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

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 $21-10 \geq X \geq 21 - 16 \implies 5 \leq X \leq 11.$
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atleast 2 dishes means  2dishes and all 3 dishes   should be (10+5)????

"at least two out of the three " both  means 2 and 3

red shaded gives both 2 and 3

so no need of again finding for 3

so ans should be only 10

21−Y  are the persons who dont eat anything

21−Y are the persons who eats at least 2.Is this correct?
@rahul yes correct
@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.

I got the same doubt.
Let A∩B∩C = x
then ( A∩B+B∩C+A∩C ), this already contains $3$x. therefore subtracting $2$x from this should result into POSITIVE value, but it is zero.
Moreover, u r right. They are asking for "atleast 2" which means ( A∩B+B∩C+A∩C ) - 2x.

Something wrong with given Data ?

@Arjun sir, plz see this.

Sir, i can't understand how this is possible?

Minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
Maximum value of Y is 21. Is this possible? No. Because 5 person eat all three item.
I did it in same way.Can some one tell,why is this approach?I dont see any flaw here/
here, we are assuming that the 21 is given as at-least 1 dishes but 21 includes (at-least 1 + none)
explain with any shortcut method ...!
15 (10+5)