At a blind taste competition a contestant is offered 3 cups : GMAT Problem Solving (PS) - Page 2
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

It is currently 10 Dec 2016, 09:22
GMAT Club Tests

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

At a blind taste competition a contestant is offered 3 cups

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

Hide Tags

Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 35950
Followers: 6863

Kudos [?]: 90114 [1] , given: 10417

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 06 Mar 2013, 01:32
1
This post received
KUDOS
Expert's post
tosattam wrote:
Not sure if my channel of thought is flawed to start with, but when it is said that it's a 'blind' taste competition, I am considering repetitions. As the contestant does not know which of the 9 cups he/she had picked up the first time, the same can possibly be repeated in the second turn and so on until the fourth.
That gives a complete different perspective to the problem.
Where am I going wrong here?


A blind taste simply means that a contestant doesn't know which samples he/she is offered to taste.
_________________

New to the Math Forum?
Please read this: 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

1 KUDOS received
Intern
Intern
avatar
Joined: 22 Dec 2012
Posts: 16
GMAT 1: 720 Q49 V39
Followers: 1

Kudos [?]: 22 [1] , given: 19

Re: Diagnostic Question [#permalink]

Show Tags

New post 09 Mar 2013, 21:37
1
This post received
KUDOS
Hi,
Let me try..
mainhoon, you are going with the logic that you select 1 cup from each sample in 3c1 x 3c1 x 3c1 ways and 1 remaining cup from 6 cups in 6c1 ways rite!
I guess the flaw with your approach is that this 6c1 is the probablity of selecting 1 item from 6 DIFFERENT items (r items from n different items is nCr)
Im this case the 6 items left are not different! They are XX,YY,ZZ types right! so you just cant use the 6c1 fomula!
This is the formula : The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of x^r in the expansion
Image

hope it helps!



mainhoon wrote:
Yes I can see that clearly I am ending up with more combinations than necessary. I was thinking if there is way to factor out those repeats the way I am doing it. I see your point in the simple example. However when I think of this, I am thinking of a 3x3 matrix where I have to make sure I get one from each row and the 4th can come from anywhere - that is the same thing as saying 2 from a row and 1 each from the remaining 2 (your and correct approach). My count differs from yours by a factor of 2 (I am 162 and you are 81). How can I rationalize this?
Intern
Intern
avatar
Joined: 10 Dec 2013
Posts: 20
Location: India
Concentration: Technology, Strategy
Schools: ISB '16 (S)
GMAT 1: 710 Q48 V38
GPA: 3.9
WE: Consulting (Consulting)
Followers: 0

Kudos [?]: 7 [0], given: 7

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 29 Jan 2014, 10:59
If we employ the counting method applied in the question http://gmatclub.com/forum/tough-p-n-c-92675.html then
Considering three different samples and contestant not to taste all three samples there would be following two conditions

1. Contestant tastes 3 cups of on sample and 1 cup of another
i.e. AAAB
The number of ways this can happen is
3C2*4!/3! = 12
3C2 = Ways to select any 2 samples out of the 3 choices
4!/3! = # of ways AAAB can be rearranged

or
2. Contestant tastes 2 cups of a sample and 2 of another
i.e. AABB
The number of ways this could happen is
3C2*4!/(2!*2!) = 18
3C2 = Ways to select any 2 samples out of the 3 choices
4!/(2!*2!) = # of ways AABB can be rearranged

Hence total number of ways by which contestant can only taste any two samples is 18 + 12 = 30

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 30/126
= 5/21

Is this approach not correct? Where am I missing ?
Manager
Manager
User avatar
Joined: 11 Jan 2014
Posts: 98
Concentration: Finance, Statistics
GMAT Date: 03-04-2014
GPA: 3.77
WE: Analyst (Retail Banking)
Followers: 2

Kudos [?]: 53 [0], given: 11

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 29 Jan 2014, 11:22
Great explanations above. A rather headache-inducing question, I admit I had to resolve to the following guessing pattern after I passed the 1:36 mark and admitted defeat. :lol:

Economist wrote:
At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. \(\frac{1}{12}\) Too low
B. \(\frac{5}{14}\) Looks reasonable
C. \(\frac{4}{9}\) Impossible, cannot be that simple
D. \(\frac{1}{2}\) Too high
E. \(\frac{2}{3}\) Too high
Intern
Intern
avatar
Joined: 10 Dec 2013
Posts: 20
Location: India
Concentration: Technology, Strategy
Schools: ISB '16 (S)
GMAT 1: 710 Q48 V38
GPA: 3.9
WE: Consulting (Consulting)
Followers: 0

Kudos [?]: 7 [0], given: 7

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 30 Jan 2014, 01:50
Quote:
If we employ the counting method applied in the question tough-p-n-c-92675.html then
Considering three different samples and contestant not to taste all three samples there would be following two conditions

1. Contestant tastes 3 cups of on sample and 1 cup of another
i.e. AAAB
The number of ways this can happen is
3C2*4!/3! = 12
3C2 = Ways to select any 2 samples out of the 3 choices
4!/3! = # of ways AAAB can be rearranged

or
2. Contestant tastes 2 cups of a sample and 2 of another
i.e. AABB
The number of ways this could happen is
3C2*4!/(2!*2!) = 18
3C2 = Ways to select any 2 samples out of the 3 choices
4!/(2!*2!) = # of ways AABB can be rearranged

Hence total number of ways by which contestant can only taste any two samples is 18 + 12 = 30

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 30/126
= 5/21

Is this approach not correct? Where am I missing ?



I have come up with another approach which gives a different result again.

Assuming the that the contestant does not taste the 3rd sample there are 2 scenarios possible

AABB and AAAB

Counting one at a time.
#of ways AABB can occur = 3C2*3C2*3C2 = 27
here,
3C2 = ways to select the 2 samples from 3
3C2 = ways to select the 2 cups from 3 cups of first sample
3C2 = ways to select 2 cups from 3 cups of second sample

# of ways AAAB can occur = 3C2*3C3*3C1 = 9
3C2 = ways to select the 2 samples from 3
3C3 = ways to select 3 cups from the 3 cups of first sample
3C1 = ways to select 1 cup from the 3 cups of second sample

Hence the total number of ways in which the contestant cannot taste any cup from third sample = 9 +27 = 36

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 36/126
= 4/14

Even after hours and hours of going through various material on Probablity and combinations I am struggling to identify a uniform approach that can be applied across problems. How would one know which approach to apply for which problem during the exams. Is there any way we can determine that? :(
Intern
Intern
avatar
Joined: 22 Feb 2014
Posts: 30
Followers: 0

Kudos [?]: 10 [0], given: 3

Reviews Badge
At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 28 Oct 2014, 22:39
# of ways of choosing 4 cups from 9 = 9c4 = 126
# of ways of choosing 4 cups such that one gets only 2 varieties of tea -

1st, 2nd and 3rd cup to be from same type = 3*2*1 per type * 3 (for 3 different types) = 27
4th cup to be from a different type (can't be from same type as none of 1 type are left) = 3*3 per type * 2 (for 2 different types) = 18

Total = 1st, 2nd, 3rd & 4th cup = 27+18 = 45

Prob (Tastes only 2 varieties) = 45/126 = 5/14 - B.

Hope this helps?
Manager
Manager
User avatar
Status: GMAT Coach
Joined: 05 Nov 2012
Posts: 115
Location: Peru
GPA: 3.98
Followers: 3

Kudos [?]: 17 [0], given: 12

At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 12 Jan 2015, 07:47
I solved it in a way that I think is more direct:
I get the probability that each cup is not a specific type of tea (type A):
1st cup, Probability it is not type A = 3/9
2nd cup, Probability it is not type A = 3/8, since we only have 8 cups left, and 3 of them are type A
3rd cup, Probability it is not type A = 3/7
4th cup, Probability it is not type A = 3/6
We do the same for types B and C (multiply by 3)
So : 3/9 * 3/8 * 3/7 * 3/6 * 3 = 5/14
_________________

Clipper Ledgard
GMAT Coach

Manager
Manager
User avatar
Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 230
Location: India
MISSION : 800
WE: Design (Manufacturing)
Followers: 5

Kudos [?]: 76 [0], given: 259

GMAT ToolKit User Premium Member
Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 24 Mar 2015, 01:45
Hi
Is the following approach not correct..?? where am wrong????

C^1_3 - # of ways to chose 1 cup out of 3 from 1st sample;
C^1_3 - # of ways to chose 1 cup out of 3 from 2nd sample;
C^1_3 - # of ways to chose 1 cup out of 3 from third sample;
C^1_6- # of ways to chose 1 cup out of 6 remaining cups;

C^4_9- # of ways to chose 4 cup out of 9 cups;

Probability = [C^1_3 x C^1_3 x C^1_3 x C^1_6] / C^4_9 = 162/126 =9/7

Hence required P = 1- 9/7 = :x
_________________

Thank you

+KUDOS

> I CAN, I WILL <

Manager
Manager
avatar
Joined: 17 Aug 2015
Posts: 101
Location: India
Concentration: Strategy, General Management
Schools: Duke '19 (II)
GMAT 1: 750 Q49 V42
GPA: 4
WE: Information Technology (Investment Banking)
Followers: 3

Kudos [?]: 46 [0], given: 338

GMAT ToolKit User
Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 07 Sep 2015, 02:25
Hi,

Can someone help me understand the question? I am having great trouble figuring out what exactly is happening here, let alone what's asked.

"3 cups of each of the 3 samples of tea" - does this mean that there are 3 types of tea, say Aasam Tea, Green Tea and Ginger Tea and a contestant is given 1 cup of each?

"in a random arrangement of 9 marked cups" - what does this mean? Are there 9 cups, each with a different "type" of tea, or are there just 9 cups that may/may not have tea? The three cups offered above - these come from these nine cups?

"each contestant tastes 4 different cups of tea" - each contestant as against "a" contestant? So what's happening here? Each tastes 4 cups, 3 of which he has been given as part of the first point above (all three of a different flavour + one he picks up on his own?)

"contestant does not taste all of the samples" - so probability that he does not taste each of aasam,green and ginger? As in, a contestant tastes either zero or one or two? What's the difference here between a contestant and each contestant?

I am completely confused! Can someone help explain what's happening with a diagram? Thanks in advance!
_________________

If you like this post, be kind and help me with Kudos!

Cheers!

Manager
Manager
avatar
Joined: 10 Mar 2014
Posts: 236
Followers: 1

Kudos [?]: 78 [0], given: 13

Premium Member
Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 18 Oct 2015, 23:04
Bunuel wrote:
Economist wrote:
At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

# \(\frac{1}{12}\)
# \(\frac{5}{14}\)
# \(\frac{4}{9}\)
# \(\frac{1}{2}\)
# \(\frac{2}{3}\)


And the good one again. +1 to Economist.

"The probability that a contestant does not taste all of the samples" means that contestant tastes only 2 samples of tea (one sample is not possible as contestant tastes 4 cups>3 of each kind).

\(\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}\).

\(C^2_3\) - # of ways to choose which 2 samples will be tasted;
\(C^4_6\) - # of ways to choose 4 cups out of 6 cups of two samples (2 samples*3 cups each = 6 cups);
\(C^4_9\) - total # of ways to choose 4 cups out of 9.

Answer: B.

Another way:

Calculate the probability of opposite event and subtract this value from 1.

Opposite event is that contestant will taste ALL 3 samples, so contestant should taste 2 cups of one sample and 1 cup from each of 2 other samples (2-1-1).

\(C^1_3\) - # of ways to choose the sample which will provide with 2 cups;
\(C^2_3\) - # of ways to chose these 2 cups from the chosen sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from second sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from third sample;
\(C^4_9\) - total # of ways to choose 4 cups out of 9.

\(P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}\).

Answer: B.

Hope it's clear.


Hi bunuel,

Could you please explain this.

C13 - # of ways to choose the sample which will provide with 2 cups; ( here we are selecting 2 cups from 3 cups right?)

Thanks.
1 KUDOS received
Manager
Manager
avatar
Joined: 13 Sep 2015
Posts: 90
Location: United States
Concentration: Social Entrepreneurship, International Business
GMAT 1: 770 Q50 V45
GPA: 3.84
Followers: 2

Kudos [?]: 9 [1] , given: 0

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 19 Oct 2015, 07:52
1
This post received
KUDOS
guess it's probability= desired result/all the possible result?

cup1: it's 1, you'll taste a sample for sure as long as you take a cup

cup2: one sample is excluded and cup 1 is excluded, 4 cups left: 4/9-1

cup3, one sample is excluded and cup 1/2 are excluded, 3 cups left:3/8-1

cup 4, one sample is excluded and cup 1/2/3 are excluded, 2 cups left: 2/7-1

so the probability is

1*4/8*3/7*2/6=1/14
1 KUDOS received
Manager
Manager
avatar
Joined: 05 Aug 2015
Posts: 60
Followers: 1

Kudos [?]: 23 [1] , given: 36

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 01 Nov 2015, 16:07
1
This post received
KUDOS
dimitri92 wrote:

i solved the question as follows. i hope my approach is correct.

let's consider that the contestant tasted all 3 flavors.
the first cup can be selected from any 9 cups. P = \(\frac{9}{9}\)
the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = \(\frac{6}{8}\)
the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = \(\frac{3}{7}\)
the fourth cup can be selected from any of the remaining 6 cups. P = \(\frac{6}{6}\)

there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

probability that the contestant tasted all flavors = \(\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}\)

required probability = \(1 - \frac{9}{14} = \frac{5}{14}\)



Hi! Can some please explain the note regarding "there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1."? I didn't understand why we need to multiple by 2.

Separately, why is my approach below wrong?

P= 1-(9*6*3*6)/(9 choose 4) is obviously wrong, but why? The denominator I understand very clearly - number of ways to choose 4 from 9 choices. The numerator is based on the reasoning that the first cup can be from any of the 9 choices, so 9 ways. The second cup has to be from any of the 6 cups (the two samples different from the first cup). The third cup has to be from any of the 3 cups (the third sample not choose by cup 1 or 2). And the last cup can be from any of the remaining 6 cups.

Thanks so much!
_________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

Intern
Intern
avatar
Joined: 26 Sep 2015
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 456

GMAT ToolKit User
Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 23 Nov 2015, 18:06
dimitri92 wrote:

i solved the question as follows. i hope my approach is correct.

let's consider that the contestant tasted all 3 flavors.
the first cup can be selected from any 9 cups. P = \(\frac{9}{9}\)
the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = \(\frac{6}{8}\)
the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = \(\frac{3}{7}\)
the fourth cup can be selected from any of the remaining 6 cups. P = \(\frac{6}{6}\)

there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

probability that the contestant tasted all flavors = \(\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}\)

required probability = \(1 - \frac{9}{14} = \frac{5}{14}\)



Hi Bunnuel,
Is this a correct solution? I am having a difficult time understanding why did he multiply by two. I would expect that through this method, you don't need to multiply to account for combinations since the probabilities for each pick already account for that. Thanks.
1 KUDOS received
Manager
Manager
avatar
Joined: 05 Aug 2015
Posts: 60
Followers: 1

Kudos [?]: 23 [1] , given: 36

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 01 Dec 2015, 21:45
1
This post received
KUDOS
happyface101 wrote:
dimitri92 wrote:

i solved the question as follows. i hope my approach is correct.

let's consider that the contestant tasted all 3 flavors.
the first cup can be selected from any 9 cups. P = \(\frac{9}{9}\)
the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = \(\frac{6}{8}\)
the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = \(\frac{3}{7}\)
the fourth cup can be selected from any of the remaining 6 cups. P = \(\frac{6}{6}\)

there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

probability that the contestant tasted all flavors = \(\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}\)

required probability = \(1 - \frac{9}{14} = \frac{5}{14}\)



Hi! Can some please explain the note regarding "there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1."? I didn't understand why we need to multiple by 2.

Separately, why is my approach below wrong?

P= 1-(9*6*3*6)/(9 choose 4) is obviously wrong, but why? The denominator I understand very clearly - number of ways to choose 4 from 9 choices. The numerator is based on the reasoning that the first cup can be from any of the 9 choices, so 9 ways. The second cup has to be from any of the 6 cups (the two samples different from the first cup). The third cup has to be from any of the 3 cups (the third sample not choose by cup 1 or 2). And the last cup can be from any of the remaining 6 cups.

Thanks so much!


BUMP my question - please help!
_________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

1 KUDOS received
Intern
Intern
User avatar
Joined: 08 Feb 2016
Posts: 46
Followers: 0

Kudos [?]: 9 [1] , given: 14

Premium Member
Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 17 Mar 2016, 23:58
1
This post received
KUDOS
The probability of tasting all 3 cups is

One cup from pool A (\(\frac{3}{9}\)) x One cup from pool B (\(\frac{3}{8}\)) x One cup from pool C (\(\frac{3}{7}\))

Since the contestant has to taste 4 cups, one possibility is picking second cup from pool A, so the probability is \(\frac{2}{6}\)

The probability of tasting in order of ABCA is \(\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6}\) ..... (1)

In how many ways can the contestant taste ABCA cups = \(\frac{4!}{2!}\) = 4 x 3 ....... (2)

The combinations ABCB & ABCC are also possible (along with ABCA) = 3 combinations ..... (3)

Since the choice of "a specific cup" from "a pool" doesn't matter here, we need NOT account for \(A_1A_2A_3, A_1A_3A_2,\) ... etc. combinations

So Probability of tasting all three cups is = (1) * (2) * (3)

= \(\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6} * 4 *3 *3\)

=\(\frac{9}{14}\)

Hence, Probability of NOT tasting all three cups = \(1- \frac{9}{14} = \frac{5}{14}\)
_________________

Once you know the answer, it is easy to justify.

Intern
Intern
avatar
Joined: 04 Apr 2016
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

New post 07 Apr 2016, 06:09
Is this an okay way to think about this problem?

In the case where we want to use the alternate solution (1-opposite), would it be okay to say the following:

The number of ways to taste all the cups would result from the following choices:

(A A) B C

- Of the 3 samples, choose 1 to be a double
- Of the 3 samples, choose 2 to be singles
- Given our choice for the double, choose 1 location of 3 possible to represent the double
- Given our choice for the singles, choose 2 locations of the 3 possible to represent the singles

Hence our total desired outcome would be: (3C1)(3C2)(3C1)(3C2) / (9C4) = 9/14.
Our answer, therefore, would be 1 - 9/14 = 5/14
Re: At a blind taste competition a contestant is offered 3 cups   [#permalink] 07 Apr 2016, 06:09

Go to page   Previous    1   2   [ 36 posts ] 

    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic A baker filled with a measuring cup with 3/4 cup water. He poured 1/2. asciijai 2 10 Jun 2016, 22:25
1 Experts publish their posts in the topic An entrepreneurship competition requires registering teams to have 3 Bunuel 4 26 May 2016, 11:33
1 Experts publish their posts in the topic Three competing juice makers conducted a blind taste test with mall Bunuel 1 06 Apr 2016, 00:14
Experts publish their posts in the topic In an intercollegiate competition that lasted for 3 days, 100 students Bunuel 3 23 Aug 2015, 12:15
11 Experts publish their posts in the topic A recipe requires 2 1/2 cups of flour 2 3/4 cups of sugar clciotola 9 17 May 2013, 07:03
Display posts from previous: Sort by

At a blind taste competition a contestant is offered 3 cups

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


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.