# Set Theory, Calendars, Clocks and Binomial Theorem

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## Sets and Unions

A factory has 80 workers and 3 machines. Each worker knows to operate atleast 1 machine. If there are 65 persons who knows to operate machine 1, 60 who knows to operate machine 2 and 55 who knows to operate machine 3,what can be the minimum number of persons who knows to operate all the three machines?
1. 15
2. 20
3. 30
4. 40

Choice B. 20

## Detailed Solution

Let’s start with taking a random value for all three category. So let’s first take 40 for the all three category. Now 65 + 60 + 55 = 180, this means there is an extra count of 180 – 80 = 100.
Now as we know that the extra count occurs in the in the exactly two area and the all three area. So let’s try put the extra count in these area
Trial 1 - Since 40 is already assumed to be in the all three area, it takes care of extra count of 40 x 2 = 80. Thus we are left with 20 as extra count which we have to place at the exactly two area
Thus in the above case our venn diagram will look as:

A close look in the figure tells us that we can further decrease the value of all the three area. A bit of logical thinking will bring us to the value 20. No value less than 20 can satisfy the conditions of the question. As there is no scope left for reallocating numbers left from one area to another in this case.
Hence the final venn diagram will look as:

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