## Saturday, September 25, 2010

### The beginning of reality.

Also done with Speed Math chapter 9, still going through chapter 10.

Spent some time in office doing questions from BTG's articles. Was fun! :) I need to get back in the groove now, still hoping to get done with GMAT by mid-November. Wish me luck guys, I neeeed it!

Happy Gmatting!

## Friday, September 24, 2010

### CP #17

**Question:**

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28

B) 46

C) 55

D) 63

E) 70

Answer:

Method 1 -The first step in this problem is to recognize that order does not matter. We must create a committee of four people, but we are not putting these people in any sort of order. Because order does not matter, we must use the combinations formula, which is n!/[k!(n-k)!], in order to determine the number of possible outcomes.

However, as is the case with most advanced combinations problems, the solution is not as simple as plugging a few numbers into the formula. Instead, we need to consider exactly what types of outcomes we are looking for.

In this case, we need to select Jane and 3 other people, none of whom are Thomas; Thomas and 3 other people, none of whom are Jane; or Jane, Thomas and 2 other people. Additionally, we need to note that this is an 'or' situation, which means that we need to add all of these possibilities together.

First, if we select Jane and three others, none of whom are Thomas, we must pick Jane plus 3 out of the remaining 6 people. Because we can pick any 3 out of the 6, we set n = 6 and k = 3. When we plug this into the formula it gives us 6!/(3!3!) = 20 (Quick review: 3! = 3 x 2 x 1, 6! = 6 x 5 x 4 x 3 x 2 x 1.)

Next we determine how many ways we can select three people, none of whom are Jane, to go along with Thomas. This is identical mathematically, thus we end up with 20 in this case too.

Lastly, we determine how many outcomes are possible in which we have Jane, Thomas and 2 others. We now select 2 out of 6, so n = 6 and k = 2. 6!/(4!2!) = 15.

All we need to do to get the answer is add up all of the possible outcomes. 20 + 20 + 15 = 55, which is the correct answer to this question

(C).

Method 2 -required combinations = total combinations - combinations excluding both Jane and Thomas.

Total combinations = 8c4 = 70

combinations excluding both Jane and Thomas = 6c4 = 15

Answer = 70-15 = 55

### CP #16

How many unique 12-letter "words" can be created using 8 A's and 4 B's such that no 2 B's are adjacent? (For example, the word "ABAABABAAABA" is permissible, but the word "AABBABABAAAA" is not permissible, since it has 2 adjacent B's)

Answer:

we have 8 A's and 4 B's and we need to figure out how many ARRANGEMENTS are possible such that no two B's are together.

First let us arrange 8 A's. As all are identical we have just one way to arrange them.

Here my advice is to actually write 8 A's on your note pad. Now see my friend we have nine places where 4 B's can go. just count.

_ A _ A _ A _ A _ A _ A _ A _ A _

these 4 places can be chosen in 126 ways. apply combination formula ie 9C4.

Now as all 4 B's are identical we don't need to worry about arranging those 4 B's ( after placing them in four of the nine places.) So, here is your answer friend. i would have further counted for the permutations if the 4 B's were not identical. But as they are identical the answer is 126.

## Sunday, September 12, 2010

### Practice Test #3...or #1? Last one was in Dec '09.

Gave the free Kaplan test this morning on the net. Got a 610 (Q39 V37). Not paying too much attention to the score. Actually found the PS questions very easy, had forgotten how to solve some "types" of questions, which is why I got a few wrong. Not a big deal, will be able to solve those easily by the time the real test arrives. Finished the section with 5 minutes to spare - timing was an issue. There were 2-3 questions of the type where I know how to solve them, just couldn't recall during the test, so I spent too much time on those questions. Finally, I guessed on those questions (which ultimately meant a massive waste of time), and got tensed so I ended up doing the next few questions super fast. But, this test is not typical of the actual test at all. I'm sure of it. A million reasons why. More on that in some other post.

During Verbal, I had to guess on last 2 questions as I ran out of time. Was slow on the section as I was getting easily distracted with the extremely boring content on all the questions. What a lame excuse! :-p A more probable reason is the lack of stamina at the moment, or the fact that I just didn't want to give a test this morning.

Just putting in my answers in the error log now...later.

## Thursday, September 9, 2010

### Is Verbal my new weakness?!

Am also done with putting in all my Diagnostic test answers in the error log. Difficult task to match all the question categories between different error logs available on the net. Need to work on it, so that I can accurately tell which areas are my weak points. At the moment, Strengthen & Weaken category of CR questions appears to be my weakest link. More on this later...

Now for some non-GMAT, non-MBA, non-work related stuff:

My boyfriend came back to town yesterday after a 9 day trip abroad. :-) That's probably how I've got so much studying done along with work! :-p

Saw a gorgeous top at Zara (Rs 3600) and an amazing dress at Forever New (Rs 5000+)...and I want them both! I know, I know, expensive...but...will probably get the Zara top this weeked...and the dress, won't really have any place to wear it to...so probably won't. But it's such a pretty dress! And it fits me sooo well! *sigh*

## Saturday, September 4, 2010

### Where's my Excellent?!

## Thursday, September 2, 2010

### Manhattan GMAT SC Guide : You're going down.

Done with Chapters 6, 7 & 8. Made good use of the holiday!