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In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]

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21 Apr 2016, 23:35

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In a marketing survey, 60 people were asked to rank three flavors of ice cream, chocolate, vanilla, and strawberry, in order of their preference. All 60 people responded, and no two flavors were ranked equally by any of the people surveyed. If 3/5 of the people ranked vanilla last, 1/10 of them ranked vanilla before chocolate, and 1/3 of them ranked vanilla before strawberry, how many people ranked vanilla first?

Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]

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21 Apr 2016, 23:42

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This question I did correctly, but still feel doubtful. Experts please help clarify.

60 people vote for 3 favorite among 4 foods: I - C - V - S The results: 36 people vote x-x-V (vanilla last) => 24 people vote Vanilla not last OR NOT VOTE FOR Vanilla

6 people vote x-V-C or V-x-C or V-C-x

20 people vote x-V-S or V-x-S or V-S-x

The only possibility of overlap between 6 and 20 people above is the case: V - S - C or V - C - S, in both case, V is the first. => The overlap = 20 + 6 - 24 = 2

But, how about NOT VOTE FOR Vanilla? What is the role of this information in my reasoning?

Thanks.
_________________

--------------------------------------------------------- Please kudos me if this helps. Thank you.

This question I did correctly, but still feel doubtful. Experts please help clarify.

60 people vote for 3 favorite among 4 foods: I - C - V - S The results: 36 people vote x-x-V (vanilla last) => 24 people vote Vanilla not last OR NOT VOTE FOR Vanilla

6 people vote x-V-C or V-x-C or V-C-x

20 people vote x-V-S or V-x-S or V-S-x

The only possibility of overlap between 6 and 20 people above is the case: V - S - C or V - C - S, in both case, V is the first. => The overlap = 20 + 6 - 24 = 2

But, how about NOT VOTE FOR Vanilla? What is the role of this information in my reasoning?

Thanks.

The Q says that all vote and they were to give their preference 1-2-3 wise, so everyone has voted and given their preference in that order.. basically there is NO oNE who has NOT voted..
_________________

In a marketing survey, 60 people were asked to rank three flavors of ice cream, chocolate, vanilla, and strawberry, in order of their preference. All 60 people responded, and no two flavors were ranked equally by any of the people surveyed. If \frac{3}{5} of the people ranked vanilla last, 1/10 of them ranked vanilla before chocolate, and 1/3 of them ranked vanilla before strawberry, how many people ranked vanilla first?

A. 2 B. 6 C. 14 D. 16 E. 24

3/5 of 60 = 36 people ranked Vanilla Last

Observation 1: i.e. 2/5 of 60 = 24 people ranked Vanilla at no. 1 or no. 2

Observation 2: and 1/10 of 60 = 6 people ranked Vanilla before Chocolate

Observation 3: and 1/3 of 60 = 20 people ranked Vanilla before Strawberry

Observation 1 is consolidation of Observation 2 and 3

i.e. there are 2 people common between Observation 2 and 3

i.e. 2 people ranked Vanilla before Chocolate and Strawberry both

Hence, Answer: Option A
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]

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22 Apr 2016, 04:31

chetan2u wrote:

johnnguyen2016 wrote:

This question I did correctly, but still feel doubtful. Experts please help clarify.

60 people vote for 3 favorite among 4 foods: I - C - V - S The results: 36 people vote x-x-V (vanilla last) => 24 people vote Vanilla not last OR NOT VOTE FOR Vanilla

6 people vote x-V-C or V-x-C or V-C-x

20 people vote x-V-S or V-x-S or V-S-x

The only possibility of overlap between 6 and 20 people above is the case: V - S - C or V - C - S, in both case, V is the first. => The overlap = 20 + 6 - 24 = 2

But, how about NOT VOTE FOR Vanilla? What is the role of this information in my reasoning?

Thanks.

The Q says that all vote and they were to give their preference 1-2-3 wise, so everyone has voted and given their preference in that order.. basically there is NO oNE who has NOT voted..

Thank you chetan2u But could you confirm my understanding that "to rank three flavors of 4 foods" = to give their preference 1-2-3 Thank you again
_________________

--------------------------------------------------------- Please kudos me if this helps. Thank you.

Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]

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01 May 2016, 07:21

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Maybe I can give an alternative but more time consuming/stupid way to solve it -

Total 60 people need to rank c, v, s 3/5 *60= 36 people rank: CSV or SCV 1/10*60 = 6 people rank: VCS or VSC or SVC 1/3*60 = 20 people rank: VCS or VSC or CVS

As we know that there are only 6 combinations for C, V, S rank - CSV, SCV, VCS, VSC, SVC & CVS. Total people add up to be 60. However, VCS and VSC are overlapped in group 2 and 3

Total people = 60 Group 1+Group 2 + Group 3 = 36+6+20 = 62 The only over lap is 2, which equals to the number for VCS + VSC

It know it's a bit lengthy and that's why I couldn't figure it out in the mock exam (this question took me ~5 minutes before I realised it.

I think the previous solution is easier ==> 60-36 = 24 means people who rank VCS as number 1 or number 2. Then 24 = 6+20 - x ===> x=2. Simple and Easy!

Unfortunately, I wasn't able to come up with this idea when I was doing the mock exam. Good luck guys!

In a marketing survey, 60 people were asked to rank three flavors of ice cream, chocolate, vanilla, and strawberry, in order of their preference. All 60 people responded, and no two flavors were ranked equally by any of the people surveyed. If 3/5 of the people ranked vanilla last, 1/10 of them ranked vanilla before chocolate, and 1/3 of them ranked vanilla before strawberry, how many people ranked vanilla first?

A. 2 B. 6 C. 14 D. 16 E. 24

It's a good question.

Total 60 people.

"3/5 of the people ranked vanilla last" For 36 people, Vanilla came last. So 24 people ranked it 1st or 2nd.

"1/10 of them ranked vanilla before chocolate" For 6 people, vanilla came before chocolate

"and 1/3 of them ranked vanilla before strawberry" For 20 people, vanilla came before strawberry

Note that this forms a sets question: 6 people ranked vanilla before chocolate, 20 people ranked vanilla before strawberry. There are 24 people who ranked vanilla before chocolate or before strawberry or before both. How many ranked vanilla before both?

Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]

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19 Jun 2016, 00:03

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I know many of you guys have posted solutions here, but here is what was most intuitive to me:

FIRST OFF Understand what you are trying to find - In this case, you are trying to find how many people rank Vanilla first. For someone to rank Vanilla first, they have to rank them before Chocolate or Strawberry. This should be the very first hint that the correct answer cannot be bigger than 1/10 * 60 = 6.

So if you are time pressured on the real test, at least you narrowed down the selections to two choices - not bad. But what now?

Solve the easier problem. From 60 people, we are already told that 3/5 of them rank Vanilla last, leaving us with 2/5 of the people not ranking vanilla last - 24 people. So what are those combinations that vanilla are not last?

A) C, V, S B) S, V, C C) C, S, V (vanilla first) D) S, C, V (vanilla first)

What are we ultimately hunting for here? The number of people that rank vanilla first: C+D Now let's represent the given info into three equations 1) A+B+C+D = 24 (2/5 of the people do not rank vanilla last) 2) A+C+D = 6 (1/10 rank vanilla before chocolate) 3) B+C+D = 20 (1/3 rank vanilla before strawberry)

Subtract equation 1 from 2, leaving us with B = 24 - 6 = 18

Sub in B into equation 3 and solve for C+D 18 + C + D = 20 C+D = 20 - 18 = 2

There you go! 2 people out of the 60 ranked Vanilla as first

In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]

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30 Jun 2016, 00:37

johng2016 wrote:

In a marketing survey, 60 people were asked to rank three flavors of ice cream, chocolate, vanilla, and strawberry, in order of their preference. All 60 people responded, and no two flavors were ranked equally by any of the people surveyed. If 3/5 of the people ranked vanilla last, 1/10 of them ranked vanilla before chocolate, and 1/3 of them ranked vanilla before strawberry, how many people ranked vanilla first?

A. 2 B. 6 C. 14 D. 16 E. 24

no two flavors were ranked equally by any of the people surveyed Does it mean -

there is no overlap for ranking (like rank 2 for S & V) and (1 for C) right?

In a marketing survey, 60 people were asked to rank three flavors of ice cream, chocolate, vanilla, and strawberry, in order of their preference. All 60 people responded, and no two flavors were ranked equally by any of the people surveyed. If 3/5 of the people ranked vanilla last, 1/10 of them ranked vanilla before chocolate, and 1/3 of them ranked vanilla before strawberry, how many people ranked vanilla first?

A. 2 B. 6 C. 14 D. 16 E. 24

no two flavors were ranked equally by any of the people surveyed Does it mean -

there is no overlap for ranking (like rank 2 for S & V) and (1 for C) right?

Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]

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08 Sep 2016, 23:08

Here is the really dumb way that I correctly "solved" this in 1:50. Not sure if the math really 'works' or I just got lucky... I just started writing numbers because I didn't know what else to do.

V last: 60*3/5 = 36 V above C = 6 V above S = 20 36 + 6 + 20 = 62

That is 2 more responses than we have people, so 2 of these responses must overlap AKA '2 people chose Vanilla ABOVE chocolate AND strawberry.'

(More further explanation, to prove that I'm right, but you don't really need this to understand how it worked) It's important to note that the only 2 categories that CAN overlap are the bottom 2 categories - "V above C" and "V above S." There is no way a person said Vanilla is better than Choc or Strawberry AND Vanilla is last place.

So maybe I just got lucky (someone smarter than me please comment? lol), I don't know, but the math takes about 30 seconds if you approach it this way.

Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]

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27 Aug 2017, 10:50

if you look at this as a simple probability problem. you have 24 as the total number of events where Vanilla is first or second. you have 6 where Vanilla>Chocolate, and 20 where Vanilla>Strawberry (not that both these numbers imply that Vanilla is ahead of "some flavor" --> i.e. vanilla is first or second).

To apply this to a simple probability problem. We have P(A) = V>C, P(B) = V>S....

now, the trick is recognize and admittedly, make the obscure connection that P(A) AND P(B), where the two conditions are both met, is when Vanilla is greater than both Chocolate and Strawberry (i.e. when Vanilla is #1, what the question is asking for)

So P(A) = 6, P(B) = 20, P(A) + P(B) - both = 24 ...... 20+26-x = 24 --> x=2.

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