What was the number of schools having facilities F2 and F4?

The correct option is 185.

Step-by-Step Explanation:

Let us define the regions of the 4-set Venn diagram representing the facilities F1, F2, F3, and F4. According to the problem constraints:
1. The total number of schools surveyed is 600.
2. 80 schools have none of the four facilities.
3. 40 schools have all four facilities.
4. The number of schools with only one facility is given as:
    • Only F1 = 25
    • Only F2 = 30
    • Only F3 = 26
    • Only F4 = 20
5. The number of schools with exactly three facilities is the same for any combination. Let this number be x. There are 4 such combinations (F1-F2-F3, F1-F2-F4, F1-F3-F4, and F2-F3-F4), each having exactly x schools.
6. 26 schools have only F2 and F3 (neither F1 nor F4).
7. Among schools with F4:
    • 24 schools have only F3 (meaning they have only F3 and F4).
    • 45 schools have only F2 (meaning they have only F2 and F4).
8. 162 schools have both F1 and F2.
9. The total number of schools with facility F2 is 313.

Step 1: Express the schools having both F1 and F2

The schools with both F1 and F2 are composed of:
• Schools with only F1 and F2
• Schools with exactly three facilities containing F1 and F2 (i.e., F1-F2-F3 and F1-F2-F4)
• Schools with all four facilities (F1-F2-F3-F4)

Using the variables defined above:

$$\text{Schools with both F1 and F2}=\text{Only (F1 \& F2)}+x+x+40=162$$

Simplifying the equation gives:

$$\text{Only (F1 \& F2)}+2x+40=162$$

$$\text{Only (F1 \& F2)}=122-2x$$

Step 2: Solve for x using the total number of schools with facility F2

The total number of schools with facility F2 is given as 313. This total is the sum of all regions that contain F2:

$$n\left(F_2\right)=\text{Only F2}+\text{Only (F1 \& F2)}+\text{Only (F2 \& F3)}+\text{Only (F2 \& F4)}+\text{Only (F1, F2 \& F3)}+\text{Only (F1, F2 \& F4)}+\text{Only (F2, F3 \& F4)}+\text{(F1, F2, F3 \& F4)}$$

Substitute the known values into the equation:

$$313=30+\left(122-2x\right)+26+45+x+x+x+40$$

Combine the constant terms and the terms involving x:

$$313=\left(30+122+26+45+40\right)+\left(-2x+3x\right)$$

$$313=263+x$$

$$x=313-263=50$$

Step 3: Calculate the number of schools having facilities F2 and F4

The schools containing both F2 and F4 lie in the following intersection regions:
• Schools with only F2 and F4 (given as 45)
• Schools with exactly three facilities containing F2 and F4 (F1-F2-F4 and F2-F3-F4, which is x = 50 for each)
• Schools with all four facilities (F1-F2-F3-F4, which is 40)

Summing these values up:

$$\text{Total schools having F2 and F4}=\text{Only (F2 \& F4)}+\text{Only (F1, F2 \& F4)}+\text{Only (F2, F3 \& F4)}+\text{(F1, F2, F3 \& F4)}$$

$$\text{Total schools having F2 and F4}=45+50+50+40=185$$

Therefore, the number of schools having facilities F2 and F4 is 185.