Archive : Quarter Wit, Quarter WisdomRSS feed
Today we will work with circles inscribed in regular polygons.
We begin by considering an equilateral triangle whose each side is of length ‘a’. Recall that every triangle has an incircle i.e. a circle can be inscribed in every triangle. The diagram given below shows the circle of radius ‘r’ inscribed in an equilateral triangle.
Last week we looked at regular and irregular polygons. Today, let’s try to understand how questions involving one figure inscribed in another are done. The most common example of a figure inscribed in another is a polygon inscribed in a circle or a circle inscribed in a polygon. Let’s see the various ways in which this can be done.
Continuing our Geometry journey, let’s discuss polygons today. Some years back, I used to often get confused in the polygon sum-of-the-interior-angles formula if I had to recall it after a gap of some months because I had seen two variations of it:
Sum of interior angles of a polygon = (n – 2)*180
Let’s go back to geometry now. We will discuss how to use diagrams to solve DS questions today. Though we discussed a DS question in a previous geometry post, we didn’t discuss how the thought process used for a DS question is different from the thought process used for a PS question. To find whether a statement is sufficient to answer the question, you should try to prove that it is not sufficient. Try to make two cases which answer the question differently using the give information. If there are two or more different answers possible, it means the given information is not enough. Let’s discuss this with the help of an official question.
If you are wondering about the absurd title of this post, just take a look at last week’s title. It will make much more sense thereafter. This post is a continuation of last week’s post where we discussed number plugging. Today, as per students’ request, we will look at the inequalities approach to the same official question. You will need to go through our inequalities post to understand the method we will use here.
Let’s take a break from Geometry today and discuss the concept of transition points. This is especially useful in questions where you are tempted to plug in values. A question often asked is: how do I know which values to plug and how do I know that I have covered the entire range in the 3-4 values I have tried? What transition points do is that they give you the ranges in which the relationships differ. All you have to do is try one value from each range. If you do, you would have figured out all the different relationships that can hold. We will discuss this concept using a GMAT Prep question. You can solve it using our discussion on inequalities too. But if number plugging is what comes first to your mind in this question, then it will be a good idea to get the transition points.
Last week, we discussed how drawing extreme diagrams can help solve Geometry questions. Today we will see how to solve another Geometry question by making diagrams. The diagram can help you understand exactly what it is that you need to do; doing it will be quite straightforward.
Let’s continue with geometry today. We would like to discuss how drawing extreme diagrams can help you solve questions. Most GMAT questions are quite intuitive and hence our non-traditional methods are perfect for them. They are not typical MATH problems per se; instead, they are logical puzzles. If you can prove why some things will not work, it means whatever is left will work.
This week, we will further build up on what we have discussed in the past two weeks. You will need to sum up everything we discussed last week in a few seconds and arrive at a conclusion and then, move on and solve the question on the basis of that conclusion. We will take you through the ‘summing up’ and ‘getting a feel for it’ process step by step so that it’s intuitive to you next time you come across this concept.
Let’s start with geometry today. It has some very interesting and intuitive concepts. We will discuss one of them today. It’s surprising how a little bit of imagination can go a long way in helping you solve questions. Let’s discuss the concept first. We will look at a question later.
Imagine a clock face. Think of the minute hand on 10. Ignore the hour hand for our discussion today. Say, the length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below (using the green and the red dotted lines). Let’s say the minute hand moves to 1. Can you say something about the lengths of the dotted black and dotted blue lines?
Another issue of assumption questions that merits discussion is the inference vs assumption confusion. On some questions, people find it hard to type the question as inference or assumption. Such questions often have the words ‘must be true’. Let’s discuss the two different cases:
There is a particular issue in assumption questions that I would like to discuss today. We discussed in our previous posts that assumptions are ‘necessary missing premises’. Many students get stuck between two options in assumption questions. The correct option is the necessary premise. The incorrect one is often a sufficient premise. Due to the sufficiency, they believe that that particular option is a stronger assumption. But the point to remember is that an assumption is only necessary for the conclusion to be true. It may not actually lead to the conclusion beyond a reasonable doubt. You only have to answer what has been asked (which is an assumption), not what you think is better to make the conclusion true.
Today, we again pay homage to the lazy bum within each one of us in our QWQW series. If you are wondering what we mean by ‘again’, check out our last two posts of the QWQW series. We have been discussing how to avoid calculations. Today let’s learn why it is advisable to avoid learning formulas too!
Last week we discussed how to solve equations with the variable in the denominator. We also said that the technique generally works for PS questions but you need to be careful while working on DS questions. Today, let’s look at the reason behind the caveat.
We have discussed before how GMAT is not a calculation intensive exam. Whenever you land on an equation which looks something like this: 60/(n – 5) – 60/n = 2, you probably think that we don’t know what we are talking about! You obviously need to cross multiply, make a quadratic and finally, solve the quadratic to get the value of n. Actually, you usually don’t need to do any of that for GMAT questions. You have an important leverage – the options. Even if the options don’t directly give you the values of n or n-5, you can use the knowledge that every GMAT question is do-able in 2 mins and that the numbers fit in beautifully well.
Today we would like to discuss a technique which is very useful in solving assumption questions. No, I am not talking about the ‘Assumption Negation Technique’ (ANT), which, by the way, is extremely useful, no doubt. The point is that ANT is explained beautifully and in detail in your book so there is no point of re-doing it here. You already know how to use it.
Today, we would like to discuss one of our own work questions. The intent is to show you how simple your calculations can get when you use the methods we discussed in the last few weeks. I couldn’t say it enough – develop a love for ratios. You will save a huge amount of time in lots of questions. If you haven’t been following the last few weeks’ posts, take a look at this link before checking out the question. Otherwise the method may not make sense to you.
This week, let’s look at some work-rate questions which use joint variation. Check out the last three posts of QWQW series if you are not comfortable with joint variation.
Question 1: A contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days will the work get over?
As promised, we will discuss inverse variation today. The concept of inverse variation is very simple – two quantities x and y vary inversely if increasing one decreases the other proportionally.
If x takes values x1, x2, x3… and y takes values y1, y2, y3 … correspondingly, then x1*y1 = x2*y2 = x3*y3 = some constant value
We can keep working on ‘pattern recognition’ questions for a long time and not run out of questions of different types on which it can be used. We hope you have understood the basic concepts involved. So let’s move on to another topic now: Variation.
Basically, variation describes the relation between two or more quantities. e.g. workers and work done, children and noise, entrepreneurs and start ups. More workers means more work done; more children means more noise; more entrepreneurs means more start ups and so on… These are examples of direct variation i.e. if one quantity increases, the other increases proportionally. Then there are quantities that have inverse variation between them e.g. workers and time taken. If there are more workers, time taken to complete a work will be less.
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.
If 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.
Last 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.
Sometimes 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.
For 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.
Last 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.
We 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.
Last 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.
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.
At 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:
People 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.
Last 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
A 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:
We 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 Applied on the GMAT
Divisibility Applied to Remainders
Today, 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?