Ratio Algebra Word Problem

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Jason76

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An inlet pipe can fill a tank in 12 hours. A second pipe can fill the tank in 15 hours. If both pipes are used, find how long it takes to fill the tank.
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\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = \dfrac{1}{x}\)

\(\displaystyle LCD = 60x\)

Multiplying it out

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)\dfrac{1}{x}\)

Multiplying this out should give (or am I doing something wrong):

\(\displaystyle \dfrac{60x^{2}}{12} + \dfrac{60x^{2}}{15} = \dfrac{60x}{x}\)

I want to make sure the above step is right before moving on.
 
An inlet pipe can fill a tank in 12 hours. A second pipe can fill the tank in 15 hours. If both pipes are used, find how long it takes to fill the tank.
Click to expand...
Jason76 said:
\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = \dfrac{1}{x}\)
Click to expand...
For what does \(\displaystyle x\) stand? How is it defined?

Thank you! ;)
 
stapel said:
For what does \(\displaystyle x\) stand? How is it defined?

Thank you! ;)
Click to expand...

I'm sorry, but that's the whole problem. There isn't any other info given.

The practice test was saying that \(\displaystyle 60x(x) = 60x\) NOT \(\displaystyle 60x^{2}\) Is this right, or aren't exponents supposed to be added?

For instance, \(\displaystyle 5x^{2}(9x^{3}) = 45x^{5}\). So wouldn't \(\displaystyle 60x^{1}(x^{1}) = 60x^{2}\) ?
 
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Jason76 said:
\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = \dfrac{1}{x}\)

\(\displaystyle LCD = 60x\)

Multiplying it out

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)\dfrac{1}{x}\)

Multiplying this out should give (or am I doing something wrong):

\(\displaystyle \dfrac{60x^{2}}{12} + \dfrac{60x^{2}}{15} = \dfrac{60x}{x}\) <------- !

I want to make sure the above step is right before moving on.
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Jason76, a main point by multiplying by the LCD is so that you cancel (reduce) so that you have no fractions: \(\displaystyle (5x)(x) + (4x)(x) = 60\) \(\displaystyle \ \ \ \) By canceling out common factors, you keep from undoing the work of multiplying the LCD by numerators of the fractions in the equation to begin with.
Jason76 said:
For instance, \(\displaystyle 5x^{2}(9x^{3}) = 45x^{5}\). So wouldn't \(\displaystyle 60x^{1}(x^{1}) = 60x^{2}\) ?
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Yes, if that were the product that was given.
 
lookagain said:
Jason76, a main point by multiplying by the LCD is so that you cancel (reduce) so that you have no fractions: \(\displaystyle (5x)(x) + (4x)(x) = 60\) \(\displaystyle \ \ \ \) By canceling out common factors, you keep from undoing the work of multiplying the LCD by numerators of the fractions in the equation to begin with.
Yes, if that were the product that was given.
Click to expand...

But isn't \(\displaystyle x\) the same as \(\displaystyle x^{1}\)?
 
An inlet pipe can fill a tank in 12 hours. A second pipe can fill the tank in 15 hours. If both pipes are used, find how long it takes to fill the tank.
Click to expand...

Jason76 said:
\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = \dfrac{1}{x}\)
Click to expand...

I'm sorry, but that's the whole problem. There isn't any other info given.
Click to expand...

So you are saying the word problem on the top and the equation given in the bottom has no connection (at least any that you know of!)?

That is a strange problem statement!!

Can you derive the equation needed to solve the word problem?
 
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Jason76 said:
An inlet pipe can fill a tank in 12 hours. A second pipe can fill the tank in 15 hours. If both pipes are used, find how long it takes to fill the tank.

\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = \dfrac{1}{x}\) Where in the world did this come from?

\(\displaystyle LCD = 60x\)

Multiplying it out

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)\dfrac{1}{x}\)

Multiplying this out should give (or am I doing something wrong):

\(\displaystyle \dfrac{60x^{2}}{12} + \dfrac{60x^{2}}{15} = \dfrac{60x}{x}\)

I want to make sure the above step is right before moving on.
Click to expand...
\(\displaystyle Let\ x = the\ volume\ of\ the\ tank.\)

\(\displaystyle So\ the\ rate\ of\ flow\ of\ the\ bigger\ pipe\ = \dfrac{x}{12}.\)

\(\displaystyle And\ the\ rate\ of\ flow\ of\ the\ smaller\ pipe\ = \dfrac{x}{15}.\)

\(\displaystyle So\ the\ rate\ of\ flow\ of\ both\ pipes\ together\ = \dfrac{x}{12} + \dfrac{x}{15} = \dfrac{5x}{5 * 12} + \dfrac{4x}{4 * 15} = \dfrac{9x}{60}.\)

So the two pipes together could fill a tank 9 times as big in 60 hours, which means together they can fill this tank in 60/9 hours or 400 minutes.

Let's check using a numeric example.

Say the tank holds 900 gallons.

So big pipe has a rate of flow of 900 gallons / 12 hours = 75 gallons per hour = 1.25 gallons / minute.

Little pipe has a rate of flow of 900 gallons / 15 hours = 60 gallons per hour = 1 gallon / minute.

Together the rate of flow is 2.25 gallons per minute. \(\displaystyle \dfrac{900\ gallons}{2.25\ gallons\ /\ minute} = \dfrac{900}{2.25}\ minutes = 400\ minutes.\)
 
Jason76 said:
I'm sorry, but that's the whole problem. There isn't any other info given.

The practice test was saying....
Click to expand...
Are you saying that what you'd posted wasn't your work and you weren't asking for us to check your work to that point? That what you'd posted was somebody else's work and you were asking us to explain it to you or correct a possible typo in some portion of it? ;)
 
Here are typical information/equations to use:

Let x = the total time in hours that is needed to complete the job when both pipes are in operation.

The smaller pipe can fill the tank in 15 hours, so it can fill 1/15 of the volume of the tank in 1 hour.
In x amount of time, it can fill x/15 of the volume of the tank.


The larger pipe can fill the tank in 12 hours, so it can fill 1/12 of the volume of the tank in 1 hour.

In x amount of time, it can fill x/12 of the volume of the tank.

Let 1 represent 100% of the volume.

Then, \(\displaystyle \ \dfrac{x}{15} + \dfrac{x}{12} = 1. \ \ \ \) The solution to that equation gives the amount of time it takes both pipes to do 100% of the job.

Or, sometimes the equation is presented as \(\displaystyle \ \dfrac{1}{15} + \dfrac{1}{12} = \dfrac{1}{x}, \ \ but \ \ that \ \ form \ \ is \ \ less \ \ intuitive \ \ to \ \ me. \)


A main idea is (time in hours)*(volume per time in hours) = the total volume.

So, the equation could also be presented as \(\displaystyle \ \ x\bigg(\dfrac{1}{15} + \dfrac{1}{12}\bigg) = 1.\)


The LCD is 60, so after multiplying each side by 60, the equation may be written as:

4x + 5x = 60.


Solve that equation for x, and you'll have the time in hours it takes both pipes to fill the tank.
 
I agree .... but

Denis said:
General:
a = time by a certain worker (12)
b = time by another worker (15)
t = time if both work together (?)

t = ab / (a + b) ← \(\displaystyle \ \dfrac{1}{t} \ = \ \dfrac{1}{a} \ + \ \dfrac{1}{b} \)

If a 3rd worker with time c is added:

t = abc / (ab + ac + bc) ← \(\displaystyle \ \dfrac{1}{t} \ = \ \dfrac{1}{a} \ + \ \dfrac{1}{b} \ + \ \dfrac{1}{c} \)

This way I find it easier to remember....
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I talked with my math professor today. Another professor wrote the problem wrong. I know the procedure on solving it, but the wrong writing of the problem threw me off. There should be 1 values, NOT x values, above the 12 and 15. But the x below the 1 is correct. Sorry guys for all the trouble, but I knew something was up with all the "x squared stuff" I was coming up with.
 
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Actually, you could keep the x values in the numerators, and get rid of the x in the denominator. I think.

Because in this college algebra textbook I was looking at, they were doing that for the same type problem. But I don't know what they were thinking when they made a problem with x values in the numerator AND a x value in the denominator (as with the original posting of the problem).

But what about 1 values in the numerator and an x value in the denominator. I DON'T think ALL of these methods could yield a correct answer.

Original posting of problem:

\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = \dfrac{1}{x}\)

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)\dfrac{1}{x}\)

Made Similar to the One in my College Algebra Textbook:

\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = 1\)

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)1\)

The Other Way

\(\displaystyle \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{1}{x}\)

\(\displaystyle (60x)\dfrac{1}{12} + (60x)\dfrac{1}{15} = (60x)\dfrac{1}{x}\)

Now, ALL 3 of these ways CANNOT yield the same answer. Agree or disagree? :rolleyes:
 
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You already posted that the original is incorrect (per your professor). Are you questioning it again? :-?
 
Jason76 said:
In that case, can the problems, minus the proven incorrect one, both yield the same answer?
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They do yield exactly the same answer. What is confusing you (I think) is that x stands for something different in the two formulations.

Correct Formulation 1

\(\displaystyle x = gallons\ in\ tank.\)

\(\displaystyle input\ rate\ of\ fast\ pipe = x / 12.\)

\(\displaystyle input\ rate\ of\ slow\ pipe = x / 15.\)

\(\displaystyle input\ rate\ of\ both\ pipes\ together = r = \dfrac{x}{12} + \dfrac{x}{15}.\)

\(\displaystyle hours\ to\ fill\ tank\ using\ both\ pipes = h.\)

\(\displaystyle But\ r = \dfrac{x}{h}= \dfrac{x}{12} + \dfrac{x}{15} = \dfrac{9x}{60} \implies h = \dfrac{60}{9}\ hours = 400\ minutes.\)

x is in the numerators because it represents gallons.

Correct Formulation 2

\(\displaystyle g = gallons\ in\ tank.\)

\(\displaystyle input\ rate\ of\ fast\ pipe = g / 12.\)

\(\displaystyle input\ rate\ of\ slow\ pipe = g / 15.\)

\(\displaystyle input\ rate\ of\ both\ pipes\ together = r = \dfrac{g}{12} + \dfrac{g}{15} = g\left(\dfrac{1}{12} + \dfrac{1}{15}\right).\)

\(\displaystyle hours\ to\ fill\ tank\ using\ both\ pipes = x.\)

\(\displaystyle But\ r = \dfrac{g}{x}= g\left(\dfrac{1}{12} + \dfrac{1}{15}\right) \implies \dfrac{1}{x} = \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{9}{60} \implies x = \dfrac{60}{9}\ hours = 400\ minutes.\)

x is in the denominator because it represents hours.
 
Looking at the 1st one.

\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = 1\)

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)1\)

\(\displaystyle \dfrac{60x^{2}}{12} + \dfrac{60x^{2}}{15} = 60x\)

\(\displaystyle 5x^{2} + 4x^{2} = 60x\)

\(\displaystyle 9x^{2} = 60x\)

Will finish soon.

Looking at the 2nd one.

\(\displaystyle \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{1}{x}\)

\(\displaystyle (60x)\dfrac{1}{12} + (60x)\dfrac{1}{15} = (60x)\dfrac{1}{x}\)

\(\displaystyle \dfrac{60x}{12} + \dfrac{60x}{15} = \dfrac{60x}{x}\)

\(\displaystyle 5x + 4x = 60\)

\(\displaystyle 9x = 60\)

\(\displaystyle x = \dfrac{60}{9}\)

\(\displaystyle x = \dfrac{20}{3}\)
 
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Jason76 said:
Looking at the 1st one.

\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = 1\)

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)1\)

\(\displaystyle \dfrac{60x^{2}}{12} + \dfrac{60x^{2}}{15} = \dfrac{60x}{x}\)

\(\displaystyle 5x^{2} + 4x^{2} = 60\)................Incorrect - should be

\(\displaystyle 5x^{2} + 4x^{2} = 60x \)

\(\displaystyle 5x + 4x = 60\)

x = 20/3

Looking at the 2nd one.

\(\displaystyle \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{1}{x}\)

\(\displaystyle (60x)\dfrac{1}{12} + (60x)\dfrac{1}{15} = (60x)\dfrac{1}{x}\)

\(\displaystyle \dfrac{60x}{12} + \dfrac{60x}{15} = \dfrac{60x}{x}\)

\(\displaystyle 5x + 4x = 60\)

\(\displaystyle 9x = 60\)

\(\displaystyle x = \dfrac{60}{9}\)

\(\displaystyle x = \dfrac{20}{3}\)
Click to expand...
.
 
We have now proven the two strategies produce the same answer of \(\displaystyle \dfrac{20}{3}\)

Looking at the 1st one.

\(\displaystyle \dfrac{x}{12} + \dfrac{x}{15} = 1\)

\(\displaystyle (60x)\dfrac{x}{12} + (60x)\dfrac{x}{15} = (60x)1\)

\(\displaystyle \dfrac{60x^{2}}{12} + \dfrac{60x^{2}}{15} = 60x\)

\(\displaystyle 5x^{2} + 4x^{2} = 60x\)

\(\displaystyle 9x^{2} = 60x\)

\(\displaystyle \dfrac{9x^{2}}{x}= \dfrac{60x}{x}\)

\(\displaystyle 9x = 60\)

\(\displaystyle x = \dfrac{60}{9}\)

\(\displaystyle x = \dfrac{20}{3}\)

Looking at the 2nd one.

\(\displaystyle \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{1}{x}\)

\(\displaystyle (60x)\dfrac{1}{12} + (60x)\dfrac{1}{15} = (60x)\dfrac{1}{x}\)

\(\displaystyle \dfrac{60x}{12} + \dfrac{60x}{15} = \dfrac{60x}{x}\)

\(\displaystyle 5x + 4x = 60\)

\(\displaystyle 9x = 60\)

\(\displaystyle x = \dfrac{60}{9}\)

\(\displaystyle x = \dfrac{20}{3}\)
 
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