To solve these questions systematically, let's analyze the problem step by step. The data provided is:

1. The ratio of students who like both Banana and Orange to those who like only Apple is 5:4.
2. Total number of students who like only one of the three fruits is 28,000.
3. Number of students who like both Apple and Banana but not Orange is 1,440 more than those who like both Apple and Orange but not Banana.
4. Total number of students who like exactly two fruits is 14,360.
5. Ratio of the number of students who like only Apple to those who like all three fruits is 6:5.
6. Students who like Banana is 225% more than students who like only Apple.
7. Total number of students who like Orange is 26,760.

The questions to solve:

Question 11: Ratio of students who like both Apple and Orange but not Banana to those who like both Apple and Banana but not Orange?

Define:

• $A$: Students who like only Apple.
• $B$: Students who like only Banana.
• $C$: Students who like only Orange.
• $AB$: Students who like both Apple and Banana but not Orange.
• $AC$: Students who like both Apple and Orange but not Banana.
• $BC$: Students who like both Banana and Orange but not Apple.
• $ABC$: Students who like all three fruits.

From the data:

1. $AB = AC + 1440$.
2. $BC + AB + AC = 14,360 - 3ABC$ (students liking exactly two fruits).

Question 12: Find the number of students who like all three fruits.

Given:

1. The ratio of $A$ to $ABC$ is $6:5$, so ( Aining given data from sympy import symbols, Eq, solve

Variables

A = symbols('A') # Students who like only Apple ABC = symbols('ABC') # Students who like all three fruits AB = symbols('AB') # Students who like both Apple and Banana but not Orange AC = symbols('AC') # Students who like both Apple and Orange but not Banana BC = symbols('BC') # Students who like both Banana and Orange but not Apple

Known values

only_one_fruit = 28000 two_fruits_total = 14360 orange_total = 26760 ratio_A_to_ABC = 6 / 5 # A : ABC banana_more_than_A = 2.25 # 225% more AB_to_AC_diff = 1440 # AB is 1440 more than AC

Equations

1. Ratio A : ABC = 6:5

eq1 = Eq(A, (6/5) * ABC)

2. B = 3.25 * A

B = 3.25 * A

3. Total number of students who like exactly two fruits

eq2 = Eq(AB + AC + BC, two_fruits_total - 3 * ABC)

4. AB = AC + 1440

eq3 = Eq(AB, AC + 1440)

5. Total number of students liking only one fruit

C = symbols('C') # Students who like only Orange eq4 = Eq(A + B + C, only_one_fruit)

Solve for A, ABC, AB, AC, BC, C

solution = solve([eq1, eq2, eq3, eq4],The initial equations are partially solved, but we need further clarification about the number of students who like exactly one fruit. The symbolic results so far depend on $C$, which is the number of students who like only Orange.

Would you like me to further compute this using an assumption or refine