CAT 1995 | Question: 177

Question

Machine M1 as well as Machine M2 can independently produce either Product P or Product Q. The times taken by machines M1 and M2 (in minutes) to produce one unit of product P and Q are given in the table below: (Each machine works 8 hours per day).

PRODUCT	M1	M2
P	10	8
Q	6	6

If the number of units of P is to be 3 times that of Q, what is the minimum idle time for maximum total units manufactured?

• A. 0 minutes 
• B. 24 minutes
• C. 1 hour 
• D. 2 hours

Answer 1

As the number of units of P is to be 3 times that of Q 

We'll be producing P by M2 

∴ In 8 hrs M2 will produce $\dfrac {{8}*{60}}{8}$ = 60 units of product P

So, we have to produce $\dfrac {60}{3}$ = 20 units of product Q in order to maintain the given condition

To produce 20 units of product Q by M1 time required = (20 * 6) min = 120 min.

∴ Remaining time of M1 can be utilized = ((8 * 60) - 120) min = 360 min

In 360 min M1 have to maintain

  the number of units of P has to be 3 times that of Q

To produce 3 units of product P, M1 needs (3*10)min = 30 min 

& to produce 1 unit of product Q, M1 needs (1*6)min = 6 min

So, in every (30 + 6)min = 36 min M1 can produce 3 units of product P and 1 unit of product Q

∴ In the remaining 360 min M1 can produce (10*3)= 30 units of product P and  (1*10) = 10 units of product Q

In order to maintain the given criterion M1 doesn't have any idle time and M2 also doesn't have any idle time