Inequality – MBA CET questions with solutions

Approach: Use every room (each extra room adds revenue) and keep triples to the minimum.

Solution:
Let s, d, t = single, double, triple rooms: s + d + t = 25, s + 2d + 3t = 55 → d + 2t = 30.
Revenue = 2000s + 3000d + 3500t = 80,000 − 500t (substituting d = 30 − 2t, s = t − 5).
s ≥ 0 → t ≥ 5; minimum t = 5 → d = 20, s = 0.
Revenue = 5 × 3500 + 20 × 3000 = 17,500 + 60,000 = ₹77,500.

Why the other options are wrong: 82,500, 87,500 and 92,500 exceed the ceiling 80,000 − 500t with t ≥ 5; 72,500 is below the maximum.

Cetking shortcut: Start with 25 singles (25 guests, ₹50,000). A second guest in a room adds ₹1,000, a third only ₹500, so give 25 rooms a second guest first (50 guests), then put the last 5 as third guests: 50,000 + 25,000 + 2,500 = ₹77,500.

Answer: B

Approach: Linear programming — check each plan against both machine-hour limits, then compare profit.

Solution:
Limits: 4S + 5D ≤ 700 (A), 6S + 10D ≤ 1250 (B).
A) 100, 60: A = 700, B = 1200 ✓ → profit 2000 + 1800 = 3800.
B) 75, 80: A = 300 + 400 = 700, B = 450 + 800 = 1250 ✓ → profit 1500 + 2400 = 3900.
C) 100, 105: A = 925 ✗.
D) 60, 90: B = 360 + 900 = 1260 ✗.
E) 50, 100: B = 300 + 1000 = 1300 ✗.
Best feasible profit = Rs. 3900 at 75 Standard + 80 Deluxe (the corner where both machines are fully used).

Cetking shortcut: In LP questions the optimum sits where both constraints are tight — solve 4S + 5D = 700 and 6S + 10D = 1250 → S = 75, D = 80.

Answer: B