Archive : Quarter Wit, Quarter Wisdom

Last week I left you with a conditional probability question. Let’s look at its solution now. This will be my last post on GMAT Combinatorics and Probability (for a while at least) until and unless you want me to take up a particular concept/question related to this topic. Next week, we will start a new topic.

Back to question at hand:

Question 2: Alex has five children. He has at least two girls (you do not know which two of her five children are girls). What is the probability that he has at least two boys too? (The probability of having a boy is 0.4 while the probability of having a girl is 0.6)

Let’s look at the concept of conditional probability in detail today. (As if the probability questions weren’t tricky enough!) But since I like to discuss advanced concepts in this blog (in addition to alternative approaches and very important fundamentals), it would not be fair on my part to end the probability discussion without a quick review of conditional probability. Let me start by tossing a question at you.

Question 1: Alex tosses a coin four times. On two of the tosses (we don’t know which two), he gets ‘Heads’. What is the probability that he gets ‘Tails’ on other two tosses?

I would like to take up a couple of questions on binomial probability today. The concepts of the topic have been covered in detail in the book so I am assuming that you know how to solve questions such as “What is the probability of getting at least 3 heads on 5 tosses of a coin?” etc. Therefore, let’s work on a couple of questions which use the binomial probability with a twist.

Question 1: Martin and Joey are playing a coin game in which each player tosses a fair coin alternately. The player who gets a ‘Heads’ first wins. The maximum number of tosses allowed in a single game for any player is 6. What is the probability that the person who tosses first will win the game?

Solution:

Let’s take another tricky probability question today and employ two different methods to solve it.

Question: Two couples and one single person occupy a row of five chairs at random. What is the probability that neither couple sits together (the husband and the wife should not occupy adjacent seats)?

(A) 1/5
(B) 1/3
(C) 3/8
(D) 2/5
(E) 1/2

Today, as requested by Pratap, we are going to take removal/replacement in mixtures. For those of you who were looking forward to some more tricky probability questions, I will make up for your disappointment next week. Meanwhile, rest assured, replacement is a very interesting, not to mention useful, concept in GMAT. So brace yourself to learn some new things today.

First of all, many “replacement” questions are nothing but the plain old mixture questions, the type we discussed in this post, with an extra step. So don’t flip out the moment you read the word “replace.” Let me show you what I mean:

I hope you have gone through the theory of probability from your book. I will not replicate that theory here but will assume that you already know it. Instead, what we will do now is take some tricky questions on probability and try and find out the various ways in which they can be solved. Hope they give you ideas and takeaways for other questions too!

Question: At the shooting range, the probability that Robert will hit the target in any one shot is 25%. If he takes four shots one after another, what is the probability that he will hit the target?

Now that we have laid the groundwork for permutations and combinations, probability will be a piece of cake. We just need to build up on what we have already learned.

The single most important concept in probability is the following:

The probability of an event A is calculated as P(A) = No. of outcomes when A occurs/Total no. of outcomes.

In this post, we will just extend the combinatorics concepts and apply them to probability. Let me explain how we will do it using some examples.

If you have been following my last few posts, I am sure you are a little wary of today’s post. They have been a little convoluted lately since we are dealing with permutations and combinations. Next, we will tackle probability but today, I am going to digress (to give you some much needed respite) and take up a simple yet interesting topic. We deal with quadratic equations on a regular basis. Tackling them effectively is pretty much one of the most basic and important skills you need for GMAT Quant. Today we will look at some relationships between the coefficients of quadratic equations and roots.

Another popular combinatorics concept deals with letters and envelopes. Let’s look at it today in some detail.

Question 1: Robin wrote 3 different letters to send to 3 different addresses. For each letter, she prepared one envelope with its correct address. If the 3 letters are to be put into the 3 envelopes at random, in how many ways can she put

(i) all three letters into the envelopes correctly?
(ii) only two letters into the envelopes correctly?
(iii) only one letter into the envelope correctly?
(iv) no letter into the envelope correctly?

Today’s post is a continuation of last week’s post and heavily refers back to it. I would suggest you to take a quick look at last week’s post again to make sense of this post. Let’s start with the variation question 1a we saw in the last post.

Today, using some examples, let’s look at different ways of distributing identical/distinct objects among people or in groups. There are some formulas which can be used in some of these cases but I will only discuss how to use the concepts we have learned so far to deal with these questions. I am not a fan of unintuitive formulas since the probability (we will come to this topic soon) that we will get to use even one of them in GMAT is quite low while the effort involved in cramming all of them is humongous. Therefore, I only want to focus on our core concepts which we can apply in various situations. Let’s start with our first example.

Question 1: In how many ways can 5 different fruits be distributed among four children? (Some children may get more than one fruit and some may get no fruits.)

Now that we have discussed both permutations and combinations independently, it’s time to look at questions that involve both. Mind you, these questions are not difficult -– they just involve both concepts. The first one is a circular arrangement question with a tiny twist. The second one requires us to make some cases. It takes a fair bit of patience to work out one case at a time and I doubt that GMAT will give you such a question since it is a little bit of a bore. (Actual GMAT questions have more entertainment value for the test maker and the test taker. They make you think and are FUN to solve) That said, it is a great question to bind together everything that we have learned till now and strengthen your understanding. Let’s start.

Let’s continue our discussion on combinations today. From the previous posts, we understand that combination is nothing but “selection.” Today we will discuss a concept that confuses a lot of people. It is similar to making committees (that we saw last week), but with a difference. Read the two questions given below:

Question 1: In how many ways can one divide 12 different chocolate bars equally among four boys?

Question 2: In how many ways can one divide 12 different chocolate bars into four stacks of 3 bars each?

Last week, we discussed the basics of combinations. Until and unless you have worked a decent bit with combinatorics in high school, the formula of combinations will not be very intuitive. We have already discussed how you can easily think of “selection” in terms of basic counting principle and un-arranging instead of the formula, if you so desire. Today, I would like to discuss some combination questions with constraints. A very common type of such questions asks you to make a committee of r people out of n people under some constraints. Let me show you what I mean with the help of some examples.

Question 1: If a committee of 3 people is to be selected from among 6 married couples such that the committee does not include two people who are married to each other, how many such committees are possible?

We will start with Combinations today. The moment we start talking about Permutations and Combinations, the first question many people ask is: “How do I know whether the given problem is a combinations problem or a permutations problem?”

My answer is: “Focus on what you have to do. Do you have to just SELECT some friends/toys/candies/candidates etc or do you have to ARRANGE them in distinct seats/among some children/in distinct positions etc too. If you have to only select, it is a combinations problem; if you have to only arrange, it is a permutations problem; if you have to first select and then arrange, it is a combinations and permutations problem. But if you are not using the formulas (nPr and nCr), you don’t have to think in terms of permutations and combinations. Just think in terms of selecting and arranging.” In the discussion below, I will start with an explanation of how we can make selections and how we can work on combinations without using the formula. We will also take a quick look at the formula and why it is what it is. Then we will move on to some examples.

With today’s post, let’s wrap up arrangements for the time being. We will discuss some complex circular arrangement constraints (which we will easily work through) today and start with combinations (i.e. picking “r” units out of “n” units) next week. Thereafter we will look at questions involving both, picking and arranging (yeah, that will be fun!).

Question 1: A group of 8 friends sit together in a circle. If A refuses to sit beside B unless C sits on the other side of A as well, how many possible seating arrangements are possible?

Solution: Let’s start with what we know. We know that the total number of ways in which 8 people can be arranged around a circular table is (8-1)! = 7!

In the last two posts, we discussed how to easily handle constraints in linear arrangements. Today we will discuss how to handle constraints in circular arrangements, which are actually even simpler to sort out. Let’s look at some examples.

Question 1: Seven people are to be seated at a round table. Andy and Bob don’t want to sit next to each other. How many seating arrangements are possible?

Solution: There are 7 people who need to be seated around a circular table. Number of arrangements in which 7 people can be seated around a circular table = (7-1)! = 6!

Today, I want to discuss the symmetry principle of linear arrangements with you. If you do not understand the symmetry principle, then it is possible that the following has happened with you:

You see a hard question and start working on it. You know that there are going to be three-four different cases. You find the number of arrangements in each case. Then, very carefully, you add them all up and get your answer. You check the answer key and behold, your answer is correct. Just for the fun of it, you turn your page to the solutions section and see that there are just two lines there which go something like this: “You can arrange 6 people in 6! ways. In half of these 6! ways, A will be ahead of B so answer is 6!/2.” and you end up feeling pretty unhappy even though you got the correct answer!

Let me first give you the solution to the question I gave you in my last post:

Question: In how many different ways (relative to each other) can 8 friends sit around a square table with 2 seats on each side of the table?

Solution: What happens when the first friend enters the room? Are all the seats same for him?

Let’s start this post with a question: In how many different ways (relative to each other) can 5 friends sit around a round table if all the seats are identical?

I guess that most of you will be able to answer it –> 4! = 24 ways

After all, you know the formula of circular arrangement which is (n-1)! But, many of you probably do not understand exactly why the formula is (n-1)! Today’s post is focused on explaining the concept of circular arrangement. If you are wondering why you need to know the theory behind the formula when all you need to do in a question is apply the formula and get the answer, here is why — you will be able to solve a straight forward 500-600 level question knowing just the formula but you will not get the 700+ level GMAT question correct. You need to understand the basics behind the formula so that you can apply it with modifications in more inventive situations. I will give you a couple of questions after discussing the theory and you will see what I mean. Right now, let’s focus on the question posed above.

After much deliberation, I have decided to start with Combinatorics and Probability this week. Why after much deliberation? Because I know that once I get into it, it will be many weeks before I get out of it. Anyway, I think it’s time we touch upon some important concepts of this vast topic. Mind you, I will stick to the GMAT-relevant sections so if you have any out-of-GMAT-scope intellectual questions, send them to me on my mail id (kbansal@veritasprep.com) and we can discuss those on the side.

The first thing I want to discuss is something we call “Basic Counting Principle” because it is useful in almost all 600-700 level questions of Combinatorics (Note here that I will avoid using the terms “Permutation” and “Combination” and the formulas associated with them since they are not necessary and make people uncomfortable). Also, many of the 700+ level questions use basic counting principle as the starting point so it’s not possible to start a discussion on combinatorics without discussing this principle first. Let’s try to understand it using an example.

Today we will continue from where we left our last post. In the last post, we discussed that of any two consecutive integers, one and only one of them will be even. Out of 20 and 21, only 20 is even. Since 2 is a factor of 20, it will not be a factor of 21. Does that make sense? Sure. Every second number will have 2 as a factor.

On the same lines, can both the consecutive numbers have 3 as a factor? Let’s take the same example – 20 and 21. 3 is a factor of 21. Can it be a factor of 20 as well? Do we even need to check? Since 21 is a multiple of 3, the previous multiple must be 3 places before (i.e., 18) and the next multiple of 3 must be 3 places ahead (i.e., 24).

We say goodbye to GMAT coordinate geometry for a while now and start today’s post with some magic tricks:

Trick 1

Step 1: Pick any two consecutive integers (Don’t tell me what they are!).

Step 2: Multiply them.

Whatever your numbers, the product is an even number! If you are wondering how I knew that, you desperately need to read this post. If you are shaking your head in disappointment, wait, I have more.

The key to doing well on GMAT Quant is to pick your standard questions, understand them really well and then try out variations of these questions. A small change in the question might force you to rethink your entire approach. The more you experiment, the more interesting your GMAT preparation will get and not to forget, the stronger your Quant will be.

Let me show you what I mean with the help of an example. Since we have been doing co-ordinate geometry for the past few weeks, let’s look at an interesting Official Guide question on coordinate geometry. Thereafter, we will try and figure out some variations of the same. We will focus more on the variations and how to think dynamically to arrive at solutions in those cases.

Continuing on our quest to master coordinate geometry, let’s look at a couple of Data Sufficiency questions today. The first question uses a great little concept. The second question has a very important takeaway that we all know but hardly ever implement.

Question 1: In the xy-coordinate plane, line l and line k intersect at the point (5, 4). Is the product of their slopes negative?

1) The product of the x-intercepts of lines l and k is positive.
2) The product of the y-intercepts of lines l and k is negative.

This week, we will take up some questions on co-ordinate geometry. Let me re-cap the relations we discussed in the last post.

Say, the equations of 2 lines are:

ax + by + c = 0

and

mx + ny + p = 0

A topic that has been steadily gaining ground in GMAT is co-ordinate geometry. First of all, I have to admit that I am not a fan of Geometry. Just something about learning the theorems and applying those to get the unknown angle/side makes me uncomfortable. It is easy to miss the big picture in some questions. That said, I adore co-ordinate geometry. I know that the moment I draw the diagram, the answer would be right there in front of my eyes. So let’s start a discussion on co-ordinate geometry this week.

Usually, GMAT deals with two dimensional figures in the XY plane. Many questions are based on points and intersecting lines. The general form of the equation of a line is ax + by = c. If you are not sure of how to draw a line given its equation, check out this post. Using an example, let’s see how this can be helpful.

This week, let’s look at some more properties of exponents and roots. Using a high level data sufficiency question, we will see how a number x is related to √x and to x^3.

Question: Is x > y?

Statement 1: √x > y
Statement 2: x^3 > y

It is one of those gorgeous questions that seem very simple at first but surprise you later.

I guess the posts on Exponents and Roots have been lifeless for the most part. If I thought they were a drag to write (I hate rules!) , I am sure you thought they were a torture to read. Nevertheless, if you find yourself wondering what to do when you see (3^2 * 9^4), you need to go through them (can’t live without the rules either!) The good news is that now that we are done with the basics, we can go on to the more “fun” concepts. And the best way to learn fun concepts is through fun questions. So let’s take a couple of problems (found them on a GMAT forum) dealing with comparisons of exponents/roots.

Let’s look at roots today. This post is chock full of mathematical symbols, so we’ve posted this week’s installment as a PDF, accessible via the link below. Click on it to read today’s post:

Roots on the GMAT

We’ve also posted an excerpt here, in this post. Enjoy!

First of all, understand that roots are a subset of exponents i.e. any number with any root can be converted to an exponent. Every rule of exponent that we have learned until now will then be applicable.

I thought I will take up “Roots” today but the rules of exponents and roots can get pretty monotonous. So let’s take a break today and do something more interesting.  I keep telling my students that GMAT questions do not involve long calculations. If you find yourself dividing a three digit number with a two digit number, it means you have missed the point of the question. Today, I will depict what I mean with the help of a couple of examples. I start with one my favorite questions from the Veritas book.

Question 1: A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant?

As promised, in this post, I will discuss the questions I gave you in the last post. Let’s apply the rules of exponents that we have learned. I will recap all the rules first and then we will proceed to the questions.

Rule 1: a^{^m} × a^ⁿ = a^{^(m + n)}

Rule 2: a^{^m} / a^ⁿ = a^{^(m – n)}

Rule 3: (a^^m)^ⁿ = a^^mn

Rule 4: For any number a, a^{^0} = 1

Let me recap the rules we discussed in the last post:

Rule 1: a^{^m} × a^ⁿ = a^(^{m + n)}

Rule 2: a^{^m} / a^ⁿ = a^(^{m – n)}

Rule 3: (a^{^m})^ⁿ = a^^mn

Rule 4: For any number a, a^⁰ = 1

No matter what score you are aiming for, knowing how to deal with exponents is crucial. Generally, questions based solely on exponents lie in the range of 600-650 and are one of the favorite topics of GMAT (but that’s just my opinion). It is also a topic which is very easy once you get the ‘hang of it’ but generates instant dislike until you don’t. I am sure many of you hate the sordid looking expressions/equations such as

(2^^a × 4 × 3^^-4 × 3^{^b} )/(3^{^4} × 2^{^2}) = 8^^-4 × 729

But of course, I don’t blame you.  Sadly, exponents and roots are basics that we should be experts in – whether we are working on Number Systems, Algebra or even Geometry! So let’s roll!

Last week, we started discussing some number properties. Let’s continue that discussion and dive into some more of those. In my opinion, it is the single most important topic on GMAT and one in which the smartest people slip easily. Think of this as a relatively easy way to earn another (or save) 20 or 30 points on your total GMAT score!

Let me show you the concept we will discuss today right away:

QUESTION: If 2^k is a factor of (10!), what is the greatest possible value of k?

Continuing the discussion of some stand alone topics, let’s discuss an important number property today. It is not only useful to know for GMAT but, if understood well, will also help you a lot during your MBA,  especially if you are keen on subjects in Finance since these subjects use a lot of ratios — e.g., Financial Leverage, P/E etc. You will often come across a situation where you will need to compare ratios. Say, you have a given ratio N/D. Now, a number ‘A’ is subtracted from both N and D. Is the new ratio (N – A)/(D – A) greater than or less than N/D? The answer depends on the original value of N/D.

The main concept is as follows:

A couple of days back, during a class, we got into a discussion on “If you have ‘n’ variables, you need at least ‘n’ equations to solve for each variable.” It centered on the cases where you cannot apply this rule of thumb. Let me discuss a couple of those cases in detail today:

1. A question such as this:

Question: What is the value of (x + y + z)?

Okay, let’s move away from “Divisibility and Remainders” on the GMAT, at least temporarily, and let’s focus our attention on another topic. If you have read some of my previous posts, I guess you know that I like to solve questions “logically.” I like to avoid making equations. Instead, I try to make myself “figure it out.”

A few days back, I came across a Time-Speed-Distance problem which was a perfect example of how you could “figure stuff out” without dealing with any equations. Actually, you can do that with a majority of GMAT questions (and save yourself loads of time!) but what was special about this question was that a couple of instructors of one of our competitor (I am not at liberty to disclose exactly who this competitor is!) had told their students that it is not possible to solve it logically. That got me thinking that perhaps, logical thinking is not as widely utilized as we at Veritas would like to believe. (I think our love for logical thinking is also apparent in the way we teach Sentence Correction!) Anyway, I thought of sharing the question and its logical solution with you! Here goes…

In this post, I would like to focus on a particular type of remainder questions and how to solve them in a particular way. For the type of questions I am going to discuss today, I like to use “Binomial Theorem.” You might be tempted to run away right now and save yourself some precious time if you are not a Math geek but wait! We will just use an application of Binomial which I will explain in as simple a language as I can think of. I am quite certain that you will be comfortable with the method if you just give it a chance.

For the past few weeks, we have been focusing on Divisibility and Remainders. There are some more ‘types’ of remainder questions. Let’s take them one by one so that by the end of it all, you are an expert in everything related to remainders. In this post I will start with a question similar to what you mind find in the Official Guide for GMAT Review.

Question: If a and b are positive integers such that a/b = 97.16, which of the following cannot be the remainder when a is divided by b?