In a society, people like three different types of cars: A, B, and C. The ratio of people who like car A to car C is $29 : 42$, and the number of people who like only car A and C together is one-fourth of the number of people who like only car B. The number of people who like only car C is 50 more than four times the number of people who like only car A and B together. Ten people like all three types of cars, which is equal to the number of people who like only car B and C together. The number of people who like only car C is twice the number of people who like only car A, and the total number of people who like car C is 210. Everyone likes at least one of the cars.

What is the percentage difference between the sum of people who like only car A and only car B together and the sum of people who like only car A and B together and only car C?

Question

In a society, people like three different types of cars: A, B, and C. The ratio of people who like car A to car C is $29 : 42$, and the number of people who like only car A and C together is one-fourth of the number of people who like only car B. The number of people who like only car C is 50 more than four times the number of people who like only car A and B together. Ten people like all three types of cars, which is equal to the number of people who like only car B and C together. The number of people who like only car C is twice the number of people who like only car A, and the total number of people who like car C is 210. Everyone likes at least one of the cars.

What is the percentage difference between the sum of people who like only car A and only car B together and the sum of people who like only car A and B together and only car C?

Accepted Answer

**Solution:**

Let's denote the number of people who like only car A as $a$, only car B as $b$, and only car C as $c$.

Given:

1. Ratio of people who like car A to car C is $29 : 42$.
   - Total people who like car A = $a + (A \& B) + (A \& C) + (A \& B \& C)$
   - Total people who like car C = $c + (A \& C) + (B \& C) + (A \& B \& C)$

2. People who like only car A and C together (i.e., $A \& C$ only) is one-fourth of people who like only car B:
   $$A \& C = \frac{1}{4} b$$

3. People who like only car C is 50 more than four times the people who like only car A and B together:
   $$c = 4 (A \& B) + 50$$

4. People who like all three cars is 10:
   $$A \& B \& C = 10$$

5. People who like only car B and C together is equal to people who like all three cars:
   $$B \& C = 10$$

6. People who like only car C is twice the people who like only car A:
   $$c = 2a$$

7. Total people who like car C is 210:
   $$c + (A \& C) + (B \& C) + (A \& B \& C) = 210$$

Let's denote:
- $x = A \& B$ (only A and B)
- $y = A \& C$ (only A and C)
- $z = B \& C$ (only B and C) = 10
- $t = A \& B \& C = 10$

From (2):
$$y = \frac{1}{4} b$$

From (3):
$$c = 4x + 50$$

From (6):
$$c = 2a$$

From (7):
$$c + y + z + t = 210$$
Substitute $y = \frac{1}{4} b$, $z = 10$, $t = 10$:
$$c + \frac{1}{4} b + 10 + 10 = 210$$
$$c + \frac{1}{4} b = 190$$

From (1), ratio of total people who like A to C is $29 : 42$:

Total A = $a + x + y + t$

Total C = $c + y + z + t = 210$

So,
$$\frac{a + x + y + t}{210} = \frac{29}{42}$$

Calculate numerator:
$$a + x + y + 10 = 210 	imes \frac{29}{42} = 210 	imes \frac{29}{42} = 5 	imes 29 = 145$$

So,
$$a + x + y = 135$$

Recall $y = \frac{1}{4} b$, so:
$$a + x + \frac{1}{4} b = 135$$

We want to find the percentage difference between:
- Sum 1: People who like only car A and only car B together = $a + b$
- Sum 2: People who like only car A and B together and only car C = $x + c$

From (3) and (6):
$$c = 4x + 50$$
$$c = 2a$$

Equate:
$$2a = 4x + 50 \ \Rightarrow 4x = 2a - 50 \ \Rightarrow x = \frac{2a - 50}{4} = \frac{a}{2} - 12.5$$

From earlier:
$$a + x + \frac{1}{4} b = 135$$
Substitute $x$:
$$a + \left(\frac{a}{2} - 12.5ight) + \frac{1}{4} b = 135$$
$$\frac{3a}{2} + \frac{1}{4} b = 147.5$$

From (7) rearranged:
$$c + \frac{1}{4} b = 190$$
Substitute $c = 2a$:
$$2a + \frac{1}{4} b = 190$$

Subtract the two equations:
$$\left(2a + \frac{1}{4} bight) - \left(\frac{3a}{2} + \frac{1}{4} bight) = 190 - 147.5$$
$$2a - \frac{3a}{2} = 42.5$$
$$\frac{4a - 3a}{2} = 42.5$$
$$\frac{a}{2} = 42.5$$
$$a = 85$$

Now find $b$ from:
$$2a + \frac{1}{4} b = 190$$
$$2 	imes 85 + \frac{1}{4} b = 190$$
$$170 + \frac{1}{4} b = 190$$
$$\frac{1}{4} b = 20$$
$$b = 80$$

Find $x$:
$$x = \frac{a}{2} - 12.5 = \frac{85}{2} - 12.5 = 42.5 - 12.5 = 30$$

Find $c$:
$$c = 2a = 2 	imes 85 = 170$$

Calculate the sums:

- Sum 1: $a + b = 85 + 80 = 165$
- Sum 2: $x + c = 30 + 170 = 200$

Calculate percentage difference of Sum 1 compared to Sum 2:

$$	ext{Percentage difference} = \frac{	ext{Sum 2} - 	ext{Sum 1}}{	ext{Sum 2}} 	imes 100 = \frac{200 - 165}{200} 	imes 100 = \frac{35}{200} 	imes 100 = 17.5\%$$

Since Sum 1 is less than Sum 2, it is 17.5\% less.

**Answer:** 17.5\%

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