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Thread: This one stumps me!

20130811, 15:55 #1
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This one stumps me!
Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming pool in 4 hours. How many hours will it take 4 large and 2 small pumps to fill the swimming pool.(We assume that all large pumps are similar and all small pumps are also similar.) I
I come up with the answer:
But the correct answer is different but my logic seems sound. Any thoughts?

20130811, 19:35 #2
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Your logic seems pretty good, but I sense that with 4 large, the time is going to be less than 2 hours.
What are we missing?

20130811, 20:26 #3
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This is the logic they give how to get the correct answer
I just don't see it!

20130811, 21:01 #4
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AH WELL...their step D suggests 4 large and 4 small, not 2 small as you indicated in your problem.
Clearly, the creator of the problem was not from PA!

20130811, 22:36 #5
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<Slap the head>
The question I posted was a copy/paste right from the web page. Totally missed the error. I'm gonna write to them and give them a piece of my mind. But what do you expect for free?
Nice pickup KW!Last edited by Maudibe; 20130811 at 22:38.

20130812, 01:53 #6
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Without reading your "spoiler" ...
Classic Algebra ... 2 unknowns need 2 equations to solve, and we have it.
The gotcha is 3rd unknown but it is common to both equations.
So the answer can be found via ratiometric of the common unknown
which we do not need to know!
Let L be # of *liters per hour* for Large pump. (Or # of gallons per hour, if that matters.)
Let s be # of *liters per hour* for Small pump.
Let P be the total liters held by the same pool (so P is the common quality).
(2L + 1s) x4 = P (1) [2 large pumps+1 small pump fill pool in 4 hours]
(1L + 3s) x4 = P (2) [1 large pump+2 small pumps fill SAME pool in 4 hours]
8L + 4s = P (3)
4L + 12s = P (4)
(3)x1, (4)x2
8L + 4s = P (3)
8L + 24s = 2P (5)
(5)(3)
20s = P
Therefore, s = (1/20)P (6)
========
Plug (6) into (3)
8L + 4xP/20 = P (7)
8L + P/5 = P
8L = P  (1/5)P =(4/5)P
Therefore, L = (1/10)P (8)
========
Or, 1 Large pump is 2x faster than small pump.
Or, 1 Larger pump equals 2 small pumps.
Now, the question:
4 Large pumps+2 small pumps, how long to fill same pool P?
4 x (1/10)P + 2 x (1/20)P
= (4/10)P + (1/10)P = (5/10)P =(1/2)P
Since
1P takes 4 hours.
Therefore,
(1/2)P takes 1/2 as much time, or 2 hours.
============
Answer: 2 hours.
============
If you don't feel comfortable with unknown common P.
You can assign a number to P.
Any number will do, as long as both equations (1) and (2) use the same number (aka P).
Easiest is P=1 (1 liter or 1 gallon).
For example, P=100,000 liters, a more common sense for a pool, if that matters.
But no difference anyway; 1/10 of 1 liter, or 1/10 of 100,000 liters, is still 1/10.]Last edited by scaisson; 20130812 at 02:13.

20130812, 02:25 #7
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Continue from my post above.
If you do not feel comfortable at arriving the answer in the above method, you can do a direct calculation, as below:
Let T hours be the time to fill the pool with 4 Large pumps and 2 Small pumps.
[4x(1/10)P + 2x(1/20)P] xT = P <4 Large pumps+2 Small pumps take T hours to fill P liters>
[(4/10)P + (1/10)P xT = P
(5/10)P xT = P
Therefore, T = 2

20130904, 09:00 #8
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Notwithstanding the error from the original site, everyone seems to have made a simple calculation very difficult...
Two large and 1 small pumps can fill a swimming pool in 4 hours...... How many hours will it take 4 large and 2 small pumps to fill the swimming pool?

20130904, 14:34 #9

20130905, 01:08 #10
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lilliebet,
Finding the needle in a hay stack ... with just a glance.
Smart and quick observation. 'Elementary', says Sherlock Holmes (or is it lilliebet?)
What if the question is: "7 big pumps and 3 small pumps." Or "3.11 big pumps and 5.73 small pumps?" It is where math helps to explain it. [Why fractional pumps? Well, the pumps are old ...]
But I prefer your 'one glance'.

20130906, 19:54 #11
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Lilliebet, do you work for a pool company or do you teach critical logic at a university?

20130906, 21:33 #12
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But, the original problem stated, at least in the explanation, it was to find:
4 large and 4 small. That's a different situation than just doubling.

20130907, 11:01 #13
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Scaisson has done it the algebraic methodical way. But look t it like this
2 large and 1 small fill the pool in 4 hours
1 large and 3 small fill the pool in 4 hours
Therefore reducing 1 large pump is the same as adding 2 small pumps since both achieve the same 4 hours to fill the pool
therefore L = 2S
Now
2L + 1 S fills the pool in 4 hours, in effect 5S fills the pool in 4 hours, therefore pool capacity = 4hrs x 5S = 20S worth
If you use 4L and 2S, that is the equivalent of 10S worth capacity (2x4L + 2S) per hour
The pool capacity is 20S
Therefore 2 hours
Then if there is a different number, like how much time for 3.2small and 4.3 large, the math is the same
4.3 x 2 + 3.2 x 1 = 11.8 total flow rate
Therefore time = 20 capacity / 11.8 flow rate = whateverLast edited by GuruPepsi; 20130907 at 11:04.

20130907, 21:21 #14
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I work in Payroll, I do maths for a living and, yes, it's fun but sometimes you just have to look for the simplest solution.

20130913, 01:42 #15
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Although GuruPepsi is only guru in Pepsi ... or drinks too much Pepsi (!),
his quickie analysis is equally good and fast.
"Elementary, so you say?"... GuruPepsi answers Sherlock.