Dark Mode

Kathleen
asked
in Set Theory & Algebra
Sep 30, 2014
recategorized
Apr 22, 2021
by Lakshman Patel RJIT

5,853 views
35 votes

Out of a group of $21$ persons, $9$ eat vegetables, $10$ eat fish and $7$ eat eggs. $5$ persons eat all three. How many persons eat at least two out of the three dishes?

33 votes

Best answer

$N(A\cup B \cup C) = N(A) + N(B) + N(C) - N(A\cap B) - N(A \cap C) - N(B\cap C) + N(A\cap B\cap C)$

Let $Y$ be the no. of persons who eat at least one item. $21 - Y$ people do not eat anything.

$Y = 9 + 10 + 7 - [N(A\cap B) + N(A\cap C) + N(B\cap C)] + 5$

$[N(A\cap B) + N(A\cap C) + N(B\cap C)] = 31 - Y$.

Now, these include the no. of persons who eat all 3 items thrice. So, excluding those, we get, no. of persons who eat at least two items (by adding the no. of persons eating EXACTLY 2 dishes and the number of persons eating all 3 dishes) as

$31 - Y - 2*5 = 21 - Y$.

The minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.

The maximum value of Y is 21. Is this possible? No. Because 5 people eat all three items. So, the no. of persons eating at most 2 items $= (9-5) + (10-5) + (7-5) = 11$. And adding 5 we get 16 people who eat at least one item.

So, our required answer is $21-10 \geq X \geq 21 - 16 \implies 5 \leq X \leq 11.$

Let $Y$ be the no. of persons who eat at least one item. $21 - Y$ people do not eat anything.

$Y = 9 + 10 + 7 - [N(A\cap B) + N(A\cap C) + N(B\cap C)] + 5$

$[N(A\cap B) + N(A\cap C) + N(B\cap C)] = 31 - Y$.

Now, these include the no. of persons who eat all 3 items thrice. So, excluding those, we get, no. of persons who eat at least two items (by adding the no. of persons eating EXACTLY 2 dishes and the number of persons eating all 3 dishes) as

$31 - Y - 2*5 = 21 - Y$.

The minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.

The maximum value of Y is 21. Is this possible? No. Because 5 people eat all three items. So, the no. of persons eating at most 2 items $= (9-5) + (10-5) + (7-5) = 11$. And adding 5 we get 16 people who eat at least one item.

So, our required answer is $21-10 \geq X \geq 21 - 16 \implies 5 \leq X \leq 11.$

22 votes

The lower bound is 5, which is very obvious, but the upper bound￼ value won't be 11, it has to be less than Or equal to 10, because if we take it to be 11, then the number of ppl eating exactly 1 item is coming -1, refer the image, and if we take the upperbound to be 10 then the ppl eating exactly 1 item is 0, which is the least it must go..

7

10 votes

@Arjun Sir ,I think the data in this question is wrong...

Let A$\cap$B$\cap$C = x

A$\cap$B+B$\cap$C+A$\cap$C contains x 3 times...

And atleast 2 means ( A$\cap$B+B$\cap$C+A$\cap$C ) - 2x

which evaluates to 0 ...which is contradict to the given data i.e.5 persons eat all three items.

Let A$\cap$B$\cap$C = x

A$\cap$B+B$\cap$C+A$\cap$C contains x 3 times...

And atleast 2 means ( A$\cap$B+B$\cap$C+A$\cap$C ) - 2x

which evaluates to 0 ...which is contradict to the given data i.e.5 persons eat all three items.

4

edited
Jun 28, 2020
by shivam001

@mehul vaidya i want to add something in your method :-

whenever you find A∪B∪C it means you have all the elements in A,B,C only once .

Let's say in a class there are 10 students out of which 3 have history , 5 have computer and 8 have science then,

**n(H∪C∪S) = n(H)+n(C)+n(S) - (n(H∩C)+n(C∩S)+n(H∩S)) + n(H∩C∩S)**

here n(H∪C∪S) :- all the students in a class who have atleast 1 subject from H/C/S. but there might be some student who do not have any of them , he/she might be having Biology or physics who knows. so don't take 10 as whole .

(n(H∩C)+n(C∩S)+n(H∩S)) :- in this case it contain all the students who have atleast 2 subjects(but it contain all 3 subjects thrice , so subtract 2*n(H∩C∩S) from this while calculating

please readhttps://gateoverflow.in/1676/gate1998-2-4 @akash.dinkar12 comment .

0

7 votes

First let us draw the generalized Venn Diagram of the situation:

As per the question the portion whose cardinality we need to find is as shown below in blue:

The required cardinality can be given as

$$N=|V\cap F| +|V\cap E|+ |F \cap E| - 2|V\cap E\cap F|$$

But we know :

$$|V\cup E\cup F|=|V|+|E|+|F|-|V\cap F| -|V\cap E|- |F \cap E| +|V\cap E\cap F|$$ $$ = 9+7+10 -|V\cap F| -|V\cap E|- |F \cap E| +|V\cap E\cap F|$$ $$ = 26 -|V\cap E\cap F| -N = 21- N \text{ [since $|V\cap E\cap F|=5$]}$$

$$\implies N=21-|V\cup E\cup F|$$

The minimum possible value of $|V\cup E\cup F|$ is $10$ as we could possibly pack the sets $V$ and $E$ in $F$ as shown below (since $F$ is having the largest cardinality among all the three sets): [This is possible because, no explicit values of $|V\cap F|$,$|V\cap E|$ or $|F \cup E|$ are given, so we are free to think of any possible sort of overlapping]

So the least value of $|V\cup E\cup F|$ is $10$.

So the max value of $N$ is $11=(21-|V\cup E\cup F|=21-10)$

But the minimum value of $N$ is $5$ since it is given that $|V\cap E\cap F|=5$, and the diagram shall be as shown:

So $$5\leq N \leq 11$$