Beginning Algebra & Rate Time Distance Problems

Beginning Algebra & Rate Time Distance Problems


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.

63 thoughts on “Beginning Algebra & Rate Time Distance Problems

  • March 27, 2013 at 4:09 am
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    You are a very good ex plainer. You have a gift my friend.

    Reply
  • March 31, 2013 at 5:11 pm
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    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

    Reply
  • April 15, 2013 at 2:40 am
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    The sound effects were very disrupting to me

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  • April 19, 2013 at 11:00 pm
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    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

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  • April 23, 2013 at 11:36 pm
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    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?

    Reply
  • April 23, 2013 at 11:37 pm
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    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!:)

    Reply
  • April 29, 2013 at 2:28 am
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    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?

    Reply
  • May 19, 2013 at 10:10 pm
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    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!

    Reply
  • May 20, 2013 at 10:26 pm
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    thank you this helped me understand more better then my teacher.. thanks for sharing

    Reply
  • May 27, 2013 at 8:33 pm
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    No wonder you need help… With that attitude you will never be successful in life.

    Reply
  • May 28, 2013 at 4:29 am
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    You just saved my ass!

    Reply
  • June 3, 2013 at 8:29 am
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    I'm not sure if this level of math will be on the asvab but for percusionary measures, why the hell not lol.

    Reply
  • August 18, 2013 at 9:51 pm
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    you forgot to put the word "EXACTLY" in the problem. Because when you read the problem you said EXACTLY the same distance.

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  • October 6, 2013 at 6:09 pm
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    9:55 can't you also just cross multiply? I believe you would get the same answer.

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  • October 31, 2013 at 8:57 pm
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    My math teacher gives us much harder problems

    Reply
  • January 7, 2014 at 8:51 pm
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    good educational video….could def do without the sound effects….very distracting 

    Reply
  • January 17, 2014 at 12:50 am
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    Thanks for the help, just needs more examples

    Reply
  • June 25, 2014 at 8:50 pm
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    nice really nice man 

    Reply
  • September 21, 2014 at 1:11 am
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    Thank you Mr. Witte. You really know how to teach. You always pay attention to the details. thanks

    Reply
  • September 24, 2014 at 6:39 am
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    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!

    Reply
  • January 30, 2015 at 3:05 am
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    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!
      

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  • March 5, 2015 at 9:32 pm
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    Great explanation! But could you do away with the silly sounds, they are distracting.

    Reply
  • March 16, 2015 at 3:10 am
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    You da real mvp. THANK YOU 

    Reply
  • June 15, 2015 at 11:44 pm
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    awsome it really helped for the asvab that im studying for, other sites wouldnt realy explain that well

    Reply
  • June 26, 2015 at 8:13 pm
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    Wow, very well done.  Thank you very much.

    Reply
  • June 29, 2015 at 4:01 am
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    Thanks for sharing and keeping it simple! I needed this.

    Reply
  • September 2, 2015 at 7:26 pm
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    Those sound effects are hella annoying but other than that great video.

    Reply
  • September 24, 2015 at 9:05 pm
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    What a GREAT help. Thank you for sharing.

    Reply
  • October 21, 2015 at 9:41 pm
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    Thank you for this helpful tutorial, however it would be a lot greater without the sound effects. But thank you for a great tutorial 🙂

    Reply
  • November 19, 2015 at 7:31 pm
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    this helped sooo much!

    Reply
  • December 6, 2015 at 6:06 pm
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    Thank you so much this helped me pass my Algebra 1 test

    Reply
  • January 7, 2016 at 4:27 am
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    I have a trig/algebra 2 test tmrw and this really helped, much better than other videos. Thank you!! :DD

    Reply
  • January 20, 2016 at 9:53 am
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    good job sir.
    aamir 00923222639668

    Reply
  • March 13, 2016 at 3:47 pm
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    This was so helpful. THANK YOU!!!

    Reply
  • March 19, 2016 at 7:43 pm
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    Again.. thanks bill!!!!

    Reply
  • September 14, 2016 at 3:19 pm
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    "Soytenly" (10:49) Ha ha ha.

    Reply
  • September 15, 2016 at 8:35 pm
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    Thank you for explaining so well where my textbook never seems to help me at all! I was stuck on this for weeks!

    Reply
  • October 3, 2016 at 11:37 pm
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    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.

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  • December 11, 2016 at 7:05 pm
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    Thank you a lots.

    Reply
  • January 16, 2017 at 10:50 pm
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    These explanations are where Honors classes kill students, life-saver video.

    Reply
  • January 22, 2017 at 7:03 am
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    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.

    Reply
  • January 28, 2017 at 11:16 pm
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    can't thank you enough. maths is always a headache for me but you made it easier for me. thank you

    Reply
  • February 13, 2017 at 2:17 am
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    thank you. You explained it so well. The sound effects are funny. I like it. 🙂

    Reply
  • May 11, 2017 at 6:52 pm
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    Hate those annoying noises.

    Reply
  • June 22, 2017 at 6:07 am
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    tnx sir it really helps me a lot

    Reply
  • September 11, 2017 at 4:21 pm
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    Great step by step formula. Hate the sound bar effects, very annoying and super distracting!!!

    Reply
  • September 12, 2017 at 1:53 pm
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    great video thanks bill!

    Reply
  • January 12, 2018 at 2:17 am
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    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.👍👌👌👌

    Reply
  • February 10, 2018 at 1:01 am
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    Thank you i have an entrance exam at a science highschool today thanks

    Reply
  • September 4, 2018 at 2:36 am
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    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!

    Reply
  • October 3, 2018 at 5:57 pm
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    Thank you so much :).

    Reply
  • October 15, 2018 at 12:15 am
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    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!!!!

    Reply
  • January 15, 2019 at 1:08 am
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    2018 anyone?

    Reply
  • January 20, 2019 at 11:12 pm
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    ECSUUUUUUUUSE MEEEEEEEEEE

    Reply
  • January 28, 2019 at 1:25 am
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    The noises… cringe

    Reply
  • February 21, 2019 at 8:36 pm
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    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.

    Reply
  • March 6, 2019 at 2:36 pm
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    Thank you .

    Reply
  • March 30, 2019 at 1:57 pm
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    Made it 2 minutes before I had to turn it off. Those sound effects are ridiculous.

    Reply
  • July 21, 2019 at 3:01 pm
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    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.

    Reply
  • August 18, 2019 at 5:47 am
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    Why do you put that kind of sound
    Not sound rather its a noise

    Reply
  • August 25, 2019 at 3:55 am
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    Great way of expressing your sense of humor. Very helpful video though

    Reply
  • October 21, 2019 at 2:40 pm
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    Difficult but I am learning

    Reply

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