A certain car averages 25 miles per gallon

[RPurewal]

courtesy of a student

A certain car averages 25 miles per gallon when driving in the city and 40 miles to the gallon when driving 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 is driven 10 miles in the city and 50 miles on the highway?
 
a)28
b)30
c)33
d)36
e)38
 
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My brain stopped working here. This doesn't seem like a difficult problem. 10 miles +50 miles=60 miles so 10/60=1/6 and 50/60=5/6. I took 25*(1/6) and 40*(5/6). This didn't get me the right answer.

[RPurewal]

no, you're right, this technique won't give you the right answer.

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 fro 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 tho 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.

[Khalid]

no, you're right, this technique won't give you the right answer.

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 fro 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 tho 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.

Ron, two questions:
1) how did you get the total of 33/20, and
2) what made you decide on the total of 60 ( was the the 10/50?)

Thanks

[RPurewal]

1) how did you get the total of 33/20

that's the sum of 2/5 and 5/4 (the numbers of gallons consumed at each step).

do you understand where the 2/5 and 5/4 come from?
if so, then the 33/20 is just the result of adding fractions, the same way you'd always add fractions.
make the common denominator: 2/5 + 5/4 = 8/20 + 25/20 = 33/20.

2) what made you decide on the total of 60 ( was the the 10/50?)

not a "decision"; that's what actually happens in the problem.
the car goes 10 miles in the city and 50 miles on the highway. that's a total of 10 + 50 = 60 miles.

[Khalid]

1) how did you get the total of 33/20

that's the sum of 2/5 and 5/4 (the numbers of gallons consumed at each step).

do you understand where the 2/5 and 5/4 come from?
if so, then the 33/20 is just the result of adding fractions, the same way you'd always add fractions.
make the common denominator: 2/5 + 5/4 = 8/20 + 25/20 = 33/20.

2) what made you decide on the total of 60 ( was the the 10/50?)

not a "decision"; that's what actually happens in the problem.
the car goes 10 miles in the city and 50 miles on the highway. that's a total of 10 + 50 = 60 miles.

Thanks Ron. I guess I am lost with the 25/20 piece?

[JonathanSchneider]

It's a hard question. You need to figure out the total miles, and the total gallons. The total miles is simple: 10 + 50 = 60. The total gallons, however, is HARD! You have to figure out the gallons used in the city and the gallons used on the highway; you have to figure these out separately.

Notice that "miles per gallon" is a ratio. You can flip this over the other way: "gallons per mile" to express the same idea. (I highly recommend writing out the units labels for problems of this type.)

You know that the car went for 10 miles in the city. You want the number of gallons used:

(10 miles) x (1 gallon / 25 miles) = 10/25 gallons = 2/5 gallon

Notice that the "miles" labels have cancelled each other out, leaving us with gallons, which is what we want. This is the beauty of writing out the labels - it helps us to make sense of the math we are doing.

Now, for the highway:

(50 miles) x (1 gallon / 40 miles) = 50/40 gallons = 5/4 gallons

Again, the "miles" labels cancel.

You now have the miles used in the two areas, so add them together:

2/5 + 5/4 = 33/20.

The final step is discussed above.

[commit.gmat]

wow. This is such a trick question. You wouldn't even realize that you made a mistake in the exam and would be so confident that you got it right under a minute.

The lesson I learned here is that 'Always write down the units along with the numbers and check whether the units make sense when you multiply two variables'.

[ashish.jere]

i took 4 minutes 30 seconds to do this problem. is it justifiable?

[RPurewal]

i took 4 minutes 30 seconds to do this problem. is it justifiable?

whoa, no. that's WAY too much time. if you took similar amounts of time on the other questions, you'd leave more than half the test blank. i don't need to tell you what would happen to your score if that were the case.

you should try NEVER to work a problem for more than 3 minutes or so.

--

of that 4:30, how much of it was "dead time"? getting stuck? staring at the problem trying to overcome inertia (or even being afraid to try alternative solution methods)?

[ashish.jere]

i took 4 minutes 30 seconds to do this problem. is it justifiable?

whoa, no. that's WAY too much time. if you took similar amounts of time on the other questions, you'd leave more than half the test blank. i don't need to tell you what would happen to your score if that were the case.

you should try NEVER to work a problem for more than 3 minutes or so.

--

of that 4:30, how much of it was "dead time"? getting stuck? staring at the problem trying to overcome inertia (or even being afraid to try alternative solution methods)?

I 've this problem.

''staring at the problem trying to overcome inertia''

so the strategy i 've been using is to pen down the data as i read.

one more point, am doing quite a few mistakes due to haphazard utilization of scratch paper. ron, could you please give some tips on that?

the strategy i have come up with is that i 've been using the scratch paper ''horizontally'' ( helps me for DS problems).

[RPurewal]

one more point, am doing quite a few mistakes due to haphazard utilization of scratch paper.

hmm
i don't really know what "haphazard utilization of scratch paper" means, so i can't really advise, but i'll take two guesses:

(1) it means "i don't write enough stuff down on the scratch paper"
--> if this is the case, then my only advice is that you should write more stuff down on the scratch paper.

(2) it means "i make algebra / arithmetic mistakes on the scratch paper"
--> if this is the case, then MAKE A FLASH CARD FOR EACH AND EVERY MISTAKE THAT YOU MAKE.
let's say that you see (x^2)^3 as part of a much larger exponent problem, and you mistakenly add the exponents to produce x^5.
then make a flash card with ONLY (x^2)^3 on the front (not the surrounding context), and the answer (x^6) on the back.
if you make these flash cards for all your mistakes, you'll soon have a deck of exactly the mistakes you made - one of the most valuable resources you can have.

[banakida]

I have an idea to help you make an educated guess for this problem.

As soon as I saw this problem, I was able to narrow the answers down between d) and e). Because the average MPG between 25 and 40 is 32.5, you can immediately eliminate a) and b) since they are less than 32. Then we know it drives 5 times more on highway than in the city, so c) is out without a doubt.

So now you have a 50/50 chance to guess. :)

[miss.t.gould]

I also encountered this problem on my GMAC practice test today. In fact, it was my last question and I was soo pressed for time that I had to guess without even reading it - not suprisingly I got it wrong.

However, in my post exam review I decided to tackle the question before scouting out an answer and came up with the following approach.

RON - I'd love it if you could please verify that this is an appropriate / correct strategy

City = 25m/gallon x 10m = 250
Highway = 40m/gallon x 50m = 2000
City + Highway / Total Actual Miles = (250 + 2000) / (10 + 50) = 2250 / 60 = 35 5/6

Therefore, the answer is closest to 36.
Effectively, I used a weighted average - although it's a little different to the technique shown above. Is this ok?

[RPurewal]

RON - I'd love it if you could please verify that this is an appropriate / correct strategy

no. if you got this to work, then it's an insanely unlikely coincidence.
i.e., don't play the lottery this month -- you already won it here.

City = 25m/gallon x 10m = 250
Highway = 40m/gallon x 50m = 2000

you definitely can't do this; you're multiplying miles/gallon x miles.
this produces numbers whose units are "miles^2 / gallon"; these are nonsense.

think about it -- what do these numbers *mean*? what actual real-life quantities are they 250 and 2000 of?

[leonitrof]

Why not keeping it simple? Meaning, how many miles will be driven and how many gallons used? Then divide the first by the second, no?

10 being 2/5 of 25 (or, for other figures, you divide the miles by the mi/gal., which gives you the gallons required), 10 miles at 25 mi./gal. means that 2/5 of a gallon will be required in the city = 0.4 gallon.

And 50 being 1.25 X 40. (or, for other figures, you do the exact same procedure that the first paragraph explains) means that 1.25 gallon will be required on the highway.

Total of miles = 60. Total of gallons = 1.65

60 miles / 1.65 gallons = 36.363636.... miles per gallon.

Voilà!