dave13 wrote:
sriharimurthy wrote:
Example 3.
Working together, printer A and printer B would finish a task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A?
Solution:
This problem is interesting because it tests not only our knowledge of the concept of word problems, but also our ability to ‘translate English to Math’
‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us that A and B combined would work at the rate of \(\frac{1}{24}\) per minute.
‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of \(\frac{1}{60}\) per minute.
At this point, it should strike you that with just this much information, it is possible to calculate the rate at which B works: Rate at which B works = \(\frac{1}{24}-\frac{1}{60}=\frac{1}{40}\).
‘B prints 5 pages a minute more than printer A’ This means that the difference between the amount of work B and A complete in one minute corresponds to 5 pages. So, let us calculate that difference. It will be \(\frac{1}{40}-\frac{1}{60}=\frac{1}{120}\)
‘How many pages does the task contain?’ If \(\frac{1}{120}\) of the job consists of 5 pages, then the 1 job will consist of \(\frac{(5*1)}{\frac{1}{120}} = 600\) pages.
pushpitkc,
Bunuel VeritasKarishma if in denominator always is given time (unless is reversed ) \(\frac{1}{120}\) then how can 120 be amount of work ? what does 5*1 mean ? and why are we dividing by 1/120 to get WORK DONE
as per FORMULA ---> WORK = RATE * TIME
pls explain
Imp: Ensure that you always have your eye on the units. When you are flipping a quantity, know why.
A and B finish a task in 24 mins. If work done has to be 1 (complete work), their rate of work is (1/24)th of the work per min. (It is usually the case that we consider work to be 1 work)
Work = Rate*Time
1 work = Rate* 24 mins
Rate = (1/24)th of work per min
Similarly, A's rate of work = (1/60)th of work per min
Since rates are additive, we get that B's rate must be (1/40)th of work per min.
So B does (1/40)th of the work in a min while B does only (1/60)th of the work. We are given that this difference of (1/40)th work and (1/60)th work is 5 pages.
So 1 complete work is 600 pages.
Now the rate of work can be expressed in terms of pages/min too.
B prints (1/40)*600 = 15 pages/min
A prints (1/60)*600 = 10 pages/min
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