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A certain car averages 25 miles per gallon when driving in [#permalink]
11 Aug 2010, 13:17

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Question Stats:

52% (02:51) correct
48% (02:09) wrong based on 348 sessions

A certain car averages 25 miles per gallon when driving in the city and 40 miles per 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 then 50 miles on the highway?

Re: More Prep questions [#permalink]
11 Aug 2010, 16:56

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A certain car averages 25 miles per gallon when driving in the city and 40 miles per 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 then 50 miles on the highway?

A. 28 B. 30 C. 33 D. 36 E. 38

Car averages x miles per gallon, means that 1 gallon is enough to drive x miles.

We are asked to find average miles per gallon (miles/gallon) --> average miles per gallon would be total miles driven divided by total gallons used (miles/gallon).

Total miles driven is 10+50=60.

As car averages 25 miles per gallon in the city for 10 miles in the city it will use 10/25=0.4 gallons; As car averages 40 miles per gallon on the highway for 50 miles on the highway it will use 50/40=1.25 gallons;

So average miles per gallon equals to \(\frac{60}{0.4+1.25}\approx{36}\)

D is closest to the correct decimal answer.... the way to solve is to find out total gallons of fuel used and total distance travelled. then divide the total distance 60 miles by X gallons.

to find X: we have 2 steps - 10/25 = 0.40 gallons is for city driving; 50/40=1.25 is for highway driving.

hence final answer = 60/1.65 which comes out to be ~36mpg. hope this is clear...

Re: Tough GMAT prep PS [#permalink]
19 Feb 2011, 10:01

For 10 miles, the fuel expended is 10/25 gallons = 0.4 gallons. Similarly for 50 miles it is 50/40 gallon = 1.25 gallons So for 60 miles total fuel spent is 1.65 gallons and hence average miles per gallon = 60/1.65 = ~36

Re: A certain car averages 25 miles per gallon when driving in [#permalink]
31 Dec 2012, 10:45

ketanth wrote:

why does weighted average method fail? answer is 37.5 miles/hr. Please explain

Your question on weighted avergae is not clear. You need to find out the fractions summing up to 1, thats the basic requirement, which is not applicable here. See my soln, and let me know if you have any question:

Km/Gallon Km Gallon neded City 25 10 10/25 = 2/5 Highway 40 50 50/40 = 5/4 Total 60 (8+25)/20 = 33/20

Mileage as asked = Total Km/Total Gallon = 60/(33/20) = 60*20/33 = 400/11 approx 38

Re: A certain car averages 25 miles per gallon when driving in [#permalink]
31 Dec 2012, 18:34

maibhihun wrote:

ketanth wrote:

why does weighted average method fail? answer is 37.5 miles/hr. Please explain

Your question on weighted avergae is not clear. You need to find out the fractions summing up to 1, thats the basic requirement, which is not applicable here. See my soln, and let me know if you have any question:

Km/Gallon Km Gallon neded City 25 10 10/25 = 2/5 Highway 40 50 50/40 = 5/4 Total 60 (8+25)/20 = 33/20

Mileage as asked = Total Km/Total Gallon = 60/(33/20) = 60*20/33 = 400/11 approx 38

Why cant we do the weighted average method: (25X10+40X50)/(50+10) = 37,5 km/gallon

I understand your logic. I am trying to understand why weighted average method logic is false.

Re: A certain car averages 25 miles per gallon when driving in [#permalink]
15 Aug 2013, 11:27

A certain car averages 25 miles per gallon when driving in the city and 40 miles per 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 then 50 miles on the highway?

Re: A certain car averages 25 miles per gallon when driving in [#permalink]
15 Aug 2013, 12:21

leafsrule99 wrote:

A certain car averages 25 miles per gallon when driving in the city and 40 miles per 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 then 50 miles on the highway?

A. 28 B. 30 C. 33 D. 36 E. 38

miles per gallon = 60/(10/25+50/40) = D _________________

Re: A certain car averages 25 miles per gallon when driving in [#permalink]
09 Apr 2014, 11:59

ketanth wrote:

maibhihun wrote:

ketanth wrote:

why does weighted average method fail? answer is 37.5 miles/hr. Please explain

Your question on weighted avergae is not clear. You need to find out the fractions summing up to 1, thats the basic requirement, which is not applicable here. See my soln, and let me know if you have any question:

Km/Gallon Km Gallon neded City 25 10 10/25 = 2/5 Highway 40 50 50/40 = 5/4 Total 60 (8+25)/20 = 33/20

Mileage as asked = Total Km/Total Gallon = 60/(33/20) = 60*20/33 = 400/11 approx 38

Why cant we do the weighted average method: (25X10+40X50)/(50+10) = 37,5 km/gallon

I understand your logic. I am trying to understand why weighted average method logic is false.

Weighted average: (n1A1+n2A2+...nxAx)/(n1+n2+...nx) for x terms. Units of n1,n2...nx should be the same as the UNITS W.R.T. which values of A1,A2...Ax are defined as rates. E.G. If Ax represents concentration in a mixture as gm/litre, units of nx should be litre and NOT gms. In the question under consideration, we have rates defined as miles/gallon, so units of nx will be in gallons NOT miles as is give in the question. To illustrate: City journey will consume: 0.4 gallons (as calculated in earlier posts) Highway journey will consume: 1.25 gallons (''-------------""--------"") Hence by weighted average formula we have: Average consumption = (0.4*25+1.25*40)/(0.4+1.25)=60/1.65=36.363636....

Re: A certain car averages 25 miles per gallon when driving in [#permalink]
10 Apr 2014, 13:29

2

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ketanth wrote:

maibhihun wrote:

ketanth wrote:

why does weighted average method fail? answer is 37.5 miles/hr. Please explain

Your question on weighted avergae is not clear. You need to find out the fractions summing up to 1, thats the basic requirement, which is not applicable here. See my soln, and let me know if you have any question:

Km/Gallon Km Gallon neded City 25 10 10/25 = 2/5 Highway 40 50 50/40 = 5/4 Total 60 (8+25)/20 = 33/20

Mileage as asked = Total Km/Total Gallon = 60/(33/20) = 60*20/33 = 400/11 approx 38

Why cant we do the weighted average method: (25X10+40X50)/(50+10) = 37,5 km/gallon

I understand your logic. I am trying to understand why weighted average method logic is false.

Weighted average: (n1A1+n2A2+...nxAx)/(n1+n2+...nx) for x terms. Units of n1,n2...nx should be the same as the UNITS W.R.T. which values of A1,A2...Ax are defined as rates. E.G. If Ax represents concentration in a mixture as gm/litre, units of nx should be litre and NOT gms. In the question under consideration, we have rates defined as miles/gallon, so units of nx will be in gallons NOT miles as is give in the question. To illustrate: City journey will consume: 0.4 gallons (as calculated in earlier posts) Highway journey will consume: 1.25 gallons (''-------------""--------"") Hence by weighted average formula we have: Average consumption = (0.4*25+1.25*40)/(0.4+1.25)=60/1.65=36.363636....[/quote]

Agreed.

Or in other words, you can say that for weighted average you need to use gallons/mile as the weights are miles

so average gallons per mile would be = (10*(1/25)+ 50*(1/40))/(10+50) = (10/25+50/40)/60

To get avg. miles/gallon we need to reverse the equation and we get 60/(10/25+50/40)...same as what you calculated by 1st method.

Re: A certain car averages 25 miles per gallon when driving in [#permalink]
05 Feb 2015, 20:30

1

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Expert's post

Quote:

A certain car averages 25 miles per gallon when driving in the city and 40 miles per 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 then 50 miles on the highway?

A. 28 B. 30 C. 33 D. 36 E. 38

I do not understand why weighted average method fails here.

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.

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