Question

In a business meeting, each person shakes hands with each other person, with the exception of Mr. L. Since Mr. L arrives after some people have left, he shakes hands only with those present. If the total number of handshakes is exactly 100, how many people left the meeting before Mr. L arrived? (Nobody shakes hands with the same person more than once.

Answer 1

It takes \(2\) persons for one handshake. If \(n\) persons attended the party "excluding" Mr. L and \(k\) of them left before arrival of Mr.L. Then \[\large \binom{n}{2} + n-k = 100\]

Answer 2

I think Mr L was also involved in shaking hands in the above equation it is like you include only the initially present people.

Answer 3

Mr L was not counted in first "n" persons' handshakes

Answer 4

By the time Mr L comes to the party, only \(n-k\) persons are present so he shakes hands with them

Answer 5

adding \(\binom{n}{2}\) and \(n-k\) gives us the total handshakes which is \(100\)

Answer 6

I apply with this but I am confused though. please check it

Answer 7

Let n be the number of people at the party.
Each person shakes hands with every other person, so each person shakes hands with (n - 1) people, a possible total of n(n - 1) handshakes.
But when person A shakes hands with person B, B also shakes hands with A, so each handshake would be counted twice.
→ number_of_handshakes = n(n - 1)/2

total number of handshakes is 100
→ n(n - 1)/2 = 100
→ n(n - 1) = 200
→ n^2 - n - 200 = 0
As 200 is negative, one factor is positive and one is negative, so we need the factor pair of 200 which has a difference of 1, namely: 20 x 10
→ (n - 20)(n + 10) = 0
→ n = 20 or -10

There cannot be a negative number of people → there are 20 people present.

Answer 8

@praxer your last line agrees with mine, but ur solution looks messy

Answer 9

yes it does. :(

Answer 10

actually no, you/me have mistake somewhere. I am getting this : \[\large \binom{n}{2} + n-k = 100 \]

\[\large \dfrac{n(n-1)}{2} + n-k = 100 \]

\[\large n(n+1)= 2(k+100) \]

Answer 11

so u get

Answer 12

can you help me swamijatin

Answer 13

in math on my question

Answer 14

@rational can you please explain your first proposition. I mean how you gave that. please explain :)

Answer 15

me

Answer 16

what is your queston @urbanmorgans

Answer 17

in math first question

Answer 18

can you help me

Answer 19

i am not getting to question

Answer 20

number of handshakes with \(n\) people is given by \(\large \binom{n}{2}\)

you familiar with this right ?

Answer 21

yes :)

Answer 22

why not it's called malinda

Answer 23

Let n be the number of people at the party.
Each person shakes hands with every other person, so each person shakes hands with (n - 1) people, a possible total of n(n - 1) handshakes.
But when person A shakes hands with person B, B also shakes hands with A, so each handshake would be counted twice.
→ number_of_handshakes = n(n - 1)/2

total number of handshakes is 100
→ n(n - 1)/2 = 100
→ n(n - 1) = 200
→ n^2 - n - 200 = 0
As 200 is negative, one factor is positive and one is negative, so we need the factor pair of 200 which has a difference of 1, namely: 20 x 10
→ (n - 20)(n + 10) = 0
→ n = 20 or -10

There cannot be a negative number of people → there are 20 people present.

Answer 24

sorry


















melinda has saved

Answer 25

\[\large n(n+1)= 2(k+100) \]

By trial and error i am getting \(n=14\) and \(k=5\)
so there are \(14\) people at the party excluding Mr L
and \(5\) ppl have left before the arrival of Mr L

Answer 26

thank you :)

Answer 27

yw