Word Problem with Solution

Introduction to the word problems with solution:

In mathematics the word problem for a recursively presented group G is the algorithmic problem of deciding whether two words represent the same element. Although it is common to speak of the word problem for the group G strictly speaking it is a presentation of the group that does or does not have solvable word problem.


Steps for solving the word problems:

Convert the phrasing into a numeric equation

that combines smaller "expressions"

Solve the equation.

Example for Word Problems with Solutions

Area/volume/perimeter problems:

A square contains area of 16 square centimeters. What is the length of every side?

Sol:

A of a square with side-length s is the formula for area:

A = s2

Here area has given, so we can insert this value into the area formula, and distinguish what it leads:

16 = s2
4 = s

Therefore, the length of every side is 4 centimeters.

Percent problem:

Jason got 30 marks in his term test. He wants to know the percentage of marks out of 40?

Sol:

Jason had the actual mark (30) and the relative mark (40). The unfamiliar in this problem is the rate or percentage. Because the report is (40) is (some percentage) of (30), in that case the variable stands for the percentage, and the equation is:

40 = (x) (30)

40 ÷ 30 = x = 1.33

Because x situate for a percentage, we want to consider translating this decimal back into a percentage:

1.33 = 133%

30 is 133% of 20.

Jason percentage is 133%.

Investment problem:

We deposit 1000 dollars into a savings yielding 6% yearly interest; we leave that money for 2 years. How much interest achieves you get at that end of the 2 years?

Sol:

In this situation, P = 1000 dollars, r = 0.06 (because we have to transfer the percent to decimal form), then the time is t = 2. From this we can get:

I = (1000) (0.06) (2) = 120

Therefore the interest is 120 dollars.

Time and distance:

John travelling in his bike with a speed of 25 m / sec. Help John to convert  his speed in km / hr.

Sol:

John's Speed = 25 m / sec

To convert the speed in km/hr multiply 25 by 18/5

Therefore Speed in km/hr =  25 * 18 / 5

John's speed is 90 km / hr.

Practice Problems on Word Problems with Solutions:

1.)     A tunnel is 160 m long. A car crosses it in 10 seconds. Find its speed?

Solution:

16 m / sec

2.) Find the volume of brick whose dimensions are 2.8 cm * 0.5 dm * 0.7 dm?

Solution:

98 CubicCentimeters.

Integrate Area of Circle

Introduction to integrate area of circle:

A circle is a locus of a point moving at a constant distance from a fixed point. The fixed point is called as center of the circle and the constant distance is called the radius of the circle. When the fixed point moves a complete rotation, the surface covered by it is the area of the circle. If you divide the circle into a number of strips, integration of all the areas of the strips gives the area of the circle.  In a rectangular coordinate system a circle is described with center at origin.

Description of Integrate Area of Circle

In a Cartesian system a circle normally symmetrical over both the rectangular axes. It means the center is at origin. If r is the radius of the circle, the equation of the circle is given by,

x2 + y2 = r2

The following graph describes the equation of a circle.


description of integrate area of circle

Therefore, at any point on the graph, the value of y is given by,

y =  ±$ \sqrt{r^2 - x^2}$

Since we are considering the first quadrant, both the values of x and y are positive. Hence let us consider only the positive value of y.Understanding Calculus problem solver is always challenging for me but thanks to all math help websites to help me out.

Area of Circle by Integration



Area of circle by integration

Consider a small strip of a rectangle in one quadrant of the circle as shown above. Let the height of the rectangle be y and its tiny width be dx. Hence area of the rectangle is,

dA = (y)(dx).

If you integrate the above, you get the area of this quadrant of the circle. That is,

A = $\int_{0}^{r}(r^2 - x^2)^\frac{1}{2}$

=[(x/2) $ \sqrt{r^2 - x^2}$ + (r2/2)sin-1(x/r)]r0

=  [(r/2)(0) + (r2/2)(Π/2)] – [0]

= (Π r2/4)

Since the circle is symmetrical, the area of this quadrant of the circle is four times the above area.

Therefore, the total area of the circle is, 4(Π r2/4) = Π r2

Polygon with Four Right Angles

Introduction to polygon with four right angles:

A polygon is defined as any plane figure which is enclosed that have 'n' straight sides or n corners. A polygon is basically a many sided plane figure. When all the sides and angles of a polygon are equal, then it is called a regular polygon.

Sum of interior angle of a polygon  = ( n - 2) × 180°

Polygon with four right angles is a closed figure, which should be a quadrilateral. Square and rectangle are the polygons with four right angles. In square and rectangle, both the interior and the external angles are right angles.

Polygon with Four Right Angles-square:

Square is a regular polygon with four sides. All the four sides of the square are equal and hence it is a regular polygon. It has four symmetric lines with equal angles.A square is a regular polygon with our right angles



Exterior angle of a Square:

Square is a four sided polygon. The exterior angle of any polygon =  `(360 ^0)/n`

Here n = 4 for square

So the exterior angle of a square = `(360 ^0)/4`  = 90°(right angle)

All the four sides of a square forms right angle.

Interior angle of a square:

The Interior angle of any polygon = = [ `(n-2)/n`  ] × 180°

Here n = 4 for square

So the interior angle of a square = [`(4-2)/4` ] * 180°

= ( `(1)/2`  ) * 180°

Interior Angle of a Square   =  90° (right angle)

Thus all the angles of the square are right angles.I like to share this geometry problems with you all through my article.

Polygon with Four Right Angles-rectangle:

Rectangle is a polygon with four sides. It has four symmetric lines with equal angles.Rectangle is a regular polygon with four right angles :

Exterior angle of a Rectangle:

Rectangle is a four sided polygon. The exterior angle of any polygon =`(360 ^0)/n`

Here n = 4 for rectangle

So the exterior angle of a rectangle = `(360 ^0)/4`  = 90°(right angle)

All the four sides of rectangle forms right angle.

Interior Angle of a Rectangle:

The Interior angle of any polygon   = [ `(n-2)/n`  ] × 180°

Here n = 4 for rectangle

So the interior angle of a rectangle  = [`(4-2)/4` ] * 180°

= ( `(1)/2`  ) * 180°

Interior angle of a rectangle        = 90° (right angle)

Solving Inequalities for Kids

Introduction:

Inequality is one of the sub topics of algebra. The inequality itself describes the meaning that it is not equal. An inequality is defined as, the size or order of two objects or about whether they are the same or not. Solving inequalities is very much similar to solving linear equations, except for one small but important change: Inequality problems are easy to solve. There are single step inequality and multi step inequality. Solving inequalities for kids involves single step solving. It is not as much as hard to solve.

Rules for Solving Inequalities

The following steps are to be followed while solving inequalities,

Step1: Adding/subtracting the constants on both sides.

Step2: Switching sides and changing the direction of the inequality sign.

Step3: Multiplying/dividing by the same number on both sides and change the symbol according to the positive and negatice sign.

Some of the notations and symbols of inequalities is given below,

The notation a < b means “a” is less than b.
The notation a > b means “a” is greater than b.
The notation a ? b means “a” is not equal to b,
The notation a = b means “a” is less than or equal to b (or, equivalently, not  greater than b)
The notation a = b means “a” is greater than or equal to b

Solving Inequalities Problem for Kids:

Example problem for kids: 1

Solve 5x + 15 > 55

Solution:

5x + 15 > 55

Subtract 15 on both sides

5x + 15 – 1 5 > 55 – 1 5

5x > 40

Divide 5 on both sides,

5x/5 > 40/5

x > 8.

The answer is x > 8.

Example problem for kids: 2

Solve: y + 28 =  +15

Solution:

y + 28 = 15

Subtract 28 on both sides,

y + 28 – 28 = 15 – 28

y  = -13

The answer is y = - 13.


Example problem for kids : 3

Solve: 2y - 28  =  -10

Solution:

2y - 28  =  -10

Add 28 on both sides,

2y - 28 + 28 = -10 + 28

2y = 18

Divide 2 on both sides,

2y/2 = 18/2

y = 9

The answer is 9.

Period of Tangent Function

Introduction to period of tangent function:

In this article, we are going to discuss about determine the period of tangent function. If all real numbers of x, the number of trigonometric function at (t + a) is equal to the number of function at x, that is f(t+a) = f(t) then we can call the function is periodic and the period is 'a'. Now we are going to see some example problems related on the period of trigonometric tangent functions.

Explanation - Period of Tangent Function :
Definition: 

The tangent function `tan x` is said to be a periodic function with period `beta` if `tan(x+beta)=tan x` . The least positive value of `beta` is known as the fundamental period of the function.  All the circular functions (trigonometrical functions) are periodic function.

For example,

`tan(x+pi) = tan x; tan(x+2pi)= tan x; tan(x+3pi)=tanx` .

Therefore,

`tan(x+npi) = tan x, n in Z` .

Here `beta = . . . -3pi, -2pi, -pi, 0, pi, 2pi, . . .` . But the fundamental period  must be least positive quantity.

Therefore `beta = pi` is the fundamental period of tangent function.

Graph:

The graph of the periods of tangent functions y = f(x) = tan x as shown in below.



The trigonometric function `f(x) = tan x` has period `pi` , because `tan(x + pi) = tan x` .

Thus tangent function `tan` is a periodic function with fundamental period `pi` . Similarly one can prove that the functions `cot x` are also periodic functions with fundamental period `pi` while `sin x, cos x, csc x` and `sec x` are periodic with fundamental period `2pi` .

Formula:

(a). A trigonometric function `f(x) = Asin(Btheta + C) + D` .Therefore,

the periods `= (360^o)/|B| = (2pi)/|B|` .

(This formulas are also used by `cos, sec,` and `csc` )

(b). A trigonometric tangent function `f(x) = Atan(Btheta + C) + D` .Therefore,

the periods `= (180^o)/|B| = (pi)/|B|` .

(This formulas are also used by `cot` ).

LCM rule for periods of trigonometric function:

The periodic trigonometric function f(x) = Ag(x) + Bh(x), when the functions g(x) and h(x) are periodic and having periods, `P_1` and `P_2` . The variable A and B are non-zero real values. Therefore, the period of function f(x) determined from the LCM of `P_1` and `P_2` .

` LCM = ` `(LCM - OF - NUMERATORS)/(HCF - OF - DENOMINATORS)` .

For example, `P_1 = (7pi)/3` , and `P_2 = (4pi)/5` .

Period = LCM = `(28pi)/(1) = 28pi` .

Note:

The LCM rule is not applicable for when the given functions are co-functions of compare to other or when the given functions are even.Is this topic How to Find the Volume of a Cylinder hard for you? Watch out for my coming posts.

Period of Tangent Function - Example :

Determine the periods of the given trigonometric tangent functions.

(a). `f(x) = 3tan(6x)` .

Solution:

Given:

`f(x) = 3tan(6x)` .

To find the periods of trigonometric function:

The trigonometric function `f(x) = tan(x)` of graph performed through a complete cycle when the angle evaluates from `0` to `pi` or equivalently the tangent function f(x) = 3tan(6x) of complete cycle performed when `x` evaluates from `0` to `(pi)/6`, that is

`6x = pi` .

`x = (pi)/6` .  [By using the above formula].

Therefore, the required periods of the trigonometric tangent functions `f(x)=3tan(6x)` is `(pi)/6` .

Interest Rate Problem

Introduction of interest rate problem:

In general, Interest is defined as the rent or charge which we pay for utilizing a particular thing or money (Principal). Most often it represents the payment for giving or using money (principal) for a period of time.  Interest cannot be charged randomly, they are charged on some rates. The rate is decided by the owner of the property and the owner only can enjoy the benefits of the money acquired as interest. The formula and example problems for finding the rate of interest when the principal and time duration is given is seen in the following sections.                 

Formula to Calculate Rate:
The General formula to calculate the simple interest is as follows,

Simple Interest, (S.I)= `(P * N * R) /100`

Where,

P = Principal amount

N = Number of years

R = Rate of Interest.

From the above formula, the formula for rate is derived as,

Rate of Interest = `(((S.I) * 100)/ (P *N))`

Example Problems to Calculate Rate:

Pro 1:   Find the rate of interest, when an interest of 300$ is obtained for a principal of 20000$, for 2 years.

Sol :   The formula for rate of interest = `(((S.I)*100)/(P * N))`

Substituting the values in the formulae,

Rate of interest  = `(300 * 100)/ (20000* 2)`

Rate of interest = `30000/ 40000`

Rate = 0.75%

Pro 2:    Find the rate of interest, when an interest of 600$ is obtained for a principal of 10000$, for 1 year.

Sol :   The formula for rate of interest = `(((S.I)*100)/(P * N))`

Substituting the values in the formulae,

Rate of interest  = `(600 * 100)/ (10000* 1)`

Rate of interest = `60000/ 10000`

Rate = 6%

Pro 3:  Find the rate of interest, when john received a interest of 450$ on investing 5000$ for two years.

Sol :   The formula for rate of interest = `(((S.I)*100)/(P * N))`

Substituting the values in the formulae,

Rate of interest  = `(450 * 100)/ (5000* 2)`

Rate of interest = `45000/ 10000`

Rate = 4.5%

Practice Problems:

Here are few practice problems to find the rate of interest,

Q 1:   Mr.John Charged 500$ for lending 13500$ for one year, calculate rate of interest charged by him.

Q 2:   Simon got 850$ credited in his account for investing 11220$ in the bank for 2 years. Calculate rate used by the bank.

Answers:

1. 3.7%

2. 3.8%

Reciprocal Math

INTRODUCTION:

RECIPROCAL MATH:

Reciprocal math is nothing but the reciprocal of a number in math. Reciprocal is any number that divides 1. If any number is consider and that number is represented as 1 divided by the considered number then such form of the number is called as reciprocal math.

Example:

If the number is 5 then the reciprocal of the number 5 is `1/5`

Reciprocal Math is mostly used in the division of a fraction number. In the case of dividing a fraction number by another fraction number then we just change the division symbol as multiplication symbol and take reciprocal math of the second number and perform multiplication. So in many cases to solve the math problem of different chapter reciprocal math is being used.

Examples on Reciprocal Math

Find the reciprocal math of the following numbers.
a)      5

b)      21

c)      99

d)      10

e)      `5/3`

f)        `2/7`

g)     ` 1/6`

Solution

The Reciprocal math of the number 5 is `1/5`

The Reciprocal math of the number 21 is `1/21`

The Reciprocal math of the number 99 is `1/99`

The Reciprocal math of the number 10 is `1/10`

The Reciprocal math of the number `5/3` is `3/5`

The Reciprocal math of the number `2/7 ` is `7/2`

The Reciprocal math of the number `1/6 ` is `6/1.` `6/1 ` can be written as 6

2.    Divide `2/3` and  `8/6`

Solution

` 2/3` / `8/6`

Keep the first fraction as it is and change the divisible sign as multiplication and find the reciprocal math of second fraction `8/6`

The reciprocal of `8/6 ` is `6/8`

`2/3` * `6/8`

Now by simplifying these we get `1/2`

3.   Perform the operation

`7/9` divide by 3

Solution

Here 3 can be written as `3/1`

`7/9` / `3/1`

Reciprocal math of` 3/1`is `1/3`

` 7/9`   / `1/3`

` 7/27`

Questions on Reciprocal Math

Find the reciprocal math of the following numbers.   a) 8   b) 1000   c) `5/7`    d) `61/11`
Divide `21/27` by` 7/9`
Answers

1.    a) The Reciprocal math of the number 8 is 18

b) The Reciprocal math of the number 1000 is `1/1000`

c) The Reciprocal math of the number `5/7` is `7/5`

d) The Reciprocal math of the number `61/11` is `11/67`

2.  1