CUET · MATHS · PYQ PAPER 2025
If a person rides his motorbike \(x km\) at 30 km per hour, he has to spend ₹ 3 per kilometer on petrol; if he rides \(y km\) at a faster speed of 40 km per hour, the petrol cost increases to ₹ 4 per kilometer. If he has ₹ 100 to spend on petrol and wishes to find the maximum distance he can travel within one hour, then linear programming problem (LPP) formulation is:
- A Maximize Distance \(( D )=x+y\) subject to the constraints \(3 x+4 y \leq 120,4 x+3 y \geq 100, x \geq 0, y \geq 0\).
- B Maximize Distance(D) \(=x+y\) subject to the constraints \(3 x+4 y \leq 70,4 x+3 y \leq 1, x \geq 0, y \geq 0\).
- C Maximize Distance(D) \(=x+y\) subject to the constraints \(3 x+4 y \leq 100,4 x+3 y \leq 120, x \geq 0, y \geq 0\).
- D Maximize Distance(D) \(=x+y\) subject to the constraints \(3 x+4 y \leq 100,4 x+3 y \leq 120, x \leq 0, y \leq 0\).
Answer & Solution
Correct Answer
(C) Maximize Distance(D) \(=x+y\) subject to the constraints \(3 x+4 y \leq 100,4 x+3 y \leq 120, x \geq 0, y \geq 0\).
Step-by-step Solution
Detailed explanation
Objective: Maximize Distance \( D = x+y \) Petrol cost constraint: \( 3x + 4y \leq 100 \) Time constraint: \( \frac{x}{30} + \frac{y}{40} \leq 1 \implies 4x + 3y \leq 120 \) Non-negativity: \( x \geq 0, y \geq 0 \) LPP formulation: Maximize Distance \((D)=x+y\) subject to…
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