Question

In a business meeting, each person shakes hands with each other person once. A person arrives after 5 people have left and he shakes hands only with those present. If the total number of handshakes is exactly 100, then the initial number of people in the party, is

• A. 12
• B. 13
• C. 14
• D. 15
• E. 16

Hint:

For quadratic equations where the constant term is around 200 (like 210), look for factors near \(\sqrt{200} \approx 14.1\). Here, \(14 \times 15 = 210\) makes the solution immediate.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem involves two sets of handshakes:
1. Handshakes among the initial \(n\) people.
2. Handshakes performed by the newcomer with the remaining people.
The sum of these handshakes equals 100.

Step 2: Key Formula or Approach:
The number of handshakes between \(x\) people is given by \(^xC_2 = \frac{x(x-1)}{2}\).

Step 3: Detailed Explanation:
Let the initial number of people be \(n\).
The number of handshakes between these \(n\) people is \(\frac{n(n-1)}{2}\).
After 5 people leave, there are \(n - 5\) people remaining.
A newcomer arrives and shakes hands with these \(n - 5\) people.
Total handshakes = (Initial handshakes) + (Newcomer handshakes) = 100.
\[ \frac{n(n-1)}{2} + (n - 5) = 100 \]
\[ \frac{n^2 - n + 2n - 10}{2} = 100 \]
\[ n^2 + n - 10 = 200 \]
\[ n^2 + n - 210 = 0 \]
Factoring the quadratic:
We need two numbers that multiply to -210 and add to 1. These are 15 and -14.
\[ (n + 15)(n - 14) = 0 \]
Since \(n\) must be positive, \(n = 14\).

Step 4: Final Answer:
The initial number of people was 14.