# What are the Weights in Weighted Averages?

We have discussed weighted averages in detail here but one thing we are yet to talk about is how you decide what the weights will be in weighted average problems. It is not always straight forward to identify the weights. For example, in a question such as this one,

While traveling from Detroit to Novi, a car averaged 10 miles per gallon, and while traveling from Novi to Lapeer, it averaged 18 miles per gallon. If the distance between Detroit and Novi is half the distance between Novi and Lapeer, what is the average miles per gallon for the entire journey?

We have two figures for mileage given here – 10 miles per gallon and 18 miles per gallon. We need to find the average mileage. So we can use the weighted average formula but what will the weights be? Will they be 1:2 since the distance between the two cities is given to be in the ratio 1:2? If you think that taking the distance to be the weights in this problem is correct, then you fell for the trap in this question.

To explain the concept, let us use a simpler example first:

When talking about average speed, what are the weights? We know that the weight given to each speed is the time for which that speed was maintained, right? Yes! But why?

Let’s review our weighted average formula:

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Time1 + Speed2*Time2)/(Time1 + Time2)
Average Speed = (Distance1 + Distance2)/(Time 1 + Time2)
Average Speed = Total Distance/Total Time

This is an accurate representation of average speed.

Now see what happens when you use distance as the weights.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2)

Speed*Distance doesn’t represent any physical quantity. So this doesn’t make sense. The units of the quantities will help you see the relation clearly.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Time1 + Speed2*Time2)/(Time1 + Time2)

Average Speed = (miles/hour * hour + miles/hour * hour)/(hour + hour)

Average Speed = (miles + miles)/(hour + hour)

Average Speed = Total miles/Total hours

What happens when you take distance as the weights?

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2)

Average Speed = (miles/hour * miles + miles/hour * miles)/(miles + miles)

miles^2/hour doesn’t represent a physical quantity and hence doesn’t make sense here. Therefore, whenever you are confused what the weights should be, look at the units.

Let’s go back to the original question now. Average required is miles per gallon. So you are trying to find the weighted average of two quantities whose units must be miles/gallon.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

The unit of Cavg, C1 and C2 is miles/gallon so w1 and w2 should be in gallons to get

miles/gallon = (miles/gallon * gallon + miles/gallon * gallon)/(gallon + gallon)

miles/gallon = Total miles/Total gallons

So how will we actually solve this question?

Question: While traveling from Detroit to Novi, a car averaged 10 miles per gallon while traveling from Novi to Lapeer, it averaged 18 miles per gallon. If the distance between Detroit and Novi is half the distance between Novi and Lapeer, what is the average miles per gallon for the entire journey?

Solution:

Let the distance between Detroit and Novi be D. So the distance between Novi and Lapeer must be 2D.

Amount of fuel used to cover distance D = D/10

Amount of fuel used to cover distance 2D = 2D/18 = D/9

So the two weights used must be D/10 and D/9

Average miles/gallon = (10*D/10 + 18*D/9)/(D/10 + D/9) = 3D*90/19D = 270/19 = 14.2 miles/gallon

Or simply, Average miles/gallon = Total miles/Total gallons = 3D/(D/10 + D/9) = 14.2 miles/gallon

Food for thought: Which one of the following can you solve?

– If a vendor sold 10 apples at a profit of 10% and 15 oranges at a profit of 20%, what was his overall profit%?

– If a vendor sold apples at a profit of 10% and oranges at a profit of 20%, what was his overall profit% if cost price of each apple was \$0.20 and the cost price of each orange was \$.06?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!