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## Algebra (all content)

### Course: Algebra (all content) > Unit 15

Lesson 3: Word problems with multiple units# Worked example: Rate problem

CCSS.Math:

Sal solves two related proportion word problems about a squirrel crossing the road and a car that approaches it. He does that using dimensional analysis. Created by Sal Khan.

## Want to join the conversation?

- I am not understanding the questions about 9 people painting 7 walls in 20 minutes so how long would it take 20 people to paint 3 walls? I don't get it.(38 votes)
- I found this explanation on Math Stack exchange, that explained it nicely: https://math.stackexchange.com/questions/2079003/rates-and-ratio-work-problem

It helped me understand why I could not formulate the probem the opposite way (walls/person) and thinking of the rate as walls/minute a person can do and that the output as x number of walls.(5 votes)

- is there other ways to solve this with out fraction like method.(6 votes)
- Yes, there is always another way

such as this way:

It takes 0.5 seconds for the car to move there (50 feet divided at 100feetpersecond is half a second)

It takes 0.75 seconds for the squirrel to move across the road(at 12 feet per second it's going to take 0.75 seconds to cross the road)

Therefore poor squirrel will die.(2 votes)

- Starting at home, Emily traveled uphill to the hardware store for 60 minutes at just 6 mph. She then traveled back home along the same path downhill at a speed of 12 mph.

What is her average speed for the entire trip from home to the hardware store and back?

I've watched the worked example rate problem video and also read the hint for this problem and I can't figure it out. Could someone explain to me in detail how to work out this type of rate problem?(6 votes)- Nichole's answer is incorrect, as the trip is shorter on the way down due to the increased speed. Thus, you must calculate a weighted average. To do this, you must first find out how far the trip is. Get this by using the first trip's travel time and speed (I will let you figure out how to do that since the video tells you how). After that, use the distance you just found and the return trip's speed to calculate the time for the second trip (again, the video shows you how to do this. It is essentially just a unit conversion). Finally, use the times to calculate a weighted average speed for the trip. In other words, instead of simply (6+12)/2, you will use the following formula, inserting the time you calculated, represented by "t" in this equation: (60*6+t*12)/(60+t). This problem is a horrible example and assumes you will know how and when to calculate weighted averages. Hopefully, my answer helps give you the outside information you need so you can focus on applying the information in this video.(3 votes)

- This video doesn't explain some problems on Rate 2.(6 votes)
- what do you need help with?(2 votes)

- Can someone PLEASE help with these type problems? I'm getting sooo frustrated on these questions. I really need help!!(6 votes)
- This video does not help me with this problem:

Starting at home, Omar traveled uphill to the gift store for 30 minutes at just 10 mph. He then traveled back home along the same path downhill at a speed of 30 mph.

How do I do this? I can't just average the two speeds, the faster one takes less time. Any suggestions?(4 votes)- You didn't say what you need to find. I assume you need to find the time it took him to go downhill.

You need to use the formula of:`distance = rate * time`

You can find the distance traveled using the info for going uphill:

30min/60 mph * 10mph = 1/2 * 10 mi=5 miles

You now know the distance uphill or downhill = 5 miles

Let T = the time to go downhill.

30mph * T = 5 miles

Divide both sides by 30 mph and you get T = 5/30 hrs = 1/6 hrs or 10 mins.

Hope this helps.(3 votes)

- I have a question about one of the practice problems.

"Starting at home, Nadia traveled uphill to the grocery store for 30 minutes at just 4 mph. She then traveled back home along the same path downhill at a speed of 12 mph."

My answer was 8mph, as the average of 12 and 4 mph is 8, but it was incorrect. How is this so? Her speed remained at 12 and 4 mph the entire journey.

In Hint 2/12, you wrote: "She traveled for a longer time uphill (since she was going slower), so we can estimate that the average speed is closer to 4 mph than 12 mph."

This contradicts what was written in the problem, because you did not clarify whether her speed was constant. When no clarification on the speed is given, we will always assume that she traveled at a constant speed.(4 votes)- The problem is that the average speed is based on time/min. So since distances are equal, 4 mph * 30 min * 1 hr/60 min gives a distance of 2 miles. We need to find the time to get home, so traveling 12 mph, we have to say 2 miles/12 mph * 60 min/hour gives 10 minutes. So she travelled 4 miles in a total of 40 minutes, 4 m/40 min * 60 min/hr = 6 m/hr as her average speed. We do see that 6 is in fact closer to 4 than to 12. Hint 2 does not contradict the constant speed, it just notes that the constant speed of 4 mph is slower than the constant speed of 12 mph.(2 votes)

- Clarify why we want seconds per feet?(5 votes)
- Because he is converting the length of the trip from feet to seconds. If feet are put in the bottom of the conversion factor they cancel out and you end up converting the length of the trip into seconds.(1 vote)

- Why can't we do "feet per second" instead of seconds per feet?(3 votes)
- If you multiply something with feet as units times something with feet per second, the units would end up as feet^2/sec, and that does not have much meaning, when you multiply units of feet times seconds/foot, the feet cancel and you are only left with seconds.(3 votes)

- In Squirrel Survival:

i don't get how he writes 9ft * 1/12 * seconds/feet

Can someone explain to me please?(0 votes)- by 1/12 (second/ feet) we get time that it takes for squirrel to run 1 feet and then multiplying it by 9 feet we get overall time to cross the road(6 votes)

## Video transcript

A squirrel is running across
the road at 12 feet per second. It needs to run 9 feet
to get across the road. How long will it take the
squirrel to run 9 feet? Round to the nearest
hundredth of a second. Fair enough. A car is 50 feet away
from the squirrel-- OK, this is a high-stakes word
problem-- driving toward it at a speed of 100
feet per second. How long will it take
the car to drive 50 feet? Round to the nearest
hundredth of a second. Will the squirrel make
it 9 feet across the road before the car gets there? So this definitely is
high stakes, at least for the squirrel. So let's answer
the first question. Let's figure out
how long will it take the squirrel to run 9 feet. So let's think about it. So the squirrel's
got to go 9 feet, and we want to figure
out how many seconds it's going to take. So would we divide or
multiply this by 12? Well, to think about that, you
could think about the units where we want to get an
answer in terms of seconds. We want to figure
out time, so it'd be great if we could multiply
this times seconds per foot. Then the feet will cancel out,
and I'll be left with seconds. Now, right over here, we're
told that the squirrel can run at 12 feet per second,
but we want seconds per foot. So the squirrel, every second,
so they go 12 feet per second, then we could also say 1
second per every 12 feet. So let's write it that way. So it's essentially
the reciprocal of this because the units are
the reciprocal of this. So, it's 1 second
for every 12 feet. Notice, all I did is I took this
information right over here, 12 feet per second, and I wrote
it as second per foot-- 12 feet for every 1 second,
1 second for every 12 feet. What's useful about this is
this will now give me the time it takes for the
squirrel in seconds. So the feet cancel
out with the feet, and I am left with 9 times
1/12, which is 9/12 seconds. And 9/12 seconds
is the same thing as 3/4 seconds, which is
the same thing as 0.75 seconds for the squirrel
to cross the street. Now let's think about the car. So now let's think
about the car. And it's the exact same logic. They tell us that the
car is 50 feet away. So the squirrel is trying
to cross the road like that, and the car is 50 feet
away coming in like that, and we want to figure out if
the squirrel will survive. So the car is 50 feet away. So it's 50 feet away. We want to figure out the
time it'll take to travel that 50 feet. Once again, we would
want it in seconds. So we would want
seconds per feet. So we would want to multiply
by seconds per foot. They give us the speed
in feet per second, 100 feet per second. And so we just have
to realize that this is 100 feet for every 1 second,
or 1/100 seconds per feet. This is once again
just this information, but we took the
reciprocal of it, because we don't
want feet per second, we want seconds per feet. And if we do that,
that cancels with that, and we're left with
50/100 seconds. So this is 50/100
is 0.50 seconds. And so now let's answer the
question, this life and death situation for the squirrel. Will the squirrel make
it 9 feet across the road before the car gets there? Well, it's going to take the
squirrel 0.75 seconds to cross, and it's going to take the
car only half a second. So the car is going to get to
where the squirrel is crossing before the squirrel
has a chance to get all the way across the road. So unfortunately for the
squirrel, the answer is no.