Oct 22 08:00 AM PDT  09:00 AM PDT Join to learn strategies for tackling the longest, wordiest examples of Counting, Sets, & Series GMAT questions Oct 22 09:00 AM PDT  10:00 AM PDT Watch & learn the Do's and Don’ts for your upcoming interview Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Oct 27 07:00 AM EDT  09:00 AM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58396

There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 04:42
Question Stats:
55% (01:23) correct 45% (01:40) wrong based on 119 sessions
HideShow timer Statistics
There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color? A. 10 B. 16 C. 23 D. 42 E. 43
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 58396

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 04:43
Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 green balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color. Answer: E.
_________________



Math Expert
Joined: 02 Aug 2009
Posts: 8004

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 05:06
Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+94+1=43
_________________



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5027
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 06:10
Bunuel wrote: Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 green balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color. Answer: E. Bunuel : Please check OA is given as 42 option D ..



Math Expert
Joined: 02 Sep 2009
Posts: 58396

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 06:12
Archit3110 wrote: Bunuel wrote: Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 green balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color. Answer: E. Bunuel : Please check OA is given as 42 option D .. _____________________ Fixed. Thank you.
_________________



Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 269

There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
Updated on: 30 Jan 2019, 07:03
chetan2u wrote: Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+94+1=43 hi do you think 38 can also be an answer, although the choice is not in the answer choice ? say first we pick all 15 green balls, then all 12 red balls, and then all 10 blue balls, so lastly we end up with only black balls remaining now, 15 + 12 + 10 +1 = 38 so, where am I getting wrong ? thanks
Originally posted by testcracker on 30 Jan 2019, 06:47.
Last edited by testcracker on 30 Jan 2019, 07:03, edited 1 time in total.



Math Expert
Joined: 02 Aug 2009
Posts: 8004

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 06:56
testcracker wrote: chetan2u wrote: Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+94+1=43 hi do you think 38 can also be an answer, although the choice is not in the answer choice ? say first we pick all 15 green balls, then all 12 red balls, and then all 10 blue balls, so lastly we end up with only black balls remaining so 15 + 12 + 10 +1 = 38 so, where am I getting wrong ? thanks The problem is with understanding the question, which can become quite perplexed. The question says the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color. The catch is ASSURED. I can pick up just 9 all and all 9 can be blacks and that will be the best scenario. But the word ASSURED means that the probability of one colour has to be 100%. These 9 could actually be combinations of all 4 colours. This means that I pick up some number blindly and then be able to claim that I have picked up atleast one set of colours, and here we will consider the worst scenario. That is I have picked up all but 4, and these 4 are one of each colour.
_________________



Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 269

There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
Updated on: 30 Jan 2019, 08:28
testcracker wrote: chetan2u wrote: Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+94+1=43 hi do you think 38 can also be an answer, although the choice is not in the answer choice ? say first we pick all 15 green balls, then all 12 red balls, and then all 10 blue balls, so lastly we end up with only black balls remaining now, 15 + 12 + 10 +1 = 38 so, where am I getting wrong ? thanks thanks a lot+ 1 I have got it
Originally posted by testcracker on 30 Jan 2019, 07:14.
Last edited by testcracker on 30 Jan 2019, 08:28, edited 1 time in total.



Manager
Joined: 17 Jun 2017
Posts: 60

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 07:46
Bunuel wrote: Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 green balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color. Answer: E. Outstanding. Thank you



Manager
Joined: 17 Jun 2018
Posts: 52
Location: Canada
GPA: 2.84
WE: Engineering (Real Estate)

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 10:02
shouldn't the question say "maximum number of balls" instead of minimum?



Math Expert
Joined: 02 Aug 2009
Posts: 8004

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 10:11
arpitkansal wrote: shouldn't the question say "maximum number of balls" instead of minimum? No, it will not be maximum. There has to be some contrast in two things.. I can say what will be the maximum number of balls that can be drawn without getting all balls of any one colour out. Here the max will become all except one of each left, so 42. But if we use maximum in present case. Maximum can be 47, as we will hav all of all colour. So here we have minimum such that one entire colour is present.. Contrast .. maximum still all not there. Minimum such that one is there. Hope it helps
_________________



Manager
Joined: 17 Jun 2018
Posts: 52
Location: Canada
GPA: 2.84
WE: Engineering (Real Estate)

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 10:55
But Chetan,the minimum could be that I draw 9 black balls simultaneously and then I would have fulfilled the condition of "have all the balls of one color"..I am still confused.



Math Expert
Joined: 02 Aug 2009
Posts: 8004

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 11:17
arpitkansal wrote: But Chetan,the minimum could be that I draw 9 black balls simultaneously and then I would have fulfilled the condition of "have all the balls of one color"..I am still confused. Arpit, the point is of being 100% sure, because there is a word ASSURED. Can you pick 9 balls without seeing and announce that all are of same colour.. No you cannot . You cannot till the time you don't take the worst scenario.. So if you pick 43, 44, 45 or 46, you will be 100% sure that you have atleast one colour completely with you and yes you can claim without hesitation. MINIMUM comes into play in these 4 numbers, so 43 is the minimum here and our answer
_________________



Manager
Joined: 17 Jun 2018
Posts: 52
Location: Canada
GPA: 2.84
WE: Engineering (Real Estate)

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 11:22
chetan2u wrote: arpitkansal wrote: But Chetan,the minimum could be that I draw 9 black balls simultaneously and then I would have fulfilled the condition of "have all the balls of one color"..I am still confused. Arpit, the point is of being 100% sure, because there is a word ASSURED. Can you pick 9 balls without seeing and announce that all are of same colour.. No you cannot . You cannot till the time you don't take the worst scenario.. So if you pick 43, 44, 45 or 46, you will be 100% sure that you have atleast one colour completely with you and yes you can claim without hesitation.MINIMUM comes into play in these 4 numbers, so 43 is the minimum here and our answer Ok..Thanks Chetan.Now I understood my mistake.



Director
Joined: 09 Aug 2017
Posts: 500

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
30 Jan 2019, 17:32
I also approached the question in same way but later I came to know my mistake. I think you would be correct if question had asked for "minimum number of balls to pick to assure you ball each color". testcracker wrote: chetan2u wrote: Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+94+1=43 hi do you think 38 can also be an answer, although the choice is not in the answer choice ? say first we pick all 15 green balls, then all 12 red balls, and then all 10 blue balls, so lastly we end up with only black balls remaining now, 15 + 12 + 10 +1 = 38 so, where am I getting wrong ? thanks



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8117
Location: United States (CA)

Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
Show Tags
01 Feb 2019, 19:01
Bunuel wrote: There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?
A. 10 B. 16 C. 23 D. 42 E. 43 The worst case scenario is that we take out 14 green balls, 11 red balls, 9 blue balls and 8 black balls. We see that we still don’t have all the balls of one color. However, if we take out one more ball, no matter what color it is, we will have all the balls of one color. Therefore, the total number of balls need to be taken out is 14 + 11 + 9 + 8 + 1 = 43. Answer: E
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.




Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball
[#permalink]
01 Feb 2019, 19:01






