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I'm studying maths as an adult, and I thought everything was going well until I hit the following activity. I have the answer in my workbook, but I just don't seem to be able to come to terms with the process.

"8 pumps working for 10 minutes raise 440 litres of water. How long will it take 6 pumps to raise 396 litres."

The process I tried was thus.

Pumps decrease at a ratio 6:8 Water drawn decreases at a ratio of 396:440

So I then try 10 * 6/8 * 396/440

I realise this is incorrect, but I'm struggling to understand the process. Any help in giving me a eureka moment would be gratefully received.

timstermatic
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  • title altered accordingly – timstermatic Nov 08 '15 at 15:52
  • After you have your eureka moment: if one and a half chickens lay one and a half eggs in one and a half days then how many eggs do nine chickens lay in nine days? – Eric Lippert Nov 08 '15 at 18:29
  • @EricLippert I'd have a go at that other than that my title included the the phrase "real world problems". The day I see half a chicken lay half an egg, I'll be back to answer :) – timstermatic Nov 08 '15 at 19:13
  • @timstermatic: In the real world, dealing with fractional quantities of indivisible goods is normal. If a company sells 18 buses a year, it may very well say, "we sell 1.5 buses per month." – Brian Nov 09 '15 at 13:58
  • @Brian this may be true, but I've yet to see any AGM of an egg company say "we sell 600.5 eggs per month" – timstermatic Nov 11 '15 at 08:53
  • @timstermatic: That's only because their volumes are high enough that they round. You normally start seeing fractions when you drill down into campaign effectiveness. That said, fractions with differing denominators are annoying to compare, so most industries consistently use factors of 10 for their denominator. E.g., 5% is shorthand for 5/100, and 5‱ is shorthand for 5/1000. One of the big exceptions is the English system of measurements, where denominators are often powers of 2. However, most of the world thinks this is dumb and uses the metric system. – Brian Nov 11 '15 at 12:40

5 Answers5

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Let's consider this:

$8$ pumps take $440L$ per $10 \min$. That means that $8$ pumps take $44L$ per $1\min$, which means that each pump takes $\frac{44}{8}L/\min$.

You want to know in how much minutes $x$, we have that $6$ pumps take $396L$. You have to multiply $6$ by $\frac{44}{8}$ to get the liters taken per minute by the $6$ pumps.

Now, you have

$$\underbrace{\left(6\cdot\frac{44}{8}\right)}_{\text{liters taken per}\,\min}\cdot \underbrace{x}_{\text{number of}\,\min }=396$$

So that it leads to

$$x=\frac{396\cdot 8}{44\cdot 6}=12$$

And it takes $12\min$ for $6$ pumps to get $396L$.

MoebiusCorzer
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When you have fewer pumps, you should not expect the time used for the entire task to be less, which is what you would get by multiplying by $6/8$ (which is less than $1$).

Instead, the number of pumps is proportional to the total rate of pumping, which is inversely proportional to the time any given task takes.

So when you change the rate, the time should be divided by the ration by which the rate changes, and your computation should be $$ 10 \div \frac{6}{8} \times \frac{396}{440} $$ which is the same as $$ 10 \times \frac{8}{6} \times \frac{396}{440} $$

hmakholm left over Monica
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  • Your first sentence is false, in general, since $396<440$. It can be misleading. Let's replace $8$ pumps by $21$ and $6$ pumps by $19$. You directly get that $19$ pumps take $\approx 9.95$ minutes to get $396L$ while it takes $10$ minutes for $21$ pumps to get $440L$. – MoebiusCorzer Nov 08 '15 at 16:18
  • @MoebiusCorzer: The change in the amount of water to pump is conceptually distinct from the change in the number of pumps, and they should be handled one a a time. The fact that you're also changing the amount of water has no bearing on what the correct way to deal with a different number of pumps is. For any given abount of water, 19 pumps will still pump it slower than 21 pumps will. – hmakholm left over Monica Nov 08 '15 at 16:19
  • Ok, I get your point, but "you should not expect the time used for the entire task to be less" can be misleading, since it is false, although the end of your sentence is correct. – MoebiusCorzer Nov 08 '15 at 16:21
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    I am accepting this answer because it addresses the process I was using and where my misunderstanding lies. That's not to say the other answers have not been useful in showing other approaches to the issue. – timstermatic Nov 08 '15 at 16:22
  • @MoebiusCorzer: It would be false if I said "you should not expect the time used for one task to be less than that used for a different task". But that's not what I'm saying -- I'm comparing the time the two different amounts of pumping capacity take for "the entire task" -- which is implicitly understood to be the same task in both the compared situations, i.e. pumping the same amount of water. – hmakholm left over Monica Nov 08 '15 at 16:23
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    @HenningMakholm Ok, you are right, I understand and it is actually without ambiguity because of the end of your sentence. Mea culpa. – MoebiusCorzer Nov 08 '15 at 16:26
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If $8$ pumps process $440$ liters of water in $10$ minutes, then how much water does one pump process per minute?

dbanet
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perhaps you can give this a thought We know: 8 pumps do 10 minutes gives 440 L, 1 pump does 10 minutes gives 55 L (one eight of 440), 6 pumps do 10 minutes give 330 L (6 times 1 pump), But we don't need 330 L, we need 396 L, so that's a factor 396/330 on the time they pump. Therefore, 6 pumps do 10*396/330 minutes give (396/330)*330 (which is 330 L) Now simplify 10*396/330 minutes

imranfat
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It's easiest to understand this as a multi-part ratio:

You get 440L using 8 pumps for 10 minutes--that translates to

440/(8*10) Liters/pump-minute = 11/2 Liters/pump-minute = 5.5 L/(p-m). This is fixed.

Your question is then easily solved...set the fixed value to the # Liters/(number of pumps*number of minutes) 11/2 = 396/(6*minutes).

t = 12 minutes, as above.

Chuck1954
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