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Math Lesson 4.2.1 - What Are Rates? Clarifying Misconceptions

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Welcome to our Math lesson on What Are Rates? Clarifying Misconceptions, this is the first lesson of our suite of math lessons covering the topic of Rates. Applications of Ratios and Rates in Practice, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

What Are Rates? Clarifying Misconceptions

One of the biggest challenges a learner has to face when dealing with rates is the similarity of rates and ratios. By definition, rate is a ratio-like structure involving two different types of quantities. Recall that a ratio is a comparison through the division of two like quantities.

A rate is usually expressed as a fraction; it is rare for the colon symbol to be used to express rates.

Rates are very common in physics. For example, velocity represents the rate of position change. More specifically, when the position of a moving object changes by 210 km in 3 hours, the rate of position change is 210 km / 3 h = 70 km/h.

At first sight, ratios and rates are similar. Indeed, some modern theories (from 2011 onwards) tend to include rates within the ratios category. This is because each ratio has an associated rate, which is equal to the value of the ratio when the second number is equal to 1. We call this special rate a unit rate. For example, when we say the ratio of boys to girls watching a football match is 35 boys to 7 girls, the unit rate is 35 / 7 = 5 boys per each girl watching the match. However, in this maths course we will consider ratios and rates as separate categories based on the distinction criterion mentioned above (ratios involve like quantities, while rates have different ones).

Recall the examples in the previous tutorial where the number of items produced in a certain period as a function of the number of workers was expressed as a ratio. In reality, those situations are closer to rates than ratios, as they involve different types of quantities (workers & items). The unit rate in those cases represent the number of items per worker produced in a day.

One of the distinctive features of rates is that in many cases they involve time. In other words, the quantities expressed as a rate where the number written in the denominator represents time are all rates. For example, in physics the rate of position change (Δx) represents the velocity v (v = Δx / t where t is the time elapsed); the rate of velocity change (Δv) gives the acceleration a (a = Δv / t); the volumetric rate of water flow R is given by R = ΔV / t, where ΔV is the volume of water flowing through the pipe, and so on.

Example 1

It takes 25 seconds to fill a 5L bucket with water. What is the flow rate of water from the tap?

Solution 1

To calculate the rate of water flow in litres per second, we must divide the water volume (in litres) by the time elapsed (in seconds). Thus, we have

Rate of water flow = Volume of water/Time elapsed
= 5 L/25 s
= 0.2 L/s

To have a clearer idea on the relationship of rates to ratios, let's consider the following example.

Example 2

A cat eats 2 kg of food in 30 days. How many kilograms of food does this cat eat in 135 days?

Solution 2

We can use two different approaches to solve this exercise.

  1. The ratio approach. In this case, we take the ratio of the number of days and then we multiply it by the amount of food the cat eats in 30 days. This is because we have the following ratio:
    30 days : 135 days = 2 kg : Total amount of food consumed
    Since the second quantity in the ratio is bigger, we can find 1:R, as explained in the previous guide. Thus,
    1/R = 135 d/30 d = 4.5
    Thus,
    Total amount of food consumed/2 kg = 4.5
    Total amount of food consumed = 2 kg × 4.5
    = 9 kg
    Therefore, the cat consumes 9 kg food in the given period.
  2. The rate approach. In this approach, we can calculate the unit rate first, i.e. the daily food consumption of the cat and then, multiply it by the total number of days. We can write:
    Unit rate = Amount of food consumed/Number of days
    = 2kg/30 days
    = 1/15 kg/day
    Then, we multiply this unit rate to the total number of days involved. Thus,
    Total amount of food consumed = Unit rate × Total number of days
    = 1/15 kg/day × 135 days
    = 135/15 kg
    = 9 kg

More Rates. Applications of Ratios and Rates in Practice Lessons and Learning Resources

Ratio and Proportion Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
4.2Rates. Applications of Ratios and Rates in Practice
Lesson IDMath Lesson TitleLessonVideo
Lesson
4.2.1What Are Rates? Clarifying Misconceptions
4.2.2Two Associated Unit Rates
4.2.3Rate of a Quantity Change
4.2.4Applications of Rates in Practice
4.2.5Applications of Ratios in Practice - The Golden Ratio and Fibonacci Numbers

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