In a survey of 90 students about their favorite meals, 63 liked Banku, 55 liked Rice, and 50 liked Fufu. 13 liked only Banku, 8 liked only Rice, and 5 liked only Fufu. If 22 students liked all three m

Let $B$ be the set of students who liked Banku, $R$ be the set of students who liked Rice, and $F$ be the set of students who liked Fufu.
Total number of students = 90.
Given:
$|B| = 63$
$|R| = 55$
$|F| = 50$
$|B \text{ only}| = 13$
$|R \text{ only}| = 8$
$|F \text{ only}| = 5$
$|B \cap R \cap F| = 22$ (students who liked all three meals)

Let $x$ be the number of students who liked Banku and Rice only ($|B \cap R \text{ only}|$).
Let $y$ be the number of students who liked Banku and Fufu only ($|B \cap F \text{ only}|$).
Let $z$ be the number of students who liked Rice and Fufu only ($|R \cap F \text{ only}|$).

The total number of students who liked Banku is the sum of those who liked Banku only, Banku and Rice only, Banku and Fufu only, and all three:
$|B| = |B \text{ only}| + x + y + |B \cap R \cap F|$
$63 = 13 + x + y + 22$
$63 = 35 + x + y \implies x + y = 28 \quad (1)$

Similarly for Rice:
$|R| = |R \text{ only}| + x + z + |B \cap R \cap F|$
$55 = 8 + x + z + 22$
$55 = 30 + x + z \implies x + z = 25 \quad (2)$

And for Fufu:
$|F| = |F \text{ only}| + y + z + |B \cap R \cap F|$
$50 = 5 + y + z + 22$
$50 = 27 + y + z \implies y + z = 23 \quad (3)$

From equation (1), $y = 28 - x$.
Substitute this into equation (3):
$(28 - x) + z = 23$
$z - x = 23 - 28$
$z - x = -5 \implies x - z = 5 \quad (4)$

Now we have a system with equations (2) and (4):
$x + z = 25 \quad (2)$
$x - z = 5 \quad (4)$

Add equation (2) and equation (4):
$(x + z) + (x - z) = 25 + 5$
$2x = 30$
$x = 15$

Substitute $x = 15$ into equation (2):
$15 + z = 25$
$z = 25 - 15$
$z = 10$

Substitute $x = 15$ into equation (1):
$15 + y = 28$
$y = 28 - 15$
$y = 13$

So, the number of students who liked:
Banku and Rice only ($x$) = 15
Banku and Fufu only ($y$) = 13
Rice and Fufu only ($z$) = 10
All three meals = 22
Banku only = 13
Rice only = 8
Fufu only = 5

a) Draw a Venn diagram to illustrate the information.
A Venn diagram would consist of three overlapping circles labeled B, R, and F. The numbers in each region would be:

• Region B only: 13
• Region R only: 8
• Region F only: 5
• Region B $\cap$ R only (intersection of B and R, outside F): 15
• Region B $\cap$ F only (intersection of B and F, outside R): 13
• Region R $\cap$ F only (intersection of R and F, outside B): 10
• Region B $\cap$ R $\cap$ F (intersection of all three): 22

b) Find the number of students who liked:

i) Banku and Rice:
This includes students who liked Banku and Rice only, and students who liked all three.
Number of students who liked Banku and Rice = $x + |B \cap R \cap F| = 15 + 22 = \boxed{37}$.

ii) only two of the meals:
This is the sum of students who liked (Banku and Rice only), (Banku and Fufu only), and (Rice and Fufu only).
Number of students who liked only two meals = $x + y + z = 15 + 13 + 10 = \boxed{38}$.

iii) none of the three meals:
First, find the total number of students who liked at least one meal:
$|B \cup R \cup F| = |B \text{ only}| + |R \text{ only}| + |F \text{ only}| + x + y + z + |B \cap R \cap F|$
$|B \cup R \cup F| = 13 + 8 + 5 + 15 + 13 + 10 + 22 = 86$.
Number of students who liked none of the meals = Total students - $|B \cup R \cup F|$
Number of students who liked none of the meals = $90 - 86 = \boxed{4}$.