Irising
Bunuel
Irising
I read quite a few solutions and they all have the following equation set up: 1/x+1/1.5x=1/4. But I have 1/x+1/1.5x=4. I don't understand why "x" is time. "x" to me is rate because the question states "constant rate of one pump was 1.5 times the constant rate of the other".
Can
Bunuel or someone else please help? Thanks!
x is an unknown and you can denote by x either time or rate, it's your choice. Doesn't really matter.
If you say x is the rate, then you get: x + 2x/3 = 1/4. My solution here uses x as the rate:
https://gmatclub.com/forum/two-water-pu ... l#p1245761If you say x is the time, then you get: 1/x + 2/(3x) = 1/4.
In any case notice that we sum the rates.
Thank you for your reply! Please bare with me as I am still trying to understand the problem. When you say "we sum rates", do you also imply that we cannot sum time? In my equation, I set x as rate. Then I thought 1/x must be time (work/rate=time). And based on that, I sum time to solve for x: 1/x+1/(1.5x)=4, time of the slow pump+time of the fast pump=total time. There must be something wrong with my logic here but I can't seem to figure it out.
Yes, we can sum rates but not times.
THEORYThere 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.
17. Work/Rate Problems
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