Sunday, July 15, 2018

106-Detailed EMI Statement maker

Purpose/Objectives

Purpose of this tool is to help users to verify EMI account statement with the bank’s EMI account statement. This tool can help users to prevent bank fraudulence if any. Features like the variable rate of interest, part payment made and skip of any EMI allow the users to analyze their loan easily.

Overview

For any loan taken, we must repay it with equated monthly installments called as EMI. This tool requires limited information like the opening balance of a loan, rate of interest, EMI amount, part payment amounts if any and the details of EMI skipped if any as input. Based on the information this tool will give the entire EMI account statement like the Opening balance of a loan, its interest, the principal amount reduced due to EMI and part payment made and closing balance for that respective month.

An educational loan of amount 3394016 is taken at an interest in 13 % pa with an EMI of 55747. EMI started from Oct-2016. Part payment of 125000 is made in Oct-2016. Then the splitting of EMI 55747 on a loan amount 3394016 at the rate of interest of 13 % per annum for Oct-2016 is 36769 (interest) and 18978 (principal amount). Here the principal amount is reduced by 143978 (125000+18978), so the new closing balance for Oct-2016 is 3250038. See the following diagram.


In Dec-2016 and Feb-2017, these two EMIs were not paid, so they are reflected in the statement. EMI of Dec-2016 was paid in May-2017 is reflected on the month May-2017. So, the total of EMI for the month May-2017 (55747) and EMI of Dec-2016 (55747) 111494 is reflected in the month of May-2017. An interest in 55747 for 5 months (Dec-16 to May-2017) at 18% interest in EMI skipped with a penalty for 427 is also reflected on this sheet for all such charges for this year.

This tool takes care of all the criteria such as:
1)      Change in EMI.
2)      Change in rate of interest.
3)      Part payment of a loan.
4)      If any EMI is skipped.
5)      The charges and interest for skipped EMI.

This interactive tool for calculating the entire EMI account statement will help all of us to verify it with the loan provider company.

Friday, July 6, 2018

105-GRE Math Test - Important Key points and formulas Part - 8

Click here for the previous Blog

Factorization (Continued):

b) Factorization of Polynomial (Continued):

3) Synthetic Division: 

Example-2: Divide 4x6-20x5+7x2-46x+55 by (x-5) using synthetic division method.

Solution:

1) Write all the terms in descending order:
     4x6-20x5+0x4+0x3+7x2-46x+55
2) Write this polynomial in the coefficient form:
     [4, -20, 0, 0, 7, -46, 55]
3) Now we will see the division using the synthetic division method.
     a) Here the divisor is (x-5) so we have x-5=0, x=5.
   b) Here the quotient is 4x6-20x5+0x4+0x3+7x2-46x+55 which can also be written as 4x6-205+7x2-46x+55. Here the remainder is 0.
    c) So we have 4x6-205+7x2-46x+55 = (x-5) (4x5+7x-11)+0.

Example-3: Divide 5x5+20x4-2x3-8x2+3x+15 by (x+4) using synthetic division method.



Solution:
1) Write all the terms in descending order:
     5x5+20x4-2x3-8x2+3x+15
2) Write this polynomial in the coefficient form:
     [5, 20, -2, -8, 3, 15]
3) Now we will see the division using the synthetic division method.
     a) Here the divisor is (x+4) so we have x+4=0, x=-4.

   b) Here the quotient is 5x5+20x4-2x3-8x2+3x+15. Here the remainder is 3.
   c) So we have 5x5+20x4-2x3-8x2+3x+15 = (x+4) (5x4-2x2+3)+3.
Next part of GRE Math Test - Important Key points and formulas will be published in the next blog.

ANIL SATPUTE

Friday, June 22, 2018

104-GRE Math Test - Important Key points and formulas Part - 7

Click here for previous blog

Factorization (Continued):

b) Factorization of Polynomial (Continued):

1) Factor Theorem: 

For a polynomial p(x), we say that (x-a) is a factor of the polynomial p(x), if we divide polynomial p(x) by (x-a), then we get remainder as 0. i.e. p(x) = q(x) (x-a) + 0 where q(x) the another polynomial called quotient (a result obtained by dividing p(x) by (x-a)).

2) Division formula in algorithm form: 

In the algorithm formula of division,  p(x) = q(x) d(x) + r(x), 
p(x) is polynomial
q(x) is quotient
d(x) is divisor
r(x) is remainder

Any polynomial can be written in the form "p(x) = q(x) d(x) + r(x)"

3) Synthetic Division: 

Few important points are to be noted to learn synthetic division.

       a) A coefficient is a real number which is the multiplicative factor of the term in the polynomial.  
Example:
1) Let the polynomial be 8x4-3x2+7x+3. First, write all the terms in descending order. So we write it as 8x4+0x3+3x2+7x+3. Now we write it in coefficient form as 8, 0, -3, 7, 3. Here, as the power of the polynomial is 4, the total number of terms of the polynomial will be 5. 
2) Let the polynomial be 7x5First, write all the terms in descending order. So we write it as 7x5+0x4+0x3+0x2+0x+0. Now we write it in coefficient form as 7, 0, 0, 0, 0, 0. Here, as the power of the polynomial is 5, the total number of terms of the polynomial will be 6. 
         b) In case of synthetic division, the divisor must of the first degree of the form (x - a). Example (x - a) = 0, so the divisor is (x - a) with x = a.

Example: Divide 3x6-6x5+7x4-18x3+13x2-9x-2 by (x-2) using synthetic division method.

Solution:
1) Write all the terms in descending order:
     3x6-6x5+7x4-18x3+13x2-9x-2

2) Write this polynomial in the coefficient form:
     [3, -6, 7, -18, 13, -9, -2]

3) Now we will see the division using the synthetic division method.

     a) Here the divisor is (x-2) so we have x-2=0, x=2.


   b) Here the quotient is 3x5+0x4+7x3-4x2+x+2 which can also be written as 3x5+7x3-4x2+x+2. here the remainder is 0.
       c) So we have 3x6-6x5+7x4-18x3+13x2-9x-2 = (x-2) (3x5+7x3-4x2+x+2)+0.

Next part of GRE Math Test - Important Key points and formulas will be published in the next blog.

ANIL SATPUTE

Friday, March 16, 2018

103-Compound interest calculating tool

Now we will see the problem of compound interest.

Formula:
A = Total amount, P = principal or amount of money deposited or borrowed, r = annual interest rate (in decimal form), n = number of times compounded per year, t = time in years.
Example 1: If you deposit $7000 into an account paying 5 % annual interest compounded bimonthly, how much money will be in the account after 6 years? What is total interest for 6 years? As per the calculation of compound interest, what is total interest per year and what is the rate of interest pcpa?

Solution:
Given:
Principal P = 7000, Rate of interest r = 0.05, n = 2 (as the interest is calculated bi-monthly) and t = 6.
Here we can use a logarithm to do these calculations. see the following steps carefully.
  












Now see the software tool for getting these results systematically. Click on the following figure and try the result for your values.
Compound interest calculating tool

Now we will see other calculation software for compound interest in the next blog.

ANIL SATPUTE

Thursday, August 24, 2017

102-Tool to calculate difference between two dates

Click on the following link:


Now we will learn very useful tool to calculate the difference between to two dates in years, months and days. Now we calculate our age today. It is a very interesting software.

Click on the following Figure:

Tool to calculate difference between two dates

Simply enter the data in Start Date and End date fields. Here your start date is smaller than the end date. It becomes very easy to calculate your age in years/Months/days. by using this software. So many times we need to calculate the difference between two dates such as "At what age I completed my Graduation" or "What is the difference between me and my mother". This is the very easy tool to calculate such difference.

 ANIL SATPUTE

Friday, August 11, 2017

101-GRE Math Test - Important Key points and formulas Part - 6

In continuation of Blog-100, we will see all the important formulas and useful statements which are to be used in Math test of GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued)

4) (x - a) (x - b) = x (x - b) - a (x - b)
                         = x 2 - b x - a x + ab
                         = x 2 - (a + b) x + ab
Generally we call x 2 as the first term, (a + b) x as the middle term and ab as the last term. 

Basic concept: 

a) Step-1: See the sign of the last term.
b) Step-2: Here it is "+" so factorize the product of the coefficient of the first term (here it is 1) and the last term in such a way that the SUM of these two factors must be the coefficient of the middle term.
c) Step-3: Get the factors.

Example-1:

Factorize: x 2 - 10 x + 21.

a) Step-1: Here sign of the last term 21 is "+"
b) Step-2: The coefficient of the first term is 1 and the last term is 21, so the product of 1 and 21 is 21. Now the factors of 21 are 3 and 7 and as the sign of the last term is "+", their addition is -3 - 7 = -10 which is the coefficient of the middle term.  
                         = x 2 - 10 x + 21
                         = x 2 - (3 + 7) x + (3 x 7)
                         = x 2 - 3 x - 7 x + (3 x 7)
                         = x (x - 3) - 7 (x - 3)
                         = (x - 3) (x - 7)
c) Step-3: So the factors of x 2 - 10 x + 21 are (x - 3) and (x - 7) 

Example-2:

Factorize: 8 x 2 - 14 x + 5.

a) Step-1: Here sign of the last term 5 is "+"
b) Step-2: The coefficient of the first term is 8 and the last term is 5, so the product of 8 and 5 is 8 X 5. Now the factors of 8 X 5 are 2, 2, 2 and 5 and as the sign of the last term is "+", so, we take two factor in such a way that their sum will be 14. Here 2, 2, 2 and 5 will give us 4 and 10. So, here addition is -4 - 10 = -14 which is the coefficient of the middle term.  

                         = 8 x 2 - 14 x + 5
                         = 8 x 2 - (4 + 10) x + 5
                         = 8 x 2 - 4 x - 10 x + 5
                         = 4 x (2 x - 1) - 5 (2 x - 1)
                         = (4 x - 5) (2 x - 1)
c) Step-3: So the factors of 8 x 2 - 14 x + 5 are (2 x - 1) and (4 x - 5).

In the next part, we will see some important formulae.

 ANIL SATPUTE

Thursday, August 10, 2017

100-GRE Math Test - Important Key points and formulas Part - 5

In continuation of Blog-99, we will see all the important formulas and useful statements which are to be used in Math test of GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued)

3) (x - a) (x + b) = x (x + b) - a (x + b)
                          = x 2 + b x - a x - ab
                          = x 2 - (a - b) x - ab
Generally we call x 2 as the first term, (a + b) x as the middle term and ab as the last term. 

Basic concept: 

a) Step-1: See the sign of the last term.
b) Step-2: Here it is "-" so factorize the product of the coefficient of the first term (here it is 1) and the last term in such a way that the DIFFERENCE of these two factors must be the coefficient of the middle term.
c) Step-3: Get the factors.

Example-1:

Factorize: x 2 - 4 x - 21.

a) Step-1: Here sign of the last term 21 is "-"
b) Step-2: The coefficient of the first term is 1 and the last term is 21, so the product of 1 and 21 is 21. Now the factors of 21 are 3 and 7 and as the sign of the last term is "-", their subtraction is 3 - 7 = -4 which is the coefficient of the middle term. (Note: Here, the coefficient of the middle term is positive so we took it as 3 - 7).  
                         = x 2 - 4 x - 21
                         = x 2 + (3 - 7) x - (3 x 7)
                         = x 2 + 3 x - 7 x - (3 x 7)
                         = x (x + 3) - 7 (x + 3)
                         = (x - 7) (x + 3)
c) Step-3: So the factors of x 2 - 4 x - 21 are (x + 3) and (x - 7) 

Example-2:

Factorize: 8 x 2 - 18 x - 5.

a) Step-1: Here sign of the last term 5 is "-"
b) Step-2: The coefficient of the first term is 8 and the last term is 5, so the product of 8 and 5 is 8 X 5. Now the factors of 8 X 5 are 2, 2, 2 and 5 and as the sign of the last term is "-", so, we take two factor in such a way that their difference will be 18. Here 2, 2, 2 and 5 will give us 2 and 20. So, here subtraction is 2 - 20 = -18 which is the coefficient of the middle term.  
                         = 8 x 2 - 18 x - 5
                         = 8 x 2 + (2 - 20) x - 5
                         = 8 x 2 + 2 x - 20 x - 5
                         = 2 x (4 x + 1) - 5 (4 x + 1)
                         = (2 x - 5) (4 x + 1) 
c) Step-3: So the factors of 8 x 2 - 18 x + 5 are (2 x - 5) and (4 x + 1).

In the next part, we will see the remaining 1 types in detail. These 1 types are given below.
4) (x - a) (x - b) = x 2 - (a + b) x + ab     

Wednesday, August 9, 2017

99-GRE Math Test - Important Key points and formulas Part - 4

In continuation of Blog-98, we will see all the important formulas and useful statements which are to be used in Math test of GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued)

2) (x + a) (x - b) = x (x - b) + a (x - b)
                          = x 2 - b x + a x - ab
                          = x 2 + a x - b x - ab
                          = x 2 + (a - b) x - ab
Generally we call x 2 as the first term, (a + b) x as the middle term and ab as the last term. 

Basic concept: 

a) Step-1: See the sign of the last term.
b) Step-2: Here it is "-" so factorize the product of the coefficient of the first term (here it is 1) and the last term in such a way that the DIFFERENCE of these two factors must be the coefficient of the middle term.
c) Step-3: Get the factors.

Example-1:

Factorize: x 2 + 4 x - 21.

a) Step-1: Here sign of the last term 21 is "-"
b) Step-2: The coefficient of the first term is 1 and the last term is 21, so the product of 1 and 21 is 21. Now the factors of 21 are 3 and 7 and as the sign of the last term is "-", their subtraction is 7 - 3 = 4 which is the coefficient of the middle term. (Note: Here, the coefficient of the middle term is positive so we took it as 7 - 3).  
                         = x 2 + 4 x - 21
                         = x 2 + (7 - 3) x - (3 x 7)
                         = x 2 + 7 x - 3 x - (3 x 7)
                         = x (x + 7) - 3 (x + 7)
                         = (x - 3) (x + 7)
c) Step-3: So the factors of x 2 + 4 x - 21 are (x - 3) and (x + 7) 

Example-2:

Factorize: 8 x 2 + 18 x - 5.

a) Step-1: Here sign of the last term 5 is "-"
b) Step-2: The coefficient of the first term is 8 and the last term is 5, so the product of 8 and 5 is 8 X 5. Now the factors of 8 X 5 are 2, 2, 2 and 5 and as the sign of the last term is "-", so, we take two factor in such a way that their difference will be 18. Here 2, 2, 2 and 5 will give us 2 and 20. So, here subtraction is 20 - 2 = 18 which is the coefficient of the middle term.  
                         = 8 x 2 + 18 x - 5
                         = 8 x 2 + (20 - 2) x - 5
                         = 8 x 2 + 20 x - 2 x - 5
                         = 4 x (2 x + 5) - 1 (2 x + 5)
                         = (2 x + 5) (4 x - 1) 
c) Step-3: So the factors of 8 x 2 + 18 x + 5 are (2 x + 5) and (4 x - 1).

In the next part, we will see the remaining 2 types in detail. These 2 types are given below.

3) (x - a) (x + b) = x 2 - (a - b) x - ab
4) (x - a) (x - b) = x 2 - (a + b) x + ab