Rate Word Problems - Problem 1
All right guys this is it, this is one of the most difficult problems you are going to see all year long and if you show your parents these problems I bet they're going to say, "I remember that problem that exact problem from algebra. This is the problem that shows up in the comic book strips because people think it's really tricky, but we're going to go through it and just try to remember distance equals rate times time.
One plane leaves New York City for LA going to 500mph. At the same time a different plane leaves LA for New York City going 650mph. If the 2 cities are approximately 3000 miles apart how far from LA are they when thy cross?
Okay so before we start this I'm going to just tell you a little bit about what is going on in the problem, like some background. Planes as you know, if you've been on an airplane, they don't always travel the same speed this is like an average sometimes you go faster, sometimes you go slower maybe there's wind or maybe they had a slow take off, whatever.
This is an average speed so it's not like it goes 500 out of the gate right it takes sometimes for it to get going. Okay other things to think about is what this looks like in the real world I'm going to draw a quick little like US map here and we'll talk about the different planes so here we go.
There's Michigan, the great lakes, New York, Florida, Texas, hi Texas, California. Okay that's not great but you get the general idea it's kind of like the US right, okay. So we have one plane that's leaving New York heading for Los Angeles going that direction and the one leaving New York City is going 500mph. At the same time this other plane leaves Los Angeles like that he is going 650 mph we have to find out how far apart from Los Angeles or how far away from Los Angeles are they when they cross that's our goal.
So it's going to be somewhere around here this guy is going slower. The New York City plane's going a little bit slower so I think my answer is going to be a little bit further than half way, because this plane, the red plane out of LA is going faster, it's going to go further. So the other thing I know is that, this whole distance is 3000 so I'm going to be keeping in mind all of these different things when 8 I go through the problem. When I'm looking at my equation distance equals rate times time I want to figure out what is the same for both airplanes. Do they travel the same distance do they go the same speed or are they on the air for the same amount of time.
Well in this problem we know they don't go the same speed; we know they have different speeds. We know they are going to be travelling different distances because they don't meet half way this thing that's the same is time. These airplanes are going to be in the air for the same amount of time. What I'm going to do then is solve the equation for time. If I were to divide both sides by rate I would get time equal to distance over rate. For both of my airplanes the time they are in the air is equals to the distance travelled divided by how fast it is.
So I have two different airplanes I'm going to call them, my red guy out of LA I'm going to call him distance 1 rate 1 and then my one out of Los Angeles or out of New York I'm going to call him distance 2 rate 2 because I have two different distances they travel and I have two different speeds for how fast they are going. That's the setup of my equation and for many students that's the most difficult part.
Now let's go through substituting some things, distance of the plane out of Los Angeles goes we don't know, but we know how fast he goes. The plane out of Los Angeles goes 650 and then the distance of the plane out of New York City goes we are not sure about yet but we know that he goes 500 mph. So far so good you guys with me? All I did is substituting the two distances, excuse me the two speeds, from the original problem 500 and 650.
Okay, next thing I'm going to think about is how I can get rid of these variables. It's hard for me to solve one equation with two variables, so instead of calling this d1 I'm going to call that 3000 take away d2 and I'm going to show you how I got that. 3000 is the entire distance between the two planes when they begin with and the amount that plane 1 travels is 3000 take away how ever far plane 2 travels. That's how I got this next piece. From here it's just a cross multiplying problem.
So we can go through and solve this pretty easily. Cross multiplying is where you multiply the diagonals like that so I'll do 500 times 300 take away d2. That's going to be equal to 650 times d2. That's the product of my diagonals.
From there I'm going to go through and solve for d2. 500 times 300 is 15 with five 0s behind it take away 500d2 is equal to 650d2. You want to add 500 of this to both sides 3, 4, 5 and I'll get 1150. Divide both sides by 1150 and I'll get the distance the second plane travels is here comes, drum roll, 1304 miles point 3 okay before I decide whether or not this is my answer I want to think about what this represents.
This represents how far plane 2 goes and in my picture plane 2 represents the guy coming out of New York right plane 2 is this guy coming out of New York. He goes 1300 miles.
I don't want how far he goes I want to know how far away from Los Angeles are we, that's what the problem asks hw far from La are they when they cross? This tells me they are 1300 miles out of New York. That's not what I want to find what I want to find is 3000 take away 1304 to tell me how far they are from Los Angeles. When I do that subtraction problem, I get 1695. That's going to be my final answer in miles make sure you label it. That tells me that when these two planes cross they are 1695 miles outside of Los Angeles.
My friends that's a difficult problem, if you get this you are like A+ level your parents and your teachers will be super impressed this is a really challenging problem but the Math wasn't that hard the hardest part was understanding the problem and setting up this first equation.
If you guys can do that you'll be an A+ student. Keep it up guys don't give up on this, if nothing else try drawing a picture try setting up some equations and see what kinds of substitutions you can do.