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$500$ students are taking one or more course out of Chemistry, Physics, and Mathematics. Registration records indicate course enrollment as follows: Chemistry $(329)$, Physics $(186)$, and Mathematics $(295)$. Chemistry and Physics $(83)$, Chemistry and Mathematics $(217)$, and Physics and Mathematics (63). How many students are taking all $3$ subjects?

  1. $37$
  2. $43$
  3. $47$
  4. $53$
in Numerical Ability by Active (2.4k points)
edited by | 89 views
Migrated from GO Electronics 6 months ago by Arjun
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Use this formula

P(AuBuC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

where we want to find P(AnBnC)

500-329-186-295+83+217+63 = 53

2 Answers

+2 votes
Best answer
Answer: $D.$ 53

Total number of students $= n(P)+n(C)+n(M)-n(P \cap C)-n(P \cap M)-n(C \cap M)+n(P \cap C \cap M)$

$\implies 500=329+295+186-217-83-63+x$
$\implies x=53.$
by Boss (15.2k points)
selected by
+1 vote

All 3 subjects =53

by Active (1.2k points)

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