3 Set Venn Diagram

No. of girls = 15

Let the no. of boys be x

No. of singers = 6

No of boys who are singers = 4

Therefore, no of girls who are singers = 2

No of dancers = 10

No of boys who are dancers = 6

Therefore, no. of girls who are dancers = 4

No. of boys who are neither singers nor dancers = x-10

No. of girls who are neither singers nor dancers = 9

Now we fill the above table,

using statements 1 and 2, we get the following table

Let the number of girls who are interested in attending a 2-day event be a and the number of girls who are dancers and are interested in 2-day event be b.

Now using statements 3 and 4, we get

$$2\le0.18x-4\le6$$

$$6\le0.18x\le10$$

0.18x should be integer for which x should be a multiple of 50, and 0.18x lies between 6 and 10; therefore, the only possible value of x is 50.

From statement 5, we can say that,

4+ 0.4a - b = 5+b +1

or, 0.4a = 2+2b

or, a = 5(1+b)

a should be a multiple of 5 as b is a whole number. So possible values of a can be 5, 10 or 15. Now, as the maximum value of b can be 4 and the maximum value of 0.4a-b can be 2, so the only possible value of a satisfying the conditions above is 5. If a= 5 then b=1.

Let n, s, d and t be the number of students who likes none of the drinks, exactly one drink, exactly 2 drinks and all three drinks, respectively.

It is given, 

n + s + d + t = 100 ...... (1)

s + 2d + 3t = 73 + 80 + 52

s + 2d + 3t = 205 ...... (2)

(2)-(1), we get

d + 2t - n = 105

Maximum value t can take is 52, i.e. t = 52, d = 1 and n = 0

Minimum value t can take is 5, i.e. t = 5, d = 95 and n = 0 (This also satisfies equation (1))

Difference = 52 - 5 = 47

The answer is option A.

Based on the information provided, we can make the following diagram;

We know that the sum of all entries in the Venn diagram should be $$150$$. Therefore,

$$111+ 75-2k + 40 - (k-x) - k + 0 = 150$$

$$76 = 4k-x$$, where $$x\geq 1$$

We must maximise the number of students who chose Physics but not Mathematics. This number is equal to $$75-2k$$. To maximise this we must minimise $$k$$. 

We have $$4k = 76+x$$ or $$k = 19 + \dfrac{x}{4}$$

Since $$x$$ and $$k$$ are both integers, $$x$$ should be divisible by $$4$$. To minimise $$k$$, we will take the minimum value of $$x$$, which is $$4$$.

Therefore, $$k = 19 + \dfrac{4}{4} = 20$$.

Thus, the maximum possible number of students who chose physics but not mathematics, is $$75-2*20 = 35$$.