I'm studying basic algebra through Khan Academy at a late age (I'm in my 30s, but what I'm learning now is probably elementary to middle school).

The reason I didn't study math when I was younger was simply because I was stupid, and even though I'm studying now, I don't think that has changed.

Anyway, the part I don't understand is

In the ratio and proportionality unit

time×speed = distance

is a formula of the form

My confusion here is that

is the process of multiplying time when speed is a fraction.

I understand this algebraically

Time×Speed = Distance is

speed = distance/time and

distance/time(speed) × time = distance

like this.

But what I'm obsessing over here is this,

if 5/6 is a speed of 5 miles per 6 hours, then when you multiply that by 12 hours.

multiplied by the number in the numerator (5×12 = 60) divided by the number in the denominator, 6.

how does that divide into 5 miles per 6 hours?

How can the number 60 in the numerator be divided by 5 miles per 6 hours?

for example, if just 2/1, like 2 miles per hour, or the number 2

2 miles x 12 hours = 24 miles, which makes sense.

What about the case I mentioned earlier?

5 miles x 12 hours = 60 miles

Does that mean 60 miles/6 hours?

But how does that equation break down to 5 miles per 6 hours? Ha....

I'm so dumb I'm not even sure how to ask the question.

I don't know if you'll understand the question.

But I'd appreciate it if someone could help me out

  • 1
    $\begingroup$ If you walk at a steady pace of 5 miles for every 6 hours, then in 12 hours you will travel 10 miles. One calculation to see this is (5 miles / 6 hours) * (12 hours) = (60 / 6) miles = 10 miles. Another way to calculate the same thing is that in the first six hours you will travel exactly 5 miles, and in the second six hours you will also travel exactly 5 miles, so in total you will travel 5 + 5 = 10 miles. In each hour you will travel exactly 5/6 of one mile. $\endgroup$
    – JBL
    Mar 25 at 14:27
  • 1
    $\begingroup$ If you go 5 miles in 6 hours, and you have 12 hours: During the first 6 of the 12 hours you go 5 miles, and then during the remaining 6 of your 12 hours you go another 5 miles. So you have in all gone 5_5=10 miles in 12 hours. $\endgroup$
    – coffeemath
    Mar 25 at 14:28
  • $\begingroup$ By the way, fractions are a genuinely difficult to understand abstraction; just because we teach them to young students doesn't mean the ideas are simple! $\endgroup$
    – JBL
    Mar 25 at 14:32
  • $\begingroup$ @coffeemath you have a crucial typo in your last equation. $\endgroup$
    – JBL
    Mar 25 at 14:32
  • $\begingroup$ I know the answer is 10, guys. I wasn't asking how to do division. I don't know how to convey to you the question in my stupid head. $\endgroup$
    – guss2222
    Mar 25 at 14:46

2 Answers 2


I think you need to distinguish the formula from the calculation.

When you say that $$ \frac 56\text{miles per hour}\times 12\text{hours}=10\text{miles} $$ then that indeed says that if your speed is such that you travel 5 miles in 6 hours, and you travel for 12 hours, then you travel 10 miles. (You don't really need a formula for that. Just read that sentence once or twice, and it should be obvious: Every 6 hours you have traveled 5 miles. If you travel for 6 hours and then 6 hours more, then the total distance must be 10 miles.)

But to get to $10$, you have to do the arithmetic. And at one point in that calculation you have the fraction $$ \frac{60}6 $$ This fraction is actually entirely divorced from any notion of time, speed or distance. There is no actual 60 anywhere in that travel.

There are several schools of thoughts on the good way to think about math as you solve problems, but for a relatively simple problem like this, I think I would personally advocate for the affectionately-named "shut up and calculate" approach. It goes something like this:

  1. Read the problem, and use what formulas and understanding you have to set up an actual calculation that will solve the problem. In your case that's $\frac 56\text{miles per hour}\times 12\text{hours}$.
  2. Forget the problem, forget the formulas and the understanding, just do the calculations you set up for yourself. In this case, that means throw away the "miles per hour", throw away the "hours", forget anything that has to do with travel, and just do $\frac 56\times 12$ the way you always do fractional arithmetic.
  3. Once you get the answer $10$, trust that everything was set up correctly, and add back the "miles" that should be there. Remember that this was a problem about travel, and put the answer "10 miles" in that context to answer the question you were actually asked.

This isn't an approach you should stick with forever. And especially when problems become more complex and have more parts to them, it could be useful to stop and contextualise the intermediate parts as well. But in this particular case I think a drive to contextualise every intermediate step is a big part of what's tripping you up.


I guess the question is also about units? When you said "5 miles x 12 hours = 60 miles", this is actually not correct, and might have led to your confusion about "60 miles/6 hours" later.

Your initial goal is to multiply a speed (given as a fraction) and a time, and you tried to first multiply the numerator and the time, which is OK:

$$\text{Distance travelled} = \frac{5\text{ miles}}{6\text{ hours}} \times 12 \text{ hours} = \frac{5\text{ miles} \times 12 \text{ hours}}{6\text{ hours}}$$

But the numerator unit is not simply "miles", and should be "miles" and "hours" multiplied:

$$\ldots = \frac{5\text{ miles} \times 12 \text{ hours}}{6\text{ hours}} = \frac{60\text{ mile-hours}}{6\text{ hours}}$$

The "$60\text{ mile-hours}$" alone is not a common physical quantity, and "mile-hour" is not a common unit. But divide that by $6\text{ hours}$, and all is good:

$$\ldots = \frac{60\text{ mile-hours}}{6\text{ hours}} = 10\text{ miles}$$

Often you might see that all intermediate units are omitted as followed, but in the background the units still exist:

$$\text{Distance travelled} = \frac{5}{6} \times 12 = \frac{5\times 12}{6} = \frac{60}{6} = 10\text{ miles}$$

Your question should be distinguished from a completely different question that involves the same numerical multiplication but different units:

If person G travels in a speed of $\frac 56$ miles per hour, and person P travels $12$ times as fast, what is P's speed?

The difference here is that the "$12$ times" has no units. You may still represent G's speed as $5\text{ miles}$ per $6\text{ hours}$:

$$\text{P's speed} = \frac{5\text{ miles}}{6\text{ hours}} \times {12}_\text{(unitless!)} = \frac{5\text{ miles} \times 12}{6\text{ hours}}$$

This time the numerator really has a unit of "miles", and the numerator alone represents the distance P would travel in $6\text{ hours}$:

$$\ldots = \frac{5\text{ miles} \times 12}{6\text{ hours}} = \frac{60\text{ miles}}{6\text{ hours}} = 10 \text{ miles per hour}$$

It also makes sense that this result shouldn't "break down to 5 miles per 6 hours", because one shouldn't expect these two (non-zero) speeds to be equal.

  • $\begingroup$ I thought to myself after I asked this question, the point of my question was this rather than units. "If you just divide the multiplied number in the numerator by '6', how does that divide by '5 miles per 6 hours'?" But I thought about it myself, and it's also in your answer. The numerator is distance x time and the denominator is time. That is, (distance × time) ÷ time. Isn't it just to think that dividing by time gives the number of multiplied distances? $\endgroup$
    – guss2222
    Mar 29 at 2:08
  • $\begingroup$ @guss2222 Anyway I was just not satisfied by two lines in your question: "2 miles x 12 hours = 24 miles" and "5 miles x 12 hours = 60 miles", and wrote a full answer about it. $\endgroup$
    – peterwhy
    Mar 29 at 2:15
  • $\begingroup$ @guss2222 You may also edit your question to clarify your thought and clean up. $\endgroup$
    – peterwhy
    Mar 29 at 2:21
  • $\begingroup$ Since the question has already been answered, I won't change it. And for what you pointed out, I don't think so anymore thanks to your answer. Thank you for taking the time to respond! $\endgroup$
    – guss2222
    Mar 29 at 2:25

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