Quant in an unorthodox way - Ankan Sengupta, FMS Delhi
FMS Delhi (2015 - 2017), CAT 2014 - 99.79 percentile
Most of us are wired to solve a problem in a conventional,formula based way as soon as we see the problem. But now a days most of the competitive exams,including CAT,wants something more from you,that is speed.Now speed comes from 2 things: clarity of concept and circumventing a regular approach, most of the times with the help of techniques that are not so widespread. The purpose of this article is about some of the unconventional methods and to share with readers the basics of solving a problem in a different way, and most of all thinking differently.
The way of doing it, use options.
A person travels equal distances with 3kmph, 4kmph and 5kmph speed. And takes 1 hour and 34 minutes to complete the journey.what’s his average speed throughout the journey ?
Ans: We can solve this one wthout calculating the total time module like the conventional one. When we try to find average speed of a journey divided in equal modules,we just take the HM of average speeds. Now you don’t even need to find the HM here. We all know AM > HM. here AM of these 3 speeds = 4 kmph.so obviously answer will be less than 4. Otion a satisfies. Here we used a known technique AM > HM in a different way to find our solution, and trust me in an exam where time matters most, these kind of techniques will come in pretty handy.
The three sides of an integral triangle are in AP. Possible length of one side is:
General approach is that we take the sides as a=x, a and a+x and then solve it algebraically. We can use pythagorean triplets intead. We know 3,4,5 is a Pythagorean triplet and this is the basic Pythagorean triplet that gives rise to an AP series.So if we multiply this triplet with any natural number we’ll get another triplet that is in AP. So from options just find which answer is divisible by either 3 or 4 or 5. Option d satisfies only. This solution would hardly take 10 seconds once you know it.
Now here comes one of my most favourite question that I solved in this way. Most of the students will leave it in the exam just by seeing it. I just solved it through the options just by logic. Let me take you through it:
In rectangle ABCD,perpendiculars DE and BF are drawn on AC.Ratio of area of triangle ADE to that of triangle EDF is 2:3.If the length of rectangle is 3cm more than it’s breadth,what is the perimeter
a) 8 + 4root10
b) 20 + 4root10
c) 16 + 4root10
d) 14 + 4 root10
General technique would incorporate complex calculations based on similarity of triangles.It would be quite tedious. My technique would be to use options again. As per as the question the perimeter=2(x+x+3=2(2x+3)),x=breadth. Now by the options we can presume x is of the form a+kroot10,both a and k are integers.now just divide all the options by 2 and subtract 3 from them.The option that’ll give an even numer(cause 2(2x+3)/2=2x+3 and subtracting 3 means 2x only) will be the answer.So the option which when divided by 2 gives an answer of the form a’+k’root10,where a’ is odd(then only after subtracting 3 a’-3 becomes even),will be the answer.only option d satisfies. This would take hardly 30 seconds to solve whereas most of the aspirants will leave it.
Will see another one using deduction.
The sum of 3/2 + 5/4+ 9/8 + 17/16 + ... to 99 terms is
a) 100 - 1/2^99
b) 101 - 1/2^99
c) 100 + 1/2^99
d) 100 + 1/2^100
Here we can go with the theory but applying some logic can save us lot of time. On a closer look we can write the given problem as
3/2 + 5/4 + ... upto n terms
a) ( n + 1 ) - 1/2^n
b) ( n + 2 ) - 1/2^n
c) ( n + 1 ) + 1/2^n
d) ( n + 1 ) - 1/2^(n+1)
for n =1, answer should be 3/2 and only option a satisfies. If more than one option satisfies then we have to check for more values with n = 2, 3 etc.. till we get our answer.
Will be ending the article with one more suggestion: sometimes knowing a few extra theorems also helps a huge lot. Mass point geometry and ladder theorems are some of the concepts that we don’t learn generally,but an acquaintance with these theorems helps a great deal while attempting problems.
2 poles,one 78 m above the ground and another 91 m above the ground are at some distance from each other.2 strings are tied,one from top of the first pole to the bottom of the second and another from top of the second to bottom of the first.Find the height of the point above ground where the 2 strings intersect.
Now general approach involves use of similar triangles property. Again it would take around 2 min to solve. But if we go for simple ladder theorem to solve, the required height h becomes, 1/h=1/78+1/91. Solving for h we’ll get 42 m. thus we can see that the problem can be solved within 20 seconds.
So techniques like these help you a lot in saving time so that you can impart more time on tougher problems or in your weaker section. A proper planning is imperative for success in a competitive exam like CAT.
So will just say keep practising and happy learning.