CAT 1995 — DILR Question 32

If M1 works at half of its normal efficiency, time taken by M1 to manufacture 1 unit of P = 20 min and Q = 12 min. For producing maximum number of units, we have to produce Q on M2 first as it takes only 6 min per piece.

Also since at least one unit of P has to be manufactured and it is more efficient to do so on M2, we would do that. So time taken to manufacture 1 unit of P on M2 = 8 min.

Hence, time remaining on M2 = (480 – 8) = 472. In this remaining time number of units of Q that can be manufactured on M2 = $\frac{472}{6}$ = 78 (only completed units taken). Now since it takes less time to manufacture Q on M1 as well, we will maximize Q on M1. Since 1 unit of number of units that can be produced = $\frac{(8\times 60)}{12}$ = 40.

Hence, the total number of units manufactured = (1 + 78 + 40) = 119 units.