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: \(24\)

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

## Other useful links

- https://www.cheenta.com/quadratic-equation-problem-amc-8-2009-problem-23/
- https://www.youtube.com/watch?v=lHgyrwyZcpw

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