we cannot add 1 to the den and num coz we don't know what the fraction would be it can be 2/5 or 4/10 or 6/15 which will give different answers every time.


Ah, ok I got it. I see, this goes in hand with a type of question that asks if it's possible to find a new ratio when something was added/subtracted to some part of current ratio.

Thank you!
VP
VP
avatar
Joined: 17 Jun 2008
Posts: 1443
Re: Probability  [#permalink]

Show Tags

New post 16 Nov 2008, 11:28
C.

Stmt2 gives number of glazed doughnuts and stmt1 gives total number of doughnuts.
VP
VP
User avatar
Joined: 05 Jul 2008
Posts: 1298
Re: Probability  [#permalink]

Show Tags

New post 16 Nov 2008, 14:44
scthakur wrote:
C.

Stmt2 gives number of glazed doughnuts and stmt1 gives total number of doughnuts.


I am not sure how stmt1 is giving the total number of DN's.

The probability of selecting a second glazed DN is 2/5. We need P of selecting the first DN.

Are you saying that the total number of DN's is 10 because 4/10 is 2/5?

I picked E because the problem does not tell us whether the first DN is put back or not. I see where you are getting at but some how I cannot explain
SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2395
Re: Probability  [#permalink]

Show Tags

New post 16 Nov 2008, 14:48
1
study wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.


Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...
so the prob of getting a glazed doughnut at first draw = 5/11
Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

VP
VP
User avatar
Joined: 05 Jul 2008
Posts: 1298
Re: Probability  [#permalink]

Show Tags

New post 16 Nov 2008, 14:53
GMAT TIGER wrote:
study wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.


Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...
so the prob of getting a glazed doughnut at first draw = 5/11
Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5


I guess I over did the problem. If the first DN was put back and the probability of drawing the second DN is 2/5 then the glazed DN's count cannot be 5
Director
Director
avatar
Joined: 14 Aug 2007
Posts: 683
Re: Probability  [#permalink]

Show Tags

New post 17 Nov 2008, 00:33
3
1
study wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.


(1)
say there are x glazed donuts, and "n" total donuts
for the first glazed donut prob = x/n
prob for second = x-1/n-1 = 2/5
But as this is ratio we can not conclude anything about x and n (it can be anythign 2/5, 10/25 etc)
insuff

(2)
we know that x = 5 but insuff. alone

together,
x-1/n-1=2/5 and x = 5
we get, 10=n-1 or n = 11

Suff to calculate prob the doughnut selected is glazed

C
SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2395
Re: Probability  [#permalink]

Show Tags

New post 17 Nov 2008, 15:07
icandy wrote:
GMAT TIGER wrote:
study wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.


Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...
so the prob of getting a glazed doughnut at first draw = 5/11
Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5


I guess I over did the problem. If the first DN was put back and the probability of drawing the second DN is 2/5 then the glazed DN's count cannot be 5


You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers.
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Director
Director
avatar
Joined: 14 Aug 2007
Posts: 683
Re: Probability  [#permalink]

Show Tags

New post 19 Nov 2008, 02:17
GMAT TIGER wrote:

You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers.


IMHO, if the question does not specifically mention replacement => no replacement. I am not sure but it just occurred to me that I have read this somewhere......
Manager
Manager
User avatar
Joined: 14 Nov 2008
Posts: 179
Schools: Stanford...Wait, I will come!!!
Re: Probability  [#permalink]

Show Tags

New post 19 Nov 2008, 03:19
alpha_plus_gamma wrote:
GMAT TIGER wrote:

You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers.


IMHO, if the question does not specifically mention replacement => no replacement. I am not sure but it just occurred to me that I have read this somewhere......


Actually, the question would have speicified that the second probability is after the replacement, then there was no value in the question. But again.. I think it is responsibility of the question maker to avoid any ambiguity to us. :)
Manager
Manager
User avatar
Joined: 10 Jul 2009
Posts: 108
Location: Ukraine, Kyiv
Re: Probability  [#permalink]

Show Tags

New post 06 Sep 2009, 02:14
IMO C too. Agree with you, guys.

I think that first statement gives the probability of selecting second doughnut without putting it back. Thus, together with statement 2 it gives us the number of total doughnuts and the number of glazed doughnuts.
_________________

Never, never, never give up

Intern
Intern
avatar
Joined: 02 Aug 2011
Posts: 43
Location: Canada
Concentration: Finance, Healthcare
GMAT 1: 750 Q50 V42
GPA: 3.4
WE: Analyst (Consulting)
Reviews Badge
Re: DS-700 level-Probability  [#permalink]

Show Tags

New post 05 Aug 2011, 14:25
(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

This does NOT mean that the first chance is 3/6 = 1/2 because if there were, for example, 4 glazed and 6 other donuts after the first draw (4/10 = 2/5 chance), then before that there were 5 glazed and 6 others (5/11 chance). Insufficient.

(2) There were originally 5 glazed doughnuts in the box.

Obviously insufficient on its own since we need to know the total number BUT, with the first one it is sufficient:

If there were 5 glazed and one was drawn, there are now 4 glazed left. There's a 2/5 chance to draw another glazed so there are 4 glazed and 6 others (10 total). So before there were 5 glazed and 6 others (5/11 chance).

C
Manager
Manager
avatar
Joined: 12 Oct 2011
Posts: 203
Re: A box of doughnuts contains glazed, cream-filled, and jelly  [#permalink]

Show Tags

New post 16 Dec 2011, 21:47
Let 'g' be the number of glazed doughnuts and let 'n' be the total number of doughnuts.
Thus, we need to find the value of (g / n).

Statement 1:
After the first glazed doughnut is selected, the number of remaining glazed doughnuts = (g - 1)
Number of remaining total doughnuts = (n - 1)
Thus this statement tells us that (g - 1)/(n - 1) = 2/5 => (5g - 5) = (2n - 2) => 5g = 2n + 3
As there are two unknowns, we cannot get a value for (g/n). INSUFFICIENT.

Statement 2: g = 5
This does not give a value for n. So INSUFFICIENT.

Combining these two statements, we have the value of 'g' from (2), which can be substituted in the equation obtained from (1) to find the value of 'n'.
Thus, we can find out (g/n). SUFFICIENT.
_________________

Consider KUDOS if you feel the effort's worth it

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50007
A box of doughnuts contains glazed, cream-filled, and jelly  [#permalink]

Show Tags

New post 20 Nov 2014, 08:43
1
2
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5. We could have total of 6 doughnuts out of which 3 were glazed or 11 doughnuts out of which 5 were glazed or 16 doughnuts out of which 7 were glazed ... Not sufficient.

(2) There were originally 5 glazed doughnuts in the box. We don't know the total number doughnuts. Not sufficient.

(1)+(2) If there were originally 5 glazed doughnuts, then after selecting 1 of them there would be 4 left, so according to (1) 4/(total - 1) = 2/5 --> total = 11 --> P(glazed) = (glazed)/(total) = 5/11. Sufficient.

Answer: C.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2657
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Premium Member Reviews Badge
Re: A box of doughnuts contains glazed, cream-filled, and jelly  [#permalink]

Show Tags

New post 31 Mar 2016, 18:28
Nsentra wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

(2) There were originally 5 glazed doughnuts in the box.



pretty straight forward question..looks like a 600-700 than a 700 lvl...

we need to find: G/(G+C+J)

1. G-1/(G+C+J-1) = 2/5
it could have been 3 G, and probability is 1/2
or it could be 5 G, and 11 total, thus probability is 5/11.
1 alone is insufficient.

2 alone is insufficient.

1+2
we know that originally, there were 5 G.
so 2x/5x => 2x=5-1 = 4, so x=2. or total there were 11 donuts, out of which 5 were G.

sufficient.
C
Manager
Manager
avatar
B
Joined: 29 Nov 2016
Posts: 53
Re: A box of doughnuts contains glazed, cream-filled, and jelly  [#permalink]

Show Tags

New post 24 Feb 2018, 10:17
Bunuel

I think I have this problem consistently. Why do we assume the probability of drawing second donut as X-1/n-1.

If the first donut is picked , then the prob of picking second shouldn't be x/n*x-1/n-1 =2/5 VeritasPrepKarishma any help would be appreciated.

Thanks
Bunuel wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5. We could have total of 6 doughnuts out of which 3 were glazed or 11 doughnuts out of which 5 were glazed or 16 doughnuts out of which 7 were glazed ... Not sufficient.

(2) There were originally 5 glazed doughnuts in the box. We don't know the total number doughnuts. Not sufficient.

(1)+(2) If there were originally 5 glazed doughnuts, then after selecting 1 of them there would be 4 left, so according to (1) 4/(total - 1) = 2/5 --> total = 11 --> P(glazed) = (glazed)/(total) = 5/11. Sufficient.

Answer: C.


Posted from my mobile device
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 6971
Re: A box of doughnuts contains glazed, cream-filled, and jelly  [#permalink]

Show Tags

New post 24 Feb 2018, 10:25
1
Mudit27021988 wrote:
Bunuel

I think I have this problem consistently. Why do we assume the probability of drawing second donut as X-1/n-1.

If the first donut is picked , then the prob of picking second shouldn't be x/n*x-1/n-1 =2/5 VeritasPrepKarishma any help would be appreciated.

Thanks
Bunuel wrote:
A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5. We could have total of 6 doughnuts out of which 3 were glazed or 11 doughnuts out of which 5 were glazed or 16 doughnuts out of which 7 were glazed ... Not sufficient.

(2) There were originally 5 glazed doughnuts in the box. We don't know the total number doughnuts. Not sufficient.

(1)+(2) If there were originally 5 glazed doughnuts, then after selecting 1 of them there would be 4 left, so according to (1) 4/(total - 1) = 2/5 --> total = 11 --> P(glazed) = (glazed)/(total) = 5/11. Sufficient.

Answer: C.


Posted from my mobile device



Hi..

If we are asked "Probability of selecting two glazed doughnuts in two picks without replacement", the answer would be correct...
But here it is ALREADY given that first selected is glazed. so now it is not a probability but a certainity.
hence we work with the remaining numbers which are
TOTAL = n-1
and glazed = g-1
so Prob will be g-1/n-1
hope it helps
_________________

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

GMAT Club Bot
Re: A box of doughnuts contains glazed, cream-filled, and jelly &nbs [#permalink] 24 Feb 2018, 10:25

Go to page    1   2    Next  [ 22 posts ] 

Display posts from previous: Sort by

A box of doughnuts contains glazed, cream-filled, and jelly

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.