Can someone please help me explaining how to convert units in problems like the one below? I've done many things, but I am very unsure of all my methods. Could you tell me what steps should I take?

$$5.66\ \frac{\mathrm{hm}}{\mathrm{min.}} \frac{\hspace{2cm}}{}\frac{\mathrm{km}}{\mathrm{hr.}}$$

Thank you.


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  • $\begingroup$ Based on Kevin's answer, which I assume is what is wanted, your teacher has made the mistake of using $m$ for minutes. $m$ usually means meters in a physics context, and certainly means meters anywhere that the $m$ in $Km$ and $Hm$ are referring to meters. $\endgroup$ – Mark Eichenlaub Apr 6 '11 at 5:11
  • 2
    $\begingroup$ this site isn't really for very specific questions about, say, one particular homework problem. We prefer questions that will be useful to the general audience of the internet (basically, anyone who comes across this looking for information on unit conversions). But I thought your question was very close to meeting that criterion, so I edited it a bit to make it more general. I hope I haven't made it any less useful to you ;) $\endgroup$ – David Z Apr 6 '11 at 6:37
  • $\begingroup$ Ah, sorry, I'll try to be more general in the future. And thank you for answering, it helped me :). I think I did a good job in the exam... it wasn't a homework ultimately! it was an unfinished practice in my notebook haha =D $\endgroup$ – Omega Apr 6 '11 at 21:38

The way you convert between units is really just multiplying by several factors of 1. But it's 1 written in a slightly unusual way. Think about this: you're probably familiar with conversion factors in the form

$$(\text{number})(\text{unit}) = (\text{other number})(\text{other unit})$$

But of course, you can divide both sides of any equation by the same thing, and the equation will continue to hold. So you can also write the conversion factor as

$$\frac{(\text{number})(\text{unit})}{(\text{other number})(\text{other unit})} = 1$$


$$\frac{(\text{other number})(\text{other unit})}{(\text{number})(\text{unit})} = 1$$

Same equation, just rearranged a bit.

Now let's see what happens when you have some value, expressed in $(\text{unit})$, that you want to convert to $(\text{other unit})$. As you know, you can always multiply anything by 1, and that doesn't change the value at all. So, you can go look up the conversion relationship between $(\text{unit})$ and $(\text{other unit})$, change it into one of the forms above, and use it like this:

$$(\text{given value})(\text{unit})\times 1 = (\text{given value})(\text{unit})\times \frac{(\text{other number})(\text{other unit})}{(\text{number})(\text{unit})}$$

Since you have $(\text{unit})$ in both the numerator and denominator, you can cancel those out. Note that it's up to you to pick the right conversion factor to use so that you can cancel out units! Anyway, you'll be left with

$$\frac{(\text{given value})(\text{other number})}{(\text{number})}(\text{other unit})$$

Everything except for $(\text{other unit})$ is just a number, so now you can actually figure out the numerical value (e.g. plug things into your calculator if necessary).

If your given value has a compound unit (one made as a combination of other units, as in your examples), then you will need to use more than one unit conversion factor - basically, you will have to multiply by 1 more than once, using a different conversion factor each time. In your example, you would have to multiply by a conversion factor to go from $\mathrm{hm}$ to $\mathrm{km}$, and another one to go from $\mathrm{min.}$ to $\mathrm{hr.}$

Here's how that applies to your example, assuming Mark Eichenlaub is right about the mistake:

  • $(\text{given value}) = 5.66$

For one part of the unit conversion:

  • $(\text{unit}) = \mathrm{hm}$
  • $(\text{other unit}) = \mathrm{km}$
  • The conversion factor is $10\text{ hm} = 1\text{ km}$

For the other part of the unit conversion:

  • $(\text{unit}) = \mathrm{min.}$
  • $(\text{other unit}) = \mathrm{hr.}$
  • The conversion factor is $60\text{ min.} = 1\text{ hr.}$

Note that the time units are in the denominator! But that doesn't really change anything - you still just need to arrange the conversion factor in the way that makes the original units cancel out.


The secret to unit conversions is to keep multiplying by 1. However, we need specially chosen values of 1.

A common value of 1 we can use is 60 min/hour or 3600 sec/hour.

Another commonly used value of 1 is 1000 m/km, and it's inverse, km/1000 m.

In your case, we need to do two conversions. hm -> km. So our chosen value of 1 is km/10hm.

The other one is min -> hr, so we use 60 min/hr.

$$5.66\ \frac{\mathrm{hm}}{\mathrm{min.}} \frac{\mathrm{km}}{\mathrm{10hm}} \frac{\mathrm{60min}}{\mathrm{hr}}=33.96\frac{\mathrm{km}}{\mathrm{hr.}}$$

Notice that the hm divides out, and the min divides out, leaving km/hr.


I think David's answer is the right one.

Here is the wrong one, but it's the way I actually think about unit conversions when I do them. (I mean "wrong" in the sense of "unofficial", not that it gives the wrong answer.)

Suppose I tell you there are 5280 feet in a mile, and ask how many feet are in two miles. You don't need to think about what to do. It's obvious there are twice as many feet in two miles as in one, so it's 2*5280 = 10,560 ft.

Suppose you want to convert 15 miles per hour to meters per second. Do it in two steps. Go from miles per hour to meters per hour, and then from meters per hour to meters per second.

There are about 1609 meters in a mile, so how many meters per hour are you going? In one hour you go 15 miles, so you go 15*1609 = 24135 meters in an hour. You have to multiply because you go a lot more meters (which are small) than miles (which are big).

Next, there are 3600 seconds in an hour. If you go 24135 meters in an hour, how many meters do you go per second? Seconds are short, while hours are long. You clearly go a much shorter distance in one second than in one hour, so divide by 3600 to get 24135/3600 = 6.7 m/s.

Also, you can always check your answer using Google like this: http://www.google.com/search?&q=15+mph+in+m/s

Or using Wolfram Alpha like this: http://www.wolframalpha.com/input/?i=15+mph+in+m%2Fs

  • $\begingroup$ In practice this is what I do too - and in fact I think this is a better way to think about it if you know what you're doing. (Don't sell yourself short by calling it "wrong" :-P) But it's more like a bag of tricks than a systematic procedure, so I don't think it's particularly well suited for someone who is just learning to do unit conversions and having a hard time with them. $\endgroup$ – David Z Apr 6 '11 at 20:29

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