## Solve this maths in 5 mins

A house wife forgot her 'ATM PIN' which is a four digit number, but luckily she remembers some hints on how to recall this 'PIN'
Here are some of the clues

1. The 1st digit is half of the 2nd
2. The sum of the 2nd and 3rd is 10
3. The 4th is equal to the 2nd plus 1
4. The sum of all digits is 23

What is her ATM PIN?

## Solution

Let the four digits PIN be   wxyz, Here 1st ,2nd .3rd and 4th digit are w,x,y and z  respectively .

### According  to 1st condition

The 1st digit is half of the 2nd i.e
w = x/2   .......................................... (1 )

### 2nd condition

The sum of the 2nd and 3rd is 10 i.e

x + y = 10

This implies   y = 10 - x  .................... ( 2 )

### 3rd condition

The 4th is equal to the 2nd plus 1    i . e

z = x + 1     .......................... ( 3 )

### According  to 4th condition

The sum of all digits is 23 . i e .

w  +  x + y + z  = 23 ............ ( 4 )

### How to find 2nd digit

1st of all we have to find the value of x .because x is related to all other equations.
Putting the values of w ( from eq1) , y (from eq 2), z( from eq3) in ( 4) we get
x/2   + x + 10 - x + x + 1 = 23,cancelling 'x'

3x/2 +11 = 23
3x/2 =23-11
3x/2 = 12

x = 8 This is our 2nd  digit

Now put the value of x in  (1)
w = 8/2
w = 4  ,This is our 1st  digit

To find 3rd digit Put the value of x in (3)
z = x + 1
z = 8 + 1
z = 9 This is our 4th  digit
To find 3rd digit Put the value of x in (2)
y = 10 - 8

## Verification

1 The 1st digit is half of the 2nd      ➡️   4  and  8 , definitely 4 is half of 8

2. The sum of the 2nd and 3rd is 10  ➡️ 8+2 =10
3. The 4th is equal to the 2nd plus 1  ➡️ 9 = 8+1
4. The sum of all digits is 23   ➡️ 4 + 8 + 2 + 9 = 23

## Conclusion

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## Application of Derivative

A piece of wire 28 cm long is to be cut into two pieces. One piece is to be made into a circle and another into a square. How should the wire be cut so that the combined area of the two figures is as small as possible?

Let the wire be cut at a distance of  x meter  from one end. Therefore then two pieces of wire be x m and (28-x) m.

## Calculate Dimension of Circle and Square

Now 1st part be turned into a square and  the 2nd part be be made into a circle.

Since 1st part of the wire is turned into square. then its perimeter will be x m.
So using formula of perimeter of square , we can calculate side of the square = x/4 m

## Calculate Areas of Circle and Square

Therefore Area of square = (x/4)(x/4) sq m

A1 = x2/16

And  when 2nd part of the wire is turned to circle, then its perimeter ( circumference ) will be 28 - x m. So using formula of perimeter of square , And if  "r" be  radius of the circle , Then
Circumference of circle =  2 π r =  (28-x)
∴  r = (28-x)/2π

We know that Area of Circle A2   = π r2

A2  π[(28-x)/2π]2

## To find value/s of x

Now to find the value of x for which this function A(x) is maximum or minimum ,put A(x) = 0

## To Test the Minimum Value of  Function

Now we have the value of "x" on which either A(x) have maximum or minimum value . To check the maximum or minimum value we have to find A''(x) as follows

So A''(x) has positive value Therefore A(x) shall have maximum value at x = 112/(π + 4)

Hence two pieces of wire should be of length x m and (28-x) m

These pieces should be of length 112/(π+4) and 28π/(π + 4)

## Verification

we can calculate the sum of these pieces , it must be 28 m

#### 1st part

112/(π+4) = 112/{(22/7)+4}=112×7/50 = 784/50

#### 2nd part

28π/(π + 4) = {28×22/7}/{(22/7)+4} = 88×7/50 = 616/50

#### Sum of Two Parts

112×7/50 + 28×7/50 = (784+616)/50

= 1400/50= 28 m

## Conclusion

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## HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY MATRIX METHOD

Discussed  short cut for inverse of matrix and solving linear equations of three variables .
Consider three linear equations of three variables .

2x + 3y - 4z = 10

3x - 2y + 4z = 12

x  -  y  +  z  =  5
Changing these equations into Matrix Form like this

AX = B
X =  A-1 B - - - - -(1)

where A is Matrix of 3×3 order , which consist of  Coefficients of  x ,y  and z respectively. X is the matrix of order 3×1 and whose elements are  variables  in given linear equations . B is the matrix of 3×1 and consist of  the constant terms on the right hand sides of all the equations such that

## How to Find A-1

To find A-1 ,we have to check the determinant value of Matrix A, if the Determinant value of Matrix A is non Zero then  A-1 Exists, otherwise not.

### How to find Determinant  Det |A|

|A| = 2(2) + 3(1) - 4(-1)

|A| = 2(2) + 3(1) - 4(-1)

|A| = 4 + 3 +4

|A| =  7
Since |A| is non Zero therefore A-1 Exists

### To find Co factors of elements  and Ad joint Matrix of A

1st of  all  put all the elements of matrix A in 3 rows and 3 columns as written in matrix A then copy the 1st and 2nd columns as  4th and 5th columns , After this we have 3×5 arrangement as shown is 1st  figure given below, Now  complete the arrangement as 5×5 by copying 1st row and 2nd row as 4th and 5th row respectively. It can be seen in 2nd figure given below.

## Find Co factors of A11  element and write it in C11   position

The element whose co factor is to be find out , is marked in red. and the co factor will be calculated by eliminated that row and column in which red coloured element is lying, Here co factor will be calculated by cross multiplication of  purple coloured four elements

## Find Co factors of  A12  element and write it in C12   position

Here co factor will be calculated by cross multiplication of  purple coloured four elements

## Find Co factors of  A13  element and write it in C13   position

Here co factor will be calculated by cross multiplication of  purple coloured four elements

## How to find values of  x , y and z

Therefore ad joint of Matrix A can be written as below

Now using the property of equality of two matrices ,
x   =   52/22
y   =   -18/11 and
z    =  -15/11

## Verification

we can check whether the values of x , y and z so calculated satisfies our system of linear equation by putting their values in one or all the given equations.

## My previous Posts

Don't forget to   read this posts

Quiz of  Mathematics For You

## Conclusion

Thanks for visiting this website and spending your valuable time to read this post regarding HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY MATRIX METHOD , short cut for inverse of matrix .If you liked this post , don't forget to   share it with your friends to benefit them also ,we shall meet in next post , till then bye and take care......

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## A Quiz Of Mathematics For You

There are Total 10 questions in all . Each question is assigned with 2 marks. Attempt all Question. Matrices question papers, determinant questions and answers, All the questions are from matrices and Determinants . Score will be Displayed at the end of Quiz.

## FINAL WORDS

Thanks for watching and responding to this quiz based on Matrices and Determinants on Mathematics.

## Appeal

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## HOW TO FIND AREA BOUNDED BY THREE LINES AND CIRCLES , AREA UNDER CURVES BY INTEGRATION METHOD

How to find common area of three lines and one circles which are  intersecting at different  points with the help of an example.

## Given Lines and Curves

Consider one circle and three lines whose equations are  given below
x - 1 )2 y2 12            ......................(1)

y = x                            .....................(2)

y = -√3 (x-2)                      .....................(3)

y = 0                                   .....................(4)

Let us draw these lines and circle in coordinate planes, We can compare the equation of circle  with standard form of circle to find  the coordinate of  centre of the circle is  (1,0)  and radius of both the circles is 1.

## How To Draw Figure 1st of all check whether these lines  intersect with circle or not . And if these lines intersect with each other or with circle then what are their coordinates of points of intersections.

## Solve (1) and (2)

Putting the value of  'y' from (2) in equation (1), we get
x - 1 )2 x2 12

x2  + 12 -2×1×x +x2 12

-2x +x2 = 0
⇒   2x(-1+x) = 0
Either  2x= 0 or (-1+x) = 0
x = 0 or x = 1
Now putting the values of x in (2) we get
x = 0 when x = 0  and y = 1 when x = 1
Therefore points of intersection of (1) and (2) are
O( 0 , 0 ) and A( 1, 1 )

## Solve (1) and (3)

Putting the value of  'y' from (3) in equation (1), we get
x - 1 )2 [-√3 (x-2)]2 12
⇒  x 2 1 2  - 2x +  3 (x-2)2 = 1

x 2 1- 2x +  3 [ x 2 +4 - 4x] =1

x 2 1- 2x +  3x 2 + 12 - 12x -1 = 0
4x 2 - 14x +12 = 0

2x 2 - 7x + 6 = 0 ,    By Factorisation Method

2x 2 - 4x - 3x  + 6 = 0

2x(x-2) -3( x - 2) = 0

(x-2)( 2x - 3) = 0

Either (x-2) = 0   or ( 2x - 3) = 0
x = 2 and x = 3/2
To find the values of y , put both the values of  'x' in (3) .i.e. in  y = -√3 (x-2)

when x = 2 ,    then  y = 0
and    x = 3/2 , then  y = √3 /2

Therefore points of intersection of (1) and (3) are
C( 2 , 0 ) and B( 3/2√3 /2 )

## How to Find Required Area

Required Area = Shaded Area = Area of ΔOAL + Area of Curve ABMLA+ Area of Δ BCM
After simplification , we get

My Previous Posts

## Final words

Thanks for visiting this website and spending your valuable time to read this post regarding how to find area bounded by three lines and circle  .If you liked this post , don't forget to   share it with your friends to benefit them also ,we shall meet in next post , till then bye and take care......

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