List of Cool Maths Tricks

Today I am going to summarize all the maths tricks that will speed up your calculation speed. I used them in CAT and IBPS exam. I am sure that these tricks will help you increase your score. I have wrote many articles in detail on these topics. Click on respective links to to get details.

List of Cool Maths Tricks

Squaring

In this simple trick we need to modify the equation and make the units digit zero. After all it is easy to multiply when units digit is zero.

For example - Find square of 43

= (43+3) × (43-3) + (3×3)
=(46×40) + 9
= (460×4) + 9
= 1840 + 9 = 1849

Read my full post on Squaring trick,

Multiplication

Vedic maths gave us the easiest method to do complex multiplications quickly. This method can be quickly explained with an example.

Multiply 62 with 32
Step 1

Step 2

Step 3

Detailed post on Multiplication technique (Multiplication of 3 digit numbers)

Read quick Multiplication technique without using complicated Vedic maths

Multiplication with 5

Simply multiply the number by 10 and then divide it by 2.
For example 99×5= 990/2= 495

Multiplication with 4

Multiplication with 9

Multiplication with 6

Multiplication with 99

Multiply the number with 100 and then minus same number from the result. Let's take an example
Multiply 32×99 = 3200 - 32 = 3168

Mixture

Questions related to mixtures can be easily solved by alligation method. By using alligation we can wide arrange of maths questions. Let me explain this with a simple example

Example

Price of wine of $60 per liter. If Samuel is adding water with and selling the mixture for $40 per liter. Profit margin remains same. What is the ratio of water and wine in the mixture.

Detailed post on Alligation method (Application on various types of questions)

Square root

Best method to find square root of large numbers is by dividing the number into parts. Let's take an example
Find square √ 1936
√ 1936 = √ 4 × √ 484 = √ 4 × √ 4 × √ 121 = 2×2×11 = 44

Simplification Techniques PDF

Time and Work

Every question in in time and work chapter can be solved easily by finding efficiency of workers or subject (such as pipes).

For example - A takes 10 days to complete a job. B takes 20 days to complete the same job. In how many days they will complete the job if they work together ?

A's efficiency = 100/10 = 10% per day
B's efficiency = 100/20 = 5% per days

A and B can do 15% of the work in a day if they work together. So they can do the whole job in 100/15 = 6.66 days or 6 days and 18 hours.

Detailed post on Time and Work chapter (Practice questions)

Profit and Loss

In case of profit 
25% of  Cost Price (1/4 of CP) = 20% of Selling Price (1/5 of SP)
Similarly 1/3 of CP  = 1/2 of SP

In case of loss
25% of  Selling Price (1/4 of SP) = 20% of Cost Price (1/5 of CP)
Similarly 1/3 of SP  = 1/2 of CP

Profit and loss practice questions

Estimation

That's the most important technique. This is not a secret that every successful candidate is using this technique during exams.

Example - 112 × 92
Simply 112 × 9 = 1008
Add a zero 10080 and then add 224 to 10080.
Answer is 10304

You need to do all the calculations in your brain. Don't use paper. You need to divide complex calculations in parts and solve it in your brain without paper. That's how toppers do complex calculations during exams.

Right now it will be difficult for you to use this method but with practice, you will be able to do any complicated calculation within seconds.


Shortcuts to getting information about the roots
-----------------------------
(1) If an equation contains all positive co-efficients of any powers of x, then it has no positive roots.[/i]
e.g. x4+3x2+2x+6=0 has no positive roots .

(2) If all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then the equation has no negative roots.
e.g. x2−x+2=0

(3) Summarizing DESCARTES rules of signs:
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).

(4)Consider the two equations
ax + by = c
dx + ey = f

Then, If ad=be=cf, then we have infinite solutions for these equations.
If ad=be≠cf , then we have no solution for these equations.
If ad≠be , then we have a unique solutions for these equations.

(5) Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i , another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real roots could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum. 

(6) If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients then the equation has no real roots in each case, except for x=0 in the second case.

(7) Besides Complex roots, even irrational roots occur in pairs. Hence if 2+root(3) is a root, then even 2-root(3) is a root . (All these are very useful in finding number of positive, negative, real, complex etc roots of an equation )

(8) |x| + |y| >= |x+y| (|| stands for absolute value or modulus ) (Useful in solving some inequations)

(9) For a cubic equation ax3+bx2+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

(10) For a biquadratic equation ax4+bx3+cx2+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


Number properties
(1) 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 .
(2) The HCF and LCM of two nos. are equal when they are equal .
(3) For any 2 numbers a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa
respectively)
(4) (GM)^2 = AM * HM
(5) For three positive numbers a, b ,c
(a+b+c) * (1/a+1/b+1/c)>=9
(6) For any positive integer n
2<= (1+1/n)^n <=3
(7) a^2+b^2+c^2 >= ab+bc+ca
If a=b=c , then the equality holds in the above.
(8) a^4+b^4+c^4+d^4 >=4abcd
(9) 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 
(10) (m+n)! is divisible by m! * n! .
(11.1)If n is even , n(n+1)(n+2) is divisible by 24
(11.2)If n is any integer , n^2 + 4 is not divisible by 4
(12) 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)
(13) 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 .
(14) 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
(15) 2<= (1+1/n)^n <=3
(16) (1+x)^n ~ (1+nx) if x<<<1
(17) |a|+|b| = |a+b| if a*b>=0 else |a|+|b| >= |a+b|

(18) In a GP (Geometric Progression?) the product of any two terms equidistant from a term is always constant .
(19)The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .

(20)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+..........)

Time and Work Formula, Shortcuts

1.     Work from Days:

If A can do a piece of work in n days, then A's 1 day's work =	1	.
	n

2.     Days from Work:

If A's 1 day's work =	1	,	then A can finish the work in n days.
	n

3.     Ratio:

If A is thrice as good a workman as B, then:

Ratio of work done by A and B = 3 : 1.

Ratio of times taken by A and B to finish a work = 1 : 3.

4.     No. of days = total work / work done in 1 day

5.     Relationship between Men and Work

More men         ------- can do ------->             More work

Less men         ------- can do ------->             Less work

6.     Relationship between Work and Time

More work        -------- takes------>               More Time

Less work         -------- takes------>               Less Time

7.     Relationship between Men and Time

More men        ------- can do in ------->          Less Time

Less men          ------- can do in ------->          More Time

8.     If M₁ persons can do W₁ work in D₁ days and M₂ persons can do W₂ work in D₂ days, then 

        

9.     If M₁ persons can do W₁ work in D₁ days for h₁ hours and M₂ persons can do W₂ work in D₂ days for h₂ hours, then

 Note:  If works are same, then M1D1h1 = M2D2h2 

10.  If A can do a work in ‘x’ days and B can do the same work in ‘y’ days, then the number of days required to complete the work if A and B work together is

11.  If A can do a work in ‘x’ days and A + B can do the same work in ‘y’ days, then the number of days required to complete the work if B works alone is

Mixtures

This is the extension of Averages which we have discussed earlier.

Formulae and shortcuts used to solve the following problems are discussed in the previous post.

Solved Problems
1. Let the cost of 2 quantities of rice be Rs.15 and Rs.19 per kg. Find the ratio of mixture which cost Rs.18 per kg?
Soln.:
          Method1:
          Let the quantity of rice of quality A (Rs.15/kg) be 'x'
          Let the quantity of rice of quality B (Rs.19/kg) be 'y'
          As per the problem, 15x+19y = 18(x+y)
          y = 3x => x/y = 1/3
          Therefore, the ratio of quantities mixed to make the quality of Rs.18 is 1:3.

          Method2: Using shortcut,
          
              Therefore, ratio is 1:3

2. Let the cost of 2 quantities of rice be Rs.15 (quality A) and Rs.19 (quality B) per kg. Find the quantity of rice of quality A that has to be mixed with 27kg of quality B to make the cost of rice as Rs.18?
Soln.:
          Method1:
          Let the quantity of rice of quality A (Rs.15/kg) be 'x'
          Let the quantity of rice of quality B (Rs.19/kg) be 'y'
          As per the problem, 15x+19y = 18(x+y)
          y = 3x => x/y = 1/3
          Therefore, the ratio of quantities mixed to make the quality of Rs.18 is 1:3.
          => 3 parts of quality B is to be mixed with 1 part of quality A
          => 27kg of quality B is to be mixed with 9kg of quality A.
          Ans: 9kgs

          Method2: Using shortcut,
          
              Therefore, ratio is 1:3
               => 3 parts of quality B is to be mixed with 1 part of quality A
               => 27kg of quality B is to be mixed with 9kg of quality A.
               Ans: 9kgs

3. Let the cost of 2 quantities of rice be Rs.15 and Rs.19 per kg. These 2 qualities of rice are mixed and sold at Rs.27 per kg of profit 50%. Find the ratio in which 2 qualities of rice mixed?
Soln.:
          Method1:
          Let the quantity of rice of quality A (Rs.15/kg) be 'x'
          Let the quantity of rice of quality B (Rs.19/kg) be 'y'
          Given Selling price, SP = 27 and the profit % is 50%
          We know that the profit%, p% = (SP-CP)/CP * 100
          => 50/100 = (27 - CP)/CP
          => CP = 18
          As per the problem, 15x+19y = 18(x+y)
          y = 3x => x/y = 1/3
          Therefore, the ratio of quantities mixed to make the quality of Rs.18 is 1:3.

          Method2: Using shortcut,
          Given Selling price, SP = 27 and the profit % is 50%
          We know that the profit%, p% = (SP-CP)/CP * 100
          => 50/100 = (27 - CP)/CP
          => CP = 18
          
              Therefore, the ratio of quantities mixed to make the quality of Rs.18 is 1:3.

4. A man purchased TV and Washing machine for Rs.30000. He sold the TV at 30% profit and washing machine at 60% profit. He makes overall profit of 50%. Then find for how much did he purchased TV and washing machine?
Soln:
          Method1:
          Let the cost price of TV be 'x'
          Let the cost price of washing machine be 'y'
          As per the problem,
          x + y = 30000 ------> Eq. 1
          SP of TV = 1.3x
          SP of washing machine = 1.6y
          1.3x + 1.6y = 30000 * 1.5
          1.3x + 1.6y = 45000 --------> Eq. 2
          On solving equations 1 and 2, x = 10000 and y = 20000
          Therefore, cost price of TV and washing machine are Rs.10000 and Rs.20000 respectively.

          Method2: Using shortcut,

          
              Therefore, ratio is 1:2. 3 parts is equivalent to 30000
              => Cost of TV is 10000 (1 part) and Cost of washing machine is 20000 (2 parts)