Question

DIRECTIONS for questions 42 and 43: Read the information below and answer the question that follows.
A truck travelled from town A to town B over several days. During the first day, it covered \(\frac{1}{p}\) of the total distance, where p is a natural number. During the second day, it travelled \(\frac{1}{q}\) of the remaining distance, where q is a natural number. During the third day, it travelled \(\frac{1}{p}\) of the distance remaining after the second day, and during the fourth day, \(\frac{1}{q}\) of the distance remaining after the third day. By the end of the fourth day, the truck had travelled \(\frac{3}{4}\) of the distance between A and B.

43. If the total distance is 100 kilometres, the minimum distance that can be covered on day 1 is ____ kilometres.

• A. 25
• B. 30
• C. 33
• D. 35

Hint:

Question 42 shows p can only be 3 or 4; the day 1 distance is 100/p, so pick the p value that gives the smaller result.

Answer 1

The Correct Option is A

Step 1: Recall the two possible cases for p and q.
From the previous question, \(\frac{(p-1)(q-1)}{pq}=\frac{1}{2}\) has only two natural number solutions: \(p=3,q=4\) or \(p=4,q=3\).

Step 2: Find the day 1 distance for each case.
The distance covered on day 1 is \(\frac{1}{p}\) of the total distance. With a total distance of 100 km:
If \(p=3\): day 1 distance \(=\frac{100}{3}\approx33.33\) km.
If \(p=4\): day 1 distance \(=\frac{100}{4}=25\) km.

Step 3: Pick the minimum.
The question asks for the minimum possible distance on day 1 across the two valid cases. Comparing 33.33 km and 25 km, the smaller value is 25 km. This rules out 30, 33 and 35 (options 2, 3 and 4), since neither valid case gives those numbers, and 25 is clearly the smaller of the two figures that are actually possible.

Final Answer:
\[ \boxed{25\text{ km}} \]