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Arithmetic  ·  I. QUANT

Arithmetic – Weights in Weighted Average

March 21, 2022

We have discussed weighted averages in detail 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:

C_{avg} = \frac{(C_1 \times w_1 + C_2 \times w_2)}{(w_1 + w_2)}

Average Speed = \frac{(Speed1 \times Time1\ +\ Speed2 \times Time2)}{(Time1\ +\ Time2)}

Average Speed = \frac{(Distance1\ +\ Distance2)}{(Time 1\ +\ Time2)}

Average Speed = \frac{Total\ Distance}{Total\ Time}

This is an accurate representation of average speed.

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

C_{avg} = \frac{(C_1 \times w_1 + C_2 \times w_2)}{(w_1 + w_2)}

Average Speed = \frac{(Speed1 \times Distance1\ +\ Speed2 \times Distance2)}{(Distance1\ +\ Distance2)}

Speed \times 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.

C_{avg} = \frac{(C_1 \times w_1 + C_2 \times w_2)}{(w_1 + w_2)}

Average Speed = \frac{(Speed1 \times Time1 + Speed2 \times Time2)}{(Time1 + Time2)}

Average Speed = \frac{(\frac{miles}{hour} \times hour + \frac{miles}{hour} \times hour)}{(hour + hour)}

Average Speed = \frac{(miles + miles)}{(hour + hour)}

Average Speed = \frac{Total\ miles}{Total\ hours}

What happens when you take distance as the weights?

C_{avg} = \frac{(C_1 \times w_1 + C_2 \times w_2)}{(w_1 + w_2)}

Average Speed = \frac{(Speed1 \times Distance1 + Speed2 \times Distance2)}{(Distance1 + Distance2)}

Average Speed = \frac{(\frac{miles}{hour} \times miles + \frac{miles}{hour} \times miles)}{(miles + miles)}

\frac{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 \frac{miles}{gallon}.

C_{avg} = \frac{(C_1 \times w_1 + C_2 \times w_2)}{(w_1 + w_2)}

The unit of C_{avg}, C_1 and C_2 is \frac{miles}{gallon} so w_1 and w_2 should be in gallons to get

\frac{miles}{gallon} = \frac{(\frac{miles}{gallon} \times gallon + \frac{miles}{gallon} \times gallon)}{(gallon + gallon)}

\frac{miles}{gallon} = \frac{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 = \frac{D}{10}

Amount of fuel used to cover distance 2D = \frac{2D}{18} = \frac{D}{9}

So the two weights used must be \frac{D}{10} and \frac{D}{9}

\frac{Average\ miles}{gallon} = \frac{ (10 \times \frac{D}{10} + 18 \times \frac{D}{9})}{(\frac{D}{10} + \frac{D}{9})} = \frac{3D \times 90}{19D} = \frac{270}{19} = 14.2 \frac{miles}{gallon}

Or simply, \frac{Average\ miles}{gallon} = \frac{Total\ miles}{Total\ gallons} = \frac{3D}{(\frac{D}{10} + \frac{D}{9})} = 14.2 \frac{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?


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