One of the classic categories of word

problems in algebra is rate, time, and distance problems. To do them we’re still

going to use the same basic steps of all word problems. So let’s review those. The

first step, as always and this is the one be people miss the most I always say, is

to look for what you’re looking for. Name what x is and remember it can only

be one thing. It’s usually next to the question mark. If you have a choice between several things, let

it be the smaller one. So named what x is and then define

everything else in terms of X that’s an object in the problem itself. Ok and i always tell people to cross out

as you do that because what’s left is going to be the equation in general. Then

we solve the equation and once you solve the equation you just know what X is. You have to, of course at the end,

answer the question. Okay. Well there’s going to be some

additional tricks, or more specific tricks if you would, with rate time and

distance problems. Generally, for steps one and two, what we

named x and everything else in terms of x, we’re going to use a chart

that looks like this. We’re going to have more than one object

moving at one time. That’s why we need a chart. So you’ll have two rows of object one

an object two and then we’ll try to define the rate,

time and distance of each of those objects in the problem. Now generally

we’re going to fill in four the six boxes from information directly given in the

problem and the other two boxes are going to get filled in by one of the

rate, time and distance formulas. I call these the hard boxes and generally those

are the two boxes we’re going to relate in step three where we write the equation. Ok, so we’ll fill in the hard boxes using

rate, time and distance formulas. Oh, I guess i have to tell you those

formulas. I bet you already know them though. [Are you pondering what I’m pondering?]

Consider if you drove 60 miles an hour, that’s a

rate, for three hours wouldn’t you have gone a hundred and

eighty miles? I think you already know that distance

equals rate, or speed, times time. What if you had to go 200 miles and you

had 4 hours to do it? How fast would you have to go per hour

anyway? 200 miles, four hours to do it, you would have

to go 50 miles each of those hours, wouldn’t you? I think you already know that rate is

distance divided by time. Well, that makes sense if distance is rate

times time we would divide distance by time to get rate. In the same way if you have to do that 200 miles and you

could only go 40 miles an hour, maybe you were on a moped or something, how long would it take you? Well, it would

take you five hours, wouldn’t it, because you know that the time it takes

you is the distance divided by how fast you go – divided by the rate. So in general, the formulas are that

distance equals rate times time, rate is distance divided by time,

and time is distance divided by rate. Now really, you don’t have to memorize those.

You can really kind of derive them if you just remember that distance is rate

times time and then solve for the one that you want. Ok let’s do one. As usual this problem is

going to have two objects of some sort moving at the same time. So, let’s use our chart. Let’s see. A passenger train travels at

80 miles an hour. A freight train travels at 30 miles an

hour. If they travel in exactly opposite directions how long will it take before they are

275 miles apart? I can’t do that one in my head so let’s use the chart. Step one: name what x is. What do they want to know

here? Well, I kind of helped you out with the color. How long. So X is time, isn’t it. ok and they’re both going the same

amount of time so I can actually and sometimes you can do this. You can

put the x in more than one place. Okay. I’ll cross that out. Now let’s

define everything else in the problem that we can get. We need to get four out

of the six boxes and we have two of them already. Well, let’s see. A passenger train travels at

80 miles an hour. There we go. His rate is 80 and I’ll cross that out. The freight

train travels at 30 miles an hour. I can put that in and cross that

out. Now I already have four out of six boxes. I fill in the other two using the

formula. Distance is rate times time so 80 times x for the passenger train and 30 times x for the freight train.

Okay and these are the two boxes and I’m going to relate in my step 3 where I

write the equation. Now the distance that the passenger train went. That’s what’s represented by 80x.

The distance that the passenger train went plus the distance that the freight

train went is going to have to equal this 275 miles because they’re going in

opposite directions and that’s going to be your equation. Let’s solve this equation. Let’s add like terms. 80x and 30x is a

110x and solve for x by dividing both sides by 110

and I get that x is 2.5 or two and a half. Now that’s x. Oh, I have to answer the question. The

question is how long before they’re 275 miles apart? How long in time? Well since the unit of

time that we used was miles per hour, I assume, I know, that the answer is 2.5 hours. And that’s your answer. Note that whatever unit goes in is going

to be the unit that comes out. Let’s try another one. Sue can ride her bike 12 miles per hour

faster than Mary can walk. Now Sue rides 95 miles in the same time Mary walks 35

miles. Well, these are some serious jocks. The question

is, what is each person’s rate? Once again, let’s start off with our

chart. We’re going to have our moving objects. In this case, Sue and Mary and

let’s define what we’re going to let x be. Now it

wants to know each person’s rate. I need to pick one of the people. I’m going to

try to pick the slower one. Well who’s going to be the slower one in

this case? The walker I assume. Mary can walk, so her rate is going to be x

and then I’ll define everything else in terms of x, ok. Let’s see what

else I know. I know that Sue can walk can ride her bike 12 miles an hour

faster than Mary can walk so I put in x + 12 for her rate. Okay and I’ll cross that out. The

distance that Sue rides is 95 miles. So I put a 95 in that position and

cross it out and the distance that Mary walks is 35 miles and I always like

to cross these things out. Ok, now we filled in four of the six

boxes. We fill in the other box with the formula, the appropriate formula for time.

Now time equals distance divided by rate. So 95 divided by or over x + 12 and Mary’s time will

be 35 over x and now what have I not crossed out? That’s what’s going to come into play

when I write the equation. Well that’s why I like crossing out.

These times, these hard boxes, it says right there are the same. Sue’s time is the same as Mary’s time

and there’s your equation. [Excuse me Professor Brainiac]

We’re going to have to solve the equation. I hope you remembered what the secret to

solving equations that have fractions in them is. We’re going to multiply both

sides by the least common denominator and the least common denominator in this

case is x and x + 12 because I’m not going to be able to cancel things that

are added. So I need that other x. Ok, if I do that, that’s going to cancel

nicely and on the left side I’ll get 95x and on the right side

I’ll distribute 35 and I’ll get 35x plu 35 times 12 or 35x + 420. Now let me solve this one, much easier.

Subtract 35x from both sides and now all I have to do is divide by 60 and I

get x is 7. Now wait a minute. Ok, let’s get this

straight. Whatever unit goes in is the unit that

comes out and what was x? I look up at my chart. Good thing I made that chart. x is

Mary’s rate. A rate is miles per hour in this case. So the answer is 7 miles per hour Mary goes 7 miles per hour. Oh and they also want to know Sue. Good thing I wrote that in the

chart too. Well Sue is x + 12. So 7 plus 12. Sue rides 19 miles an hour and that

makes sense doesn’t it and that’s about as hard as it gets. Ok, remember that when you want to solve

a fractional equation – they’re going to occur a lot in these type of problems – you are going to

multiply by the LCD. Okay. Go try that homework.

You are a very good ex plainer. You have a gift my friend.

Greetings! Your tutorials are excellent and so very helpful – Thank you! I would just like to add a note about calculating time – When calculating time it is important to express your answer in meaningful terms.

As an example: If John drove 30 miles at a constant rate of 45 MPH, how long did it take? You wouldn't say it took him .6667 hours (30/45) You would want to take .6667 and multiply it by 60 to express the answer in minutes. Thus, (30/45) = .6667 X 60 = 40 minutes. Hope this was helpful

The sound effects were very disrupting to me

when you divide on paper do you get 2.49 or is my math wrong if thats the case should have said round up to nearest

Shut Up! Can you at least thank this person for trying to help who ever needs help with Algebra by trying to make it more fun and easy like him?

Oh yeah and thank you very much for this video it kinda helped me and now I am trying to get the hang of thank you!:)

Here's one I can't solve: A plane is flying from point A to point B against a strong headwind of 50 km/h. The plane takes 1/2h longer for this flight than it would take in calm air. if the distance between the two points is 1200km, what is the plane's speed in calm air?

5:47 is Jeff Spicolli!!!! lol, Mr. Witte. After returning to college after 20 some years, I have found this video very helpful, thank you so much!

thank you this helped me understand more better then my teacher.. thanks for sharing

No wonder you need help… With that attitude you will never be successful in life.

You just saved my ass!

I'm not sure if this level of math will be on the asvab but for percusionary measures, why the hell not lol.

you forgot to put the word "EXACTLY" in the problem. Because when you read the problem you said EXACTLY the same distance.

9:55 can't you also just cross multiply? I believe you would get the same answer.

My math teacher gives us much harder problems

good educational video….could def do without the sound effects….very distracting

Thanks for the help, just needs more examples

nice really nice man

Thank you Mr. Witte. You really know how to teach. You always pay attention to the details. thanks

i learned this last year but never understood and tomorrow im having a quiz and i almost fail, but thanks to you now I get it! Thank you!

This is was definitely a great help! I am taking my ASVAB on monday and I have been struggling with these problems and every resource I have turned to has not helped me, you are awesome! Great to see this video, thank you so so much!

Great explanation! But could you do away with the silly sounds, they are distracting.

You da real mvp. THANK YOU

wow.

awsome it really helped for the asvab that im studying for, other sites wouldnt realy explain that well

Wow, very well done. Thank you very much.

Thanks for sharing and keeping it simple! I needed this.

Those sound effects are hella annoying but other than that great video.

What a GREAT help. Thank you for sharing.

Thank you for this helpful tutorial, however it would be a lot greater without the sound effects. But thank you for a great tutorial 🙂

this helped sooo much!

Thank you so much this helped me pass my Algebra 1 test

I have a trig/algebra 2 test tmrw and this really helped, much better than other videos. Thank you!! :DD

good job sir.

aamir 00923222639668

This was so helpful. THANK YOU!!!

Again.. thanks bill!!!!

"Soytenly" (10:49) Ha ha ha.

Thank you for explaining so well where my textbook never seems to help me at all! I was stuck on this for weeks!

For the last equations, with the 95x and 35x, aren't you suppose to add those variables to get 130x. Then you divide 130x to 420.

Thank you a lots.

These explanations are where Honors classes kill students, life-saver video.

Great video. You explain things clearly, and the sound effects added a nice touch of humor. However, I am confused about something…

I watched some of YourMathGal's Youtube tutorials about Rate-Time-Distance word problems. She said that any time you have 2 unknown numbers ('x') adding up to something, you must express one of those unknown numbers as the resulting number minus x.

If I take her advice and apply it to your first word problem, the Passenger Train's time is 'x' and the Freight Train's time is '274 – x.'

To find out the distance traveled by the Passenger Train, I must multiply 80 by x, which equals 80x. To find out the distance traveled by the Freight Train, I must multiply 30 by 274 – x, which equals 8,250 – 30x.

I must then solve the equation using these values: 80x + 8,250 – 30x = 275. When I combine the x terms, I get 50x + 8,250 = 275. When I subtract 8,250 from both sides of the equation (to isolate the x value on the left side of the equation), I get 50x = =7,975. When I divide both sides of the equation by 50 (to get the value of just 'x'), I get x = -159.5.

That number (-159.5) is a far cry from the number you got (2.5). I don't understand why her advice screwed up my answer to your word problem.

can't thank you enough. maths is always a headache for me but you made it easier for me. thank you

thank you. You explained it so well. The sound effects are funny. I like it. 🙂

Hate those annoying noises.

tnx sir it really helps me a lot

Great step by step formula. Hate the sound bar effects, very annoying and super distracting!!!

great video thanks bill!

Video was alright I did understand 1part of it which helped me a lot but like c'mon now why would u add those noises and your gonna say something like if I don't like the vid with noises I have to do something but I watched I'm not bout to watch it again them sounds affects are annoying (no offense) but all in all your video was alright pretty decent.👍👌👌👌

Thank you i have an entrance exam at a science highschool today thanks

very clear and helpful video. My mind started wandering and the annoying sound effects brought my attention back so I didn't have to scrub backwards. lol ty again!

Thank you so much :).

I like the sound effects because they're funny and this math is boring as hell…thank you for the tricks, it helped so much!!!!

2018 anyone?

ECSUUUUUUUUSE MEEEEEEEEEE

The noises… cringe

I really wanted to watch this video but those sound effect just distracted me. i hope you consider doing a new one without them because you break it down well.

Thank you .

Made it 2 minutes before I had to turn it off. Those sound effects are ridiculous.

It’s bad enough to be frustrated by algebra, the. You have to put up with the background stupidness. Take off the background noises, they’re dumb.

Why do you put that kind of sound

Not sound rather its a noise

Great way of expressing your sense of humor. Very helpful video though

Difficult but I am learning