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Joined: 08 May 2017
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A group of people is seated at a table at which a toast is made. Follo  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 25% (02:42) correct 75% (02:36) wrong based on 100 sessions

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A group of people is seated at a table at which a toast is made. Following the toast, each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people, how many people are seated at the table?

1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.
Math Expert V
Joined: 02 Aug 2009
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A group of people is seated at a table at which a toast is made. Follo  [#permalink]

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1
Graina wrote:
A group of people is seated at a table at which a toast is made. Following the toast, each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people, how many people are seated at the table?

1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.

let there be n people, so basically we are asked to find 2 out of n people.
nC2..

let us see each choice now...
1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.
Now, the strength becomes n-2, so ways you can choose 2 out of n-2 = (n-2)C2
So, $$17\leq{nC2-(n-2)C2}\leq{19}$$.
let us find (n)C2-(n-2)C2 => $$\frac{n!}{(n-2)!2!}-\frac{(n-2)!}{(n-2-2)!2!}=\frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=\frac{n^2-n-n^2+5n-6}{2}=2n-3$$..
so $$17\leq{2n-3}\leq{19}=20\leq{2n}\leq{22}$$=$$10\leq{n}\leq{11}$$..
so n can be 10 or 11
insuff..

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.
Now, the strength becomes n-2, so ways you can choose 2 out of n-2 = (n-2)C2
So, $$18\leq{nC2-(n-2)C2}\leq{21}$$.
let us find nC2-(n-2)C2 => $$\frac{n!}{(n-2)!2!}-\frac{(n-2)!}{(n-2-2)!2!}=\frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=\frac{n^2-n-n^2+5n-6}{2}=2n-3$$..
so $$18\leq{2n-3}\leq{21}=21\leq{2n}\leq{24}$$=$$10.5\leq{n}\leq{12}$$..
so n can be 11 or 12
insuff..

Combined..
n can be 11..
sufficient..

C

You can also work with taking different values of n..
find 8C2-6C2 and then 9C2-7C2 and so on..
example 9C2-7C2 = 9*4-7*3=36-21=15, so increase n by 1..
10C2-8C2=5*9-4*7=45-28=17..
11C2-9C2=11*5-9*4=55-36=19..
so n as 10 and 11 are correct.

As we can see with increase of 1 in n, the possible value of n increases by 2..
so, if 10C2-8C2=17, 11C2-9C2=19, 12C2-11C2=21..
so statement I gives us 17 and 19 and statement II gives us 19 and 21..
C
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Re: A group of people is seated at a table at which a toast is made. Follo  [#permalink]

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Graina wrote:
A group of people is seated at a table at which a toast is made. Following the toast, each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people, how many people are seated at the table?

1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.

Can some one please explain, "each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people" ?
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Joined: 21 Oct 2018
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Re: A group of people is seated at a table at which a toast is made. Follo  [#permalink]

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jashandeep2332 wrote:
Graina wrote:
A group of people is seated at a table at which a toast is made. Following the toast, each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people, how many people are seated at the table?

1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.

Can some one please explain, "each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people" ?

to explain above statement , for example take 5 persons seated and named as a,b,c,d,e

first person a can clink with b,c,d,e .so on total there will be 4 clinks
in the same way, second person can clink with b,c,e on total there will be 3 clinks... so on

Manager  S
Joined: 09 Jul 2018
Posts: 63
Re: A group of people is seated at a table at which a toast is made. Follo  [#permalink]

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chetan2u wrote:
Graina wrote:
A group of people is seated at a table at which a toast is made. Following the toast, each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people, how many people are seated at the table?

1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.

let there be n people, so basically we are asked to find 2 out of n people.
nC2..

let us see each choice now...
1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.
Now, the strength becomes n-2, so ways you can choose 2 out of n-2 = (n-2)C2
So, $$17\leq{(n-2)C2-nC2}\leq{19}$$.
let us find (n-2)C2-nC2 => $$\frac{n!}{(n-2)!2!}-\frac{(n-2)!}{(n-2-2)!2!}=\frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=\frac{n^2-n-n^2+5n-6}{2}=2n-3$$..
so $$17\leq{2n-3}\leq{19}=20\leq{2n}\leq{22}$$=$$10\leq{n}\leq{11}$$..
so n can be 10 or 11
insuff..

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.
Now, the strength becomes n-2, so ways you can choose 2 out of n-2 = (n-2)C2
So, $$18\leq{(n-2)C2-nC2}\leq{21}$$.
let us find (n-2)C2-nC2 => $$\frac{n!}{(n-2)!2!}-\frac{(n-2)!}{(n-2-2)!2!}=\frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=\frac{n^2-n-n^2+5n-6}{2}=2n-3$$..
so $$18\leq{2n-3}\leq{21}=21\leq{2n}\leq{24}$$=$$10.5\leq{n}\leq{12}$$..
so n can be 11 or 12
insuff..

Combined..
n can be 11..
sufficient..

C

You can also work with taking different values of n..
find 8C2-6C2 and then 9C2-7C2 and so on..
example 9C2-7C2 = 9*4-7*3=36-21=15, so increase n by 1..
10C2-8C2=5*9-4*7=45-28=17..
11C2-9C2=11*5-9*4=55-36=19..
so n as 10 and 11 are correct.

As we can see with increase of 1 in n, the possible value of n increases by 2..
so, if 10C2-8C2=17, 11C2-9C2=19, 12C2-11C2=21..
so statement I gives us 17 and 19 and statement II gives us 19 and 21..
C

can u please explain this equation $$17\leq{(n-2)C2-nC2}\leq{19}$$.
why its (n-2)C2-nC2?
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Posts: 1417
Location: India
Re: A group of people is seated at a table at which a toast is made. Follo  [#permalink]

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1
ritu1009 wrote:
chetan2u wrote:
Graina wrote:
A group of people is seated at a table at which a toast is made. Following the toast, each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people, how many people are seated at the table?

1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.

let there be n people, so basically we are asked to find 2 out of n people.
nC2..

let us see each choice now...
1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.
Now, the strength becomes n-2, so ways you can choose 2 out of n-2 = (n-2)C2
So, $$17\leq{(n-2)C2-nC2}\leq{19}$$.
let us find (n-2)C2-nC2 => $$\frac{n!}{(n-2)!2!}-\frac{(n-2)!}{(n-2-2)!2!}=\frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=\frac{n^2-n-n^2+5n-6}{2}=2n-3$$..
so $$17\leq{2n-3}\leq{19}=20\leq{2n}\leq{22}$$=$$10\leq{n}\leq{11}$$..
so n can be 10 or 11
insuff..

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.
Now, the strength becomes n-2, so ways you can choose 2 out of n-2 = (n-2)C2
So, $$18\leq{(n-2)C2-nC2}\leq{21}$$.
let us find (n-2)C2-nC2 => $$\frac{n!}{(n-2)!2!}-\frac{(n-2)!}{(n-2-2)!2!}=\frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=\frac{n^2-n-n^2+5n-6}{2}=2n-3$$..
so $$18\leq{2n-3}\leq{21}=21\leq{2n}\leq{24}$$=$$10.5\leq{n}\leq{12}$$..
so n can be 11 or 12
insuff..

Combined..
n can be 11..
sufficient..

C

You can also work with taking different values of n..
find 8C2-6C2 and then 9C2-7C2 and so on..
example 9C2-7C2 = 9*4-7*3=36-21=15, so increase n by 1..
10C2-8C2=5*9-4*7=45-28=17..
11C2-9C2=11*5-9*4=55-36=19..
so n as 10 and 11 are correct.

As we can see with increase of 1 in n, the possible value of n increases by 2..
so, if 10C2-8C2=17, 11C2-9C2=19, 12C2-11C2=21..
so statement I gives us 17 and 19 and statement II gives us 19 and 21..
C

can u please explain this equation $$17\leq{(n-2)C2-nC2}\leq{19}$$.
why its (n-2)C2-nC2?

Hello

If there are total 'n' people, number of clinks produced will be nC2.
If there are total 'n-2' people, number of clinks produced will be (n-2)C2.

First statement says that in case of 2 fewer people, the 'difference in number of clinks produced' will be between 17 and 19 inclusive.
And what is the difference in number of clinks produced ? It will be nC2 - (n-2)C2 only.

Thats why Chetan has made the equation:
17 <= [nC2 - (n-2)C2] <= 19

(I think just by typo Chetan has written (n-2)C2 - nC2.. instead he should have written nC2 - (n-2)C2.
But his calculation is completely fine, and all else that he has written is also perfect).
Intern  B
Joined: 28 Sep 2018
Posts: 6
Re: A group of people is seated at a table at which a toast is made. Follo  [#permalink]

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This was my thought process without the need of any calculations

1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

--let the current people be n+2,
--now statement 1 says
17<=n/2<=19
=> 34<=n<=38
=> 34<= current_people <=38 ----------------------II

Thus current people = n+2
=> 36<= n+2 <= 40

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.
--let the current people be n+2,
--now statement 1 says
18<= n/2 <=21
=> 36<= n <=42

Thus current people = n+2
=> 38<= n+2 <= 44
=> 38<= current_people <= 44 ----------------- I

Combining I and II

current_people = 38
Hence C
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Posts: 156
GMAT 1: 610 Q48 V25 GMAT 2: 690 Q50 V32 GMAT 3: 710 Q50 V36 Re: A group of people is seated at a table at which a toast is made. Follo  [#permalink]

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chetan2u wrote:
Graina wrote:
A group of people is seated at a table at which a toast is made. Following the toast, each person must clink glasses exactly once with each of the other people at the table. If each clink is produced by the glasses of only two people, how many people are seated at the table?

1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.

let there be n people, so basically we are asked to find 2 out of n people.
nC2..

let us see each choice now...
1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.
Now, the strength becomes n-2, so ways you can choose 2 out of n-2 = (n-2)C2
So, $$17\leq{nC2-(n-2)C2}\leq{19}$$.
let us find (n)C2-(n-2)C2 => $$\frac{n!}{(n-2)!2!}-\frac{(n-2)!}{(n-2-2)!2!}=\frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=\frac{n^2-n-n^2+5n-6}{2}=2n-3$$..
so $$17\leq{2n-3}\leq{19}=20\leq{2n}\leq{22}$$=$$10\leq{n}\leq{11}$$..
so n can be 10 or 11
insuff..

2. If two fewer people were seated at the table, there would be at least 18 but no more than 21 fewer clinks.
Now, the strength becomes n-2, so ways you can choose 2 out of n-2 = (n-2)C2
So, $$18\leq{nC2-(n-2)C2}\leq{21}$$.
let us find nC2-(n-2)C2 => $$\frac{n!}{(n-2)!2!}-\frac{(n-2)!}{(n-2-2)!2!}=\frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=\frac{n^2-n-n^2+5n-6}{2}=2n-3$$..
so $$18\leq{2n-3}\leq{21}=21\leq{2n}\leq{24}$$=$$10.5\leq{n}\leq{12}$$..
so n can be 11 or 12
insuff..

Combined..
n can be 11..
sufficient..

C

You can also work with taking different values of n..
find 8C2-6C2 and then 9C2-7C2 and so on..
example 9C2-7C2 = 9*4-7*3=36-21=15, so increase n by 1..
10C2-8C2=5*9-4*7=45-28=17..
11C2-9C2=11*5-9*4=55-36=19..
so n as 10 and 11 are correct.

As we can see with increase of 1 in n, the possible value of n increases by 2..
so, if 10C2-8C2=17, 11C2-9C2=19, 12C2-11C2=21..
so statement I gives us 17 and 19 and statement II gives us 19 and 21..
C

HI chetan2u
1. If two fewer people were seated at the table, there would be at least 17 but no more than 19 fewer clinks.

I cannot understand this statement.
According to me, it means:
if there are n people
then it becomes n-2

So 17<n-2c2<19
if there are 2 fewer people seated at the table, there would min 17 and max of 19.
Why are you saying that th DIFFERENCE will be min 17 and max 19.
you could simply also say that with 2 fewer people, the clinks shall be min 17 and max 19.

Please tell me if I am going wrong t=somewhere.

Regards Re: A group of people is seated at a table at which a toast is made. Follo   [#permalink] 12 Dec 2018, 03:14
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