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

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Intern
Joined: 28 Jun 2008
Posts: 39
A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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13 Jun 2009, 14:04
11
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Difficulty:

85% (hard)

Question Stats:

60% (02:28) correct 40% (02:37) wrong based on 313 sessions

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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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8810
Location: Pune, India

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09 Jun 2011, 19:25
2
3
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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Karishma
Veritas Prep GMAT Instructor

##### General Discussion
Manager
Joined: 28 Jan 2004
Posts: 197
Location: India

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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
Senior Manager
Joined: 20 Mar 2008
Posts: 422

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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.
Manager
Joined: 28 Jan 2004
Posts: 197
Location: India

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14 Jun 2009, 22:02
Can Jivana pls. expalin how the gas consumed is 1.25 gallons ?
Manager
Joined: 26 Feb 2010
Posts: 75
Location: Argentina

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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
Manager
Joined: 16 Feb 2010
Posts: 135

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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
Intern
Joined: 18 May 2011
Posts: 12

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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
Manager
Joined: 08 Sep 2010
Posts: 129

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09 Jun 2011, 20:04
1
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
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Intern
Joined: 18 May 2011
Posts: 12

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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
Manager
Joined: 12 Oct 2009
Posts: 145

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10 Jun 2011, 10:26
Excellent approach Karishma !!!
Current Student
Joined: 26 May 2005
Posts: 505

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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: 8810
Location: Pune, India

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10 Jun 2011, 11:00
1
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

Manager
Joined: 22 Jul 2014
Posts: 53
Concentration: Operations, General Management
GMAT 1: 660 Q47 V34
GMAT 2: 710 Q49 V38
GPA: 3.27
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

Hail the Messiah, Lionel Andres Messi Cuccitini
VP
Joined: 07 Dec 2014
Posts: 1155
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
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Posts: 9464
Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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13 Mar 2018, 09:50
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Re: A certain car averages 25 miles per gallon of gasoline when &nbs [#permalink] 13 Mar 2018, 09:50
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