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A Prototype fuel-efficient car (P-Car) is estimated to get 80% more

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aparnaharish
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A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   08 Aug 2013, 01:21
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A Prototype fuel-efficient car (P-Car) is estimated to get 80% more miles per gallon of gasoline than does a traditional fuel-efficient car (T-Car). However, the P-Car requires a special type of gasoline that costs 20% more per gallon than does the gasoline used by a T-Car. If the two cars are driven the same distance, what percent less than the money spent on gasoline for the T-Car is the money spent on gasoline for the P-Car?

A) 16 2/3%
B) 33 1/3 %
C) 50%
D) 60%
E) 66 2/3 %
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B
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blueseas
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   Updated on: 08 Aug 2013, 01:34
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aparnaharish wrote:
A Prototype fuel-efficient car (P-Car) is estimated to get 80% more miles per gallon of gasoline
than does a traditional fuel-efficient car (T-Car). However, the P-Car requires a special type
of gasoline that costs 20% more per gallon than does the gasoline used by a T-Car. If the two
cars are driven the same distance, what percent less than the money spent on gasoline for the
T-Car is the money spent on gasoline for the P-Car?

A) 16 2/3%
B) 33 1/3 %
C) 50%
D) 60%
E) 66 2/3 %


let say T car travel 100 miles with 100 $ gasoline
therefore according to quest
P car will travell 180(80%more) mile and gasoline cost will be 120(20%more)

therfore p car will travel \(100\) miles in \((120/180)*100\) $= \(200/3\) $

difference of money for both cars travelling for 100 miles = 100-(200/3) = 100/3

so percentage less will be ((100/3)/100)*100 (DENOMINATOR IS 100 as cost for T was 100 $...and multiplying by 100 in order to take out the percent value)
= 33 1/3

hence B

Originally posted by blueseas on 08 Aug 2013, 01:33.
Last edited by blueseas on 08 Aug 2013, 01:34, edited 1 time in total.
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   08 Aug 2013, 01:33
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Assumption

Miles/Gallon
T = 100
P = 180 ( 80% more)

$/gallon
T = 100
P = 120 (20% more)

Miles
100 for both

Cost = (Miles X $/ Gallon)/ M/gallon

T = 100
P = 66.67

Hence 100 - 66.67 = 33.3 = 33 1/3%

Ans B

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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   08 Aug 2013, 01:37
\(miles=gallons*MPG\).

\(G_p*MPG_p=G_t*MPG_t\) with the data we know that \(G_p*1.8=G_t\), \(\frac{G_p}{G_t}=\frac{1}{1.8}\).

We know also that \(price_p=1.2*price_t\)so the cost is (price*number of gallons)\(\frac{P}{T}=\frac{Price_p}{Price_t}*\frac{G_p}{G_t}=\frac{1.2}{1.8}=\frac{2}{3}\) or 1/3 (33.3%) less.
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   09 Aug 2013, 00:24
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aparnaharish wrote:
A Prototype fuel-efficient car (P-Car) is estimated to get 80% more miles per gallon of gasoline
than does a traditional fuel-efficient car (T-Car). However, the P-Car requires a special type
of gasoline that costs 20% more per gallon than does the gasoline used by a T-Car. If the two
cars are driven the same distance, what percent less than the money spent on gasoline for the
T-Car is the money spent on gasoline for the P-Car?

A) 16 2/3%
B) 33 1/3 %
C) 50%
D) 60%
E) 66 2/3 %


Or you can plug in numbers:

P-Car is estimated to get 80% more miles per gallon of gasoline than does a traditional fuel-efficient car (T-Car) - So if T-Car gets 10 miles per gallon, P-Car gets 18 miles per gallon.

P-Car requires a special type of gasoline that costs 20% more per gallon than does the gasoline used by a T-Car - So if T-Car gasoline costs $1 per gallon, P-Car gasoline costs $1.2 per gallon.

T-Car runs 10 miles in $1 i.e. 180 miles in $18
P-Car runs 18 miles in $1.2 i.e. 180 miles in $12

P-Car costs (6/18)*100 = 33.33% less
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   09 Aug 2013, 02:46
aparnaharish wrote:
A Prototype fuel-efficient car (P-Car) is estimated to get 80% more miles per gallon of gasoline
than does a traditional fuel-efficient car (T-Car). However, the P-Car requires a special type
of gasoline that costs 20% more per gallon than does the gasoline used by a T-Car. If the two
cars are driven the same distance, what percent less than the money spent on gasoline for the
T-Car is the money spent on gasoline for the P-Car?

A) 16 2/3%
B) 33 1/3 %
C) 50%
D) 60%
E) 66 2/3 %

..........
solution:
(1-2/3)/1 * 100 = 33.33%
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   12 Aug 2013, 19:12
VeritasPrepKarishma wrote:
aparnaharish wrote:
A Prototype fuel-efficient car (P-Car) is estimated to get 80% more miles per gallon of gasoline
than does a traditional fuel-efficient car (T-Car). However, the P-Car requires a special type
of gasoline that costs 20% more per gallon than does the gasoline used by a T-Car. If the two
cars are driven the same distance, what percent less than the money spent on gasoline for the
T-Car is the money spent on gasoline for the P-Car?

A) 16 2/3%
B) 33 1/3 %
C) 50%
D) 60%
E) 66 2/3 %


Or you can plug in numbers:

P-Car is estimated to get 80% more miles per gallon of gasoline than does a traditional fuel-efficient car (T-Car) - So if T-Car gets 10 miles per gallon, P-Car gets 18 miles per gallon.

P-Car requires a special type of gasoline that costs 20% more per gallon than does the gasoline used by a T-Car - So if T-Car gasoline costs $1 per gallon, P-Car gasoline costs $1.2 per gallon.

T-Car runs 10 miles in $1 i.e. 180 miles in $18
P-Car runs 18 miles in $1.2 i.e. 180 miles in $12

P-Car costs (6/18)*100 = 33.33% less


Hi VeritasPrepKarishma,

I solved exactly the same way as you did, however I am unable to calculate the percent difference.
Car T - 10 miles and car P - 18 miles. I assumed the cost for one gallon as 20 for T and 24 for P . Hence for a distance of 100 miles, car T uses $ 200 and car P uses $400/3. But when I find the percent difference which is (200-400/3)/ 200 = 33%
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   12 Aug 2013, 21:32
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Diipz wrote:

Hi VeritasPrepKarishma,

I solved exactly the same way as you did, however I am unable to calculate the percent difference.
Car T - 10 miles and car P - 18 miles. I assumed the cost for one gallon as 20 for T and 24 for P . Hence for a distance of 100 miles, car T uses $ 200 and car P uses $400/3. But when I find the percent difference which is (200-400/3)/ 200 = 33%


I don't see any problem here. You got the % difference as 1/3 which is 33.33%. This is the correct answer.
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   30 May 2014, 13:34
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Let's say that T car is driven at 10mpg while P car is driven at 18mpg.

Now, then Let's also assume that T car's gallon costs 10 bucks wihle P car gallon costs 12 bucks.

Finally, let's assume that the total distance is 180. Therefore P car will spend 120 and T car will spend 180. Therefore the difference is 60/180=33.3%.

Answer is B

Hope this helps
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink] New post   01 Feb 2024, 04:36
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Re: A Prototype fuel-efficient car (P-Car) is estimated to get 80% more [#permalink]
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