# Shortcut methods to solve IIT JEE 2012 and AIEEE 2012 Problems

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[NOTE:  Try the given problems with  the conventional methods first and then look into the shortcut methods given.This makes it evident to you, the lesser  labour involved, in comparison to the conventional methods.]

In the competitive exams like the IIT/AIEEE/CET it is important to utilize every single micro second effectively.The way in which you arrive at the solution doesn’t matter. A person solving the problem fully and thereby getting the answer,is given the same marks as the person merely guessing the answer correctly.Hence it is necessary to try and use as many short cut methods as possible in order to solve more problems in less time,correctly.This section stresses on different shortcut methods which can be used to solve any kind of problem IN LESS THAN A QUARTER OF A MINUTE:

1.    Find the value of 36C3

A. 7282                                      B.7140

C. 6954                                      D.8326

Yeah,you got it right. The answer is 7140.

IF this is how you did it, you wasted approximately 45 seconds.

36C3= (36*35*34)/(3*2*1)
= ( 6*35*34)
on calculating
=7140
did you actually calculate the value of 6*35*34???
and then get the answer as 7140.

But was this necessary???Did you have to waste almost a minute when you could have got the correct answer in less than 15 seconds.

Yes,look at the options.Each option has the last digit different.
Hence if you find the last digit of the product it is sufficient.

The last digit of ( 6*35*34) is 0 which consumes only a few seconds.In the options only the option ‘b’ has the last digit as 0.Hence you can directly choose 7140 as the answer.

NOW, IF YOU ARE LIKE MOST OF THE STUDENTS,YOU WOULD HAVE WASTED ALMOST A MINUTE IN COMING TO THE SOLUTION,WHICH WAS ABSOLUTELY UNNECCESSARY.

BY USING THIS MINUTE YOU WOULD HAVE SAVED ALMOST A MINUTE WHICH YOU CAN UTILIZE FOR SOLVING OTHER PROBLEMS.

THIS IS WHERE SHORTCUT METHODS COME INTO HANDY

ABOVE DISCUSSED IS  THE “LAST DIGIT METHOD”.

This is a very simple problem and you may wonder why this has been included here.In exams,seldom would you find such a simple problem.However such a problem can be a part of another problem,as discussed later on in this section.

2.    Find the value of
1/2!+1/4!+1/6!+…….                            (AIEEE 2004)

A.(e^2-1) / 2                               B.(e^2-2) / e

C.(e^2-1) / 2e                             D.(e-1)^2 / 2e

This is also an irritating problem which requires a sound knowledge of the exponential series.This problem is tough to solve,and almost impossible, in less than a minute if you go by the normal method.

But if you examine the series carefully you may write it as,

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 can be approximated by taking only the first 2 terms because the succeeding terms become negligible.

thus

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

Now lets check the options one by one

A .(e^2-1) / 2

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

= 3.5

Similarly

B .(e^2-2) / e   =(8-2)/2.71 ~= 2 which too cant be the answer

C.(e^2-1) / 2e  =(8-1)/(2.71*2) > 1

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

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

This way of solving problems may appear a bit tedious in the beginning.But mark my words,once you solve more problems like these,you get accustomed to these methods, and can solve them easily.

THIS IS CALLED “THE METHOD OF APPROXIMATIONS”.

3.The co efficient of x^n in the expansion of

(1+x)(1-x)^n is                                                                       (AIEEE 2004)

A.  (n-1)                                   B. (-1)^(n-1) * n

C.  (-1)^(n-1) * (n-1)^2          D. (-1)^n * (1-n)

Whenever you see terms like m,n etc it is highly advisable that you put  particular values for m,n.
Here let us put n=0

Then the question becomes,find the co efficient of x^0 (which is 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.(n-1)
= 0-1
= -1   therefore A cant be the answer

B.(-1)^(n-1) * n
=(-1)^(-1) * 0
= 0    Hence B cant be the answer.

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

Since A,B,C dont qualify to be the answers,we can directly deduce that D is the correct answer.However lets check if this is true.

D.  (-1)^n * (1-n)
= (-1)^0 * (1-0)
=  1
This is exactly what we should get if the option is right.Hence D is the correct answer.

THIS IS CALLED “THE METHOD OF SUBSTITUTION”.

4. 3 particles,each of mass M are situated at the vertices of an
equilateral triangle of the side a.The only forces acting on
the particles are their mutual gravitational forces.It is
desired that each particle move in a circle,while maintaining
the original mutual separation a.Then the initial velocity
that should be given to each particle is

From the last sentence of the question it is clear that we are expected to find the value of initial velocity (the word velocity being more important here).Observing the options carefuly we notice that  each of the options have different dimensions (you will know what i mean,if you have studied the chapter on ‘Units and dimensions’ in Physics).But since we need to find  the value of initial velocity which should after all have the dimension [L1  T-1] ( irrespective of whether it is inital or final or whatever else ,so long as it is a velocity).Now check in the options to see which option has the dimension [L1 T-1].

Quantity Dimension

a (length)                                    [M0 L1 T0]

G (Gravitational constant)        [M-1 L3 T-2]

M (mass)                                     [M1 L0 T0]

2 ∏ ,3                                           [M0 L0 T0]

a) [M0 L-1/2 T1 ]      not=  [M0 L1 T-1]

b) [M0 L-3/2 T1 ] not=  [M0 L1 T-1]

c) [M0 L-1/2 T-1 ] not=  [M0 L1 T-1]

d) [M0 L1 T-1 ] =  [M0 L1 T-1]

Therefore  option d is the correct answer.

Look,how an otherwise challenging and a tedious problem can be solved in less than half a minute.Such problems aren’t very uncommon in the exams like IIT,AIEEE.CET s .You can always manage to find a couple of problems having such weak options  in most of the exams.Hence this method is a valuable tool for the exams.But this should be taken with a pinch of salt.Dont try this method on questions with  all or more than one options having the same dimensions. However if atleast one option has a different dimension then this method can be used to eliminate some options.

5.The sum of the first n terms of the series

1^2+2 . 2^2 + 3^2 +2 . 4^2 +5^2+2 . 6^2 + …………. is

n(n+1)^2/2 when n is even.

When n is odd the sum is           (AIEEE 2004)

A.3n(n+1)/2                             B.[n(n+1)/2]^2

C.n(n+1)^2/4                           D.n^2*(n+1)/2

Without even bothering to see what the question actually is let us put a value for n,

ie; n=1 (note that this value should be odd)

Now the required sum = 1^2      (the first term of the series only)
=  1

Now lets check in the options

A.3n(n+1)/2
=3*1(1+1)/2

B.[n(n+1)/2]^2
=[1(2)/2]^2
= 1

C.n(n+1)^2/4

=1

D.n^2*(n+1)/2
=1

Now we see that the options B,C,D all give us the correct answer.But only one of them is correct.Hence we need to put another value for n.Lets put n=3 and check.

The required sum is  1+8+9=18

A.3n(n+1)/2

B.[n(n+1)/2]^2
=[3(4)/2]^2
= 36      therefore not the answer

C.n(n+1)^2/4