A certain car averages 25 miles per gallon of gasoline when : GMAT Problem Solving (PS)
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# A certain car averages 25 miles per gallon of gasoline when

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A certain car averages 25 miles per gallon of gasoline when [#permalink]

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13 Jun 2009, 14:04
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A certain car averages 25 miles per gallon of gasoline when driven 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
[Reveal] Spoiler: OA
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13 Jun 2009, 19:14
D.
City -
25 miles in 1 gallon
1 miles in 1/25 gallon or .04 gallons.
10 miles in .04*10 = .4 gallons

Highway -
40 miles in 1 gallon.

gas consumed by car = .4(for city) + 1(for highway) = 1.4 gallons
distance travelled = 50 miles
so avearge/gallon = 50/1.4 = 35.7
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13 Jun 2009, 21:19
mdfrahim wrote:
D.
City -
25 miles in 1 gallon
1 miles in 1/25 gallon or .04 gallons.
10 miles in .04*10 = .4 gallons

Highway -
40 miles in 1 gallon.

gas consumed by car = .4(for city) + 1(for highway) = 1.4 gallons
distance travelled = 50 miles
so avearge/gallon = 50/1.4 = 35.7

I think it should be 1.25 gallons for the 50 miles on the highway, so total of 60 miles in 1.65 gallons, with approx. 36.36 miles / gallon output. The answer doesn't change however.
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14 Jun 2009, 22:02
Can Jivana pls. expalin how the gas consumed is 1.25 gallons ?
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13 May 2010, 06:46
mdfrahim wrote:
Can Jivana pls. expalin how the gas consumed is 1.25 gallons ?

40m ______ 1g
50m ______ x = 50/40 = 5/4 = 1.25
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14 May 2010, 11:27
simple avg problem
40 miles ----- 1 G
therefore 1 mile = 1/40 G
so for 50 miles (1/40) 50= 5/4 G
similerly
25 miles = 1G
therefore 1 mile = 1/25 g
so 10 miles= (1/25)*10 = 2/5
now total distance = 60 mile hence the avg = 60/((5/4) + (2/5))
hence 36 ans
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09 Jun 2011, 02:25
Here is my reasoning and please feel free to correct me if im wrong.

the formula is (weight1)(datapoint1) + (weight2) (data point2) / sum of weights.

Weight 1 = 25
Weight 2= 40

Datapoint 1= 10
Data point 2= 50

25*10 + 40*50/ ( 25+40)

250+2000/65
2250/65 =36.5

D
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09 Jun 2011, 19:25
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Carol680 wrote:
Here is my reasoning and please feel free to correct me if im wrong.

the formula is (weight1)(datapoint1) + (weight2) (data point2) / sum of weights.

Weight 1 = 25
Weight 2= 40

Datapoint 1= 10
Data point 2= 50

25*10 + 40*50/ ( 25+40)

250+2000/65
2250/65 =36.5

D

Actually, you need to find the average miles/gallon so your 'datapoints' will be 25 and 40 miles/gallon. Weights will be the number of gallons in each case, not number of miles so 10 and 50 are not even the weights. If you get confused when deciding what the weights will be, just look at the denominator of the units of the quantity you are averaging. Since you want to find the average miles/gallon, your weights should be 'gallons' (because average miles/gallon will be total miles/total gallons... the formula has 'sum of weights' as the denominator so 'sum of weights' should be 'total gallons')
Number of gallons used in each case = 10/25 and 50/40

Now you can use the formula:
[25*(10/25) + 40*(50/40)]/[10/25 + 50/40] = 36.36 miles/gallon
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 08 Sep 2010 Posts: 166 Followers: 0 Kudos [?]: 24 [1] , given: 18 Re: average rate problem [#permalink] ### Show Tags 09 Jun 2011, 20:04 1 This post received KUDOS Ans= D City: 25 miles - 1 gallon so for 10 miles = 10/25 gallons Highway: 40 miles - 1 gallon so for 50 miles = 50/40 gallons Total Distance = 10 + 50 = 60 Total Gallons = 10/25 + 50/40 = 66/40 So avg = 60/1 * 40/60 = approx 36 _________________ My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. If you like my explanations award kudos. Intern Joined: 18 May 2011 Posts: 17 Followers: 0 Kudos [?]: 1 [0], given: 13 Re: average rate problem [#permalink] ### Show Tags 10 Jun 2011, 01:15 VeritasPrepKarishma wrote: Carol680 wrote: Here is my reasoning and please feel free to correct me if im wrong. the formula is (weight1)(datapoint1) + (weight2) (data point2) / sum of weights. Weight 1 = 25 Weight 2= 40 Datapoint 1= 10 Data point 2= 50 25*10 + 40*50/ ( 25+40) 250+2000/65 2250/65 =36.5 D Actually, you need to find the average miles/gallon so your 'datapoints' will be 25 and 40 miles/gallon. Weights will be the number of gallons in each case, not number of miles so 10 and 50 are not even the weights. If you get confused when deciding what the weights will be, just look at the denominator of the units of the quantity you are averaging. Since you want to find the average miles/gallon, your weights should be 'gallons' (because average miles/gallon will be total miles/total gallons... the formula has 'sum of weights' as the denominator so 'sum of weights' should be 'total gallons') Number of gallons used in each case = 10/25 and 50/40 Now you can use the formula: [25*(10/25) + 40*(50/40)]/[10/25 + 50/40] = 36.36 miles/gallon Thank you very much for the clarification +1 Senior Manager Joined: 12 Oct 2009 Posts: 268 Schools: Columbia, INSEAD, RSM, LBS Followers: 5 Kudos [?]: 159 [0], given: 4 Re: average rate problem [#permalink] ### Show Tags 10 Jun 2011, 10:26 Excellent approach Karishma !!! Current Student Joined: 26 May 2005 Posts: 565 Followers: 18 Kudos [?]: 207 [0], given: 13 Re: average rate problem [#permalink] ### Show Tags 10 Jun 2011, 10:42 VeritasPrepKarishma wrote: Carol680 wrote: Here is my reasoning and please feel free to correct me if im wrong. the formula is (weight1)(datapoint1) + (weight2) (data point2) / sum of weights. Weight 1 = 25 Weight 2= 40 Datapoint 1= 10 Data point 2= 50 25*10 + 40*50/ ( 25+40) 250+2000/65 2250/65 =36.5 D Actually, you need to find the average miles/gallon so your 'datapoints' will be 25 and 40 miles/gallon. Weights will be the number of gallons in each case, not number of miles so 10 and 50 are not even the weights. If you get confused when deciding what the weights will be, just look at the denominator of the units of the quantity you are averaging. Since you want to find the average miles/gallon, your weights should be 'gallons' (because average miles/gallon will be total miles/total gallons... the formula has 'sum of weights' as the denominator so 'sum of weights' should be 'total gallons') Number of gallons used in each case = 10/25 and 50/40 Now you can use the formula: [25*(10/25) + 40*(50/40)]/[10/25 + 50/40] = 36.36 miles/gallon Hi Karishma, Was wondering if I can ue the following approach city mileage = 25 now we are going 10 miles = 40% highway = 40 now = 50 = 125% total mileage = (10+50)/165% = 36.36 Thanks Sudhir Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7183 Location: Pune, India Followers: 2165 Kudos [?]: 14011 [1] , given: 222 Re: average rate problem [#permalink] ### Show Tags 10 Jun 2011, 11:00 1 This post received KUDOS Expert's post sudhir18n wrote: VeritasPrepKarishma wrote: Carol680 wrote: Here is my reasoning and please feel free to correct me if im wrong. the formula is (weight1)(datapoint1) + (weight2) (data point2) / sum of weights. Weight 1 = 25 Weight 2= 40 Datapoint 1= 10 Data point 2= 50 25*10 + 40*50/ ( 25+40) 250+2000/65 2250/65 =36.5 D Actually, you need to find the average miles/gallon so your 'datapoints' will be 25 and 40 miles/gallon. Weights will be the number of gallons in each case, not number of miles so 10 and 50 are not even the weights. If you get confused when deciding what the weights will be, just look at the denominator of the units of the quantity you are averaging. Since you want to find the average miles/gallon, your weights should be 'gallons' (because average miles/gallon will be total miles/total gallons... the formula has 'sum of weights' as the denominator so 'sum of weights' should be 'total gallons') Number of gallons used in each case = 10/25 and 50/40 Now you can use the formula: [25*(10/25) + 40*(50/40)]/[10/25 + 50/40] = 36.36 miles/gallon Hi Karishma, Was wondering if I can ue the following approach city mileage = 25 now we are going 10 miles = 40% highway = 40 now = 50 = 125% total mileage = (10+50)/165% = 36.36 Thanks Sudhir Yes, this is fine but make sure you use the units so that you know exactly what you are doing. The following is what I mean: city mileage = 25 miles/gallon now we are going 10 miles = 40% = 0.4 gallons used highway mileage = 40 miles/gallon now 50 miles = 125% = 1.25 gallons used total mileage = Total Distance / Total amount of fuel used (10+50) miles/1.65 gallons = 36.36 miles/gallon _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A certain car averages 25 miles per gallon of gasoline when [#permalink]

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09 Sep 2015, 09:18
netrix wrote:
mdfrahim wrote:
Can Jivana pls. expalin how the gas consumed is 1.25 gallons ?

40m ______ 1g
50m ______ x = 50/40 = 5/4 = 1.25

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Re: A certain car averages 25 miles per gallon of gasoline when [#permalink]

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09 Sep 2016, 12:05
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A certain car averages 25 miles per gallon of gasoline when [#permalink]

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09 Sep 2016, 14:38
A certain car averages 25 miles per gallon of gasoline when driven 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

10*1/25 gpm+50*1/40 gpm=1.65 total gallons
60 total miles/1.65 total gallons=36.4 average mpg
D. 36
A certain car averages 25 miles per gallon of gasoline when   [#permalink] 09 Sep 2016, 14:38
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