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# Car X averages 25.0 miles per gallon of gasoline and Car Y

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Math Expert
Joined: 02 Sep 2009
Posts: 43785
Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]

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24 Sep 2012, 04:02
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Difficulty:

15% (low)

Question Stats:

81% (01:19) correct 19% (01:38) wrong based on 1417 sessions

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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
[Reveal] Spoiler: OA

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Math Expert
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Posts: 43785
Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink]

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24 Sep 2012, 04:02
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SOLUTION

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

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 1,000-480=520 gallons.

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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, 04:06
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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, 04:20
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12000(25-12)/(25-12) = 520
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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, 07:51
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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, 09:14
1
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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

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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, 06:06
1
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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, 20:54
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, 21:53
1
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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, 08:35
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, 20:30
4
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Expert's post
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 30 May 2008 Posts: 73 Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink] ### Show Tags 24 Jul 2013, 21:18 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 Manager Joined: 12 Jan 2013 Posts: 216 Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink] ### Show Tags 13 Jan 2014, 02:52 1 This post received KUDOS 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. Senior Manager Joined: 18 Aug 2014 Posts: 300 Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink] ### Show Tags 24 Oct 2015, 14:51 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. _________________ Please help me find my lost Kudo's bird Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7934 Location: Pune, India Re: Car X averages 25.0 miles per gallon of gasoline and Car Y [#permalink] ### Show Tags 25 Oct 2015, 22:20 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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, 04:19
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.

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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, 08:29
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....

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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, 01:51
Here is what i did in this Question.

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 1000-480=> 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, 06:30
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, 11:04
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

Approximate difference in gallons used = 1000 - 480 = 520

[Reveal] Spoiler:
C

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Re: Car X averages 25.0 miles per gallon of gasoline and Car Y   [#permalink] 02 Feb 2018, 11:04
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