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: Note that the day 1 distance depends only on p.
The day 1 distance is $\frac{100}{p}$. A bigger value of $p$ gives a smaller distance, since we are dividing 100 by a bigger number. So to get the minimum day 1 distance, pick the larger of the two valid values of $p$.

Step 2: List the valid values of p.
Question 42 showed that $(p,q)$ can only be $(3,4)$ or $(4,3)$, so $p$ is either 3 or 4. The larger value is $p=4$.

Step 3: Compute the distance for the larger p.
\[ \text{day 1 distance} = \frac{100}{4} = 25 \text{ km} \]
As a check, with $p=3$, day 1 distance is $\frac{100}{3}\approx33.3$ km, which is bigger than 25 km, confirming that 25 km is indeed the smaller of the two possible values.

Step 4: Final Answer.
The minimum distance covered on day 1 is 25 km, so option 1 is correct.
\[ \boxed{25\text{ km}} \]