Question
A company manufactures two products, A and B. The
contribution per unit for A is ₹50 and for B is ₹60. Each unit of A requires 4 machine hours, and B requires 6 machine hours. Total machine hours available = 1,400. Fixed costs = ₹4,000. If the company can sell a maximum of 200 units of each, what is the maximum profit possible under machine hour constraints?Solution
Contribution per hour: A = 50/4 = ₹12.5, B = 60/6 = ₹10 → A has higher contribution per hour so first of all the machine hours will be utilized in manufacturing of product A. Max possible units = A: 200 units × 4 = 800 hrs Remaining = 1,400 – 800 = 600 hrs → B = 600/6 = 100 units Total contribution = (200×50) + (100 ×60) = ₹10,000 + ₹6,000 = ₹16,000 Profit = Contribution – Fixed = ₹16,000 – ₹4,000 = ₹12,000
100, 75, 59, 55, 46, 45
147 490 707 831 895 930
16, 27, 44, 63, 98, 139
Find the odd number out in the following set and state the reason:
131, 151, 171, 191, 211
Find the wrong number in the given number series.
2, 6, 12, 22, 30, 421320, 1352, 1390, 1436, 1488, 1548
Find the wrong number in the given number series.
16, 24, 36, 54, 81, 120.5
850, 849, 841, 814, 750, 688
7, 78, 324, 900,1855, 3128
66, 115, 34, 155, 14, 211