A truck travelled from town A to town B over several days. During the first day, it covered 1/p of the total distance, where p is a natural number. During the second day, it travelled 1/q of the remaining distance, where q is a natural number. During the third day, it travelled 1/p of the distance remaining after the second day, and during the fourth day, 1/q of the distance remaining after third day. By the end of the fourth day the truck had travelled 3/4 of the distance between A and B.
If the total distance is 100 kilometres, the minimum distance that can be covered on day 1 is ......... kilometres.

Options

A. 25
B. 30
C. 33
D. 35

Answer

To solve the problem, we need to analyze the distances covered by the truck over the four days based on the fractions given. Let's denote the total distance from town A to town B as \( D = 100 \) km.

### Step-by-Step Solution:

1. **Distance Covered on Day 1:**
The truck covers \( \frac{1}{p} \) of the total distance on the first day.
\[ \text{Distance on Day 1} = \frac{1}{p} \times D = \frac{100}{p} \]

2. **Remaining Distance After Day 1:**
After the first day, the remaining distance is:
\[ \text{Remaining Distance after Day 1} = D - \text{Distance on Day 1} = 100 - \frac{100}{p} = 100 \left(1 - \frac{1}{p}\right) = 100 \left(\frac{p-1}{p}\right) \]

3. **Distance Covered on Day 2:**
On the second day, the truck travels \( \frac{1}{q} \) of the remaining distance:
\[ \text{Distance on Day 2} = \frac{1}{q} \times \left(100 \left(\frac{p-1}{p}\right)\right) = \frac{100(p-1)}{pq} \]

4. **Remaining Distance After Day 2:**
The remaining distance after the second day is:
\[ \text{Remaining Distance after Day 2} = 100 \left(\frac{p-1}{p}\right) - \frac{100(p-1)}{pq} = 100 \left(\frac{p-1}{p}\right) \left(1 - \frac{1}{q}\right) = 100 \left(\frac{p-1}{p}\right) \left(\frac{q-1}{q}\right) \]

5. **Distance Covered on Day 3:**
On the third day, the truck covers \( \frac{1}{p} \) of the remaining distance:
\[ \text{Distance on Day 3} = \frac{1}{p} \times \left(100 \left(\frac{p-1}{p}\right) \left(\frac{q-1}{q}\right)\right) = \frac{100(p-1)(q-1)}{p^2q} \]

6. **Remaining Distance After Day 3:**
The remaining distance after the third day is:
\[ \text{Remaining Distance after Day 3} = 100 \left(\frac{p-1}{p}\right) \left(\frac{q-1}{q}\right) - \frac{100(p-1)(q-1)}{p^2q} = 100 \left(\frac{p-1}{p}\right) \left(\frac{q-1}{q}\right) \left(1 - \frac{1}{p}\right) = 100 \left(\frac{p-1}{p}\right) \left(\frac{q-1}{q}\right) \left(\frac{p-1}{p}\right) \]

7. **Distance Covered on Day 4:**
On the fourth day, the truck covers \( \frac{1}{q} \) of the remaining distance:
\[ \text{Distance on Day 4} = \frac{1}{q} \times \left(100 \left(\frac{p-1}{p}\right) \left(\frac{q-1}{q}\right) \left(\frac{p-1}{p}\right)\right) = \frac{100(p-1)^2(q-1)}{p^2q^2} \]

8. **Total Distance Covered After Four Days:**
The total distance covered after four days is:
\[ \text{Total Distance} = \text{Distance on Day 1} + \text{Distance on Day 2} + \text{Distance on Day 3} + \text{Distance on Day 4} \]
Setting this equal to \( \frac{3}{4} \times D = 75 \) km, we can solve for \( p \) and \( q \).

9. **Finding Minimum Distance on Day 1:**
To minimize the distance covered on Day 1, we need to maximize \( p \). The maximum value of \( p \) that keeps \( \frac{100}{p} \) a natural number is 4 (since \( p \) must be a natural number). Thus:
\[ \text{Minimum Distance on Day 1} = \frac{100}{4} = 25 \text{ km} \]

### Final Answer:
The minimum distance that can be covered on Day 1 is **25 km**.