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Answer to Q34

At a party, everyone shook hands with everybody else. There were 4005 handshakes. How many people were at the party?

The number of handshakes between people is always the sum of 1+2+3+...+(n-1). n = number of people. Say there are 5 people, handshake between them will be 1+2+3+4 = 10 (try that with your friends 😀)

The sum of 1 to n of any number is (n(n+1))÷2. For example 1+2+3+4+5 = 15 and (5(5+1))÷2 = (5*6)÷2 = 15

So, to find the number of people at the party

(n(n+1))÷2 = 4005

Cross multiplying:
n(n+1) = 4005x2 = 8010

Expanding the bracket:
n² + n = 8010
n² + n - 8010 = 0

Solving that using the Almighty Formula (I don't think I need to go into that details)

n = 90 and -89

Ofcourse there can be negative people at the party so answer is 90 people

Answer: 90

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515 viewsAug 16, 2019 at 14:52

Answer to Q34

At a party, everyone shook hands with everybody else. There were 4005 handshakes. How many people were at the party?

The number of handshakes between people is always the sum of 1+2+3+...+(n-1). n = number of people. Say there are 5 people, handshake between them will be 1+2+3+4 = 10 (try that with your friends 😀)

The sum of 1 to n of any number is (n(n+1))÷2. For example 1+2+3+4+5 = 15 and (5(5+1))÷2 = (5*6)÷2 = 15

So, to find the number of people at the party

(n(n+1))÷2 = 4005

Cross multiplying:
n(n+1) = 4005x2 = 8010

Expanding the bracket:
n² + n = 8010
n² + n - 8010 = 0

Solving that using the Almighty Formula (I don't think I need to go into that details)

n = 90 and -89

Ofcourse there can be negative people at the party so answer is 90 people

Answer: 90

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