Acre of Land

Acre of land - Introduction:

The acre is a unit of area in a numeral of systems, with the regal and United States normal systems.Generally the used acres nowadays are the worldwide acre and, in the U.S., the survey acre. The general use of the acre is to determine tracts of land.

One acre is equal to

1 acre = 4,840 square yards

1 acre = 43,560 square feet

Acre of Land - Examples:

Acre of land - Example 1:

To build colleges in 25 acres how much of the square yards and the square feet are there in land square yards and square feet.

Solution:

Step 1:

x acre          = x * 4,840 square yards

x acre          = x * 43,560 square feet

Step 2:

25 acres = 25 * 4840 square yards = 121 000 square yards

25 acres = 25 * 43560 square feet = 1089 000 square feet

Answer:

Therefore 121 000 square yards (or) 1089000 square feet’s land are needed to build a college.

Acre of land - Example 2:

To build church in 15 acres how much of the square yards and the square feet are there in land square yards and square feet.

Solution:

Step 1:

x acre          = x * 4,840 square yards

x acre          = x * 43,560 square feet

Step 2:

15 acres = 15 * 4840 square yards = 72 600 square yards

15 acres = 15 * 43560 square feet = 653 400 square feet

Answer:

Therefore 72 600 square yards (or) 653 400 square feet’s land are needed to build a church.

Acre of Land - more Examples:

Acre of land - Example 1:

To build schools in 10 acres how much of the square yards and the square feet are there in land square yards and square feet.

Solution:

Step 1:

x acre          = x * 4,840 square yards

x acre          = x * 43,560 square feet

Step 2:

10 acres = 10 * 4840 square yards = 48400 square yards

10 acres = 10 * 43560 square feet = 435600 square feet

Answer:

Therefore 48400 square yards (or) 435600 square feet’s land are needed to build a school.

Acre of land - Example 2:

To build temple in 5 acres how much of the square yards and the square feet are there in land square yards and square feet.

Solution:

Step 1:

x acre          = x * 4,840 square yards

x acre          = x * 43,560 square feet

Step 2:

5 acres = 5 * 4840 square yards = 24 200 square yards

5 acres = 5 * 43560 square feet = 217 800 square feet

Answer:

Therefore 24 200 square yards (or) 217 800 square feet’s land are needed to build a temple.

Different Situation

Introduction :

In math, different situation questions are nothing but the questions involving the different situations. The situations may be past, current or future. In general, we can observe the different situation based questions via word problems. In this article different situation, we are going to discuss few questions based on different situations.

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Example Problems for Different Situation Questions:

The example problems for different situation questions are as follows:

Example 1:

A library consists of 600 books. Mike bought 80 more books for the library.  At present, how many books are there in the library?

Solution:

Number of books already in the library = 600

Number of books that Mike bought   =  80

Number of books in the library          = 600 + 80

= 680

Therefore, there are 680 books in the library.

Example 2:

Chris sold 1200 meatballs on Monday. He sold 600 more meatballs on Tuesday than on Monday. On Tuesday, how many meatballs did Chris sell?

Solution:

Number of meatballs on Monday = 1200

Steve sold 600 more meatballs on Tuesday

Number of meatballs on Tuesday = 1200 + 600

= 1800

Example 3:

There are 750 seats in a theater. At present, 500 seats are occupied. Calculate the percentage of seats that are occupied.

Solution:

Total number of seats   =  750

Occupied seats   =  500

Percentage of seats occupied  =  `<< 500 / 750>>`   x 100

= `<< 50/75>>`   x 100

=  66.67


Practice Problems for Different Situation Questions:

The practice problems for different situation questions are as follows:

1) A library consists of 2000 books. Milton bought 120 more books for the library.  At present, how many books are there in the library?

Answer: 2120 books

2) Kalvin sold 1310 meatballs on Monday. He sold 100 more meatballs on Tuesday than on Monday. On Tuesday, how many meatballs did Kalvin sell?

Answer: 1410 meatballs.

3) There are 450 seats in a theatre. At present, 410 seats are occupied. Calculate the percentage of seats that are occupied.

Answer: 91.1 percent.

Symbol for less than or Equal To

Introduction :

In math, the inequality shows the major task. In arithmetic it has many signs like greater than, less than, greater than or equal to and less than or equal to. In this list of symbols less than equal to symbol is `<=` . This less than equal to symbol specifies that, the term which is left hand side of the inequality is less than or equal to the term which is right of the inequality. In a number line it is mentioned as the closed dot on a number line and the arrow mark which is pointed to the left side of the dotted number.

Symbols and Rules:

Symbol – Symbol for less than or equal to.

`<=` This symbol is the specification of less than or equal to sign.
Rules – Symbol for less than or equal to:

When multiply the inequality term less than equal to sign by negative sign then the negative sign becomes positive and the positive sign becomes negative and the less than equal to sign becomes greater than equal to symbol.
When divide the inequality term less than sign by negative sign then the negative sign becomes positive and the positive sign becomes negative and the less than equal to symbol becomes greater than equal to sign.

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Example Problems – less than Equal to Symbol:

Example 1 – Symbol for less than or equal to:

Find the value of x from the inequality 19a+17`<=` 23

Solution:

Given, 19a+17`<=` 23

Now add the above inequality by -17 on both sides.

19a+17`<=` 23

-17 `<=` -17

----------------------

19a `<=` 6

Now divide by 19 on both sides so, a`<=` 6/19

Example 2 – Symbol for less than or equal to:

Find the value of y from the inequality -5b+83 `>=` 3.

Solution:

Given, -5b+83 `>=` 3.

Now add -83 on both sides

-5b+83`>=` 3

-83  `gt=` -83

---------------

-5b     `>=` -80

Now multiply the above term -5b `>=` -80 by negative sign.

- (-5b `>=` -80) is 5b `<=` 80.

The answer is b`<=` 16.

Example 3 – Symbol for less than or equal to:

Find the value of y from the inequality -m+6 `>=` 4.

Solution:

Given, -m+6 `>=` 4.

Now add -6 on both sides

-m+6 `>=` 4

-6 `>=` -6

---------------

-m      `>=` -2

Now multiply the above inequality -m `>=` -2 by negative sign.

- (-m `>=` -2) becomes m`<=` 2.

m`<=` 2 is the answer.

Example 4 – Symbol for less than or equal to:

Find the value of y from the inequality -n+5 `>=` 5.

Solution:

Given, -n+5 `>=` 5.

Now add -5 on both sides

-n+5 `>=` 5

-5  `>=` -5

---------------

-n      `>=` -0

Now multiply the above inequality -n`>=` -0 by negative sign.

- (-n `>=` -0) is n `<=` 0.

n`<=` 0 is the simplified form of given inequality.

Five Steps of Problem Solving

Introduction to five steps of problem solving:

Here we are going to see the article as five steps of problem solving method.In problem solving concepts we have to do follwing steps.

First understand the problem
Find what should be calculated
Use require formula
Form the equation and plug the data.
Simplify and get the result
let us see the five steps of problem solving.

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Factoring- Five Steps of Problem Solving:

Factor x^2+8x+6=0

Solution:

Step 1:

it is in the standard form of the quadratic equation Ax2+bx+c=0

Step 2:

here A=1, B=8 and c=6, the formula for the quadratic equation is

x= `(-b+- sqrt((b^2-4ac)))/ (2a) `

Step 3:

plug a, b and c values in the above formula

So it will be look like that below,

x= `(-6+- sqrt((6^2-4xx1 xx 8)))/ (2xx1) `X= `"(-6+- sqrt((36-32)))/ (2)`

Step 4:

if we solve these equation means we get the two root valueX= `"(-6+ sqrt(4))/ (2)`X=`"(-6- sqrt(4))/ (2)`

Step 5:

so the answer is x=-2 and x=-4

Geometry- five steps of problem solving:

Find the area and perimeter of the rectangle whose length is 6cm and width is 8cm?

Solution:

Step 1:

Length of the rectangle is 6cm and width is 8cm

Step2:

Formula for area of the rectangle is ==>  length* width

Step3:

Plug the values of length and width in the above formula
Area of the rectangle => 6x8=48cm2

Step 4:

Perimeter of the rectangle formula is 2(length + breath)

Step5:

Plug the values of length and width in the above formula

Perimeter = 2(6+8) =2(14) =28cm

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Fractions - Five Steps of Problem Solving

Add `5/8+3/2`

Solution:

Step 1:

Here the denominators of the fraction are different such as 8 and 2 so we take the greatest common factor of 8 and 2

Step 2:

Here the greatest common factor is 8 it will be divided by the denominator of each fraction (8 is divided by 8 so we get 1) so we have to multiply both the numerator and denominator by 1 so we get the faction like that` (5xx1)/ (8xx1) =5/8`

Step3:

Similarly we do that in the second fraction the denominator of that fraction is 2 so 8 will be divided by 2 means we get the value as 4

Step 4:

Multiply 4 on both the numerator and denominator so the second fraction looks like that below

`(3xx4)/ (2xx4) =12/8`

Step 5:

So the two fractions have the same denominator value now we do the operation as addition

`5/8+12/8=17/8`

Problem Solving Training

Introduction to problem solving training

In this article we are giving problem solving training and we can understand how to solving problems. It is very helping you to improve your problem solving skills. In mathematical terms we can see different types of problem solving. Here you can learn math terms, Exam practices test, problem solving and online test and our tutor will helps you and you can get free online tutor. Let us see Problem solving training. Please express your views of this topic An Equation with no Solution by commenting on blog.

Problem Solving:

Let us see few problems and their solving methods.

Problem solving training 1:

Solve:  `12 / 6 + 5 / 3`

Solution:

Above problem is showing fraction addition. So we should solve this problem in fraction addition operation.

Step 1: `12 /6 + 5 / 3`

Here numerator values are same, but denominators are different. So we should take LCM then only we can add both values. (We can take LCM if denominators are different).

Step 2:  `12 / 6 + 5 / 3`

Take LCM 6, 3 (therefore LCM is 6)

we have to change denominators values like 6.

Step 3: `(12*1)/(6*1) = 12/6 , (5*2)/ (3*2) = 10 / 6`

`12/6 + 10 / 6`

Now denominators are same so add both values

Step 4: `12 / 6 + 10 / 6`

`22 / 6`

Therefore `12/ 6 + 5 / 3 = 22 / 6.`

Problem solving training 2:

Solve:    20 ___ 5   = 25

Solution:

Step 1: Given 20 ___ 5   = 25.

Step2: here we find the symbols which are need for this operation.

Step 3: if we put the - (minus) symbols like 20 - 5 = 15 we can get 15.

So minus operation is not accept

Step4: + (plus) is correct operation for this problem.

20 + 5 = 25

Step 5: Therefore + (plus) symbol is making the number sentences true.

Problem solving training 3:

Divide the two fractions `40 -: 1/ 5`

Solution:

Above problem is showing fraction division. So we should solve this problem in fraction division operation.

Step 1: given` 40 -: 1/ 5`

Step 2: It denoted by `40 -: 1/5`

The right hand side denominator will be change like as 5/1 so

=   `40 * 5/1`

= `200`

Step 3: Therefore answer is 200. Is this topic formula chart for math hard for you? Watch out for my coming posts.

Practices Problems:

1) Solve this fraction `40/ 20`      answer: 2

2) Add `2 / 3 + 3/ 2 `                      answer: `13/6`

3) Find missing number 4 , 8  12 , ___ , 20 , 24 , ____ , 32      answer: 16 , 28

Problem Based Learning Math

Introduction :

Math is used throughout the whole world that has fundamental tool in various fields that include natural science, engineering, medicine, and the social sciences. Mathematics is the learning of quantity, arrangement, space, and change. Math seeks out patterns that originate the new conjecture, and ascertain truth by precise deduction from properly selected axioms and definitions.

Example Problems for Problem Based Learning Math:

Problem based learning math – Example: 1

Find a function that has choral on the slip between the lines `y=-x+3, y=-x-3` that takes the values -50 and 10 on the lower and upper lines.

Solution:

Guess `\phi(x,y) = Ax + By + C`

Find the values of `A, B, C.`

`\phi(3,0) = 3A+C=10`

`\phi(-3,0) = -3A+C=-50`

`10 = -50 + 6A`

`A = 10`

`\phi(0,3) = 3B+C=10`

`\phi(0,-3) = -3B+C=-50`

`B = 10`

`\phi(3,0) = 30 + C = 10, C = -20`

The solution is

`\phi(x,y) = 10x + 10y - 20`

Problem based learning math – Example: 2

Find the partial fraction decomposition of `\frac{4z+4}{z(z-1)(z-2)^2}.`

Solution:

`\frac{4z+4}{z(z-1)(z-2)^2} = \frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2} + \frac{D}{(z-2)^2}`

`4z+4=A(z-1)(z-2)^2 + Bz(z-2)^2 + Cz(z-1)(z-2) + D z(z-1)`

Plug in z=0 to get A=-1

Plug in z=1 to get B=8

Plug in z=2 to get D=6

Differentiate both sides of the equation once with respect to z and plug in z=2 to get C=-7.

Finally

`\frac{4z+4}{z(z-1)(z-2)^2} = \frac{-1}{z} + \frac{8}{z-1} - \frac{7}{z-2} + \frac{6}{(z-2)^2}`

Problem based learning math – Example: 3

If `u(x,y) = e^x\sin y`   find `f(x,y) = u(x,y) + i v(x,y)`   and check if it satisfies the Cauchy-Riemann equations.

Solution:

The Cauchy-Riemann equations are `u_x=v_y, v_x=-u_y.`

`u_x = e^x\sin y, u_y = e^x\cos y`

`v_y = e^x\sin y`

`v =-e^x\cos y+g(x)`

`v_x =-e^x\cos y+g'(x)`

For the CR equations to hold, we must have `g'(x)=0` so that `g(x)=c\isin{R}.`

`f(x,y) = e^x\sin y + i(-e^x\cos y + c)`

`=e^x(\sin y-i\cos y) + ic`

`=-i e^x(\cos y + \frac{1}{-i}\sin y) + ic`

` =-i e^x e^{iy} + ic = -i e^z+ic`

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Practice Problems for Problem Based Learning Math:

1. Show that if `\phi(x,y)`   is harmonic then `\phi_x - i \phi_y`   is analytic.

2. Find a function that is choral on the vertical slip from x = 1 to 2 and equals 20 and 30 at x = 1 and 2.

`Answer: \phi(x,y) = 10x+10 `

Surface Area of Prisms and Cylinders

Introduction about prism and cylinder:

Prism:

In geometry, an n-sided prism is a polyhedron made of an n-sided polygonal base, a translated copy, and n faces joining corresponding sides. Thus these joining faces are parallelograms

Cylinder:

A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder.

(Source – Wikipedia)


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Formula Used to Find the Surface Area of the Prism and Cylinder:

Rectangular prism:

Surface area of the rectangular prism (A) = 2(wh + lw + lh) square units

w – Width

h – Height

l – Length

Triangular prism:

Total Surface Area of the triangular prism (T.S.A)

T.S.A = L.S.A + 2 x Base Area

Total surface area T.S.A = (P x h + 2 A) sq. units

P - Perimeter

A - Area of the base

h - Height of the prism

Cylinder:

Surface area of cylinder (SA) = 2 p r^2 + 2 p r h square units

r – Radius

h – Height

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Surface Area of Prisms and Cylinders - Example Problems:

1. The rectangular prism has the length 25 cm, width 13cm and height 10 cm. find the surface area of the rectangular prism.

Solution:

Given:

Length (l) = 25 cm

Width (w) = 13 cm

Height (h) = 10 cm

Formula to find the surface area of the rectangular prism:

Surface area (A) = 2(wh + lw + lh) square units

= 2 (13 x 10 + 25 x 13 + 25 x 10)

= 2 (130 + 325 + 250)

= 2 (705)

= 1410

Surface area of the rectangular prism (A) = 1410 cm 2

2. The triangular prism has the base side length 3 cm, 4cm and 5 cm. its height is 10 cm. find total surface area of prism.

Solution:

Given:

Three side length of triangular prism = 3 cm, 4 cm, 5 cm

Height of the prism = 10 cm

Base of the triangle = 3 cm

Height of the triangle = 4 cm

The lateral surface area of the prism = p x h square units

= (3 + 4 + 5) x 10

=12x10
L.S.A =120 cm2

Now the area of the bases, A = 1/2 bh square units

h is the height of the triangle

= 1/2 x 3 x 4

A= 6 cm2

The total surface area of the prism = p h + 2 A square units

= 120 + 2 x 6

Total surface area of the prism  =  1440 cm2

3. The cylinder has radius r = 3 cm, h= 12 cm. Find the surface area of cylinder.

Solution:

Given:

r= 3 cm

h=12 cm

Geometric Formula:

The surface area of the cylinder  = 2 p r^2 + 2 p r h square units

= 2 x 3.14 x 32 + 2 x 3.14 x 3 x 12

= 56.52 + 226.08

The surface area of the cylinder = 282.6 cm2