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18 Jun 2013, 23:24
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I remember having seen a formula for rates
\(\frac{A*B}{A+B}\) but I can't remember what it is for (if anything at all). Can someone clarify this for me?
Thanks
\(\frac{A*B}{A+B}\) but I can't remember what it is for (if anything at all). Can someone clarify this for me?
Thanks
19 Jun 2013, 02:16
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Skag55
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.
Some work problems with solutions:
https://gmatclub.com/forum/time-n-work- ... cal%20rate
https://gmatclub.com/forum/facing-probl ... reciprocal
https://gmatclub.com/forum/what-am-i-do ... reciprocal
https://gmatclub.com/forum/gmat-prep-ps ... cal%20rate
https://gmatclub.com/forum/questions-fr ... cal%20rate
https://gmatclub.com/forum/a-good-one-9 ... hilit=rate
https://gmatclub.com/forum/solution-req ... ate%20done
https://gmatclub.com/forum/work-problem ... ate%20done
https://gmatclub.com/forum/hours-to-typ ... ation.%20R
Theory on work/rate problems: https://gmatclub.com/forum/work-word-pr ... 87357.html
All DS work/rate problems to practice: https://gmatclub.com/forum/search.php?s ... &tag_id=46
All PS work/rate problems to practice: https://gmatclub.com/forum/search.php?s ... &tag_id=66
All DS work/rate problems to practice: https://gmatclub.com/forum/search.php?s ... &tag_id=46
All PS work/rate problems to practice: https://gmatclub.com/forum/search.php?s ... &tag_id=66
Hope it helps.
19 Jun 2013, 01:09
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Skag55
hi,
If A can finish a work in X time and B can finish the
same work in Y time then both of them together can
finish that work in (X*Y)/ (X+Y) time.
If A can finish a work in X time and A & B together can
finish the same work in S time then B can finish that
work in (XS)/(X-S) time.
If A can finish a work in X time and B in Y time and C in
Z time then all of them working together will finish the
work in (XYZ)/ (XY +YZ +XZ) time
kudos if it helped
