Cistern

B

BigGlenntheHeavy

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\(\displaystyle A \ cistern \ can \ be \ filled \ by \ any \ one \ of \ three \ pipes.\)

\(\displaystyle The \ largest \ pipe \ can \ fill \ it \ alone \ in \ 6 \ hours.\)

\(\displaystyle The \ largest \ and \ smallest \ together \ can \ fill \ it \ in \ 5 \ hours,\)

\(\displaystyle and \ the \ two \ smaller \ pipes \ can \ fill \ it \ in \ 12 \ hours.\)

\(\displaystyle How \ long \ would \ it \ take \ to \ fill \ the \ cistern \ if \ all \ three \ pipes \ were \ used?\)
 
largest : speed = a gph : cistern volume = 6a gallons
2nd largest : speed = b gph
smallest : speed = c gph

6a / (a + c) = 5 ; c = a / 5

6a / (b + c) = 12 ; 6a / (b + a/5) = 12 ; b = 3a / 10

hours if all working = 6a / (a+b+c) = 6a / (a + 3a/10 + a/5) = 60a / 15a = 4
 
\(\displaystyle Denis, \ this \ is \ how \ I \ did \ it \ sticking \ to \ the \ formula:\)

\(\displaystyle Rate \ X \ Time \ = \ Fraction \ of \ Work \ Done\)

\(\displaystyle Label \ pipes \ A \ (largest), \ B \ (medium), \ and \ C \ (smallest).\)

\(\displaystyle We \ are \ given \ that \ the \ rate \ of \ A \ = \ 1/6, \ hence \ if \ we \ find \ the \ rate \ of \ B \ and \ C, \ we \ are\)

\(\displaystyle in \ business.\)

\(\displaystyle \frac{5}{6}+\frac{5}{C} \ = \ 1 \ \implies \ C's \ rate \ is \ \frac{1}{30},\)

\(\displaystyle and \ \frac{12}{B}+\frac{12}{30} \ = \ 1 \ \implies \ B's \ rate \ = \ \frac{1}{20}\)

\(\displaystyle Ergo, \ \frac{x}{6}+\frac{x}{20}+\frac{x}{30} \ = \ 1 \ \implies \ x \ = \ 4 \ hrs. \ QED\)
 
Yes, agree (of course) Glenn.

However, as an introduction to this type of problem to students, you'd need to "show" a bit like:
ok you sob's:
1: in 1 hour, A does 1/6 of job: since we know he does job in 6 hours
2: so, after 5 hours working with C, A has done 5/6 of job ... got that?
3: so, it's evident that C has done 1 - 5/6 = 1/6 of job
4: so, in 1 hour, C does (1/6)/5 = 1/30 of job
...and likewise for B...

From that, you then "lead" to the "rate*time = fraction of work" formula.

And it can be made "clearer to students" by using the resulting 60 gallons capacity of cistern:
A: 60/6 = 10
A+C: 60/5 = 12
B+C: 60/12 = 5
B: 60/3 = 20
C: 60/2 = 30

I'm saying all this because I noticed that this kind of problem is often posted by students.
I also noticed that students have difficult time "seeing" that when "fractions of job" are used,
the "job" then equals 1.
 
\(\displaystyle Denis, \ all \ I \ can \ say \ is \ I \ damned \ if \ I \ do \ and \ I \ damned \ if \ I \ don't,\)

\(\displaystyle ergo, \ no \ mystique \ here, \ what \ you \ see \ is \ what \ you \ get.\)

\(\displaystyle Note: \ However, \ I \ am \ always \ open \ for \ constructive \ criticism, \ so \ thanks \ for \ your \ input.\)
 
BigGlenntheHeavy said:
\(\displaystyle Note: \ However, \ I \ am \ always \ open \ for \ constructive \ criticism, \ so \ thanks \ for \ your \ input.\)
Click to expand...
Intended no criticism, constructive or non-constructive.
Your solution is "perfecto".
I was just "typing out loud"!
 
For those who have worked out many of these, generally speaking, if it takes one person A units of time and another person B units of time to complete a specific task working alone, the time it takes them both to complete the task working together is T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.

You might have also deduced that the equivalant expression involving 3 people working alone and together which results in T = ABC/(AB + AC + BC).

Letting A be the largest pipe, C be the smallest and B be the one in between, we have:

A = 6
AC/(A + C) = 5 and
BC /(B + C) = 12
Substituting A = 6 into AC/(A + C) = 5 yields C = 30
Substituting A = 6 into BC/(B + C) = 12 yields B =20

Therefore, the time for all three to fill the tank becomes T = 6(20)30/[6(20) + 6(30) + 20(30)] = 4 hours.

Just a thought.
 
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