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tricky math question

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madcowudub
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tricky math question [#permalink] New post   05 Aug 2007, 01:20
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?

the answer is 5. I'm not sure how this answer was reached. please help.



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



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Re: tricky math question [#permalink] New post   05 Aug 2007, 06:42
madcowudub wrote:
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?

the answer is 5. I'm not sure how this answer was reached. please help.


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)

so the answer is 5
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Re: tricky math question [#permalink] New post   24 Jan 2008, 19:41
good explanation! it's hard figuring this out under time constraint....
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Re: tricky math question [#permalink] New post   24 Jan 2008, 20:56
madcowudub wrote:
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?

the answer is 5. I'm not sure how this answer was reached. please help.


(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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Re: tricky math question [#permalink]
24 Jan 2008, 20:56
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