Hard Fraction Problems

Hard fraction problems:

This articles discusses hard fraction problems solving. A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (`1/2` , `5/8` , `3/4` etc.) and which consist of a numerator and a denominator. (Source – Wikipedia)

Hard fraction problems solve for proper fraction, improper fraction and complex fraction problems.

Example for Hard fraction problems:

Example 1:
Subtract the fractions `4/5` – `3/4`

Solution:
The denominator (bottom number) is different so we have to take least common denominator (lcd).
LCD = 5 x 4 = 20
`(4 xx 4)/ (5 xx 4)` = `16/20` and `(3 xx 5)/ (4 xx 5)` = `15/20`
`16/20 ` – `15/20`
The denominators are equals
So subtracting the numerator directly = `(16-15)/20`
Simplify the above equation we get = `1/20`
Therefore the final answer is` 1/20`

Example 2:
Subtract the mixed fractions for given fractions,` 4 7/5` – `5 8/5`

Solution:
The given two mixed fractions are `4 7/5` – `5 7/5`
We need convert to mixed fraction to improper fraction `27/5` – `33/5`
The same denominators of the two fractions, so
                                         = `27/5` – `33/5`
Subtract the numerators the 27 and 33 = 27 - 33 = - 6.
The same denominator is 5.
                                         = `-6/5`
The subtract fraction solution is -`6/5` .

Example 3:
Multiply the mixed fractions for given two fraction,` 4 2/4` x `5 2/6`

Solution:
The given two mixed fractions are `4 2/4` x `5 2/6`
We need convert to mixed fraction to improper fraction `18/4` x `32/6`
Multiply the numerators the 18 and 32 = 18 x 32 = 576.
Multiply the denominators the 4 and 6 = 4 x 6 = 24
                                         =` 576/24`
The multiply fraction solution is 24

Example 4:
Convert `16/ (5/4)` to a simple fraction and reduce.

Solution:
The given complex fraction `16/ (5/4)`
Can be written as `16/1 -: 5/4`
First we have to take the reciprocal of the 2^nd number, and then multiply with the second one
Reciprocal of `5/4 ` is `4/5`
`16/1` x `4/5`
Multiply the numerator and denominator
`(16 xx 4) / (1 xx 5)` = `64/5`
Therefore complex fraction solution is `64/5`

My forthcoming post is on cbse course for class 11, class 11 cbse sample papers will give you more understanding about Algebra

Practice for hard fraction problems:

Problem 1: Convert `5/ (5/4)` to a simple fraction and reduce.
Solution: 4.
Problem 2: Adding the mixed fractions for given two fraction,` 4 2/3` + `5 2/3`
Solution: `31/3`
Problem 3: Subtract the mixed fractions for given fractions,` 4 7/3` – `5 8/3`

Solution:`-4/3`

Written Translation

Introduction:
In word problem we can use the English words for translating words into mathematical expressions. In the translation we can use the math equation. This is used in the math problem to convert in the fairly simple problems. But we can’t do the figuring problem. This translating method is help to solve the word problems.

1. Working clearly will help you think clearly, and
2. Figuring out what you require will help you convert your final answer back into English.

Written translation - Translation methods:

• Addition
• Subtraction
• Multiplication
• Division
• Power

Written translation - Addition
Words that mean adding

words that mean adding	example
the sum of	the sum of a number and 6	y + 6
more than	6 more than a number	y + 6
plus	a number plus two	y + 2
added to	6 added to a number	y + 6
increased by	a number increased by 3	y + 3

Written translation - Subtraction

Words that mean subtracting

words that mean subtracting	example
less than	6 less than a number	y - 6
less	6 less a number	6 - y
minus	6 minus a number	6 - y
subtract from	subtract 6 from a number	y - 6
reduced by	a number reduced by 6	y - 6
decreased by	a number decreased by 6	y - 6
diminished by	6 diminished by a number	6 - y

Written translation - Multiplication

Words that mean multiplying

Words that mean multiplying	example
multiply	multiply a number by 3	3y
multiplied by	a number multiplied by 3	3y
times	4 times a number	4y
double	double a number	2y
twice	twice a number	2y
triple	triple a number	3y

Written translation - Division

Words that mean dividing

Words that mean dividing	example
divided by	a number divided by 6	y/6
divide into	divide 6 into a number	y/6
the quotient of	the quotient of a number and 3	y/3
half of	half of a number	y/2
one third of	one third of a number	y/3

Power

Word that mean rising to the n^th power

words that mean raising to the nth power	example
a number raised to the nth power	a number raised to the fifth power	y6
raise a number to the nth power	raise a number to the fifth power	y6
the nth power of a number	the fifth power of a number	y6
squared	a number squared	y2
cubed	a number cubed	y3

Binomial Theory

In algebra, a binomial theory is nothing but the study of binomial. The binomial is defined as the polynomial with sum of two monomial terms. Some of the binomial term is given as,The binomial also consist of distributions, coefficient, variables etc.some of the binomial terms are given as,

3x + 1 = 0

4x² + 5 = 0

9x³ + 15 = 0

The difference of binomial a2 − b2 is the product of two other binomials:

a2 − b2 = (a + b) (a − b).



Properties of binomial:

A sum of binomial coefficients of an exponent (a + b) n is equal to 2 n.

(1 + 1) n = 2 n

A sum of binomial coefficients of even term is equal to a sum of binomial coefficients of odd terms, and it is equal to

2n-1

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

Examples

Example1: find the value of 5²- 3² and prove it?

Solution:

The difference of binomial a2 − b2 is the product of two other binomials:

a2 − b2 = (a + b) (a − b).

52 − 32 = (5 + 3) (5 − 3).

= (8)(2)

= 16.      ---------- (1)

Proof:

a2 − b2 = 52 − 32

= 25 – 9

= 16       ---------- (2)

From 1 and 2 it proved.

Example 2: find the value of (3x + 1) and (2x + 3)?

Solution:

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

(3x + 1)(2x + 3) = 3*2x2 + 3*3x + 1*2x + 1*3.

= 6x2 + 9x + 2x + 3.

Therefore, (3x + 1) (2x + 3) = 6x2 + 9x + 2x + 3

Example 3: Prove that (a + b) ³ = 8?

Solution:

From the property of binomial,

A sum of binomial coefficients of an exponent (a + b) n is equal to 2 n.

(1 + 1) n = 2 n

(a + b) ³ = (1 + 1) 3 = 2 3

= 8

Hence it is proved.

Algebra is widely used in day to day activities watch out for my forthcoming posts on taylor series cos and cbse previous year question papers class 12. I am sure they will be helpful.

Practice problem:

Problem1: find the value of 9²- 4² and prove it?

Answer is 65

Problem2: find the value of (2x + 1) and (x + 3)?

Answer is 2x2 + 6x + 1x + 3.

Solving Sums Integrals

Integration is an important concept in mathematics and, together with differentiation, is one of the two main operations in calculus. The term integral may also refer to the notion of antiderivative, a function F whose derivative is the given function f. In this case it is called an indefinite integral. Integral can be classified as definite and indefinite integral. (Source: Wikipedia)

I like to share this Table of Derivatives and Integrals with you all through my article.

Examples problems for solving sums integrals

Solving sums integral problem 1:

Integrate the given function ∫ (234x2 + 132x4 - 5x) dx.

Solution:

Given ∫ (234x2 + 132x4 - 5x) dx

Integrate the given function with respect to x, we get

∫ (234x2 + 132x4 - 5x) dx = ∫ 234x2 dx + ∫ 132x4 dx - ∫ 5x dx.

= 234 (x3 / 3) + 132 (x5 / 5) - 5 (x2 / 2) + c.

= 78x3 + `(132 / 5)` x5 - `(5 / 2)` x2 + c.

Answer:

The final solution is  78x3 + `(132 / 5)` x5 - `(5 / 2)` x2 + c.

Solving sums integral problem 2`:`

Find the value of the integration

`int_2^5(x^6)dx`

Solution:

Integrate the given function with respect to x, we get

`int_2^5(x^6)dx`  = `(x^7 / 7)`52

Substitute the lower and upper limits, we get

= `((5^7 / 7) - (2^7/ 7))`

= `((78125 / 7) - (128 / 7))`

= `(77997 / 7)`

Answer:

The final answer is `(77997 / 7)`

Solving sums integral problem 3:

Integration using algebraic rational function ∫ 7dx / (17x + 37)

Solution:

Using integrable function method,

Given function is ∫ `(7dx) / (17x + 37)`

Formula:

∫ [L / (ax + c)] dx = (L / a) log (ax + c)

From given, L = 7, a = 17, and c = 37

Integrate the given equation with respect to x, we get

= `(7 / 17)` log (17x + 37)

Answer:

The final answer is `(7 / 17)` log (17x + 37).

Practice problems for solving sums integrals

Solving sums integral problem 1:

Integrate the given function using integrable function ∫ `(14 / (11x + 12))` dx

Answer:

The final answer is `(14 / 11)` log (11x +12)

Solving sums integral problem 2:

Integrate the given function using integrable function ∫ `(15 / (21x + 92))` dx

Answer:

The final answer is `(15 / 21)` log (21x + 92)

My forthcoming post is on cbse class 10 syllabus will give you more understanding about Algebra.

Solving sums integral problem 3:

Integrate the given function ∫ (7.9x2 - 12.9x) dx

Answer:

The final answer is `(7.9 / 3)` x3 - `(12.9 / 2)` x2

Difference of Two Means

Mean is defined as the average for the total number of values given in the data set. It is the sum made between the given data set and it is divided it by the total number of values given in the data set. Tis is called as the mean for the given data set. This can also be called as arithametic mean or sample mean. The difference between two mean is nothing but the taking difference for the two sets of data (mu_1 , mu_2 ) and calculating the difference for that mean.

Difference of two mean  = mu_1 - mu_2

Where as

mu_1 is the mean value for the first set of data.
mu_2 is the mean value for the second set of data.

Mean is calculated by the way,
mu = (Sum of all the values given) / (Total Number of values)


Steps for calculating the difference of two means:

Get the two sets of data for calculating the mean.
Measure the sum for the first set of data and divide it by the total number of data's given. This is the mean for the first data set
Measure the sum for the second set of data and divide it by the total number of data's given. This is the mean for the second data set.
Now measure the difference of the two means from the mean calculated.



Difference of two means - Example Problems:

Difference of two means - Problem 1:

Find the difference of two means from the given two data set. 2, 3, 4, 5, 6, 7 and 1, 2, 3, 4, 5, 6

Solution:

Mean for the first set of the data given

mu_1 = (2+3+4+5+6+7) / 6

= 27 / 6

=  4.5

Mean for the second set of the data given

mu_2 = (1+2+3+4+5+6) / 6

=  21 / 6

mu_2    = 3.5

Difference of the two means is given by

Difference of two mean = mu_1 - mu_2

= 4.5 - 3.5

Difference of two mean = 1

Difference of two means - Problem 2:

Find the difference of two means from the given two data set. 24, 23, 24, 25, 26, 27 and 11, 13, 13, 14, 15, 16

Solution:

Mean for the first set of the data given

mu_1 = (24+23+24+25+26+27) / 6

= 149 / 6

=  24.8333333

Mean for the second set of the data given

mu_2 = (11+13+13+14+15+16) / 6

=  82 / 6

mu_2    = 13.6666667

Algebra is widely used in day to day activities watch out for my forthcoming posts on icse syllabus and cbse books. I am sure they will be helpful.

Difference of the two means is given by

Difference of two mean = mu_1 - mu_2

= 24.8333333 - 13.6666667

Difference of two mean  = 11.1666666

Is a Cube a Polygon

Cube is not a polygon,because cube is a three dimensional shaped figure .But polygon is a two dimensional object. Generally polygon must be flat, plane figure and it’s made up of line segment. Here we are going to study about cube and polygon shape and its example problems.


Shape of cube:

Cube is a regular solid three-dimensional figure it has six square faces .it has 12 edges of equal length and 8 vertices it is otherwise called as regular hexahedron.

Shape of polygon:

It will be triangle, quadrilateral, pentagon, hexagon, octagon, heptagon…etc. All are two dimensional shapes.

Example problems for cube:

Example: 1

Find the volume of the cube with side length 6 meter.

Solution:

We know that volume of the cube is a 3

Given a = 6

Therefore a3 = (6)3

= 6 * 6 * 6

= 216 meter cube

Example: 2

Find the surface area of the cube each side length of a cube is 16 feet.

Solution:

We know that surface area of cube is,

A = 6a2

Here the given  a = 16 feet

Substitute the a value in the above formula we get

A= 6*162

162 = 16*16 = 256

Therefore the area of the cube is

= 6*256

=1536 feet square

The surface area of the given cube is 1536 feet square.

Example problems for polygon:

Example: 1

Find the perimeter of the polygon which is hexagon shape with side length is 9 feet

Solution:

We know that perimeter of the hexagon is =6*a

a represents the side length

Therefore perimeter = 6 *9

= 54

Perimeter of the given hexagon is 54 feet

My forthcoming post is on syllabus of class x cbse and tamil nadu text book will give you more understanding about Algebra.

Example: 2

Find the perimeter of the regular pentagon with side length is 8 meter

Solution:

We know that pentagon is one of a polygon and formula for finding the perimeter is 5 *a

= 5 * 8

= 40

Perimeter of the pentagon is = 40 meter

Unit of Measurement for Volume

The volume is defined as the space occupied by any object in three dimensions. There are various units for the measurement of the volume of the objects. The most commonly used is the standard international unit of measurement, the cubic meter unit. But there are other units of volume measurement. In this article we will see them in detail.


Unit of measurement for volume:

The various units of volume measurement are related with the standard international unit of volume measurement the cubic meter. With the relations the various units can also be related to one other. The various units of volume measurement and their relations are,

1 cubic meter = 1000 L = 264.2 gallons

1 cubic meter = 35.31 ft3 = 1.308 yd3

1 gallon = 0.1337 ft3 = 3.785 L

1 cubic feet = 7.481 gallons = 0.0283 m3

1 cubic yard = 27 ft3 = 202 gallons = 0.7646 m3 = 764.6 L

1 imperial barrel = 163.7 L = 6.10 m3

The above relations can be used for the conversion between the various units above by relating with each other.

Example problems on unit of measurement for volume:

1. A container volume is measured to be 15000 liters. Convert the volume into m3 and gallons.

Solution:

1 cubic meter = 1000 L

1 liter = `1/1000` m3

15000 liter = `15000*(1/1000)` m3

15000 liter = 15 m3

1000 liter = 264.2 gallons

15000 liters = 15*264.2 gallons

15000 liters = 3963 gallons

2. A water tank can store 20.5 m3 of water. Convert the volume into yd3 and gallons.

Solution:

1 cubic meter = 1.308 yd3

20.5 cubic meters = 20.5 * 1.308 yd3

20.5 cubic meters = 26.8 yd3

1 cubic meter = 264.2 gallons

20.5 cubic meter = 20.5*264.2 gallons

20.5 cubic meter = 5416 gallons

Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiplying Mixed Number Fractions and polynomial function degree. I am sure they will be helpful.

3. Convert 1.35 Barrel into cubic meters.

Solution:

1 imperial barrel = 6.10 m3

1.35 imperial barrel = 1.35*6.1 m3

1.35 imperial barrel = 8.23 m3
Practice problems on unit of measurement for volume:

1. Convert the volume of 560 gallons into ft3 and liters.

Answer: 74.9 ft3 and 2119.6 L

2. Convert the volume of 2.5 yd3 into liters.

Answer: 1911.5