If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Algebra (all content)

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

Lesson 3: Word problems with multiple units

# Worked example: Rate problem

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.
• is there other ways to solve this with out fraction like method.
• 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.
• 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?
• 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.
• This video doesn't explain some problems on Rate 2.
• what do you need help with?
• Can someone PLEASE help with these type problems? I'm getting sooo frustrated on these questions. I really need help!!
• 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?
• 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.
• 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.
• 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.
• Clarify why we want seconds per feet?
• 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?