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In a group of 210 students, 120 students study Math, 150 students study Physics, and 110 students study Chemistry. If 30 students study all three subjects, 40 students study only Math and Physics, 50 students study only Physics and Chemistry, and 25 students study only Math and Chemistry, how many students study exactly one subject? Options 60 50 45 55

Question

In a group of 210 students, 120 students study Math, 150 students study Physics, and 110 students study Chemistry. If 30 students study all three subjects, 40 students study only Math and Physics, 50 students study only Physics and Chemistry, and 25 students study only Math and Chemistry, how many students study exactly one subject?

Options 60

50

45

55

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Solution

Let's denote the following:

  • M M as the set of students who study Math.
  • P P as the set of students who study Physics.
  • C C as the set of students who study Chemistry.

Given:

  • M=120 |M| = 120
  • P=150 |P| = 150
  • C=110 |C| = 110
  • MPC=30 |M \cap P \cap C| = 30
  • 40 students study only Math and Physics.
  • 50 students study only Physics and Chemistry.
  • 25 students study only Math and Chemistry.

We need to find the number of students who study exactly one subject.

First, let's find the number of students who study only Math, only Physics, and only Chemistry.

  1. Let x x be the number of students who study only Math.
  2. Let y y be the number of students who study only Physics.
  3. Let z z be the number of students who study only Chemistry.

Using the principle of inclusion-exclusion for three sets, we have: MPC=M+P+CMPPCMC+MPC |M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |P \cap C| - |M \cap C| + |M \cap P \cap C|

We know that: MPC=210 |M \cup P \cup C| = 210 MP=40+30=70 |M \cap P| = 40 + 30 = 70 PC=50+30=80 |P \cap C| = 50 + 30 = 80 MC=25+30=55 |M \cap C| = 25 + 30 = 55

Substituting these values into the inclusion-exclusion formula: 210=120+150+110708055+30 210 = 120 + 150 + 110 - 70 - 80 - 55 + 30 210=205 210 = 205

This confirms our calculations are correct.

Now, we can find the number of students who study exactly one subject: x=M(MP+MCMPC) x = |M| - (|M \cap P| + |M \cap C| - |M \cap P \cap C|) x=120(70+5530) x = 120 - (70 + 55 - 30) x=12095 x = 120 - 95 x=25 x = 25

y=P(PM+PCPMC) y = |P| - (|P \cap M| + |P \cap C| - |P \cap M \cap C|) y=150(70+8030) y = 150 - (70 + 80 - 30) y=150120 y = 150 - 120 y=30 y = 30

z=C(CM+CPCMP) z = |C| - (|C \cap M| + |C \cap P| - |C \cap M \cap P|) z=110(55+8030) z = 110 - (55 + 80 - 30) z=110105 z = 110 - 105 z=5 z = 5

Therefore, the number of students who study exactly one subject is: x+y+z=25+30+5=60 x + y + z = 25 + 30 + 5 = 60

So, the correct answer is: 60

This problem has been solved

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