Practice Test for Basic Math

Introduction to mathematics:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. There is debate over whether mathematical objects such as numbers and points exist naturally or are human creations. (Source: Wikipedia)

Example problems of practice test for basic math

Basic math test example problem 1:

Add the given two values 35 and 84.

Solution:

Given numbers are 35 and 84

(35 + 84) = 35
84  ( + )
____
119
____

Answer:

The final answer is 119

Basic math test example problem 2:

Find the area of the triangle with the base length is 20 m and its height is 12 m.

Solution:

Given base length (b) = 20 m and height (h) = 12 m

Area of the triangle = `(1 / 2)` * (base length) * (height)

= `(1 / 2)` * 20 m * 12 m

= 120 m^2

Answer:

The final answer is 120 m^2

Basic math test example problem 3:

The sum of the two numbers is 23. Smaller number is three less than that of larger number. Find out the two number values.

Solution:

Let us consider,

x is the larger number and y is the smaller number

Given, sum of the two numbers is 23

x + y = 23 ------- (1)

Smaller number is three less than larger number, we get

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

Substitute equation 2 in equation 1, we get

2x - 3 = 23

After simplifying, we get

x = 13

Substitute the value of x in equation 2, we get

y = 10

Finally, that two numbers are x = 13, y = 10

Answer:

The final answer is x = 13, y = `10`

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Practice problems of practice test for basic math

Basic math test practice problem 1:

Subtract 45 and 34

Answer:

The final answer is 11

Basic math test practice problem 2:

Multiply the given values 31 and 10

Answer:

The final answer is 310

Basic math test practice problem 3:

Solve: x = 2x - 7

Answer:

The final answer is x = 7

Basic math test practice problem 4:

Solve:

3y - 10 = 2

Answer:

The final answer is y = 4

Basic Algebra Calculations

Introduction to basic algebra calculations:

Basic 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. The part of basic algebra calculations called elementary algebra calculations is often part of the curriculum in secondary education and introduces the concept of variables representing numbers. The variables are manipulated using the rules of operations that apply to numbers, such as addition. (Source: Wikipedia).

Examples for basic algebra calculations:

Example 1 for basic algebra calculations:

Find the value for x in (x+2) +(x-22) +(x+45) =0.

Solution:

The given expression is(x+2) +(x-22) +(x+45) =0.

In this above expression first add the constants separately and then add the variable x.

(x+2) +(x-22) +(x+45) = (x+ x+ x) + (2-22+45)

(x+ x+ x) + (2-22+45) =3x+25

3x+ 25=0

3x =-25

x=-25/3

The value for x in (x+2) +(x-22) +(x+45) =0 is -25/3.

Example 2 for basic algebra calculations:

Find the value for the quadratic equation x ^2+6x +8=0.

Solution:

The given quadratic equation is x ^2+6x +8=0.

We have to find the roots for the above quadratic equation.

This can be solved by the factoring by middle term.

A=co-efficient 0f x ^2=1

B= co- efficient of x=6

C= constant=8

We split the 6 as (2+4), and then only we get 2 x4 =8.

x ^2+6x +8= x ^2+2x+4x +8

Take x as common in first two terms and 4 as common in next two terms.

x ^2+6x +8= x(x+2)+4(x+2)

x ^2+6x +8= (x+4) (x+2)

(x+4) (x+2) =0

x+4 =0 and x+2 =0

In the equation x+4 =0, subtract 4 on both sides.

x+4 =0

x+4-4 =0-4

x=-4

In the equation x+2 =0, subtract 2 on both sides.

x+2 =0

x+2-2 =0-2

x=-2

The roots for the equation x ^2+6x +8=0 is x=-2,-4.

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Practice problem for basic algebra calculations:

Find the value of x for the equation (x+12)+( x-22)=0
Answer: x=5.

Find the roots for the quadratic equation x ^2+ 11x+30=0.
Answer: x=-5,-6.

Basic Proportionality Theorem

Introduction of basic proportionality theorem:
The Basic proportionality theorem is the theorem associated to triangle's property. This theorem says that if a line intersects two sides of one triangle and is parallel to the third side, then it divides the first two sides proportionally. That’s the Basic Proportionality Theorem shows the relation between the triangle and a straight line which bisects the two sides and parallel to another side.

Explanatory of basic proportionality theorem:

If a line is drawn parallel to the triangle, thus the line intersect two sides of the triangle parallel to the other (third) side in distinct points, then it cuts(divide) the two sides proportionally in the same ratio. This is known as proportionality theorem. It is otherwise called as Thales theorem.

Now we are going to prove the theorem and illustrative the statements

Given: A triangle LMN, S and T are any other two points on PQ and PR respectively such that ST parallel to QR.

To prove:

LS / SM = PT / TN

Construction:

Join MT and NS. From TU perpendicular to LM.

Proof:

In triangle LMN

ST is parallel to MN

And

If LS/MS = k (Where k is constant)

Thus LT / NT is also equal to k

Or we can say

LS / MS = LT / NT

We are given LMN,

ST is parallel to MN.

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Basic proportionality theorem:

Proportionality Theorem says that, A line parallel to one side of a triangle divides the other two sides into parts of equal proportion.

DATA:

In triangle ABC, a line PQ drawn parallel to BC cuts AB and AC at P and Q respectively.

To prove:

AP/ PB = AQ /QC

Construction:

Let the point P divide AB in the ratio of l: m where l and m are natural numbers. Divide AP into 'l' and PB into 'm' equal parts. Through all these points on AB, draw lines parallel to BC to cut AC.

Proof:

Statement                                                                           Reason

1. AP is cut in l equal parts and lines are drawn through                    Construction

these points parallel to BC.

2. Therefore, AQ is cut l equal parts.                                                        Intercept  theorem

3. Similarly,                                                                                                    Statement 1

QC is cut in m equal parts.

4. AP / PB = 1 /m                                                                                          Construction

5. AQ / QC = 1/ m                                                                                         Statement 2 and 3

6. Therefore AP / PB = AQ / QC                                                                 Statement 4 and 5.

Simplify Monomials

An algebraic expression consisting of only one term is called a monomial. It can be a product/quotient of variable and numerals or only variables or even constant terms.

Some of the examples are 3x^2, -p/2, 4xy, 63, -7z etc. Operations addition and subtraction cannot be used to represent a mono form as they are used when there are more than one term in an algebraic expression also monomials cannot have negative exponents. For instance 5-y, 7x+3, ab-3 are not mono forms.

The sum of the exponents of all the variables would be the degree of the same.
The degree of such a constant term is zero. To simplify monomials we would be using law of exponents, multiplication of mono forms and division of mono forms accordingly wherever required.

The order of operations is also followed while solving
The combination of these expressions which are given inside the parenthesis have to be simplified first.
If we have a power of an exponential algebraic expression then law of exponents is used. For example, given an expression (3a2b3)2 can be simplified by using the law of multiplying the exponents [(am)n=amn] giving 3a4b6

Multiplication of Monomials:

To multiply the given expression the exponents of the like terms are to be added using the law of exponents, am.an=am+n
Example: Given an expression 3x2y4. 4xy3 then we simplify applying the law of exponents am.an=am+n, we get 3.4 x2+1y4+3= 12x3y7
Example: -3(m2n).(-4mn3)
Coefficients are multiplied first and then the variable part
(-3.-4)(m2.m1)(n1.n3)=(12)(m2+1)(n1+3)= 12m3n4

Dividing monomials:

While dividing exponential expressions, the exponents of like terms are subtracted
Example: 9x3y4/3x2y this expression can be simplified applying the law of exponents am/an=am-n
We get, 3x3-2y4-1= 3xy3

Let us now learn the steps involved in solving monomials with the help of few examples as given below,
Example: -18a^3b^2c^5/3ab^2c
Here first the numeral part is simplified, -18/3= -6
Each of the like terms are simplified using the law of exponents,
a^3/a= a^3-1= a^2
b^2/b^2 = b^2-2=b0=1
c^5/c= c^5-1=c^4
Writing all the terms together we get, -6a^2c^4 which is the simplified form

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Example: Simplify (3p^2q3/3p) . (9p^2q^2/q^5)
First step would be to multiply the numerators and add the exponents of like terms
Numerator= (3.9)(p^2+2.q3+2) = 27p4q5
Then multiply the denominators and add the exponents of like terms if any
Denominator= (3.1)(p1q^5)  [when the exponent is not given it is considered as 1]
Now divide the coefficients of the expression if possible
27/3 = 9
Next step would be to subtract the exponents of like terms
P4-1q5-5=p3.q0=p3.1=p3
The final answer is (3p^2q^3/3p).(9p^2q^2/q^5)= 9p^3

Basic Algebra For

Introduction of Basic Algebra For:
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. Together with geometry, analysis, topology, combinatorics, and number theory, algebra is one of the main branches of pure mathematics. The part of algebra called elementary algebra.

Source – Wikipedia.

Basic Algebra for Addition and Subtraction:

The followings are some of the examples for basic algebra of addition and subtraction.

Addition of fractions:  `4/9 ` +` 4/9` .

Solution:

From the fraction denominators are same.

By adding the numerator and same as in the denominator:

= `4/9` +` 4/9`

=` (4 +4)/9`

=` 8/9`

By Simplify and adding the fraction:

= `8/9`

= `4/3`

Subtraction of fractions: ` 2/6` - `4/6` .

Solution:

From the fraction denominators are same.

By subtract the numerator and same as in the denominator:

= `2/6 ` - `4/6`

= `(2- 4)/ 6`

=` -2/6`

By Simplify and subtract the fraction:

=` -2/6`

= `-1/3`

Algebraic expression of addition: 10x + 7y + 3x + 5a.

Solution:

The given expressions are 10x + 7y + 3x + 5a

The like terms in expression are 10x and -3x. We cannot do anything the 7y or 5a.

Then group the terms and add.

= (10x + 3x) + 5a + 7y

= 5a + 13x + 7y

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Basic Algebra for Multiplication and Division:

The followings are some of the examples for basic algebra of multiplication and division.

Multiply the integers: 5 × 9.

Solution:

Given integers are (5) × (9)

= |5| × |9|

= 5 × 9

= 45 is the solution for the given integers.

Multiply the integers: 4 × (-5).

Solution:

Given integers are 4 × (-5)

= 4 × -5

= 4 × -5

= -20 is the solution for the given integers.

Divide: 8a^2 – 40 by 8

Solution:

The given polynomials are 8a^2 – 40 ÷ 8

= `(8a^2-40)/8`

By dividing them we get

= `(8a^2)/ 8` * `-40/8`

= a^2 - 5

Divide: 12a^2 – 24 by 4

Solution:

The given polynomials are 12a^2 – 24 ÷ 4

= `(12a^2-24)/4`

By dividing them we get

= `(12a^2)/4` *` -24/4`

= 3a^2 - 6

Grade 7 Math Powers

Introduction of grade 7 math powers:

Grade 7 math powers are nothing but the repeated multiplication of the particular value upto the numbers in the powers. The grade 7 math powers are made to have the multiplication, addition, subtraction etc. The grade 7 math powers are the way which shows the powers of the particular term leads to the multiple of the particular value. The math power saves the time for the multiplication process.

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Grade 7 math powers:


The grade 7 math power is involved in the number sense. The powers that involves in the number sense that gives the power values on the particular term. The grade 7 math powers are made to express through the scientific notations such as power values. The math powers are made to have the range of the numerical values of the particular value.

The grade 7 math powers are simple to compute the answers let us see the simple examples for the grade 7 math powers. The multiplication of the same factor is mentioned as the power of the terms. The math powers are done along the way which has the products of the factor which represents on the right top of the factor.


Example problems for grade 7 math powers:


Example 1:

Calculate 2 power 1. That is `2^(1)` = ?

Solution:

`2^(1)` = 2.

Repeated multiplication of two upto one time.

Example 2:

Calculate 2 power 2. That is `2^(2)` = ?

Solution:

`2^(2)` = 2 * 2 = 4.

Repeated multiplication of two upto two times.

Example 3:

Calculate 2 power 3. That is `2^(3)` = ?

Solution:

`2^(3)` = 2 * 2 * 2 = 4 * 2 = 8.

Repeated multiplication of two upto two times.

These are the example problems on grade 7 math powers.

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Practice problems for grade 7 math powers:


Practice problem help you to get thorough in the concept of powers.

Problem 1:

Calculate 2 power 7. That is calculate `2^(7)`

Answer:

128

Problem 2:

Calculate 2 power 8. That is calculate `2^(8)`

Answer:

256.

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Help in 7th Grade Math

Introduction to Help in 7th grade math:

In this article we discuss about help in 7th grade math solving problems. In this article we are going to discuss for 7th grade topics for help. In everyday life we are using mathematical concepts often. 7th grade math covers following chapters

Numbers
Measures
Algebra
Geometry
Handling data.
The some help in 7th grade solving problems are given below.

Example problems for help in 7th grade math:

Example 1:

Addition of integers: ( + 5) + ( + 7) =?

Solution:

The sum of positive integers is a positive integer obtained by adding the two integers.

(+5) + (+7) = +12

Solution is 12

Example 2:

Subtraction of integers:

(+5) – (+3) =?

Solution:

To subtract an integer from another integer, add the additive inverse of the second number to the first number.

(+5) – (+3) = 5 – 3 = 2

Solution is 2

Example 3:

Multiplication of integers:

(+5) * (+8) =?

Solution:

The product of two positive integers is a positive integer.

(+5) * (+8) = 5 * 8 = 40

Solution is 40

Example 4:

Division of integers:

(+10)/(+5) =?

Solution:

Positive integer / positive integer = positive value

(+10)/ (+5) = 10/5 = 2

Solution is 2

Example 5:

John inverse RS 6000 for 60 months. Hear received the simple interest of RS 1500. Find the rate of interest.

Solution:

One year = 12 months

Five year = 60 months

Principal, p = RS 6000

Number of years, n = 5

Simple interest, i = RS 1500

Rate of interest, r =?

Rate of interest, r = 100 * i/ p * n

100 * 1500 / 6000 * 5 = 5

Rate of interest 5 %

Example 6:

Find the area and perimeter of a rectangle with length 7.5 cm and breadth 4.4cm.

Solution:

Given length = 7.5 cm

Breadth = 4.4 cm

Area of a rectangle = l * b square unit

= 7.5 * 4.4 square unit

= 33

Area = 33 cm2

Perimeter of rectangle = 2 (l * b) unit

= 2 ( 7.5 * 4.4 )

= 2 * 11.9

= 23.8

Perimeter = 23.8 cm

Example 7:

Solve 4x + 5 = 25

Solution:

Given expression 4x + 5 = 25

Subtract 5 on both sides

4x + 5 – 5 = 25 – 5

4x = 20

Divide 4 on both sides

4x/4 = 20/ 4

x = 5

Solution is 5

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Practice problems and solutions for help in 7th grade math:


Problem 1: Addition of integer (+4) + (+2) =?

Solution: 6

Problem 2: Subtraction of integer (+8) – (+3) =?

Solution: 5

Problem 3: Multiplication of integer (+5) * (+5) =?

Solution: 25

Problem 4: Division of integer (+8) / (+4) =?

Solution: 2

Problem 5: Find the area of a rectangle with length 7 cm and breadth 4cm.

Solution: 28cm2

Problem 6: Solve 5x + 2 = 37

Solution: 7

How To Do Basic Long Division

INTRODUCTION FOR HOW TO DO BASIC LONG DIVISION:

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: 24 ÷ 12 = 2. Here 24 is the dividend, 12 is the divisor, and 2 is the quotient.  Here we are going to see how we are going to divide a number using long division method.

The two definitions :

the number that is to be divided into is called as the dividend
The number which divides the dividend is called as the divisor


The steps followed to do long division are:

Division , Multiplication, Subtraction, and bring down.

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Example problem to show how to do basic long division:


Ex 1: Solve 3264 ÷ 2.

2)3265(1632
2
12
12
6
6
5
4
1
Steps to show how to do basic long division.

Step – 1: In the given problem the dividend is 3264 and the divisor is 2

Step – 2: Now we are going to divide 2364 by 2, we take the first number of the divisor. Since 3 is less than 2 we can stop with 3

Step – 3: The divisor 2 is multiplied with the quotient 1 and gives the result or 2, so when we subtract 3 with 2 we get the remainder of 1.

Step – 4: The remainder we have is 1; one will not go with 2 so we bring down 2.

Step – 5: Now 2 is multiplied with 6 to get 12. When 12 subtracted with 12 we get 0 so we bring 6 down.

Step – 6: When 2 multiplied with 3 we will get 6 so when 6 subtracted with 6 we will get 0. Now we bring 5 down

Step – 7: When two multiplied with two we will get four. And the remainder is one.

Step – 8: we cannot proceed further. So we stop with this. The remainder is 1 and the quotient is 1632. This is how we do basic long division.

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Practice problems for how to do basic long division:


Pro -1: solve 564 by 4

Pro -2: solve 840 by 10.

Ans 1) 141 and 2) 84

Cross Multiplying in Math

Introduction to cross multiplying in math:

Normal multiplying:

In math,The basic operation of multiplying is addition.

For example: 5 × 3 = 5 + 5 + 5 = 15

By multiplying the whole numbers we can also multiply fractions as well.

For example 5 × 3½ = 5 + 5 + 5 + (half of 5) = 17.5

Cross multiplying:

In math,The cross multiplying is defined as the following form

a/b= c/d

by cross multiplying , we get

a*d= b*c

ad= bc.

Generally the cross multiplying is in the fraction form.


Example problems for cross multiplying in math


There are some rules which very useful for cross multiplying :

1)

The denominator of the fraction should be equal to zero but both the there may be chance for both fractions can equal to zero. This is one of the such easiest way to reduce the fraction in cross multiplying for both the fractions values.

Example:

3/2 = 4/a

Rule 3 represents as

a = (4*2) / 3 ( by fractions cross multiplying we get)

a = 8/3

Cross multiplying working process:

The cross multiplying is mainly used in the process of comparing the two set of equations. Let we see some sample problems for cross multiplying in math.

The first car A has travelled to Q distance at P speed

And the second card B has travelled  to R distance at S speed

We can make the proportions as like

P/Q=R/S

Now we need to solve the distance value for B

So,                                                 S= (Q*R)/P

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Practice problems for cross multiplying:


Example 1:

2/a-3/b find the solution by using cross multiplication

Solution:

Given: 2/a – 3/b

2b/ab+ 3a/ba               ( by cross multiplying, , we get)
= (2b-3a)/ab

Answer  to the given problem is = (2b-3a)/ab

Example 2:

5/6 + 6/5

Solution:

Given : 5/6 + 6/5

=5*5/6*5 + 6*6/ 6*5 ( by cross multiplying , we get)

=25/30+36/30

=61/30.

Practice problems for cross mutiplying in math:

try to solve these practice problems:

1) solve :   3/x-4/y =?

2)find  :  24/3 + 13/4 =?