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In a class of 400 students, 180 read English, 371 read Sanskrit, 270 read Hindi, at least how many students read all the three?

No other information is given about how many students reading any two of the given subjects. I want to use $\#(A\cup B \cup C)$ formula but I don't have sufficient information about $\#(A\cap B)$ or $\#(B \cap C)$ or $\#(A\cap C).$ I don't have other ideas for dealing with this problem. How can I solve this problem?

BruceET
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tejasree
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  • I want to use n(AuBuC) formula but i dont have sufficient information about n(AnB) or n(BnC) or n(AnC). I dont have other idea in which the the problem can be dealt @YukiJ – tejasree May 16 '18 at 09:34
  • yes, i have practised problems on de morgan's laws. @BruceET – tejasree May 16 '18 at 09:34

3 Answers3

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$400 - 371 = 29$ students do not read sanskrit.

$270$ students read Hindi. At most $29$ of them do not read Sanskrit. So at least $270 - 29 = 241$ students read both Hindi and Sanskrit.

So at most $400 - 241 = 159$ students don't read both.

$180$ read english and at most $159$ of the don't read both sanskrit and hindi. So at least $180 - 159 = 21$ of the students who read english read both sanskrit and hindi.

So at the very least $21$ students read all three.

....

To do it by exclusion/inclusion.

$|E \cup S \cup H| = 400$

$|E \cup S \cup H|= |E|+ |S| + |H| - (|E\cap S| + |E \cap H| + |S \cap H|) + |E \cap S \cap H|=180 + 371 + 270 - (|E\cap S| + |E \cap H| + |S \cap H|) + |E \cap S \cap H|$

So $|E \cap S \cap H| = (|E\cup S| + |E \cup H| + |S \cup H| )- 421$.

Now $400 \ge |E\cup H| = |E| + |H| - |E\cap H|= 180+270- |E\cap H|$ so $|E\cap H|\ge 50$.

$400\ge |E\cap S| = 180 + 371 + |E\cap S|$ so $|E\cap S| \ge 151$

$400\ge |H\cap S| = 270 + 371 + |H\cap S|$ so $|H\cap S| \ge 241$

So $|E \cup S \cup H| = (|E\cap S| + |E \cap H| + |S \cap H| )- 421\ge 50 + 151 + 241 - 421 = 21$

tejasree
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fleablood
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Let $A$ be the set of people who don't read English, $B$ the set of people who don't read Sanskrit, $C$ the set of people who don't read Hindi. The number of people who don't read all three languages is $$\#(A\cup B\cup C)\le\#(A)+\#(B)+\#(C)=220+29+130=379,$$ so the number of people who do read all three languages is at least $400-379=\boxed{21}.$

bof
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Hint: You have a room with 400 benches, each of which can hold up to 3 people, and 180+371+270 people must find a seat -- how many benches are full at a minimum?

You can think people enter the room one by one, and never sit on a bench if there's a bench with more space available. How many people must enter to have one person on each bench? How many more to have two? After that point, every additional person fills a bench!

Anonymous
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