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
To find the Next number in the given series.
342 510 726 996 1326 ?
...19 8 ? -14 63 - 36
...If 16 15 26 69 260 x,
Find 75% of x+ x
124 174 211 ? 254 264
...315 146 267 ? 235 210
A series is 2100, 3431, 2431, 3160, 2648, 2991
If another series 1728, __, __, __, __, p, follows the same pattern as the given number series, th...
Study the given pattern carefully and select the number that can replace the question mark (?) in it.
4 7 6
15 ? 21
44 68 60
9 4.5 4.5 9 36 ?
18 29 51 84 128 182
5 4 29 24 185 ?
