Quarter Wit, Quarter Wisdom: Pattern Recognition – Part II

Quarter Wit, Quarter Wisdom
Today’s post comes from Karishma, a Veritas Prep GMAT instructor. Before reading, be sure to check out Part I from last week!

Last week we saw how to use pattern recognition. Today, let’s take up another question in which this concept will help us. Mind you, there are various ways of solving a question. Most questions we solve using pattern recognition can be solved using another method. But pattern recognition is a method we can use in various cases. It is something that comes to our aid when we forget everything else. If you don’t know from where to start on a question, try to give some values to the variables. You might see a pattern. You may not ‘know’ something. Even then, you can ‘figure out’ the answer because GMAT is not a test of your knowledge; it is a test of your wits. It is a test of whether you can keep your cool when faced with the unknown and use whatever you know to solve the question.

Let’s look at a question now.
Continue reading “Quarter Wit, Quarter Wisdom: Pattern Recognition – Part II”

Quarter Wit, Quarter Wisdom: Pattern Recognition

Quarter Wit, Quarter WisdomIf you are hoping for a 700+ in GMAT, you need to develop the ability to recognize patterns. GMAT does not test advanced concepts but you can certainly get advanced questions on simple concepts. For such questions, the ability to quickly observe patterns can come in quite handy. We will discuss a complicated question today which can be easily solved by observing the pattern.

Question: If a and b are distinct integers and a^b = b^a, how many solutions does the ordered pair (a, b) have? Continue reading “Quarter Wit, Quarter Wisdom: Pattern Recognition”

Quarter Wit Quarter Wisdom: GCF and LCM of Fractions

Quarter Wit, Quarter WisdomLast week we discussed some concepts of GCF. Today we will talk about GCF and LCM of fractions.

LCM of two or more fractions is given by: LCM of numerators/GCF of denominators

GCF of two or more fractions is given by: GCF of numerators/LCM of denominators

Why do we calculate LCM and GCF of fractions in this way? Let’s look at the algebraic explanation first. Then we will look at a more intuitive reason.

Algebraic Approach:

Consider 2 fractions a/b and c/d in their lowest form, their LCM is Ln/Ld and GCF is Gn/Gd, also in their lowest forms.
Let’s work on figuring out the LCM first.

LCM is a multiple of both the numbers so Ln/Ld must be divisible by a/b. This implies (Ln/Ld)/(a/b) is an integer. We can re-write this as:

Ln*b/Ld*a is an integer.

Since a/b and Ln/Ld are in their lowest forms, Ln must be divisible by a; also, b must be divisible by Ld (because a and b have no common factors and Ln and Ld have no common factors).

Using the same logic, Ln must be divisible by c; also d must be divisible by Ld.

Ln, the numerator of LCM, must be divisible by both a and c and hence should be the LCM of a and c, the numerators. Ln cannot be just any multiple of a and c; it must be the lowest common multiple so that Ln/Ld is the lowest multiple of the two fractions.

b and d both must be divisible by Ld, the denominator of LCM, and hence Ld must be their highest common factor. Mind you, it cannot be just any common factor; it needs to be the highest common factor so that Ln/Ld is the lowest multiple possible.

This is why LCM of two or more fractions is given by: LCM of numerators/GCF of denominators.

Using similar reasoning, you can figure out why we find GCF of fractions the way we do.

Now let me give you some feelers. They are more important than the algebraic explanation above. They build intuition.

Intuitive Approach:

Let me remind you first that LCM is the lowest common multiple. It is that smallest number which is a multiple of both the given numbers.

Say, we have two fractions: 1/4 and 1/2. What is their LCM? It’s 1/2 because 1/2 is the smallest fraction which is a multiple of both 1/2 and 1/4. It will be easier to understand in this way:

1/2 = 2/4. (Fractions with the same denominator are comparable.)

LCM of 2/4 and 1/4 will obviously be 2/4.

If this is still tricky to see, think about their equivalents in decimal form:

1/2 = 0.50 and 1/4 = 0.25. You can see that 0.50 is the lowest common multiple they have.

Let’s look at GCF now.

What is GCF of two fractions? It is that greatest factor which is common between the two fractions. Again, let’s take 1/2 and 1/4. What is the greatest common factor between them?

Think of the numbers as 2/4 and 1/4. The greatest common factor between them is 1/4.

(Note that 1/2 and 1/4 are both divisible by other factors too e.g. 1/8, 1/24 etc but 1/4 is the greatest such common factor)

Now think, what will be the LCM of 2/3 and 1/8?

We know that 2/3 = 16/24 and 1/8 = 3/24.
LCM = 16*3/24 = 48/24 = 2

LCM is a fraction greater than both the fractions or equal to one or both of them (when both fractions are equal). When you take the LCM of the numerator and GCF of the denominator, you are making a fraction greater than (or equal to) the numbers.

Also, what will be the GCF of 2/3 and 1/8?

We know that 2/3 = 16/24 and 1/8 = 3/24.

GCF = 1/24

GCF is a fraction smaller than both the fractions or equal to one or both of them (when both fractions are equal). When you take the GCF of the numerator and LCM of the denominator, you are making a fraction smaller than (or equal to) the numbers.

We hope the concept of GCF and LCM of fractions makes sense to you now.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Some GCF Concepts

Quarter Wit, Quarter WisdomSometimes students come up looking for explanations of concepts they come across in books. Actually, in Quant, you can establish innumerable inferences from the theory of any topic. The point is that you should be comfortable with the theory. You should be able to deduce your own inferences from your understanding of the topic. If you come across some so-called rules, you should be able to say why they hold. Let’s discuss a couple of such rules from number properties regarding GCF (greatest common factor). Many of you might read them for the first time. Stop and think why they must hold.

Rule 1: Consecutive multiples of ‘x’ have a GCF of ‘x’

Explanation: What do we mean by consecutive multiples of x? They are the consecutive terms in the multiplication table of x. For example, 4x and 5x are consecutive multiples of x. So are 18x and 19x…

What will be the greatest common factor of 18x and 19x? We know that x is their common factor. Do 18 and 19 have any common factors (except 1)? No. So greatest common factor will be x. Take any two consecutive numbers. They will have no common factors except 1. Hence, if we have two consecutive factors of x, their GCF will always be x.

For more on common factors of consecutive numbers, check:
http://www.veritasprep.com/blog/2011/09 … c-or-math/
http://www.veritasprep.com/blog/2011/09 … h-part-ii/

Can you derive some of your own ‘rules’ based on this now? Let’s give you some ideas:

Two consecutive integers have GCF of 1.

Two consecutive odd multiples of x have GCF of x.

Rule 2: The G.C.F of two distinct numbers cannot be larger than the difference between the two numbers.

Explanation: GCF is a factor of both the numbers. Say, the GCF of two distinct numbers is x. This means the two numbers are mx and nx where m and n have no common factor. What can be the smallest difference between m and n? m and n cannot be equal since the numbers are distinct. The smallest difference between them can be 1 i.e. they can be consecutive numbers. In that case, the difference between mx and nx will be x which is equal to the GCF. If m and n are not consecutive integers, the difference between them will be much larger than x. The difference between mx and nx cannot be less than x.
Say, GCF of two numbers is 6. The numbers can be 12 and 18 (GCF = 6) or 12 and 30 (GCF = 6) etc but they cannot be 12 and 16 since both numbers must have 6 as a factor. So after a multiple of 6, the other multiple of 6 must be at least 6 away.

Let’s look at a third party question based on these concepts now.

Question 1: What is the greatest common factor of x and y?

Statement 1: Both x and y are divisible by 4.
Statement 2: x – y = 4

Solution:

Statement 1: Both x and y are divisible by 4

We know that 4 is a factor of both x and y. But is it the highest common factor? We do not know. There could be another factor common between x and y and hence highest common factor could be greater than 4. e.g. 4 and 16 have 4 as the highest common factor but 12 and 36 have 12 as the highest common factor though both pairs have 4 as a common factor.

Statement 2: x – y = 4

We know that x and y differ by 4. So their GCF cannot be greater than 4 (as discussed above). The GCF could be any of 1/2/4 e.g. 7 and 11 have GCF of 1 while 2 and 6 have GCF of 2.

Taking both statements together: From statement 1, we know that x and y have 4 as a common factor. From statement 2, we know that x and y have one of 1/2/4 as highest common factor. Hence 4 is the highest common factor.

Answer (C)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Have a Game Plan

Quarter Wit, Quarter WisdomFor the past few weeks, we have been discussing conditional statements. Let’s switch back to Quant today. I have been meaning to discuss a question for a while. We can easily solve it by plugging in the right values. The only issue is in figuring out the right values quickly. The point we are going to discuss is that there has to be a plan.

Question 1: Six countries in a certain region sent a total of 75 representatives to an international congress. No two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives?

Statement 1: One of the six countries sent 41 representatives to the congress.
Statement 2: Country A sent fewer than 12 representatives to the congress. Continue reading “Quarter Wit, Quarter Wisdom: Have a Game Plan”

Quarter Wit, Quarter Wisdom: Necessary Conditions

Quarter Wit, Quarter WisdomLast week we discussed a very tricky CR question based on conditional statements. This week, we would like to discuss another CR question based on necessary conditions. Note that you don’t need to be given ‘only if’ or ‘only when’ to mark a necessary condition. The wording of the statement could imply it. You need to keep a keen eye to figure out necessary and sufficient conditions.

Question: Successfully launching a new product for supermarket sale requires either that supermarkets give prominent shelf space to the product or that plenty of consumers who have not tried it seek it out. Consumers will seek out a new product only if it is extensively advertised, either on television or in the press. One way for a manufacturer to obtain prominent shelf space for a new product is to promote it in trade journals.
Continue reading “Quarter Wit, Quarter Wisdom: Necessary Conditions”

Quarter Wit, Quarter Wisdom: A Question on Conditional Statements

Quarter Wit, Quarter WisdomWe hope that you have understood conditional sentences we discussed in the last post. The concept is very important and you will come across questions using this concept often. Now, let’s discuss the GMAT question we gave you last week.

Question: A newborn kangaroo, or joey, is born after a short gestation period of only 39 days. At this stage, the joey’s hind limbs are not well developed, but its forelimbs are well developed, so that it can climb from the cloaca into its mother’s pouch for further development. The recent discovery that ancient marsupial lions were also born with only their forelimbs developed supports the hypothesis that newborn marsupial lions must also have needed to climb into their mothers’ pouches.
Continue reading “Quarter Wit, Quarter Wisdom: A Question on Conditional Statements”

Quarter Wit, Quarter Wisdom: Conditional Statements

Quarter Wit, Quarter WisdomLast week we discussed a Critical Reasoning question in detail. Today, I want to discuss a very important concept of CR — analyzing a conditional statement. You will often encounter these even though they may not be in the exact same format as the one we will discuss below. We will discuss the basic framework and then we will look at questions where this concept will be very helpful. Mind you, without this framework, it can get a little tricky to wrap your head around these questions.

Statement 1:

If you trouble your teacher, you will be punished.

What does this imply? It implies that ‘troubling your teacher’ is a sufficient condition to get punished. If you trouble, you will get punished.
Continue reading “Quarter Wit, Quarter Wisdom: Conditional Statements”

Quarter Wit, Quarter Wisdom: Evaluate the Conclusion

Quarter Wit, Quarter WisdomAt the risk of generalizing from a relatively small sample, let me say that people who are good at Quant, tend to be good at Critical Reasoning in Verbal. I certainly cannot comment about their SC and RC prowess but they are either good at CR right from the start or improve dramatically after just a couple of our sessions. The reason for this is very simple – CR is more like Quant than like Verbal. CR is very mathematical. You need to keep in mind some basic rules and based on those, you can easily crack the most difficult of problems. There is only one catch – don’t get distracted by options put there to distract you!

Today I would like to discuss a great CR question from our book. It upsets a lot of students even though it is simple – just like a GMAT question is supposed to be. Here is the question:
Continue reading “Quarter Wit, Quarter Wisdom: Evaluate the Conclusion”

Quarter Wit, Quarter Wisdom: Rates Revisited

Quarter Wit, Quarter WisdomPeople often complain about getting stuck in work-rate problems. Hence, I would like to take some 700+ level questions on rate today. I have discussed the basic concepts of work-rate (using ratios) in a previous post:

Cracking the Work Rate Problems

You  might want to go through that post before you set out to work on these problems. Ensure that you are very comfortable with the relation: Work = Rate*Time and its implications: If rate doubles, work done doubles too if the time remains constant; if one work is done, rate = 1/time etc. Thorough understanding of these implications is fundamental to ‘reasoning out’ the answer.
Continue reading “Quarter Wit, Quarter Wisdom: Rates Revisited”

Quarter Wit, Quarter Wisdom: Questions on Factorials

Quarter Wit, Quarter WisdomLast week we discussed factorials – how we can take something common when we have factorials in some equations. Today let’s discuss a couple of questions based on factorials. They look intimidating but they are pretty simple. Factorial is all about multiplication and hence there is a high probability that you will be able to take something common and cancel something. These techniques reduce our work significantly. Hence, seeing a factorial in a question should bring a smile to your face!

Question 1: Given that x, y and z are positive integers, is y!/x! an integer?

Statement 1: (x + y)(x – y) = z! + 1

Statement 2: x + y = 121
Continue reading “Quarter Wit, Quarter Wisdom: Questions on Factorials”

Quarter Wit, Quarter Wisdom: Managing Factorials in Equations

Quarter Wit, Quarter WisdomA concept we have not yet covered in this series is factorials (though we used some factorials in the post Power in Factorials). Let’s first discuss the basics of factorials. Once we do, we will see that most factorial expressions can be easily solved using a single method: taking common!

First of all, what is (n!)?

n! = 1*2*3*4*5*6*…*(n – 2)*(n -1)*n

Let’s take some examples:
Continue reading “Quarter Wit, Quarter Wisdom: Managing Factorials in Equations”

Quarter Wit, Quarter Wisdom: Successive Division

Quarter Wit, Quarter WisdomWe discussed divisibility and remainders many weeks ago. Today, we will use those concepts and discuss another type of question – successive division. But before we do, you need to go through the previous related posts on division if you haven’t read them already:

Divisibility Unraveled

Divisibility Applied on the GMAT

Divisibility Applied to Remainders
Continue reading “Quarter Wit, Quarter Wisdom: Successive Division”

Quarter Wit, Quarter Wisdom: Working on Getting the Full Picture Again!

Quarter Wit, Quarter WisdomToday, we will continue our discussion on why it is important to understand the workings behind seemingly miraculous shortcuts. We will use another example from probability.

Question 1: A bag contains 4 white balls, 2 black balls & 3 red balls. One by one three balls are drawn out with replacement (i.e. a ball is drawn and then put back. Thereafter, another ball is drawn). What is the probability that the third ball is red?

Solution:

The question is simple, isn’t it? When you draw balls with replacement, the probability stays the same at the beginning of every cycle. On first draw, the probability of drawing a red ball is 3/9 (since there are 3 red balls and 9 balls in all). When you draw the ball and put in back, the probability of drawing a red ball again stays the same i.e. 3/9 (since again there are 3 red balls and 9 balls in all). The situation at the beginning of every draw is the same.
Continue reading “Quarter Wit, Quarter Wisdom: Working on Getting the Full Picture Again!”

Quarter Wit, Quarter Wisdom: Get the Full Picture – Part II

Quarter Wit, Quarter WisdomLet’s refer back to last week’s post. We discussed why it is important to fully understand what you are doing and why you are doing it, especially while using an innovative method. We talked about it using a combinatorics example.

Let’s revisit it here:

Question 1: What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

We discussed the ways of obtaining various sums. The regular way of obtaining a sum of 8 is enumerating all the possibilities. An innovative way was using 7C2 (as discussed last week).
Continue reading “Quarter Wit, Quarter Wisdom: Get the Full Picture – Part II”

Quarter Wit, Quarter Wisdom: Get the Full Picture

Quarter Wit, Quarter WisdomToday we will discuss why ‘understanding’ rather than just ‘learning’ a concept is important. Most questions can be solved using different methods. Sometimes, a particular method seems really easy and quick and we tend to ‘learn’ it without actually knowing why we are doing what we are doing. We need to understand the strengths and the weaknesses of the method before we use it. Let me elaborate with an example.

Question: What is the probability that you will get a sum of 8 when you throw three dice simultaneously?

When you throw three dice simultaneously, you can obtain a total sum ranging from 3 (1 on each die) to 18 (6 on each die). Let’s consider all the possible sums that we can obtain:

Sum of 3: This can happen in only 1 way. Each die shows 1 in this case

Sum of 4: This can happen in 3 ways. One die shows 2 and the other two show 1 each. Any of the three dice could show 2 so there are a total of 3 ways of getting a sum of 4.

Sum of 5: There are different ways of obtaining a sum of 5:

1, 1, 3 – Any of the three dice could show 3 so there are a total of 3 ways of obtaining 5 in this way.

1, 2, 2 – Any of the three dice could show 1 so there are a total of 3 ways of obtaining 5 in this way.

Total number of ways of obtaining a sum of 5 = 3 + 3 = 6

Sum of 6: There are different ways of obtaining a sum of 6:

1, 1, 4 – Any of the three dice could show 4 so there are a total of 3 ways of obtaining 6 in this way.

1, 2, 3 – These 3 numbers could be arranged among the 3 dice in 3! ways (using basic counting principle). There are a total of 3! = 6 ways of obtaining 6 in this way.

2, 2, 2 – This can happen in only one way. All dice show 2.

Total number of ways of obtaining a sum of 6 = 3 + 6 + 1 = 10

Similarly, we can find the number of ways in which all other sums can be obtained. Below I will list the number of ways in each case.

Sum of 3: 1 way                          Sum of 18: 1 way

Sum of 4: 3 ways                        Sum of 17: 3 ways

Sum of 5: 6 ways                        Sum of 16: 6 ways

Sum of 6: 10 ways                     Sum of 15: 10 ways

Sum of 7: 15 ways                    Sum of 14: 15 ways

Sum of 8: 21 ways                    Sum of 13: 21 ways

Sum of 9: 25 ways                    Sum of 12: 25 ways

Sum of 10: 27 ways                 Sum of 11: 27 ways

Notice the symmetry here. There is only 1 way of obtaining 3 and only one way of obtaining 18. Similarly, there are 3 ways in which you can obtain 4 and 3 ways in which you can obtain 17. Is it a co-incidence? No. The second half of the column can be obtained by replacing 1 by 6, 2 by 5, 3 by 4, 4 by 3, 5 by 2 and 6 by 1. The number of ways of obtaining a sum is symmetrical about the center.

We have simplified one aspect of this problem. If we need to find the number of ways of obtaining 15, we can instead find the number of ways of obtaining a sum of 6 (which is psychologically easier to handle). Now the problem is whether there is an easier way of obtaining the number of ways in which you can get a sum of 6 or do we have to enumerate it in every case.

You might have come across something like this:

Sum of 3: 2C2 = 1 way

Sum of 4: 3C2 = 3 ways

Sum of 5: 4C2 = 6 ways

Sum of 6: 5C2 = 10 ways

Sum of 7: 6C2 = 15 ways

Sum of 8: 7C2 = 21 ways

Perfect till now! Matches the numbers we obtained above. It seems like a good method to use instead of writing down all the cases, doesn’t it? But what happens further on?

Sum of 9: 8C2 = 28 ways

Sum of 10: 9C2 = 36 ways

These don’t match! The key to understanding this is to understand why it works in the above given cases. First review this post.

Focus on method II of question 2. Notice how you divide n identical objects among m distinct groups. Let’s take the example of a sum of 7. You have to divide 7 among 3 dice such that each die must have at least 1 (no die face can show 0). First step is to take 3 out of the 7 and give one each to the three dice. Now you have 4 left to distribute among 3 distinct groups such that it is possible that some groups may get none of the four. Think of partitioning 4 in 3 groups. This can be done in (4+2)!/4!*2! = 6C2 ways (check out the given link if you do not understand this)

This is how you obtain 6C2 for the sum of 7.

The concept works perfectly till the sum of 8. Thereafter it fails. Think why. I will explain it next week.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Three Overlapping Sets

Today, let’s take a look at how to use Venn diagrams to solve questions involving three overlapping sets.

First, let me show you what the three overlapping sets diagram looks like.

Notice that the total comprises of the elements that do not fall in any of the three sets and the elements that are a part of at least one of the three sets.

The elements falling in the red, yellow or green region (region a, b or c) fall in only one set. The elements falling in region d, e or f fall in exactly two sets and the elements falling in region g fall in all three sets.

Now some quick questions to get a clear picture:

Question 1: Which regions represent the elements that belong to at least 2 sets?

Answer 1: d + e + f + g

Question 2: Which regions represent the elements that belong to at least 1 set?

Answer 2: a + b + c + d + e + f + g = Total – None

Question 3: Which regions represent the elements that belong to at most 2 sets?

Answer 3: None + a + b + c + d + e + f = Total – g

Hope there are no doubts up till now. Let’s look at a question to see how to apply these concepts.

Question: Three table runners have a combined area of 200 square inches. By overlapping the runners to cover 80% of a table of area 175 square inches, the area that is covered by exactly two layers of runner is 24 square inches. What is the area of the table that is covered with three layers of runner?

(A) 18 square inches
(B) 20 square inches
(C) 24 square inches
(D) 28 square inches
(E) 30 square inches

Solution: Let’s first try to understand what exactly is given to us. The area of all the runners is equal to 200 square inches.

Runner 1 + Runner 2 + Runner 3 = 200

In our diagram, this area is represented by
(a + d + g + e) + (b + d + g + f) + (c + e + g + f) = 200

(We need to find the value of g i.e. the area of the table that is covered with three layers of runner.)

Area of table covered is only 80% of 175 i.e. only 140 square inches. This means that if each section is counted only once, the total area covered is 140 square inches.
a + b + c + d + e + f + g = 140
So the overlapping regions are obtained by subtracting second equation from the first. We get d + e + f + 2g = 60
But d + e + f (area with exactly two layers of runner) = 24
So 2g = 60 – 24 = 36

g = 18 square inches

Note that you don’t need to make all these equations and can directly jump to d + e + f + 2g = 60. We wrote these equations down only for clarity. It is a matter of thinking vs solving. If we think more, we have to solve less. Let’s see how.

Combined area of runners is 200 square inches while area of table they cover is only 140 square inches. So what does the extra 60 square inches of runner do? It covers another runner!
Wherever there are two runners overlapping, one runner is not covering the table but just another runner. Wherever there are three runners overlapping, two runners are not covering the table but just the third runner at the bottom.
So can we say that (d + e + f) represents the area where one runner is covering another runner and g is the area where two runners are covering another runner?
Put another way, can we say d + e + f + 2g = 60?
We know that d + e + f = 24 giving us g = 18 square inches
This entire ‘thinking process’ takes ten seconds once you are comfortable with it and your answer would be out in about 30 sec!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: A Sets Question that Upsets Many

Quarter Wit, Quarter WisdomWe hope last week’s discussion improved your understanding of sets and showed you how you can come across some tricky sets questions even though the concepts seem very simple. Today, let’s further build up on what we learned last week with the help of an example.

Example: A group of people were given 2 puzzles. 79% people solved puzzle X and 89% people solved puzzle Y. What is the maximum  and minimum percentage of people who could have solved both the puzzles?
Continue reading “Quarter Wit, Quarter Wisdom: A Sets Question that Upsets Many”

Quarter Wit, Quarter Wisdom: Nuances of Sets

Quarter Wit, Quarter WisdomWe will start with sets today. Your Veritas Prep GMAT book explains you the basics of sets very well so I am not going to get into those. If you have gone through the concepts, you know that we can use Venn diagrams to solve the sets questions.

First, let’s look at why we should focus on terminology in sets question. Thereafter, we will put up a very nice question from our very own book which is simple but takes down many people (just like a typical GMAT question):

Say, there are a total of 100 people in a housing society. There are two clubs close to the society – A and B. You are given that of the 100 people of the housing society, 60 people are members of club A and 50 people are members of club B.
Continue reading “Quarter Wit, Quarter Wisdom: Nuances of Sets”

Quarter Wit, Quarter Wisdom: Working like Clockwork

Quarter Wit, Quarter WisdomIn the last few weeks, we have gone through the concepts of relative speed. You might be surprised to know that we can use the same concept to solve some clock problems too. The reason some clock problem can be tricky is that the hour hand and the minute hand move simultaneously so handling them separately is not easy. In such questions you can easily use relative speed i.e. speed of the minute hand relative to the hour hand. Let’s try to understand this with the help of an example.

Example: In a circular clock, the minute hand is the radius of the circle. At what time is the smaller angle between the minute hand and the hour hand of the clock not divisible by 10?
Continue reading “Quarter Wit, Quarter Wisdom: Working like Clockwork”

Quarter Wit, Quarter Wisdom: Taking the Official GMAT Prep Test

Quarter Wit, Quarter WisdomToday’s post of this series is not based on Quant section of the GMAT. Instead, we will try to answer the often asked question: When should I take the two GMAT prep tests available on mba.com?

Let’s first discuss a hypothetical situation: Say, your goal is to beat a particular opponent called Mr. Dude in a game of chess in a competition being held three months from now. Over three months, you can play four friendly matches against him on a day of your choice. The results of the friendly matches are immaterial and will not be made known to any third party. What will be your strategy?
Continue reading “Quarter Wit, Quarter Wisdom: Taking the Official GMAT Prep Test”

Quarter Wit, Quarter Wisdom: Moving in Circles

Quarter Wit, Quarter WisdomHopefully, you are a little comfortable with the relative speed concept now. The concept can come in handy in some circular motion questions too. Today, we will show you how you can use the fundamentals we learned in the last few weeks to solve questions involving moving in a circle. Let’s try a couple of GMAT questions to put what we learned to use:

Question 1: Two cars run in opposite directions on a circular path. Car A travels at a rate of 4? miles per hour and car B runs at a rate of 6? miles per hour. If the path has a radius of 8 miles and both the cars start from point S at the same time, how long, in hours, after the cars depart will they meet again for the first time after leaving?
Continue reading “Quarter Wit, Quarter Wisdom: Moving in Circles”

Quarter Wit, Quarter Wisdom: Some Tricky Relative Speed Concepts

Quarter Wit, Quarter WisdomAs 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.

Question 1: Cities A and B are 20 miles apart. From both of these cities, simultaneously, two people start walking toward each other at a constant speed of 2 miles/hr. At the same time, a dog leaves city B and runs at a constant speed of 5 miles/hr toward city A. When it reaches the person from city A, it immediately turns around and runs back to the person from city B. When it reaches the person from city B, it turns around and runs back to the person from city A. It keeps doing so until the two people meet. How many miles did the dog run?

(A)   15 miles

(B)   20 miles

(C)   25 miles

(D)   30 miles

(E)    35  miles

Solution: Looks really tricky, doesn’t it? Actually it is really easy once you look at it from the right perspective. It would take you less than a minute to solve almost any 700+ level GMAT question if you can figure out the most optimum method to solve it. The point is – how long will it take you to figure out the most optimum method? Take a minute to think what you would do in this question.

We know the dog’s speed. We need to know the distance it has run. Directly calculating distance is a little complicated since we need to consider the distance covered by the two men too. Instead, if we can figure out the time for which the dog ran, we can easily calculate the distance. For how long did the dog run? He started when the two men started from their respective cities and stopped when the two men met. Therefore, if we can find the time taken by the men to cover the 20 miles, we will get the time for which the dog ran.

This is where the relative speed concept comes in handy.

Total distance between the two men = 20 miles

Relative speed of the men with respect to each other = 2 + 2 mph (since they are travelling in opposite directions)

Time taken by the men to meet = 20/4 = 5 hrs

Therefore, the dog also ran for 5 hrs.

In 5 hrs, the dog must have run 5*5 = 25 miles

Answer (C)

I hope you see that the right perspective simplifies the question immensely. Let’s look at another 700+ level question.

Question 2: A man cycling along the road at a constant speed noticed that every 12 minutes a bus overtakes him and every 6 minutes he meets an oncoming bus. If all buses move at the same constant speed and leave the bus station at fixed time intervals, what is the time interval between consecutive buses?

(A) 5 minutes

(B) 6 minutes

(C) 8 minutes

(D) 9 minutes

(E) 10 minutes

Solution: Again, the question is a little tricky but once you understand how to tackle it, it takes less than a minute.

Buses are coming from opposite directions. A bus overtakes the man every 12 minutes i.e. a bus moving in the same direction as the man overtakes him every 12 mins. To clarify: say, at constant intervals, buses leave a bus station located from where the man left and travel on the same road as the man. Since they are faster, they overtake the man. The man noticed that a bus overtakes him every 12 mins. Obviously then, they must be leaving at constant intervals. Also, he meets a bus coming from the opposite direction every 6 mins. So buses must be leaving from a bus station located at the opposite end of the road at constant intervals.

I hope the problem is clear to you. Now let’s try to work out the solution.

Say the cyclist is stationary at a point. Buses are coming from opposite directions (same speed, same time interval). A bus will meet the cyclist every t minutes from either direction. This ‘t minutes’ must be the time interval between consecutive buses. Let’s say, a bus from each direction just met him. After t minutes, 2 more buses from opposite directions will meet him and so on… We need to find ‘t’.

Now the cyclist starts moving at speed c. His speed relative to the bus going in the same direction becomes b – c. His speed relative to the bus from the opposite direction becomes b + c. This is the reason that the time interval between two buses is different for the opposite directions. Time interval is in the ratio 12:6. Then, the ratio of the relative speed in the two cases must be inverse i.e. 6:12

(b-c):(b+c) = 6:12 which gives you c = (1/3)b

This means that the bus travelling at a relative speed which is 2/3rd of its usual speed (b-c = 2b/3) takes 12 minutes to meet the man. If it were travelling at its usual speed, it would have taken 12*(2/3) = 8 mins to meet the man. This 8 mins is the value of ‘t’ i.e. the time interval between buses.

Answer (C)

You might need to go through the question a few times before you fully understand it. It will also be helpful to draw a diagram and see what the situation looks like.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Questions on Speeding

Quarter Wit, Quarter WisdomWe 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.

Question 1: A man walking at a constant rate of 4 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 20 miles per hour. The woman stops to wait for the man 5 minutes after passing him, while the man continues to walk at his constant rate. How many minutes must the woman wait until the man catches up?
Continue reading “Quarter Wit, Quarter Wisdom: Questions on Speeding”

Quarter Wit, Quarter Wisdom: Speeding, Relatively

Quarter Wit, Quarter WisdomLet’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)

So the first question is – what is relative speed? To understand this, let’s first consider ‘speed’. When your car’s speedometer shows 60 mph, what does it mean? It means that you are traveling at a speed of 60 mph relative to Earth. For an observer in space, your speed could be different. It would be the resultant of your speed and Earth’s speed. If this is hard to imagine, think of a plane flying in the sky. Say, it is flying at a speed of 500 mph. What happens if a strong wind starts blowing in the same direction as the plane? The wind’s speed gets added to the plane’s speed and the plane travels even faster. So anyway, the speed we usually talk about is relative to Earth.

It is sometimes useful to consider the speed of an object relative to another object. Say, two people are sitting next to each other in the same train which is moving at a speed of 100 mph. Are they moving relative to each other? No. Their speed relative to each other is 0. Ofcourse, relative to the Earth, they are moving at the same speed as the train i.e. 100 mph. When working on relative speed questions, just focus on the two objects we are talking about and how fast they are moving relative to each other. Their actual speed may not be of much relevance to us.

Let’s consider a few cases related to relative speed. Say, there are two people, A and B.

Case 1: A is standing still and B is moving at a speed of 5 mph.

What is B’s speed relative to A? B is moving away from A at a rate of 5 miles every hour. So B’s speed relative to A is 5 mph. What is A’s speed relative to B? Is it 0? No! A’s speed relative to Earth is 0. A’s speed relative to B is 5 mph since distance between A and B is increasing at a rate of 5 mph. Confused? When we say ‘relative to B’, we assume that B is stationary. Since distance between the two is increasing at a rate of 5 mph, we say A’s speed relative to B is 5 mph.

Case 2: A is walking due east at a speed of 5 mph and B is walking due east at a speed of 2 mph.

What is A’s speed relative to B? Distance between A and B is increasing by 3 miles every hour. So A’s speed relative to B is only 3 mph, not 5 mph. Think of it this way – A is moving fast but not as fast when compared with B since B is also moving. Similarly, B’s speed relative to A is also 3 mph. Notice that when they move in the same direction, their relative speed is the difference of their speeds.

Case 3: A is walking due east at a speed of 5 mph and B is walking due west, away from A, at a speed of 2 mph.

What is A’s speed relative to B? Distance between A and B is increasing by 7 miles every hour. So A’s speed relative to B is 7 mph, not 5 mph. Think of it this way – A is moving fast and even faster as compared with B since B is moving in the opposite direction. Similarly, B’s speed relative to A is also 7 mph. Notice that when they move in the opposite directions, their relative speed is the sum of their speeds. It doesn’t matter whether they are moving toward each other or away from each other.

Let’s look at some easy questions on relative speed to cement the concepts. We will look at some tough nuts next week.

Question 1: Train A starts from station A traveling at 30 miles per hour toward station B. At the same time, on a parallel track, train ? leaves station B at 40 miles per hour toward station A. When the two trains meet, how far is train A from station B if the distance between stations A and B is 700 miles?
(A) 140
(B) 240
(C) 300
(D) 340
(E) 400

Solution: Both the trains start at the same time from stations that are 700 miles away from each other. This means that in the beginning, distance between them is 700 miles. When they meet, they have together covered the entire 700 miles. Relative to each other, their speed is 30 + 40 = 70 mph.

Time for which they travel till they meet = 700/70 = 10 hrs

Train A covered 30*10 = 300 miles. This means, it is 400 miles away from station B. Answer (E)

Meanwhile, train B covered 40*10 = 400 miles.

Let’s look at a variation of this question now.

Question 2: Train A leaves the station and travels at 30 miles per hour. Three hours later, train ? leaves the same station traveling in the same direction at 40 miles per hour on a parallel track. How far from the station was train A overtaken by train B?
(A) 90
(B) 180
(C) 200
(D) 300
(E) 360

Solution: Train A travels at 30 mph for 3 hrs and covers 30*3 = 90 miles in this time. This is when train B leaves the station. Now both the trains are running in the same direction and their relative speed is 40 – 30 = 10 mph. This means that train B covers an extra 10 miles every hour. Since the initial distance between the two trains is 90 miles, it takes train B ( 90/10 = ) 9 hrs to catch up with train A. In 9 hrs, train B must have traveled 40*9 = 360 miles. Hence, train A must also be 360 miles away from the station.

Answer (E)

You can easily do these questions using variables. The relative speed concept is not a must. But, you definitely end up saving some time if you use relative speed. You don’t take any variables and don’t make any equations. Next week, we will look at some tougher relative speed questions.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches GMAT prep for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Some Inequalities, Mods and Sets

Quarter Wit, Quarter WisdomToday, 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!

Anyway, enough of introduction! Let’s get to the question now.

Question: If x/|x| < x, which of the following must be true about x?

(A)   x > 1

(B)   x > -1

(C)   |x|< 1

(D)   |x| = 1

(E)    |x|^2 > 1

Solution:

First thing we do is tackle the mod. We know that |x| is just the absolute value of x.

So, x/|x| can take only 2 values: 1 or -1
If x is positive, x/|x| = 1 e.g. if x = 4, then 4/|4| = 1
If x is negative, x/|x| = -1 e.g. if x = -4, then -4/|-4| = -4/4 = -1
x cannot be 0 because we cannot have 0 in the denominator of an expression.

Now let’s work on the inequality.

x > x/|x| implies x > 1 if x is positive or x > -1 if x is negative.
Hence, for this inequality to hold, either x > 1 (when x is positive) or -1 < x< 0 (when x is negative)

x can take many values e.g. -1/3, -4/5, 2, 5, 10 etc.

Now think – which of the following MUST BE TRUE about every value that x can take?

(A)   x > 1

or

(B)   x > -1

I hope that you agree that x > 1 doesn’t hold for every possible value of x whereas x > -1 holds for every possible value of x. Mind you, every value greater than -1 need not be a possible value of x.

This concept might need some more work. Let me explain with another example.

Forget this question for a minute. Say instead you have this question:

Example 1: x > 2 and x < 7. What integral values can x take?
I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line.

The top arrow shows x > 2 and the bottom arrow shows x < 7. You see that the overlapping area includes 3, 4, 5 and 6. That is the region that satisfies both the inequalities.

Now consider this:

Example 2: x > 2 or x > 5. What integral values can x take?
Let’s draw that number line again.

So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. the overlapping part) as was the case in example 1. OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 …

Now go back to this question. The solution is a one liner.

x/|x| is either 1 or -1.
So x > 1 or x > -1
So which values can x take? All values included in at least one of the sets. Therefore, x > -1.

Note that the confusion lies only between the first two options. The other three options are rejected outright.

(C)   |x|< 1 implies -1 < x < 1. Definitely doesn’t hold.

(D)   |x| = 1 implies x = 1 or -1. Definitely doesn’t hold.

(E)    |x|^2 > 1 implies either x < -1 or x > 1. Definitely doesn’t hold.

So what do you say? Are you convinced that the answer is (B)?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Questions on Inequalities

Quarter Wit, Quarter WisdomNow 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).

Question 1: Is n between 0 and 1?
Statement 1: n^2 is less than n
Statement 2: n^3 is greater than 0
Continue reading “Quarter Wit, Quarter Wisdom: Questions on Inequalities”

Quarter Wit, Quarter Wisdom: Inequalities with Complications – Part II

Quarter Wit, Quarter WisdomLast 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.

Complication No 3: Even powers

(x  – a)^2(x – b)(x – c)(x – d) > 0

How would you handle even powers of factors? Say, if the question has a factor (x – a) which has an even power, would you still plot ‘a’ on the number line? No, you will just ignore that factor while making the number line! Why? This is so because this factor will never be negative. It will be either 0 (in case the inequality includes the equality sign e.g. (x  – a)^2(x – b)(x – c)(x – d) ? 0) or it will be positive. Therefore, a factor with an even power acts just like a positive constant. Then, does it mean factors with even powers have no role to play at all? No, it doesn’t. While writing out the range, they could impact the final answer. We will discuss this in more detail using an example later on.

e.g. (x – 4)(x – 2)(x + 8)^4 < 0

You don’t need to plot -8 here. Since (x + 8)^4 can never be negative, it doesn’t change the sign of the expression.

The expression will be negative in the range 2 < x < 4.

Complication No 4: Odd powers

(x  – a)^3(x – b)(x – c)(x – d) > 0

What will you do in case of odd powers? Notice that in the last two posts, we have handled questions where the power of all the factors is 1. 1 is an odd power. So when you have any other odd power, you will handle it the same way. You can assume that the odd power is equal to 1 and proceed as usual. This is so because the sign of (x – a) will be the same as the sign of (x – a)^3.

e.g. (x – 4)(x – 2)(x + 1)^3 < 0

The expression will be negative in the range x < -1 or 2 < x < 4.

Now let’s look at a question involving both these complications.

Question: Find the range of x for which the given inequality holds.

[(2x – 5) * (6 – x)^3]/x^2 ? 0

Solution:

I hope you agree that it doesn’t matter whether the factors are multiplied or divided. We are only concerned with the sign of the factors.

[2(x – 5/2) * (-1)(x – 6)^3]/x^2 ? 0 (take 2 common)

[2(x – 5/2) * (x – 6)^3]/(x – 0)^2 ? 0 (multiply both sides by -1)

Now let’s draw the number line. We don’t need to plot 0 since (x – 0) has an even power.

We want to find the range where the expression is positive. The required range is x ? 5/2 or x ? 6. But we are missing something here. x ? 5/2 implies that all values less than 5/2 are acceptable but note that x cannot be 0 since x^2 is in the denominator. Hence the acceptable range is x ? 5/2 or x ? 6 but x ? 0.

When you have the equal to sign, you have to be careful about the way you choose your range.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Inequalities with Complications – Part I

Quarter Wit, Quarter WisdomLast 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.

Complication No. 1: (a – x)(x – b)(x – c)(x – d) > 0

We want our inequality to be of the form (x – a), not (a – x) because according to the logic we discussed last week, when x is greater than a, we want this factor to be positive. The manipulation involved is pretty simple: (a – x) = -(x – a)

So we get: – (x – a)(x – b)(x – c)(x – d) > 0

But how do we handle the negative sign in the beginning of the expression? We want the values of x for which the negative of this expression should be positive. Therefore, we basically want the value of x for which this expression itself (without the negative sign in the beginning) is negative.

We can manipulate the inequality to (x – a)(x – b)(x – c)(x – d) < 0

Or simply, multiply – (x – a)(x – b)(x – c)(x – d) > 0 by -1 on both sides. The inequality sign flips and you get (x – a)(x – b)(x – c)(x – d) < 0

e.g. Given: (4 – x)(2 – x)(-9 – x) < 0

We can re-write this as –(x – 4)(2 – x)(-9 – x) < 0

(x – 4)(x – 2)(-9 – x) < 0

-(x – 4)(x – 2)(x + 9) < 0

(x – 4)(x – 2)(x – (-9)) > 0  (multiplying both sides by -1)

Now the inequality is in the desired form.

Complication No 2: (mx – a)(x – b)(x – c)(x – d) > 0 (where m is a positive constant)

How do we bring (mx – a) to the form (x – k)? By taking m common!

(mx – a) = m(x – a/m)

The constant does not affect the sign of the expression so we don’t have to worry about it.

e.g. Given: (2x – 3)(x – 4) < 0

We can re-write this as 2(x – 3/2)(x – 4) < 0

When considering the values of x for which the expression is negative, 2 has no role to play since it is just a positive constant.

Now let’s look at a question involving both these complications.

Question 1: Find the range of x for which the given inequality holds.

-2x^3 + 17x^2 – 30x > 0

Solution:

Given: -2x^3 + 17x^2 – 30x > 0

x(-2x^2 + 17x – 30) > 0  (taking x common)

x(2x – 5)(6 – x) > 0 (factoring the quadratic)

2x(x – 5/2)(-1)(x – 6) > 0 (take 2 common)

2(x – 0)(x – 5/2)(x – 6) < 0 (multiply both sides by -1)

This inequality is in the required form. Let’s draw it on the number line.

We are looking for negative value of the expression. Look at the ranges where we have the negative sign.

The ranges where the expression gives us negative values are 5/2 < x < 6 and x < 0.

Hence, the inequality is satisfied if x lies in the range 5/2 < x < 6 or in the range x < 0.

Plug in some values lying in these ranges to confirm.

Next week, we will look at some more variations which can be brought into this form.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Inequalities with Multiple Factors

Quarter Wit, Quarter WisdomStudents 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:
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Quarter Wit, Quarter Wisdom: Some Tricky Standard Deviation Questions

Quarter Wit, Quarter WisdomLast 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.
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Quarter Wit, Quarter Wisdom: Dealing with Standard Deviation – Part II

Quarter Wit, Quarter WisdomThis 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.
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Quarter Wit, Quarter Wisdom: Dealing with Standard Deviation

Quarter Wit, Quarter WisdomWe 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
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Quarter Wit, Quarter Wisdom: Mean Questions on Median

Quarter Wit, Quarter WisdomAs 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
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Quarter Wit, Quarter Wisdom: Application of Arithmetic Means

Quarter Wit, Quarter WisdomLast 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?
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Quarter Wit, Quarter Wisdom: Finding Arithmetic Mean Using Deviations

Quarter Wit, Quarter WisdomToday’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.
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Quarter Wit, Quarter Wisdom: Some Mean Questions!

Quarter Wit, Quarter WisdomI 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.
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Quarter Wit, Quarter Wisdom: The Meaning of Arithmetic Mean

Quarter Wit, Quarter WisdomLet’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.
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Quarter Wit Quarter Wisdom: Blundering through Calculations

Quarter Wit, Quarter WisdomI 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.

Obviously, the two exams – GMAT and CFA are very different and one of the differences is that on the CFA exam, we are allowed to use a financial calculator. Now, if you have read some of my posts before, I guess you will agree that I am a devoted ‘logical solution’ fan. To increase speed, I advice my students to try to solve all quant questions orally – to sit without a pen and paper and see if you can force yourself to figure out the answer without writing a word. I solve most of the GMAT relevant questions orally so it is certainly possible. If a question makes me get up and get my pen and paper and then, if I can solve the question within two minutes, I consider it a very tricky but good GMAT question.
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