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Question

In a university, out of 100 students 15 offered Mathematics only; 12 offered statistics only; 8 offered Physics only; 40 offered Physics and Mathematics; 20 offered Physics and Statistics; 10 offered Mathematics and Statistics; 65 offered Physics. What is the number of students who did not offer any of the three subjects ?

The correct answer is
1

Identify Given Information

Total number of students = 100.

  • Students offering Mathematics only = 15
  • Students offering Statistics only = 12
  • Students offering Physics only = 8
  • Students offering Physics and Mathematics = 40 (This represents $|P \cap M|$)
  • Students offering Physics and Statistics = 20 (This represents $|P \cap S|$)
  • Students offering Mathematics and Statistics = 10 (This represents $|M \cap S|$)
  • Students offering Physics = 65 (This represents $|P|$)

We need to find the number of students who did not offer any of the three subjects.

Calculate Intersection of All Three Subjects

Let $x$ be the number of students offering all three subjects (Mathematics, Statistics, and Physics), i.e., $x = |M \cap S \cap P|$.

The number of students offering Physics ($|P|$) can be broken down:

$|P| = (\text{Physics only}) + (\text{Physics and Maths only}) + (\text{Physics and Stats only}) + (\text{Physics, Maths, and Stats})$

We know:

  • Physics only = 8
  • Physics and Maths only = $|P \cap M| - |M \cap S \cap P| = 40 - x$
  • Physics and Stats only = $|P \cap S| - |M \cap S \cap P| = 20 - x$
  • Physics, Maths, and Stats = $x$

Substitute these into the equation for $|P|$:

$65 = 8 + (40 - x) + (20 - x) + x$

$65 = 8 + 40 - x + 20 - x + x$

$65 = 68 - x$

$x = 68 - 65$

$x = 3$

So, 3 students offered all three subjects.

Determine Total Offering At Least One Subject

Now, calculate the number of students in each distinct category:

  • Mathematics only = 15
  • Statistics only = 12
  • Physics only = 8
  • Physics and Mathematics only = $40 - 3 = 37$
  • Physics and Statistics only = $20 - 3 = 17$
  • Mathematics and Statistics only = $10 - 3 = 7$
  • Physics, Mathematics, and Statistics = 3

The total number of students offering at least one subject is the sum of these distinct groups:

$|M \cup S \cup P| = 15 + 12 + 8 + 37 + 17 + 7 + 3$

$|M \cup S \cup P| = 99$

Find Number Offering No Subjects

The number of students who did not offer any of the three subjects is the total number of students minus the number of students offering at least one subject:

Number offering none = Total students - $|M \cup S \cup P|$

Number offering none = $100 - 99$

Number offering none = 1

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