8th Grade Problem Solving

Introduction to 8th grade problem solving:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Here we are going to see about 8th grade problem solving and its example problems.                                                                                      Source: Wikipedia

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Example for 8th grade problem solving:


8th grade problem solving example: 1

Solve 3x + 3 = 12

Solution:

Given that 3x + 3 = 12

Subtract the -3 on both sides

3x + 3 – 3 = 12 – 3

3x = 9

Divide both side using 3

`(3x) / 3` = `9 / 3`

x = 3

The solution is x = 3.

8th grade problem solving example: 2

Solve 4x + 3y = 9

5x + 3y = 5

Solution:

Given that   4x + 3y = 9 ------------- (1)

5x + 3y = 5-------------- (2)

Subtract the first equation and second equation

4x + 3y = 9

(`-` )  5x + 3y = 5

_______________

-x = 4

x = - 4

x = - 4 take the value and substitute in equation (1).

4x + 3y = 9 ------------- (1)

x = -4

4(-4) + 3y = 9

-16 + 3y = 9

Add the +16 on both sides

+16 – 16 + 3y = 9 + 16

3y = 25

Divide using 3 on both sides

` (3y) / 3` = `25 / 3`

y = 8.33

The solution is x = -4

y = 8.33

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Example for 8th grade problem solving:


8th grade problem solving example: 3

Solve 6x + 4y = 2

7x + 5y = 3

Solution:

Given that   6x + 4y = 2 ------------- (1)

7x + 5y = 3 ------------- (2)

From these two equations we cannot cancel particular variable

So we need to change this equation at least one variable equal on both equations

(1)   * 5 =   30x + 20y = 10

(2)   * 4 =   28x + 20y = 12

From this equation we can cancel one variable.

30x + 20y = 10

( - ) 28x + 20y = 12

Subtract you can get the value

2x = - 2

Divide using 2 on both sides

x = - 2 / 2

x = -1

x = - 1 take the value and substitute in equation (1).

6x + 4y = 2 ------------- (1)

6(-1) + 4y = 2

-6 + 4y = 2

Add +6 on both sides

+6 – 6 + 4y = 2 + 6

4y = 8

Divide using 4 on both sides

`y / 4` = `8 / 4`

y = 2

The solution is x = -1

y = 2

Logic Problem Solving

Introduction of solving logic problems:

Logic (from the Greek λογική logikē)[1] is the study of reasoning.[2] Logic is used in most intellectual activity, but is studied primarily in the disciplines of philosophy, mathematics, and computer science. Logic examines general forms which arguments may take, which forms are valid, and which are fallacies. It is one kind of critical thinking. In philosophy, the study of logic falls in the area of epistemology, which asks: "How do we know what we know?" In mathematics, it is the study of valid inferences within some formal language.[3]

-Source Wikipedia

The solving logic problems are given as follow:


Logic problem solving examples:


The examples of logic problem solving are given as follow:

Problem 1:

Roger can mow the lawn in 60 minutes and Sam can mow the lawn in 80 minutes. How long will it take for them to mow the lawn together?

Solution:

Step 1: Let assume variables:

Assume y = Time to mow lawn together

Step 2: Using the formula:

1/s1+1/s2 =1/sa

1/60 + 1/80 = 1/y

Step 3: Solving the equation

Take LCM

60 x 80 =480

To multiply each sides with 480

1/60+1/80 = 1/y

480/60+480/80 = 480/y

8+6 =480/y

14y = 480

Y=480/14 = 34.28

Answer is: The time taken for each of them to mow the lawn together is 34.28 minutes.

Problem 2:

John, Richard and Alexander can complete painting the bench in 3 hours. If John does the job alone he can complete it in 6 hours. If Richard does the job alone he can complete it in 7 hours. How long will it take for Alexander to complete the job alone?

Solution:

Step 1: Let assume variables:

Assume y = time taken by Alexander

Step 2: Using the formula:

1/6 + 1/7 +1/y = 1/3

Step 3: Solving the equation

To multiply both sides with 42y

1/6+1/7+1/y = 1/3

42y/6+42y/7+42/y = 42y/3

7y+6y+42 = 14y

14y-13y = 42

Y = 42

Answer is: The time taken for Alexander to paint the bench alone is 42 hours.


Logic problem solving practice problem:


The logic problem solving practice problems are given as follow:

Problem 3:

C tank can be filled by pipe C in 4 hours and by pipe D in 6 hours. When the tank is full, it can be drained by pipe E in 5 hours. if the tank is initially empty and all 3 pipes are open, how many hours will it take to fill up the tank?

Solution:

Step 1: Let assume variables:

Assume y = time taken to fill up the tank

Step 2: Using the formula:

Pipe C drains the water it is subtracted.

¼+1/6-1/5 =1/y

Step 3: Solving the equation

Take LCM

Multiply both sides with 120

¼+1/6-1/5 = 1/y

120/4+120/6 -120/5 =120/y

30+20-24 = 120/y

26y = 120

Y= 120/26 = 4

Answer: The time taken to fill the tank is 4 x 8/13 hours.

Word Problems in Algebra with Solution

Introduction of word problems in algebra with solution:

Algebra is a very important terms in math. We have different concepts in algebra. Word problems are also one of the concepts in algebra. Word problems are commonly difficult to understand for kids. Basically word problems are very easy .If we understand the word problem clearly, we can easily find the answer.Here we are going to discuss with some word problems with solution.

Addition and subtraction word problems in algebra:


Example 1:

In a city there are 3,942 men, 4,2 14 women and 6,231 children. What is the population of the city ?

Solution:

Step 1: Men in the city             =3,942

Step 2: Women in the city        =4,214

Step 3: Children’s in the city    = 6,231.

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

Step 4: Total                             = 14,387.


Subtraction word problem:

Example 2:

Natalie has 53 marbles. Sam has 105 marbles and lost 8 when his bag of marbles dropped. Find the difference between the number of marbles that Sam has left and Natalie’s.

Solution:

Step 1: We know that Natalie has 53 marbles.

Step 2: Sam has 105 marbles and lost 8 marbles.

Step 3: Therefore same has 105 marbles – 8 marbles – Natalie had 53 marbles.

Step 4: Now we need to subtract 105-8-53 = 44 marbles.

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Multiplication and division word problems in algebra.


Example 3:

There are 9 desks in a class room and each desk has 6 legs. What is the total number of the legs of the desks?

Solution:

Step 1: We know that we have 9 desks in a class room. And each desk has 6 legs .

Step 2: Now we need to find total number of the legs of the desks.

Step 3: Here we need to multiply 9 times 6 = 54

Step 4: Therefore, the result is 54 legs.


Division word problems:

Example 4:

5 people can sit on a bench. How many benches are required for 45 people?

Solution:

Step 1: We know that, 5 people can sit on a bench.

Step 2: Now we need to find how many benches are required for 45 people.

Step 3: Here we need to divide 45/ 5 = 9 benches.

Step 4: Therefore 9 benches are required for 45 people.

Solving Math Eqations

Introduction to Math equations:

In math, Equation is a condition on a variable. Here the condition is meant that two expressions should have equal value. In an equation, we note that at least one of the two expressions must contain the variable.

There is always an equality sign in an equation in math. That is the equality sign looks that the value of the expression to the left of the sign must equal to the value of the expression to the right of the sign. Only equality sign is used in an equation.

Here the equation remains the same, if the expression on the left side and on the right side is interchanged.

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Solving Example Problems on Math equations:


Problem 1:

Solve x in this math equation:  x – 7 = 14

Solution:

The given equation is x – 7 = 14.

Solving x,

Add 7 to each side in the equation.

x – 7 + 7 = 14 + 7

x  = 21.

To check:

x – 7 = 14

21 – 7 = 14

14 = 14 .

Problem 2:

Solve x in this math equation:  3m + 7 = 1

Solution:

The given equation is 3m + 7 = 1.

Solving m,

Subtract from 7 to each side in the equation.

3m + 7 - 7 = 1 – 7

3m = - 6

Divide by 3 to each side.

`(3m)/3` = `-6/3`

m  = - 2.

To check:

3m + 7 = 1

3(-2) + 7 = 1

-6 + 7 = 1

1 = 1 .

Problem 3:

Solve the math equation : 5r = 20

Solution:

The given equation is 5r = 20

Solving r,

Divide by 5 to each side.

`(5r)/5` = `20/5`

r = 4.

To check:

5r = 20

5*4 = 20

20 = 20.

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Example word problem on math equations:


Problem 4:

Peter’s father’s age is 5 years more than three times Peter’s age. Peter’s father is 44 years old. Find the math equation to find Peter’s age?

Solution:

We do not know Peter’s age. So let us take peter’s age is A years.

Three times Peter’s age is 3A years.

Peter’s father’s age is 5 years more than 3A. that is Peter’s father is 3A + 5 years old.

Peter’s father age is 44 years old.

Therefore,

The equation is 3A + 5 = 44

Solving A,

Now subtract 5 from each side.

3A + 5 – 5 = 44 – 5

3A = 39

Divide by 3 to each side.

`(3A)/3` = `39/3`

A = 13.

So Peter’s age is 13 years old.

7th and 8th Grade Math

7th and 8th grade math:

In this article we discuss about solving problems in 7th and 8th grade math. In this article we are going to discuss for solving problems in 7th and 8th grade math topics. In our day to day we are using arithmetical concepts often. Solving problems in 7th and 8th grade math covers following chapters

Numbers
Measures
Algebra
Geometry

The solving problems in 7th and 8th grade math topics are given below.

Example problems in 7th and 8th grade math:

Example 1:

Simplify the given expression `((-80) * (3))/ (- 2)^2`

Solution:

Given expression `(-80) * (3)/ (- 2)^2`

`(-80) * (3) / (4)` =` - (80 * 3)/ (4)` =  -60

Solution is -60.

Example 2:

Simplify the given expression 8 + (5 – 2) =?

Solution:

Given expression 8+ (5-2)

8 + (5 – 2) = 8+3 = 11

Solution is 11.

Example 3:

Davit sells a house through a broker for Rs.5, 00,000 paying a brokerage of 2%. Find the amount of brokerage.

Solution:

The selling price of the house =Rs 500000

Rate of brokerage                     = 2%

Amount of brokerage                 = 2/100 * 500000

= 10000

The amount of brokerage         = Rs 10000

Example 4:

Radius of a sector is 10 cm. Its arc length is 18 cm. Find its perimeter.

Solution:

Given, r = 10cm and l = 18 cm

P = l+2r units.

= 18 + 2 * 10 cm.

= 18+20 = 38cm.

P = 38cm.

The perimeter is 38cm.

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Example 5:


Find the volume of the right prism whose are of the base is 350 cm2 and height is 24 cm.

Solution:

Given that area of the base, A = 350 cm2 and height (h) of the prism = 24 cm.

Volume of the right prism = Area of the base * height cu units.

= A h

= 350 * 24 = 8400

Volume = 8400 cm3

Example 6:

Find the circumference of a circle whose radius is 7cm.

Solution:

Radius r = 7cm

Circumference, C = 2 pi r

= 2 * 22/7 * 7

= 44

Circumference of the circle = 44cm.

Example 7:

One third of a certain number is 9. Find the number.

Solution:

Let the unknown number be m. One third of m is 1/3 * m

But 1/3 m = 9

m = 9 * 3 (by rule 3)

m = 27

The number is 27.

Example 8:

Sum of 3 consecutive odd numbers is 51. Find the numbers.

Solution:

Framing the equation:

Let the first odd numbers be x.

Then the second and third odd numbers are (x+2) and (x+4)

There sum is 51.

x+(x+2)+(x+4) = 51

x+x+2+x+4 = 51

3x + 6 = 51(adding the like terms)

3x+6 = 51 is the required equation.

Solving the equation:

3x+6 = 51

3x = 51 – 6 (by rule 1)

= 45

x = 45 * 1/3 (by rule 3)

x = 15

The consecutive odd numbers are x, (x=2), (x=4)

15, (15+2) and (15+4)

The required numbers are 15, 17 and 19.

Solving Solution Set

Introduction:

Information, lettering, any other article restricted in a set. Example, the solving solution elements of the set {p, q, r} are the letters p, q, and r.

These are used to solving the set problems.

Type 1: Empty set

Type 2: Equivalent sets

Type 3: Singleton set

Type 4: Universal Set

Type 5: Subset

Type 6: Proper Subset

Type 7: Power Set

We are leaving to learn in feature about the laws of set operations, Relations and Functions.


Examples:


1. Identify finite and infinite sets from the following in solving solution set:

(i) {All schools in Tamil Nadu}.

(ii) N.

(iii) The set of all prime numbers.

Solution:

(i) Every part of schools in Tamil Nadu can be counted one by one and we come to an end in the solution counting process. So the set {All schools in Tamil Nadu} is a finite set.

(ii) N = {1, 2, 3,…}. at what time we count up rudiments of N one by one as 1 for 1, 2 for 2, 3 for 3, 4 for 4, we are not able

to come to an end in the counting process. ? the set N is an infinite set.

(iii) When we write the prime numbers one by one as 2, 3, 5, 7, 11, 13, 17 and so on, we are unable to come to an

end in the counting process. ? the set of all solving prime numbers are an infinite set.

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More Problems:

Represent the following solving sets in Rule Form:


(i) The set of all natural numbers less than 6.


(ii) The set of vowels in English alphabet.


(iii) The set of the numbers 2, 4, 6, … .


Solution: (i) A natural number less than 6 can be described by the statement’s ? N, x < 6.

?the set is { x | x ? N, x < 6}.


(ii) A vowel in English alphabets can be described by the statement: x is a vowel in English alphabet.


? the set is {x | x is a vowel in English alphabet}.


(iii) A number x of the form 2, 4, 6, … can be described by the statement:x = 2n,n ? N.

? the set is { x | x = 2n, n ? N}.

Study Solution Set

Definition

A solution set is the set of values that satisfy a given set of equations or inequalities.

For example, for a set {fi} of polynomials over a ring R, the solution set is the subset of R on which the polynomials all vanish (evaluate to 0).i

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Expressing the Requirements


a solution set is a set of possible values that a variable can take on in order to satisfy a given set of conditions.

When you say that the solution set of two parallel lines.  I assume you mean the intersection between two parallel lines y=m*x+b and y=m*x+c. That is, we seek m*x+b = m*x+c.

Expressing this requirement in the terms of a polynomial f(x) =0, where f(x) =b-c, first, we observe that this has no solution unless b=c. i.e., unless the two parallel lines are in fact one and the same. In this case, the variable x is could not constrained and we have an infinite set of solutions since f(x)=b-c=0 is independent of x. In general, however, there is no solution to f(x)=b-c=0 when b is different from c. No x-value in R will satisfy this condition. This means two parallel lines do not intersect to the irrespective of the x value.



The solution set of the single equation f(x) = x is the set {0}.

1. The solution set of the single equation f(x) = x is the set {0}.

2. For any non-zero polynomial f can be over the complex numbers in one variable, the solution set is made up of finitely many points.

3. However, for that the complex polynomial in more than one variable the solution set has no isolated points.

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If the solution set is empty, then there are no such xi such that

f(x0,...,xn) = c

Becomes true for a given c.

He solution set of an equation in the form ax + by = c with a, b, and c real-valued constants, this forms a line in the vector space R^2. However, it cannot always be easy to graphically depict solutions sets – for example, the solution set to an equation in the form ax + by + cz + dw = k (with a, b, c, d, and k real-valued constants) is a hyper plane