Try this beautiful problem Based on Set Theory useful for ISI B.Stat Entrance.

Set Theory| ISI B.Stat Entrance | Problem 53


There were  41 candidates in an examination and each candidate was examined in algebra, geometry and calculus. It was found that 12 candidates failed in algebra, 7 failed in geometry and 8 failed in calculus , 2 in geometry  and calculus , 3 in calculus and algebra , 6 in algebra and geometry, whereas  only  1 failed in one of the three  subjects. Then, find  the number of candidates  who passed in all three subjects?

  • \(24\)
  • \(26\)
  • \(28\)

Key Concepts


SET

Algebra

Cardinal number

Check the Answer


But try the problem first…

Answer: \(24\)

Source
Suggested Reading

TOMATO, Problem 53

Challenges and Thrills in Pre College Mathematics

Try with Hints


First hint

use set theory concept

Can you now finish the problem ……….

Second Hint

we assume that A ,G,C be the sets of the students who failed on algebra ,geometry and calculus respectively.

Find the complement of \(N(A \cup G \cup C)\)

can you finish the problem……..

Final Step

Let S be the set of total students i.e N(s) = 41

we assume that A ,G,C be the sets of the students who failed on algebra ,geometry and calculus respectively.

Therefore

N(A)=12

N(G)=7

N(C)=8

and

\(N(A \cap G \cap C)\)=1

\(N(G \cap C)\) =2 ,

\(N(C\cap A)\) =3

\(N(A\cap G)\) =6

\(N(A \cup G \cup C)\)=\(N(A) + N(G) +N(C) -N(G \cap C) -N (C \cap A) -N(C \cap A) + N(A \cap G \cap C)\)=12 + 7+8-2-6-3+1=17

Therefore complement of \(N(A \cup G \cup C)\) =41-17=24

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