To have Machine B work the least, we have to maximise the capacity of Machine A, hence let A work for 8 hours.

In 8 hours, Machine A produces 9000x8=72000. Total left = 100000-72000 = 28000 to be produced by B.

So to produce 28000 B needs 28000/7000 hours. = 4 hours

Not sure if this is the best method though. I thought long and hard before arriving at this method. What about you

Angie between each hour (eg. 2 to 3, 7 to 8) = 360/12 = 30deg
In one minute, the second hand moves by 360/60=6deg
In one minute, the hour hand moves by 30/60=0.5deg
Set up 6t=30+0.5t where t is time in minute travelled by both hands
Solve t=5 5/11 min =5.5min approx. Ans B

Thanks for all responses. Testluv, how do you go about thinking of 4C2? 4C2 means "Ther are 4 seats. Choose 2 to be seated by 2 students, the order of the students doesn't matter". But the order does matter so it should be 4P2. 4P2 = 12 as is 4C2 * 2 = 12 Both reasoning are correct? Is the...

4 seats and 2 students permutation

How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students? 2 3 4 6 12 What's wrong with my method? Number of ways = Total ways without restrictions - no. of ways where 2 students are next to each other = 4!/2! - 3!/2! =...

Thank you Stuart. That was bad on my part. Can't believe I didn't spot that even after I kept repeating the steps. It's cleared now. Thank you.

Hi Can someone explain to me what is wrong with my method. To get 3 person without a married couple in it, 1) First, select any 3 couples out of 4 from which the committee will be selected. There are 4C3 ways to do it. 2) Each couple can only contribute one member. Therefore, there are 2C1 * 3 ways ...