Question 157 Problem Solving 2018 GMAT Official Guide
Question 157 Problem Solving 2018 GMAT Official Guide
Video explanation: A certain experimental mathematics program was tried out…
Comments
Georgesays
Hi, I understand your way to solve the problem, but could we discuss this question in logic? I kinda can’t work it out.
Let’s assume that, we let 37 teachers teach the first 37 classes(now each teacher teaches at least 1 class), then we select 27 teachers to teach the rest 27 classes. Now we have 10 teachers who teach only 1 class and 27 teachers who teach 2 classes. So n=0.
We can narrow the answer to A and B.
Again, let’s assume that 37 teachers teach the first 37 classes. We then let all 37 teachers teach these 37 classes again, which means all teachers have taught at least 2 classes. Now we have 27 classes that have not yet been taught. Then we let 27 teachers to teach these 27 classes. So we have 27 teachers teaching 3 classes, and 10 teachers teaching 2 classes. If so, the n could be 27!
I think it’s ok because the question just said each teacher teaches at least 1 class, so according to my second assumption, the fact that each teacher teaches at least 2 classes is not going against the question’s condition(I make your x=0).
So could you please point out the logic flaw above? Thanks!
The first part of your argument that gives n=0 is correct.
The second part of your argument is where the problem lies. Since each teacher has to teach at least 1 class, we will assign 1 class to each of the 37 teachers. This takes care of the 37 classes out of the total of 64 classes that need to be assigned to the 37 teachers. Because we are trying to maximize the number of teachers who will teach 3 classes, we will reduce the number that will teach 2 classes. We have 27 classes remaining, we will split this into 13 groups of 2 and 1 extra class left. We then assign the 13 groups of 2 classes to 13 of the 37 teachers, and this means we will have 13 teachers that will teach 3 classes, 1 teacher who will teach 2 classes, and 23 teachers who will teach only 1 class. I hope this makes sense.
Hi, I understand your way to solve the problem, but could we discuss this question in logic? I kinda can’t work it out.
Let’s assume that, we let 37 teachers teach the first 37 classes(now each teacher teaches at least 1 class), then we select 27 teachers to teach the rest 27 classes. Now we have 10 teachers who teach only 1 class and 27 teachers who teach 2 classes. So n=0.
We can narrow the answer to A and B.
Again, let’s assume that 37 teachers teach the first 37 classes. We then let all 37 teachers teach these 37 classes again, which means all teachers have taught at least 2 classes. Now we have 27 classes that have not yet been taught. Then we let 27 teachers to teach these 27 classes. So we have 27 teachers teaching 3 classes, and 10 teachers teaching 2 classes. If so, the n could be 27!
I think it’s ok because the question just said each teacher teaches at least 1 class, so according to my second assumption, the fact that each teacher teaches at least 2 classes is not going against the question’s condition(I make your x=0).
So could you please point out the logic flaw above? Thanks!
George,
The first part of your argument that gives n=0 is correct.
The second part of your argument is where the problem lies. Since each teacher has to teach at least 1 class, we will assign 1 class to each of the 37 teachers. This takes care of the 37 classes out of the total of 64 classes that need to be assigned to the 37 teachers. Because we are trying to maximize the number of teachers who will teach 3 classes, we will reduce the number that will teach 2 classes. We have 27 classes remaining, we will split this into 13 groups of 2 and 1 extra class left. We then assign the 13 groups of 2 classes to 13 of the 37 teachers, and this means we will have 13 teachers that will teach 3 classes, 1 teacher who will teach 2 classes, and 23 teachers who will teach only 1 class. I hope this makes sense.