Re: Pumps A, B, and C operate at their respective constant rates. Pumps A
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27 Jul 2022, 14:05
Generally, if we are told that:
A hours is needed for worker A (pump A etc.) to complete the job --> the rate of A=1A;
B hours is needed for worker B (pump B etc.) to complete the job --> the rate of B=1B;
C hours is needed for worker C (pump C etc.) to complete the job --> the rate of C=1C;
You can see that TIME to complete one job=Reciprocal of rate. eg 6 hours needed to complete one job (time) --> 1/6 of the job done in 1 hour (rate).
Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance.
Time*Rate=Distance
Time*Rate=Job
Also note that we can easily sum the rates:
If we are told that A is completing one job in 2 hours and B in 3 hours, thus A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together.
Time needed for A and B working simultaneously to complete the job=A∗BA+B hours, which is reciprocal of the sum of their respective rates. (General formula for calculating the time needed for two workers working simultaneously to complete one job).
Time needed for A and C working simultaneously to complete the job=A∗CA+C hours.
Time needed for B and C working simultaneously to complete the job=B∗CB+C hours.
General formula for calculating the time needed for THREE workers working simultaneously to complete one job is: A∗B∗CAB+AC+BC hours. Which is reciprocal of the sum of their respective rates: 1A+1B+1C.
We have three equations and three unknowns:
1. 1A+1B=56
2. 1A+1C=23
3. 1B+1C=12
Now the long way is just to calculate individually three unknowns A, B and C from three equations we have. But as we just need the reciprocal of the sum of relative rates of A, B and C, knowing the sum of 1A+1B+1C=AB+AC+BCABC would be fine, we just take the reciprocal of it and bingo, it would be just the value we wanted.
If we sum the three equations we'll get:
2∗1A+2∗1B+2∗1C=56+23+12=2
1A+1B+1C=1, now we just need to take reciprocal of 1, which is 1.
So the time needed for A, B, and C working simultaneously to complete 1 job is 1 hour.
Hope it helps.