Solved Algebra 1 Word Problems

Introduction solved algebra 1 word problems:

The term word problems are often used to refer at  the mathematical exercise where significant background information on the problem is presented as text rather than in mathematical notation. As word problems are often be involve in a narrative of some sort, they are occasionally also referred to as story problems and may vary in the amount of language used. Let us see about the solved algebra 1 word problems. (Source - Wikipedia).


Examples in solved algebra 1 word problems


Example  1 :

The truck starts from Florida at 10 P.M. and reaches  Indusland at 2.25 A.M. The length between the two places is 170km. If the time of the stoppages at 25 min, find the speed of the truck ?
Solution: Distance = 170 Km

No. of hours between 10 P.M. and   2. 25   A.M.

= 4 hrs 25 min

Time of stoppage = 25 min

Time Taken = 4 hrs 25 min – 25 min

= 4 hrs

Distance
Speed= ___________
Time
170
=    ______
4
The speed of the truck is = 42.5 KM.

Example 2:

Alvin saved 3897 and Fernando saved 4869 more than Alvin. How much did they saved overall?
Solution:

Step 1: Calculate how much Alvin saved.

Alvin = 3897

Fernando = 4869 more than Styrene

3897 + 4869   = 8766

Alvin saved 8766

Step 2: Calculate how much they have saved overall.

8766+ 3897 = 12663

They saved 12663 overall.

Example  3 :

The sum of two consecutive numbers is 211. Generate the next two numbers.
Solution

Let x be one number and x + 1 be another number (since both are consecutive numbers).

So the equation is

x + (x+1) = 211

2x + 1  = 211

2x =211 -1

2x = 210

x = 105

The sum of two consecutive 105 and 106 .


One more examples in solved algebra 1 word problems


Simmons has 125 computers and 132 Pointing . How many computers do they have?
Solution:

Let Z = Total number of computers
The sum of 125 computers and 132 Pointing is equal to the total number of Computers. It assumes the problem into an equating form.
Z = 125 + 132

Solve this equation.
Let Z = Total number of Computers

Z = 257 .

There are 257 total numbers of computers.

The above solved problems are deals with algebra concepts .

What is 9th Grade Math

Introduction to what is 9th grade math:

In mathematics, ninth (9) grade math students learn number of skills and to workout the practice problem. It is more helpful for students to improve our practice skills. In year 9 math is based on 9th grade math. The 9th grade student’s math practice does on regular basis. In 9th grade math is comprise of general math application, algebra and geometry etc.

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Lessons in what is 9th grade math:-


In following basic topics involves the what is 9th grade math

Arithmetic
Commercial mathematics
Algebra
Geometry
Statistics
Trigonometry

Arithmetic

In 9th grade math arithmetic is oldest one it is performed normal operations like addition, subtraction, multiplication and division.

Commercial mathematics

In 9th grade math commercial mathematics is learn ratio and proportion, percentage, compound interest, banking. It is more helpful to higher studies and entry in business.

Algebra

In 9th grade math Algebra is fundamental arithmetic and learns number system, indices, radicals, polynomials, and factorization, linear equations, Quadratic equations and number patterns.

Geometry

In 9th grade math geometry in objects and learns lines and Angles, Congruence of Triangles, Concurrent Lines, Quadrilaterals, Similarity of Triangles, Circles, Angles in a Circle and Cyclic Quadrilateral, Secants , Tangents and their Properties, Constructions, Coordinate Geometry.

Statistics

In 9th grade math statistics is normally used the mean and median and learn data and their representation, graphical representation of data, measures of central tendency, Introduction to Probability

Trigonometry

In 9th grade math trigonometry means study the angles and triangles learn the introduction to trigonometry, trigonometric ratios of some special angles

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Example Problems for what is 9th grade math:-


Problem 1:-

Solve the algebra equation |-2x + 2| -3 = -3.

Solution:-

The equation to solve is given by.

|-2x + 2| -3 = -3

Add 3 to both sides of the equation and simplify.

|-2x + 2| = 0

|-2x + 2| is equal to 0 if -2x + 2 = 0. Solve for x to obtain x = 1

Problem 2:-

Find the equation of the line that passes through the points (-1 , -1) and (-1 , 2).

Solution:-

To find the equation of the line through the points (-1 , -1) and (-1 , 2), we first use the slope m.

m = `(y2 - y1) / (x2 - x1)` = `(2-(-1)) / (-1-(-1))` = `3 / 0`

The slope line is perpendicular to the x axis and its equation has the form x = constant. Since both points are equal to x coordinates -1, the equation is given by:
x = -1

How Do You Divide Math

INTRODUCTION FOR HOW DO YOU DIVIDE IN MATH:

Division is the arithmetical operation for finding how many times a number is in another number. This is a one kind of operations in basic arithmetic. Division is nothing but the repeated reduction. For example: 12 ÷ 2 = 6. Here 12 is the dividend, 2 is the divisor, and 6 is the quotient.  Here we are going to see how to divide.

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How do you divide in math:


Solve 6598 ÷ 2.

2) 6598 ( 3299
6
5
4
19
18
18
18
0
Steps to show how to do division.

Step – 1: In the given problem the dividend is 6598 and the divisor is 2

Step – 2: Now we are going to divide 6598 by 2, we take the first number of the divisor. Since 6 is less than 2 we can stop with 6

Step – 3: The divisor 2 is multiplied with the quotient 3 and gives the result or 6, so when we 6 subtract  with 6 we get the remainder of 0.

Step – 4: The remainder we have is 0;  so we bring down 5.

Step – 5: Now 2 is multiplied with 2 to get 4. When 4 subtracted with 5 we get 1 so we bring 9 down.

Step – 6: When 2 is multiplied with 9 we will get 18 so when 18 subtracted with 19 we will get 1. Now we bring 8 down

Step – 7: When two multiplied with Nine we will get 19 . And the remainder is 0.

Step – 8: The remainder is 0 and the quotient is 3299. This is how we divide in math

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Practice problem to how do you divide in math:


Problem -1: solve 986 by 2

Problem -2: solve 333 by 3.

Answers

Answer for Problem -1 = 493

Answer for Problem -2 = 111

Different Math Properties

Introduction about Different Math properties:

The mathematics has the number of different properties based on the operations of a number. The3 basic properties are Associative property, commutative property, Distributive property and the identity property. Other than this the mathematics has some other different properties. They are Property of closure, inverse, equality. The Different types of special properties are Reflexive, symmetric, transitive, Comparision etc... Now we are going to learn about the different math properties with examples.

Different basic math properties:

Associative property:

Addition and multiplication is satisfies the associative property. We can group a numbers in any type but the answer is does not change. Here we can do the operation first inside the parenthesis. But the order of numbers cannot be changed.

Example:

(l+m)+n =l+(m+n)        =      (6+2)+3  =  6+(2+3)   here the answer is same for both sides as 11.

(l. m) . n  = l. (m.n)      =       (6.2).3   =  6.(2.3) =36

Commutative property:

Here without changing the result we can change the order for those numbers. It satisfied for both addition and multiplication. But not satisfies the subtraction.

Example:

L+m =  m+l         =    4+9 =9+4 =13

l.m   =  m.l          =     4.9 = 9.4  =36

Distributive property:

It is the combination of addition and multiplications. Here two numbers are added inside the parenthesis and multiplied with some number.

Example:

L *(m+n)  =  l*m + l *n

3*(2+3)   =   3*2+3*3  =6+9 =15.

Identity property:

It is the special types of math property. If any identity combine with another number the result will be the same.

Example:

0+l  = l+0

1*l  = l  =l*1

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Some other different math properties:


Reflexive:

If anything is similar to its equal twin.

Example :

m = m.

Symmetric:

If anything is turn over its sides of the identical sign.

Example:

L= m & m=L

Transitive:

If two numbers are equal to the third number then those two numbers are equal.

Special kind of different property:

Properties of Multiplication.
Identity for additive and multiplicative.
Property of opposites and reciprocals.
Multiplication for 0 properties.
Multiplicative for (-1) property.
Cancellation property for addition and multiplication.
Definition of subtraction, division and exponents.
Substitution and comparison’s property.

Algebra ii Math Solutions

Introduction:

The invention of algebra starts with the Greek mathematician Plato. In general, the term algebra defines the constants and variables. The applications of algebra are widely used in day to day life. In general, algebra is classified into algebra 1, algebra ii and college algebra. Algebra ii is advanced topic when compared to algebra i. In this article, we are going to see about algebra ii math solutions.

Algebra ii math solutions examples:


Example 1:

3x + 6y + 9z = 42

9x + 3y + 6z = 33

6x + 9y + 3z = 33,

Solution:

The given equations are,

3x + 6y + 9z = 42 ---------- {i)

9x + 3y + 6z = 33 ---------- (ii)

6x + 9y + 3z = 33 ---------- (iii)

Let’s take first two equations and solve it,

2* equ (i) => 6x + 12y + 18z = 84

3*equ (ii) =>27x + 9y + 18z = 99 (subtract)

-21x + 3y        = -15 equation (iv)

Solve equation (ii) and (iii)

=> 9x + 3y + 6z    = 33

2*equ (iii) =>12x +18y + 6z = 66 (subtract)

-3x -15y          = -33 equation (v)

Equate equation 4 and 5,

-21x + 3y = -15

-3x   -15y = -33

5*equ(iv) => -105x + 15y = -75

=>   -3x    -  15y = -33   (Add)

-108x       = -108

-108x = -108

Divide 108 on both sides,

108x/108 = 108/108

x = 1

Plugging x in equation (iv)

-21(1) +3y = -15

-21 + 3y = -15

Add 21 on both sides,

-21 + 21 + 3y = -15 + 21

3y = 6

Divide 3 on both sides,

3y/3 = 6/3

y = 2

Plugging x and y values in first equation,

3(1) + 6(2) + 9z = 42

3 + 12 + 9z = 42

15 + 9z = 42

Subtract 15 on both sides,

15 + 9z – 15 = 42 – 15

9z = 27

Divide 3 on both sides,

9z/3 = 27/3

z = 3,

The solutions are x = 1, y = 2, z = 3.

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More examples on algebra ii math solutions


Example 2:

Solve: 5x + 9y > 9

Solution:

Arrange the given inequality in slope intercept form,

5x + 9y = 9

Subtract 5x on both sides,

5x + 9y – 5x = 9 – 5x

9y = -5x + 9

Divide 9 on both sides,

9y/9 = -5x/9 + 9/9

y = -0.56x + 1

Put y = 0

0 = -0.56x + 1

Subtract 1 on both sides,

0 – 1 = 0.56x + 1 – 1

-1 = 0.56x

Divide 0.56 on both sides,

-1/0.56 = 0.56x/0.56

-1.78 = x

Put x = 0, to find x.

y = 0 + 1

y = 1

The solutions are x= -0.56, y =1.

Sat Math Practice Problems

Introduction:

Scholastic Assessment Test or Sat reasoning test is a test which is used to get admissions in US universities and colleges. Sat test deals with quantitative questions and English skills. Sat test is conducted for 3 hrs and 45 minutes. It consists of three parts; they are i) Critical reasoning ii) Math aptitude and iii) Writing.  Math section of Sat questions deal with quantitative questions and logical reasoning questions. Math questions are given with multiple choices. Students should Practice math problems to get good scores in Sat test.

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Sat Math Problems:


Example 1:

There are 240 balls, 4 times as many are brown, and the rest is black. How many are green and how many are black colors?

Solution:

There are 4 times as many brown balls as there is black color,

Hence, there must be 4 brown for each 1 of black color.

Given that 4 + 1 = 5,

Therefore,

240/5 = 48

On solving this we get, 48 sets of balls with 4 greens and one of black color in each set.

The total is then,

48 * 4 = 192 green and 48 of black color

Answer Check:

Number of green balls = 192 balls

Remaining black balls = 48

Total balls = 192 + 48

= 240 balls.



Example 2:

Mani drives from his house to hills 150 miles away, and at the end of the day drives home. If Mani drives at a standard speed of 50 miles per hour, how long does Mani takes to drive the round trip?

Solution:

Here Mani takes 150 miles to reach the destination,

The total distance covered by Mani during the round trip = 150 + 150

= 300 miles

Mani drives at an average of 50 miles per hour.

Using the formula

Distance = speed x time

300 = 50 x X

Divide 50 on both sides,

300/50 = 50X/50

6 = X

Mani takes 6 hrs to complete the round trip.

Answer is 6 hrs.


Sat Math Practice Problems:


Practice Problems1:

In a class of 78 students 41 are taking Tamil, 22 are taking English and 9 students are taking both Tamil and English. How many students are not enrolled in any of the course?

A) 10

B) 15

C) 24

D) 34

Answer: C

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Practice Problems 2:

Six years ago rani was X times as old as raji was. If rani is now 17 years old, how old is raji now in terms of X?

A) 12/X + 6

B) 11/X + 6

C) 17X

D) 18/X

Answer: B

Practice Problems 3:

Which of the following numbers can be used to demonstrate that all prime numbers are not odd?

A) 2

B) 5

C) 11

D) 13.

Answer: A

AP Solved Problems

Introduction for arithmetic progression:
A   Arithmetic progression is a sequence of numbers in that every term except the first is attained by adding a permanent number to the immediately before term. This Permanent number is known as the common difference. For an example in the series 2, 4, 6, 8, 10... Each term apart from the first is attained by adding 2 to the previous term. The progressions are called arithmetic progression.

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Arithmetic progression (AP) solved problems:


Solved problem 1: Find out the arithmetic progression (AP) for common difference then the next 3 terms of the arithmetic progression is 1, 4, 7...

Solution:

The common difference of AP  d = 4 – 1 = 3
General form of AP are  a, a + d,a + 2d, a + 3d ....

Next three terms are a + 3d, a + 4d,a + 5d

a + 3d = 1 + 3(3) = 1 + 9 = 10

a + 4d = 1 + 4(3) = 1 +12 = 13

a + 5d = 1 + 5(3) = 1 + 15 = 16

Therefore the next 3 terms are 10, 13, and 16.

Solved problem 2: Find out the arithmetic progression (AP) for 12th term of an A.P. 6, 2, –2...

Solution:

Assume the arithmetic progression in the form a, a + d, a + 2d ...
where, a = 6, d = 2 – 6 = –4, n = 12
tn = a + (n–1) d
t12 = 6 + (12 – 1) (–4) = 6 + (11 * –4) = 6 – 44 = – 38
The 12th term is –38

Solved problem 3: Find out the number of integers between 60 and 600 that are divisible by 9.

Solution:
Let us take the 1st number divisible by 9 among 60 and 600 is 63. The last number divisible by 9 that is 594. 594 is less than 600.
The series is 63, 72, 81... 594.

Here AP is 594.
Where, a = 63, d = 72 – 63 = 9

tn= 594 `=>`   a + (n –1) d = 594
`=>` 63 + (n–1) 9 = 594 `=>` (n–1) 9 = 594 – 63 = 531
`=>` n – 1 = 59 `=>` n = 60
Among 60 and 600 which are divisible by 9 is 60 integers.

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Arithmetic progression (AP) practice solved Problems:


1. If a bank clerk is fixed in the salary 3200 – 85 – 4900, when will he achieve his maximum?

Sol: The clerk will achieve his maximum salary in his 21st year of service.

2. The nth term of a series is 7n – 3. Demonstrate that it is an AP and find out the 1st term and the common difference.

Sol: 1st term is 4, Common difference is 7.

3. Find the angles of a triangle whch are in arithmetic progression. If it is greatest angle equals the calculation of the other two, find out the angles.

Sol: The angles of the triangles are 30°, 60°, and 90°