A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?
A. 74
B. 75
C. 76
D. 77
E. 78
PS
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Let:abhasjha wrote:A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?
A. 74
B. 75
C. 76
D. 77
E. 78
Class A have 4 students, each scoring 65 points, for a total of 260 points.
Class B have 6 students, each scoring 80 points, for a total of 480 points.
Class C have 5 students, each scoring 77 points, for a total of 385 points.
Average score for all 15 students = (260+480+385)/15 = 75.
The correct answer is B.
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A slightly different approach:abhasjha wrote:A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?
A. 74
B. 75
C. 76
D. 77
E. 78
This is a weighted averages question, so we can use the following formula:
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...
For more information on weighted averages, you can watch this free GMAT Prep Now video: https://www.gmatprepnow.com/module/gmat- ... ics?id=805
Ratio of classes A, B, C = 4 : 6 : 5, respectively
4 + 6 + 5 = 15
So, class A represents 4/15 of the TOTAL population
Class B represents 6/15 of the TOTAL population
Class C represents 5/15 of the TOTAL population
So, weighted average of 3 groups combined = (4/15)(65) + (6/15)(80) + (5/15)(77)
= 260/15 + 480/15 + 385/15
= 1125/15
= 75
= B
Cheers,
Brent
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Hi abhasjha,
Mitch's solution is likely the easiest way for most Test Takers to go about answering this question (and it's exactly how I would solve it). You can use the answer choices against the prompt though and use some algebra "concepts" to figure out which is correct.
Since we're calculating an average, the "net above average" must "cancel out" the "net below average" of the numbers.
eg. the average of 1, 2 and 6 is 3.
The 1 is "2 below" the average,
the 2 is "1 below" the average, and
the 6 is 3 "above" the average.
The net "3 below" cancels out the net "3 above" and we end up with the average.
With the info in the prompt, we know that we essentially have 4 "65s", 6 "80s" and 5 "77s"
If we start with answer C: 76....
The 65s are each "11 below" --> 4(11) = 44 below
The 80s are each "4 above" --> 6(4) = 24 above
The 77s are each "1 above" --> 5(1) = 5 above
Here we have 44 below and 29 above; these numbers don't cancel out, so 76 CANNOT be the average. Since we have more "below", the average must be SMALLER.
With answer B: 75....
The 65s are each "10 below" --> 4(10) = 40 below
The 80s are each "5 above" --> 6(5) = 30 above
The 77s are each "2 above" --> 5(2) = 10 above
Here, the 40 below cancels out the 40 above, so 75 MUST be the average.
Final Answer: B
To be fair, I wouldn't go about solving this problem in this fashion, but it does serve as an interesting concept for how groups of numbers can affect an average.
GMAT assassins aren't born, they're made,
Rich
Mitch's solution is likely the easiest way for most Test Takers to go about answering this question (and it's exactly how I would solve it). You can use the answer choices against the prompt though and use some algebra "concepts" to figure out which is correct.
Since we're calculating an average, the "net above average" must "cancel out" the "net below average" of the numbers.
eg. the average of 1, 2 and 6 is 3.
The 1 is "2 below" the average,
the 2 is "1 below" the average, and
the 6 is 3 "above" the average.
The net "3 below" cancels out the net "3 above" and we end up with the average.
With the info in the prompt, we know that we essentially have 4 "65s", 6 "80s" and 5 "77s"
If we start with answer C: 76....
The 65s are each "11 below" --> 4(11) = 44 below
The 80s are each "4 above" --> 6(4) = 24 above
The 77s are each "1 above" --> 5(1) = 5 above
Here we have 44 below and 29 above; these numbers don't cancel out, so 76 CANNOT be the average. Since we have more "below", the average must be SMALLER.
With answer B: 75....
The 65s are each "10 below" --> 4(10) = 40 below
The 80s are each "5 above" --> 6(5) = 30 above
The 77s are each "2 above" --> 5(2) = 10 above
Here, the 40 below cancels out the 40 above, so 75 MUST be the average.
Final Answer: B
To be fair, I wouldn't go about solving this problem in this fashion, but it does serve as an interesting concept for how groups of numbers can affect an average.
GMAT assassins aren't born, they're made,
Rich
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Alligation is a great approach for determining the ratio of two ingredients in a mixture composed solely of these two ingredients.abhasjha wrote:Mitch ,
since this is a weighted avg problem - is it possible to solve the question using method of mxiture and alligation too?
Alligation should not be used for a mixture composed of 3 ingredients.
For the reason, check my post here:
https://www.beatthegmat.com/allegation-t69552.html
For those unfamiliar with alligation, check my post here:
https://www.beatthegmat.com/ratios-fract ... 15365.html
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As a tutor, I don't simply teach you how I would approach problems.
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