b2b quants shortcut booklet

Back 2 Basics –Handbook of Shortcuts and Tricks in Quants and English

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POWER CYCLE

The last digit of a number of the form ab falls in a particular sequence or order depending on the

unit digit of the number (a) and the power the number is raised to (b). The power cycle of a

number thus depends on its unit digit.

Consider the power cycle of 2

21 =2, 2

5 =32

22 =4 2

6 =64

23 =8 2

7 =128

24 =16 2

8 =256

As it can be observed, the unit digit gets repeated after every 4th

power of 2. Hence, we can say

that 2 has a power cycle of 2,4,8,6 with frequency 4.

This means that, a number of the form

24k+1

will have the last digit as 2

24k+2


24k+3


24k+4

will have the last digit as 6 (where k=0, 1, 2, 3…)

This is applicable not only for 2, but for all numbers ending in 2.

Therefore to find the last digit of a number raised to any power, we just need to know the power

cycle of digits from 0 to 9, which are given below -

Q1. Find the remainder when (3454372)73

is divided by 5?

Solution:

1. Express power in the form of 4k+x, where x=1, 2, 3, 4. In this case, 73 = 4K+1.

2. Take the power cycle of unit digit 2 which is 2, 4, 8, 6. As power is of the form 4K+1, take the 1st

digit in the cycle, which is 2. For any number, if the number is divided by 5, the remainder will be

units digit divided by 5. In our case, the unit digit of (3454372)73

is 2. Hence, remainder when

(3454372)73

is divided by 5 is 2.

Unit Digit Power Cycle Frequency

0 0 1

1 1 1

2 2, 4, 8, 6 4

3 3, 9, 7, 1 4

4 4, 6 2

5 5 1

6 6 1

7 7, 9, 3, 1 4

8 8, 4, 2, 6 4

9 9, 1 2


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Sometimes, we may encounter a question which is in terms of variables, where we should substitute

values to get the answer in fastest possible time –

Q2. Find the unit digit of 73^4n

Solution: Put n = 1, the problem becomes 73^4

which is 781

. Since 81 = 4K+1, the answer is first digit in the

power cycle of 7, which is 7.

Technique to find last 2 digits of any expression of the form ab

TYPE METHOD EXAMPLES

1) Numbers ending in

1

The last digit is always 1.

The 2nd last digit = product of

tens digit of base * unit digit of

the power.

In 5756

; 5 is the tens digit of base and

6 is the units digit of the power

1) 2167

= __41 (2 *7 =__4)

2) 4187

=__81(4 *7 = __8)

3) 1261167

=__21 (6 * 7 =__2)

4) 31124

= __21(3 *4 = __2)

2) Number ending

with 5

Last two digits: always 25 or 75 e.g. 155534

= _25

3) Numbers ending in

3,7,9

Change the power so that the

base ends with 1 and then use

the same technique as that for

numbers ending with 1.

e.g. 34, 7

4 and 9

2 all end in 1.

e.g. 17288

= (174)

72 (taking power 4

as 74 ends in 1.

(172* 17

2)

72

= (_89*_89)72

(as last 2 digits of

172=89)

= ( _21)72

(as last 2 digits of

89*89=21)

Answer = __41 ( as 2*2=4)

4) For even numbers

(2,4,6,8)

Use the pattern of the number

210

= 1024 i.e.

*210

raised to even power ends with

76 and

*210

raised to odd power ends with

24

e.g. 2788

= (210

)78

* 28

=

__76 * __56 = __56


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It is also important to note that,

1. 76 multiplied by the last 2 digits of any power of 2 will end in the same last 2 digits

E.g. 76*04 = 04, 76*08 = 08, 76*16 = 16, 76*32 = 32

2. The last two digits of, x2, (50-x)

2, (50+x)

2 , (100-x)

2 will always be the same. For example last

2 digits of 122, 38

2, 62

2, 88

2, 112

2 …. will all be the same.

Also, last two digits of 112 = 39

2 = 61

2 = 89

2 = 111

2 =139

2 = 161

2=189

2

3. To find the squares of numbers from 30-70 we can use the following method

e.g To find 412

Step 1: Difference from 25 will be first 2 digits, (41-25) = 16

Step 2: Square of the difference from 50 will be last 2 digits, (50 – 41)2 = 81 Answer = 1681 .

Similarly, To find 432

Step1: Difference from 25 will be first 2 digits = 18

Step 2: Square of the difference from 50 will be last 2 digits = 49 Answer = 1849

4. Combining all these techniques we can find the last 2 digits for any number because every

even number can be written as 2* an odd number (except for the powers of 2)

Minimum of All regions in an N set Venn diagram Let us understand this concept with the help of an example.

Q. In a survey conducted among 100 men in a company, 100 men use brand A, 75 use brand B,

80 use brand C, 90 use brand D & 60 use brand E of the same product. What is the minimum

possible number of men using all the 5 brands, if all the 100 men use at least one of these

brands?

Solution: This is a 5 set Venn diagram question which is impractical to represent using

conventional methods. We can solve such questions by the following method-

Step 1: Take the difference of each set from 100 = (100-100) = 0, (100-75) = 25,

(100-80)= 20, (100-90) = 10, (100-60) = 40.

Step 2: Add the differences together = 0+25+20+10+40 = 95.

Step 3: Subtract this sum from 100, (100-95) = 5 (answer)


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Similar to different grouping in Permutation and Combination

All questions in P&C can be categorized into 4 types –

1) Similar to Different (S�D)

2) Different to Similar (D�S)

3) Similar to Similar (S�S)

4) Different to Different (D�D)

Here, we will have a brief look at the first category; i.e. Similar to Different, which is the

number of ways of dividing ‘n’ identical (similar) things into ‘r’ distinct (different) groups

S��D No limit questions

The concept of S�D can be easily understood with the help of following example.

Q. In how many ways can you divide 12 identical books to 3 different persons?

This condition can be expressed as equation A+B+C = 8. Where A, B, C are 3 different persons.

The question is also equivalent to finding the whole no. solutions for equation A+B+C = 8.

We classify this question as a NO limit question because there is no upper or lower limit defined

on the number of books that a person can get. Any person can get 0 books or even all 8 books.

Let us solve this question by using a very innovative technique, which is very easy to understand

and master. We call it the ZERO-ONE technique. Let us represent the 8 books by zeros as

follows-

Now to divide the books represented by zeros above among 3 persons, we need 2 lines. If we

want to give 2 books to person A, 3 books to person B and 3 books to person C, we can do it as

follows –

A B C

Now suppose if we want to give 0 books to 1st

person A, all 8 books to 2nd

person B and 0 books

to 3rd

person C, we can do it as follows –

A B C


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Now suppose if we want to give 1st

person 0 books, 2nd

person also 0 books and 3rd

person all 8

books, we can do it as follows –

A B C

So to divide 8 books to 3 persons, we need 2 lines. To divide entities among N persons, we

need N-1 lines. We call these lines as 1 in ZERO-ONE method.

So now, we have a total of 10 entities (8 zeros + 2 ones). From this, the number

of ways of selecting 2 ones from 10 entities is 10

C2 = (10 !) / (2! * 8!) = 45 ways

(answer)

The only P&C concept you need to know in this case is the formula NCR.

S��D Lower Limit questions

Now suppose we have a restriction that groups cannot be empty. i.e. in the above question,

each person should get at least 1 book.

Example: In how many ways can you distribute 8 books to 3 persons where each person should

get at least 1 book?

Solution: In this case, we distribute 1 book to each person at the beginning itself. So we are left

with 5 books only (8-3 = 5) and we have to divide them among 3 persons in such a ways that

each person get 0 or more books. Hence, this becomes a no limit question now.

A + B + C = 5 (No limit)

Hence, the answer by ZERO-ONE method in this case is 7C2 = 21 (5 zeros and 2 ones).


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Q. There are four balls to be put in five boxes where each box can accommodate any number

of balls. In how many ways can one do this if Balls are similar and boxes are different (S��D)

1) 275 2) 70 c) 120 d) 19

When the balls are similar and the boxes are different, it’s a S�D question

A+B+C+D+E=4, where A,B,C,D,E are the different boxes. The number of ways of selection and

distribution = 8C4 = 70 (answer)

GRAPHICAL DIVISION IN GEOMETRY Let’s look at a technique which will help you solve a geometry question in no time.

ABCD is a square and E and F are the midpoints of AB and BC respectively.

Find the ratio of Area( ABCD): Area(DEF)?

FIGURE A FIGURE B

Lets divide the figure using dotted lines as shown in Figure B. Area of ABCD=100%. Area

AEDG=50%. Then Area in shaded region 1(AED)= 25%. Similarly, DCFH=50%. Area in shaded

region 2(DCF) = 25%.

Now Area of EOFB = 25%. .Area of shaded region 3(BEF)=12.5%. Total area outside triangle=

62.5%. Area inside triangle= 100-62.5=37.5%. required ratio = 100/37.5 = 8:3


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ASSUMPTION method

There are more than 9 questions out of 25, both in CAT 2007 and CAT 2008 where we can apply

this technique. This involves assuming simple values for the variables in the questions, and

substituting in answer options based on those values. Assumption helps to tremendously

speedup the process of evaluating the answer as shown below.

Q. k & 2k2 are the two roots of the equation x

2 – px + q. Find q + 4q

2 + 6pq

a) q2 b) p

3 c) 0 d) 2p

3

Solution: Assume an equation with roots 1&2 (k=1)

=>p (sum of roots)= 3 and q(product of roots)= 2 .

Substitute in q + 4q2 + 6pq = 54. Look in the answer options for 54 on substituting values of p=3

and q=2. we get 2p3 = 54 � Ans = 2p

3

Q. Let ‘x’ be the arithmetic mean and y,z be the two geometric means between any two

positive numbers. The value of (y3 + z

3 ) / (xyz) is -

a) 2 b) 3 c) 1/2 d) 3/2

Solutions:

Assume a GP 1 2 4 8 implies 2 GMs between 1 and 8, i.e. y=2 and z=4.

Arithmetic Mean, x =(8+1)/2 = 4.5 .Substitute in (y + z ) / (xyz) Answer = (2 +4 )/(2x4x4.5) = 2.

3 3 3 3

NOTE:- Assume a GP 1 1 1 1 then x=1, y=1 z=1. Answer on substitution=2, which will make

the calculation even faster, half of the problems in Algebra can be solved using assumption. This

is not direct substitution. In the next example see how you can use the same technique in an

equation question.

SOME ACTUAL QUESTIONS FROM CAT-2007 SOLVED USING THIS TECHNIQUE

Q. Consider the set S={2,3,4……2n+1), where n is a positive integer larger than 2007. Define X

as the average of odd integers in S and Y as the average of the even integers in S. What is the

value of X-Y?

a) 1 b) n/2 c)(n+1)/2n d) 2008 e) 0

The question is independent of n. There is no number series, which behaves differently for 1st

N

numbers and is different for numbers after that. Therefore, N > 2007 is insignificant, which is

shown below.

Take n=2. Then S= {2,3,4,5). X= 4 and Y=3. X-Y =1, Take n=3. then S={2,3,4,5,6,7}. X=5 and Y=4. X-

Y=1. Hence, you can directly mark the answer option (a) .

You can solve the question in less than 60 seconds.


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Q. Let S be the set of all pairs (i,j) where 1=i<j=n and n=4. Any two distinct members of

S are called friends, if they have one constituent of the pairs in common and “enemies”

otherwise. For example, if n=4, then S={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}. Here (1,2), (1,3) are

friends but (1,4), (2,3) are enemies. For general n, consider any two members of S that are

friends. How many other members of S will be common friends of both these members?

a) 2n-6 b) n/2(n-3) c) n-2 d) (n2-7n+16)/2

e) (n2-5n+8)/2

We will start with n=5, then only two options (c) and (d) satisfies the condition

Now take n =6, let’s take two members (1,2) and (1,3). Then their common friends will be

(1,4)(1,5),(1,6) and (2,3). i.e. four common friends. So substitute n=6 among answer options and

check where you get answer 4.Only option (c) satisfies and hence, answer = c) n-2.

There are more questions in past CAT and XAT which can be solved with the same techniques.

The only requirements are some basic conceptual knowledge in subject, and right strategies to

attack such question, which can be easily mastered by anyone with some practice.

Tricks and strategies for solving Reading Comprehension questions in English

section

Reading Comprehension(RC) questions can be categorized in the following types –

i. Main theme questions ii. Database questions (description question)

iii. Structure questions iv. Inference (interpolation) questions

v. Hypothetical (extrapolation) questions vi. Tone based questions

Tricks and strategy for Main theme questions –

Answer for main theme questions can be found in –

i. 1st

paragraph 1st

line ii. Last paragraph last line

iii. 1st

paragraph last line iv. 2nd

paragraph first line

Working Principles –

1. If the content of 1st

paragraph does not match with the content of last paragraph, the 1st

paragraph gives background information. Therefore, the main theme lies in 2nd

paragraph 1st

line or last paragraph last line.

2. Main theme lines are usually associated with structure words.

Do’s and Don’ts for Main theme questions –

1. Answer option can never be too specific not be too generic

2. Examples can never be a part of main there

3. Don’t add value to what the author has stated. Cut-Copy-Paste or restate

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