Shortcuts for Engineering Entrance Exam

Every year hundred thousands of students appear for entrance level examination in pursuit of bright and promising career. Meritorious trumps the mediocre by overwhelming margins. The usual difference between two candidates separated by several ranking positions is decided by one right dot on the OMR sheet. Hence, it’s imperative that not only a candidate must tick the right answer without a doubt but also finishes his paper on time. The ranking is based on marks scored in all three sections, so it is important to attempt all sections to clear the minimum sectional cut off or else failing to comply with this condition in any of the section may deem you disqualified or will put your ranking down to abyss.

Time is of the Essence

Time and tide waits for none. Have you ever pondered why is that an overwhelming number of questions are given in the question paper that is already so cramped up with the sand of time. It is because the demand of sharp and flexible professionals is much more than an all knowledgeable, arrant, double battery individual. A sharp person will apply jacks and Vedic Arithmetic tricks to do a question in 5 or 8 seconds which would be stretched to all most a minute by Mr Straight line. This saved time would give a greater capital to Mr Sharp to reach to maximum of the questions in the allotted time than Mr Straight Line could ever dream to come across. Hence, speed and accuracy is directly proportional to great rankings.

Tricks to make the Correct Ticks

Mathematical shortcuts have always been prevalent for quick calculation since time infinite. The usage of calculator made this tricks obsolete but in modern and competitive time, they have made their way back to the expertise of smart and sharp. To explain a few we’ll lead with example and then explanation of it.

Last Digit Method

It can be used in the following problem as\

1. Find the value of 36C3
A. 6532
B. 8246
C. 7534
D. 7140

The conventional way to solve this problem is

36C3=(36X35X34)/(3X2X1)
=(6X35X34) Multiplying 6,35 and 34 will take forever
=7140. By then correct answer would have taken the time that could have been given two or more questions.

The smarter way to do this question is by applying the last digit method. On multiplying 36X35X34 it can be foreseen that whatever may be the answer the last digit will always be zero. The the last digit of the sol. of 6X5X4 is 0.  The last digit 0 comes in the answer sequence (D) as 7140 thus being the only answer among the set of options. Orally you just have saved 3/4th of a minute by doing it in 15 secs or less.
This is Last Digit Method.

Substitution Method

It can be used in the following problem as
The co efficient of x^n in the expansion of (1+x)(1-x)^n is
A. (-1)^(n-1)* (n-1)^2
B. (n-1)
C. (-1)^(n-1) * n
D. (-1)^n * (1-n)

Dealing with variable like m,n etc it is wise to assign values to these m,n.
Here, let us say n=0

The question now stands as, find the co efficient of x^0 (that’s 1) in the expansion
of (1+x)(1-x)^0=1+x

the co efficient of 1 in this expression is 1.Hence the correct option should lead us to the answer 1.

Let’s check

A. (-1)^(n-1) * (n-1)^2
=(-1)^(-1)  * (-1) ^2
= -1   Thus A is not the answer.

B.(n-1)
= 0-1
= -1   therefore B can’t be the answer

C.(-1)^(n-1) * n
=(-1)^(-1) * 0
= 0    Hence C can’t be the answer.

Since A, B, C stand void to be the answers, we can directly deduce that D is the correct answer. However let’s check if this is true.

D.  (-1)^n * (1-n)
= (-1)^0 * (1-0)
=  1
It’s stands in accordance with our estimate. Hence D is the correct answer.

The Method of Approximations

It can be used in the following problem as
Find the value of
1/2! +1/4! +1/6! +…….

A. (e^2-1) / 2e
B. (e^2-1) / 2
C. (e^2-2) / e
D. (e-1)^2 / 2e
 
Solving the above problem by method of approximation
1/2! +1/4! +1/6! +……. = 1/2 + 1/4.3.2.1 +1/6.5.4.3.2.1 +……..

=   0.5 + 1/24 +1/120 +……..

=   0.5 + 0.05 (approx) + 0.01 (approx) +….

This series is approximated by taking only the first 2 terms because the succeeding terms become negligible.
So,

1/2!+1/4!+1/6!+……. =  0.5 + 0.05    (approximately)

Now let’s dry run the options one by one
A.(e^2-1) / 2e  =(8-1)/(2.71*2) > 1
It can’t be the answer

B .(e^2-1) / 2

= (2.71^2 -1) / 2     take 2.71^2 ~= 8 (i.e.: a little less than 3^2)

= 3.5
which can’t be the answer.

Similarly

C .(e^2-2) / e = (8-2)/2.71 ~= 2
which can’t be the answer as well

D.(e-1)^2 / 2e  =(2.71-1)^2/(2*2.71)  < 1

We note that  the option D appears closer to the answer. Hence the answer is D.

The above methods may go against the habitual scheme of things but with practice and logic it comes more naturally than the tiring conventional practices.

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