Shortcut methods for solving CAT questions by Patrick Dsouza, CAT 100 Percentiler (CAT 07, 09, 16) - Part 2

  • 100 Percentile - CAT 07, 09, 16 | XAT 16, 17 : Rank 1 | 99.99 Percentile - CAT15 | 99.98 Percentile - CAT08,11,14 | Rank 1 - CET11,12

    The perception with a lot of students with regards to CAT is that CAT is for Engineers. A non engineer finds it difficult as there is a lot of Math in CAT. Most of the students tend to give up on CAT as they can’t find how to break the Math conundrum. They learn formulae by rote and try to use the formulae in different sums in the Mocks only to realize that their scores are not improving.

    The whole concept of learning formulae by rote for CAT is flawed. CAT does not check how good you are at remembering formulae and for most of the sums in CAT you cannot apply formulae directly. Rather, CAT checks whether you can logically think through sums. Its not about how much You study from Preparatory material but how smartly you can apply what you have learnt.

    If you look at the previous CAT papers, most of the sums can be solved without the use of any high funda formulae. The whole thing about learning Euler’s Formula, Chinese Theorem, etc is basically useless. Rote learning formulae just makes you focus on formulae and you try to wonder which formulae to use for the sums. Frankly speaking I do not know any of these high funda formulae. I don’t know what is Euler’s formula or Chinese Theorem or most of the formulae which are used by various classes to solve their sums in the Mock CATs. As a matter of fact, some of the formulae which I encounter in the Mocks of these papers are new to me and I do not bother to read it, but still in the given time I could get 78 marks in Math section in CAT07 and 90 marks in Math section in CAT08 both the years scoring 100%ile in the Math section. Understand this - that CAT does not check on your mathematical skill but on your reasoning skills. Let’s try and understand how using formulae or equations can make you slow

    Example: Lets take a question from CAT08

    A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?

    (1) 2 ≤ x ≤ 6 (b) 5 ≤ x ≤ 8 (c) 9 ≤ x ≤ 12 (d) 11 ≤ x ≤ 14 (e) 13 ≤ x ≤ 18


    Usually students who are used to formulae and equation will tend to use x as the initial amount of rice and try to form an equation to solve the sum. A better way would be to go reverse way. Say the last person has no rice left after he was given half the rice plus half a kg of rice. So he would have (0+1/2) x 2 = 1Kg before the last customer came in. Similarly before the second customer had come the shopkeeper would have (1 + ½) x 2 = 3. Before the first customer came in the shopkeeper would have (3 + ½) x 2 = 7kgs which is the answer.

    Another question from CAT08:

    Find the sum √(1+ 1/1^2 + 1/2^2) + √(1 + 1/2^2 + 1/3^2) + ….. + √(1 + 1/2007^2 + 1/2008^2)

    (1) 2008 – 1/2008 (2) 2007 – 1/2007 (3) 2007 – 1/2008 (4) 2008 – 1/2007 (e) 2008 – 1/2009


    It is almost unsolvable by using equations. But use the first term √(1+ 1/1^2 + 1/2^2) and we get 3/2 = 2 – ½. The denominator of last number of first term is 2. Take two terms √(1+ 1/1^2 + 1/2^2) + √(1 + 1/2^2 + 1/3^2) = 3/2 + 7/6 = 8/3 = 3 – 1/3. The denominator of the last number of second term is 3. So the answer has to be 2008 – 1/2008.

    Solving sums in these ways is not difficult and CAT does have similar type repeated.

    Take, for example, CAT05:


    1 x 2 x 3 x … x n for integer n ≥ 1. If p = 1! + (2 x 2!) + (3 x 3!) + … + (10 x 10!), then p + 2 when divided by 11! Leaves a remainder of:

    (a) 10 (b) 0 (c) 7 (d) 1


    This sum is similar to the previous sum. If p = 1! Then p + 2 = 3 when divided by 2! Gives a remainder 1. If p = 1! + (2 x 2!) = 5. then p + 2 = 7 when divided by 3! Gives a remainder 1. The result is the same for p = 1! + (2 x 2!) + (3 x 3!). So the answer for the given sum should be 1.

    Another example


    X = √(4 + √(4 - √(4 + √(4 ….)))) to infinity. Then x equals (CAT05)

    (a) 3 (b) (√13 – 1)/2 (c) (√13+ 1)/2 (d) √13


    We know that √4 – anything will be less than 2. So we get √ 4 + (less than 2). Which should be more than 2 but less than 2.5. The only option that satisfied is c option with √13 is approximately 3.5. So (3.5 + 1)/2 = 2.25. The same sum can be solved by formulae but that will take time.

    Another similar sum


    What is the value of y?


    (a) (√13 + 3)/2 (b) (√13-2)/2 (c) (√15 + 3)/2 (d) (√15 -3)/2


    Y = 1 / 2 + something. So the answer has to be less than half. Check the options. Only option less than half is d option where √15 is less than 4 lets say 3.8. So (3.8 – 3)/2 = 0.4.


    In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE : EB = 1: 2, and DF is perpendicular to MN such that NL : LM = 1: 2. The 1ength of DH in cm is (CAT05)


    (a) 2√2 – 1 (b) (2√2 -1)/2 (c) (3√2 - 1)/2 (d) (2√2 -1)/3


    In this case AE is 1. Also DH is less than AE so it has to be less than 1. From options value of (b) 0.9 and that of (d) is 0.6. But if you draw to scale you will realize that DH is almost similar to AE and cant be 0.6. So the answer has to be (b).


    A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?




    Radius of bigger Circle is 2. So diameter of the smaller circle is less than 1. So radius of smaller circle is less than 0.5. From options (a) is 0.2 and (d) is 0.4. So the answer is (d) as it can’t be as small as 0.2.

    What should be done to improve Math?

    • Keep away from difficult formulae. The less formulae you know the better it is. Obviously you need to know basic formulae like 1 + 2 + 3 + … = n(n+1)/2 or (a+b)^2 = a^2 + 2ab + b^2. But don’t get into any new formulae you have not heard of. It will only slow you down.

    • Go back to all the mocks you have solved and go through the questions again to search for better ways of solving. If you find it write it down next to the sum so that when you go through the paper again you could remember the methods you used. Also once before the exam go and check the shortcut methods you used to solve the sums. This would ensure that it is fresh in your mind when you solve the paper.

    • Continue writing the usual Mock Papers and analyze it. Try to get better ways of solving it than that provided by explanatory answers.

    • Go through the 5 previous CAT papers and see if you could get methods to solve the questions there. Understand this that CAT does not repeat questions, but the methods used to solve CAT sums can be repetitive as shown in the examples.


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