Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jul 1311:00 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Don’t settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1206:30 AM PDT
-08:30 AM PDT
Join our webinar to master GMAT RC Main Point Questions! Learn effective strategies to decipher passage themes and purposes. Senior Verbal Expert- Sunita Singhvi will equip you with tools to navigate complexities and avoid common traps set by the GMAT.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Jul 1611:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
Jul 1911:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment
01 Jan 2024, 06:24
Kudos
Bookmarks
nazii
Dear KarishmaB
By your approach (weighed average) the answer would be 37.5, which is much closer to answer choice 38, which is a wrong answer choice, how we can avoid such mistake? or can you tell when we can find this approach is better or the approach provided by Bunuel? Because t first I used weght average for this question and mistakenly chose 38 as answer, by looking at Bunuel's approach I saw it's totally different!
'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.
By your approach (weighed average) the answer would be 37.5, which is much closer to answer choice 38, which is a wrong answer choice, how we can avoid such mistake? or can you tell when we can find this approach is better or the approach provided by Bunuel? Because t first I used weght average for this question and mistakenly chose 38 as answer, by looking at Bunuel's approach I saw it's totally different!
KarishmaB
khushboochhabra
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??
(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.
Note that the post you have quoted responds to exactly this query. Before we decide to use weighted averages, we need to fully understand what the weights will be in this situation. First check this post: https://anaprep.com/arithmetic-weights- ... d-average/
Now look at the explanation I gave in my post above: https://gmatclub.com/forum/a-certain-ca ... l#p1079109
11 May 2025, 11:29
Kudos
Bookmarks
Thanks for the detailed explanation, Ana.
Even I made the mistake of assuming this was a weighted average problem.
'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.
Here is a post on Weights in Weighted Avg:
https://anaprep.com/arithmetic-weights-in-weighted-average/
Even I made the mistake of assuming this was a weighted average problem.
KarishmaB
khushboochhabra
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??
(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.
Here is a post on Weights in Weighted Avg:
https://anaprep.com/arithmetic-weights-in-weighted-average/
