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average rate problem [#permalink]
 13 Jun 2009, 15:04
Question Stats:
28% (02:37) correct
71% (01:53) wrong based on 0 sessions
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
Answer: D.
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Re: average rate problem [#permalink]
13 Jun 2009, 20:14
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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Re: average rate problem [#permalink]
13 Jun 2009, 22:19
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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Re: average rate problem [#permalink]
14 Jun 2009, 23:02
Can Jivana pls. expalin how the gas consumed is 1.25 gallons ?
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Re: average rate problem [#permalink]
13 May 2010, 07:46
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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Re: average rate problem [#permalink]
14 May 2010, 12:27
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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Re: average rate problem [#permalink]
09 Jun 2011, 03:25
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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Re: average rate problem [#permalink]
09 Jun 2011, 20:25
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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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Re: average rate problem [#permalink]
09 Jun 2011, 21:04
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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Re: average rate problem [#permalink]
10 Jun 2011, 02:15
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
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Re: average rate problem [#permalink]
10 Jun 2011, 11:26
Excellent approach Karishma !!!
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Re: average rate problem [#permalink]
10 Jun 2011, 11:42
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
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Re: average rate problem [#permalink]
10 Jun 2011, 12:00
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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
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Re: average rate problem
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10 Jun 2011, 12:00
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