Skip to main content

Number system

Number system (Basic)

1. Addition: The mother of all calculation. Performing addition is no big deal for anyone but doing it at a pace of 5 to 10 times faster, which is required in competitive exams like CAT is surely a heck of work. 
Following method also called as number-line method discussed below enables doing addition without carry-over and with proper practice increases addition capacity multi folds.

Suppose we have to add 78 and 86, 


#How to progress? Start with addition of two 2 digit numbers, then try adding multiple 2 digit number and finally multiple digits numbers. 

2. Subtraction: As subtraction is extension of addition, the trick is just an extension of method discussed above.

Suppose we have to subtract 38 from 72, treat it as what we have to add to 38 to get 72. Again number line system comes to rescue. 

             

First we find the number which have to be added to 38 toe get the nearest number greater than 38 having same unit place as 72. So we have to add 4 to yield 42 having same unit place as 72. Now we add 30 to 42 to get 72. Hence 72-38=34. 

#How to progress? Start with subtracting of two 2 digit numbers, then proceed in subtracting  multiple digits numbers.

3. Multiplication: In simple terms multiplication is addition of same number many times. In general most of the competitive exams have multiplication involving numbers less than 1000. Apart from conventional multiplication technique taught in our school days, there are four other methods which can considerably reduce time taken in multiplication. 
i) Using squares: a²-b² = (a+b)(a-b). Eg: 18 x 22 = (20-2)(20+2)=20²-2²=400-4=396
   This method is not so useful under the following cases:
  •    When obtaining/computing square are difficult.
  •    Multiplying two numbers which are far apart.
  •    Multiplying two very large numbers (3 digit or 4 digit)
ii) Multiplying number close to 100 or 1000: Suppose we have to multiply 94 x 96. Following steps will be used -

  • Difference of both the number from the nearest power of 10 is figured out.                                                        
  • Th last two digit of product (94 x 96) will be the product of difference i.e. (-6)x(-4)  = 24   
  • Initial digits are obtained by cross adding difference i.e. either 94-4 or 96-6 both resulting in 90 as shown below:                                                                                                                               
                       
  • So the final product is 9024. Following are couple of more such examples, 102 x 103 and 104 x 92.  Look carefully the 2nd example i.e. 104 x 92                                                                                           
  • This method however becomes inconvenient when the numbers are far away from the power of 10. Thus adopt this method only if both the numbers are between 80 and 120. 
iii) Using addition to multiply: Th method is one of the simplest method and we will understand it with couple of examples:
    83 x 32 = 83 x 30 + 83 x 2 = 2490 + 166 = 2656. The advantage of this method is that at no point of time we doing anything more than single digit multiplication. Another example,
      77 x 48 = 77*40 + 77*8 = 77*40 + 70*8 + 7*8
iv) Use of percentage to multiply: Use of percentage is the most powerful tool required for carrying out all kinds multiplication one comes across in aptitude exams.
Suppose we have to multiply 43 and 78. First we compute 43% of 78 and then multiply the result by 100. 43% of 78 = 10% of 78 + 10% of 78 + 10% of 78 + 10% of 78 + 1% of 78 + 1% of 78 + 1% of 78= 7.8 + 7.8+7.8+7.8+0.78+0.78+0.78.
Considering only integral part, 7x4=28. Coming to fractional part, 0.8+0.8+0.8+0.8+0.78+0.78+0.78=5.54.
Hence 43% of 78 = 33.54 and 43*78=3354.
Although this method looks time consuming, but after practice and having sufficient command on addition, the total time taken will be much less than conventional multiplication procedure.

Comments

  1. tittle of article/post is missing. Please update it.

    ReplyDelete

Post a Comment

Popular posts from this blog

SSC Practice Problem Solution

SSC Practice Problem Solution In this post, we will discuss solution of  SSC Practice Problem . Question 1  The efficiency of (A + B) = 100/24 = (25/6)% The efficiency of (A + B + C) = 100/8 = (25/2)% The efficiency of C = 25/2 – 25/6 = (50/6)% Hence, C can alone finish this job in = 100/(50/6) = 12 days Answer: 12 days option (1) Question 2  Original price = Rs. 6000; Price after discount = 6000 – 1200 = Rs. 4800; Price after raising service contract = 4800 + 480 = Rs. 5280 Answer: Rs. 5280 option (2) Question 3  Suppose the received money by A, B, and C is respectively 5x, 6x, and 9x. 5x = 450; => x = 90; Hence, the total money = 20x = 20*90 = Rs. 1800; Answer: Rs. 1800 option (2) Question 4 Suppose the Cost price of the bag= Rs. x; Hence, x + 0.15x = 230; => x = Rs. 200; Selling price after selling on 20% = 200 + 20% of 200 = Rs. 240 Answer: Rs. 240 option (3) Question 5  The required percentage ...

Average

Average In this post, We will learn about average.  Average of any collection of data is the sum of data divide by number of data. Average = (Sum of observations /  Number of observations) Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.  problems The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is (CAT 2018) As specified in the question mean of x and y and z is 80 x+y+z=3*80 mean of x, y, z,u,v is 75 x+y+z+u+v=75*5 2u=x+y 2v=y+z x+y+z+((x+y+y+z)/2)=75*5 y+3(x+y+z)=75*10 y=750-720 y=30 since x+y+z=240 y=30 so x+z=210 since x>=z so minimum value of x=210/2 = 105 Problem2: In first 5 inings sehwag has scored at average of 40. In sixth ining He had scored 52 what is His average after 6th ining. one simple approach is getting total score divided by no of inin...

Complex Number

Complex Number While solving the equation of the form  a x 2 + b x + c = 0 , the roots of the equations can take three forms which are as follows: Two Distinct Real Roots Similar Root No Real roots (Complex Roots) The introduction of complex numbers in the 16th century made it possible to solve the equation  x 2 + 1 = 0 . The roots of the equation are of the form  x = ± − 1 ‾ ‾ ‾ √  and no real roots exist. Thus, with the introduction of complex numbers, we have Imaginary roots. We denote  − 1 ‾ ‾ ‾ √  with the symbol i, where i denotes Iota (Imaginary number). An equation of the form z= a+ib, where a and b are real numbers, is defined to be a complex number. The real part is denoted by Re z = a and the imaginary part is denoted by Im z = b. Algebraic Operation on Complex numbers: Addition of two complex numbers Subtraction of two complex number Multiplication of two complex number Division of two complex number Power of Iota (i) ...

Logarithm

Logarithm When a x   = N , then we say that x = logarithm of N to the base a and write it as x = log a N . In simple words, it represents the power to which a number must be raised. Let me expand on that by giving a simpler example. If we are asked, what would be the result if ‘a’ is multiplied with itself ‘b’ times; then your answer would be x = a*a*a*…. b (times). This can also be written as a^b. This is also known as ‘a raised to the power of b’ If we are asked, which number multiplied with itself ‘b’ times, will result in a; then you are asked for the value of x such that x*x*x*… b(times) = a => x^b = a => x = a^(1/b) This is also known as ‘bth root of a’ If swe are asked, how many times should you multiply ‘a’ with itself to get ‘b’, that is where the concept of logarithm comes into the picture. You are asked for the value of ‘x’ such that a*a*a…. x (times) = b => x = Log  a  b This is also known as ‘ Log b to the base a’ ...