Appendix 2: Mathematical Modelling — Practice Quiz
Madhya Pradesh Board · Class 12 · Mathematics
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A tank contains 1000 litres of brine with 250 g of salt per litre. Brine containing 200 g of salt per litre flows in at 25 litres per minute, and the mixture flows out at the same rate. What is the amount of salt in the tank at time t minutes?
A manufacturer has the profit model P = (s - b)x - a. If fixed cost a = ₹12000, selling price s = ₹18 per unit, variable cost b = ₹10 per unit, and 3000 units are produced and sold, what is the profit?
For the profit model P = (s - b)x - a, suppose dP/dx = 12 and b = ₹9 per unit. What is the selling price s?
A tower has eye height h = 1.5 m, horizontal distance l = 40 m, and angle of elevation α = 45°. What is the height H of the tower?
Sample Questions
For the inaccessible-foot tower model, the observer’s eye height is h = 1.8 m and the angle of depression β = 30°. What is the horizontal distance l?
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1.8√3 m
1) Use l = h cotβ. 2) For β = 30°, cot30° = √3. 3) So l = 1.8 × √3. 4) Therefore l = 1.8√3 m.
For the medicine production LPP, the feasible region has vertices O(0,0), P(20000,0), Q(20000,6000), R(10500,34500), S(5000,40000), and T(0,40000). What is the maximum value of Z = 8x + 7y?
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₹325500
1) Evaluate Z at each vertex. 2) At R(10500,34500), Z = 8(10500) + 7(34500). 3) 8(10500) = 84000 and 7(34500) = 241500. 4) Sum = 325500. 5) This is the maximum value.
A company has cost model C = a + bx with a = ₹5000 and b = ₹12 per unit, and income model I = sx with s = ₹20 per unit. What is the profit when x = 2500 units are sold?
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₹15000
1) Use P = (s - b)x - a. 2) Here s - b = 20 - 12 = 8. 3) Multiply: 8 × 2500 = 20000. 4) Subtract fixed cost: 20000 - 5000 = 15000. 5) Profit = ₹15000.
In the brine model, y(t) = 200 + 50e^{-t/40}. What is the salt amount after 40 minutes?
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200 + 50/e kg
1) Substitute t = 40 into y(t) = 200 + 50e^{-t/40}. 2) Then exponent becomes -40/40 = -1. 3) So y(40) = 200 + 50e^{-1}. 4) This is 200 + 50/e kg.
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