We are told that X takes 12 hours to finish a certain job, not 2/3 of the job. So:
12 hours = 1 job, not 2/3 of the job.
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Hi SivaKumarP,

The formula is an extension of the basic concept of work rate problem. Let me explain it with an example.

Assume two persons A & B working alone who complete a work \('W'\) in \(5\) and \(6\) days respectively. We need to find the time taken by A & B working together to complete the same work \(W\).

We know the amount of work to be done (i.e. \(W\)) and are asked to find the time taken. From the work rate equation Work = Rate * Time, the only variable remaining is the rate, so let's find out the rates of persons A & B and put in the equation to calculate the time taken.

Rates of work done by A & B
Work done by A = \(W\)
Time taken by A to do W amount of work = \(5\) days

Since Rate = Work/time

Hence rate of work done by A = \(\frac{W}{5}\)

Similarly rate of work done by B = \(\frac{W}{6}\)

Putting the values of rates of A & B in the Work Rate equation to find out the time taken when both A & B work together:

Work = Rate * Time

\(W = (\frac{W}{5} + \frac{W}{6}) * t\) (since A & B would work for the same time 't' to complete the work when working together)

\(W = \frac{11W}{30} * t\) i.e. \(t = \frac{30}{11}\) days = \(\frac{(5*6)}{(5+6)}\) days

Time taken when A & B work together to do the same work = Time taken by A * Time taken by B/(Time taken by A + Time taken by B)


On similar lines, if we need to find the work done by A & B working together for 30 days, we can calculate it using the rate of work of A & B working together which is \(\frac{11W}{30}\).

Work done by A & B working together for 30 days = \(\frac{11W}{30} * 30 = 11W = 6W + 5W\) i.e. work done by A & B respectively working alone for 30 days.

Work done by A & B together in time t = Work done by A in time t + Work done by B in time t


Hope its clear!

Regards
Harsh
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STEP BY STEP:

4 machines, 2 hours to complete 2 jobs;
1 machine, 8 hours to complete 2 jobs;
1 machine, 4 hours to complete 1 job;
3 machines, 4/3 hours to complete 1 job;
3 machines, 4 hours to complete 3 jobs.
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Rate*Time = Job done.

(4*rate of one machine)*2 = 2;

rate of one machine = 1/4 job/hour.

Rate*Time = Job done.

(3*rate of one machine)*x = 3;

(3*1/4)*x = 3;

x = 4 hours.
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CONCEPT: \(\frac{(Machine_Power * Time)}{Work} = Constant\)

i.e. \(\frac{(M_1 * T_1)}{W_1} = \frac{(M_2 * T_2)}{W_2}\)

i.e. \(\frac{(4 * 2))}{2} = \frac{(3 * T_2)}{3}\)

i.e. \(T_2 = 4\)
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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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