Free Mock CAT Test Answer Keys

Below are answer keys for free mock CAT test:

Full Mock CAT Answer Keys 


1. d 26. a 51. c  76. d
2. a 27. d 52. c 77. c
3. a 28. c 53. a 78. d
4. c 29. a 54. d 79. a
5. b 30. a 55. d 80. a
6. b 31. d 56. a 81. a
7. c 32. d 57. d 82. c
8. b 33. a 58. c 83. c
9. c 34. a 59. d 84. b
10. c 35. d 60. d  85. d
11. a 36. c 61. c 86. d
12. b 37. 5 62. d 87. c
13. c 38. c 63. b 88. a
14. d 39. c 64. b 89. b
15. a 40. b 65. a 90. b
16. a 41. d 66. b 91. d
17. d 42. a 67. a 92. b
18. c 43. a 68. b 93. d
19. c 44. a 69. b 94. b
20. b 45. b 70. c 95. d
21. d 46. d 71. d 96. d
22. c 47. d 72. b 97. c
23. a 48. c 73. a 98. b
24. d 49. a 74. a 99. b
25. a 50. b 75. d 100. a

Mini Mock CAT Answer Keys 


1. d 11 c 21 d 31 c




2. a 12. a 22 c 32 a




3. b 13 b 23 b 33 c




4. c 14 b 24 b 34 b




5. a 15 c 25 b 35 d




6. d 16 b 26 d 36 b




7. c 17 a 27 d 37 b




8. a 18 a 28 a 38 c




9. d 19 b 29 d 39 d




10. d 20 d 30 c 40 d

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Answer Keys for the questions would be published on 8th September 2014 by 8 AM

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Free Online Mock CAT Test Instructions

Instructions:
1. You are allotted 80 minutes to answer 40 questions
2. You are advised to take this exam at one stretch
3. Once you start the exam you cannot pause it in between
4. You are advised to have pen and paper for rough work and calculations
5. As this mock test is for your benefit, you are requested to attempt the paper judiciously without cheating ;)
6. A CAT aspirant with 60-65+ score in this test could expect 85-90+ percentile in CAT
7. The mock test will display your score on completion of test i.e. when you click on Submit button or the allotted time of 80 minutes is over
8. There might be slight discrepancy in the time displayed by two of the timers on the slower networks, but be assured allotted 80 minutes is calculated appropriately

Marks:
1. Each correct answer will give you 3 marks
2. There is penalty of 1 marks for each incorrect answer
3. Maximum Marks for the exam is 40*3= 120 Marks

Click on the below link to start the test

Note:

Answer Keys for the questions is available on the blog.

Though we have tried our best to keep this test error free and enjoyable, still if you have any issues kindly forgive us and share the details of the same on our Facebook Page so that we could fix and make sure you have a better experience in future.

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Free Online Mock CAT

Free Mini Mock CAT Details:

Date: Saturday 5th October 2013
Time: 9:30 AM On wards 
Test URL would be available after 9:30 AM and shared on our Facebook Page or at below Page:

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Number of Questions: 40 Questions 
Time Allotted: 100 minutes

Answers to the questions would be published on the same day

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SNAP 2013 Admissions Test Notification for Symbiosis Institutes

The admission process for MBA programmes at Symbiosis International University (SIU) is simplified. After SNAP shortlisted candidates will need to appear for only ONE Personal Interaction (PI) and Written Ability Test (WAT) for admission to MBA programmes at 13 constituent institutes of Symbiosis.

Below is SNAP 2013 notification and important dates.

Below is the structure of SNAP 2013:
SNAP TEST STRUCTURE
What is the SNAP TEST STRUCTURE?

The SNAP test pattern is as follows :

Subject

Total Marks
1. General English : Reading Comprehension, Verbal Reasoning, Verbal Ability40
2. Quantitative, Data Interpretation & Data Sufficiency40
3. General Awareness : General Knowledge, Current Affairs, Business Scenario40
4. Analytics & Logical Reasoning60
Total180
Each wrong answer attracts 25% negative marks.

Which MBA Stream to Choose - Which MBA Specialization should you go for?


While preparing for MBA exams aspirants tend to think about the different specializations of MBA and are confused which MBA stream should they opt for.  

This is one of the most critical decisions for all MBA aspirants. Opting for a specialization depends mostly on your interests, personality and also somewhat on your profile. Below are some of the key point which could help you in choosing the right MBA stream and making the right decision .

1. Marketing/Sales MBA stream is good for people who are outgoing in nature, are good at convincing others and posses excellent negotiation skills
2. Finance MBA stream is good for those who are good in interpretation, have good mathematical aptitude and are excellent at analysis
3. IT MBA stream is for those who want to go ahead in their IT career at much higher pace  and want to manage IT related projects.
4. HR MBA stream is for those who are good at understanding people mindset & bonds easily with people, should be excellent in organizing events and also good at co-ordination.


You should also go through the below key attributes expected out of professionals in various specializations of MBA:

Key Qualities of Marketing and Sales Executives:

-A Marketing or Sales Manager must have very very good communication skills and must be persuasive
-Must be very adaptive and tech savvy to gauge the ever changing market
-Must be Customer focused and should have ability to quickly understand the customer needs
-Must have excellent analytical ability to understand the market and the competitors
-Must have very strong business fundamentals to drive revenue and customers needs
-Must have passion for the brand and the business 
A good Sales manager needs to be like Chris Gardner's Character played by Will Smith in The Pursuit of Happyness

-Should support teams with valuable, targeted programs to foster business growth
-Should be collaborative, visionary and a thoughtful leader. Must be good at building business relationships with customers as well as leaders in all levels with in the company
-Should have ability to plan and execute business strategy in the details and the big picture.
-Must be result oriented and must have excellent negotiation skills.



Key Qualities of Finance Executives:

-A finance manager must have an analytical and logical approach to problem solving
-Must have command of both technology and finance issues
-Must posses excellent communications skills, both verbally and orally
-Must develop the best strategy to efficiently fund the company goals and do strategic planning
-Must have ability to quickly adapt and understand business processes
-Must posses good all-round IT skills
-Must have ability to work on their own initiative
-Must be Motivated and ambitious




Key Qualities of HR Executives:

-An HR Manager should be approachable to employees and bond well with them.
-Should communicate effectively, both verbally and orally and should have good influencing & convincing skills
-Should be good in administering policies
-HR managers should support the company to ensure they are optimizing the cost of operation of the company by hiring right people with effective salary negotiation. They also need to support the employee by ensuring their career growth and job satisfaction.

-They should actively listen to the employees.
-Define vision and values of the company and percolate the same to employees
-They should have ability to to cope with change and operating difficulties
-Skilled at motivating and inspiring others. 
-Should be good in Conflict Management, Problem Solving & Change Management


Please do post share your views and comments below.

CAT Free Online Test - Permutation Answers

Below are detailed answers for each of the questions asked in test, the explanations given below would further strengthen your understanding of Permutation & Combinations topic. 

Please do provide your comments and share this free running blog with CAT aspirants via Facebook & online groups, we desperately need your support.

Q1. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Answer:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.


Q2.In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Answer:
Required number of ways           = (8C5 x 10C6)
= (8C3 x 10C4)
=           8 x 7 x 6           x           10 x 9 x 8 x 7   
3 x 2 x 1            4 x 3 x 2 x 1
= 11760.


Q3.In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together??
Answer:
Let 4 girls be one unit and now there are 6 units in all.
They can be arranged in 6! ways.
In each of these arrangements 4 girls can be arranged in 4! ways.
=> Total number of arrangements in which girls are always together
=6!*4!=720*24= 17280


Q4. 12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?
For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is 12C4=495
. Therefore, we can draw 495 quadrilaterals.​


Q5. The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and at least 4 bowlers?
We are to choose 11 players including 1 wicket keeper and 4 bowlers
or, 1 wicket keeper and 5 bowlers.
Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players
=2C1*5C4*9C6=840
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players
=2C1*5C5*9C5=252
=> Total number of ways of selecting the team:
=840+252= 1092


Q6. How many factors of (2^4)×(5^3)×(7^4) are odd numbers?
Any factor of this number should be of the form 2a×3b×5c.
For the factor to be an odd number, a should be 0.
b can take values 0,1, 2,3, and c can take values 0, 1, 2,3, 4.
Total number of odd factors =4×5= 20


Q7. How many factors of (2^5)×(3^6)×(5^2) are perfect squares?
Any factor of this number should be of the form 2a×3b×5c.
For the factor to be a perfect square a, b, c have to be even.
a can take values 0, 2, 4. b can take values 0,2, 4, 6 and c can take values 0,2.
Total number of perfect squares =3×4×2= 24

Q8. In how many ways can eight directors, the vice-chairman and chairman of a firm be seated at a round table, if the chairman has to sit between the vice-chairman and the director?
In a circular arrangement problem, we always fix one position and the look at ways of arranging the rest with respect to the fixed position.

Case 1 - seating of chairman - In this question, we fix the position of the Chairman. Thus the Chairman can be seated in 1 way only.

Case 2 - seating of vice-chairman - The vice-chairman can be to the left or right of the chariman. Thus the vice chairman can be seated in 2 ways.

Case 3 - seating of the 8 directors - The rest of the 8 directors can be seated in 8 positions in 8! ways. This is because the first director can be seated in andy of 8 positions in 8 ways. And then the second director can be seated in any of the remaning 7 positions in 7 ways and so on, thus giving the total number of ways of seating the 8 director as 8 * 7 * 6* 5 * 4 * 3 * 2 * 1, i.e. 8! ways.

Thus the desired answer is when case 1 and 2 and 3 happen together, which is 1 * 2 * 8! = 2 * 8! ways.


Q9. In how many ways can 15 people be seated around two round tables with seating capacities of 7 and 8 people?
‘n' objects can be arranged around a circle in (n - 1)!.

If arranging these 'n' objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.
i.e., number of arrangements = (n-1)!/2.
You can choose the 7 people to sit in the first table in 15C7 ways.
After selecting 7 people for the table that can seat 7 people, they can be seated in (7-1)! = 6!.
The remaining 8 people can be made to sit around the second circular table in (8-1)! = 7! Ways.

Hence, total number of ways 15C8 * 6! * 7!


Q10. Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel from A to E?
The bus from A to B can be selected in 3 ways.
The bus from B to C can be selected in 4 ways.
The bus from C to D can be selected in 2 ways.
The bus from D to E can be selected in 3 ways.
So, by the General Counting Principle, one can travel from A to E in 3*4*2*3= 72 ways


Q11. How many words can be formed by using 4 letters at a time of word "SURPRISE" (words may be meaningful or meaningless according to dictionary)
(1) No. of words having all letters different = 6*5*4*3 = 360
(2) No. of words having any one letter repeated
=(2*1)*(5*4*3*2*1/3*2*1*2*1)*(4*3*2*1/2*1) = 240
(3) No. of words having two letter repeated = (1)*(4*3*2*1/2*1*2*1) = 6
Total words = 360+240+6 = 606



Permutation-Combinations Workshop for CAT

Factorial
Let n be a positive integer. Then, factorial n, denoted n! is defined as:
n! = n(n - 1)(n - 2) ... 3.2.1.
Solved Examples:
i. We define 0! = 1.
ii. 1! = 1.
iii. 4! = (4 x 3 x 2 x 1) = 24.
iv. 5! = (5 x 4 x 3 x 2 x 1) = 120.


Difference between Permutation & Combination
In English we use the word "combination" loosely, without thinking if the sequence/order of things is important or not. Examples:



"My mock-tail is a combination of  pineapple juice, grapes juice and lime juice" . 
In this case we don't care what order the juices are mixed in, they could also be "grapes juice, pineapple juice and lime juice" or "lime juice, grapes juice and pineapple juice", its the same mock-tail.





" 5178 is my ATM Pin combination ". 
Now in this case we do care for the order. 5187 will not work, nor will 5718, 5781, 1857, 1875 etc. The only sequence which will work is 5178.


In mathematics:
If the order doesn't matter, it is a Combination.
If the order does matter it is a Permutation.
=>     A Permutation is an ordered Combination.


Permutation - The order of arrangement matters
In mathematics permutation means act of rearranging (permuting) objects or values. In simple words: "the different arrangements of a given number of things by taking some or all at a time, are called permutations."
Consider following three fruits which needs to be arranged in a straight line on a thin bench:












What are the different ways one can arrange these 3 fruits on table?
Lets number these fruits as 1. Apple, 2. Mango and 3. Banana.













All possible arrangements of these 3 fruits are:







The number of different arrangements as you could see above is 6 or 3! = 3 • 2 • 1

-The number of permutations of n distinct objects is n×(n − 1)×(n − 2)×⋯×1, which is commonly denoted as "n factorial" and written "n!".
-All possible arrangements of letters of a word is a permutation of its letters. 
-For (a,b,c) or abc number of distinct objects are 3, hence permutation made with the letters a, b, c taking all at a time are 3! = 3*2 = 6 => (abc, acb, bac, bca, cab, cba)



Important Concepts on Permutation:

1. Taken r items at a time
Number of all permutations of n things, taken r at a time, is given by:
    
 nPr = n(n - 1)(n - 2) ... (n - r + 1) = n!/(n-r)! 


Solved Example: 
Question 1: How many different 3-digit numerals can be made from the digits  4, 5, 6, 7, 8 if a digit can appear just once in a numeral?
Answer:  We need to find permutation of 5 distinct numbers taken 3 at a time.
Applying formula for nPr => 5P3 = 5!/(5-3)! = (5*4*3*2*1)/(2*1) = 5*4*3 = 60

 
2 a. Identical Items in arrangement
If x out of n items are identical, then the number of different permutations of the n items is:  n!/x! 
Solved Example: 
Question 2: How many ways can you arrange the letters of the word 'twist'? 
Answer:  We need to find permutation of 5 letters(i,s,t,t,w) of which 2 are identical(t,t).
Applying n!/x! for identical items => 5!/2! = 60

 
2 b. Multiple Identical Items in arrangement
If a set of n items contains p identical items, q identical items, and r identical items etc.., then the total number of different permutations of n objects is:  n!/(p!.q!.r!...) 
Solved Example: 
Question 3: How many can you arrange the letters in the word 'MISSISSIPPI''? 
Answer:  We need to find permutation of 5 letters(I,I,I,I,M,P,P,S,S,S,S) of which 4Is, 2 Ps and 4Ss are identical.
Applying n!/(p!.q!.r!..) for multiple identical items => 11!/(4!.2!.4!) = 34,650

 
3. Conditional/Restrictive Permutation 
There are situations where we need to find out possible arrangements  by keeping some of conditions/restrictions in mind. 
Solved Example: 
Question 4:  Using the letters in the word  " square ", tell how many 6-letter arrangements, with no repetitions, are possible if the  first letter is a vowel? 
Answer:  When working with "arrangements", it is often helpful to put lines down to represent the locations of the items. 
       For this problem, six "locations" are needed for 6-letter arrangements.
       _____  •   _____  •  _____  •  _____  •  _____  •  _____

The first locations must be a vowel (u, a, e).  There are three ways to fill the first location.
       _3_ •  _____ •  _____ •  _____ •  _____ •  _____

After the vowel has been placed in the first location, there are 5 letters left to be arranged in the remaining five spaces.

      _3_ • _5_  • _4_  • _3_  •  _2__1_     or
                        
         3  5P5    =     3  120  =  360

 
4. Circular Arrangements
To understand circular arrangement better below illustration comparing it with linear arrangement would be useful:

1. Lets consider 4 people sitting around a round table and 4 people sitting on a linear bench

2.  Shifting each of the 4 people sitting around round table in clockwise direction we get following arrangements:
Looking at the above illustration we could say that if 4 persons are sitting at a round table, then they can be shifted four times, but these four arrangements will be the same, because the sequence of P1, P2, P3, P4 is same.

3. Now we will compare this with linear arrangement where 4 people P1, P2, P3 & P4 are sitting on a bench.  Shifting each of the 4 people sitting in a row on a bench all the 4 arrangement formed will be different:
 All the above 4 sequences are unique and are different arrangements: 
[P1, P2, P3, P4] , [P4, P1, P2, P3], [P3, P4, P1, P2], [P2, P3, P4, P1]

Thus you can see how results in a circular arrangement differs from linear one.



Number of circular-permutations of ‘n’ different things taken ‘n’ at a time
(a)  When clockwise & anti-clockwise order are considered different
If clockwise and anti clock-wise orders are considered different, then total number of circular-permutations is given by (n-1)!

(b)  When clockwise & anti-clockwise order are not considered different 
If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by  (n-1)!/2!

Solved Example: 
Question 5: If 6 people are going to sitting at a round table, how many different ways can the group of 6 sit
Answer:  Applying formula (n-1)! = (6-1)! = 120

Question 6:If 6 people are going to sitting at a round table, but Raj will not sit next to Simran, how many different ways can the group of 6 sit? 
Answer:
First Approach
a. Total circular permutations = (6-1)! = 5! = 120. 
b. Ways in which Raj and Simran sit together = 2! * 4! = 2*24 = 48 
Required ways = Total - Together = 120 - 48 = 72. 

Second Approach
a. We have total of 6 places. Fix Simran . Now Raj can't sit at either seat beside her. So number of places where Sam can sit = 5-2 = 3. 
For the other 4 people we can arrange them in 4! ways in 4 seats. 
So total ways = 3 * 4! = 72. 



Number of circular-permutations of ‘n’ different things taken ‘r’ at a time

(c) When clockwise & anti-clockwise order are considered different

If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations  =   nPr /r


(d) When clockwise & anti-clockwise order are not considered different

If clock-wise and anti-clockwise orders are taken as not different, then total number of circular – permutation =  nPr/2r



Solved Example:
Question 7: How many necklace of 12 beads each can be made from 18 beads of different colors ? 
Answer:  Here clock-wise and anti-clockwise arrangement s are same.
Applying formula:  nPr/2r
Hence total number of circular–permutations = 18P12/2x12
=18!/(6! x 24)


 
Combinations: The order of arrangement doesn't matters
In mathematics a combination is a way of selecting several things out of a larger group, where  order does not matter.

Number of combinations 

The number of all combinations of n things, taken r at a time is:


nCr = n(n - 1)(n - 2) ... to r factors/r! = n!/(r!(n-r)!)


Important:

Solved Example: 
Question 8: In a class of 10 students, how many ways can a club of 4 students be formed? 
Answer: Applying nCr formula:
=> 10C4 = 10!/(4!*6!) = 21

Question 9: How many factors of 2^4 * 5^3 * 7^4 are odd numbers? ? 
AnswerAny factor of this number would be of the form 2^a * 3^b * 5^c. For the factor to be an odd number, a should be 0. b can take values 0,1, 2,3, and c can take values 0, 1, 2,3, 4. Total number of odd factors = 4 * 5 = 20 


Take Online Test on Permutation-Combinations to evaulate yourself 
Test Consists of 11 Questions
Time Allotted is 16 minutes
Total Mraks is 33
Average cut Off Score 25 out 33

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