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# tricky math question

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Intern
Joined: 02 Mar 2007
Posts: 27

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05 Aug 2007, 01:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I found this question online.

At a club, there can be a number of 10-40 people. If 3 people sit at one table, then the remainding people can fit with 4 per table. If every 3 people sit on one table, then the remainding people can fit 5 per table. The question is: if 6 people sit per table, one table will not be able to be full, how many people can sit on that one table that isn't full?

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VP
Joined: 28 Mar 2006
Posts: 1367

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05 Aug 2007, 06:42
I found this question online.

At a club, there can be a number of 10-40 people. If 3 people sit at one table, then the remainding people can fit with 4 per table. If every 3 people sit on one table, then the remainding people can fit 5 per table. The question is: if 6 people sit per table, one table will not be able to be full, how many people can sit on that one table that isn't full?

There are 10-40 people

1st condition is 4*a+3 (where a is the # of people sitting/table)
2nd condition is 5*b+3 (where b is the # of people sitting/table)

But bothe trhe total are equal

4*a+3 = 5*b+3

This can happen

when a=5 and b=4
OR
when a=10 and b=8 (but this cannot happen 'cos the max # of people is 40 and we cannot exeed that #)

4*10+3 = 43 (so ruled out)

So there are only 23 people

Now coming to the 3rd condition
6*c+d = 23 where d is the spill over people on the last table left

we have c=1 d=17
c=2 and d=11
c=3 and d=5

here d cannot be 6 or more (since this table has <6)

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Intern
Joined: 02 Mar 2007
Posts: 27

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24 Jan 2008, 19:41
good explanation! it's hard figuring this out under time constraint....

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CEO
Joined: 29 Mar 2007
Posts: 2554

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24 Jan 2008, 20:56
I found this question online.

At a club, there can be a number of 10-40 people. If 3 people sit at one table, then the remainding people can fit with 4 per table. If every 3 people sit on one table, then the remainding people can fit 5 per table. The question is: if 6 people sit per table, one table will not be able to be full, how many people can sit on that one table that isn't full?

(x-3)/4= an integer (x-5)/3= an integer

(x-y)/6 = an integer.

X is btwn 10-40 people... From here i dunno what to do but try numbers

15-3 = 12/4 = integer 12-5/3 no this doesnt work

23-3/4 --> integer 23-5 -> 18/3 is an integer, This maybe it.

23-y/6 is an integer if y is 5 then this works. So y must be 5.

Took bout 2 1/2 min, not impressive, but least to the right answer.

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VP
Joined: 22 Oct 2006
Posts: 1437

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Schools: Chicago Booth '11

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25 Jan 2008, 15:17
1
KUDOS
Instead of picking numbers I set both conditions to X and set them equal

4X + 3 = 3X + 5

This gives me X = 2

total number is 11 which satisfies condition > 10 so this works out.

Now if 6 People sit we have

6(X) + Remainder = 11

Since X cant be anything greater than 1, because it would go over 11, we set X=1 and remainder will be 5.

Kudos [?]: 196 [1], given: 12

Re: tricky math question   [#permalink] 25 Jan 2008, 15:17
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