The chart above shows the distribution of letter grades for three groups of students on one test. Point values are assigned to the letter grades so that A = 4 points, B = 3 points, C = 2 points, and D = 1 point. Which of the following is closest to the average (arithmetic mean) of the grades of all the students combined? A. 2.1 B. 2.5 C. 2.9 D. 3.1 E. 3.3
(B) We start by determining the number of students that received each grade: 3 + 5 + 4 = 12 grades of 1. 5 + 7 + 7 = 19 grades of 2. 7 + 9 +10 = 26 grades of 3. 3 + 5 + 2 = 10 grades of 4.
All together, there are 12 + 19 + 26 + 10 = 67 students. The total number of points earned by all of the students is: (12 × 1) + (19 × 2) + (26 × 3) + (10 × 4) = 12 + 38 + 78 + 40 = 168.
The average for all of the students is then: 168/67 ≈ 2.5
The correct answer is B.
Alternate Method: This is a case where we can get to the answer without doing many calculations if we take a step back and look at the information in the figure. The students who received A’s basically offset the students who received D’s, with a few more students receiving D’s. So the average of these two groups should be approximately: (1 + 4)/2 = 2.5.
The same holds true of the students who received B’s and C’s. There were a few more students who received B’s, so the average grade in these two groups will be a little closer to a 3 than to 2, but as a rough approximation, it should be approximately: (2 + 3)/2 = 2.5. So then the overall average must be close to 2.5. Looking at our choices, we see 2.5 and next largest choice is 2.9, which is too large.
Again, the closest answer choice is 2.5. ---------- I did not understand the question could u please elaborate. which groups are we talking about?
I don't really understand the reasoning behind the shorter method.
First of all, I'd like to note, that this shortcut is quite advanced one. It can be used here only if you see it as a glimpse and are very familiar with it. So if you don't get it right away - it's better and faster to use the general method, which is provided in the explanation.
1. We have some quantity of D-marks (denote by x), each corresponds to "1 point". We have about the same quantity of A-marks (x as well), each corresponds to "4 points". Their average will be (x × 1 + x × 4)/(2x) = (1 + 4)/2 for ANY x.
Here is the tricky moment #1. You need to feel this "about the same quantity". What can the difference be: 2, 10, 50? It depends on the quantities. Let's say 98 of 1s and 100 of 4s are about the same, but 4 of 1s and 2 of 4s are NOT. 98 of 0.00001s and 100 of 2s are NOT good as well.
2. We did the same for C-marks and B-marks and then we've combined the results, because they were "about the same".
Here is the tricky moment #2. We've skipped all the rest of the calculations, because the points were "about the same" in the both case. If they were different, let's say 2 and 3, this shortcut would NOT work. One group could have contained 100 marks (3 average), while the other one could have contained 10 marks (2 average). The combined average is NOT (2 + 3)/2 = 2.5, but it IS (3 × 100 + 2 × 10)/110 ≈ 2,9.
ADVICE: If you don't see a shortcut, or not sure if it can be applied, then use the general method - it will deliver the desired results. If you're low on time and see a shortcut, but not so sure if it can be applied - your decision depends on how much time you have left and how sure you are about your shortcut.
USEFUL FACT: A relevant fact about average values, that can lead to a shortcut.
An average value always lies between the maximum and minimum values. E.g. we have some quantity of 2s and 4s - the average value will be between 2 and 4. If there're more 4s - it will be closer to 4. If there're more 2s - it will be closer to 2.
The answers 2.8 and 2.9 are very close to each other, so it is not convincing that based on visual determination it is possible to say that 2.8 is more likely than 2.9. This solution lacks smarter way of solving the problem.
Users browsing this forum: No registered users and 0 guests
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum
GMAT(TM) and GMAT CAT (TM) are registered trademarks of the Graduate Management Admission Council(TM). The Graduate Management Admission Council(TM) does not endorse, nor is affiliated in any way with the owner or any content of this site.