Saturday, December 1, 2018

107-GRE Math-9-Important Key points and formulas

In continuation of part - 8, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued):

3) Synthetic Division: 

Example-4: Divide 4x4+(13/3)x3-(23/3)x2+11x-(18/3) by (x-2/3) using the synthetic division method.

Solution:

1) Write all the terms in descending order:
     4x4+(13/3)x3-(23/3)x2+11x-(18/3)
2) Write this polynomial in the coefficient form:
     [4, 13/3, -23/3, 11, -18/3]
3) Now we will see the division using the synthetic division method.
     a) Here the divisor is (x-2/3) so we have x-2/3=0, x=2/3.
b) Here the quotient is 4x3+7x2-3x+9 and the remainder is 0.
c) So we have 4x4+(13/3)x3-(23/3)x2+11x-(18/3) = (x-2/3) (4x3+7x2-3x+9)+0.

Example-5: Divide 3x5+(9/5)x4-2x-(6/5) by (x+3/) using synthetic division method.


Solution:
1) Write all the terms in descending order:
     3x5+(9/5)x4+0x3+0x2-2x-(6/5)
2) Write this polynomial in the coefficient form:
     [3, 9/5, 0, 0, -2, 6/5]
3) Now we will see the division using the synthetic division method.
     a) Here the divisor is (x+(3/5)) so we have x+3/5=0, x=-3/5.

b) Here the quotient is 3x4-2 and the remainder is 0.
c) So we have 3x5+(9/5)x4-2x-(6/5) = (x+3/5) (3x4-2)+0.

In the next part, we will see a few examples and some essential formulae.

ANIL SATPUTE

Sunday, July 15, 2018

106-Detailed EMI Statement maker

Purpose/Objectives

The purpose of this tool is to help users to verify their 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 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, part payment made, and the closing balance for that respective month.

An educational loan of amount 3394016 is taken at an interest of 13 % pa with an EMI of 55747. EMI started in 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 and is reflected in the month of May 2017. So, the total EMI for the month of 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-8-Important Key points and formulas

In continuation of part - 7, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

Factorization (Continued):

b) Factorization of Polynomial (Continued):

3) Synthetic Division: 

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

In the next part, we will see a few examples and some essential formulae.

Friday, June 22, 2018

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

In continuation of part - 6, we will see all the important formulas and useful statements which are to be used in the Math test GRE.

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 the remainder as 0. i.e. p(x) = q(x) (x-a) + 0 where q(x) 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: 

A 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 the case of synthetic division, the divisor must be 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 the 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.

In the next part, we will see a few examples and some essential formulae.

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, and 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 the total interest for 6 years? As per the compound interest calculation, 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.


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

ANIL SATPUTE