Shortcut methods for solving CAT questions by Patrick Dsouza, CAT 100 Percentiler (CAT 07, 09, 16)
patrick_dsouza last edited by zabeer
There is a lot of buzz on shortcuts for solving various questions in CAT. It is one thing to know shortcuts and another to use it in the exam. I have heard from a lot of students who tell me that after the paper I can solve the sums or think of how to solve a sum in a better way, but in the actual mocks or exams it does not click. The reason for this is that one is not habituated to think in the right way in the exam. One of the basic problems with getting shortcuts is our education system. Through out the school and college, we have been given marks for the equation / formula and the steps we put while solving the sums. So whenever we see a sum we try to look out for which equation / formula to apply rather than how to solve it faster. This is made worse by the coaching classes which focus so much on equation based solving (because that is the best way that all the students understand), that the approach to solving in any alternate way is done only once the syllabus is complete when it is too late to develop new methods or one finds it difficult to inculcate new ways of solving sums.
While solving a particular sum, you can follow the following steps:
- Firstly try solving the sum mentally, without using pen and paper.
- If you do not get the method in 2 minutes then take a pen and try solving.
- If you still do not get then go through the solution and try to understand it.
- Once you have understood the solution, come back to the sum and try to solve in an alternate method.
Lets take a simple sum to understand how to find alternate methods:
N men can do a piece of work in 8 days. The same work can be done by (N+4) men in 6 days. Find the values of N.
Method 1: Normal method of solving this would be number of man days have to be equal.
8 x N = 6 x (N+4)
N = 12.
Students usually get happy that they have got the answer and move ahead. But that does not help them to improve on CAT. What is important for CAT is to get different methods.
So we go back to the question and think, is there any other way of solving
Method 2: Use options – substitute values. We see that if we substitute 12 we get 12 x 8 = (12+4) x 6
Method 3: Ratios – days is inversely proportional to number of men. So if days is 6/8 or 3/4 th then men will become 4/3 . Also the difference is 4 – so it should be 16 and 12.
Method 4: In 8 x N = 6 x (N+4) -- > Now 6 is a multiple of 3. So left side should also be a multiple of 3. Since is not a multiple of 3, it should be N. Only one option that is multiple of 3.
What we are doing out here is finding multiple solutions to the same sum that helps to build in flexibility to solve the sums. This helps when the students are taught shortcuts. Lot of students- are so used to learning the mathematical way of solving that it becomes too late and difficult to unlearn old methods and learn new methods.
Let me give you another example. I had given the following sum from Nishit Sinha book to couple of students who had scored 94%ile + in the CAT previous year.
If 29 goats can feed on a field of uniformly grown grass in 7 days or 25 goats can feed on the same field in 9 days, how many goats will feed in 6 days?
They started by forming equation for the first part for 7 days. Formed another equation for second part for 9 days and tried to solve it. And still after 2 minutes I was not able to get the answer. I informed them that the approach was completely un-CAT like. So I explained to them that while solving any question first understand the question what is asked. Here the question would be simple but for the condition of uniformly grown grass. Next look at the options to see if you can use them. At first glance you do not get any hint. So go back to the question and focus on the issue. So to find the amount by which grass is grown, I can find the total grass in 7 days which is 29x7 = 203 goat days and the total grass in 9 days which is 25x9 = 225 goat days. So in 2 days the grass grown is 225-203 = 22 goat days. Which means the growth is 11 goat days every day. So in 6 days there will be 203-11 = 192 goat days which can be consumed by 192/6 = 32 goat days.
So only with a small change in the way you look at the question you can solve the sum in a much faster way. The issue is, the students on an everyday basis use equation to solve all the sums that are available that it is difficult for them to change the way they solve the sums. Every forum that I see has the same method of solving. They just include new formulae like the Chinese Theorem and some others which I have never heard of. Understand one thing that CAT does not check on how many formulae you know or how great are you at Math. The focus of CAT is to understand whether given a situation and options available are you able to find a solution in a simple way, and believe me almost all the sums in CAT can be solved without using equation and it can be done at a much faster pace. In CAT 14 where there were 50 problems in the first section, I was able to attempt all 50 Math section question much before the time limit and had time to go back and check the questions which I was not sure of. I managed to score a 99.99%ile in section 1 (with a 99.98 overall) and not a 100%ile which I have done quite a few times in the past.
Also while solving a sum, if you get a shortcut method, don’t just be happy with it. Strive to see if there is a better method to solve the same sum.
Lets take an example:
a, b and c are three positive numbers and s = (a + b + c) / 2
If (s -a) : (s – b) : (s – c) = 1:7:4, the ratio a : b : c =
(a) 8 : 10 : 6
(b) 9 : 4 : 12
(c) 7 : 8 : 9
(d) 11 : 5 : 8
One shortcut way is to substitute the options. If we substitute d option. Take a=11, b=5, c=8 we will get s=12 and substituting in (s -a) : (s – b) : (s – c) we see that we get the values.
But if you look at it closely you will realize you need not even solve it. In the ratio since s-b is the largest, so b has to be the smallest. We can say either (b) or (d) is the answer. Again since s-a is the smallest so a should be the largest. We can say that (d) has to be the answer.
Options are one of the keys to get the shortcut methods in most of the sums. Therefore it is said that you have read a sum completely only when you have also read the options.
Also all the shortcut methods may not be restricted to options. Lets take the following sum:
In a particular week the average number of people who visited the Taj Mahal is 40. If we exclude the holidays then the average is increased by 16. Further if we exclude also the day on which the maximum number of 112 people visited the Taj Mahal, then the average becomes 42. The number of holidays in the week is
(d) data insufficient
In this sum we one shortcut way is to substitute simple values and check. So if the average people who visited Taj Mahal was 40 and there are 7 days a week, so the total number of people who visited Taj Mahal should be 7x40 = 280. Now excluding holidays the average becomes 40+16=56. We realize for the average to be 56 the number of days should be 280/56 = 5. Now to confirm, if 112 people are excluded then 280-112 = 168 people visited in 4 days which is 42. This satisfies. But there could be a shorter way to solve this. As the total people is constant (including and excluding holidays), days will be inversely proportional to average people. More the days less the average. So when the ratio of average is 40:56, then the ratio of days will be 56:40 which is 7:5. So if initially there were 7 days, excluding holidays there will be 5 days. So 2 days will be holiday in a week.
P1, P2 and P3 are three consecutive prime numbers and P1 × P2 × P3 = 190747. What is the value of P1 + P2 + P3?
Here if we see options we see that the options are above 150. So the numbers should be above 50. Look at prime numbers above 50. We get 53, 59, 61. Multiply and we get the answer. So sum is 173.
A function f(x) is defined for real values of x as:
Easiest value to substitute is 0. But x=0 will give you a square root of negative number. So option b ruled out as there is 0. Then if we substitute 5 we see we get log to base 0 which is not possible. So x = 5 not possible so a and d options not possible. So answer is b.
In the regular hexagon shown below, what is the ratio of the area of the smaller circle to that of the bigger circle?
a. 3 : 7+2√3
b. 3 : 7+√3
c. 3 :16+ 4√3
d. 3 : 7 + 4√3
First check if the diagrams are drawn to scale. This diagram is. Also if you see the diameter of the larger circle is approximately double of that of smaller circle. So we see the options. We can say that option a which is 3:10.4 and option d which is 3:13.8 is the closest. Going back to the sum we see that it should be slightly more than 2, so the answer should be d option.
The radius of the cross-sections of pipes P1 and P2 are 7 m and 14 m respectively. Water flows through P1 at a constant rate of 10 m/s and it can alone fill a tank in 2 hours. If P1 is used as the inlet pipe and P2 as the outlet pipe then together they fill the tank in 4 hours. What is the rate of water flow (in m/s) through P2?
For the tank to fill in 2 hours it has to be filled in 10m/s, so for it to be filled in 4 hours it should be filled in 5m/s. 10m/s is coming from inlet pipe. So 5m/s should go out from outlet pipe. As the ratio of the radius is 1:2 so the ratio of the cross section area has to be 1:4. That means the same water has to go from 4 times the cross section area. So the speed will be 1/4th which is 5/4 = 1.25m/s.
10 straight lines, no two of which are parallel and no three of which pass through any common point, are drawn on a plane. The total number of regions (including finite and infinite regions) into which the plane would be divided by the lines is
(d) Not unique
When I had given this sum to one of my student, he immediately gave me a formula for it. Which is n(n+1)/2 + 1. But I was not aware of the formula and in the exam there will be a lot of sums where you may not be aware of formulas. Usually the exam paper is set in such a way where you cannot use formulae. So here you can form a pattern. So learn to solve sums without formula. If you draw 1 line 2 regions are formed. If you draw 2 line 4 regions are formed. If you draw 3 lines we see 7 regions are formed. So if you take the pattern forward where difference in numbers is increasing by 1 we get the answer as 55.
In a triangle ABC, right angled at B, a median BE and an angle bisector BD are drawn. The lengths of DE, AD and EC in the same order, are in Arithmetic progression. If the length of AC is 10 cms and AB < BC, then what is the length (in cm) of BC?
(a) 6 (b) 4 (c) 2rt5 (d) 4rt5
Look at the options and substitute BC as 6. In that case AB becomes 8 (6–8–10 is a triplet) and AB < BC does not satisfy. So we know the answer should be more than 6 cms and the only option is 4rt5.
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.
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.
Rajiv last edited by
Awesome work sir.
If possible please post more question solving techniques like it .
Secondly, sir which books did you refer for your cat prep.