9K * 8 + 7K * t = 100K => t = 4. Tricky.

Machine A working atmost 8 hours can produce = 9000*8 = 72000
Remaining Pencils to be produced = 100000-72000 = 28000
Machine B can produce 7000 pencils in one hour
Time taken by B to produce remaining pencils = 28000/7000 = 4 hours.
Least amount of time machine B must operate = 4 hours.

KItuz wrote:Machine A produces pencils at a constant rate of 9000 pencils per hour and machine B produces pencils at a constant rate of 7000 pencils per hour. If two machines together must produce 100000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time in hours that machine B must operate?

a 4
b 4 2/3
c 5 1/3
d 6 1/4

I thought answer was D but Correct answer is A.

Any clue how?

Absolutely the same approach as neya -

The following statement seems a bit confusing - If two machines together must produce 100000 pencils


We can never assume that both the machines are working simultaneously at the same time , coz we are asked to find out the least amount of time in hours that machine B must operate...

To minimize the time taken by B we must try to use A to the maximum extent possible ( ie. 8 hours ) to minimize B...