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Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Sep 2012, 05:02
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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SOLUTIONCar X averages 25.0 miles per gallon of gasoline and Car Y averages 11.9 miles per gallon. If each car is driven 12,000 miles, approximately how many more gallons of gasoline will Car Y use than Car X ?(A) 320 (B) 480 (C) 520 (D) 730 (E) 920 Car X will use \(\frac{12,000}{25}=480\) gallons of gasoline for 12,000 miles; Car Y will use \(\frac{12,000}{(\frac{119}{10})}=12,000*\frac{10}{119}\approx{12,000*\frac{10}{120}}=1,000\) gallons of gasoline for 12,000 miles. The difference is 1,000480=520 gallons. Answer: C.
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Sep 2012, 05:06
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For simplicity we can assum e 11.9 to b 12, so answer would b 1200/12 1200/25
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Sep 2012, 05:20
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12000(2512)/(2512) = 520 Answer C
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Sep 2012, 08:51
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Bunuel wrote: Car X averages 25.0 miles per gallon of gasoline and Car Y averages 11.9 miles per gallon. If each car is driven 12,000 miles, approximately how many more gallons of gasoline will Car Y use than Car X ?
(A) 320 (B) 480 (C) 520 (D) 730 (E) 920
X: 25 miles  in  1 gallon 12000 miles  in  12000/25 = 480 gallons Y: 11.9 miles  in  1 gallon 12000 miles in  12000 / 11.9 gallons since Q is asking appox value and all options are not very close to each other, so we can assume 11.9 as 12 So, 12000/12 = 1000 gallons Hence gasoline used by Y  by X = 1000  480 = 520 Hence C
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Sep 2012, 10:14
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Just think distance = rate * time 12,000 miles = 25mpg * # gallons 12,000 miles = 11.9mpg * # gallons
Similar idea as distance = r*t
So we want to compare # of gallons used.
# gallons X = 12,000 / 25 # gallons Y = 12,000 / 11.9
12,000/11.9  12,000/25
~12,000/12  12,000/100 *4 1,000  120*4 = 1,000  480 = 520
(C)



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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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28 Sep 2012, 07:06
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Using R*T=D chart we can arrive to a solution pretty quickly R T D car X 25 12000 Car Y 11.9 12000 here is much faster to use 12 instead of 11.9 (also the stem says us the word approximately, whenever we see this word we can rounde some value with attenton) Car X > T = 12000/25 = 480 Car Y > T = 12000/12= 1000 The difference lead us to C NOTE: here the trap answer is to think to do 25  12 = 13 and then 12000 / 13 = 923.07......that approximately is E
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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23 Jul 2013, 21:54
carcass wrote: Using R*T=D chart we can arrive to a solution pretty quickly
R T D
car X 25 12000
Car Y 11.9 12000
here is much faster to use 12 instead of 11.9 (also the stem says us the word approximately, whenever we see this word we can rounde some value with attenton)
Car X > T = 12000/25 = 480
Car Y > T = 12000/12= 1000
The difference lead us to C
NOTE: here the trap answer is to think to do 25  12 = 13 and then 12000 / 13 = 923.07......that approximately is E Can someone please explain why it is wrong to use 25  12 = 13 then use 12,000 / 13 ?



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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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23 Jul 2013, 22:53
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As far as the difficulty, I'd rate it as a 550. catty2004 wrote: carcass wrote: Using R*T=D chart we can arrive to a solution pretty quickly
R T D
car X 25 12000
Car Y 11.9 12000
here is much faster to use 12 instead of 11.9 (also the stem says us the word approximately, whenever we see this word we can rounde some value with attenton)
Car X > T = 12000/25 = 480
Car Y > T = 12000/12= 1000
The difference lead us to C
NOTE: here the trap answer is to think to do 25  12 = 13 and then 12000 / 13 = 923.07......that approximately is E Can someone please explain why it is wrong to use 25  12 = 13 then use 12,000 / 13 ? 13 is closer to 12 than 25. So your answer is approximately the number of gallons used by car Y. The MPH of Car X is 2.08 times that of Car Y. For the same distance traveled Car Y uses 2.08 times more gas than Car X. Since we are dealing with multiples, you can not subtract at the beginning of the problem. You have to use \(Distance (in miles)/Miles per Gallon = Gallons Used\) for each car, then subtract the difference between each car to obtain the difference in Gallons used per car.
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Jul 2013, 09:35
MzJavert wrote: As far as the difficulty, I'd rate it as a 550. catty2004 wrote: carcass wrote: Using R*T=D chart we can arrive to a solution pretty quickly
R T D
car X 25 12000
Car Y 11.9 12000
here is much faster to use 12 instead of 11.9 (also the stem says us the word approximately, whenever we see this word we can rounde some value with attenton)
Car X > T = 12000/25 = 480
Car Y > T = 12000/12= 1000
The difference lead us to C
NOTE: here the trap answer is to think to do 25  12 = 13 and then 12000 / 13 = 923.07......that approximately is E Can someone please explain why it is wrong to use 25  12 = 13 then use 12,000 / 13 ? 13 is closer to 12 than 25. So your answer is approximately the number of gallons used by car Y. The MPH of Car X is 2.08 times that of Car Y. For the same distance traveled Car Y uses 2.08 times more gas than Car X. Since we are dealing with multiples, you can not subtract at the beginning of the problem. You have to use \(Distance (in miles)/Miles per Gallon = Gallons Used\) for each car, then subtract the difference between each car to obtain the difference in Gallons used per car. Thanks for the explanation, but im not sure why using 25 12 = 13 is equivalent of approximating car Y's distance traveled??



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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Jul 2013, 21:30
catty2004 wrote: Thanks for the explanation, but im not sure why using 25 12 = 13 is equivalent of approximating car Y's distance traveled??
Responding to a pm: Notice what it is given to you: Car X runs 25 miles in 1 gallon of fuel. Car Y runs approx 12 miles in 1 gallon of fuel (since options are not close to each other, we can easily approximate 11.9 to 12. If options are very close to each other, you need to calculate a more accurate answer) So Car X runs an extra 13 miles in one gallon of fuel. Now what does the term 12000/13 represent? Is it the extra fuel used by car Y? No. This will be the fuel used by a car which runs 12000 miles at an average of 13 miles in one gallon. But neither car X nor car Y does that. Car X needs 12000/25 = 480 gallons of fuel Car Y needs 12000/12 = 1000 gallons of fuel What we need to do here is \(\frac{12000}{12}  \frac{12000}{25}\). I hope this is obvious to you that this is not equal to \(\frac{12000}{(25  12)}\) When the denominators are different, we cannot combine the fractions. We can combine them in case the denominators are the same. Since 25 and 12 are in the denominator of the quantity we need, we cannot combine them.
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Jul 2013, 22:18
VeritasPrepKarishma wrote: catty2004 wrote: Thanks for the explanation, but im not sure why using 25 12 = 13 is equivalent of approximating car Y's distance traveled??
Responding to a pm: Notice what it is given to you: Car X runs 25 miles in 1 gallon of fuel. Car Y runs approx 12 miles in 1 gallon of fuel (since options are not close to each other, we can easily approximate 11.9 to 12. If options are very close to each other, you need to calculate a more accurate answer) So Car X runs an extra 13 miles in one gallon of fuel. Now what does the term 12000/13 represent? Is it the extra fuel used by car Y? No. This will be the fuel used by a car which runs 12000 miles at an average of 13 miles in one gallon. But neither car X nor car Y does that. Car X needs 12000/25 = 480 gallons of fuel Car Y needs 12000/12 = 1000 gallons of fuel What we need to do here is \(\frac{12000}{12}  \frac{12000}{25}\). I hope this is obvious to you that this is not equal to \(\frac{12000}{(25  12)}\) When the denominators are different, we cannot combine the fractions. We can combine them in case the denominators are the same. Since 25 and 12 are in the denominator of the quantity we need, we cannot combine them. Thank you! No longer confused!! I thought by doing 25  11.9 is easier than doing long division of 12k/12  12k/25



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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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Bunuel wrote: Car X averages 25.0 miles per gallon of gasoline and Car Y averages 11.9 miles per gallon. If each car is driven 12,000 miles, approximately how many more gallons of gasoline will Car Y use than Car X ? (A) 320 (B) 480 (C) 520 (D) 730 (E) 920 Practice Questions Question: 49 Page: 158 Difficulty: 600 Basically, for X we have approximately \(\frac{12*1000}{12*2}\) = 500 (though less than 500, because 24 < 25) And for Y we have approximately \(\frac{12000}{12}\) = 1000 (though more than 1000, because 12 > 11.9) So, the subtraction needs to be slightly higher than 500, so our answer is C.



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Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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24 Oct 2015, 15:51
Wondering if this is too lazy/simple? Instead of dividing 12000/25, I just did the approximation of 11.9 to 12  so quick dividing 12000/12 brings you to 1000  and just went with an extra approximation saying 12 practically half of 25 so half of 1000 = 500, and since I rounded up the nearest rounded up answer is 520. It seems like more steps when you write it out but it seemed faster as you can just do that in your head but I'm wondering if, while I arrived at the same answer, this is safe to do a second approximation like that or if on the GMAT that might come back to bite me on problems like these.
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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25 Oct 2015, 23:20
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redfield wrote: Wondering if this is too lazy/simple?
Instead of dividing 12000/25, I just did the approximation of 11.9 to 12  so quick dividing 12000/12 brings you to 1000  and just went with an extra approximation saying
12 practically half of 25 so half of 1000 = 500, and since I rounded up the nearest rounded up answer is 520.
It seems like more steps when you write it out but it seemed faster as you can just do that in your head but I'm wondering if, while I arrived at the same answer, this is safe to do a second approximation like that or if on the GMAT that might come back to bite me on problems like these. Approximation is fine if the options are far apart (as in this case). But keep in mind that whenever you approximate, keep the direction in mind. What I mean by that is that if you are going to say that 12 is practically half of 25, so half of 1000 is 500, you need to keep in mind that 12 is "less than" half of 25 so when you divide 12000 by 25, you will get something less than half i.e. less than 500. So when you subtract "something less than 500" from 1000, you will get "something more than 500" and that is the reason your answer is 520. You stumbled in your logic when you said "since I rounded up...". You did round up 25 but it "divides" 12000 so the answer goes down. But thereafter, you need to subtract it from 1000 and that's why you finally "round up"! I hope that makes sense.
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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30 May 2016, 05:19
Bunuel wrote: Car X averages 25.0 miles per gallon of gasoline and Car Y averages 11.9 miles per gallon. If each car is driven 12,000 miles, approximately how many more gallons of gasoline will Car Y use than Car X ?
(A) 320 (B) 480 (C) 520 (D) 730 (E) 920 To solve this problem we will need to use the formula: rate x gallons = distance, which means: gallons = distance/rate We need to calculate the number of gallons used by car X and the number of gallons used by car Y. Let’s start with car X. For car X we know the following:rate = 25 distance = 12,000 gallons = 12,000/25 = 480 gallons For car Y we know the following:rate = 11.9 Since we are told we can approximate, we can round up the rate of Car Y to 12. distance = 12,000 gallons = 12,000/12 = 1,000 Thus, we can say that car Y will use approximately 1,000 – 480 = 520 more gallons of gasoline than car X. The answer is C.
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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30 May 2016, 09:29
Bunuel wrote: Car X averages 25.0 miles per gallon of gasoline and Car Y averages 11.9 miles per gallon. If each car is driven 12,000 miles, approximately how many more gallons of gasoline will Car Y use than Car X ? (A) 320 (B) 480 (C) 520 (D) 730 (E) 920 Practice Questions Question: 49 Page: 158 Difficulty: 600 X uses 12000/12 =>480 gallons Y uses 12000/12 => 1000 gallons So, Car Y uses 520 gallons of gasoline ( 1000  480 ) gallons.... Answer will be (C)
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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04 Dec 2016, 02:51
Here is what i did in this Question. Its all about approximation here.
Car x => 25 miles in 1 gallon 1 mile in \(\frac{1}{25}\) gallon 12000 miles in\(\frac{1}{25} * 12000 => 480\) gallons
Car y 11.9miles in 1 gallon 1 mile in \(\frac{1}{11.9}\) gallons 12000 miles in \(\frac{1}{11.9}*12000 =>1000\) gallons (Note the approximation of 11.9 to 12) Hence the difference is 1000480=> 520 Gallons Hence C
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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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27 Jan 2018, 07:30
question is asking approximately how much .. So car x = 12000/25 = 480 car y = 12000/11.9 ==> 11.9 is approx. 12 => 12000/12 = 1000
hence difference is 520. hence c



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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]
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02 Feb 2018, 12:04
Hi All, This question is ultimately about division. We're told how many miles per gallon each car averages and the distance that each car travels, so we can determine how many gallons of gas each car uses. Car X: 12,000/25 = 480 gallons Car Y 12,000/(about 12) = about 1,000 gallons Approximate difference in gallons used = 1000  480 = 520 Final Answer: GMAT assassins aren't born, they're made, Rich
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