Articles

HANDA KA FUNDA

Articles for Quant My posts on various websites Ravi Handa

2012

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Hi, I am Ravi Handa. I have been teaching Maths for CAT and other competitive exams for more than 6 years. During this period I have contributed to Pagalguy.com, Minglebox.com, Rediff.com and a host of other websites. I thought it would be beneficial to students if I compiled all the articles in one place. This PDF / ebook is the realization of that thought. I have prepared a very detailed set of videos on Maths / Quantitative Aptitude for competitive exams which is available for sale. The course is available here Use the coupon code ARTBOOK to get extra discount of 10% (Offer valid till 31st August, 2012) Your support of buying the course is what makes these ebooks / videos possible. Cheers, Ravi Handa Email: [email protected] Twitter: @ravihanda

PS: If you notice any typos, grammar errors, calculation mistakes, please send me an email.

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INDEX

1. Basic Applications of Remainder Theorem 2. Cyclicity of Remainders 3. Winning Silver – Figuring out the second last digit 4. Divide & Conquer – Divisibility Rules 5. Time Speed and Distance – Equations Stay Away 6. Dealing with Factorials 7. Box of Chocolates(P & C) 8. Interpretation of Data (DI) 9. Logical Reasoning Explained (LR) 10. Figuring out Remainders 11. Remainders Reloaded 12. Games and Tournaments 13. Games and Tournaments – Part 2 14. Set Theory 15. Geometry Funda Part-I 16. Geometry Funda Part-II 17. Functions 18. WizIQ Course Details

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1. Basic Applications of Remainder Theorem In my previous post, we discussed the cyclical nature of the remainders when an is divided by d. In this post, we will see how problems on finding out the remainder can be broken down into smaller parts.

Funda 1: Remainder of a sum when it is being divided is going to be the same as the sum of the individual remainders.



𝑹𝒆𝒎

𝒂+𝒃+𝒄… 𝒅

= 𝑹𝒆𝒎

𝒂 𝒅

+ 𝑹𝒆𝒎

𝒃 𝒅

+ 𝑹𝒆𝒎

𝒄 𝒅

…

Let us look at an example for this case: Eg: Find out the remainder when (79+80+81) is divided by 7. If we add it up first, we get the sum as 240 and the remainder as 2 as shown below: 𝑅𝑒𝑚

79 + 80 + 81 240 = 𝑅𝑒𝑚 = 2 7 7

However, it would be easier to find out the individual remainders of 79, 80 & 81; which come out to be 2, 3 & 4 respectively and adding them up later to get the answer. This process is shown below: 𝑅𝑒𝑚

79 + 80 + 81 2+3+4 9 = 𝑅𝑒𝑚 = 𝑅𝑒𝑚 =2 7 7 7

I hope you would agree that the second method is easier. May be the difference in difficulty level is not highlighted here. Let us look at another idea on the same lines.

Funda 2: Remainder of a product when it is being divided is going to be the same as the product of the individual remainders.



𝑹𝒆𝒎

𝒂∗𝒃∗𝒄… 𝒅

= 𝑹𝒆𝒎

𝒂 𝒅

∗ 𝑹𝒆𝒎

𝒃 𝒅

∗ 𝑹𝒆𝒎

𝒄 𝒅

…

Let us look at an example for this case: Eg: Find out the remainder when (79 x 80 x 81) is divided by 7. If we multiply it first, we get the product as 511920 and the remainder as 3 as shown below:

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𝑅𝑒𝑚

79 ∗ 80 ∗ 81 511920 = 𝑅𝑒𝑚 =3 7 7

However, it would be easier to find out the individual remainders of 79, 80 & 81; which come out to be 2, 3 & 4 respectively and multiply them to get 24, which will eventually lead to the remainder of 3. This process is shown below: 𝑅𝑒𝑚

79 ∗ 80 ∗ 81 2∗3∗4 24 = 𝑅𝑒𝑚 = 𝑅𝑒𝑚 = 3 7 7 7

I guess there is no doubt in this question that the second method is easier. To be honest, I would take more time to just find out the product of (79 x 80 x 81) than to solve the entire question. That is the reason I recommend breaking down the problem into smaller parts. Funda 3: Negative Remainders – When the absolute value of the -ive remainder is lesser than the absolute value of the positive remainder, it is recommended that you consider a multiple greater than the divisor. When 7 is divided by 4, the remainder can be considered as 3 or -1. When 18 is divided by 7, the remainder can be considered as 4 or -3. When 689 is divided b 23, the remainder can be considered as 22 or -1. As you can see from above, the calculations would reduce drastically in the third case if you consider a negative remainder. As a tip, in remainder questions, you should always think of multiples or powers which can lead to a remainder of 1 or 1. Till now, the examples I have taken are too simple to be asked in CAT or for that matter any other MBA entrance exam. Let us look at an example that uses all the above-mentioned ideas and is of a slightly higher difficulty level.

Eg: Find out the remainder when 83261 is divided by 17. First of all we need to break down 83261 into smaller parts.



𝑅𝑒𝑚

83 261 17

= 𝑅𝑒𝑚

83∗83∗83…261 𝑡𝑖𝑚𝑒𝑠 17

83

We know that 𝑅𝑒𝑚 17 = 15 𝑜𝑟 − 2. It would be easier if consider the remainder as -2 because our calculations would be lesser. So essentially, our question reduces to:



𝑅𝑒𝑚

−2 ∗ −2 ∗ −2 …261 𝑡𝑖𝑚𝑒𝑠 17

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= 𝑅𝑒𝑚

−2 261 17

= −𝑅𝑒𝑚

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2261 17

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Now, referring to the tip I gave above, think of a power of 2 that would give a remainder of 1 or -1 from 17. 24 is 16 and would give a remainder of -1 from 17. We have a 2261 here. We will have to break it down to (2 260 x 2) so that we can convert it to a power of 16. This step is shown below:



−𝑅𝑒𝑚



−𝑅𝑒𝑚

2261 17

= − 𝑅𝑒𝑚

−1 65 ∗2 17

2260 ∗2 17

= −𝑅𝑒𝑚

= −𝑅𝑒𝑚

−1∗2 17

16 65 ∗2 17

=𝟐

I highly recommend that in this question and other questions of this type, you should verify the answer from Wolfram Alpha. Hope you found this article useful. I look forward to your suggestions for future posts.

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2. Cyclicity of Remainders In this post I would like to discuss some of the really fundamental ideas that can be used to solve questions based on remainders. If you have just started your preparation for CAT 2012, you might find this article helpful. On the other hand, if you are looking for some advance stuff, I suggest that you check out some of my posts from last year on the same topic. First of all, 𝑅𝑒𝑚

𝑎𝑛 = 0 𝑡𝑜 𝑑 − 1 𝑑

What I am trying to say above is that if you divide a^n by d, the remainder can be any value from 0 to d-1. Not only that, if you keep on increasing the value of ‘n’, you would notice that the remainders are cyclical in nature. What I am trying to say is that the pattern of remainders would repeat. Let me try to clarify this with an example: 4^1 divided by 9, leaves a remainder of 4. 4^2 divided by 9, leaves a remainder of 7. {Rem(16/9) = 7} 4^3 divided by 9, leaves a remainder of 1. {Rem (64/9) = 1} 4^4 divided by 9, leaves a remainder of 4. {Rem (256/9) = 4} 4^5 divided by 9, leaves a remainder of 4. {Rem (1024/9) = 7} 4^6 divided by 9, leaves a remainder of 4. {Rem (4096/9) = 1} 4^(3k+1) leaves a remainder of 4 4^(3k+2) leaves a remainder of 7 4^3k leaves a remainder of 1 As you can see from above that the remainder when 4^n is divided by 9 is cyclical in nature i.e. the remainders obtained are 4,7,1, 4,7,1, 4,7,1 … and so on. They will always follow the same pattern. Funda 1: a^n when divided by d, will always give remainders which will have a pattern and will move in cycles of ‘r’ such that r is less than or equal to d. With the help of the above idea, you can solve a large number of remainder questions. All you need to do is to figure out the cycle / pattern in which the remainders are moving, and it will lead you to the answer. Let us look at an example to illustrate the above idea. Eg: What will be the remainder when 4^143 is divided by 9. http://bit.ly/handacatcourse

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From the calculations that I did in the beginning of the post, I know that: Remainders of 4^n when divided by 9, move in a cycle of 3. So, I need to express 143 = 3k + x and that would lead to the answer. I know that 143 = 141 + 2 {141 is divisible by 3} So, my answer would be the 2nd value in the list, which is 7. In the questions where you have to find out the remainder of a^n by d, as a rule you can follow this process: Step 1: Find out the cycle of remainders when a^n is divided by d and make a list of those values. Step 2: Find out the cyclicity, say ‘r’ Step 3: Find out the remainder when the power is divided by the cyclicity i.e. Rem[n/r] = p Step 4: The answer would be the pth value in the list. {If p = 0, it would be the last value in the list} Funda 2: While trying to find the cycle / pattern of remainders when a^n is divided by d, just multiply the previous remainder with ‘a’ to get the next value. If you notice in the example mentioned in the beginning of this post, I have calculated 4^5 and 4^6 and then found out the remainder. As you might have realized by now that it is a long and tedious process. But the good part is – you can avoid that tedious process by just multiplying the previous remainder. In that example instead of calculating 4^5 and then dividing by 9, I could have just multiplied the previous remainder, which was 4 with 4 to get 16, which would have directly given me a remainder of 7. Got confused? Well, let us look at a new example. Eg: Find out the cyclicity of remainders when 3^n is divided by 11. 𝑅𝑒𝑚 𝑅𝑒𝑚 𝑅𝑒𝑚

31 11 32 11 33 11

=3 =9 = 𝑅𝑒𝑚

27 11

=5

As you can see that till here there is no problem in calculating the remainders. 34

𝑅𝑒𝑚 11 = 5 ∗ 3 = 15 = 4 {In this case instead of using 3^4 = 81, I took the previous remainder, which was 5 and multiplied it with 3 to get 15, which lead to my current remainder = 4} 𝑅𝑒𝑚

35 11

= 4 ∗ 3 = 12 = 1

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{In this case instead of using 3^5 = 243, I took the previous remainder, which was 4 and multiplied it with 3 to get 12, which lead to my current remainder = 1} 𝑅𝑒𝑚

36

=1∗3 =3

11

{In this case instead of using 3^6 = 729, I took the previous remainder, which was 1 and multiplied it with 3to get my current remainder = 3} As you might have noticed that the remainder 3 repeated itself and so the cycle / pattern of remainders as -> 3,9,5,4,1 and the cyclicity as 5. Let us try and solve a slightly more complicated problem with this idea. Eg: Find out the remainder when 𝟑𝟐𝟑𝟐 𝑅𝑒𝑚

32 32 7

32

= 𝑅𝑒𝑚

4 32

𝟑𝟐

is divided by 7.

32

7

Step 1: Find out the cycle / pattern of remainders when 4^n is divided by 7. Rem [4^1 / 7 ] = 4 Rem [4^2 / 7 ] = 2 Rem [4^3 / 7 ] = 1 So, the cycle/pattern is 4,2,1. Step 2: The cyclicity is 3. Step 3: Rem [Power/Cyclicity] = Rem [32^32 / 3] = (-1)^32 = 1 Step 4: The answer is the 1st value in the list, which is 4. I hope you found this post useful. Suggestions for future posts are more than welcome.

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3. Winning Silver – Figuring out second last digit Nike caused controversy with its advertising campaign during the 1996 Olympics by using the slogan, "You Don't Win Silver — You Lose Gold." Nike's use of this slogan drew harsh criticism from many former Olympic Silver medalists. In a way, it did undermine the importance of the second position but in Math things are often very different. Figuring out the second last digit is often tougher than figuring out the last digit. It is unlikely but definitely not impossible that in CAT you get a straightforward question that asks you to find out the second last digit of a number (abc pqr). It did happen in CAT 2008. In few cases, you will be able to do it by forming a cycle and observing the pattern. Those will be the easier cases. Read on if you wish to do the same for the not so easy cases. The question becomes really simple if the last digit in abc pqr is 0 or 5 because if it 0, second last digit will be 0 and if it is 5, second last digit will be 2 or 7 (which can be easily figured out by observing the cyclicity). All the other questions can be divided in two broad categories: a) Last digit is odd b) Last digit is even I recommend that before using any of the concepts given below, you should try and see if a pattern exists. Let us consider our number is abcpqr where a,b,c,p,q and r are digits and c is not 0 or 5. Concept 1: What to do when the last digit is odd? The second last digit always depends on the last two digits of the number so anything before that can be easily neglected. We first convert the number in such a way that the last digit of the base becomes 1. Second last digit of the number will simply be: Last digit of (Second last digit of base) X (Last digit of power) Let us look at few examples Eg 1a: Second last digit of 3791768 = Last digit of 9×8 = 2 Eg 1b: Second last digit of 1739768 = Second last digit of 39768 = Second last digit of Second Last digit of 1521384 = Last digit of 2×4 = 8 Eg 1c: Second last digit of 9317768 = Second last digit of 17768 = Second last digit of (174) 192 = Second last digit of (…21) 192 = Last digit of 2 x 2 = 4

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Concept 2: What to do when the last digit is even? The second last digit always depends on the last two digits of the number so anything before that can be easily neglected. We need to remember the following ideas:  **2 raised to power 10 will always end in 24.  24 raised to an even power will always end in 76 and to an odd power will always end in 24.  76 raised to any power will always end in 76. Now we can use these to find out the second last digit. We reduce the number in such a way that the last two digits of the base become 76. Eg 2a: Second last digit of 1372482  Second last digit of 72482  Second last digit of 72480 x 722  Second last digit of (7210) 48 x (**84)  Second last digit of 2448 x (**84)  Second last digit of 76 x 84  Second last digit of 6384 = 8 Eg 2b: Second last digit of 48307 = (483) 102 x 48 = (****92) 102 x 48  Second last digit of 92100 x 922 x 48 = 76 x (**64) x 48  Second last digit of (****72) = 7 Eg 2c: Second last digit of 15484 = Second last digit of (54) 84  Second last digit of (545) 16 x 544 = (***24) 16 x (542) 2  Second last digit of 76 x (2916) 2  Second last digit of 76 x 56  Second last digit of 4256 = 5 I hope that after reading this post you will be at ease in figuring out the second last digit. I also hope that you will not mind winning silver medals either.

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4. Divide & Conquer – Divisibility Rules The concept of ‘divide and conquer’, derived from the Latin phrase ‘Divide etimpera’, was put into use effectively by everyone from Caesar to Napoleon to The British in India. Even Gaddafi tried using the same but as current events show us – he wasn’t very effective. Dividing rather divisibility rules to be specific can come in really handy at times in solving problems based on Number Systems. The standard rules which nearly all of us are very comfortable with are the ones for 2n and 5n. For these all that one needs to do is look at the last ‘n’ digits of the number. If the last ‘n’ digits of a number are divisible by 2 n or 5n, then the number is divisible by 2 n or 5n and vice versa. For details about other numbers, I suggest that you read on. Funda 1: For checking divisibility by ‘p’, which is of the format of 10n – 1, sum of blocks of size ‘n’ needs to be checked (Blocks should be considered from the least significant digit ie the right side). If the sum is divisible by p, then the number is divisible by p. Eg 1.1 Check if a number (N = abcdefgh) is divisible by 9  9 is 101 – 1  Sum of digits is done 1 at a time = a + b + c + d + e + f + g + h = X  If X is divisible by 9, N is divisible by 9  Also, N is divisible by all factors of 9. Hence the same test works for 3. Eg 1.2 Check if a number (N = abcdefgh) is divisible by 99  99 is 102 – 1  Sum of digits is done 2 at a time = ab + cd + ef + gh = X  If X is divisible by 99, N is divisible by 99  Also, N is divisible by all factors of 99. Hence the same test works for 9, 11 and others. Eg 1.3 Check if a number (N = abcdefgh) is divisible by 999  999 is 103 – 1  Sum of digits is done 3 at a time = ab + cde + fgh = X  If X is divisible by 999, N is divisible by 999  Also, N is divisible by all factors of 999. Hence the same test works for 27, 37 and others. Funda 2: For checking divisibility by ‘p’, which is of the format of 10 n + 1, alternating sum of blocks of size ‘n’ needs to be checked (Blocks should be considered from the least significant digit ie the right side). If the alternating sum is divisible by p, then the number is divisible by p.

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(Alternating Sum: Sum of a given set of numbers with alternating + and – signs. Since we are using it to just check the divisibility, the order in which + and – signs are used is of no importance.) Eg 1.1 Check if a number (N = abcdefgh) is divisible by 11  11 is 101 + 1  Alternating sum of digits is done 1 at a time = a - b + c - d + e - f + g - h = X  If X is divisible by 11, N is divisible by 11 Eg 1.2 Check if a number (N = abcdefgh) is divisible by 101  101 is 102 + 1  Alternating sum of digits is done 2 at a time = ab - cd + ef -gh = X  If X is divisible by 101, N is divisible by 101 Eg 1.3 Check if a number (N = abcdefgh) is divisible by 1001  1001 is 103 + 1  Sum of digits is done 3 at a time = ab - cde +fgh = X  If X is divisible by 1001, N is divisible by 1001  Also, N is divisible by all factors of 1001. Hence the same test works for 7, 11, 13 and others. Funda 3: Osculator / seed number method For checking divisibility by ‘p’, Step 1: Figure out an equation such that 𝑝 ∗ 𝑛 = 10𝑚 ± 1 If we have this equation, the osculator / seed number for ‘p’ will be ∓ 𝑚. (-m in case of 10m+1 and +m in case of 10m – 1) Step 2: Remove the last digit and multiply it with the seed number. Step 3: Add the product with the number that is left after removing the last digit. Step 4: Repeat Steps 2 and 3 till you get to a number which you can easily check that whether or not it is divisible by p. Eg: Check whether 131537 is divisible by 19 or not.  19*1 = 10*2 – 1 (Seed number is +2)  131537 → 13153 + 7 ∗ 2 = 13167 → 1316 + 7 ∗ 2 = 1330 → 133 + 0 ∗ 2 = 133  133 is divisible by 19  131537 is divisible by 19 I hope that these divisibility rules will enable you to divide and conquer few of the Number Systems problems that you encounter during your preparation.

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5. Time Speed & Distance Equations – Stay Away! I guess my first fascination with problems of Time, Speed and Distance began when I first saw Heena. An important part of the storyline, if you can call it that, saw Rishi Kapoor floating from India to Pakistan without drowning. I remember arguing with my friends that if could float for that long – he could swim back to India as well. My friends nullified the argument by saying: Speed River> Speed Rishi Kapoor I know that the reference is a little dated for most readers of this post, but Zeba Bhaktiyar made me look beyond reason. In this post we will discuss some of the ideas that have helped me solve TSD problems without forming too many equations. Funda 1: Average Speed We know that the average speed in a journey is given by (Total Distance Covered) / (Total Time Taken); but there are few special cases which might help in solving questions 



If the distance covered is constant(𝑑1 = 𝑑2 = 𝑑3 … . = 𝑑𝑛 ) in each part of the journey, then the average speed is Harmonic Mean of the values. 𝑛 SpeedAvg = 1/𝑠 + 1/𝑠 + 1/𝑠 ….1/𝑠 1

2

3

𝑛

If the time taken is constant (𝑡1 = 𝑡2 = 𝑡3 … . = 𝑡𝑛 ), in each part of the journey then the average speed is Arithmetic Mean of the values. 𝑠 + 𝑠 + 𝑠 ….𝑠 SpeedAvg = 1 2 𝑛 3 𝑛

Funda 2: Using Progressions (Arithmetic & Harmonic) In many questions, you come across a situation when a person is going from Point A to Point B at various speeds and taking various times. We know that if distance is constant, speed and time are inversely proportional to each other. But, this information can also be used to deduce these two facts –  

If the various speeds which are mentioned are in AP, then the correspondingtimes taken will be in HP. If the various speeds which are mentioned are in HP, then the corresponding times taken will be in AP.

Let us use these ideas to solve couple of quant questions. Eg 2.1 Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30,40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start? [CAT 2006]

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Note: Arun Barun Kiranmala is a 1968 Bangladeshi film. Now you can guess what inspires CAT question setters. Here is a song from the film. Solution As you can see that the speeds are in HP, so we can say that the times taken will be in AP. Time difference between Arun and Barun is 2 hours, so the time difference between Barun and Kiranbala will also be 2 hours. Hence, Kiranbala started 4 hours after Arun. Eg 2.2 Rishi Kapoor can swim a certain course against the river flow in 84 minutes; he can swim the same course with the river flow in 9 minutes less than he can swim in still water. How long would he take to swimthe course with the river flow? Solution Let us say Speed of the Rishi Kapoor in still water is ‘RK’ and Speed of the river is ‘R’. Hence, Rishi Kapoor’s speeds against the river flow, in still water and with the river flow are: RK – R, RK and RK +R. As you can see, they are in AP. Hence, the corresponding times taken will be in HP. Let us say that the time taken to row down with the stream is ‘t’, then 84, t+9 and t are in HP. So, 2 ∗ 84 ∗ 𝑡 𝑡+9 = 84 + 𝑡  𝑡 2 + 93𝑡 + 756 = 168𝑡  𝑡 2 − 75𝑡 + 756 = 0  𝑡 = 63 𝑜𝑟 12. Funda 3: Special Case Let us say that two bodies ‘a’ & ‘b’ start at the same time from two points P & Q towards each other and meet at a point R in between. After meeting at R, ‘a’ takes t a time to reach its destination (Q) and ‘b’ takes t b time to reach its destination (P); then:

𝑆𝑎 = 𝑆𝑏

𝑡𝑏 𝑡𝑎

Also, the time taken by ‘a’ & ‘b’ to meet (i.e. to reach point R from P & Q respectively) is given by: 𝑡 = 𝑡𝑎 ∗ 𝑡𝑏 Note: The same formulae will be valid if two bodies ‘a’ & ‘b’ start at different times from two points P & Q towards each other. They meet at a point R in between after travelling for ta and tb time respectively. After meeting, they take the same amount of time (‘t’) to reach their respective destinations (Q & P).

I hope that these ideas will help you reduce the number of equations that you form while solving TSD problems if not completely eliminate them. http://bit.ly/handacatcourse

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6. I got the Power – Dealing with Factorials We all know what factorials (n!) are. They look friendly and helpful but looks can be deceiving, as many quant problems have taught us. Probably it is because that Factorials are simple looking creatures, most students prefer attempting questions based on them rather than on Permutation & Combination or Probability. I will cover P&C and Probability at a later date but in today’s post I would like to discuss some fundas related to factorials, which as a matter of fact form the basis of a large number of P&C and Probability problems. Some of the factorials that might speed up your calculation are: 0! = 1; 1! = 1; 2! = 2; 3! = 6; 4! = 24; 5! = 120; 6! = 720; 7! = 5040. Funda 1: Rightmost non-zero digit of n!or R(n!) R(n!) = Last Digit of [ 2a x R(a!) x R(b!) ] where n = 5a + b Eg1.1: What is the rightmost non-zero digit of 37! ?  R (37!) = Last Digit of [ 27 x R (7!) x R (2!) ]  R (37!) = Last Digit of [ 8 x 4 x 2 ] = 4 Eg 1.2: What is the rightmost non-zero digit of 134! ?  R (134!) = Last Digit of [ 226 x R (26!) x R (4!) ]  R (134!) = Last Digit of [ 4 x R (26!) x 4 ] We need to find out R (26!) = Last Digit of [ 25 x R (5!) x R (1!) ] = Last digit of [ 2 x2x1]=4  R (134!) = Last Digit of [ 4 x 4 x 4 ] = 4 Funda 2: Power of a prime ‘p’ in a factorial (n!) The biggest power of a prime ‘p’ that divides n! (or in other words, the power of prime ‘p’ in n!)is given by the sum of quotients obtained by successive division of ‘n’ by p. Eg 2.1: What is the highest power of 7 that divides 1342!  [1342 / 7] = 191  [191 / 7] = 27  [27 / 7] = 3  Power of 7 = 191 + 27 + 3 = 221 Eg 2.2: What is the highest power of 6 that divides 134! ? As 6 is not a prime number, we will divide it into its prime factors. 3 is the bigger prime, so its power will be the limiting factor. Hence, we need to find out the power of 3 in 134!  [134/3] = 44  [44/3] = 14  [14/3] = 4 http://bit.ly/handacatcourse

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 [4/3] = 1  Power of 3 in 134! = 44 + 14 + 4 + 1 = 63 Eg 2.3: What is the highest power of 9 that divides 134! ? As 9 is not a prime number, we will divide it into its prime factors. 9 is actually 32. The number of 3s available is 63, so the number of 9s available will be [63/2] = 31. Highest power of 9 that divides 134! is 31. Highest power of 18 and 36 will also be 31. Highest power of 27 will be [63/3] = 21. Note: To find out the highest power of a composite number, always try and find out which number (or prime number) will become the limiting factor. Use that to calculate your answer. In most cases you can just look at a number and say that which one of its prime factors will be the limiting factor. If it is not obvious, then you may need to find it out for two of the prime factors. The above method can be used for doing the same. Funda 3: Number of ending zeroes in a factorial (n!) Number of zeroes is given by the sum of the quotients obtained by successive division of ‘n’ by 5. This is actually an extension of Funda 1. Number of ending zeroes is nothing else but the number of times n! is divisible by 10 or in other words, the highest power of 10 that divides n!. 10 is not a prime number and its prime factors are 2 and 5. ‘5’ becomes the limiting factor and leads to the above-mentioned idea. Eg 3.1: What is the number of ending zeroes in 134! ?  [134/5] = 26  [26/5] = 5  [5/5] = 1  Number of ending zeroes = 26 + 5 + 1 = 32 I hope that this gets you started with factorials and you might start singing this song.

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7. Box of Chocolates As an astute man Mr. Gump once said, “Life is like a box of chocolates. You never know what you gonna get.” The Permutations & Combinations that life presents us daily is baffling and probably it is because of that inherent fear of choices and cases we get intimidated by such questions in the exam. I understand that P & C is one of the dreaded topics but I hope that once you understand the fundas given below, your fear will reduce. Funda 1: De-arrangement If ‘n’ distinct items are arranged in a row, then the number of ways they can be rearranged such that none of them occupies its original position is: 𝑛!

1 1 1 1 −1 − + − +⋯ 0! 1! 2! 3! 𝑛!

𝑛

Note: De-arrangement of 1 object is not possible. Dearr(2) = 1; Dearr(3) = 2; Dearr(4) =12 – 4 + 1 = 9; Dearr(5) = 60 – 20 + 5 – 1 = 44 Eg1.1: A person has eightletters and eightaddressed envelopes corresponding to those letters.In how many ways can he put the letters in the envelopes such that exactly 5 of them get delivered correctly? Solution: At first, select the five letters that get delivered correctly. That can be done in 8C5 ways. Now, the other three must get delivered to the wrong address. That can be done in Dearr(3) = 2 ways. So, total ways is 2 x 8C5 = 2 x 56 = 112 ways. Funda 2: Partitioning ‘n’ identical items in ‘r’ distinct groups ‘n’ distinct objects in ‘r’ distinct groups

No restrictions: No group empty: No restrictions:

n+r-1 n-1

Cr-1 Cr-1

rn

Arrangement in a group is important:

𝑛 +𝑟−1 ! 𝑟−1 !

Note: Other than standard distribution / partitioning problems, these ideas can be used to solve questions in which number of solutions are asked. Eg 2.1: How many solutions are there to the equation a + b + c = 100; given that a) a, b and c are whole numbers. b) a, b and c are natural numbers. Solution: Case a) is identical to a case in which 100 identical chocolates are being distributed in three kids a, b and c. It is possible that one kid gets all the http://bit.ly/handacatcourse

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chocolates. In this case, we will use the formula for distributing ‘n’ identical items in ‘r’ distinct groups where n = 100 and r = 3. So, it can be done in 102C2 ways. Case b) is identical to a case in which 100 identical chocolates are being distributed in three kids a, b and c. Every kid must get at least one chocolate. In this case, we will use the formula for distributing ‘n’ identical items in ‘r’ distinct groups where no group is empty and n = 100 and r = 3. So, it can be done in 99C2 ways. Eg 2.2:In how many ways can you distribute 5 rings in a) 4 boxes. b) 4 fingers. Solution: First of all we need to identify the difference between distributing in boxes and distributing in 4 fingers. The distinction is that in case of fingers, unlike boxes, the order in which rings are placed matters. In Case a; Ring 1 can go in any of the four boxes, so it has four choices. Ring 2 can also go in any of the four boxes, so it has four choices. Similarly for Ring 3, Ring 4 and Ring 5; there are 4 choices each. So, the total number of ways of distribution is = 4 x 4 x 4 x 4 x 4 = 45. This is essentially how the formula rn is derived. In Case b; Ring 1 can go in any of the four fingers, so it has 4 choices. Ring 2 can go in any of the four fingers but it has five choices. There is a finger, say F3, which contains the ring R1. Now, on F3, R2 has two choices – it can go above R1 or below R1. So, the total number of choices for R2 is 5. Ring 3 can go in any of the four fingers but it now has 6 choices. Ring 4 can go in any of the four fingers but it will now have 7 choices. Ring 5 can go in any of the four fingers but it will now have 8 choices. 8! So, the total number of way of distribution of rings is = 4 x 5 x 6 x 7 x 8 = 3! This is essentially how the formula

𝑛+𝑟−1 ! 𝑟−1 !

is derived.

Funda 3: Number of ways of arranging ‘n’ items, out of which ‘p’ are alike, ‘q’ are alike and ‘r’ are alike given that p + q + r = n == Number of ways of distributing ‘n’ distinct items, in groups of size ‘p’, ‘q’ and ‘r’ given that p + q + r = n == 𝑛! 𝑝! 𝑞! 𝑟!

I hope that this would help you solve problems in the exam. May be the chocolate you end up getting is a Bournville. May be you would have earned it.

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8. Interpretation of Dreams…Oops! Data Data Interpretation questions typically have large amount of data given in the form of tables, pie charts, line graphs or some non-conventional format. The questions are calculation heavy and typically test your approximation abilities. A very large number of these questions check your ability to compare or calculate fractions and percentages. If you sit down to actually calculate the answer, you would end up spending more time than required. Here are few ideas that you can use for approximation. Funda 1 – Calculating (Approximating) Fractions When trying to calculate (approximate) a fraction ‘p/q’, add a value to the denominator and a corresponding value to the numerator before calculating (approximating). 1789

Example: What is the value of 762 Now for the denominator, either we take it close to 750 or to 800. Let’s see how it happens in both cases. Clearly the answer is between 2 and 3, so for adding values / subtracting values from the denominator or the numerator, I will consider a factor of 2.5 Case 1: 762 is 12 above 750, so I will subtract 12 from the denominator. Keeping the factor of 2.5 in mind, I will subtract 25 from the numerator. 1789 −25 1763 4 7.052 My new fraction is 762 −12 = 750 = 1763 ∗ 3000 = 3 = 2.350666 Actual answer is 2.34776. As you can see, with very little effort involved in approximation, we got really close to the actual answer. Case 2: 762 is 38 below 800, so I will add 38 to the denominator. Keeping the factor of 2.5 in mind, I will add 95 to the numerator. 1789 +95 1884 My new fraction is 762 +38 = 800 = 2.355 As you can see, even this is close to the answer. The previous one was closer because the magnitude of approximation done in the previous case was lesser. Funda 2 – Comparing Fractions If you add the same number to the numerator and denominator of a proper fraction, the value of the proper fraction increases. If you add the same number to the numerator and denominator of a proper fraction, the value of the improper fraction decreases. Note: You can remember this by keeping in mind that 1 2 3 4 3 4 5 6 < < < … 𝑎𝑛𝑑 > > > … 2 3 4 5 2 3 4 5 117 128 223

Example: Arrange the following in increasing order 229 , 239 , 449 Let’s first compare 117/229 & 128/239.

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If we add 11 to the numerator and the denominator of the first proper fraction, the resulting proper fraction would be 128/240, which will be bigger than the original. We know that 128/240 is smaller than 128/239, as the latter has a lower base. So, 117/229 < 128/240 < 128/239  117/229 < 128/239 Now let’s compare 117/229 and 223/449. If we add 11 to the numerator and the denominator of the second proper fraction, the resulting proper fraction would be 234/460, which will be bigger than the original. If we double the numerator and denominator of the first proper fraction, the resulting proper fraction would be 234/458. We know that 234/460 is smaller than 234/458, as the latter has a lower base. So, 223/449 < 234/460< 234/458  223/449 < 117/229 Using the above two results, we can say that 223/449 < 117/229 < 128/239 Note: This question can be solved much simply by just looking at the numbers and approximately comparing them with ½. I used this long explanation to illustrate the funda given above. #Given below are few other shortcuts that might come in handy in calculations. Funda 3 – Percentage Growth If the percentage growth rate is ‘r’ for a period of ‘t’ years, the overall growth rate 𝑡 𝑡−1 is approximately: 𝑟𝑡 + 2 𝑟 2 Note: Derived from the Binomial theorem, this approximation technique works best, when the value of r is small. If the rate goes above 10%, then this approximation technique gives bad results. Also, if the rate is 5% then r = 0.05; if the rate is 7.2% then r = 0.072. Funda 4 –Comparing Powers Given that natural numbers a > b > 1, a b will always be less than ba Note: There are only two exceptions to this funda. I hope someone in the comments will point them out. As I end this post, I am also wondering if “Approximation of Dreams… Oops! Data” would have been a better title.

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9. Look Through The Glass – Logical Reasoning ‘If it was so, it might be; and if it were so, it would be; but as it isn't, it ain't. That's logic.’ – Tweedledee in Lewis Caroll’sThrough the Looking Glass. If the above line confused you, trust me – you are not alone. Even God can vanish in a puff of logic. To know how, you can probably jump to the end of this post. To those who choose not to skip – let us discuss few common types of Logical Reasoning problems. Type 1: Cube problems A cube is given with anedge of unit ‘N’. It is painted on all faces. It is cut into smaller cubes of edge of unit ‘n’. How many cubes will have ‘x’ faces painted? In these types of questions, the first thing that we need to figure out is the number of smaller cubes. For this, we look at one particular edge of the big cube and figure out how many smaller cubes can fit into this. It will be N/n. So, the number of smaller cubes will

𝑁 3 𝑛

.

A cube has 6 faces and none of the smaller cubes will have all faces painted. As a matter of fact, none of the smaller cubes will have even 5 or 4 faces painted. The maximum number of faces, which will be painted on a smaller cube, will be 3. This will happen only in the case of the smaller cubes that emerge from the corners of the big cube. So, number of smaller cubes with 3 faces painted = 8 (Always) For 2 faces to be painted, we will have to consider the smaller cubes that emerge from the edges of the big cube (leaving out the corners). So, the smaller cubes on 𝑁−2𝑛 every edge will be 𝑛 . There are 12 edges in a cube. So, number of smaller cubes with 2 faces painted =

𝑁−2𝑛 𝑛

∗ 12

For 1 face to be painted, we will have to consider the smaller cubes that emerge from the face of the big cube (leaving out the corners and the edges). So, the smaller cubes on every face will be

𝑁−2𝑛 2 𝑛

. There are 6 faces in a cube.

So, number of smaller cubes with 1 face painted =

𝑁−2𝑛 2 𝑛

∗6

For no face to be painted, we will have to consider the smaller cubes that emerge from the inside of the big cube (leaving out the outer surface which was painted). Imagine this as taking a knife and cutting a slice of width ‘n’ from every face of the cube. You will be left with a smaller cube with an edge of ‘N-2n’. Number of smaller cubes that you can make from the resulting cube is So, number of smaller cubes with 0 face painted =

𝑁−2𝑛 3

𝑁−2𝑛 3 𝑛

𝑛

Let us take an example to elucidate this type of problem. http://bit.ly/handacatcourse

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Eg 1: A painted cube is given with an edge of 15 cm. Smaller cubes are cut out from it with an edge of 3 cm each. How many cubes will have 3 faces painted, 2 faces painted, 1 face painted and no face painted. Solution: Total number of smaller cubes = (15/5)3 = 125 3 faces painted = 8 cubes. 2 faces painted: Consider an edge of size 15 cm. We have removed the corners that take away 3 cm from each corner of the edge. Now our edge is of 9 cm. 3 cubes of 3 cm each can come from it. There are 12 edges. So, there will be 3 x 12 = 36 cubes. 1 face painted: Consider a face. If we have removed 3 cm from each edge of the face, we will be left with a square of side 9 cm or area 81 sq cm. There can be 9 smaller squares that can be formed on that face with an area of 9 sq cm each. These 9 will be the cubes which will have 1 face painted. There are 6 faces. So, there will be 9 x 6 = 54 cubes. No face painted: Cut slices of 3 cm each from each face of the cube. We will be left with a smaller cube of edge 9 cm. Number of smaller cubes that can be formed from it is (9/3)3 = 27. So, 27 cubes will have no faces painted. You can use this to verify the formulas above and also note that 8 + 36 + 54 + 27 = 125. This means that there is no need to find out all four using the formula, just find any three of them and the other would emerge by using the total. In an exam, this might save you some valuable time. Type 2: Matchstick Game You are playing a matchstick game with Mr. Bond. There are ‘n’ matchsticks on a table. On a player’s turn, he can pick any number of matchsticks upto ‘p’ (p is typically quite smaller than ‘n’). Whosoever picks the last matchstick loses the game. It is your turn first. How many matchsticks should you pick (assuming that you are smart and will play to win) that you will always win? First remove 1 matchstick from consideration, as that would be the matchstick that Mr. Bond will pick and lose the game. 𝑛−1 Find out Remainder 𝑝+1 = 𝑞 You should pick ‘q’ matchsticks in the first turn. After that if Mr. Bond picks ‘r’ sticks, you should pick ‘p+1-r’ sticks and you will win the game. Once again, let us take an example. Eg 2: There are 105 matchsticks on a table and a player can pick any number of matchsticks from 1 to 10. The person who picks the last matchstick loses the game. You are playing the game against Mr. Bond and it is your turn first. How many matchsticks should you pick in the first turn such that you always win the game?

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105 −1

Solution: You should pick Remainder 10+1 = 5 matchsticks to win the game. Let us look at few scenarios, in which you have picked 5 sticks and there are 100 sticks left on the table. It is Mr. Bond’s turn now. Round ID Round 1 Round 2 Round 3 Round 4 Round 5 Round 6 Round 7 Round 8 Round 9

Mr. Bond Picks 5 8 7 4 10 8 1 2 9

Sticks Left 100 – 5 = 95 89 – 8 = 81 78 – 7 = 71 67 – 4 = 63 56 – 10 = 46 45 – 8 = 37 34 – 1 = 33 23 – 2 =21 12 – 9 = 3

You Pick 10 + 1 – 5 = 6 10 + 1 – 8 = 3 10 + 1 – 7 = 4 10 + 1 – 4 = 7 10 + 1 – 10 = 1 10 + 1 – 8 = 3 10 + 1 – 1 = 10 10 + 1 – 2 = 9 10 + 1 – 9 = 2

Sticks Left 95 – 6 = 89 81 – 3 = 78 71 – 4 = 67 63 – 7 = 56 46 – 1 = 45 37 – 3 = 34 33 – 10 = 23 21 – 9 = 12 3–2=1

As only 1 stick is left, Mr. Bond will have to pick it and lose the game. I recommend you to try out such scenarios with a friend. Nothing validates a concept more than a real-life implementation, especially if it is on a bet.  For those who are still wondering – what did just happen (as I did when I first read this concept), I suggest you pick up Hitchhiker’s Guide to the Galaxy and read about how God vanished in a puff of logic.

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10. Figuring Out Remainders Figuring out the last digit is the same as figuring out the remainder of a number when divided by 10, but I guess you already know that. Figuring out the last two digits is the same as figuring out the remainder of a number when divided by 100. Even if you don't already know how to do that, you can always read this post. However, if you wish to figure the remainder when the divisor is not 10 or 100, I suggest you read on. Funda 1: Basic idea of remainders can be used to solve complicated problems. 𝑅𝑒𝑚

𝑎 ∗𝑏∗𝑐… 𝑎 𝑏 𝑐 = 𝑅𝑒𝑚 ∗ 𝑅𝑒𝑚 ∗ 𝑅𝑒𝑚 … 𝑑 𝑑 𝑑 𝑑 𝑎+𝑏+𝑐… 𝑎 𝑏 𝑐 = 𝑅𝑒𝑚 + 𝑅𝑒𝑚 + 𝑅𝑒𝑚 … 𝑑 𝑑 𝑑 𝑑

𝑅𝑒𝑚

There is nothing special or unique about this idea. At first glance it seems like something really obvious. But it is its usage that makes it special and helpful in questions related to remainders. Let us look at couple of examples to see how this can be used effectively. In the first example we will see the idea that will work in cases of a b and in the second example 𝑐 we will see the idea that will work in case of 𝑎𝑏 Example 1: Find out the remainder when 2525 is divided by 7. 𝑅𝑒𝑚  𝑅𝑒𝑚  𝑅𝑒𝑚

2525 25 25 25 = 𝑅𝑒𝑚 ∗ 𝑅𝑒𝑚 ∗ 𝑅𝑒𝑚 … (25 𝑡𝑖𝑚𝑒𝑠) 7 7 7 7

25 25 7 25 25 7

= 4 ∗ 4 ∗ 4 … 25 𝑡𝑖𝑚𝑒𝑠 = 425 = 424 ∗ 4 = 43 = 648 ∗ 4 = 𝑅𝑒𝑚

64

8

7

Example 2: Find out the remainder when 2526 27

8

∗4

∗ 4 = 18 ∗ 4 = 4

27

is divided by 7. 27

2526 426 𝑅𝑒𝑚 = 𝑅𝑒𝑚 7 7 3 We know that 4 gives a remainder of 1, when divided by 7. So, if we have 43k, it will give a remainder of 1 when divided by 7 If we have 43k+1, it will give a remainder of 4 when divided by 7 If we have 43k+2, it will give a remainder of 2 when divided by 7 So, we need to reduce 2627as 3k or 3k+1 or 3k+2. If we can do that, we will know the answer. So our task has now been reduced to figuring out 𝑅𝑒𝑚 𝑅𝑒𝑚

26 27 3

= 𝑅𝑒𝑚

−1 27 3

26 27 3

= −1 = 2

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 𝑅𝑒𝑚

25 26

27

= 𝑅𝑒𝑚

7

4 26

27

7

4 3𝑘 +2

= 𝑅𝑒𝑚

7

=2

Note: As you can see in solving this example, we have used the concept of negative remainder. In some cases, using the negative remainder can reduce your calculations significantly. It is recommended that you practice some questions using negative remainders instead of positive ones. Funda 2: While trying to find out the remainder, if the dividend (M) and the divisor (N) have a factor (k) in common; then  Cancel out the common factor  Find out the remainder from the remaining fraction  Multiply the resulting remainder with the common factor to get the actual remainder In equation format, this can be written as: 𝑀 𝑘𝑚 𝑚 𝑅𝑒𝑚 𝑁 = 𝑅𝑒𝑚 𝑘𝑛 = 𝑘 𝑅𝑒𝑚 𝑛 Example: Find out the remainder when 4 15 is divided by 28. 415 4 ∗ 414 414 43𝑘 +2 𝑅𝑒𝑚 = 𝑅𝑒𝑚 = 4 ∗ 𝑅𝑒𝑚 = 4 ∗ 𝑅𝑒𝑚 = 4∗2= 8 28 4∗7 7 7 Funda 3: While trying to find out the remainder, if the divisor can be broken down into smaller co-prime factors; then  𝑅𝑒𝑚

𝑀 𝑁

 Let 𝑅𝑒𝑚 𝑹𝒆𝒎

𝑴 𝑵

= 𝑅𝑒𝑚 𝑀 𝑎

𝑀

{HCF (a,b) = 1}

𝑎 ∗𝑏

= 𝑟1 & 𝑅𝑒𝑚

𝑀 𝑏

= 𝑟2

= 𝒂𝒙𝒓𝟐 + 𝒃𝒚𝒓𝟏 {Such that ax+by = 1}

Note: If you wish to read more about it and how it happens, I suggest you read about the Chinese Remainder Theorem. Example: Find out the remainder when 715 is divided by 15. 𝑅𝑒𝑚

715 15

= 𝑅𝑒𝑚

 𝑅𝑒𝑚  𝑅𝑒𝑚

715 3 715 15

715 3∗5

= 1 & 𝑅𝑒𝑚

715 5

= 𝑅𝑒𝑚

= 3∗𝑥∗3+5∗𝑦∗1

215 5

=3 {Such that 3x+5y=1}

 Valid values are x = -3 and y = 2  𝑅𝑒𝑚

715 15

= 9𝑥 + 5𝑦 = −27 + 10 = −17 ≡ 13

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By using the above fundas, solving remainder problems will get a little easier. But if you are thinking, that this is all you need to know to solve remainder problems in CAT – I beg to differ. Great mathematicians like Euler, Fermat&Wilson developed some theorems that come in handy while solving remainder questions. We will discuss these in my next post.

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11. Remainders Reloaded In previous posts, we have already discussed how to find out the last two digits and basic ideas of remainders. However, there are theorems by Euler, Fermat & Wilson that make calculation of remainders easier. Let’s have a look at them. Funda 1 – Euler: Number of numbers which are less than N = 𝑎𝑝 ∗ 𝑏𝑞 ∗ 𝑐 𝑟 and co-prime to it are 1

1

 ∅ 𝑁 = 𝑁 1−𝑎

1

1−𝑏

1−𝑐

If M and N are co-prime ie HCF(M,N) = 1  𝑅𝑒𝑚

𝑀∅ 𝑁 𝑁

=1

A very common mistake that students tend to make while using Euler’s Theorem in solving questions is that they forget M and N have to be co-prime to each other. There is another set of students (like me in college) who don’t even understand what to do with the theorem or how to use it solving questions. Let us look at couple of examples in which Euler’s Theorem is used. Note: ∅ 𝑁 is also known as Euler’s Totient Function. Example 1: 𝑅𝑒𝑚

750 90

=? 1

1

 ∅ 90 = 90 1 − 2 1

2

1−3

1

1−5

4

 ∅ 90 = 90 ∗ 2 ∗ 3 ∗ 5 = 24  𝑅𝑒𝑚  𝑅𝑒𝑚

724

= 1 = 𝑅𝑒𝑚

90 750

= 𝑅𝑒𝑚

90

72 90

748 90

∗ 𝑅𝑒𝑚

748 90

= 49 ∗ 1 = 49

Funda 2 – Fermat: If ‘p’ is a prime number and ‘a’ and ‘p’ are co-primes  𝑅𝑒𝑚  𝑅𝑒𝑚

𝑎𝑝 𝑝

=𝑎

𝑎 𝑝 −1 𝑝

=1

 (ap – a) will be divisible by p. If you notice, the three statements above are saying the exact same thing but in a different way. It is important to keep all three in mind because sometimes it http://bit.ly/handacatcourse

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becomes a little difficult to analyze which interpretation of Fermat’s little theorem is to be used. A simple illustration of this would be: 107 106 𝑅𝑒𝑚 = 10𝑂𝑅 𝑅𝑒𝑚 = 1 𝑜𝑟 107 − 10 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑏𝑙𝑒 𝑏𝑦 7 7 7 We can check it by noticing that 107 = 10000000 = 9999990 + 10 = 142857 ∗ 7 + 10 Another way that you can remember Fermat’s Little Theorem (I am not joking, that is the official name – check this) is that it is a special case of Euler’s Theorem where ‘N’ is a prime number. Because, if ‘N’ is prime then∅ 𝑁 or the Euler’s Totient Function will always be (N-1) Funda 3: Wilson Sometimes people find the history behind Wilson’s theorem to be more interesting than the theorem itself. Actually, it was know to the great Muslim polymath Alhazen approximately seven and a half centuries before John Wilson was born. Alhazen, being the great scientist that he was, never bothered to prove it and tried to regulate the floods in river Nile. After being ordered by Al-Hakim bi-Amr Allah, the sixth ruler of the Fatimid caliphate, to carry out this operation, Alhazen quickly perceived the impossibility of what he was attempting to do, and retired from engineering. Fearing for his life, he feigned madnessand was placed under house arrest, during and after which he devoted himself to his scientific work until his death. The English mathematician, John Wilson, stated it in the 18 th century but he could not prove it either. Actually Wilson was a student of Edward Waring, who announced the theorem in 1770. None of them could prove it. Lagrange proved it in 1771. There is evidence that Leibniz was also aware of the result a century earlier, but he never published it. I think I will end the history lesson here and continue with the mathematical part. For a prime number ‘p’ 𝑅𝑒𝑚

𝑝−1 ! = 𝑝−1 𝑝

Another related result to the Wilson Theorem is: 𝑅𝑒𝑚

𝑝−2 ! 𝑝

= 1

Example: 10! 9! = 10 & 𝑅𝑒𝑚 = 1 11 11 82! 81! 𝑅𝑒𝑚 = 82& 𝑅𝑒𝑚 = 1 83 83 1008! 1007! 𝑅𝑒𝑚 = 1008& 𝑅𝑒𝑚 = 1 1009 1009 𝑅𝑒𝑚

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Note: I have checked the related result for primes up to 120 and found it to be valid. I could not find a proof for it that I could understand. Do note that the key part of the previous sentence is not ‘find a proof for it’ but ‘that I could understand’. May be one of you can help me out in comments. I also recommend that while trying these ideas or any other remainder questions, keep Wolfram Alpha open.

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12. Games & Tournaments In Logical Reasoning very often we encounter problems based on games of tournaments. The first thing that as a CAT taker you need to realize is that such tournament based format offers the examiner a multitude of options. So, there cannot be a set formula for solving such kind of questions. However, if you look at the CAT papers of past few years – a pattern seems to emerge. Let us discuss couple of them. Type 1: The questions are typically in a set where the data will be either in the standard tabular format or a format which you would never find on Cricinfo or for that matter any other ESPN website. The ‘different for the sake of being different’ format essentially tests a CAT taker’s ability to infer data in newer formats. An example of this would be: Each diagram communicates the number of runs scored by the three top scores from India, where K, R, S, V, and Y represent Kaif, Rahul, Saurav, Virender, and Yuvraj respectively. The % in each diagram denotes the percentage of total score that was scored by the top three Indian scorers in that game.

You can get the complete set of questions from here. I will not get into the detail of solving this particular set. Once you interpret the information, the questions are really simple. The catch in this question (this type of questions) is to interpret the given data. Let us look at this information in a different way:

Runs by Kaif Runs by Rahul Runs by Saurav Runs by Virender Runs by Yuvraj Runs by Top 3 Top 3 as a %age of total Total runs http://bit.ly/handacatcourse

Pakistan 28

South Africa 51 49 75

130 40 90%

Australia 55 50 87

70%

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As you can see, the triangular format is no different from providing the same information to the student in a table. It is just a little more intimidating in a pressure situation – and that intimidation is exactly what you should avoid. With some very simple addition and calculation you will be solve this problem set. Bottom line: Even if it takes a couple of minutes, it is best to represent information in a format that you are comfortable with. Type 2: For some reason, Tennis appears to be a favorite among exam setters. Actually Tennis does offer some very interesting possibilities – such as seeds, an unconventional scoring and the knockout feature. Knockouts are inherent in the sport of Tennis and hence used frequently by exam setters. Note: In a knockout tournament, No. of matches = No. of players – 1 Let us look at few ideas related to questions on seeded players. Let’s say in a tournament there are ‘n’ players and they are seeded (ranked) from 1 to n. Typically this ‘n’ is a power of 2 like 32 or 64 or 128. In the first round the highest seeded player plays the lowest seeded player, the second highest seeded player plays the second lowest seeded player and so on. To put it into perspective: Round 1 – Match 1 – Seed 1 Vs Seed n Round 1 – Match 2 – Seed 2 Vs Seed (n-1) Round 1 – Match 3 – Seed 3 Vs Seed (n-2) . . . Round 1 – Match n/2 – Seed n/2 Vs Seed n/2 + 1 In the second round, winner of Match 1 plays winner of Match n/2; winner of Match 2 plays winner of Match n/2 – 1 and so on. In this kind of questions, an ‘upset’ comes into the picture which essentially means that a lower seeded plays beat a higher seeded player. The questions are typically of the format: Ques: Who will play match 36 in Round 1? Ans: It will be played between the 36th highest seed and the 36th lowest seed. The 36th lowest seed can be sometimes difficult to figure out but you can figure it out easily by calculating (n+1) – 36. Note: The rth match in Round 1 will be played between Seed ‘r’ and Seed ‘n+1-r’ Ques: If there are no upsets, then in Round 2 who will play the 5 th match? Ans: One way of solving this question would be figuring out the winners of Round 1 and then figuring out the 5th from the top and the bottom. http://bit.ly/handacatcourse

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If there is no upset, then seed 5 will be there. The other player would be (n/2+1 – 5) Note: If there are no upsets, then the rth match in the pth round will be played between – 𝑛 Seed ‘r’ and Seed 2𝑝 −1 + 1 − 𝑟 Ques: Who will meet Seed 37 in the Quarterfinals of a tournament in which 64 players are taking part?Other than Seed 37’s matches, there were no other upsets. We first need to analyze which round would be the quarterfinal: Round 1 (32 matches), Round 2 (16 matches), Round 3 (8 matches – prequarter), Round 4 (4 matches – quarterfinals). In Round 1, Seed 37 must have defeated 64 + 1 – 37 = 28 In Round 2, Seed 37 played the match that Seed 28 would have played. Seed 28 would have played against Seed 32 + 1 – 28 = Seed 5 In Round 3(pre-quarters), Seed 37 played the match that Seed 5 would have played. Seed 5 would have played against Seed 16 + 1 – 5 = Seed 12 and won it. In Round 4 (quarterfinals), Seed 37 would meet the player that Seed 5 would have met. Seed 5 would have met 8+1-5 = 4. Hence, Seed 37 will meet Seed 4 in the quarterfinals. As a matter of fact, even the above solution is not the most optimal one. Because once you realize that Seed 37 defeated Seed 5, he would keep meeting the opponents that Seed 5 would have met. I think I have taken enough of your time with this lengthy post but this time would be well spent if such a question appears in CAT or some other management entrance exam.

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13. Games & Tournaments – Part 2 In my previous post, we discussed about two of the popular types of questions when it comes to games and tournament questions. So, if you are looking for questions on new types of data representation or questions based on seeding in a tennis tournament, probably you should read that. However, there is another popular type of questions with respect to Games & Tournaments and that is – Football / Hockey tournament questions in which we have to find out Goals scores, winners, ties, etc. In such tournaments, all competitors play a fixed number of matches. Points are awarded for wins / draws / losses. Then an overall ranking is decided by total points or average points per match. Sometimes other factors such as goals scored / goals faced also come into the picture to resolves ties in ranking. Let us look at a question from CAT 2000. (Full set of questions here) Question: “Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament was conducted in two stages. In the first stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of first stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprised several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated. The team that remains undefeated in the second stage is declared the winner and claims the Gold Cup. The tournament rules such that each match results in a winner and a loser with no possibility of a tie. In the first stage, a team earns one point for each win and no points for a loss. At the end of the first stage, teams in each group are ranked on the basis of total points to determine the qualifiers advancing to the next stage. Ties are resolved by a series of complex tie-breaking rules so that exactly four teams form each group advanced to the next stage.” Now questions were asked on – Total number of matches, minimum number of wins required for a team to guarantee advance (or possible advance) to next stage, maximum number of matches that a team can win in the first stage without advancing, etc. In first stage, teams are divided into two groups of 8 teams each. There they play a match against everyone exactly one ie 8C2 matches in every group. So 2*8C2 = 56 matches for the first stage. In second stage, there are 8 teams in a knockout stage. There will be one winner, so 8 – 1 = 7 So, total number of matches is 56 + 7 = 63 For a team to advance to the second stage, it should be among the top 4 in its group. Total points on stake in a group is the same as the total number of http://bit.ly/handacatcourse

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matches which is 8C2 = 28. To guarantee advance, it can have 3 teams with the same or more points. There can be 5 teams with 5 wins or 5 points. So, 5 wins is not good enough to ensure a birth in round 2. However, 6 wins will guarantee its advance. This also tells us that a team might have 5 wins but still not advance. To figure out the minimum wins required to possibly advance, let us look at the method for ‘n’ teams. n/2 – 1 teams should win maximum no. of matches. n/2 + 1 teams should have exactly the same number of wins. So in this question, the top 3 teams can have a maximum of 7 + 6 + 5 = 18 points. All other teams (5) have a combined score of 28 – 18 = 10 points. Their individual score is 2 points each and one of these five teams will advance to second stage. So, minimum wins required to advance is 2. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Let us look at another type of question in which we are given a table and we have to fill it. Given below is a random table at the end of hockey tournament. For each win two points were awarded and for a draw one point was given. We also know that the South Africa – Spain match was a draw. No two teams have the exact same count for Win/Draw/Loss and Australia has won more matches than Spain. Figure out the result of every match from the table given below: Team Name India Pakistan Australia Spain South Africa

Played Won

Draw

Lost 0

Points 6 6 3 3

Each team has played 4 matches. A team can get a score of 6 in two ways: 3 Wins and 1 loss or 2 wins and 2 draws. India did not lose, so it will have 2 wins and 2 draws whereas on the other hand Pakistan will have 3 wins. No. of matches played will be 5C2 = 10 The total no. of points at stake is 20. South Africa has the left-over points which is 2. We also know that the South Africa – Spain match was a draw.

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So, now our table looks like this: Team Name India Pakistan Australia Spain South Africa

Played Won 4 2 4 3 4 4 4

Draw 2 0

Lost 0 1

Min(1) Min(1)

Points 6 6 3 3 2

2 Points can be achieved by 1 Win, 0 Draw, 3 Loss OR 2 Draw, 2 Loss. We know that both Spain & South Africa have at least 1 Draw. This means that South Africa’s 2 points are by 2 Draw, 2 Loss Team Name India Pakistan Australia Spain South Africa

Played 4 4 4 4 4

Won 2 3

Draw 2 0

Lost 0 1

0

Min(1) 2 2

Points 6 6 3 3 2

3 Points can be achieved by 1 Win, 1 Draw, 2 Loss OR 3 Draw, 1 Loss. As no two teams have the same Win/Draw/Loss count, one of the above applies to Australia whereas the other one applies to Spain. As Australia has won more matches, it will get the 1 Win, 1 Draw, 2 Loss. Our final table will look like this: Team Name India Pakistan Australia Spain South Africa

Played 4 4 4 4 4

Won 2 3 1 0 0

Draw 2 0 1 3 2

Lost 0 1 2 1 2

Points 6 6 3 3 2

Now, let us try and analyze the match results for the 10 matches (in no special order): Pakistan has won 3 and lost 1. Pakistan cannot win against India as India did not lose a match. So, Match 1 – India beat Pakistan Match 2 – Pakistan beat Australia Match 3 – Pakistan beat Spain Match 4 – Pakistan beat South Africa We also know the result of South Africa Vs Spain Match 5 – South Africa & Spain drew the match. http://bit.ly/handacatcourse

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Spain has lost against Pakistan and it needs to draw all other matches. Match 6 – India & Spain drew the match Match 7 – Australia & Spain drew the match Australia cannot draw another match as it has only 1 draw. It cannot win against India as India has no losses. So, it must have lost against India and the win must have come against the remaining team ie South Africa. Match 8 – India beat Australia Match 9 – Australia beat South Africa The only match remaining between India & South Africa must have been a draw as India scored wins against Pakistan and Australia. Match 10 – India & South Africa drew the match >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Phew!! I hope that you lasted this long without actually playing the game of “Banging head against the wall”.

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14. Set Theory – Maxima & Minima The conceptsof Set Theory are applicable not only in Quant / DI / LR but they can be used to solve syllogism questions as well. Let us first understand the basics of the Venn Diagram before we move on to the concept of maximum and minimum. A large number of students get confused in this so I have listed out each area separately. A venn diagram is used to visually represent the relationship between various sets. What do each of the areas in the figure represent?

I – only A; II – A and B but not C; III – Only B; IV – A and C but not B; V – A and B and C; VI – B and C but not A; VII – Only C n(A∪B∪C) = n(A) + n(B) + n(C) — n(A∩B) — n(A∩C) - n(B∩C) + n(A∩B∩C) As per the above diagram, n(A) = I + II + IV + V n(B) = II + III + V + VI n(C) = IV + V + VI + VII n(A∩B) = II + V n(B∩C) = V + VI n(C∩A) = IV + V n(A∩B∩C) = V n(A∪B∪C) = I + II + III + IV + V + VI + VII Note: While doing such questions, it is advisable that you take the least no. of variables to fill up the empty space. As a practice if n(A∩B∩C) is missing, take that as ‘x’ and proceed. http://bit.ly/handacatcourse

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For Maximum and Minimum of values, the key point to note is: If you allot the a value to the intersection, it will get added to all the indivudal sets but will bring down the total. Example: In a survey it was found that 80% like tea whereas 70% like coffee. What is the maximum and minimum number of those who like both? Ans: First thing to note is that no information is mentioned about the people who don’t like either of them. So that value is flexible and can change. n(tea) = 80 n(coffee) = 70 n(total) = 100 {This includes those who like neither.} n(tea U coffee) = ??? {We don’t know this value and it is flexible} If we want to maximize those who like both, we have to maximize the value in the intersection. So, we have to minimize the value of the union. n(tea ∩ coffee)max = 70 {It is limited by the higher of the two values} In this case, our venn diagram will look something like this (Red is tea & Purple is coffee):

In this case 20% of people like neither tea nor coffee.

If we want to minimize those who like both, we have to minimize the value in the intersection. So, we have to maximize the value of the union. We know that the maximum possible value of the union ie n(tea U coffee) = 100 So, we need to figure out the surplus n(tea) + n(coffee) = 80 + 70 = 150. The surplus is = 150 – 100 = 50 So, the value of the intersection = value of the surplus = 50 This could have also been obtained by the formula n(a U b) = n(a) + n(b) – n (a ∩ b)

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In this case, our venn diagram will look something like this (Red is tea & Purple is coffee):

In this case, there is no one who likes neither coffee nor tea. Let us look at a slightly more complicated problem when we have to deal with three sets and the value of union of the sets is fixed. Example: In a survey it was found that 40 % like tea, 50 % like coffee and 60 % like milk. Every person likes at least one of the three items – tea / coffee / milk. What are the maximum and minimum possible values of those who like all three? Solution: Currently our Venn Diagram looks like this:

First of all, we need to figure out the surplus. Surplus = 40 +50 + 60 – 100 = 50 The surplus should be taken care of by adding to the intersection of all three or any of the two. If we want to maximize those who like all three, we need to maximize the intersection of all three. Adding ‘1’ to the intersection of all three takes care of a surplus of ‘2’. To take care of a surplus of 50, we need to make n (tea ∩ coffee ∩milk) = 25 Note: If the union of the sets was not fixed i.e. the line ‘Every person likes at least one of the three items – tea / coffee / milk’ was not given in the question then the answer would have been 40.

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Our venn diagram will now look like this:

If we want to minimize those who like all three, we need to minimize the intersection of all three. But we have to take care of the surplus of 50. We can do that adding them to the intersection of any two of them. Adding ‘1’ to the intersection of two sets, takes care of a surplus of ‘1’. So, n (tea ∩ coffee ∩milk)min = 0 We can take care of the surplus 50 in many ways by adding them in any order to the intersection of two sets. Three of those many ways are given below in Venn Diagrams.

I hope that this will help you in solving problems related to Set Theory.

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15. Geometry Funda Part-I I got a lot of feedback via comments on my previous posts that people are looking for a post on Geometry. I have been avoiding it for sometime because of two main reasons: a) It is not one of my strongareas. b) It takes a lot of time to draw the diagrams that are sometimes required to explain the fundas. With Diwali and CAT approaching I realized that I could not procrastinate any more because Geometry, as you would agree, is a very important part of CAT-prep. I have compiled a list of fundas that you might find helpful in solving CAT level questions. I am splitting those in two posts so that one post does not become too long / intimidating. In this post, we will discuss Geometry fundas related to lines, triangles, parallelograms, trapeziums, polygons, etc. You might find some of them very simple or ideas that are obvious to you. If that is the case, be glad that your prep is up to the mark. If not, then be glad you got them in time. (Yes – I am inspired by two-face) Funda 1:

The ratio of intercepts formed by a transversal intersecting three parallel lines is equal to the ratio of corresponding intercepts formed by any other transversal. 

𝑎 𝑏

𝑐

𝑒

=𝑑 =𝑓

Funda 2: Centroid and Incenter will always lie inside the triangle. About the other points: -

For an acute angled triangle, the Circumcenter and the Orthocenter will lie inside the triangle. For an obtuse angled triangle, the Circumcenter and the Orthocenter will lie outside the triangle. For a right-angled triangle,the Circumcenter will lie at the midpoint of the hypotenuse and the Orthocenter will lie at the vertex at which the angle is 90°.

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Funda 3: The orthocenter, centroid, and circumcenter always lie on the same line known as Euler Line. - The orthocenter is twice as far from the centroid as the circumcenter is. - If the triangle is Isosceles then the incenter lies on the same line. - If the triangle is equilateral, all four are the same point. Funda 4: Appolonius’ Theorem {AD is the median}

AB2 + AC2 = 2 (AD2 + BD2) Funda 5: For cyclic quadrilaterals –

Area = 𝑠 − 𝑎 𝑠 − 𝑏 𝑠 − 𝑐 (𝑠 − 𝑑) where s is the semi perimeter 𝑠 = Also, Sum of product of opposite sides = Product of diagonals  ac + bd = PR x QS

𝑎 +𝑏+𝑐+𝑑 2

Funda 6:

If a circle can be inscribed in a quadrilateral, its area is given by = 𝑎𝑏𝑐𝑑 http://bit.ly/handacatcourse

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Funda 7: Parallelograms 

   

A parallelogram inscribed in a circle is always a Rectangle. A parallelogram circumscribed about a circle is always a Rhombus. So, a parallelogram that can be circumscribed about a circle and in which a circle can be inscribed will be a Square. Each diagonal divides a parallelogram in two triangles of equal area. Sum of squares of diagonals = Sum of squares of four sides AC2 + BD2 = AB2 + BC2 + CD2 + DA2 A Rectangle is formed by intersection of the four angle bisectors of a parallelogram. From all quadrilaterals with a given area, the square has the least perimeter. For all quadrilaterals with a given perimeter, the square has the greatest area.

Funda 8: Trapeziums 

 

Sum of the squares of the length of the diagonals = Sum of squares of lateral sides + 2 Product of bases. AC2 + BD2 = AD2 + BC2 + 2 x AB x CD If a trapezium is inscribed in a circle, it has to be an isosceles trapezium. If a circle can be inscribed in a trapezium, Sum of parallel sides = Sum of lateral sides.

Funda 9:  

A regular hexagon can be considered as a combination of six equilateral triangles. All regular polygons can be considered as a combination of ‘n’ isosceles triangles.

I will wrap up this post here. In my next and final post on Geometry we will discuss fundas related to circles (specifically – common tangents), solid figures, mensuration and co-ordinate geometry.

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16. Geometry Funda Part-II This is my second post on PG with respect to Geometry Fundas. We discussed lines, triangles, parallelograms, trapeziums, polygons, etc. in my previous post. Let us look at few of the fundas / formulae in Geometry that are often neglected by students and can fetch some crucial marks in the exam. Funda 1: Angle made by Secants

θ = ½ [ m(Arc AC) – m(Arc BD) ]

θ = ½ [ m(Arc AC) + m(Arc BD) ]

In both these cases, PA * PB = PC * PD

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Funda 2: Common Tangents

Two Circles

No. of Common Tangents

Distance Between Centers (d)

One is completely inside other

0

r1 + r2

Length of the Direct Common Tangents (DCT)  AD = BC = 𝑑 2 − (𝑟1 − 𝑟2 )2 Length of the Transverse Common Tangent (TCT)  RT = SU = 𝑑 2 − 𝑟1 + 𝑟2 2 Note: The two centers(O and O’), point of intersection of DCTs (P)and point of intersection of TCTs (Q) are collinear. Q divides OO’ in the ratio r1 : r2 internally whearea P divides OO’ in the ratio r1 : r2 externally.

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Funda 3:Solids  If a sphere is inscribed in a cube of side a, the radius of the sphere will be a/2. If a sphere is circumscribed about a cube of side a, the radius of the sphere will be 3 a /2.  If a largest possible sphere is inscribed in a cylinder of radius ‘a’ and height h, its radius r will be o r = h/2 {If 2a > h} o r=a {If 2a < h}  If a largest possible sphere is inscribed in a cone of radius r and slant height equal to 2r, then the radius of sphere = r/ 3  If a cube is inscribed in a hemisphere of radius r, then the edge of the cube = r

2 3

Funda 4: Co-ordinate Geometry  The X axis divides the line joining P(x1,y1) and Q(x2,y2) in the ratio of y1 : y2  The Y axis divides the line joining P(x1,y1) and Q(x2,y2) in the ratio of x1 : x2  If we know three points A(x1,y1), B(x2,y2 ) and C(x3,y3) of a parallelogram, the fourth point is given by o (x1 + x3 – x2, y1 + y3 – y2) With this I will like to wrap up this post on Geometry. Best of Luck to all of you for CAT!

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17 –Functions XAT’s Quant is always a little bit on the tougher side. It is said that the paper would be do-able and the level of difficulty will see a dip. That does not mean that the difficulty level would suddenly drop to the standard of elementary mathematics. XAT traditionally focuses more on topics like functions, probability, permutation & combination, etc. more than the CAT exam. In this post we will discuss some basic tips about functions and how graphs of functions change. Let us see what the function y = f(x) = x3 + 7 looks like:

When we do y = -f (x) we will have to reflect the graph in the x-axis. Given below is what it is going to look like. As you can see, this image would be obtained if we had put a mirror on the X-axis.

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When we do y = f(-x) = -x3 + 7 we will have to reflect the graph in the y-axis. Given below is what it is going to look like. As you can see, this image would be obtained if we had put a mirror on the Y-Axis

We also know that for even functions, f(x) = f(-x), so their graphs would be identical in nature. We can also say that a function is even if its mirror image in the Y-axis is identical to the original. Given below are the graphs of even functionscos(x) and cos(-x)

We also know that for odd functions, f(-x) = - f (x) The graph of an odd function has rotational symmetry with respect to the origin, meaning that its graph remains unchanged after rotation of 180 degrees about the origin. This means that if you reflect the graph of an odd function first in the X-axis and then in the Y-axis, the resultant graph would be same as the identical.

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Let us check this out: y = f(x) = x + 5 sin(x)

To find out its reflection in the X-axis, we will need to y = - f(x) = - [x + 5 sin(x)]

As you can see, if you reflect the above graph in Y-axis, you will get back the original. In the modifications discussed above, we talked about reflection about the X-axis or the Y-axis. However, there can be other modifications as well, in which the graph shifts up, down, left or right. Let us look at those. If y = f(x), the graph of y = f(x) + c (where c is a constant) will be the graph of y = f(x) shifted c units upwards (in the direction of the y-axis). If y = f(x), the graph of y = f(x) - c (where c is a constant) will be the graph of y = f(x) shifted c units downwards (in the direction of the y-axis). If we consider f(x) = x2, given below are the graphs of f(x), f(x) + 20 and f(x) – 10. As you can see, the red graph is shifted 20 units upwards and the orange graph is shifted 10 units downwards from the original blue graph.

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If y = f(x), the graph of y = f(x + c) will be the graph of y = f(x) shifted c units to the left. If y = f(x), the graph of y = f(x – c) will be the graph of y = f(x) shifted c units to the right. If we consider f(x) = x2, given below are the graphs of f(x), f(x+5) and f(x-3). As you can see, the red graph is shifted 5 units left and the orange graph is shifted 3 units right from the original blue graph.

The graph of y = af(x) is a stretch scale factor a in the y-axis. This is because all the y-values become 'a' times bigger.

The graph of f(ax) is also a stretch. This time the multiple affects the x-values. (Everything happens 'a' times quicker.) Therefore: The graph of f(ax) is a stretch scale factor 1/a in the x-axis.

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Another modification which happens is in the case of y = |f(x)| In this case, whatever portion is below the X-axis gets reflected in the x-axis. Check the examples below:

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Given below is the graph of f(x) = |Sin(x)| to clarify it further.

I guess I will wrap it up here and hope this would help you with your functions problems.

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Course Details & Structure The course will contain:  Access 60+ recorded videos explaining shortcuts & tricks for CAT (sample: http://www.youtube.com/ravihanda ) Complete details are given below.  LIVE doubt clearance sessions(sample: http://www.wiziq.com/onlineclass/929122-number-systems)  A bunch of PDFs like Past years papers, GK Papers, etc.

Use the coupon code ARTBOOK to get extra discount of 10% (Offer valid till 31st August, 2012)

Your support of buying the course is what makes these ebooks / videos possible. If you want to know more, have any queries – feel free to call me on 09765142632 or email me ([email protected])

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VIDEO DETAILS: Number System

Arithmetic

Algebra

Geometry

Modern Maths

Basics of Numbers Divisibilty Rules Divisibility & Factors Finding out last digit Finding out number of zeroes Finding power of a number in a factorial Remainder Theorem Basics Cyclicity of Remainders Remainder - Advanced Base Systems - Basics Base Systems - Converting Bases Base Systems - Advanced Averages Mean, Median, Mode Percentages Simple Interest, Compound Interest Profit & Loss Alligations Mixtures Time Speed Distance Time & Work Basics of Algebra Inequalities Quadratic Equations Polynomial Equations Inequalities of higher order Basics of Geometry Triangles Trigonometry Quadrilaterals Polygons Circles Solids Co-ordinate Geometry Set Theory Binomial Theorem Permutation & Combination Probability Sequence & Series Functions

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1 3 2 1 1 1 1 1 2 1 1 1 1 1 3 1 2 1 1 6 2

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