CUET · MATHS · PYQ PAPER 2025
A manufacturing unit makes two models, 'classic' and 'supreme' of the scooter.
Each piece of the classic model requires 9 labour hours for assembling and 1 labour hour for finishing.
Each piece of supreme model requires 12 labour hours for assembling and 3 labour hour for finishing.
Maximum labour hours available for assembling and finishing are 180 and 30 respectively.
Profit is Rs. 10000 on each classic piece and Rs. 15000 on each supreme piece.
Which option describes the LPP to maximize profit Z (where x and y are classic and supreme pieces)?
- A Max \(Z=10 x+15 y\), subject to \(3 x+4 y \le\) \(180, x+3 y \le\) \(30, x \ge 0, y \ge 0\).
- B \(\operatorname{Max} Z=10000 x+15000 y\), subject to \(9 x+y \le\) \(180,12 x+3 y \le \) \(30, x \ge\) \(0, y \ge 0\).
- C \(\operatorname{Max} Z=10000 x+15000 y\), subject to \(3 x+4 y \ge 180\), \(12 x+3 y \ge\) \(30, x \ge 0, y \ge 0\).
- D \(\operatorname{Max} Z=10000 x+15000 y\), subject to \(3 x+4 y \le\) \(60, x+3 y \le\) \(30, x \ge 0, y \ge 0\).
Answer & Solution
Correct Answer
(D) \(\operatorname{Max} Z=10000 x+15000 y\), subject to \(3 x+4 y \le\) \(60, x+3 y \le\) \(30, x \ge 0, y \ge 0\).
Step-by-step Solution
Detailed explanation
Maximize Profit: \(Z = 10000x + 15000y\) Subject to: Assembling hours: \(9x + 12y \le 180 \Rightarrow 3x + 4y \le 60\) Finishing hours: \(x + 3y \le 30\) Non-negativity: \(x \ge 0, y \ge 0\)
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