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# Intro to rates

Sal introduces rates using examples like 35 miles per hour and 10 dollars per hour.

## Want to join the conversation?

• What is the easiest method for solving rates • What is the difference between a rate and a unit rate mathematically? • Just like a unit fraction when the numerator is 1, representing 1 piece of the whole, a unit rate is when the rate is representing 1 quantity of the rate. Take this problem as an example: Jake worked at a bakery. He earns \$80 every 4 hours. How much does he earn every hour? The ratio of hours to money is 4:80, so to get the unit rate we have to divide by four which gives us . He earns \$20 per hour. So, in this problem the unit rate was . Hope this helped!
(sorry about the , it isn’t meant to be a benchmark)
• what is the difference between rates and unit rates? •  A rate is a special ratio in which the two terms are in different units. For example, if a 12-ounce can of corn costs 69¢, the rate is 69¢ for 12 ounces. ... When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates.
sorry that is so long to read and you will probably think that is boring! ;)
• Because the rate in which you are doing something changes, do you take the average? • HI! I'm confused. I don't understand what the difference is between *ratio and rates.* Aren't they basically the same thing? • The following table shows how many multiplication problems 4 students were able to solve in different amounts of time. For example, Murray solved 58 problems in 2 minutes.
Student Problems Minutes
Murray 58 2
Kris 78 3
Logan 84 3
Taylor 92 4 Which student solved problems at a rate of 28 per minute? • What is the easiest way to do this (and a 5th grader can understand it)🤔?? • This is the way I learned it, and I hope this helps. Say you are going on a vacation. You know that it takes two hours to go 50 miles, and that you will be traveling 120 miles. You need to figure out how long it will take you to go 120 miles. It takes 2 hours for 50 miles, so you would write it as a fraction. 2/50. You don't know the amount of time it will take you to go 120 miles, so the time is your variable, x. x/120.
Next, you would set it up in an equation.
2/50 = x/120
Now you need to pair the 2 with the 50 and the 120 with the x. In every equation like this, you will always pair the first number with the variable and the second number with the third number. Now the equation looks like this:
50x = 2 x 120. All you have to do now is multiply the right side and divide it by the coefficient (the number next to the variable)
2 x 120 = 240
240 divided by 50 = 4.8
This gets rid of the 50, leaving you with x = 4.8.
It takes 4.8 hours to go 120 miles.
• How would I solve this?: A bus leaves Oak City at AM traveling at a speed of 45 mph. 25 minutes later, a car leaves Oak City in the same direction as the bus at 60 mph. At what time would the car catch up with the bus?

A) AM
B) AM
C) AM
D) AM
E) AM • This is the way I would do it:

Let's say "t" is the number of minutes the bus drove until the car caught up.

Distance bus drove: 45 mph * t
Distance car drove: 60 mph * (t - 25 mins)

The distance the bus drove at time "t" is the same as the distance the car drove, because they start at the same location.
==> 45 * t = 60 * (t - 25)
<==> 45t = 60t - 1500
<==> 1500 = 60t - 45t = 15t
<==> 100 = t

The car caught up after 100 minutes. Apply that to the starting time am:
• What is the most common or useful method • Is it possible to simplify rates? Because I found you can simplify ratios. So, since the ideas are so similar you should probably be able to do some of the same stuff with them, right? 