A certain car averages 25 miles per gallon of gasoline when driven in

Average miles per gallon is given by: Total distance traveled/Total gallons used.

Total distance traveled is given to us and is equal to 10+50 = 60 miles

Total gallons used can be determined by reverse working the formula listed above.

Number of gallons in the city = Total distance / Average = \(\frac{10}{25}\)

Number of gallons in the highway = Total distance / Average = \(\frac{50}{40}\)

Total number of gallons = Sum of the above two quantities.

So the average overall

\(\frac{60}{\frac{10}{25} + \frac{5}{4}} = \frac{60}{\frac{2}{5} + \frac{5}{4}} = \\
\frac{60}{\frac{33}{20}} = \frac{1200}{33} = \frac{400}{11}\)

If you solve this you get 36.3636. So the answer is 36.

Hi !

You need average miles/ gallon with the given data

Total Miles traveled => 10+50 = 60

City Fuel Consumption => 1 gallon - 25 miles => 10 Miles - 10/25 gallons => 0.4 gallons
Highway Fuel Consumption => 1 gallon - 40 miles => 50 Miles - 50/40 gallons => 1.25 gallons

Now the tricky part which even I got wrong in my first timed attempt

Average fuel consumption (City + Highway) = 60 miles / (0.4+1.25) = 60/1.65 = 36.6 miles / gallon

divide 60 / 15 = 40 but since 1.65 is greater than 15 the fraction has to be smaller to its either 38 / 36

38 is just too close so eliminate

Hope this helps

My take on this -

whenever I see this type of question, I make sure what the question is asking me to solve...and this makes the question easier to solve-

Here, its asking - Total distance travelled/ number of gallons used.-----------------(1)

Now start calculating individual entity.

Total distance = 50+10 = 60

Total galoons - This is the tricky part.

Generally, I solve by using unitary method.

such as - 25 miles ---- 1 gallon
1 mile = 1/25
10 miles = 1/25*10

Similarly calculate for Highway.

Try solving for highway and put the values in equation 1. You will get the answer.

Thanks
H

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??

Actually we are asked to find the average, but the average of miles per gallon (miles/gallon). So, you should have total miles driven in numerator and total gallons used in denominator (miles/gallon). Now, ask yourself: what do you have in numerator and denominator?

Bunnel can you comment on this

Total gallons consumed = 25*10 + 40*50 = 2250
Total miles driven = 50

Avg Mileage in MPG = 2250/50 = 37.5 .. hence close to 38. I picked E pls help me understand

Total miles driven is 10+50=60, not 50.
Total gallons used is 0.4+1.25=1.65, not 2250 (!!!). How, could we need 2250 gallons for only 60 miles?

The average miles per gallon (miles/gallons) equals 60/1.65=~36.

Complete solution here: a-certain-car-averages-25-miles-per-gallon-when-driving-in-98942.html#p762784

Hope it's clear.

I just went through this thread and I have a doubt. Applying Karishma's method of using units, in this case, we get
(25*10+40*50)/(10+ 50) . The units are (miles/g * miles)/ miles = miles/gallon. Please point out the mistake in my logic.

We need total miles in the numerator and gallons in the denominator.

(C1*W1 + C2*W2) should make "miles".
(W1 + W2) should make "gallons".

Methods don't fail. You might need to manipulate them in particular circumstances.
You need to understand that whatever is given cannot be the weight. For example, when trying to find the average profit, the weight must be cost price, not number of items or selling price. Similarly, when calculating average speed, weight must be time, not distance. In the same way, here the weight must be amount of fuel used, not distance traveled.
How do you figure out what the weight will be in each case? The units help you in figuring that out.

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/12 ... -averages/

1. Since mileage is varying according to city or highway, average is calculated by total miles traveled / total gallons used
2. Total miles traveled=60
3. total gallons used = gallons used in city + gallons used in highway= 10/25 + 50/40= 33/20
4. So average is 60/(33/20)=36(approx)

We can use the formula:

average = total distance/total gallons

average = 60/(10/25 + 50/40)

average = 60/(2/5 + 5/4)

average = 60/(8/20 + 25/20)

average = 60/(33/20)

average = 1200/33 ≈ 36

Answer: D

BACKGROUND:

this is one of the test's most time-worn trick questions: a weighted average that seems as though it's going to work, but is presented with units that make the answer counterintuitive. the most common variation on this theme is the 'round trip problem', in which someone travels to and from at different speeds. for instance, say you travel a 120-mile round trip (60 miles each way), at 30mph on the way there and 20mph on the way back. is the average speed 25mph? no way. you have to calculate the average speed the old-fashioned way: find the total distance (120 miles) and divide by total time (2 hours there + 3 hours back = 5 hours), for an average speed of 120/5 = 24mph. the most important realization for you to make here is that the two trips take different TIMES, and that speeds have TIMES as their denominators. if the two legs of the journey each took the same amount of TIME (which would make one of them longer than the other), then you could calculate an average speed in the way you normally create averages of everything else.

SO WHAT DOES THIS HAVE TO DO WITH THE ABOVE PROBLEM?

same deal. you're trying to find average miles PER GALLON, but the legs of the journey are given in miles (not gallons). therefore, you can't calculate the average using the simple method for weighted averages. instead, you have to do the problem the old-fashioned way again:
* figure the total miles (60 miles)
* figure the total gallons (10/25 = 2/5 = 0.4 gallons city, 50/40 = 5/4 = 1.25 gallons highway, for a total of 33/20 = 1.65 gallons)
* divide: 60 miles / 1.65 gallons = 60 miles / (33/20) gallons which is about 36 miles/gallon, by long division.
Correct answer is (D).

Posted from my mobile device

This question looks hard but it’s not. By following a simple trick , we cane easily get the answer.
Step 1:
In city - 1 gallon gives 25 miles. So, 1 mile consumes 0.04 gallons of gasoline. For 10 miles, it takes 0.4 gallons
Step 2:
On highway, 1 gallon gives 40 miles. So, 1 mile consumes 0.025 gallons of gasoline. For 50 miles, it takes 1.25 gallons
Now comes the final step :
Count step 1 and step 2: For 60 miles - 1.65 gallons. Now, you can determine that 1 gallon of gasoline gives 36 miles of mileage.

Like my post if you find it useful .

#believeinurself

Posted from my mobile device

My 2 cents on this:

Tricky question to apply, but here's how I did it:

We're given the number of miles that a car can drive per gallon in a city and highway

We are also given the distance traveled to city and hwy (Not the two statements above mean very different things. Miles per gallon is not the same as distance travelled)

So, Miles/Miles per gallon --> 10 miles / 25 miles per gallon --> gives us 0.4 gallons and similarly, 50miles/40miles per gallons gives us 1.25 gallons/

OK. lets move on --> we are asked to find the number of miles per gallon that the car averages when it is driven 10 miles in the city and then 50 miles on the highway

So avg --> Total Distance traveled/ Total gallons consumed --> 10+50/0.4 + 1.25 --> approx 36

Ans D.


--

Hang in there folks! Stay motivated.

10 miles in the city at 25mpg is 0.4 gallons.
50 miles on the highway at 40mpg is 1.25 gallons.
Totals are 60 miles and 1.65 gallons.

\(\frac{60}{1.65} = \frac{6000}{165} = \frac{1200}{33} = \frac{3600}{99}\)

That's 36.

Answer choice D.

Dear KarishmaB
By your approach (weighed average) the answer would be 37.5, which is much closer to answer choice 38, which is a wrong answer choice, how we can avoid such mistake? or can you tell when we can find this approach is better or the approach provided by Bunuel? Because t first I used weght average for this question and mistakenly chose 38 as answer, by looking at Bunuel's approach I saw it's totally different!