For a school trip, buses and minivans are to be hired. A bus seats 50 students and costs Rs. 6000, and a minivan seats 20 students and costs Rs. 3000. The school can spend at most Rs. 48000, and only 10 drivers are available, one for each vehicle. The greatest number of students that can be taken on the trip is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 400
Let x buses and y minivans be hired. Then
Maximise \(Z = 50x + 20y\)
subject to
\(6000x + 3000y \le 48000\), that is \(2x + y \le 16\) (money)
\(x + y \le 10\) (drivers)
\(x \ge 0,\ y \ge 0\)
The two lines meet where, subtracting,
\(x = 6,\ y = 4\)
The corner points are (0, 0), (8, 0), (6, 4) and (0, 10). Values of Z:
\(Z(0, 0) = 0\)
\(Z(8, 0) = 400\)
\(Z(6, 4) = 300 + 80 = 380\)
\(Z(0, 10) = 200\)
The maximum is at the corner (8, 0): 8 buses take 400 students. The optimum need not be at the point where the two constraint lines meet, so every corner must be checked.
Why the other options are incorrect
Option (1) 380 — Incorrect. This is Z at (6, 4), which is less than Z(8, 0) = 400.
Option (2) 200 — Incorrect. This is Z at (0, 10), with minivans only.
Option (3) 420 — Incorrect. It is more than the largest corner value 400; for example 8 buses and 1 minivan would cost Rs. 51000.