Fraction Math Word Problems

Introduction to fraction math word problems:

Numbers can be represented in different forms like decimals, fractions in the mathematics. Fraction is used to represent the values accurately. Fractions consist of two parts, the numerator and denominator values and it can undergo all the basic mathematical operations. Here we will prepare word problems for fractions. In this article we will illustrate about fraction math word problems.

Fraction Math Word Problems:

Fraction math word problems simply involves in preparing word problems for fractions. Some sample problems are given as follows,

Example problem 1- Fraction math word problems

In the car showroom we have 50 cars; one- fifth of the car is Lenovo. Find out how many Lenovo cars present in the car showroom?

Solution:

`1 / 5` of 50 cars are Lenovo.

To find the number of Lenovo,

Divide 50 ÷ 5 = 10

10 Lenovo are` 1 / 5` of 50.

10 Lenovo are in the car showroom.

Example problem 2:

Of the soft drinks sold yesterday at Kevin’s supermarket, `2/4` was Pepsi and another `3 /4` was coke. Find out at what fraction of the soft drinks sold was either Pepsi or coke?

Solution:

Fraction of Pepsi sold = `2/4`

Fraction of coke sold = `3/4`

Fraction of soft drinks sold was either Pepsi or coke =?

Add the Fraction of Pepsi and coke to get the fractions for either Pepsi or coke

= `2 / 4` + `3 / 4`

Here the denominators are same, so add the numerators.

= `5 / 4`

Fraction of soft drinks sold was either Pepsi or coke is` 5/4` .

Example problem 3:

Quen prepared some cookies in the morning. She used` 5 / 6 ` of a cup of flour, and `3 / 6 ` cup of sugar for the preparation of cookie. How much more flour did Quen use than sugar?

Solution:

Amount or quantity of flour used = `5/ 6`

Amount or quantity of sugar used = `3 / 6`

Amount or quantity of flour used than sugar =?

By subtracting the amount of sugar from the flour, we can get the amount of flour used.

=  ` 5/ 6` –` 3 / 6`

=` 2 / 6`

Amount or quantity of flour used than sugar = `1 / 3.`

Example problem 4:

Ben owns 20 acres of farmland where he planted flowers on `1/6` of the land. On how many acres of land does Ben plant flowers?

Solution:

Ben grows flowers on `1/6` of 20 acres of land.

Acres of land that Ben grow flowers =?

=      20  ` xx`   `1 / 6`

=   `10 / 3`

Ben plants flowers on `10 /3 ` of an acre of land.I have recently faced lot of problem while learning how to calculate simple interest, But thank to online resources of math which helped me to learn myself easily on net.

Practice Problem: Fraction Math Word Problems

Practice problem 1 - Fraction math word problems

Quen prepared some cookies in the morning. She used `8 / 6` of a cup of flour, and `7 / 6` cup of sugar for the preparation of cookie. How much more flour did Quen use than sugar?

Answer: `1/ 6`

Practice problem 2 - Fraction math word problems

Of the soft drinks sold yesterday at Kevin’s supermarket, `7/10` was Pepsi and another `4 /10` was coke. Find out at what fraction of the soft drinks sold was either Pepsi or coke?

Answer:   `11/10`

Twice Differentiable Function

Introduction for twice differentiable function:

The twice differentiable function explains the process which takes double time differentiation on the operations of differentiation equations on the different functions like algebraic function, trigonometric functions, logarithmic functions, exponential functions etc..   In this article we are going to deal with the different functions with different variables to differentiate for the complex equations. If we have the constant in the second time differentiation then its results in the zero value.

Examples to Explain Twice Differentiable Function

Review on the twice differentiable function for the equation in the form of function f(x) = `x^2 + 3x + 10`
Solution:

The given function is f(x) = `x^2 + 3x + 10`

Differentiate the above function

`f(x)` = `2x + 3 + 0`

`f(x)` = `2x + 3`

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `2 + 0`

`f'(x)` = `2 `    is the required solution obtained after twice differentiation.

Review on the twice differentiable function for the equation in the form of function `f(x)` = `x^3 - x^2 + 3x + 10`
Solution:

The given function is` f(x) ` =  `x^3 - x^2 + 3x + 10`

Differentiate the above function

`f(x)` = `3x^2 - 2x + 3 + 0`

`f(x)` = `3x^2 - 2x + 3 `

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `6x + 2 + 0`

`f'(x)` = `6x + 2 `    is the required solution obtained after twice differentiation.

Review on the twice differentiable function for the equation in the form of function `f(x)` = `5x^2 + x^3 - x^2 + 3x + 10`
Solution:

The given function is` f(x) ` =  `5x^2 + x^3 - x^2 + 3x + 10`

This equation is rewritten as

` f(x) ` =  `x^3 + 3x^2 + 3x + 10`

Differentiate the above function

`f(x)` = `3x^2 + 6x + 3 + 0`

`f(x)` = `3x^2 + 6x + 3 `

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `6x + 6 + 0 `

`f'(x)` = `6x + 6 `    is the required solution obtained after twice differentiation.Understanding statistics help is always challenging for me but thanks to all math help websites to help me out.

Problems to Explain Twice Differentiable Function

Review on the twice differentiable function for the equation in the form of function f(x) = `x^2 + 3x + sin x`
Solution:

The given function is f(x) = `x^2 + 3x + sin x`

Differentiate the above function

`f(x)` = `2x + 3 + cos x`

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `2 + 0 - sin x`

`f'(x)` = `2 - sin x`        is the required solution obtained after twice differentiation.

Review on the twice differentiable function for the equation in the form of function f(x) = `x^2 + 3x + sin x + cos x`
Solution:

The given function is f(x) = `x^2 + 3x + sin x + cos x`

Differentiate the above function

`f(x)` = `2x + 3 + cos x - sin x`

Again differentiate (twice differentiation)  the above function f(x),

`f'(x)` = `2 + 0 - sin x - cos x`

`f'(x)` = `2 - sin x - cos x`        is the required solution obtained after twice differentiation.

Solving Making Box and Whisker Plots

Introduction to making box and whisker plots:

Box and whisker plots is one of the important concept in math.  This method is used to split the data into different number of quartiles.  For this we can calculate the values of median and upper quartile and lower quartile range values.  In this topic we are going to seen about how to solving and making a box and whisker plots problems with suitable diagrams.

Description about Making a Box and Whisker Plots

Median:

First we can solve the median for the given set of values.  So we can arrange the numbers in to the format of ascending or else descending order. Then count the number of values if it is odd middle value is median or else find the mean for the middle two values.

Lower quartile part:

The group of values before the median is noted as Lower quartile part.  Again we can measure the median for this LQR values.

Upper quartile part:

The group of values after the median is noted as upper quartile part.  Again we can find the median for this UQR values.

Number line:

Draw a number line with particular scale level.  Then mark the given values and shade the LQR and UQR and median values.  These gives the following four quartiles.

Lowest value to LQR value
LQR value to Median value
Median value to UQR value
UQR value to greatest value

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Problems on Making a Box and Whisker Plots

Solving and making a box and whisker plots for the first 10 multiples of 20

Solution:

The first 10 multiples of 20 are 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

Solving for Median:

Ascending order: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

Number of values: 10

10 is even number so median is a mean of 100 and 120.  That is 110

Median = 110

Solving for LQR:

Lower quartile Part: 20, 40, 60, 80, 100

Number of values: 5

5 is a odd number so median is center number 60

LQR = 60

Solving for UQR:

Upper quartile Part: 120, 140, 160, 180, 200

Number of values: 5

5 is a odd number so median is center number 160

UQR = 160

Number line:



Quartiles:

20 to 60
60 to 110
110 to 160
160 to 200

Practice Prime Numbers

Introduction – Practice prime numbers:

A prime number is a natural number. This has exactly divisible by one and itself only. If any one number divisible by more than these two factors that is not a prime number. First twenty-five prime numbers are listed below up to 100.

First twenty-prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.The only one even prime number is 2. Because other than 2 all even numbers will divisible by two. 

Finding Prime Numbers – Practice Prime Numbers:

Finding prime number less than two digits are very easy. If it is three digits we have to follow some rules.

For example:
Determine the number 281 is prime number or not?
Step 1:

Find the square root for the given number. That is 281 = 16.76
Then we have to round- up the decimal value16.76 as 17.
Step 2:

Mention all the prime number up to 17. That is 2, 3, 5, 7, 11, 13 and 17.
Step 3:
Now divide the given number by all obtained prime numbers. The number is not divisible by any number. So the given number is a prime number.

Example Problems – Practice Prime Numbers:

Problem 1:

Determine the number 150 to 155 is prime number or not?

Solution:

150 – Factors for 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75 and 150. So 150 is not a prime number.

151 – Factors for 151 are 1 and 151 only. So 151 is a prime number.

152 – Factors for 152 are 1, 2, 4, 8, 19, 38, 76 and 152. So 152 is not a prime number.

153 – Factors for 153 are 1, 3, 9, 17, 51 and 153. So 153 is not a prime number.

154 – Factors for 154 are 1, 2, 7, 11, 14, 22, 77 and 154. So 154 is not a prime number.

155 – Factors for 155 are 1, 5, 31 and 155. So 155 is not a prime number. I like to share this prime factorization of 72 with you all through my article.

Problem 2:

Determine the number 307 is prime number or not?

Solution:

The given number 307 has two factors only. Those factors 307 are 1 and 307 only. So the given number 307 is considered as a prime number.           

Problem 3:

Determine the number 330 is prime number or not?

Solution:

The number 330 has more than two factors. Those factors are 1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55, 66, 110, 165 and 330. Hence the given number 330 is not a prime number.

Practice problems – practice prime numbers:

Problem 1:

Determine the number 593 is prime number or not?

Solution:

The given number 593 is a prime number.

Problem 2

Determine the number 641 is prime number or not?

Solution:

The given number 641 is a prime number.

Problem 3

Determine the number 432 is prime number or not?

Solution:

The given number 432 not a prime number.

Methods of Analyzing Data

Introduction to methods of analyzing data:

Any collection of information in the form of giving the required information is called data. Methods of analyzing data are used to compare the collection of data. Data analysis method Graphs mostly help to analysis the various types of database. For example methods are statistics, bar graphs, histogram graphs, pie charts, and line graphs are used to easily analysis the data. In this article, we are going to see about methods of analyzing data.

Example Problem for Methods of Analyzing Data:

1. Solve the statistics data analyzing using the given data set, find the mean, mode, median, and range of the given data set.

59, 57, 56, 59, 54, 55, and 66

Solution:

Rearrange the data set for ascending order.

54, 55, 56, 57, 59, 59, 66

Mean:

Mean is the sum of data divided by the number of data in the given data set.

Mean = (sum of data)/ (number of data)

Mean = `406/7`

= 58

Mode:

Mode is the most common value in given data set.

Most common value is 59.

Therefore, mode is 59.

Median:

Median is a middle number else we have two middle value means; we find the average of their two values is median. We find it can be arranged in ascending order.  

Median:

In this problem we have a middle number.

Therefore, Median is the middle number 57.

Range:

Range is difference between greatest and least values in the given data set.

Range = maximum value – minimum value

= 66 - 54.

Range = 12.

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More Example Problem for Methods of Analyzing Data:

1. Solve and create a bar chart for data analyzing using the given data set, worker name with their salary, given below.

Worker name

Salary in dollar($)

Aldo

4500

Dillan

6000

Joan

5500

Philip

7000

Alec

5000

Steps to draw bar charts and learn examples                                        

Step 1:

First, we need to take a paper and draw the horizontal line and vertical line. In bar chart, horizontal line called as x-axis and another line that vertical line called as y-axis.

Step 2:                                                                                

The horizontal axis mention the information that ‘worker name’ and in the vertical axis, we take the other information namely ‘salary in dollar ($)’.

Bar chart:



The above bar graph analyzed the given data set that which one is high range and low range. This is the needed bar chart for given data set.

Pre Calculus Problems Exam

Introduction to pre calculus problems exam:

Pre calculus is the advanced form of school algebra and its exam. Actually, pre algebra gives us a brief introduction to algebra while Pre Calculus exam explores some of the important topic about calculus. Pre calculus includes sets, real numbers, complex numbers solving inequalities and equations, properties of functions, composite functions, polynomial functions, rational functions, etc.

Sample Pre Calculus Problems Exam:

Pre calculus problem 1:

Find the perpendicular bisector equation of the line which passes through (4, -2) and (6, 6).

Solution;

The perpendicular bisector is given by

P = (x1 + x2) / 2, (y1 + y2) / 2

P = ((4 + 6) / 2, (-2 + 6) / 2).

= (5, 2)

The slope of the line segment (m1) is given by

m1 = (- 2 - 6) / (4 - 6)

= 4

So the slope of the perpendicular bisector m2 will be

m2 = -1 / m1

= - 1/4.

Therefore, the equation for perpendicular bisector is,

y – 2 = - 1 / 4(x - 5)

y = (- 1 / 4) x + (5 / 4) + 2

y = (-1 / 4) x + 13 / 2

Pre calculus problem 2:

Find an equation of the line which passes through (2, 6) and is parallel to the line 4x - 3y = 24

Solution:

4x - 3y= 24,

Solve the above last equation for y

we get,                    y = (4/3) x - 8

So this line has the slope m = 4/ 3.

The parallel in has its slope is also 4 / 3,

So the equation becomes

y- (-6) = (4/3) (x - (-2))

y+6 = (4/3) (x+2)

y = (4/3) x + 8/3 - 6

y = (4/3) x - 10/3

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Pre Calculus Problems Exam-test Problems:

Exam problem 1:

Find the perpendicular bisector equation of the line which passing through the points (6, -3) and (2, 5).

Answer: y = 1 / 2x -1

Exam problem 2:

Find an equation of the line which passes through (4, -3) and is perpendicular to the line 2x – 5x = 20.

Answer: y = - 5 / 2x +7

Exam problem 3:

Find an equation of the circle with center C (7, -5) which is tangent to the x-axis.

Answer: (x-7)2 + (y +5)2 = 25

Solving One and Two Step Equations

Introduction to solving one and two step equations:
Solving one and two step equations is the problems for systems of linear equations. Solving one step equation is the problem that involves the single step of solving the equation and finding the values of the variables. Solving two step equations is the problem that involves the two step of solving the equation and finding the values of the variables.Let us see sample problems for solving one and two step equations.

Solving One and Two Step Equations:

Solving one step equations:

To solve a one-step equation, we must get the variable alone on one side of the equation. This can be done by using inverse operations. Addition and Subtraction (or adding the opposite) are inverse operations. Multiplication and Division are example of inverse operations. Whatever you do to one side of an equation, you must do to the other side (like a balance scale).

Example1:

1) Simplify  `X/5` = 15

`X/5` ·5 = 15 · 5

X = 75

2) Simplify  X – 12 = 18

X – 12 + 12 = 18 +12

X – 12 + 12 = 18 +12

X =30

3)  Simplify   3X =9

`(3X) / X` = 9 /3

`(3X) / X ` = 9 /3

X =3

Example 2:

x +1 = 7  , 1 is being added to the variable, so we must subtract 1 (or add -1) from (to) both sides

x = 6

Example 3:       

6 = y - 5 , 5 is being subtracted from the variable, so we must add 5 to both sides

y =11

Example 4:

y - 1/5 = 1/10 , 1/5 is being subtracted from the variable, so we must add 1/5 to both sides

y = 3/10

Example 5:

-20.2+ 4.1= n - 38.1

n = 22

First we must add -20.2 and 4.1, 38.1 is being subtracted from the variable, so we must add 38.1 to both sides

Example 6:

5 - x =12 , because the 5 is positive, we must subtract 5 (or add -5) to both sides, then the opposite of x is 7, so x is -7

-x = 7

x = -7

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Solving One and Two Step Equations:

Solving Two-Step Equations - Getting variable alone

1. Remove constant by adding opposite to both sides.

2. If variable is being divided by a number, multiply both sides by the number. If variable is being multiplied by a number, divide both sides by the number (remember dividing by a fraction is the same as multiplying by the reciprocal).

Examples:

1) Simplify     `X/3` + 3 = 12

`X/3` + 3 – 3 = 12 – 3

`X/3` + 3 – 3 = 12 – 3

`X/3` = 9

`X/3` = 9

`X/3` · 3 = 9 ·3

X =  `9*3`

X = 27

2) Simplify     4X – 5 = 6

4X – 5 + 5 = 6 + 5

4X – 5 + 5 = 11

4X = 11                  

`(4X)/4` = `11/4`

X = `11/4`

X = 2.75.