Finding the Mean - Elementary Math

The list below shows the names of students and the number of merits each student received during the school year:

James received 20 merits
Ann received 45 merits
Alex received 37 merits
Jennifer received 56 merits
Aidan received 27 merits

What was the mean number of merits received?

a. 37
b. 25
c. 43
d. 57

Answer : a

Explanation:
The mean value of any set of numbers indicates the average of those numbers. The average of the numbers is calculated by adding all the numbers and dividing them by how many times they occurred. For this example it will be as follows:
20 + 45 + 37 + 56 + 27 = 185
185 / 5 = 37
For additional information on this subject go to : How to Find the Mean

©Copyright 2007.Najwa S. Hirn. All rights reserved.

Distributive Property - College Level

Multiply the following:

Explanation:

The term distribute means to multiply an outer term which is usually outside a paranthesis by all the other tems that are within the paranthesis. We start this one step at a time as follows:

must be distributed into the equation. In order to distribute, multiply the outside term by every term inside the paranthesis. Therefore we end up with:

Remeber to add the exponents when you are multiplying the same varaibles as follows:

The final answer will be:

© Copyright 2007.Najwa S. Hirn. All rights reserved.

Solving Systems Of Equations - High School

Sam sells two different deals of Apples and Bananas to produce stands. First deal contains 30 pounds Bananas and 40 pound Apples for a total cost of $360. The other package contains 15 pounds Banana and 10 pound Apples for $130.00. The following system of equations represent this relationship where A represents the cost of one dozen Apples and O represents the cost of one dozen Oranges:
30 B + 40 A = 360
15 B + 10 A = 130
Solve the system of equation and find the cost of one pound Banana (B) and the cost of one pound Apples (A) in dollars.

Explanation:
In order to solve the above equations with two variables A and B we must first get rid of one variable and solve in terms of the other. I will be using the method of addition and subtraction of the two equations in order to eliminate one variable and solve for the other. I will then substitute the known variable into the equation and solve for the final missing variable. Lets go:

To be able to subtract the second equation from the first, I must have change the second equation and have one of my variables either O or A equal the same coefficient as in the first equation. I have chosen to eliminate O. To do this, I must multiply the second equation by 2 . Here goes:
30 B+ 40 A = 360
(2) 15 B + 10 A = 130
So now we have :
30 B+ 40 A = 360
30 B + 20 A = 260
Now we can subtract the second equation from the first
20 A = 100
In order to solve for A, divide both sides by 20
A = 100 / 20 = $5
The cost of one pound Apples (A) = $5
To find the cost of one pound Banana (B) then substitute in the first equation :
30 B+ 40 (5) = 360
30 B + 200 = 360

Subtract 200 from both sides :
30 B +200 - 200 = 360-200
30 B = 160

Divide both sides by 30 to find B :
30 / 30 B = 160 / 30
B = $5.333

The cost of one dozen Bananas = $5.3

In order to check our work we can substitute these numbers in one of the original equations and find out if the right hand of the equation equals what is on the left. So lets check with the second equation:
30 ($5.333) + 40 ($5) = $360

$160 + $200 = $360 It checks OK

©Copyright 2007.Najwa S. Hirn. All rights reserved.

Solving equations for a variable - Middle Level

James is filling his new swimming pool with water. For his first try, he filled 10 gallons less than half of the maximum volume that the pool can hold. The equation below represents this relationship:
t= ½ v – 10
Where t is the weight of his first set and v is the maximum volume that the pool can hold. If the volume of his first try is 115 gallons, what is the maximum gallons that he can fill?

a. 65.7 gallons
b. 82.5 gallons
c. 175.0 gallons
d. 250.0 gallons

Answer : d

Explanation :
Lets try to solve the equation by asking : what is it we are trying to find out?
We realize that this linear question is asking us to find out what the maximum volume is, which means to find what v is. This means that we are trying to solve this equation for the variable v.
In order to do that we have to plug in the knowns into the equation to be able to find the unkown.
We know that:
t = 115 so lets start:
115 = ½ v - 10
now we have to isolate the variable by grouping the numbers on one side of the equation.
In order to do that we must add 10 to each side so here goes:
115 + 10 = ½ v – 10 + 10
The 10’s cancel on the right side and we are left with :
125 = ½ v
In order to solve for v we must get rid of the ½. To do that we must multiply each side of the equation by 2, so lets try it:
125 (2) = ½ (2) v
Notice that the 2's cancel each other out on the right side and we are left with:
250 = v
We have now solved for v.
To check our work lets substitute the value we found for v in the original equation to find out if the right side of the equation equals the left side.
115 = ½ (250) – 10
115 = 125 – 10
115 = 115 It checks ok
For additional information about solving linear equation try this webpage:
Solving Linear Equations

©Copyright 2007.Najwa S. Hirn. All rights reserved.

Division and Number recognition - Elementary Math

Introduction

Welcome back to school kids. Many are excited but to those who fear and hate math, a start of a new school year brings on added stress and pressure.

I am creating this blog to reach out to students and give them just an extra little help in math. I will post one math problem daily taken from grade levels ranging from elementary school to college level and write a detailed explanation about how to solve it. Participation from students, teachers and parents in the form of comments on this blog is welcomed and appreciated. I can help you more if you leave me feedback on what you need.

I hold a Bachelors degree in Engineering Technology so I have had my share of math courses. I do admit that I truly love the subject. I have tutored it privately and at college for the past 17 years. My biggest reward stems from seeing a struggling student succeed in his or her studies.

It is my pleasure to be here today to try to bring that joy to all the students who fear and hate math. Lets begin our journey.

A little History

I thought that it would be appropriate to explain the history behind the name of this blog. I wanted to use a unique and authentic name that reflects math. After some search, I found out that Ahmes, who was born in Egypt at about 1680 B.C. was a scribe who wrote the oldest known mathematical document titled as “The Rhind” Papyrus. The Front of this document contains division by of 2 by the odd numbers 3 to 101 in unit fractions and the numbers 1 to 9 by 10. The reverse of the document has 87 problems on the four operations, solution of equations, progressions, volumes of granaries, and the two-third rule etc.
For additional information on this document visit this article: Rhind Mathematical Papyrus

Elementary school level

Students attending art class must work in teams of 4 students each. The following list gives the number of students in each class.

First grade has 15 students
Second grade has 9 students
Third grade has 20 students
Fourth grade has 11 students

Which class has exactly enough students to work in art class with no students left over?

a. First grade
b. Second grade
c. Third grade
d. Fourth grade

Answer : c

Explanation :
This is a division problem. The question wants you to divide the number of students in each grade by four. The class that has a number of students that divide without any left over students will be the class that can work in teams of 4 each. So lets try it:

First grade has 15 students : 15 divided by 4 = 3.75 (there is a remainder)
Second grade has 9 students : 9 divided by 4 = 2.25 (there is a remainder)
Third grade has 20 students : 20 divided by 4 = 5 (no remainder)
Fourth grade has 11 students : 11 divided by 4 = 2.75 (there is a remainder)

If you need some additional help on understanding long division with a remainder try this website: Long Division with Remainders

©Copyright 2007.Najwa S. Hirn. All rights reserved.