Archive : Quarter Wit, Quarter Wisdom

As promised, we tackle some 700+ level questions today. Mind you, these questions are not your typical GMAT type questions. The reason we are discussing them is that they look mind boggling but are easily workable when the concepts of relative speed are used. They give insights that help you understand relative speed. Once you are good with the concepts, you can solve most of the relative speed based questions easily.

We discussed the concepts of relative speed in GMAT questions last week. This week, we will work on using those concepts to solve questions. The questions we take today will be 600-700 level. I intend to take the 700+ level questions next week (don’t want to scare you away just yet!). Let’s get going now.

Let’s look at the concept of relative speed today. A good understanding of relative speed can be very useful in some questions. If you don’t use relative speed in these GMAT questions, you can still solve them but they would be rather painful to work through (they might need multiple variables and you know my policy on variables – use one, if you must)

Today, let’s look at a question that involves inequalities and modulus and is best understood using the concept of sets. It is not a difficult question but it is still very tricky. You could easily get it right the first time around but if you get it wrong, it could take someone many trials before he/she is able to convince you of the right answer. Even after I write a whole post on it, I wouldn’t be surprised if I see “but I still don’t get it” in the comments below!

Now that we have covered some variations that arise in inequalities in GMAT problems, let’s look at some questions to consolidate the learning.

We will first take up a relatively easy OG question and then a relatively tougher question which looks harder than it is because of the use of mods in the options (even though, we don’t really need to deal with the mods at all).

Last week, we discussed a couple of variations of inequality questions with many factors. Let’s now look at some more complications that we should know how to handle.

Last week we learned how to handle inequalities with many factors i.e. inequalities of the form (x – a)(x – b)(x – c)(x – d) > 0. This week, let’s see what happens in cases where the inequality is not of this form but can be manipulated and converted to this form. We will look at how to handle various complications.

Students often wonder why ‘x(x-3) < 0’ doesn’t imply ‘x < 0 or (x – 3) < 0’. In this post, we will discuss why and we will see what it actually implies. Also, we will look at how we can handle such questions quickly.

When you see ‘< 0’ or ‘> 0’, read it as ‘negative’ or ‘positive’ respectively. It will help you think clearly.

So the question we are considering today is:

Last week we promised you a couple of tricky standard deviation (SD) GMAT questions. We start with a 600-700 level question and then look at a 700 – 800 level one.

Question 1: During an experiment, some water was removed from each of the 8 water tanks. If the standard deviation of the volumes of water in the tanks at the beginning of the experiment was 20 gallons, what was the standard deviation of the volumes of water in the tanks at the end of the experiment?

Statement 1: For each tank, 40% of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.

Statement 2: The average volume of water in the tanks at the end of the experiment was 80 gallons.

This week, we pick from where we left last week. Let’s discuss the last 3 cases first.

Question: Which set, S or T, has higher SD?

Case 5: S = {1, 3, 5} or T = {1, 3, 3, 5}

The standard deviation (SD) of T will be less than the SD of S. Why? The mean of 1, 3 and 5 is 3. If you add another 3 to the list, the mean stays the same and the sum of the squared deviations is also the same but the number of elements increases. Hence, the SD decreases.

We will work our way through the concepts of Standard Deviation (SD) today. Let’s take a look at how you calculate standard deviation first:



Ai – The numbers in the list

Aavg – Arithmetic mean of the list

n – Number of numbers in the list

Let’s discuss the idea of “range” today. It is simply the difference between the smallest and the greatest number in a set. Consider the following examples:

Range of {2, 6, 10, 25, 50} is 50 – 2 = 48

Range of {-20, 100, 80, 30, 600} is 600 – (-20) = 620

and so on…

As promised, we discuss medians today! Conceptually, the median is very simple. It is just the middle number. Arrange all the numbers in increasing/decreasing order and the number you get right in the middle, is the median. So it is quite straight forward when you have odd number of numbers since you have a “middle” number. What about the case when you have even number of numbers? In that case, it is just the average of the two middle numbers.

Median of [2, 5, 10] is 5

Median of [3, 78, 102, 500] is (78 + 102)/2 = 90

Last week we discussed arithmetic means of arithmetic progressions in GMAT math problems. Today, let’s see those concepts in action.

Question 1: If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?

Today’s post is again focused on arithmetic mean. Let’s start our discussion by considering the case of arithmetic mean of an arithmetic progression.

We will start with an example. What is the mean of 43, 44, 45, 46, 47? (Hint: If you are thinking about adding the numbers, that’s not the way I want you to go.)

As we discussed in our previous posts, arithmetic mean is the number that can represent/replace all the numbers of the sequence. Notice in this sequence, 44 is one less than 45 and 46 is one more than 45. So essentially, two 45s can replace both 44 and 46. Similarly, 43 is 2 less than 45 and 47 is 2 more than 45 so two 45s can replace both these numbers too.

I hope the theory of arithmetic mean we discussed last week is clear to you. Let’s see the theory in action today. I will pick some mean questions from various sources (Official Guide, GMAT prep tests, etc.) and we will try to use the concepts we learned last week to solve them.

Let’s start with a simple question.

Let’s start today with statistics – mean, median, mode, range and standard deviation. The topics are simple but the fun lies in the questions. Some questions on these topics can be extremely tricky especially those dealing with median, range and standard deviation. Anyway, we will tackle mean today.

So what do you mean by the arithmetic mean of some observations? I guess most of you will reply that it is the ‘Sum of Observations/Total number of observations’. But that is how you calculate mean. My question is ‘what is mean?’ Loosely, arithmetic mean is the number that represents all the observations. Say, if I know that the mean age of a group is 10, I would guess that the age of Robbie, who is a part of that group, is 10. Of course Robbie’s actual age could be anything but the best guess would be 10.

I have my CFA level II exam in June and I am certainly looking at disaster (but that’s not what I am going to discuss today). While studying for it yesterday, I did something really stupid and that gave me an insight on ‘mind matters’. That is what I want to talk about today but I will have to give you some background to make my point clearer.

Last week, at the end of the Geometric Progression (GP) post, I gave you a question to figure out. I hope some of you did try it. Today we will discuss the question in detail and look at two different approaches – one without using GP formula (we discussed this approach in a previous post) and another with the formula. As I said before, you can solve every sequence question on GMAT without using the formulas we are discussing. We are still investing time in these formulas so that we can save some in the actual exam. Let me show you how.

Question 3: For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by (-1)^n*2^(-n). If N is the sum of the first 200 terms in the sequence, then N is:

Let’s look at geometric progressions (GP) now. Before I start, let me point out that GMAT is unlikely to give you a statement which looks like this: “If S denotes a geometric progression whose first term is… ” GMAT will not test your knowledge of GP (i.e. you don’t really need to learn the formulas of the sum of n terms of a GP or sum of infinite terms of a GP etc) though it may give you a sequence which is a geometric progression and ask you questions on it. You will be able to solve the question without using the formulas but recognizing a GP can help you deal with such questions in an efficient manner. That is the reason we are discussing GPs today.

For those of you who are wondering what exactly a GP is, let me begin by giving you the definition.

Last week, we looked at some basic formulas related to Arithmetic Progressions. This week, we will look at a particular (and related) type of Arithmetic Progression — Consecutive Integers.

Look at the following three sequences:

S1 = 3, 4, 5, 6, 7

S2 = -1, 0, 1, 2, 3, 4, 5, 6, 7

S3 = 1, 2, 3, 4, 5, 6, 7, 8, 9

All of them are APs of consecutive integers so every formula we looked at last week is applicable here.

Today’s topic is Arithmetic Progression (AP). An AP is a sequence of numbers such that the difference between the consecutive terms is constant.

For example:

2, 4, 6, 8…

-1, 10, 21, 32…

4, 3, 2, 1, 0, -1, -2, -3

-1/2, -3/2, -5/2, -7/2…

and so on.

Today, I will take the question I gave you in the last post (and that is all we will tackle today!) It is a question from GMAT prep test so it is quite indicative of the tricky questions you might get on actual GMAT. Solving the question takes less than two minutes since the calculations required are negligible. However, if you start calculating the actual value, you could end up spending many painful minutes before giving up. I implore you to always remember that GMAT does not give you calculation intensive questions. Since they do not provide you with an HP12C, there will always be a logical solution -– you will just need to think a little harder! Let’s get going then…

Question 3: For every integer m from 1 to 10 inclusive, the mth term of a certain sequence is given by [(-1)^(m+1)]*[(1/2)^m]. If T is the sum of the first 10 terms in the sequence, then T is

Today, let’s dig into sequences on the GMAT. Let’s first understand what a sequence is (from Wikipedia):

A sequence is an ordered list of objects. The number of terms it contains (possibly infinite) is called the length of the sequence. Unlike a set, order matters in a sequence, and exactly the same elements can appear multiple times at different positions in the sequence. Since order matters, (A, B, C) and (B, C, A) are two different sequences. (A series is the sum of the terms of a sequence but we will not deal with series today.)

There are some special sequences e.g. arithmetic progressions and geometric progressions. We will deal with these in subsequent weeks. Today we will look at some generic sequence questions and will learn how to approach them. I will start with a very basic question. Mind you, most sequence questions will be higher level questions since sequence questions look complicated (even though they are very straight forward, believe me!). Let me show you using some questions from external sources:

Some of the trickiest questions in GMAT are based on positive/negative bases and powers. Today, let’s look at some of their properties. First thing you must understand is that if the base is positive, it will stay positive no matter what the power . a^n is equal to a*a*a* … (n times). Since only positive numbers are multiplied with each other, the product will always be positive. (We cannot say the same thing about negative bases but let’s ignore them in today’s post.)

For example, 5^n must be positive no matter what the value of n.
The base ‘a’, which is positive, can belong to one of the two ranges – ‘Greater than 1’ or ‘Between 0 and 1’ (or it can be equal to 1). Let’s see what happens to a^n is each case.

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.