Word Problems 5th Ed. Chapter 5 pg108 Data Sufficiency

[GMATman13]

Here is the problem on page 108 in the 5th edition that I am having trouble with:

The average number of students per class at School X is 25 and the average number
of students per class at School Y is 33. Is the average number of students per
class for both schools combined less than 29 ?

(1) There are 12 classes in School X.
(2)There are more classes in School X than in School Y.

I reasoned that I need to find the variables for this equation:

(Total Students in School X + Total Students in School Y) / (Total number of classes in School X + Total number of classes in school Y)


I selected C.

First I reasoned that if school x has an average of 25 students per class then the equation must me

Average per class for school x = (number of total students in school x) / (number of classes in school x)

So:

With Statement 1, I re-arranged the formula to give me the total number of students for school X, by knowing how many classes were in school x total

12*25 = 300 Students in School X

Now with Statement 2 It says the number of classes in school Y are less than school x so that means it has to be an integer and the highest possible value based on this condition is 11.

So I now can calculate the students in school Y

Classes of school y * Average per class = Students in School Y

11*33 = 363

Now with this information I can solve for the average students per class combined for school x and school y

(Total students in school X + total students in school Y) / (Total number of classes in school x and y)

(300+ 363) / (12+11) = 28.8 students per class

which is less than 29

..-------------------

However after reading the description in the book I am confused because it doesnt take the number of students in the schools into account.

[RonPurewal]

I don't have the book in front of me, so there may or may not be any overlap between my explanation and the book's.

I'm not going to treat statement 2 here, since you've correctly ascertained that it's insufficient. The issue is statement 1.

If you know how "weighted averages" work, then statement 1 is a quick study.
The deal is this:
1/
If there are the same number of classes in school X as in school Y, then the average number of students per class will be exactly 29.
2/
Therefore, since school X is the school with the smaller classes, if school X has more classes, then the average will be less than 29.

Fact #1 (that the average is 29 if the #s of classes are the same) isn't hard to prove algebraically. In that case, let "n" be the number of classes at each school.
Then the overall average is
(total # students) / (total # classes)
= (25n + 33n) / (n + n)
= 29.

If you prove that, then you might be able to figure out the deal here conceptually, without further calculations. I.e., if having the same # of classes gives an average of 29, then adding more small classes (= adding more classes to school x) will bring the average down below 29.
... In the same way that, if your current exam average in a class is 80%, then scoring 75% on the next exam will bring that average down.
You probably have more intuition about this kind of stuff than you might think, because you've probably thought a lot about school grade averages over the years. See whether you can import that intuition into this exam.

[RonPurewal]

If you don't have the intuition described in the previous post, then you can write an inequality for the question, and then simplify it.
Say there are "x" number of classes at school X, and "y" number of classes at school Y. Then the average number of students per class is
(total # students) / (total # classes)
= (25x + 33y) / (x + y)
So the inequality in the question is...
Is (25x + 33y) / (x + y) < 29?
Is 25x + 33y < 29x + 29y ?
Is 4y < 4x ?
Is y < x ?

Hey! Look at that. Knowing that x > y is exactly the same as knowing that the answer to the question is "yes".

[RonPurewal]

As for this...

However after reading the description in the book I am confused because it doesnt take the number of students in the schools into account.

Yep. That's actually the whole point.
Almost all DS problems (except for those at very low performance levels) will allow you to find what you want to find, WITHOUT finding other things.

In other words, if you find out too much information in a problem like this one, you should actually be suspicious of that thought process.
Especially on these kinds of problems -- dealing with averages, medians, and the like -- the DS section is absolutely (in)famous for giving situations in which you can find the average without finding any specifics about the things you're taking an average of.
Be forewarned.

[coolparthi]

If you don't have the intuition described in the previous post, then you can write an inequality for the question, and then simplify it.
Say there are "x" number of classes at school X, and "y" number of classes at school Y. Then the average number of students per class is
(total # students) / (total # classes)
= (25x + 33y) / (x + y)
So the inequality in the question is...
Is (25x + 33y) / (x + y) < 29?
Is 25x + 33y < 29x + 29y ?
Is 4y < 4x ?
Is y < x ?

Hey! Look at that. Knowing that x > y is exactly the same as knowing that the answer to the question is "yes".

Ron,
This helps. Thank you! I love setting up of equations method, that is how my mind is used to..
Thank you!

[RonPurewal]

You're welcome -- but, when you study, you should try NOT to do too much of "what your mind is used to".

If you only practice the things with which you're already comfortable, then you aren't going to see much improvement on this exam. You should spend almost all of your review time on methods that you don't like as much.