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.

42. The value of p + q is:

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

Track the fraction of distance left after each day; after four days the leftover fraction is the square of \(\frac{(p-1)(q-1)}{pq}\), and it must equal \(\frac{1}{4}\).

Answer 1

The Correct Option is D

Step 1: Set up the remaining distance after each day.
Let the total distance from A to B be 1 unit. On day 1 the truck covers \(\frac{1}{p}\) of it, so the fraction left after day 1 is \(1-\frac{1}{p}=\frac{p-1}{p}\).

Step 2: Apply the same idea for days 2, 3 and 4.
On day 2 the truck covers \(\frac{1}{q}\) of what remains, so the fraction left after day 2 is \(\frac{p-1}{p}\cdot\frac{q-1}{q}\).
On day 3 it again covers \(\frac{1}{p}\) of what is left, so the fraction remaining after day 3 is \(\frac{(p-1)^2}{p^2}\cdot\frac{q-1}{q}\).
On day 4 it covers \(\frac{1}{q}\) of that, so the fraction remaining after day 4 is \(\frac{(p-1)^2(q-1)^2}{p^2q^2}\), which is the square of \(\frac{(p-1)(q-1)}{pq}\).

Step 3: Use the given information.
By the end of day 4 the truck has covered \(\frac{3}{4}\) of the distance, so what remains is \(\frac{1}{4}\).
\[ \left[\frac{(p-1)(q-1)}{pq}\right]^2=\frac{1}{4} \]
Since p and q are natural numbers, \((p-1)(q-1)\) and \(pq\) are both zero or positive, so we take the positive square root:
\[ \frac{(p-1)(q-1)}{pq}=\frac{1}{2} \implies 2(p-1)(q-1)=pq \]

Step 4: Simplify to a factorable form.
Expanding, \(2pq-2p-2q+2=pq\), so \(pq-2p-2q+4=2\), which factors neatly as:
\[ (p-2)(q-2)=2 \]

Step 5: Find natural number solutions.
Since p and q are natural numbers, \(p-2\) and \(q-2\) are integers whose product is 2. The only pairing that keeps both p and q as natural numbers is \(p-2=1, q-2=2\) or \(p-2=2, q-2=1\), giving \((p,q)=(3,4)\) or \((4,3)\). A pairing like \(p-2=-1, q-2=-2\) would force \(q=0\), which is not a natural number, so it is rejected. This is why sums such as 4, 5, 6 or 8 (options 1, 2, 3, and the eliminated option 5) never occur; only 7 is possible.

Final Answer:
Either way, \(p+q=3+4=7\).
\[ \boxed{p+q=7} \]