VeritasPrepKarishma wrote:

khushboochhabra wrote:

I got Bunuel's approach.. but was wondering if weighed average should work here that is

(1*25+5*40)/6 = 37.5 ~ 38

Seems it doesnt work here.. But shouldn't it??

'Distances traveled' (i.e. ratio of 1:5) cannot be the weights here to find the average mileage. The weights have to be 'number of gallons'.

If we change the question and make it:

A certain car averages 25 miles per gallon of gasoline when driven in the city and 40 miles per gallon when driven on the highway. According to these rates, which of the following is closest to the number of miles per gallon that the car averages when it uses 10 gallons in the city and then 50 gallons on the highway

Now you can use (1*25+5*40)/6

Why?

Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled or should they be gallons of fuel used...), look at the units.

Average required is \(\frac{miles}{gallon}\). So you are trying to find the weighted average of two quantities whose units must be \(\frac{miles}{gallon}\).

\(C_{avg} = \frac{C_1*W_1 + C_2*W_2}{{W_1 + W_2}}\)

\(C_{avg}, C_1, C_2 - \frac{miles}{gallon}\)

So \(W_1\) and \(W_2\) should be in gallon to get:

\(\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)\)

Only if weights are in gallons, do we get 'Total miles' in the numerator and 'Total gallons' in the denominator.

We know that Average miles/gallon = Total miles/Total gallons

Takeaway: The weights have to be the denominator units of the average.So what do we do in this question?

What is the distance travelled (i.e.total miles)?

(10+50) = 60

What is the total gallons used?

Fuel used in the city = 10/25

Fuel used to go from B to C = 50/40

So Average miles/gallon = (60)/(10/25 + 50/40) = apprx 36

You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon.

VeritasPrepKarishma Your clarification seems impressive as if unit is not same the result will be something different than actual

So first we need to check what is given . If it is miles per gallon then weighted average can only be calculated when amount of fuel used in given.

So, to solve such problem I want to tweak the approach to make the solution aligned with weighted average.

Now since we know that the no. of miles i.e. distance is given and we can't take weighted average method, lets try to make the given rate i.e. miles per galon favorable to distance by reversing it and then using weighted average and then again reversing it to get my answer. All the reversing process will go in mind and we will make full use of Karishma's approach.

So. Car average in city = 25 miles per gallon

Lets reverse it to understand it as car will use 1/25 gallon per mile in city.

Car average on highway = 40 miles per gallon

Lets reverse it to understand it as car will use 1/40 gallon per mile on highway.

Also distance traveled in city is given as 10 miles and on highway is 50 miles.

So, here we can use the weighted average now .

Avg. = (1/25*10 + 1/40* 50 )/(10+50) =( 0.4 +1.25 )/60 = 1.65/60 = 11/400 gallon per mile.

Lets reverse it again to get the answer.

Hence car average = 400/11 = 36.6 miles per gallon.

So, basically I have gone a step further using Karishma's approach to understand the problem, avoid mistake regarding weighted average and then solving it the same way by tweaking it in the mid way.

Hope that helps...

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