'Work' Word Problems Made Easy

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

Can someone help with this question?
If it takes 4 machines, working at the same constant rate, 2 hours to complete 2 jobs, how long will it take 3 machines, working at the same constant rate, to complete 3 jobs?"

The answer is 4 hours but I have no idea how to arrive at the answer, hopefully someone can help.

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.

Bunuel - Could you possibly structure that in an algebraic manner?

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.

You can follow the method mentioned by Bunuel above .

The only thing you need to remember is that the work rate problems are similar to distance-speed-time wherein Distance is similar to job, speed is similar to rate and time is constant.

Thus, as distance = speed X time

Job or work = rate x time

Additionally, as with more machines, the total time required will reduce for the same amount of work, combined rates of 'n' machines will be n*rate of 1 machine.

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. \(T_2 = 4\)

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

Can someone please explain this in Example 2:
‘In how much time does Y finish his work?’ If Y completes 1/120 of the work in 1 hour, then he will complete 1/3 of the work in 40 hours.

VeritasKarishma

Can you pls explain me why in statement 2 we are dividing combined time with individual time and also if there is any easier method of solving this

Thanks in advance!

A group of 5 craftsmen, working together at the same rate, can finish a job in 3 hours. How long will it take a group of 4 apprentices working together to do the same job?
(1) Each apprentice works at 2/3 the rate of a craftsman.
(2) The 5 craftsmen and the 4 apprentices working together will take 45/23 hours to finish the job.

Time taken by 5 craftsmen (group 1) = 3 hrs
Rate of work of group 1 = 1/3

Time taken by 5 craftsmen (group 1) + 4 apprentices (group 2) = 45/23 hours
Rate of work of group1 + group 2 = 23/45 hrs

Rate of work of group 2 = 23/45 - 1/3 = 8/45
Time taken by group 2 = 45/8 hrs

6 men can finish a piece of work in 2 hours. How long will 9 men take to finish the work if they can work only at 4/5th the capacity of the initial workers?

You have 3/2 as many men, working at 4/5 the capacity. Multiplying the two ratios gets us 6/5. Since we have 6/5 as much “workforce”, the job will take just 5/6 as long. 5/6 of two hours is an hour and 2/3, or an hour and 40 minutes.



Posted from my mobile device

Nice shortcut, I have utilized proportion:
\(\frac{#of workers}{Rate1}\) = \(\frac{#New of workers}{Rate2}\)
in which rates can be similar or different.

Thus,
\(\frac{7}{3/8 }\) = \(\frac{2}{1/time}\)

\(\frac{7*8}{3 }\) = \(\frac{2*time}{1}\)

\(\frac{7*8}{3*2 }\) = time =\( \frac{28}{3}\) = 9.3h

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