Answer :
Oh, I like this problem! Let's see what information we have here!
He burns, 7 cal/min when bicycling at 11 miles/hour
He burns, 11.75 cal/min when bicycling at 15 miles/hour
There is one notable problem here! The problem is in bold, and underlined. We have different units of time. We'll need to convert the cal/min into cal/hour. (Keep in mind we could instead convert the mph into mpmin, but that would lead to using much larger numbers making the problem a bit harder)
7.00 cal/min * (60 min / 1 hour) = 420 cal/hour
11.75 cal/min * (60 min / 1 hour) = 705 cal/hour
Let me quickly explain how I got the following conversion rate of 60min/1 hour. (Specifically how I knew it should be multiplied by 60 and not divided. Keep in mind that while in this problem it's obvious in others it isn't, so it'd be a good idea to know this!)
It'd be a good idea to do what I'm about to tell you to help you visualize this. Take a piece of paper, and write this:
7 cal
1 min
With a line between it. (Basically you are making a fraction... so 7/1)
In a problem where you want to convert to a different unit you will need to cancel that unit out. Canceling a unit out is very similar to canceling a variable out of the problem, or say canceling out 3/3 because it's 1. You simply need it to be over itself. So you look at the position of the variable you want to change and put it in the numerator or denominator of the conversion fraction depending on where it is currently. (If it's in the numerator then the conversion fraction will have it in the denominator!)
Once you find the place you need to put your current units you then put the unit you want to convert to!
(7 cal / 1 min) * (X min / Y hour)
You'll have to remember the conversion rates yourself, but if you ever end up converting to certain units like Kilogram, or a Microgram. The smaller unit is almost always the one that will have a value other than 1. (Remember that conversion rates always have a ratio of 1 to some number. Making them easy to work with. In the case of minutes to hours. There are 60 minutes in 1 hour) Once you work out the problem you'll "replace" the unit you canceled out with the one you found. In this case it is the hours.
(7 cal / 1 min) * (60 min / 1 hour) = 7 * 60 = 420 cal/hours.
420 cal/hour = 11.00 mph
705 cal/hour = 15.00 mph
In the following problem I'll drop the units for now to make it less cramped, but I'd recommend you do not do this. I'm just doing it to make it easier to understand for you, because it is much more difficult to absorb information on a screen from someone else than what you yourself have figured out and have written out.
(Every number should have cal/hour)
420x + 705y = 600
x represents the number of hours biking at 11.00 mph.
y represents the number of hours biking at 15.00 mph.
We stated above that each of the cal/hour have a number which they are equal to. We can use this to create another equation, which will look very similar with the "same" variables used.
I realized this after I got the answer that I overlooked one simple part of the problem that makes this a whole lot easier! My apologies if this frustrates you, but it is a part of math.
In this problem we need a way to substitute in that 1 hour is our time limit. So we need an equation that denotes 1 hour.
x + y = 1
Remember x and y both represent our time. So this is simply stating x and y added together will equal 1 (hour).
Solve for x.
x = 1 - y
Substitute in for x.
420x + 705y = 600
420 (1 - y) + 705y = 600
420 - 420y + 705y = 600
420 + 285y = 600
285y = 180
y = 180/285 = 0.6316
Use the fraction when substituting the value into solve, because the decimal is rounded!.
420x + 705 (180/285) = 600
420x = 600 - 705 (180/285)
x = (600 - 705 (180/285)) / 420
x = 0.3684
So our ACTUAL answer is he will bike 11.00 mph for 0.3684 hours, and 15.00 mph for 0.6316 hours!
He burns, 7 cal/min when bicycling at 11 miles/hour
He burns, 11.75 cal/min when bicycling at 15 miles/hour
There is one notable problem here! The problem is in bold, and underlined. We have different units of time. We'll need to convert the cal/min into cal/hour. (Keep in mind we could instead convert the mph into mpmin, but that would lead to using much larger numbers making the problem a bit harder)
7.00 cal/min * (60 min / 1 hour) = 420 cal/hour
11.75 cal/min * (60 min / 1 hour) = 705 cal/hour
Let me quickly explain how I got the following conversion rate of 60min/1 hour. (Specifically how I knew it should be multiplied by 60 and not divided. Keep in mind that while in this problem it's obvious in others it isn't, so it'd be a good idea to know this!)
It'd be a good idea to do what I'm about to tell you to help you visualize this. Take a piece of paper, and write this:
7 cal
1 min
With a line between it. (Basically you are making a fraction... so 7/1)
In a problem where you want to convert to a different unit you will need to cancel that unit out. Canceling a unit out is very similar to canceling a variable out of the problem, or say canceling out 3/3 because it's 1. You simply need it to be over itself. So you look at the position of the variable you want to change and put it in the numerator or denominator of the conversion fraction depending on where it is currently. (If it's in the numerator then the conversion fraction will have it in the denominator!)
Once you find the place you need to put your current units you then put the unit you want to convert to!
(7 cal / 1 min) * (X min / Y hour)
You'll have to remember the conversion rates yourself, but if you ever end up converting to certain units like Kilogram, or a Microgram. The smaller unit is almost always the one that will have a value other than 1. (Remember that conversion rates always have a ratio of 1 to some number. Making them easy to work with. In the case of minutes to hours. There are 60 minutes in 1 hour) Once you work out the problem you'll "replace" the unit you canceled out with the one you found. In this case it is the hours.
(7 cal / 1 min) * (60 min / 1 hour) = 7 * 60 = 420 cal/hours.
420 cal/hour = 11.00 mph
705 cal/hour = 15.00 mph
In the following problem I'll drop the units for now to make it less cramped, but I'd recommend you do not do this. I'm just doing it to make it easier to understand for you, because it is much more difficult to absorb information on a screen from someone else than what you yourself have figured out and have written out.
(Every number should have cal/hour)
420x + 705y = 600
x represents the number of hours biking at 11.00 mph.
y represents the number of hours biking at 15.00 mph.
We stated above that each of the cal/hour have a number which they are equal to. We can use this to create another equation, which will look very similar with the "same" variables used.
I realized this after I got the answer that I overlooked one simple part of the problem that makes this a whole lot easier! My apologies if this frustrates you, but it is a part of math.
In this problem we need a way to substitute in that 1 hour is our time limit. So we need an equation that denotes 1 hour.
x + y = 1
Remember x and y both represent our time. So this is simply stating x and y added together will equal 1 (hour).
Solve for x.
x = 1 - y
Substitute in for x.
420x + 705y = 600
420 (1 - y) + 705y = 600
420 - 420y + 705y = 600
420 + 285y = 600
285y = 180
y = 180/285 = 0.6316
Use the fraction when substituting the value into solve, because the decimal is rounded!.
420x + 705 (180/285) = 600
420x = 600 - 705 (180/285)
x = (600 - 705 (180/285)) / 420
x = 0.3684
So our ACTUAL answer is he will bike 11.00 mph for 0.3684 hours, and 15.00 mph for 0.6316 hours!