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# There is a club of car-owners consisting of 100 members. 93

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Senior Manager
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There is a club of car-owners consisting of 100 members. 93 [#permalink]  12 Feb 2004, 19:00
There is a club of car-owners consisting of 100 members.

93 persons own BMW
90 persons own Ferrari
81 persons own Porsche
75 persons own Aston Martin
62 persons own Lamborghini

How many people own all the cars?
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shubhangi

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[#permalink]  14 Feb 2004, 01:20
You should be able to see that you cannot answer this question given the information in the question. Perhaps a more suitable question is:

Given the above information, what is the minimum number of people that own all 5 cars?
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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[#permalink]  14 Feb 2004, 04:05
Shooting in the dark. Is it 2?
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[#permalink]  14 Feb 2004, 05:06
rakesh, agree if you mean 62
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[#permalink]  14 Feb 2004, 08:17
Ans is 1 .. dont know how??
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shubhangi

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[#permalink]  14 Feb 2004, 15:01
2 was close enough to 1.
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[#permalink]  15 Feb 2004, 02:32
I was able to get 1 but due to unavailibility of net was not able to post it. My hard Luck!!!!!!
Well to explain it is a bit difficult. But the way i that i found out the total as 401. So i thought that lets all have 4 cars. So we are remaining with 1. Hence atleast 1 person need to have more then 4 cars or all the 5 cars.
Initially i thought that its not logically correct but if you reframe it to 3 cars and 10 persons ans then on a paper draw all the combinations u will know that it works!!!!!!!!
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[#permalink]  23 Feb 2004, 07:13
Okay. Everyone seems to be focusing on "1" as the answer. Let's apply a sanity check.

Lets says that 62 people own all five cars. Of the 38 remaining people. 10 own BMV, 7 own Ferrari, and 21 own both. Now everyone has at least one car. We can divide up the remainder Porsches and AMs among those 38 people any way we want since there are no Lamborghinis left.

Hence, given the infomation in the question, IT IS POSSIBLE for as many as 62 people to own all of the cars.

So why is the answer "1"?

I still maintain that the question must be reworded.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
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Kudos [?]: 52 [0], given: 0

[#permalink]  23 Feb 2004, 07:30
AkamaiBrah wrote:
You should be able to see that you cannot answer this question given the information in the question. Perhaps a more suitable question is:

Given the above information, what is the minimum number of people that own all 5 cars?

This question seems difficult because setting up a Venn diagram is impossible with more than 3 variable (unless you are one of those freaky math or physics majors that think in 4 and 5-D space-time)

Assuming that the question is reworded as above, here is a simple step by step way to get to the solution:

There are 93 beamers and 90 ferraris. Hence, there are at a minimum 93+90-100 or 83 people with both cars. Agree? (you can set up a venn diagram and come up with this equation quite easily -- leave that as an exercise).

There are at a minimum 83 people with BMW and Ferr and 81 Porsche drivers. Hence there are at a minimum 83 - 81 - 100 or 64 people who drive all 3 cars.

There are at a minimum 64 owning B, F, and P. There are 75 people driving AMs, hence there are at a minimum 64 + 75 - 100 or 39 people driving B,F, P, and AM.

There are at least 39 people driving B,F,P, and AM and 62 people driving Ls. Hence there are at least 62+39-100 = 1 person driving all 5 cars.

QED
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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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[#permalink]  11 Mar 2004, 02:44
Akamai, if the question is reworded then i think my appraoch is absolutely correct.Infact i solved this by assuming that we need to find the minimum number of person.

Can you just confirm on this.
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[#permalink]  11 Mar 2004, 11:52
Guys,

My 2 cents. I think the answer should be 1.

First time got it without analysing but after checking that u ppl. will not accept without proper reasoning , pondered over it for a minute.

here is what i think:

7 do not own A
10 do not own B
19 do not own C
25 do not own D
38 do not own E

So 99 do not own one or the other.
Hence, one person owns all of them.

cheers
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[#permalink]  12 Mar 2004, 02:18
Virtual wrote:
Guys,

My 2 cents. I think the answer should be 1.

First time got it without analysing but after checking that u ppl. will not accept without proper reasoning , pondered over it for a minute.

here is what i think:

7 do not own A
10 do not own B
19 do not own C
25 do not own D
38 do not own E

So 99 do not own one or the other.
Hence, one person owns all of them.

cheers

Virtual, you know your permise do not hold true, if i say 7 do not own A constitues in 38 who do not owns E.

Question needs to reword either.

Dharmin

Dharmin
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[#permalink]  12 Mar 2004, 04:13
What do u guys says on my logic??????
Had the total been 402...there would have been atleast 2 people who would have had all the 5 cars!!!!!!!!!

Just check and let me know......
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[#permalink]  12 Mar 2004, 11:13
Dharmin,

I agree.

Guys,

In that case, will the question be like this :

What is the minimum number of people who own all the cars?
The answer to which should be 1.

For exact/max. number, date is insufficient.

cheers.
[#permalink] 12 Mar 2004, 11:13
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# There is a club of car-owners consisting of 100 members. 93

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