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16 Sep 2014, 01:53
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B
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Difficulty:
65%
(hard)
Question Stats:
56% (02:12) correct
44%
(02:08)
wrong
based on 86
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A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13
B. 13.5
C. 14
D. 14.5
E. 15
A. 13
B. 13.5
C. 14
D. 14.5
E. 15
16 Sep 2014, 01:53
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Official Solution:
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13
B. 13.5
C. 14
D. 14.5
E. 15
The fastest way to solve this problem is to pick smart numbers for the two distances, so that you'll get integer numbers of gallons. The distance from A to B should be a multiple of 12, and the distance from B to C should be a multiple of 18. With a little trial and error, we can find suitable numbers.
A-B: 72 miles -- the car burns 6 gallons.
B-C: 36 miles -- the car burns 2 gallons.
In total, the car goes 108 miles on 8 gallons, so the average fuel efficiency is \(\frac{108}{8} = 13.5\) miles per gallon.
You can check a different set of numbers:
A-B: 36 miles -- the car burns 3 gallons.
B-C: 18 miles -- the car burns 1 gallon.
In total, the car goes 54 miles on 4 gallons, so again, the average is 13.5 miles per gallon.
This value may not be what you expected: maybe you thought that the answer would be 14, which is a weighted average of 12 and 18, weighted 2:1 toward the 12. However, that weighting is faulty. If you want to weight two ratios or rates (such as miles PER gallons), then you must weight by the denominator (gallons), NOT by the numerator (miles). As we saw by picking numbers, the gallons used on each stage of the trip wind up in a 3:1 ratio, and the weighted average of 12 and 18 (weighted 3:1) is 13.5.
Answer: B
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?
A. 13
B. 13.5
C. 14
D. 14.5
E. 15
The fastest way to solve this problem is to pick smart numbers for the two distances, so that you'll get integer numbers of gallons. The distance from A to B should be a multiple of 12, and the distance from B to C should be a multiple of 18. With a little trial and error, we can find suitable numbers.
A-B: 72 miles -- the car burns 6 gallons.
B-C: 36 miles -- the car burns 2 gallons.
In total, the car goes 108 miles on 8 gallons, so the average fuel efficiency is \(\frac{108}{8} = 13.5\) miles per gallon.
You can check a different set of numbers:
A-B: 36 miles -- the car burns 3 gallons.
B-C: 18 miles -- the car burns 1 gallon.
In total, the car goes 54 miles on 4 gallons, so again, the average is 13.5 miles per gallon.
This value may not be what you expected: maybe you thought that the answer would be 14, which is a weighted average of 12 and 18, weighted 2:1 toward the 12. However, that weighting is faulty. If you want to weight two ratios or rates (such as miles PER gallons), then you must weight by the denominator (gallons), NOT by the numerator (miles). As we saw by picking numbers, the gallons used on each stage of the trip wind up in a 3:1 ratio, and the weighted average of 12 and 18 (weighted 3:1) is 13.5.
Answer: B
27 Oct 2014, 01:12
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Bunuel I wanted to ask you this..actually I also tried to solve through weighted average and got the correct answer by mistake, my calculation are:
[ (12*3)+18]/4 = 13.5. Then I realized that I did a mistake but got the right answer by chance.
I tried to figure out how this gave me the correct answer but failed. So my question is if we can make some alterations in weighted average for such cases or picking numbers should be the best??
[ (12*3)+18]/4 = 13.5. Then I realized that I did a mistake but got the right answer by chance.
I tried to figure out how this gave me the correct answer but failed. So my question is if we can make some alterations in weighted average for such cases or picking numbers should be the best??
