Friday, September 16, 2011

How to calculate day on a given date

Day of the Week:
January has 31 days. It means that every date in February will be 3 days later than the same date in January(28 is 4 weeks exactly). The below table is calculated in such a way. Remember this table which will help you to calculate.
January0
February3
March3
April6
May1
June4
July6
August2
September5
October0
November3
December5

Step1: Ask for the Date. Ex: 1st June 1988
Step2: Number of the month on the list, June is 4.
Step3: Take the date of the month, that is 01
Step4: Take the last 2 digits of the year, that is 88.
Step5: Find out the number of leap years. Divide the last 2 digits of the year by 4, 88 divide by 4 is 22.
Step6: Now add all the 4 numbers: 4 + 01 + 88 + 22 = 115.
Step7: Divide 115 by 7 = 16 remainder 3.
The reminder tells you the day.
Sunday0
Monday1
Tuesday2
Wednesday3
Thursday4
Friday5
Saturday6
Answer: Wednesday

Wednesday, September 14, 2011

EQUATION SOLVING METHOD


x2 + 2x + 7             x + 2
__________ =        _____
x2+ 3x + 5               x + 3

In the conventional method we proceed as
x2 + 2x + 7             x + 2
__________ =       _____
x2 + 3x + 5             x + 3

(x + 3) (x2 + 2x + 7) = (x + 2) (x2 + 3x + 5)
x3 + 2x2 + 7x + 3x2 + 6x + 21 = x3 + 3x2 + 5x + 2x2 + 6x + 10
x3 + 5x2 + 13x + 21 = x3 + 5x2 + 11x + 10

Canceling like terms on both sides
13x + 21 = 11x + 10
13x – 11x = 10 – 21
2x = -11
x = -11 / 2


Now we solve the problem using anatyayoreva formula.
x2 + 2x + 7          x + 2
__________ =     _____
x2 + 3x + 5         x + 3


Consider
x2 + 2x + 7              x + 2
__________ =           _____
x2+ 3x + 5               x + 3

Observe that
x2 + 2x             x (x + 2)                     x + 2
______ =          ________         =      _____
x2+ 3x              x (x + 3)                     x + 3

This is according to the condition in the sutra. Hence from the sutra
x + 2              7
_____    =     __
x + 3              5

5x + 10 = 7x + 21
7x – 5x = -21 + 10
2x = -11
x = -11 / 2

HCF of two algebraic expression


To find the Highest Common Factor i.e. H.C.F. of algebraic expressions, the

factorization method and process of continuous division are in practice in the
conventional system. We now apply' Lopana - Sthapana' Sutra, the 'Sankalana
vyavakalanakam' process and the 'Adyamadya' rule to find out the H.C.F in a
more easy and elegant way.


Question: Find the H.C.F. of x2 + 5x + 4 and x2 + 7x + 6.

1. Factorization method:


x2 + 5x + 4 = (x + 4) (x + 1)
x2 + 7x + 6 = (x + 6) (x + 1)

H.C.F. is ( x + 1 ).


2. Continuous division process.

x2 + 5x + 4) x2 + 7x + 6 ( 1
                     x2+ 5x + 4
                    ___________
                              2x + 2 ) x2 + 5x + 4 ( ½x
                                            x2+   x
                                           __________
                                                     4x + 4) 2x + 2 ( ½
                                                                    2x + 2
                                                                   ______
                                                                         0

Thus 4x + 4 i.e., ( x + 1 ) is H.C.F.



3. Lopana - Sthapana process i.e. elimination and retention or alternate
destruction of the highest and the lowest powers is as below:

i.e.,, (x + 1) is H.C.F

Cube of a number


Cubing of Numbers:

Example : Find the cube of the number 106.
We proceed as follows:

i) For 106, Base is 100. The surplus is 6. Here we add double of the surplus i.e. 106+12 = 118.
(Recall in squaring, we directly add the surplus)
This makes the left-hand -most part of the answer.
i.e. answer proceeds like 118 / - - - - -

ii) Put down the new surplus i.e. 118-100=18 multiplied by the initial surplus i.e. 6=108.


Since base is 100, we write 108 in carried over form 108 i.e. .

As this is middle portion of the answer, the answer proceeds like
118 / 108 /....

iii) Write down the cube of initial surplus i.e. 63 = 216 as the last portion
i.e. right hand side last portion of the answer.

Since base is 100, write 216 as 216 as 2 is to be carried over. Answer is 118 / 108 / 216
Now proceeding from right to left and adjusting the carried over, we get the
answer
119 / 10 / 16 = 1191016.



let us Consider 1063.

i) The base is 100 and excess is 6. In this context we double the excess and then add.
i.e. 106 + 12 = 118. ( 2 X 6 =12 ) This becomes the left - hand - most portion of the cube.
i.e. 1063 = 118 / - - - -


ii) Multiply the new excess by the initial excess i.e. 18 x 6 = 108 (excess of 118 is 18)
Now this forms the middle portion of the product of course 1 is carried over, 08
in the middle.
i.e. 1063 = 118 / 08 / - - - - -


iii) The last portion of the product is cube of the initial excess.
i.e. 63 = 216.
16 in the last portion and 2 carried over.
i.e. 1063 = 118 / 081 / 16 = 1191016

Example:


Find 10023:
 
i) Base = 1000. Excess = 2. Left-hand-most portion of the cube becomes 1002+(2x2)=1006.


ii) New excess x initial excess = 6 x 2 = 12. Thus 012 forms the middle portion of the cube.

iii) Cube of initial excess = 23 = 8.

So the last portion is 008.

Thus 10023 = 1006 / 012 / 008 = 1006012008.


Find 943.

i) Base = 100, deficit = -6. Left-hand-most portion of the cube becomes 94+(2x6)
=94-12=82.

ii) New deficit x initial deficit = -(100-82)x(-6)=-18x-6=108
Thus middle potion of the cube = 08 and 1 is carried over.

iii) Cube of initial deficit = (-6)3 = -216
                                 __                    __
Now 943 = 82 / 08 / 16 = 83 / 06 / 16
                         _
                   1    2
                                  = 83 / 05 / (100 – 16)
                                  = 830584.




Equation of a line passing through two points


Question:Find the equation of the line passing through the points (9,7) and (5,2).

Solution:
Step1: Put the difference of the y - coordinates as the x - coefficient and vice -versa.
i.e. x coefficient = 7 - 2 = 5
y coefficient = 9 - 5 = 4.
Thus L.H.S of equation is 5x - 4y.


Step 2: The constant term (R.H.S) is obtained by substituting the co-ordinates of either of the given points in
L.H.S (obtained through step-1)
i.e. R.H.S of the equation is

5(9) - 4(7) = 45 - 28 = 17
or 5(5) - 4(2) = 25 - 8 = 17.


Thus the equation is 5x - 4y = 17.


Example: Find the equation of the line passing through (2, -3) and (4,-7).

Step 1 : x[-3-(-7)] –y[2-4] = 4x + 2y.
Step 2 : 4(2) + 2(-3) = 8 –6 = 2.
Step 3 : Equation is 4x + 2y =2 or 2x +y = 1.

Numbers ending with 5




Now we relate the formula to the ‘squaring of numbers ending in 5’.

Consider the example 252

Here the number is 25. We have to find out the square of the number. For the
number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more than
the previous one', that is, 2+1=3. The Sutra, in this context, gives the procedure
'to multiply the previous digit 2 by one more than itself, that is, by 3'. It becomes
the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S (right hand
side) of the result is 52, that is, 25.


352= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;
652= 6 X 7 / 25 = 4225;
1052= 10 X 11/25 = 11025;