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

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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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johnnguyen2016 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 \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,

Originally posted by GMATinsight on 22 Apr 2016, 04:21.
Last edited by GMATinsight on 21 Sep 2020, 21:04, edited 1 time in total.
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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This question I did correctly, but still feel doubtful.

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.
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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johnnguyen2016 wrote:
This question I did correctly, but still feel doubtful.

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..
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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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!
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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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
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I think I have a simpler solution, can anyone confirm if it accurate?

1/10 ranked V before C, and 1/3 ranked V before S, and you want the amount that ranked V before C AND S, can't you just multiply the two?

Therefore: 1/10*1/3=1/30
Multiply this with the total group (60) to get the correct answer of 2.

This approach does not use the 3/5 information, so I'm not sure if this is a good approach. Can anyone confirm?
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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Hi Experts,

It would be great if anyone can verify my approach here.
This question popped as my 7th questions on GMAT Prep QP2 and I got it wrong T___T

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

Total = Vanilla before Chocolate + Vanilla before Strawberry - Vanilla before Both + Vanilla before neither
1 = 1/10 + 1/3 - Vanilla before Both + 3/5
1 - 3/5 = 13/30 - Vanilla before Both
2/3 = 13/30 - Vanilla before Both
Vanilla before Both = 1/30.

1/30 of 60 = 2.
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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The sum of the number of people who ranked vanilla should be 60.
Here, 36(third) + 6(before chocolate) + 20(before strawberry) = 62.
The extra count is the number of people who rated vanilla before both chocolate and strawberry i.e. who ranked vanilla first.
This means, that 62 – 60 = 2 people ranked vanilla first.
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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

Let C represent chocolate, V represent vanilla, and C represent strawberry

Let a = the number of people who ranked the flavors (from best to worst) as V-C-S
Let b = the number of people who ranked the flavors (from best to worst) as V-S-C
Let c = the number of people who ranked the flavors (from best to worst) as C-S-V
Let d = the number of people who ranked the flavors (from best to worst) as C-V-S
Let e = the number of people who ranked the flavors (from best to worst) as S-C-V
Let f = the number of people who ranked the flavors (from best to worst) as S-V-C

We know that a + b + c + d + e + f = 60 [since there are 60 people in the survey]

3/5 of the people ranked vanilla last
This refers to people who had any of the following rankings: C-S-V (c) or S-C-V (e)
3/5 of 60 = 36
So, we can write: c + e = 36

1/10 of the people ranked vanilla before chocolate
This refers to people who had any of the following rankings: V-C-S (a), V-S-C (b) or S-V-C (f)
1/10 of 60 = 6
So, we can write: a + b + f = 6

1/3 of the people ranked vanilla before strawberry
This refers to people who had any of the following rankings: V-C-S (a), V-S-C (b) or C-V-S (d)
1/3 of 60 = 20
So, we can write: a + b + d = 20

How many people ranked vanilla first?
This refers to people who had any of the following rankings: V-C-S (a) or V-S-C (b)
So, our GOAL is to find the value of a + b

So, we have the following equations:
c + e = 36
a + b + f = 6
a + b + d = 20
ADD them all to get: 2a + 2b + c + d + e + f = 62
We also know that a + b + c + d + e + f = 60

If we SUBTRACT the bottom equation from the top equation, we get: a + b = 2

Cheers,
Brent
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"1/10 of them ranked vanilla before chocolate"

so 9/10 of then ranked vanilla AFTER chocolate. SO 54 ranked vanilla AFTER chocolate. 6 of them have vanilla before chocolate, but we don't know if vanilla is 1st or second for now. We can deduce the answer has to be 6 or 2.

36 ranked vanilla last so these are part of the 54 who ranked vanilla after chocolate. There must be 54-36 who ranked vanilla after chocolate but did not rank it last. So the order must be c,v,s....There are 1-(1/3) or 40 who ranked vanilla after strawberry. These are part of the 36 who ranked vanilla last so 4 must rank vanilla after strawberry but must not have it be last. The order must be s,v,c. So 36+18+4=58 DO NOT RANK VANILLA FIRST. The remaining 2 people must rank vanilla first.
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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Hi,

If we read the entire question all its asking is in how many ways can we Write V first

So either its VSC or VCS

Or think of it like this
So we have 3 Slots where First Slot is Selected in 1 way ( we can select only Vanilla)

First Slot can be filled 1 Way
Second Slot we can Select i out of 2 so 2 ways
Third slot can be Selected in 1 way after the second slot is filled

So we have 1 * 2 *1 = 2 Ways

So we can have vanilla 2 ways in first place .
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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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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

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Re: In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
Can someone confirm if this is accurate or not?

free water wrote:
I think I have a simpler solution, can anyone confirm if it accurate?

1/10 ranked V before C, and 1/3 ranked V before S, and you want the amount that ranked V before C AND S, can't you just multiply the two?

Therefore: 1/10*1/3=1/30
Multiply this with the total group (60) to get the correct answer of 2.

This approach does not use the 3/5 information, so I'm not sure if this is a good approach. Can anyone confirm?
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If you are a fan of visualization, the idea of using a triple venn diagram or a double matrix should pop up in your head after reading this question.

Notice how the question actually asks us how many people rate Vanilla before Chocolate AND Strawberry.
And notice that Vanilla can be either before or after Chocolate. Vanilla can be either before or after Strawberry.
We actually have a double matrix scenario!

Just take a look at the attached photo. There are only 2 people who have rated Vanilla before Chocolate AND Strawberry.

This is by far the easiest way to solve this question for us prone to visual understanding.

PS: When doing this during the test you should of course use short abbreviations like BC (before chocolate), AS (after strawberry), and T (total). And you do not need to draw that whole matrix in boxes, just draw a few lines to from a grid. This takes less than 5 seconds.
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In a marketing survey, 60 people were asked to rank three flavors of i [#permalink]
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Attached is a visual that should help.

Also note that 20+36+6 = 62, so there must be an overlap of (62 - 60) = 2 in VSC and VCS, which are each counted twice.
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