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

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

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25 Feb 2012, 19:00
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Difficulty:

85% (hard)

Question Stats:

55% (02:29) correct 45% (02:29) wrong based on 765 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 driven 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

Struggling on this one too. Any idea how to solve?

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Re: Miles in a gallon  [#permalink]

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25 Feb 2012, 22:57
15
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enigma123 wrote:
A certain car averages 25 miles per gallon of gasoline when driven in the city and 40 miles per gallon when driven 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

Struggling on this one too. Any idea how to solve?

Car averages x miles per gallon, means that 1 gallon is enough to drive x miles.

We are asked to find average miles per gallon (miles/gallon) --> average miles per gallon would be total miles driven divided by total gallons used (miles/gallon).

Total miles driven is 10+50=60.

As car averages 25 miles per gallon in the city for 10 miles in the city it will use 10/25=0.4 gallons;
As car averages 40 miles per gallon on the highway for 50 miles on the highway it will use 50/40=1.25 gallons;

So average miles per gallon equals to $$\frac{10+50}{0.4+1.25}\approx{36}$$

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Re: Miles in a gallon  [#permalink]

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25 Feb 2012, 21:55
2
Hi !

You need average miles/ gallon with the given data

Total Miles traveled => 10+50 = 60

City Fuel Consumption => 1 gallon - 25 miles => 10 Miles - 10/25 gallons => 0.4 gallons
Highway Fuel Consumption => 1 gallon - 40 miles => 50 Miles - 50/40 gallons => 1.25 gallons

Now the tricky part which even I got wrong in my first timed attempt

Average fuel consumption (City + Highway) = 60 miles / (0.4+1.25) = 60/1.65 = 36.6 miles / gallon

divide 60 / 15 = 40 but since 1.65 is greater than 15 the fraction has to be smaller to its either 38 / 36

38 is just too close so eliminate

Hope this helps
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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07 Apr 2012, 02:39
1
My take on this -

whenever I see this type of question, I make sure what the question is asking me to solve...and this makes the question easier to solve-

Here, its asking - Total distance travelled/ number of gallons used.-----------------(1)

Now start calculating individual entity.

Total distance = 50+10 = 60

Total galoons - This is the tricky part.

Generally, I solve by using unitary method.

such as - 25 miles ---- 1 gallon
1 mile = 1/25
10 miles = 1/25*10

Similarly calculate for Highway.

Try solving for highway and put the values in equation 1. You will get the answer.

Thanks
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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26 Apr 2012, 17:51
I got Bunuel's approach.. but was wondering if weighed average should work here that is

(1*25+5*40)/6 = 37.5 ~ 38 Seems it doesnt work here.. But shouldn't it??
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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26 Apr 2012, 23:09
1
khushboochhabra wrote:
I got Bunuel's approach.. but was wondering if weighed average should work here that is

(1*25+5*40)/6 = 37.5 ~ 38 Seems it doesnt work here.. But shouldn't it??

Actually we are asked to find the average, but the average of miles per gallon (miles/gallon). So, you should have total miles driven in numerator and total gallons used in denominator (miles/gallon). Now, ask yourself: what do you have in numerator and denominator?
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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27 Apr 2012, 06:34
8
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khushboochhabra wrote:
I got Bunuel's approach.. but was wondering if weighed average should work here that is

(1*25+5*40)/6 = 37.5 ~ 38 Seems it doesnt work here.. But shouldn't it??

'Distances traveled' (i.e. ratio of 1:5) cannot be the weights here to find the average mileage. The weights have to be 'number of gallons'.

If we change the question and make it:
A certain car averages 25 miles per gallon of gasoline when driven in the city and 40 miles per gallon when driven 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 uses 10 gallons in the city and then 50 gallons on the highway

Now you can use (1*25+5*40)/6

Why?

Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled or should they be gallons of fuel used...), look at the units.

Average required is $$\frac{miles}{gallon}$$. So you are trying to find the weighted average of two quantities whose units must be $$\frac{miles}{gallon}$$.

$$C_{avg} = \frac{C_1*W_1 + C_2*W_2}{{W_1 + W_2}}$$

$$C_{avg}, C_1, C_2 - \frac{miles}{gallon}$$

So $$W_1$$ and $$W_2$$ should be in gallon to get:

$$\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)$$

Only if weights are in gallons, do we get 'Total miles' in the numerator and 'Total gallons' in the denominator.

We know that Average miles/gallon = Total miles/Total gallons

[highlight]Takeaway: The weights have to be the denominator units of the average.[/highlight]

So what do we do in this question?

What is the distance travelled (i.e.total miles)?
(10+50) = 60

What is the total gallons used?
Fuel used in the city = 10/25
Fuel used to go from B to C = 50/40

So Average miles/gallon = (60)/(10/25 + 50/40) = apprx 36

You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon.
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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27 Apr 2012, 09:59
Thanks VeritasPrepKarishma. I was having issues with understanding weights in averages. This cleared it up.
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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28 Aug 2013, 01:36
I just went through this thread and I have a doubt. Applying Karishma's method of using units, in this case, we get
(25*10+40*50)/(10+ 50) . The units are (miles/g * miles)/ miles = miles/gallon. Please point out the mistake in my logic.
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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28 Aug 2013, 04:07
3
I just went through this thread and I have a doubt. Applying Karishma's method of using units, in this case, we get
(25*10+40*50)/(10+ 50) . The units are (miles/g * miles)/ miles = miles/gallon. Please point out the mistake in my logic.

We need total miles in the numerator and gallons in the denominator.

(C1*W1 + C2*W2) should make "miles".
(W1 + W2) should make "gallons".
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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18 Sep 2013, 11:30
enigma123 wrote:
A certain car averages 25 miles per gallon of gasoline when driven in the city and 40 miles per gallon when driven 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

Struggling on this one too. Any idea how to solve?

Sol:

10 miles in city while it gives 25miles/gallon in city, so 10 miles will consume : 10/25 =>2/5 gallon

10 miles on highway while it gives 40 miles/gallon on highway, so 10 miles will consume : 10/40 =>1/4 gallon

40 miles on high way whilte it gives 40 miles/gallon on highway, so 40 miles will consume: 40/40 =>1 gallon

Total gallons : 1+ (2/5)+ (1/4) = 33/20

Total distance travelled is (10+50)= 60

so avg is 60/(33/20) ==>36
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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19 Sep 2013, 03:46
i took 4:37 seconds to solve this question...
Did i spend too much time ?
I actually commited a calculation mistake in between. Even if I wouldn't have done that it would take me at least 3mins.
I see the average of correct and wrond both to be 1:50 (approx)
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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25 Dec 2015, 11:13
total miles=60
total gallons=10/25+50/40=33/20
total miles/total gallons=36.4 average mpg
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A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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18 Nov 2016, 08:11
enigma123 wrote:
A certain car averages 25 miles per gallon of gasoline when driven in the city and 40 miles per gallon when driven 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

Struggling on this one too. Any idea how to solve?

well...a pretty straight forward one...instead of RTD chart, we have a RGD one...where G stands for gallons.

1st thing to do, find how much gas the car consumed during the first 10 miles in the city: 10/25 = 2/5
2nd thing to do, find how much gas the car consumed during the 50 miles on the highway: 50/40 = 5/4

total gas consumed: 2/5 + 5/4 = 33/20

now...we have the total distance: 10+50 = 60 miles
we have the total gas consumed: 33/20 gallons
the average consumption is: 60 : 33/20 or 60*20/33 = 20*20/11 = 400/11 -> let's approximate...this takes only few seconds..
400/12 = 100/3 = 33.33
400/10 = 40
so it must be between 33.33 and 40. A, B, and C are out right away.
now...11 * 30 = 330, and we are left with another 70...in 70, 11 fits only 6 times and slightly more.
therefore, the correct answer must be 30+6=36.

D

total time to solve - under 1min50sec.
a relatively easy question, compared to other motion related ones...
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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18 Nov 2016, 09:34
enigma123 wrote:
A certain car averages 25 miles per gallon of gasoline when driven in the city and 40 miles per gallon when driven 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

Struggling on this one too. Any idea how to solve?

City -

Mileage = 25 m/g
Distance covered is 10 m

So, Fuel required is 10/25 => 0.40

Highway -

Mileage = 40 m/g
Distance covered is 50 m

So, Fuel required is 50/40 =>1.25

Total Miles driven in City + Highway is 60 miles
Total Fuel used in City + Highway is 1.65

Mileage = $$\frac{60}{1.65} = 36.36$$

Hence, answer will be (D) 36

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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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29 Jul 2017, 04:30
VeritasPrepKarishma wrote:
khushboochhabra wrote:
I got Bunuel's approach.. but was wondering if weighed average should work here that is

(1*25+5*40)/6 = 37.5 ~ 38 Seems it doesnt work here.. But shouldn't it??

'Distances traveled' (i.e. ratio of 1:5) cannot be the weights here to find the average mileage. The weights have to be 'number of gallons'.

If we change the question and make it:
A certain car averages 25 miles per gallon of gasoline when driven in the city and 40 miles per gallon when driven 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 uses 10 gallons in the city and then 50 gallons on the highway

Now you can use (1*25+5*40)/6

Why?

Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled or should they be gallons of fuel used...), look at the units.

Average required is $$\frac{miles}{gallon}$$. So you are trying to find the weighted average of two quantities whose units must be $$\frac{miles}{gallon}$$.

$$C_{avg} = \frac{C_1*W_1 + C_2*W_2}{{W_1 + W_2}}$$

$$C_{avg}, C_1, C_2 - \frac{miles}{gallon}$$

So $$W_1$$ and $$W_2$$ should be in gallon to get:

$$\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)$$

Only if weights are in gallons, do we get 'Total miles' in the numerator and 'Total gallons' in the denominator.

We know that Average miles/gallon = Total miles/Total gallons

Takeaway: The weights have to be the denominator units of the average.

So what do we do in this question?

What is the distance travelled (i.e.total miles)?
(10+50) = 60

What is the total gallons used?
Fuel used in the city = 10/25
Fuel used to go from B to C = 50/40

So Average miles/gallon = (60)/(10/25 + 50/40) = apprx 36

You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon.

VeritasPrepKarishma Your clarification seems impressive as if unit is not same the result will be something different than actual
So first we need to check what is given . If it is miles per gallon then weighted average can only be calculated when amount of fuel used in given.

So, to solve such problem I want to tweak the approach to make the solution aligned with weighted average.

Now since we know that the no. of miles i.e. distance is given and we can't take weighted average method, lets try to make the given rate i.e. miles per galon favorable to distance by reversing it and then using weighted average and then again reversing it to get my answer. All the reversing process will go in mind and we will make full use of Karishma's approach.

So. Car average in city = 25 miles per gallon
Lets reverse it to understand it as car will use 1/25 gallon per mile in city.

Car average on highway = 40 miles per gallon
Lets reverse it to understand it as car will use 1/40 gallon per mile on highway.

Also distance traveled in city is given as 10 miles and on highway is 50 miles.

So, here we can use the weighted average now .
Avg. = (1/25*10 + 1/40* 50 )/(10+50) =( 0.4 +1.25 )/60 = 1.65/60 = 11/400 gallon per mile.

Lets reverse it again to get the answer.
Hence car average = 400/11 = 36.6 miles per gallon.

So, basically I have gone a step further using Karishma's approach to understand the problem, avoid mistake regarding weighted average and then solving it the same way by tweaking it in the mid way.

Hope that helps...
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Re: A certain car averages 25 miles per gallon of gasoline when  [#permalink]

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31 Jul 2018, 01:05
Case 1:
25 miles --> 1 gallon
1 mile --> 1/25 gallon
10 miles --> 1/25*10 --> 2/5 gallon used

Case 2:
1 mile --> 1/40
50 miles --> 1/40*50 = 5/4

Average => total d/total mileage --> 60/(2/5 + 5/4) -> 60/(20/33)--> 400/11 approx = 33
Re: A certain car averages 25 miles per gallon of gasoline when &nbs [#permalink] 31 Jul 2018, 01:05
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