CAT

Are You
Preparing For
CAT 2018 ?
Problem Solving Techniques Booklet

By Ashank Dubey CAT 100%iler

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QUANT TECHNIQUES BY Ashank Dubey

INDEX

(1) CONSTANT PRODUCT RULE
(2) POWER CYCLE
(3) “MINIMUM OF ALL” REGIONS IN VENN DIAGRAMS
(4) LAST 2 DIGITS TECHNIQUE
(5) SIMLIAR TO DIFFERENT GROUPING (P&C)
(6) APPLICATION OF FACTORIALS
(7) GRAPHICAL DIVISION FOR GEOMETRY
(8) ASSUMPTION METHOD

(1) CONSTANT PRODUCT RULE (1/x) & 1/(x+1)

This rule can be applied when we have two parameters whose product is constant, or in other words, when they are inversely
proportional to each other.
e.g.) Time x Speed = Distance, Price x consumption = Expenditure, Length x breadth = Area

A 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other, Parameter if the parameters are
inversely proportional.
Let’s see the application with the following examples

1. A man travels from his home to office at 4 km/hr and reaches his office 20 min late. If the speed had been 6 km/hr he would
have reached 10 min early. Find the distance from his home to office?
Solution: Assume original speed= 4km/hr. Percentage increase in speed from 4 to 6= 50% or ½. 1/2 increase in speed will result
in 1/3 decrease in original time=30 minutes.(from given data). Original time= 90 minutes= 1.5 hours! Answer is Distance=
4x1.5=6 km

2. A 20% increase in price of sugar. Find the % decrease in consumption a family should adopt so that the expenditure remains
constant
Solution: Here price x consumption= expenditure (constant)
Using the technique, 20% (1/5) increase results in 16.66% (1/6) decrease in consumption. Answer=16.66%

(2) POWER CYCLE

b
The last digit of a number of the form a 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
1 5
2 =2, 2 =32
2 6
2 =4 2 =64
3 7
2 =8 2 =128
4 8
2 =16 2 =256
th
As it can be observed, the unit digit gets repeated after every 4 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
4k+1
2 will have the last digit as 2
4k+2
2 will have the last digit as 4
4k+3
2 will have the last digit as 8
4k+4
2 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



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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

For example
75
1. Find the remainder when 3 is divided by 5.
(a) Express power in the form, 4k+x where x=1, 2, 3, 4. In this case 75 = 4k+3.
(b) Take the power cycle of 3 which is 3,9,7,1. Since the form is 4k+3, take the third digit in the cycle, which is 7
Any number divided by 5, the remainder will be that of the unit digit divided by 5. Hence the remainder is 2.
Sometimes, you may get a question in the term of variables, where you need to substitute values to get the answer in the fastest
way possible.
For example,

3^4n
2. Find the unit digit of 7
3^4 81.
Put n=1, the problem reduces to 7 , which is 7
Since 81=4k+1, take the first digit in the power cycle of 7, which is 7.

1957 1982
3. What is the first non zero integer from the right in 8330 + 8370 ?
(a) 3 (b) 1 (c) 9 (d) none of these
Solution:
1982
8370 will end with more number of zeroes so we need to consider only the first part. Rightmost non-zero integer of the
1957 1957
expression will be = unit digit of 833 = unit digit of 3
Since, 1957 = 4k+1, take the first digit in the power cycle of 3, which is 3.

1! + 2! + 3! + …..+ 13! 1! + 2! + 3!....+ 28! 1! + 2! + 3! + …....+ 32! 1! + 2! + 3! + …....+ 67!
4. If N = (13) + (28) + (32) + (67) then the unit digit of N is
(a) 4 (b) 8 (c) 2 (d) none of these
Solution:
Based on Power Cycle
After 4! Every number is of the form 4k+4 , here in all four terms power becomes 4k+1. So taking the first digit from the power
cycles of 3,8,2, and 7 we will get the unit digit as i.e 3+8+2+7 = 4. Ans = 0

3) Useful technique to find the last 2 digits of any expression of the form ab

Depending on the last digit of the number in question, we can find the last two digits of that number.
We can classify the technique to be applied into 4 categories.

TYPE METHOD EXAMPLES
67
1) Numbers ending in 1 The last digit is always 1. 1) 21 =__41(2 x 7=__4)

nd 87
The 2 last digit = product of tens 2) 41 =__ 81( 4 x 7= __8)
digit of base * unit digit of the
167
power. 3) 1261 =__21 ( 6 x 7= __2)
67
In 21 ; 2 is the tens digit of base and
124
7 is the unit digit of power 4) 31 =__ 21( 3 x 4=__2)

34
2) Number ending with 5 e.g.) 1555 = __ 25
Tens power Unit Second
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digit digit last

of digit
base
Odd Odd 5 7
Odd Even 5 2
Even Odd 5 2
Even Even 5 2
288 4 72 4

3) Numbers ending in 3, 7, 9 Change the power so that the base e.g.) 17 =(17 ) (taking power 4 as 7
ends with 1 and then use the same will end in 1.
2 2 72
technique as for those numbers (17 x17 )
ending with 1.

4 4 2
eg) 3 , 7 &9 all will end in1.
72 2
=( _89*_89) (as last 2 digits of 17 =89)
72
= ( _21) (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 1024
10 788 10 78 8
=2 i.e. e.g.) 2 = (2 ) x 2 =
10
x 2 raised to even power ends with
76 and
10
x 2 raised to odd power ends with
24.
__76 x __56 = __56.

It is also important to note that,

" 76 multiplied by the last 2 digits of any power of 2 will end in the same last 2 digits
E.g. 76 x 04 = 04, 76 x 08 = 08, 76 x 16 = 16, 76 x 32 = 32

2 2 2 2 2 2 2 2 2
" The last two digits of, x , (50-x) , (50+x) , (100-x) will always be the same. For example last 2 digits of 12 , 38 , 62 , 88 , 112 ….
will all be the same.
2 2 2 2 2 2 2 2
Also, last two digits of 11 =39 =61 =89 =111 =139 =161 =189 and so on

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

2
To find 41
Step1: Difference from 25 will be first 2 digits = 16
Step 2: Square of the difference from 50 will be last 2 digits = 81. Answer = 1681.

2
To find 43
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

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

4) “MINIMUM OF ALL” REGIONS IN VENN DIAGRAMS

1. 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?

Sum of the difference from 100 = (100-100) + (100-75) + (100-80) + (100-90) + (100-60) = 95
Again take the difference from 100 = 5 (answer)
This method will be explained in detail in the next booklet.
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5) SIMILAR TO DIFFERENT GROUPING IN PERMUTATION & COMBINATION

All questions in Permutation and Combination fall into 4 categories, and if you master these 4 categories, you can understand all
concepts in P&C easily.
(i) Similar to Different
(ii) Different to Similar
(iii) Similar to Similar
(iv) Different to Different
In this booklet, we will look at the first category; i.e. Similar to Different, which entails the number of ways of dividing ‘n’
identical (similar) things into ‘r’ distinct (different) groups

(a) NO LIMIT QUESTIONS
Let me explain this with an example. Suppose I have 10 identical chocolates to be divided among 3 people. The 10 chocolates
need to be distributed into 3 parts where a part can have zero or more chocolates.

So let us represent chocolates by blue balls. The straight red lines are used to divide them into parts. So you can see that for
dividing into 3 parts, you need only two lines.

st nd rd
Suppose you want to give 1 person 1 chocolate, 2 3 chocolates and 3 6 chocolates. Then you can show it as:

Suppose you want to give one person 1 chocolate, another person 6 chocolates and another one 3, then it can be represented
as:

Now if first person gets 0, second gets 1 and third gets 9 chocolates then it can be represented as:

Now suppose you want to give first person 0, second also 0 and third all of 10 then you can show it like:

So, for dividing 10 identical chocolates among 3 persons you can assume to have 12 (10 balls +2 sticks) things among which ten
are identical and rest 2 are same and of one kind. So number of ways in which you can distribute ten chocolates among 3
people is the same in which you can arrange 12 things among which 10 are identical and of one kind while 2 are identical and of
one kind which can be done in 12!/(10!2!)

The above situation is same as finding the number of positive integral solutions of a + b + c = 10. a, b, c is the number of
chocolates given to different persons.
st
Or else I can also say how many integral points lie on the line a + b + c = 10 in the 1 quadrant. In both the cases the answer is
12
C2.

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(b) LOWER LIMIT QUESTIONS

Now suppose we have a restriction that the groups cannot be empty i.e. in the above example all 3 persons should get at least
1.
You have to divide ten chocolates among 3 persons so that each gets at least one. So in the start only give them one each. This
you will do in just 1 way as all the chocolates are identical. Now, you are left with 7 chocolates and you have to divide them
among 3 people in such that way that each gets 0 or more. You can do this easily as explained above using balls and sticks.
Number of ways = 9! /(7!2!).

The above situation is same as finding the number of positive natural number solutions of a + b + c = 10. a, b, c is the number
of chocolates given to different persons.
st
Or else I can also say how many integral natural points lie on the line a + b + c = 10 in the 1 quadrant.
9
In both the cases the answer is C2.


Now suppose I change the question and say that now you have to divide 10 chocolates among 3 persons in such a way that one
gets at least 1, second at least 2 and third at least 3.
It’s as simple as the last one. First full fill the required condition.
st
Give 1 person 1, second 2 and third 3 and then divide the left 4 (10–1–2-3) chocolates among those 3 in such a way that each
gets at least 1.
6
This is same as arranging 4 balls and 2 sticks which can be done in C2 ways.

The above situation is same as finding the number of positive integral solutions of
a + b + c = 10 such that a ≥ 1, b ≥ 2, c ≥ 3. a, b, c is the number of chocolates given to different persons.
9
In this case the answer is C2.

2. Rajesh went to the market to buy 18 fruits in all. If there were mangoes, bananas, apples and oranges for sale then in how
many ways could Rajesh buy at least one fruit of each kind?
17 18 21 21
(a) C3 (b) C4 (c) C3 (d) C4
Solution:
This is a Grouping type 1 Similar to Distinct question, with a lower limit condition.
M + B + A + O=18
Remove one from each group, therefore 4 is subtracted from both sides. The problem changes to
M+B+A+O=14
Using our shortcut, The answer is the arrangement of 14 balls and 3 sticks
17
i.e. C3

3. 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:
(a) Balls are similar and boxes are different
(1) 275 (2) 70 (3) 120 (4) 19
Solution: When the balls are similar and the boxes are different, it’s a grouping type 1 question
8
A+B+C+D+E=4, where A,B,C,D,E are the different boxes. The number of ways of selection and distribution= C4 = 70

4. The number of non-negative integral solutions of A + B + C ≤ 10
(a) 84 (b) 286 (c) 220 (d) none of these
By non-negative integral solutions, the conditions imply that we can have 0 and natural number values for A, B, C, and D
To remove the sign ≤ add another dummy variable x4. The problem changes to
A + B + C + D = 10
This is an example of grouping type1 (Similar to Distinct)
It is the arrangement of 10 balls and 3 sticks.
13
Using the shortcut of balls and sticks, Therefore the answer is C3 = 286

6) APPLICATION OF FACTORIALS
A thorough understanding of Factorials is important because they play a pivotal role not only in understanding concepts in
Numbers but also other important topics like Permutation and Combination

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Definition of Factorial! N! = 1x2x3x…x(n-1)xn

E.g. 1) 5!= 1x2x3x4x5=120 e.g. 2) 3!=1x2x3 =6

Let us now look at the application of Factorials

(I) Highest power in a factorial or in a product
Questions based on highest power in a factorial are seen year after year in CAT. Questions based on this can be categorized
based on the nature of the number (prime or composite) whose highest power we are finding in the factorial, i.e

(ii) Highest power of a prime number in a factorial:
To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

1. The highest power of 5 in 100!
Solution:
100/5=20; 20/5=4; Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24
ALTERNATIVE METHOD
2 2
100/5 + 100/5 =20+4=24 (We take up to 5 as it is the highest power of 5 which is less than 100)

(iii) Highest power of a composite number in factorial
Factorize the number into primes. Find the highest power of all the prime numbers in that factorial using the previous method.
Take the least power.

2. To find the highest power of 10 in 100!
Solution: Factorize 10=5*2.
1. Highest power of 5 in 100! =24 2. Highest power of 2 in 100! =97
Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are available. So take the
lesser i.e. 24 is the answer.

3. Highest power of 12 in 100!
2 2
Solution: 12=2 *3. Find the highest power of 2 and 3 in 100!
First find out the highest power of 2.
Listing out the quotients:100/2 = 50; 50/2 = 25; 25/2 = 12; 12/2 = 6; 6/2 = 3; 3/2 = 1
2 2
Highest power of 2 = 50 + 25 + 12 + 6 + 3 + 1 = 97. So highest power of 2 = 48 (out of 97 2’s only 48 can make 2 )
Now for the highest power of 3. 100/3 = 33; 33/3 = 11; 11/3 = 3; 3/3 = 1;
Highest power of 3 = 48, Highest power of 12 = 48

(iv) Number of zeroes in the end of a factorial or a product
Finding the number of zeroes forms the base concept for a number of application questions. In base 10, number of zeros in the
end depends on the number of 10s; i.e. effectively, on the number of 5s
" In base N, number of zeroes in the end highest power of N in that product

4. Find the number of zeroes in 13! In base 10
Solution: We need to effectively find the highest power of 10 in 13! = Highest power of 5 in 13! As this power will be lesser.
13/5=2

5. Find the number of zeroes in the end of 15! in base 12.
2
Solution: Highest power of 12 in 15! =highest power of 2 *3 in 15! =Highest power of 3 in 15! = 5

(V) Number of factors of any factorial
There is a technique, which can be used to find the number of factors in a factorial

6. Find the factors of 12!
10 5 2
STEP 1: Prime factorize 12! i.e. find out the highest power of all prime factors till 12 ( i.e. 2,3,5,7,11). 12! = 2 x 3 x 5 x 7 x 11
STEP2: Use the formula
m n
N=a *b (a, b are the prime factors). Then number of factors= (m+1)(n+1)
The number of factors= (10+1)(5+1)(2+1)(1+1)(1+1) =792. Answer=792
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(VI) Right most non zero integer in a factorial

7. What is the last non – zero digit of 20!
Solution: Using above property we can write:
1 x 2 x3 x 4 x5 x6 x7 x8 x9 x10 = 4 x
11 x 12 x13 x 14 x15 x16 x17 x18 x19 x20 = 4 x
So last non – zero digit of 20! = x last non – zero digit of 4! = last non – zero digit of 6 x 4 = 4

APPLICATION QUESTION BASED ON FACTORIAL

8. How many natural numbers are there such that their factorials are ending with 5 zeroes?
10! is 1x2x3x4x(5)x6x7x8x9x(2x5). From this we can see that highest power of 5 till 10! is 2. Continuing like this, 10!-14!, highest
power of 5 will be 2. The next 5 will be obtained at 15 = (5*3).
Therefore, from 15! To 19! - The highest power of 5 will be 3.
20!-24! – Highest Power = 4, In 25, we are getting one extra five, as 25 = 5x5. Therefore, 25! to 29!, we will get highest power of
5 as 6. The answer to the question is therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.

7) USE OF GRAPHICAL DIVISION IN GEOMETRY

Let’s look at a technique which will help you solve a geometry question in no time
1. ABCD is a square and E and F are the midpoints of AB and BC respectively. Find the ratio of Area (ABCD) : Area (DEF)?








Let’s 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

8) ASSUMPTION method

There are at least 10 questions out of 25 in CAT 2007 where you 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.

2 2 2
1. k & 2k are the two roots of the equation x – px + q. Find q + 4q + 6pq =
2 3 3
(a) q (b) p (c) 0 (d) 2p
Solution: Assume an equation with roots 1&2 (k=1)
2
=>p (sum of roots)= 3 and q(product of roots)=2. Substitute in q + 4q + 6pq = 54. Look in the answer options for 54 on
3 3
substituting values of p=3 and q=2. we get 2p = 54. => Answer = 2p .
3 3
2. Let ‘x’ be the arithmetic mean and y,z be the two geometric means between any two positive numbers. The value of (y + z ) /
(xyz) is
(a) 2 (b) 3 (c) 1/2 (d) 3/2
Solution: 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 e.g.: see how you can
use the same technique in an equation question.

#SOME ACTUAL QUESTIONS FROM CAT SOLVED USING THIS TECHNIQUE

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1. 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
Solution: The question is independent of n, 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.

2. 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 these members?
2 2
(a) 2n-6 (b) n/2(n-3) (c) n-2 (d) (n -7n+16)/2 e) (n -5n+8)/2
Solution: We will start with n=5, then you will get two options (c) and (d).
Now take n =6, lets 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).

There were more questions which could be solved using similar strategies. The methods given above clearly show that for
someone with good conceptual knowledge and right strategies the quant section is a cakewalk.

………………………………………………………………………………………………………………………………………………….....................................
About the trainers of Ashank Dubey

" Ashank is unarguably India's best Quant trainer and 100%iler of CAT – 16 (Overall) & CAT – 15 (Qunat). He has
trained numerous CAT toppers ever since the CAT went online, including 99.9%tilers of CAT 2015 – Shrikanth,
Rushil, Sandeep, Abhilasha, Prakhar, Pratik, Shivani to name a few.

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