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Five Ways To Drastically Improve Your Quant Skills

Most of the MBA entrance examinations require you to be extremely quick, agile and efficient a problem-solving. The MBA aspirants spend several hours mastering the ways to speed up their problem-solving abilities. The usual tools for speeding up problem-solving are : finding short-cuts, memorizing all the formula, learning tricks to compute faster, etc. These tools do have a role to play, but they do nothing to enhance your problem-solving abilities.

Try out these tips and see your problem-solving abilities soar up instantly.

Tip No. 1 The best way to speed up calculations is to eliminate the need of calculations.

Most problems in CAT/GMAT are actually meant to test your ability to look at a complex scenario and find simple solutions to it. You are wannabe managers, right ? A manager is not somebody who is a mathematical genius, but somebody who is proficient at efficient utilization of resources. In most problems, a bulk of calculation can be avoided by using logical arguments. This is the best way to speed up calculations avoid the need for calculations. Before rushing on to make a calculation stop to see if there are logical explanations that can resolve things to a great degree ? You will be surprised, that in most cases, this actually turns out to be so. The questions for these examinations are carefully crafted not to test whether you can find a cube-root, but to find out whether you can find the inherent patterns in the problem and reveal the underlying simplicity. Consider the following problem : S is a 6-digit number that begins with 1. When the digit "1" is moved from the left-most to the right-most place, the resultant number is three times the original number. The sum of the digits of S is :

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Try it out for a while before you proceed. There are several ways to solve this problem some involving some trial-and-error, some involving some solid calculations. Most of them involve finding the value of S and then adding the digits to find the sum. Now read through the problem statement again thoroughly s l o o o o w w w l y and mark all the important points in the problem statements. What we have to find is the sum of the digits of S. Can we do something directly about the sum of the digits of S ? Note that the digits of S remains the same, only the order changes. Therefore, the sum of the digits of S and the second number (which is 3S) are the same. Now, since 3S is a multiple of 3, what can you say about the sum of digits of S ? It has to be a multiple of 3. What can you say about the sum of digits of S ? It is the same as the sum of digits of 3S, isn't it ? Therefore, the sum of digits of S is a multiple of 3. Therefore, S is a multiple of 3. Therefore, 3S is a multiple of 9. Therefore, the sum of digits of 3S is a multiple of 9. Therefore, the sum of digits of S is a multiple of 9. Can you find the answer now ? Although this argument takes a lot of space to write and a lot of time to read, it would take a few seconds for you to go through the entire argument and get to the right answer. What does it take to come up with solutions like these ? The answer is extreeeeeeeeeeemely simple and is disclosed in the next tip.

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Tip No. 2 The most important step in solving a problem is to "study" the problem statement - NOT read it , NOT skim through it, but study it thoroughly chew it, digest it, assimilate it.

Every second that you spend assimilating the problem statement will save you minutes when you start solving it. Most of the time we just read through a problem and rush through to the solution. The biggest excuse for this approach is the lack of time. We just have seconds to solve each problem, right ? Ironically, this would be your biggest undoing. A skillful, efficient problem-solver follows an entirely different approach. He enjoys exploring the problem, looking at it through different perspectives, and while doing so an interesting, elegant solution emerges which saves him several minutes of hard-work in getting the answer. Yes, yes, I can hear you scream Where is the time ? We have to solve the problems in seconds. I know. Read the following statement again : Every second that you spend assimilating the problem statement will save you minutes when you start solving it. Consider the following problem : Ram and Shyam start from a point A, move to B which is 5 Kms away from A, and then travels back to A. Ram starts at 9 AM, traveling at 5 Km/hr while Shyam starts at 9:45 AM, traveling at 10 Km/hr. When do they meet first ? Try the problem for a while before proceeding further. There is a general tendency to start forming equations about the problem statement and trying to solve them. If you followed that approach, you would realize the complexity with the problem you don't know whether they were moving in the same direction or in the opposite direction when they meet first, and because of this, you cannot really form a simple equation to represent the above problem. Let us dig into the problem further.

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Let us explore what Ram does. He starts from the point A at 9 AM at 5 Km/Hr. Therefore, he reaches B at 10 AM and back to A at 11 AM. What about Mr Shyam ? He starts from the point A at 9:45 AM at 10 Km/Hr. Therefore, he reaches B at 10:15 AM and back to A at 10:45 AM. What happens at 10 AM ? At 10 AM, Ram has started moving from point B back to point A at 5 Km/Hr At 10 AM, Shyam is 2.5 Kms away from B and is moving towards it at 10 Km/Hr Now we have a new problem to solve : At 10 AM, A and B are 2.5 Kms apart, they are moving in the opposite directions at 5 Km/Hr and 10 Km/Hr respectively. When do they meet ?

Did the original problem really have so much information in it ? Yeeeeesssssss --- only if you go down deep enough.

As an exercise, try finding the time when Ram and Shyam meet for the second time.

It is IMPOSSIBLE to get to solutions like these, if you tend to rush through to the solution using the most familiar ways known to you.

Similarly, it is POSSIBLE for EVERYONE (yes, and that includes YOU) to be able to come up with solutions like these, provided you spend sufficient time with the problem statement, the options given. Being inefficient and ineffective at problem-solving is a simple 3-step process : 1) Read through the problem fast 2) Start solving the problem 3) Congratulations if you still got a great solution. Otherwise, happy merry-go-rounding.

Being efficient and effective at problem-solving is even simpler :

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1) Chew, Digest and Assimilate the problem statement entirely. 2) Watch the solution emerge.

Tip No 3. Do not carry all the luggage on your head.

Imagine if you were traveling in a train with your suitcase on your head, move to your seat and sit down comfortably. Now it feels so lazy to take off that damn suitcase off your head. So, you leave it right there anyway what harm is it going to do ? Funny, isn't it ? But that's exactly what most of us do while solving a problem !!! storing every single piece of information in our heads is more painful (and ridiculous) than traveling on a train with the luggage on our heads. We tend to keep the problem statement in mind, all the data for the problem in the mind, we perform the steps mentally and store the intermediate results in the mind every bit of data that you are "trying" to "remember" is hindering your thought process. Any extra luggage on your head is going to make your journey that much uncomfortable. Well, the solution is easy. Do not be lazy and do not be presumptuous (O, I can remember all that !!!) . Write down all the bit of information in front of you with sufficient clarify. Represent the problem well. Draw diagrams if necessary. Mark everything that you know on the diagram. If you make any inferences, mark that too. At any point, whatever you have known so far about the problem transfer that out of your head to a piece of paper in front on you. Keep your mind FREE FREE, so that it can think, be creative and solve the problem, rather than working hard at just keeping a tab on all the darn data. Consider the following problem : An equilateral triangle ABE is drawn inside the square ABCD. Find the angle CED.

Now suppose we have drawn the following diagram :

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So, we have this square ABCD and the triangle AEB. Try solving the problem with just this diagram in front of you. Try for a while before proceeding forward.

We have a square and an equilateral triangle so we have got lots of equal sides ( and equal angles) to manage. Can you figure out which all sides are equal and which all angles are equal ? Not in your head - Mark them ALL on the diagram.

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For most students, it is impossible NOT to solve this problem beyond this point. You just can't help seeing that BE = EC and so angles BEC = BCE,. the angle EBC is 30 degrees and the rest of the argument follows simply. Did that help ? For most students, just this single step "Transferring every bit of information from your head to a piece of paper in front of you" creates the difference between solving the problem and not reaching about it anywhere. Genius could sometimes be very simple isn't it ?

Tip No. 4 There is a scope for "speeding up" at every single step of the solution. Every second counts.

For most students, the toughest part of the process of solving is the first step getting the right "idea" about the problem. And once they have it, they rush through to the end of the solution for their life, without waiting for a breather.

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Remember every step of the solution is a sub-problem in itself and there is ample scope to speed up every single step of the solution. You need a "habit" and a probing "eye" for areas to simplify. Unless you have mastered this habit, you will be performing a number of steps which are utmost wasteful. Consider multiplying 11 x 14 on step 2 and later dividing that result by 14 to get 11 on step 5.

How would you want to speed this up by learning how to multiply and divide faster, or by simply eliminating these two unnecessary steps ?

The "eye" for speeding up has to be deeply ingrained, so that you can automatically spot the hot areas that can be speedened up. Consider the following problem : Find the remainder of 2^3015 with 17. The general solution of the problem would be to observe that 2^4 = 16 gives a remainder of -1 with 17. Hence, 2^8 would give a remainder of 1 with 17. Therefore, because of this cyclicity of 8, you would express 3015 = 8 x 376 + 7 Hence 2^(3015) = 2^(8 x 376) * 2^(7). Now, the first term would give a remainder 1 with 17, so we can ignore it.

2^7 = 128 which gives a remainder 9 with 17. Pretty simple, isn'it ? Could we have done it better ? Observe that all we need to worry about is the remainder that 3015 leaves with 8. Can we do it faster than the way we did it before ? We know that 3000 is a multiple of 8, as any number that ends with 3 zeroes is a multiple of 8.

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So, 3015 would give the same remainder with 8 as 15 , which is 7. Time taken for this process : 1 second There are hundreds of ways you can speed up calculating certain parts of a problem. Watch our for a possible speed-up at every single step and you will save several minutes.

Tip No 5 Watch out for simplicity

There is a simple test for whether a solution is good enough is it simple ? All elegant solutions have an inherent simplicity and beauty to it. If you catch yourself doing several unnecessary computations and complicated arguments, rest assured the method is ineffective. There has to be a better method than that. This is a natural corollary of Tip No 2 study the problem carefully, watch out for simplifying patterns. Consider the following problem : Perform the following binary addition : 1 + 11 + 111 + 1111 + 11111 (all these numbers are binary). There are several "quick-fix" solutions to these : i) ii) convert these numbers to decimal, add them and convert the result back to binary perform the brute-force binary addition.

However, the solution to this could be much simpler. Consider how would the problem look slightly differently and be extremely simple to solve ? Would 1 + 10 + 100 + 1000 + 100000 be simple enough ? What, then, makes the given problem complicated ? Can we find some simplifying patterns in it ?

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The CAT Lounge What if we could represent the last three numbers as 111 = (1000 1) 1111 = (10000 1) 11111 = (100000 1) Then the sum would be 1 + 11 + (1000 1) + (10000 1) + (100000 1) = 1 + 11 + (111000 11) = 1 + 111000 = 111001

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(Note that 3 in decimal is 11 in binary)

There are patterns like these hidden throughout the problem statements only if you can pause, look and identify the gold. Consider another example : The sum of the `n' terms of an arithmetic progression is (an + bn^2). Find the common difference between the terms. The most straightforward way to solve this problem is to use the formula for the `n' terms. Consider this : The sum of 1 term of the AP is the same as the first term of the AP , which is ( a + b). The sum of 2 terms of the AP is the sum of the first and the second terms and it equals (2a + 4b). So, what is the second term ? (2a + 4b) = (a + b) = a + 3b. So, we know the first term, we know the second term, can we find the common difference between the terms ?

There is sufficient simplicity in each problem, waiting to reveal itself. Don't insult the beauty by ignoring it and rushing past through it. Stop. Relax. Enjoy the beauty and your journey will be much simpler.

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These tips, however simple or obvious they look, can help you increase your problemsolving abilities many many times. Each tip would make your problem-solving much easier and exciting. As you master these, you would start to enjoy Quant and develop a natural flair for problem-solving.

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