how many gallons of gasoline will the car use in 5 hours

eddy2017

eddy2017

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Hi, I need help in how to set this up and why, the logic behind it. How do I instantly recognize that is the path I should take?.

A car travels 200 miles at an average speed of 55 miles per hour and has a gas mileage of 22 miles per gallon of gasoline. If the car continues to travel at the same speed and continues getting the same mileage per gallon of gasoline, how many gallons of gasoline will the car use in 5 hours?

My try:

The car travels 200 miles.
it is going at 55 miles per hour
it uses up 1 gallon every 22 miles
x = the number of gallons of gas used /burned in 5 hours.

Please, I need help with this. I don't want you to give me the answer. I need you to teach me how to think logically to get better at word problems. I do struggle with them.

Thanks a lot,
eddy
 
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It appears to me that you know some algebra. So

[MATH]x = \dfrac{\text {gallons}}{\text {hour}}.[/MATH]
You know that

[MATH]55 = \dfrac{\text {miles}}{\text {hour}}[/MATH]
[MATH]22 = \dfrac{\text {miles}}{\text {gallon}}.[/MATH]
Using those two known fractions, how can you compute x, which does not involve miles and has gallons over hours. Obviously addition will not work. How about subtraction? What other operations come to mind?

Once you know x, how will you get your final answer?

(This approach is called dimensional analysis. It is very useful when dealing with ratios involving units.)
 
One gallon of gas every 22 miles is 1/22 "gallon per mile".

Driving 5 hours at 55 miles/hour, you will go (5 hours)(55 miles/hour)= 275 miles. AT 1/22 gallon per mile that will require 275/22= 12.5 gallons of gas.
 
eddy2017 said:
Hi, I need help in how to set this up and why, the logic behind it. How do I instantly recognize that is the path I should take?.

A car travels 200 miles at an average speed of 55 miles per hour and has a gas mileage of 22 miles per gallon of gasoline. If the car continues to travel at the same speed and continues getting the same mileage per gallon of gasoline, how many gallons of gasoline will the car use in 5 hours?

My try:

The car travels 200 miles.
it is going at 55 miles per hour
it uses up 1 gallon every 22 miles
x = the number of gallons of gas used /burned in 5 hours.

Please, I need help with this. I don't want to to give me the answer. I need you to teach me how to think logically to get btter at word problems. I do struggle with them.

Thanks a lot,
eddy
Click to expand...
First calculate

how far did the car travel in 5 hours = 55 mile/hr * 5 hr = 275 mile

How much gas is used to travel 275 miles = (275 mile) / (22 mile/gal.) ......... continue......
 
Thanks a lot!!!.
Now, it is going to be easy, I think, to continue.
Following JeffM reasoning:
x= gallons/hour
55mph= miles/hour
55mph= x/hours
55mph = x miles/5hr
solve for x
x= 55 mph x 5hr (of travel)
x= 275 miles (total of miles driven)

So,
22mph (car fuel efficiency)= total of miles driven/gallons
22mph= 275/x ( I still don't know how many gallons the car will use)

Solving for x

275/22 = x
x =12.5 gallons needed to go for five hours.

Is this good?
I want to thank you all,
Jeff set it up, but Hall and Sub, thank you very much because I was stuck as how to continue after Jeff's explanation.
:thumbup:
 
You got the correct answer, which is GREAT.

What I was hoping that you would see is

[MATH]\dfrac{\text {miles}}{\text {hour}} \div \dfrac{\text {miles}}{\text {gallon}} = \dfrac{\cancel { \text {miles}}}{\text {hour}} \times \dfrac{\text {gallon}}{\cancel { \text {miles}}} = \dfrac{\text {gallons}}{\text {hour}} = x.[/MATH]
[MATH]\dfrac{\text {miles}}{\text {hour}} \div \dfrac{\text {miles}}{\text {gallon}} = 55 \div 22 = \dfrac{5}{2} = 2.5.[/MATH]
So to drive 5 hours, the gallons you need are

[MATH]5x = 5 * 2.5 = 12.5.[/MATH]
 
Jeff, could I ask you to explain the setup that you did for me?. when you wrote you were hoping me to see that setup.
I would really like to have it explained by you. And I know it will fix it better in my mind. you used dimensional analysis which is something I studied a little but, but still fail to see it up there in your setup.
Thanks a lot
 
Dimensional analysis is a mechanical way to organize your thoughts.

First, remember that [MATH]a = \dfrac{a}{1}.[/MATH]
Now in this problem you are given

[MATH]55 = \dfrac{\text {miles}}{\text{hour}},[/MATH]
[MATH]22 = \dfrac{\text {miles}}{\text {gallon}}, \text { and}[/MATH]
[MATH]5 = \text {hours.}[/MATH]
And you want an answer in gallons.

Now there is a short cut when you have numerous ratios. You treat the ratios as fractions and cancel units.

You want an answer in gallons. So [MATH]x \text { gallons} = \dfrac{x \text { gallons}}{1}.[/MATH]
You want gallons in the numerator. But you have nothing with gallons in the numerator although you have something with gallons in the denominator. Numerators and denominators switch places when you divide by a fraction. So our calculation will include

[MATH] \dfrac{\text {something}}{\dfrac{22 \text { miles}}{\text {gallon}}} = \text {something} * \dfrac{\text {gallons}}{22 \text { miles}}.[/MATH]
How do we get rid of miles in the denominator? If we multiply by something with miles in the numerator, miles will cancel. And we do have a fraction with miles in the denominator. Thus

[MATH]\dfrac{55 \cancel {\text { miles}}}{\text {hour}} * \dfrac{\text {gallon}}{22 \cancel {\text { miles}}} = \dfrac{5 \text { gallons}}{2 \text { hour}}.[/MATH]
Now we need to get hours out of the denominator because we want an answer in gallons. The way to do that is to multiply by some number of hours.

[MATH]5 \text { hours} \implies \dfrac{5 \text { hours}}{1} \implies \dfrac{5 \cancel {\text { hours}}}{1} * \dfrac{5 \text { gallons}}{2 \cancel {\text { hours}}} = \dfrac{25}{2} \text { gallons} = 12.5 \text { gallons}.[/MATH]
All we were doing is figuring out how to cancel units to get where we wanted to go. No real thought.

Because of the lack of thought, we should check our answer for reasonableness. How far would we go in 5 hours at 55 miles per hour? 275 miles. How many gallons does that take at 22 miles per gallon. How many times does 22 go into 275? 12.5 So our answer looks good. The value of dimensional analysis is that it's very easy to do cancellations.

[MATH][/MATH]
 
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In the solution of the problem above only one aspect of the process of dimensional analysis above was used - balancing dimensions. That aspect is kind of "thought-less".

There are many aspects of dimensional analysis that require lot of thought and induce inventive "aha.." moments. I have to say that many laws of fluid-dynamics were first postulated through "dimensional analysis". Those postulates became law after some length time when the "intuition" of dimensional analysis was supported by experiments and "inductive logic". Of course "that" dimensional analysis is slightly different from "cancelling units". But cancelling units is always an important step of the process.

In the types of problems above, dimensional analysis can guide the "next step" to take.

One of the things to remember while checking answers with dimensions - if the dimensions do not match, "something" is definitely wrong.

However, if the dimensions match (balanced) - the solution could still be wrong (reasonable or not).
 
As always, Subhotosh has helpful thoughts. Personally, I'd be reluctant to use a discovery in physics as an example of the value of dimensional analysis because of the critical importance of replicable empirical validation to laws in the physical sciences. But Popper's distinction between the psychology of discovery and the logic of discovery means that my reluctance is not fundamental.

In any case, I really like SK's point that a calculation resulting in the correct units does not guarantee that the calculation is correct. That is a more profound way to justify my comment that you must ask whether the answer achieved by dimensional analysis makes sense. But even more important is his point that any calculation resulting in incorrect units is guaranteed to be wrong.

I still believe that dimensional analysis itself, the process of deciding to multiply or divide by cancelling unwanted units, is essentially mindless. Although there can be some thought involved in setting up the mechanism in the first place, that thought does not address the important question of whether the mechanism will generate a valid answer.
 
As I see it, dimensional analysis is based on an assumption of proportionality. If everything in a problem is proportional (in the simplest possible way, usually), then it is applicable; and this is common in the sorts of problems for which it is recommended. But far too often students are introduced to it in such a way that the assumptions are left unstated, and the method appears to be magical. They are not taught to first judge whether proportionality is likely (not to mention proving it is actually valid). This is, in fact, true of many applications of proportion. Blindly assuming it is bad math, because it is math misapplied.

This is why I like to approach such problems one piece at a time (though I might be writing the same chain of fractions as I do so). So, rather than think, "we need to cancel miles from the denominator," I would be thinking, "so far we have the number of gallons used per mile [assumed to be proportional, though it isn't really, if speed and driving conditions vary]; now we need to relate miles to hours [also proportional, because the speed is constant], ..."

In science, on the other hand, you might be saying, "the units suggest that it's possible that this is proportional to that; let's find out if it is."
 
Dr.Peterson said:
As I see it, dimensional analysis is based on an assumption of proportionality. If everything in a problem is proportional (in the simplest possible way, usually), then it is applicable; and this is common in the sorts of problems for which it is recommended. But far too often students are introduced to it in such a way that the assumptions are left unstated, and the method appears to be magical. They are not taught to first judge whether proportionality is likely (not to mention proving it is actually valid). This is, in fact, true of many applications of proportion. Blindly assuming it is bad math, because it is math misapplied.

This is why I like to approach such problems one piece at a time (though I might be writing the same chain of fractions as I do so). So, rather than think, "we need to cancel miles from the denominator," I would be thinking, "so far we have the number of gallons used per mile [assumed to be proportional, though it isn't really, if speed and driving conditions vary]; now we need to relate miles to hours [also proportional, because the speed is constant], ..."

In science, on the other hand, you might be saying, "the units suggest that it's possible that this is proportional to that; let's find out if it is."
Click to expand...
My sentiments exactly ......
 
Jeff, thank you and thank you all for your help and commnets. the sicussin got a little bit ot techy for me, but I enjoyed anyways.:D
Just one question. I suppose that the 5gallons/2hours comes from the reduction of 55 gallons/22 miles.
Am I right?.
 
eddy2017 said:
Jeff, thank you and thank you all for your help and commnets. the sicussin got a little bit ot techy for me, but I enjoyed anyways.:D
Just one question. I suppose that the 5gallons/2hours comes from the reduction of 55 gallons/22 miles.
Am I right?.
Click to expand...
You are indeed!
 
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