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i want shortcuts in quantitative aptitude plz send me the related shortcuts to it

I suggested you to study the book R.S. AGGRAWAL. this book is the best book for this subject . another many books age as follows-----











best of luck

Tricks and
Tips !!

See the way to Go out of this issue is Only practice, Don't think much about Shortcut tricks because some times they do fails.

For Example if you Want to solve 50 maths/aptitude problems in just 25 mins then Yes you can but Not through Tricks but through Practice.

Try to solve as many problems you can ! Because If you try hard for practice they familiar Questions gets answer in few seconds in the Exams.

But if you Go for Short cuts then they need some manipulations to Solve them..

I must say Go for practice and Hard work, You will really succeed in your Aim.

As during practice it will take time, But Excessive practice of same kind of questions make you More confidant and Passionate..

Good luck and Hope you will crack the Exam..

No special short cut tricks exist to solve quants questions. If your concepts are clear you can make your own short cuts. What I mean is very simple fact. If you have sound understanding of different concepts, have a lot of practice, you will be able to solve the problem quickly skipping some steps doing some mental maths. So, if your goal is to solve problems quickly and efficiently, there is no way out there except practice and hard working.
It is sometimes hyped that one can learn a lot of short cut tricks after joining coaching classes. Remember it is only a hype. May be for one or two questions solving by some special methods are possible but it is only a few. Short cuts comes in handy only when your fundamentals are strong. To make fundamentals strong you are required to practice a lot.
So, start practice. There is no short cut for success.
Regards
Uttam Nandi

Dear Friend,

To find the number of factors of a given number, express the number as a product of powers of prime numbers.

In this case, 48 can be written as 16 * 3 = (24 * 3)

Now, increment the power of each of the prime numbers by 1 and multiply the result.

In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)

Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 - 2 = 8 factors.
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The sum of first n natural numbers = n (n+1)/2

The sum of squares of first n natural numbers is n (n+1)(2n+1)/6

The sum of first n even numbers= n (n+1)

The sum of first n odd numbers= n^2

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To find the squares of numbers near numbers of which squares are known

To find 41^2 , Add 40+41 to 1600 =1681

To find 59^2 , Subtract 60^2-(60+59) =3481
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If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.
eg: x^4+3x^2+2x+6=0 has no positive roots .
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For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) .
Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)
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For a cubic equation ax^3+bx^2+cx+d=o

sum of the roots = - b/a
sum of the product of the roots taken two at a time = c/a
product of the roots = -d/a
----- ----- ----- ----- -----
For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0

sum of the roots = - b/a
sum of the product of the roots taken three at a time = c/a
sum of the product of the roots taken two at a time = -d/a
product of the roots = e/a
----- ----- ----- ----- -----
If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if
x=y(=k/2). The maximum product is then (k^2)/4
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If for two numbers x*y=k(=constant), then their SUM is MINIMUM if
x=y(=root(k)). The minimum sum is then 2*root(k) .
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|x| + |y| >= |x+y| (|| stands for absolute value or modulus )
(Useful in solving some inequations)
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Product of any two numbers = Product of their HCF and LCM .
Hence product of two numbers = LCM of the numbers if they are prime to each other

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For any regular polygon , the sum of the exterior angles is equal to 360 degrees
hence measure of any external angle is equal to 360/n. ( where n is the number of sides)

For any regular polygon , the sum of interior angles =(n-2)180 degrees

So measure of one angle in

Square =90
Pentagon =108
Hexagon =120
Heptagon =128.5
Octagon =135
Nonagon =140
Decagon = 144

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If any parallelogram can be inscribed in a circle , it must be a rectangle.

----- ----- ----- ----- -----

If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:e oblique sides equal).

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For an isosceles trapezium , sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .
----- ----- ----- ----- -----

Area of a regular hexagon : root(3)*3/2*(side)*(side)
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For any 2 numbers a>b

a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively)

(GM)^2 = AM * HM
----- ----- ----- ----- -----

For three positive numbers a, b ,c

(a+b+c) * (1/a+1/b+1/c)>=9

----- ----- ----- ----- -----
For any positive integer n

2<= (1+1/n)^n <=3

----- ----- ----- ----- -----

a^2+b^2+c^2 >= ab+bc+ca
If a=b=c , then the equality holds in the above.

a^4+b^4+c^4+d^4 >=4abcd
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(n!)^2 > n^n (! for factorial)
----- ----- ----- ----- -----

If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s will be maximum
if a/p = b/q = c/r = d/s .
----- ----- ----- ----- -----

Consider the two equations

a1x+b1y=c1
a2x+b2y=c2

Then ,
If a1/a2 = b1/b2 = c1/c2 , then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2 , then we have no solution for these equations.(<> means not equal to )
If a1/a2 <> b1/b2 , then we have a unique solutions for these equations..
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For any quadrilateral whose diagonals intersect at right angles , the area of the quadrilateral is
0.5*d1*d2, where d1,d2 are the lenghts of the diagonals.
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Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is ,
the minute hand describes 6 degrees /minute
the hour hand describes 1/2 degrees /minute .

Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .

The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight.
(This can be derived from the above) .
----- ----- ----- ----- -----
If n is even , n(n+1)(n+2) is divisible by 24

If n is any integer , n^2 + 4 is not divisible by 4

----- ----- ----- ----- -----
Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for
[(a+e)/2,(b+f)/2] =[ (c+g)/2 , (d+h)/2]

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Area of a triangle
1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where s=a+b+c/2
=a*b*c/(4*R) where R is the CIRCUMRADIUS of the triangle = r*s ,where r is the inradius of the triangle .

In any triangle
a=b*CosC + c*CosB
b=c*CosA + a*CosC
c=a*CosB + b*CosA
----- ----- ----- ----- -----

If a1/b1 = a2/b2 = a3/b3 = .............. , then each ratio is equal to
(k1*a1+ k2*a2+k3*a3+..............) / (k1*b1+ k2*b2+k3*b3+..............) , which is also equal to
(a1+a2+a3+............./b1+b2+b3+..........)
----- ----- ----- ----- -----

(7)In any triangle
a/SinA = b/SinB =c/SinC=2R , where R is the circumradius
----- ----- ----- ----- -----x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples .For example (17-14=3 will be a multiple of 17^3 - 14^3)
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e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
2 < e < 3
----- ----- ----- ----- -----

log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [ Note the alternating sign . .Also note that the ogarithm is with respect to base e ]
----- ----- ----- ----- -----

In a GP the product of any two terms equidistant from a term is always constant .
----- ----- ----- ----- -----

For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2
----- ----- ----- ----- -----

For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the internal opposite angle.

(m+n)! is divisible by m! * n! .
----- ----- ----- ----- -----
If a quadrilateral circumscribes a circle , the sum of a pair of opposite sides is equal to the sum of the other pair .
----- ----- ----- ----- -----

The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .
----- ----- ----- ----- -----
The equation whose roots are the reciprocal of the roots of the equation ax^2+bx+c is cx^2+bx+a
----- ----- ----- ----- -----
The coordinates of the centroid of a triangle with vertices (a,b) (c,d) (e,f)
is((a+c+e)/3 , (b+d+f)/3) .
----- ----- ----- ----- -----
The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 .
----- ----- ----- ----- -----

Area of a parallelogram = base * height
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APPOLLONIUS THEOREM:

In a triangle , if AD be the median to the side BC , then
AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .

----- ----- ----- ----- -----

for similar cones , ratio of radii = ratio of their bases.

The HCF and LCM of two nos. are equal when they are equal .

----- ----- ----- ----- -----

Volume of a pyramid = 1/3 * base area * height
----- ----- ----- ----- -----

In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.

----- ----- ----- ----- -----

In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.
----- ----- ----- ----- -----
The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.
----- ----- ----- ----- -----

Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1
----- ----- ----- ----- -----

|a|+|b| = |a+b| if a*b>=0
else |a|+|b| >= |a+b|
----- ----- ----- ----- -----

2<= (1+1/n)^n <=3
----- ----- ----- ----- -----
WINE and WATER formula:

If Q be the volume of a vessel
q qty of a mixture of water and wine be removed each time from a mixture
n be the number of times this operation be done
and A be the final qty of wine in the mixture

then ,
A/Q = (1-q/Q)^n
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Area of a hexagon = root(3) * 3 * (side)^2
----- ----- ----- ----- -----

(1+x)^n ~ (1+nx) if x<<<1
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Some pythagorean triplets:

3,4,5 (3^2=4+5)
5,12,13 (5^2=12+13)
7,24,25 (7^2=24+25)
8,15,17 (8^2 / 2 = 15+17 )
9,40,41 (9^2=40+41)
11,60,61 (11^2=60+61)
12,35,37 (12^2 / 2 = 35+37)
16,63,65 (16^2 /2 = 63+65)
20,21,29(EXCEPTION)
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Appolonius theorem could be applied to the 4 triangles formed in a parallelogram.
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Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * height
where median is the line joining the midpoints of the oblique sides.
----- ----- ----- ----- -----++++

when a three digit number is reversed and the difference of these two numbers is taken , the middle number is always 9 and the sum of the other two numbers is always 9 .
----- ----- ----- ----- -----
ANy function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .
----- ----- ----- ----- -----

Let W be any point inside a rectangle ABCD .
Then
WD^2 + WB^2 = WC^2 + WA^2

Let a be the side of an equilateral triangle . then if three circles be drawn inside
this triangle touching each other then each's radius = a/(2*(root(3)+1))
----- ----- ----- ----- -----++++

Let 'x' be certain base in which the representation of a number is 'abcd' , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d
----- ----- ----- ----- -----
when you multiply each side of the inequality by -1, you have to reverse the direction of the inequality.
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To find the squares of numbers from 50 to 59

For 5X^2 , use the formulae

(5X)^2 = 5^2 +X / X^2

Eg ; (55^2) = 25+5 /25
=3025
(56)^2 = 25+6/36
=3136
(59)^2 = 25+9/81
=3481
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many of u must b aware of this formula, but the ppl who don't know it must b useful for them.
a+b+(ab/100)

this is used for succesive discounts types of sums.
like 1999 population increses by 10% and then in 2000 by 5%
so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999

and if there is a decrease then it will be preceeded by a -ve sign and likeiwse

Dear Aspirant,

I provide you lot of short cuts.but you are not remembering all short cut
Short cuts not useful for Exams.Because In Exam fear ,we forget all shortcuts.

Show My better Advise and Tips for you..

TIPS

I provide here some valuable information here,Those are very useful for to Solve the answers with in time.

You have to learn subject 100% you can do any bit with in time.

If you want to do all the questions with in time.You should learn TABLES 1 to 20

And you learn Basics in every chapter and fundamental rules also

And practice all the questions from different branded materials.

above three tips leads to you solve any question with in time.

Use full books for you do the fast in Exams is Quicker maths-MS tyra BSC publications.

All the best for you

Practice is main hint for solve any question with in time.

As the time alloted in quantitative aptitude is less there are chances that you might answer a question wrong in case a statement is twisted
The best way is to practice a lot on these type of questions because there you have race against time
You need to be quick in calculations and it comes only when you practice a lot for it

practice a new set of exam papers everyday.it will help in maintaining time.

Originally Posted by Unregistered

i want shortcuts in quantitative aptitude plz send me the related shortcuts to it

Hi, please don't rely on shortcuts, they never help you in the life. Shortcuts are not always right and flawless.

Be clear in your fundamentals and get into hard practice, you will be successful in life.

Best of luck. Feel free to ask if doubts remain.

Dear Aspirant,
TIPS AND TRICKS:
There is a trick called practise makes perfect. So,by frequently practicing a questions it will be easier to solve.
Prepare for all these types of questions.
1) Reasoning
2) Aptitude
3) Logical thinking
4) English etc.,

Also you can take an online test in this websites for the aptitude and computer related questions.
www.indiabix.com

All the best.

Hi,


There is no shortcut for success.you have to study hard for success.

you have to practiec more and more question for good command.

But if you Go for Short cuts then they need some manipulations to Solve them..

I must say Go for practice and Hard work, You will really succeed in your Aim.

As during practice it will take time, But Excessive practice of same kind of questions make you More confidant and Passionate..

hi
shortcuts in quantitative aptitude has advantages & disadvantages
in my opinion try to solve as many problem you ,then you will get practise to solve the anewer .

practise make man perfect

Originally Posted by Unregistered

i want shortcuts in quantitative aptitude plz send me the related shortcuts to it

all shorsirtcuts pls sir..

The reasoning questions can be solved more accurately by practicing as more as you can do. Since the only trick to solve reasoning question accurately is practice as much problems as you can do.

There are some particular tricks for doing the aptitude questions they are not fixed you can make them by your own practice and solve as many questions as you can.

For this purpose you can refer the book written by R.S. Agarwal. famous book writer for quantitative aptitude. Main thing that you must keep in your mind is that practice always to control time the questions are not very hard but you have to do them in particular time.

Please any one post some important basic formulas in every type of question for the quantitative aptitude test.....

how to solve apti question in mimimum time .ailso send previous paper of bank po exam to my [email protected]

See the way to Go out of this issue is Only practice, Don't think much about Shortcut tricks because some times they do fails.

For Example if you Want to solve 50 maths/aptitude problems in just 25 mins then Yes you can but Not through Tricks but through Practice.

Try to solve as many problems you can ! Because If you try hard for practice they familiar Questions gets answer in few seconds in the Exams.

But if you Go for Short cuts then they need some manipulations to Solve them..

I must say Go for practice and Hard work, You will really succeed in your Aim.

As during practice it will take time, But Excessive practice of same kind of questions make you More confidant and Passionate..

Following are suggestions for Quantitative aptitude

Improving speed of calculation

Do put in some efforts to improve your calculation speed. For attempting a question, we can divide time into two parts, to grasp the question and to calculate it further. Fast calculation will definitely save your time which you can then allocate to other questions.


Handling traps

Many a times there are language traps laid by the paper setter. Even simple questions confuse you with their tricky language which is a deliberate attempt to affect your understanding.

Tricks and
Tips !!

See the way to Go out of this issue is Only practice, Don't think much about Shortcut tricks because some times they do fails.

For Example if you Want to solve 50 maths/aptitude problems in just 25 mins then Yes you can but Not through Tricks but through Practice.

Try to solve as many problems you can ! Because If you try hard for practice they familiar Questions gets answer in few seconds in the Exams.

But if you Go for Short cuts then they need some manipulations to Solve them..

I must say Go for practice and Hard work, You will really succeed in your Aim.

Good luck.................

which book will be better for the preparation for open-matignou>