Venn Diagrams: A group of 102 students took exams in...

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judocallin02

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A group of 102 students took examinations in Chinese, English, and Mathematics. Among them, 92 passed Chinese, 75 English, and 63 Mathematics; at most 65 passed Chinese and English, at most 54 Chinese and Mathematics, and at most 48 English and Mathematics. Find the largest possible number of students who could have passed all three subjects.

I tried making a Venn Diagram but I can't seem to complete it. All I was able to do is to put the cardinalities of the given set intersections in the venn diagram.


I need help. Please


Thanks...
 
Re: Venn Diagrams

judocallin02 said:
A group of 102 students took examinations in Chinese, English, and Mathematics. Among them, 92 passed Chinese, 75 English, and 63 Mathematics; at most 65 passed Chinese and English, at most 54 Chinese and Mathematics, and at most 48 English and Mathematics. Find the largest possible number of students who could have passed all three subjects.

I tried making a Venn Diagram but I can't seem to complete it. All I was able to do is to put the cardinalities of the given set intersections in the venn diagram.


I need help. Please


Thanks...
Click to expand...

You don't need Venn diagram for this problem.

Just logically think through - people who passed Chinese + Math + English

must have passed English + Mathematics too!!
 
Re: Venn Diagrams

judocallin02 said:
Just to clarify,

so the answer will be 102 or 48? <<< You tell me!!! Why ? Why do you have "or"?
Click to expand...
 
Re: Venn Diagrams

Hello, judocallin02!

A group of 102 students took examinations in Chinese, English, and Mathematics.

Among them:
. . 92 passed Chinese
. . 75 English
. . 63 Mathematics
. . at most 65 passed Chinese and English
. . at most 54 Chinese and Mathematics
. . at most 48 English and Mathematics.

Find the largest possible number of students who could have passed all three subjects.
Click to expand...

\(\displaystyle \text{You should be familiar with this formula: }\;n(A \cup B ) \;=\;n(A) + n(B) - n(A \cap B)\)

\(\displaystyle \text{Are you familiar with the three-set version of the formula?}\)
. . \(\displaystyle n(A\;\cup\;B\;\cup\;C) \;=\; n(A) \;+\; n(B) \;+\; n(C) \;-\; n(A\;\cap\;B) \;-\;n(A\;\cap\;C)\;-\;n(B\;\cap\;C)\;+\;n(A\;\cap\;B\;\cap\;C)\)


\(\displaystyle \text{We have:}\)

\(\displaystyle \underbrace{n(C \cup E \cup M)}_{102} \;=\;\underbrace{n(C)}_{92} + \underbrace{n(E)}_{75} + \underbrace{n(M)}_{63} - \underbrace{n(C\cap E)}_{65} - \underbrace{n(C\cap M)}_{54} - \underbrace{n(E\cap M)}_{48} +\; n(C\;\cap\;E\;\cap\;M)\)

. - - . . . . .\(\displaystyle 102 \;=\;63 + n(C \cap E \cap M)\)

\(\displaystyle n(C \cap E \cap M) \;=\;39\)


\(\displaystyle \text{Therefore, at most 39 students passed all three subjects.}\)

 
Re: Venn Diagrams

soroban said:
Hello, judocallin02!

A group of 102 students took examinations in Chinese, English, and Mathematics.

Among them:
. . 92 passed Chinese
. . 75 English
. . 63 Mathematics
. . at most 65 passed Chinese and English
. . at most 54 Chinese and Mathematics
. . at most 48 English and Mathematics.

Find the largest possible number of students who could have passed all three subjects.
Click to expand...

\(\displaystyle \text{You should be familiar with this formula: }\;n(A \cup B ) \;=\;n(A) + n(B) - n(A \cap B)\)

\(\displaystyle \text{Are you familiar with the three-set version of the formula?}\)
. . \(\displaystyle n(A\;\cup\;B\;\cup\;C) \;=\; n(A) \;+\; n(B) \;+\; n(C) \;-\; n(A\;\cap\;B) \;-\;n(A\;\cap\;C)\;-\;n(B\;\cap\;C)\;+\;n(A\;\cap\;B\;\cap\;C)\)


\(\displaystyle \text{We have:}\)

\(\displaystyle \underbrace{n(C \cup E \cup M)}_{102} \;=\;\underbrace{n(C)}_{92} + \underbrace{n(E)}_{75} + \underbrace{n(M)}_{63} - \underbrace{n(C\cap E)}_{65} - \underbrace{n(C\cap M)}_{54} - \underbrace{n(E\cap M)}_{48} +\; n(C\;\cap\;E\;\cap\;M)\)

. - - . . . . .\(\displaystyle 102 \;=\;63 + n(C \cap E \cap M)\)

\(\displaystyle n(C \cap E \cap M) \;=\;39\)


\(\displaystyle \text{Therefore, at most 39 students passed all three subjects.}\)
Click to expand...

I have a problem with this problem statement.

When the -ve numbers ( e. g.\(\displaystyle n(\cap E \cap M)\)) are at most(= maximum) - then the resulting positive number (\(\displaystyle n(C \cap E \cap M)\)) would be minimum (= at least) or smallest.

The largest # to pass all three would be the minimum of those maximum numbers.
 
Sorry, what I mean by "or" - is the answer would be 48 or is the answer would be 102?


The minimun of those maximum numbers is 34 which is also the maximun possible number of students who could have passed all those subjects. Is that what you mean?
 
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