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

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?

- \(24\)
- \(26\)
- \(28\)

SET

Algebra

Cardinal number

But try the problem first...

Answer: \(24\)

Source

Suggested Reading

TOMATO, Problem 53

Challenges and Thrills in Pre College Mathematics

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

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

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