M13-28

THEORY
There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

\(time*speed=distance\) <--> \(time*rate=job \ done\). For example when we are told that a man can do a certain job in 3 hours we can write: \(3*rate=1\) --> \(rate=\frac{1}{3}\) job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then \(5*(2*rate)=1\) --> so rate of 1 printer is \(rate=\frac{1}{10}\) job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then \(3*(2*rate)=12\) --> so rate of 1 printer is \(rate=2\) pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is \(rate_a=\frac{job}{time}=\frac{1}{2}\) job/hour and B's rate is \(rate_b=\frac{job}{time}=\frac{1}{3}\) job/hour. Combined rate of A and B working simultaneously would be \(rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\) job/hour, which means that they will complete \(\frac{5}{6}\) job in one hour working together.

3. For multiple entities: \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}\), where \(T\) is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}\), where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that \(t_1\) and \(t_2\) are the respective individual times needed for \(A\) and \(B\) workers (pumps, ...) to complete the job, then time needed for \(A\) and \(B\) working simultaneously to complete the job equals to \(T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}\) hours, which is reciprocal of the sum of their respective rates (\(\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}\)).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

\(T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}\) hours.



Hope this helps

I understand the solution given.
Just have one query.

R * 1 hr = 20 page => the rate is 20 page/1hour or 20 p/hr
I took individual rates as 1/x + 1/y. Why is this incorrect?

Since time is 1 - the rates should be reciprocal of it. hence 1/x and 1/y.
Please tell me what is it that I'm missing.

x and y in the solution provided are respective RATES of pages per hour of John and Jack, while in your solutions, x and y are respective TIMES in hour in which John and Jack could type 1 page. This is not incorrect but since we need the ratio of rates it's better to denote the rates by x and y from the beginning. Still if you proceed with your approach you'd get the correct answer:

1/x + 1/y = 20 and 1/x + 1/y*1.25 = 22 gives x = 1/12 hours and y = 1/8 hours. Thus rates would be 12 pages per hour and 8 pages per hour. Ratio = 8/12 = 2/3.

Hope it's clear.

Bunuel, could you please expand the algebra? I can´t figure out how to solve the two equations

1/x + 1/y = 20 and 1/x + 1/y*1.25 = 22

Thank you.

(i) \(\frac{1}{x} + \frac{1}{y} = 20\)

(ii) \(\frac{1}{x} + \frac{1}{y}*1.25 = 22\)

Subtract (i) from (ii):

\(\frac{1}{y}*0.25=2\);

\(y = \frac{1}{8}\).

Plug back \(y = \frac{1}{8}\) in (i) to get \(y = \frac{1}{12}\).

Hope it helps.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

So if 25% = 2 ie (22-20)
then 100% = 8

20-8 = 12
Therefore 8:12
ie 2:3

25% of Jack's original speed is equivalent to (22-20) 2 pages. That means originally Jack was typing 8 pages ((100/25)*2) among the 20 pages.