CAT 2011: Last minute tips to crack the exam

With just few days left for CAT 2011, students must be very stressed out. However, now is the time to sit back, relax and revise. In order to help you crack the exam, Learnhub spoke to IIM alumni from across the country and asked them their last-minute strategies that helped them to score that desirable percentile. Let us divide various sections of the CAT2011 exam and examine expert tips on how to tackle these sections effectively-:

1. CAT 2011 Quantitative Ability and Data Interpretation: The key to solve this section is speed and accuracy. One should have a thorough knowledge of mathematical concepts and have short cut to answers. Abhishek Jha, an alumnus of IIM Lucknow, says, "For data interpretation questions, one has to be very comfortable with numbers, charts, and tables and be very good with mental calculations." He further suggests that one should try and practice a data interpretation problem everyday, reading and analysing stories which consist of graphs and pie charts in business sections of newspapers. In order to improve your speed, it is very essential to take mock tests. However, experts warn that one should take only mock test per day.
   
   On the exam day, try to answer the caselets with which one is familiar with, and one can spend up to 4 minutes in analysing a caselet. Experts suggest that if after four minutes there is no progress in understanding then it would be wise to leave the questions. Another advise for the exam day is to avoid guess work as CAT has 1/3 negative marking. In this section, if you're able to attempt 16-19 questions with 90 per cent accuracy is enough to secure 97 percentile and above in this section.
2. CAT 2011 Verbal Ability and Logical Reasoning: As we all know that verbal ability consists of two sections namely reading comprehension and English usage. The only way to do well in reading comprehension is to increase your speed of reading. This can be done by practising the habit of focusing on keywords that convey the meaning of a sentence. On how one should work on enhancing his Logical Reasoning, Jha says, "One should create a chart or a diagram of the question to get clarity. The aspirant should read questions very carefully as these questions are intended to check your ability to manage ambiguous data. One should avoid to assume any information that is not given in the question." Experts also suggest that it is always advisable to answer those questions you're most comfortable in. Around 15 minutes should be spent on the verbal ability section. One should allot five minutes on reading each comprehension passage and logical reasoning caselet to understand the same. Even after spending five minutes if you fail to understand the comprehension passage then you should move on to the next.
3. Time Management: The total examination time is 140 minutes and examinees will have 70 minutes to solve 30 questions in each section. The test will also contain a 15-minute tutorial. Experts feel that since each section is of about 60 minutes duration, so one must keep ten minutes as a buffer. A couple of minutes should be spent at the beginning of every section to understand the number, type and nature of questions. When we spoke to some faculty member of coaching institutes, they specified that you should allocate equal time to all the questions and one should not spend more than four minutes in any question. You can spend up to eight minutes on a Reading Comprehension passage or Data Interpretation caselet assuming it consist about 3 questions.

Last-Minute Tips-:

1. Never make the mistake of attempting the questions without looking at all four options first
2. Try to focus on your speed and accuracy
3. Go into the exam with an open mind and expect the unexpected
4. Begin with your strongest section and attempt the weakest section in the middle
5. Do not be nervous and don't start with any new concept now

Quant Shortcuts

Quant Shortcuts

Finding number of Factors

To find the number of factors of a given number, express the number as a product of powers of prime numbers.
In this case, 48 can be written as 16 * 3 = (24 * 3)

Now, increment the power of each of the prime numbers by 1 and multiply the result.

In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)

Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 – 2 = 8 factors.



Sum of n natural numbers

-> The sum of first n natural numbers = n (n+1)/2

-> The sum of squares of first n natural numbers is n (n+1)(2n+1)/6

-> The sum of first n even numbers= n (n+1)

-> The sum of first n odd numbers= n^2



Finding Squares of numbers

To find the squares of numbers near numbers of which squares are known

To find 41^2 , Add 40+41 to 1600 =1681

To find 59^2 , Subtract 60^2-(60+59) =3481



Finding number of Positive Roots

If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.

Eg: x^4+3x^2+2x+6=0 has no positive roots .



Finding number of Imaginary Roots

For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) .
Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)



Reciprocal Roots

The equation whose roots are the reciprocal of the roots of the equation ax^2+bx+c is cx^2+bx+a



Roots

Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1



Finding Sum of the roots

For a cubic equation ax^3+bx^2+cx+d=o sum of the roots = - b/a sum of the product of the roots taken two at a time = c/a product of the roots = -d/a

For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0 sum of the roots = - b/a sum of the product of the roots taken three at a time = c/a sum of the product of the roots taken two at a time = -d/a product of the roots = e/a



Maximum/Minimum

-> If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if x=y(=k/2). The maximum product is then (k^2)/4

-> If for two numbers x*y=k(=constant), then their SUM is MINIMUM if x=y(=root(k)). The minimum sum is then 2*root(k) .



Inequalties

-> x + y >= x+y ( stands for absolute value or modulus ) (Useful in solving some inequations)

-> a+b=a+b if a*b>=0 else a+b >= a+b

-> 2<= (1+1/n)^n <=3 -> (1+x)^n ~ (1+nx) if x<<<1> When you multiply each side of the inequality by -1, you have to reverse the direction of the inequality.



Product Vs HCF-LCM 

Product of any two numbers = Product of their HCF and LCM . Hence product of two numbers = LCM of the numbers if they are prime to each other


AM GM HM

For any 2 numbers a>b a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively) (GM)^2 = AM * HM



Sum of Exterior Angles


For any regular polygon , the sum of the exterior angles is equal to 360 degrees hence measure of any external angle is equal to 360/n. ( where n is the number of sides)

For any regular polygon , the sum of interior angles =(n-2)180 degrees

So measure of one angle in

Square-----=90
Pentagon--=108
Hexagon---=120
Heptagon--=128.5
Octagon---=135
Nonagon--=140
Decagon--=144


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Problems on clocks


Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is , the minute hand describes 6 degrees /minute the hour hand describes 1/2 degrees /minute . Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .
The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight. (This can be derived from the above) .
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Co-ordinates
Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for [(a+e)/2,(b+f)/2] =[ (c+g)/2 , (d+h)/2]

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Ratio
If a1/b1 = a2/b2 = a3/b3 = .............. , then each ratio is equal to (k1*a1+ k2*a2+k3*a3+..............) / (k1*b1+ k2*b2+k3*b3+..............) , which is also equal to (a1+a2+a3+............./b1+b2+b3+..........)
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Finding multiples

x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples .For example (17-14=3 will be a multiple of 17^3 - 14^3)
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Exponents
e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity 2 <>GP
-> In a GP the product of any two terms equidistant from a term is always constant .

-> The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .



Mixtures

If Q be the volume of a vessel q qty of a mixture of water and wine be removed each time from a mixture n be the number of times this operation be done and A be the final qty of wine in the mixture then ,
A/Q = (1-q/Q)^n

Some Pythagorean triplets:

3,4,5----------(3^2=4+5)
5,12,13--------(5^2=12+13)
7,24,25--------(7^2=24+25)
8,15,17--------(8^2 / 2 = 15+17 )
9,40,41--------(9^2=40+41)
11,60,61-------(11^2=60+61)
12,35,37-------(12^2 / 2 = 35+37)
16,63,65-------(16^2 /2 = 63+65)
20,21,29-------(EXCEPTION)----------------------------------------------------------


Appolonius theorem

Appolonius theorem could be applied to the 4 triangles formed in a parallelogram.



Function
Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .



Finding Squares

To find the squares of numbers from 50 to 59

For 5X^2 , use the formulae

(5X)^2 = 5^2 +X / X^2

Eg ; (55^2) = 25+5 /25
=3025
(56)^2 = 25+6/36
=3136
(59)^2 = 25+9/81
=3481



Successive Discounts

Formula for successive discounts
a+b+(ab/100)
This is used for succesive discounts types of sums.like 1999 population increses by 10% and then in 2000 by 5% so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999 and if there is a decrease then it will be preceeded by a -ve sign and likewise.



Rules of Logarithms:
-> loga(M)=y if and only if M=ay

-> loga(MN)=loga(M)+loga(N)

-> loga(M/N)=loga(M)-loga(N)

-> loga(Mp)=p*loga(M)

-> loga(1)=0-> loga(ap)=p

-> log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [ Note the alternating sign . .Also note that the ogarithm is with respect to base e ]

Divisibility rules

Divisibility rules:

-> A number is divisible by 2 if and only if the last digit is divisible by 2.


-> A number is divisible by 3 if and only if the sum of the digits is divisible by 3.


-> A number is divisible by 4 if and only if the last 2 digits is a number divisible by 4.


-> A number is divisible by 5 if and only if the last digit is divisible by 5.


-> A number is divisible by 6 if and only if it is divisible by 2 and 3.


-> A number is divisible by 8 if and only if the last 3 digits is a number divisible by 8.


-> A number is divisible by 9 if and only if the sum of the digits is divisible by 9.


-> A number is divisible by 10n if and only if the number ends in n zeros.


-> A number is divisible by 11 iff the sum of every other digit minus the sum of the rest of the digits is divisible by 11.


-> To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.


-> If n is even , n(n+1)(n+2) is divisible by 24