shasadou wrote:

Hi Bunuel

Can we solve this using weighted average approach? I am actually stuck at some point: so we derive that the faster machine B takes 40 minutes to complete the job. That is 20 minutes faster (can we come to solution from here somehow given that in 20 minutes B will print 100 pages more) that A alone.

Then I draw the weighted average line and get that: a/b = 16/36 = 4/9. I am trying to plug in further a/(a+5) = 9/4. But I dont get the answer and get stuck. Can you help please?

hi,

I am not sure if it can be done by weighted average, as it is more to do with allegations and mixture and the averages..

It is not the case here..

I'll just tell you two methods here..

1) both finish the work in 24 minutes..

so A does 24/60=2/5 work ..

therefore B does 1-2/5=3/5 work..Also in 24 minutes B can print 5*24=120 pages..

so 120=(3/5-2/5) of total work..

so 1/5 of total work = 120..

or total work = 600 pages..2) 1/a + 1/b = 1/24..

1/60 + 1/b = 1/24..

b=40 minutes..

total work = 24(A+B)=24(A+A+5)=48A+120..

A's work individually= 60A..

so 60A=48A+120..

A=10..

so total pages = 60A=600..