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Chairs

by maihuna » Wed Dec 23, 2009 3:13 am
In the convention halls chairs are evenly placed around a circular table. If the chairs are counted clockwise chair number 3 is exactly opposite chair number 18. How many chairs are in the conference hall.

30
31
32
33
34
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by Lattefah84 » Wed Dec 23, 2009 3:58 am
maihuna wrote:In the convention halls chairs are evenly placed around a circular table. If the chairs are counted clockwise chair number 3 is exactly opposite chair number 18. How many chairs are in the conference hall.

30
31
32
33
34
I'll never pass the gmat
:cry:
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by munaf » Wed Dec 23, 2009 3:59 am
IMO A-30

No 3 is exactly opposite to 18 so there are total 16 chairs including no 3 and no 18 on one side of the circle.So 14 more chairs are to the other side since the chairs are placed symmetrically.So 16+14=30 total chairs.

Guys correct me if i am wrong.

OA plz
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by maihuna » Wed Dec 23, 2009 8:36 am
munaf wrote:IMO A-30

No 3 is exactly opposite to 18 so there are total 16 chairs including no 3 and no 18 on one side of the circle.So 14 more chairs are to the other side since the chairs are placed symmetrically.So 16+14=30 total chairs.

Guys correct me if i am wrong.

OA plz
Let it be 30, good reasoning enjoy.
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by linkinpark » Wed Dec 23, 2009 9:16 am
Lattefah84 wrote:
maihuna wrote:In the convention halls chairs are evenly placed around a circular table. If the chairs are counted clockwise chair number 3 is exactly opposite chair number 18. How many chairs are in the conference hall.

30
31
32
33
34
I'll never pass the gmat
:cry:
don't be so pessimist, GMAT isn't harder than conquering Mt.Everest :) don't jump to hard questions directly, create foundation on basic stuff. Be positive.
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by Stuart@KaplanGMAT » Wed Dec 23, 2009 6:05 pm
munaf wrote:IMO A-30

No 3 is exactly opposite to 18 so there are total 16 chairs including no 3 and no 18 on one side of the circle.So 14 more chairs are to the other side since the chairs are placed symmetrically.So 16+14=30 total chairs.

Guys correct me if i am wrong.

OA plz
That's definitely one way to solve; here's another way to think about it:

there are 14 numbers between 3 and 18 (i.e. 4, 5, 6, ..., 17). To keep the circle halved, there must also be 14 chairs between 18 and 3.

So, we have 14 + 14 + chair 3 + chair 18 = 30 chairs.
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by viidyasagar » Thu Dec 24, 2009 11:08 pm
Not to undermine Stuart's method......but i thought of the sum in the following manner and it helped me solve the sum in 10 sec...

Think of a round dial clock hanging on the wall..... think of any number and its diametric opposite....for instance 2 and its diametric opposite 8

2 is the 10th minute on the clock and 8 has 20 minutes to go to reach centre number 12..... when 10 and 20 are added we get 30...which is 60 minutes (total no of minutes in an hour) divided by 2... this holds true for any minute on the clock....

hence the equation for this sum is as follows.....let the total number of chairs be x....(x corresponds to 60 minutes)

3 + (x-18) = x/2...solve for x, x= 30....

This explanation may make the problem more tedious...but if we can imagine a clock then the answer really presents itself

tx
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by harsh.gupta.175 » Fri Dec 25, 2009 6:05 pm
Hi All,

All solutions are great. And i suppose the Solution from Stuart would take the least time.

Here is how i solved the problem.

1. If there are 30 chairs then the distance between any two chairs is c/30 where c is the cirsumference of the circle.
2. Now lets assume the first chair as the starting point(origin) of a traversal around the circle or 0.
3. The length of the arc between the third chair and the origin(0) should be equal to the length of the arc between mid circle (c/2) and the 18th chair.[Since these chairs are 180 derees apart)
4. length of the first arc = (3-1) c/30 - 0(origin) =2c/30=c/15
5. length of the second arc = (18-1)c/30 -c/2(midway)= 17c/30 - c/2=2c/30=c/15

Hence 30 is the correct answer.
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by harsh.gupta.175 » Fri Dec 25, 2009 6:06 pm
Hi All,

All solutions are great. And i suppose the Solution from Stuart would take the least time.

Here is how i solved the problem.

1. If there are 30 chairs then the distance between any two chairs is c/30 where c is the cirsumference of the circle.
2. Now lets assume the first chair as the starting point(origin) of a traversal around the circle or 0.
3. The length of the arc between the third chair and the origin(0) should be equal to the length of the arc between mid circle (c/2) and the 18th chair.[Since these chairs are 180 derees apart)
4. length of the first arc = (3-1) c/30 - 0(origin) =2c/30=c/15
5. length of the second arc = (18-1)c/30 -c/2(midway)= 17c/30 - c/2=2c/30=c/15

Hence 30 is the correct answer.
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by harsh.gupta.175 » Fri Dec 25, 2009 6:06 pm
Hi All,

All solutions are great. And i suppose the Solution from Stuart would take the least time.

Here is how i solved the problem.

1. If there are 30 chairs then the distance between any two chairs is c/30 where c is the cirsumference of the circle.
2. Now lets assume the first chair as the starting point(origin) of a traversal around the circle or 0.
3. The length of the arc between the third chair and the origin(0) should be equal to the length of the arc between mid circle (c/2) and the 18th chair.[Since these chairs are 180 derees apart)
4. length of the first arc = (3-1) c/30 - 0(origin) =2c/30=c/15
5. length of the second arc = (18-1)c/30 -c/2(midway)= 17c/30 - c/2=2c/30=c/15

Hence 30 is the correct answer.
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by harsh.gupta.175 » Fri Dec 25, 2009 6:24 pm
Hi All,

All solutions are great. And i suppose the Solution from Stuart would take the least time.

Here is how i solved the problem.

1. If there are 30 chairs then the distance between any two chairs is c/30 where c is the cirsumference of the circle.
2. Now lets assume the first chair as the starting point(origin) of a traversal around the circle or 0.
3. The length of the arc between the third chair and the origin(0) should be equal to the length of the arc between mid circle (c/2) and the 18th chair.[Since these chairs are 180 derees apart)
4. length of the first arc = (3-1) c/30 - 0(origin) =2c/30=c/15
5. length of the second arc = (18-1)c/30 -c/2(midway)= 17c/30 - c/2=2c/30=c/15

Hence 30 is the correct answer.
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