What was the number of schools having only facilities F1 and F4?

Sixteen patients in a hospital must undergo a blood test for a disease. It is known that exactly one of them has the disease. The hospital has only eight testing kits and has decided to pool blood samples of patients into eight vials for the tests. The patients are numbered 1 through 16, and the vials are labelled A, B, C, D, E, F, G, and H. The following table shows the vials into which each patient’s blood sample is distributed.

If a patient has the disease, then each vial containing his/her blood sample will test positive. If a vial tests positive, one of the patients whose blood samples were mixed in the vial has the disease. If a vial tests negative, then none of the patients whose blood samples were mixed in the vial has the disease.

The correct answer is 20.

Note on the Provided Image and Question Text:
The attached image shows a patient-vial distribution table mapping 16 patients (such as Patient 1 in vials B, D, F, H and Patient 16 in vials A, C, E, G) to 8 vials (A through H). However, the main question asks: "What was the number of schools having only facilities F1 and F4?", which belongs to a classic 4-set Venn diagram set from the CAT 2020 Slot 3 examination involving 600 schools and their Online Teaching-Learning Process (OTLP) facilities. Below is the step-by-step mathematical derivation for this problem.

1. Defining the Variables for the 4-Set Venn Diagram:
Let the four facilities be:
F1: Own software for OTLP
F2: Trained teachers for OTLP
F3: Training materials for OTLP
F4: All students having laptops

Let the number of schools having only one facility be:
Only F1 = 25, Only F2 = 30, Only F3 = 26, Only F4 = 20.
Total schools having exactly one facility is:
$n\left(1\right)=25+30+26+20=101$

Let the number of schools having exactly three facilities be the same for all combinations. Let this be $y$. Since there are 4 combinations of exactly three facilities, the total number of schools having exactly three facilities is:
$n\left(3\right)=4y$

The number of schools having all four facilities is given as:
$n\left(4\right)=40$

The number of schools having none of the facilities is 80. Thus, the total number of schools having at least one facility is:
$600-80=520$

Therefore, the number of schools having exactly two facilities, denoted as $n\left(2\right)$, can be calculated from the sum of all regions:
$n\left(1\right)+n\left(2\right)+n\left(3\right)+n\left(4\right)=520$
$101+n\left(2\right)+4y+40=520$
$n\left(2\right)+4y=379$

2. Solving for the Exact Overlaps:
Let $x_{ij}$ represent the number of schools with only facilities $F_i$ and $F_j$.
- We are given that 26 schools had only F2 and F3, so:
$x_{23}=26$
- Among schools having F4, 24 had only F3 and 45 had only F2:
$x_{34}=24$
$x_{24}=45$
- We are given that 162 schools had both F1 and F2:
$x_{12}+y+y+40=162$
$x_{12}+2y=122$
$x_{12}=122-2y$

- We are given that 313 schools had F2:
$\text{Total F2}=30+x_{12}+x_{23}+x_{24}+3y+40=313$
$30+x_{12}+26+45+3y+40=313$
$141+x_{12}+3y=313$
$x_{12}+3y=172$

Substituting $x_{12}=122-2y$ into the equation:
$(122-2y)+3y=172$
$122+y=172$
$y=50$

With $y=50$, we have:
$x_{12}=122-2\left(50\right)=22$
And the total number of schools with exactly three facilities is:
$n\left(3\right)=4y=200$

Substitute this back into the equation for $n\left(2\right)$:
$n\left(2\right)+200=379$
$n\left(2\right)=179$

Since $n\left(2\right)$ is the sum of all double intersections:
$x_{12}+x_{13}+x_{14}+x_{23}+x_{24}+x_{34}=179$
$22+x_{13}+x_{14}+26+45+24=179$
$117+x_{13}+x_{14}=179$
$x_{13}+x_{14}=62$

3. Using the Symmetry Condition to Find the Final Answer:
We are given that the number of schools having F1 is equal to the number of schools having F4.
Total schools having F1:
$25+x_{12}+x_{13}+x_{14}+3y+40$
$=25+22+x_{13}+x_{14}+150+40=237+x_{13}+x_{14}$

Total schools having F4:
$20+x_{14}+x_{24}+x_{34}+3y+40$
$=20+x_{14}+45+24+150+40=279+x_{14}$

Setting them equal to one another:
$237+x_{13}+x_{14}=279+x_{14}$
$237+x_{13}=279$
$x_{13}=42$

Using $x_{13}+x_{14}=62$:
$42+x_{14}=62$
$x_{14}=20$

Thus, the number of schools having only facilities F1 and F4 is 20.