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

It is currently 26 Sep 2018, 07:57

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

M26-22

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

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49544
M26-22  [#permalink]

Show Tags

New post 16 Sep 2014, 01:25
1
22
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

55% (01:57) correct 45% (02:02) wrong based on 192 sessions

HideShow timer Statistics

There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445

_________________

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

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49544
Re M26-22  [#permalink]

Show Tags

New post 16 Sep 2014, 01:25
1
7
Official Solution:

There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445


Total ways to select 8 marbles out of \(7+5=12\) is \(C^8_{12}\);

Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\);

Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble remain in the jar is \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).


Answer: D
_________________

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

Manager
Manager
User avatar
Joined: 19 Dec 2015
Posts: 111
Location: United States
GMAT 1: 720 Q50 V38
GPA: 3.8
WE: Information Technology (Computer Software)
Re: M26-22  [#permalink]

Show Tags

New post 09 Mar 2016, 21:05
1
Bunuel,

This is a duplicate question. Please refer to M13-37.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49544
Re: M26-22  [#permalink]

Show Tags

New post 09 Mar 2016, 22:49
Manager
Manager
avatar
B
Joined: 17 Sep 2014
Posts: 148
Location: India
Concentration: Operations, Strategy
GMAT 1: 710 Q49 V38
GPA: 3.65
WE: Engineering (Manufacturing)
Reviews Badge
Re: M26-22  [#permalink]

Show Tags

New post 11 Oct 2016, 20:54
Bunuel wrote:
Official Solution:

There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445


Total ways to select 8 marbles out of \(7+5=12\) is \(C^8_{12}\);

Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\);

Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble remain in the jar is \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).


Answer: D


The question says at least one red marble AND at least one blue marble.
But you add the possibilities (5+35). Shouldn't it be multiplied ?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49544
Re: M26-22  [#permalink]

Show Tags

New post 12 Oct 2016, 00:09
avdgmat4777 wrote:
Bunuel wrote:
Official Solution:

There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445


Total ways to select 8 marbles out of \(7+5=12\) is \(C^8_{12}\);

Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\);

Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble remain in the jar is \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).


Answer: D


The question says at least one red marble AND at least one blue marble.
But you add the possibilities (5+35). Shouldn't it be multiplied ?


There are 2 separate cases when there can be 0 red or 0 blue left:
If we pick 7 red (out of 7) and 1 blue.
If we pick 5 blue (out of 5) and 3 red.

These cases are separate, they cannot happen together so we add.
_________________

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

Intern
Intern
avatar
B
Joined: 07 Oct 2015
Posts: 7
Premium Member
Re: M26-22  [#permalink]

Show Tags

New post 30 Nov 2016, 02:57
Total ways of selecting 8 marbles=495
Marbles that remain in the jar= 1R and 3B or 3R and 1B= 35/2 *2= 35. In this case shouldn't the answer be 460?
Am I missing something?
Retired Moderator
avatar
P
Joined: 04 Aug 2016
Posts: 544
Location: India
Concentration: Leadership, Strategy
GPA: 4
WE: Engineering (Telecommunications)
Premium Member
Re: M26-22  [#permalink]

Show Tags

New post 06 Dec 2016, 23:04
I took the long way :(

Applicable Cases: 4B and 4 Red, 3 Blue and 5 Red, 2 Blue and 6 Red - Remaining two cases with 1B and 7 Red and 5 B and 3 Red Discarded


175 + 210+70 =455
Intern
Intern
avatar
B
Joined: 11 Oct 2017
Posts: 11
Reviews Badge
Re: M26-22  [#permalink]

Show Tags

New post 13 Jan 2018, 11:41
Bunuel
Are the same colored marbles identical?
I didn't understand how there would be 5 ways to choose a blue marble if they are all identical? I thought there should be 1 way only if the red marbles are identical to each other and blue identical to the other blue.

thank you :)
Intern
Intern
avatar
B
Joined: 03 Sep 2017
Posts: 15
Reviews Badge
Re: M26-22  [#permalink]

Show Tags

New post 23 May 2018, 14:54
Another approach:

The selecting/de-arranging method (without C formulas).

Total ways to choose 12 balls.
12*11*10*9*8*7*6*5 / 8! -> 495
the divided by 8! is because we are ''de-arranging'' since the order doesn't matter.

Ways to choose all red:
RRRRRRRB
By formulas would be 7*6*5*4*3*2*1*5 / 7!*1! = 5
7! and 1! same de-arrange concept.

Ways to choose all blue
BBBBBRRR = 5*4*3*2*1*7*6*5 / 5!*3! = 35
5! and 3! same de-arrange concept.


495-35-5 = 455.
Intern
Intern
avatar
B
Joined: 29 May 2016
Posts: 10
Re: M26-22  [#permalink]

Show Tags

New post 16 Aug 2018, 10:55
The question prompt states:"at least one red marble AND at least one blue marble." Yet, the answer seems to consider one red marble OR one blue marble.
Intern
Intern
avatar
B
Joined: 26 Jun 2018
Posts: 7
Re: M26-22  [#permalink]

Show Tags

New post 16 Aug 2018, 23:00
3ksnikhil wrote:
Total ways of selecting 8 marbles=495
Marbles that remain in the jar= 1R and 3B or 3R and 1B= 35/2 *2= 35. In this case shouldn't the answer be 460?
Am I missing something?

2R and 2B is what your'e missing.
Manager
Manager
User avatar
B
Joined: 23 May 2018
Posts: 112
Location: Pakistan
GMAT 1: 770 Q48 V50
GPA: 3.4
Re: M26-22  [#permalink]

Show Tags

New post 18 Aug 2018, 03:27
Xin Cho wrote:
The question prompt states:"at least one red marble AND at least one blue marble." Yet, the answer seems to consider one red marble OR one blue marble.


Because when you take out the possibility of there being either zero Reds or zero Blues from all available options, it gives you all the other ways (in numbers) where after taking out 8 marbles, there is at least 1 Red and at least 1 Blue left in the jar.
_________________

If you can dream it, you can do it.

Kudos are appreciated.

Manager
Manager
User avatar
B
Joined: 23 May 2018
Posts: 112
Location: Pakistan
GMAT 1: 770 Q48 V50
GPA: 3.4
Re: M26-22  [#permalink]

Show Tags

New post 18 Aug 2018, 03:27
vitorcbarbieri wrote:
Another approach:

The selecting/de-arranging method (without C formulas).

Total ways to choose 12 balls.
12*11*10*9*8*7*6*5 / 8! -> 495
the divided by 8! is because we are ''de-arranging'' since the order doesn't matter.

Ways to choose all red:
RRRRRRRB
By formulas would be 7*6*5*4*3*2*1*5 / 7!*1! = 5
7! and 1! same de-arrange concept.

Ways to choose all blue
BBBBBRRR = 5*4*3*2*1*7*6*5 / 5!*3! = 35
5! and 3! same de-arrange concept.


495-35-5 = 455.


This is the same thing as the C formula.
_________________

If you can dream it, you can do it.

Kudos are appreciated.

Intern
Intern
avatar
B
Joined: 12 Jan 2017
Posts: 36
CAT Tests
M26-22  [#permalink]

Show Tags

New post 11 Sep 2018, 12:48
Bunuel wrote:
Official Solution:

There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble remain in the jar?

A. 460
B. 490
C. 493
D. 455
E. 445


Total ways to select 8 marbles out of \(7+5=12\) is \(C^8_{12}\);

Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\);

Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);

Hence ways to select 8 marbles so that at least one red marble and at least one blue marble remain in the jar is \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).


Answer: D


Just to explain this a bit more:

Problem says there are 12 marbles total, 7 red 5 blue.
Problem wants to select 8, so that there are at least 1 red and 1 blue not selected.

When you see "at least" (or "at most") this is a sign that you want to consider doing P(A) - P(not A) to get your answer.
So consider how many ways there are to select 8 marbles out of 12, with no restrictions. This is 12C8, or 12!/4!8! = 495.

Then think about what the problem wants. What is the opposite of the situation in which you have at least one red and at least one blue? The opposite is you have no red and no blue. That is the opposite of at least 1 red and at least 1 blue.

With that in mind, think about how you would get into those situations... and note that these are two separate (OR) situations.
To have zero red left, you need to pick 7 red and 1 blue.
To have zero blue left, you need to pick 5 blue and 3 red.

Convert these both to math...
Zero red = 7C7*5C1 = 7!/7! * 5!/1!4! = 5
Zero blue = 5C5 * 7C3 = 5!/5! * 7!/3!4! =35
Add the two together and you get 40.

Now remember P(A)-P(not A) = answer.
Do the math:
495 - 40 = 455

455 is the answer!
GMAT Club Bot
M26-22 &nbs [#permalink] 11 Sep 2018, 12:48
Display posts from previous: Sort by

M26-22

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

Moderators: chetan2u, Bunuel

Events & Promotions

PREV
NEXT


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®.