Percentage in Grade VI

Hello Friends, in today's math help online session we are going to learn about Percentage. Percentage is a simple way of expressing a number in the ratio form or as a fraction of 100. It is denoted by a symbol %, known as the percentage sign.
For Example 45% is read as forty- five percent, and is equal to 45/100 or 0.45.
It is usually used to represent how large or small a quantity is with respect to the another one. For example, an increase of Rs. 0.15 on a price of Rs. 2.50 is an increase by a fraction of 0.15/2.50 = 0.06. Expressed in percentage form, this is therefore a 6% increase.
What we should keep in mind is that the value of percentage always lies in between zero and one. If the percentage value is greater than 1.00 or 100% then it is not a valid result.
The fundamental concept to remember when performing calculations with percentages is that the percent symbol can be treated as being equivalent to the pure number constant 1 / 100 = 0.01, for example 35% of 300 can be written as(35/100) × 300 = 105 which is our result.
We take another example like if you are having 1250 apples and you want to find the percentage of a single apple, then you will take it as 1/ 1250 x 100% = 0.08%. So if now you have given one apple to your friend then you have reduced 0.08% from the total of 100%. Now what if you gave 100 apples to your friend then the percentage would be 0.08% x 100 = 8%, which is our result.
Let's solve some Examples for better understanding.

1. Solve  50% of 30.
   

30 x 50/ 100 = ½ x 30 = 15.

2. Solve 13 % of 98.
   

13/100 x 98 = 12.74.

3. Solve 60% of all university students are female. There are 2400 female students. How many students are in the university?
   

2400 = 60% x n; n = 2400/ (60/100) = 4000.
We also have increase and decrease in percentage, which is of utmost importance.
Sometimes we say like 10% increase or 10% fall, so this is usually done with initial quantity of the value.
So some examples of percentage change would be -

1. An increase of 100% in a quantity means that the final amount is 200% of the initial amount (100% of initial + 100% of increase = 200% of initial); in other words, the quantity has doubled.
   
2. An increase of 800% means the final amount is 9 times the original (100% + 800% = 900% = 9 times as large).
   
3. A decrease of 60% means the final amount is 40% of the original (100% − 60% = 40%).
   
4. A decrease of 100% means the final amount is zero (100% − 100% = 0%).
   

So now you would be able to understand what percentage is and how to solve the problems based on the increase and decrease of some quantity. 

In upcoming posts we will discuss about Prime Factorization in Grade VI. Visit our website for information on syllabus for class 12th ICSE

Math Blog on Grade VI

Hello Friends, in today's session we are going to learn about the three most important properties of mathematics. The three properties are Commutative, Associative and Distributive Property which are of great importance to us. They help us in making a very complex problem a simple one.

Let's just take them one by one and see what these properties say:

Distributive property – in a very simple manner the property say that multiplication distributes over addition. Like if we do it in a general manner it will be a ( b + c ) = a b + a c. In numerical form it can be written as 2 ( 3 + 4 ) = 2 x 3 + 2 x 4. If in any question you are asked to used the distributive property then what you have to do is just take something common from the parenthesis or factorize it.

is this true 2 ( x + y ) = 2 x + 2 y.

To check this question we take the Left Hand Side and apply the distributive property, now after applying the property we can see that the LHS becomes 2 x + 2 y, so it is true.

Solve 4 x – 8.

To make this in simpler form, we look at the problem and we see that the term 4 is common from the above the expression, so now by applying distributive property the output is 4 ( x – 2 ), so this is our answer. But whether the property applies over subtraction also or not.

So now in place of x + y we have x – y, then what to do?

So we can write the term x – y as x + (-y), so now you can easily apply the property on it.

Now we can say that the multiplication distributes over subtraction also.

Associative property is usually used to associate or form groups.

For Example a + ( b + c ) = ( a + b ) + c or a ( b x c) = ( a x b ) c.

Numerically it can be written as 2 + (3 + 4) = (2 + 3) + 4 or (3×4) = (2×3)4.

The commutative property is used to commute or move around the numbers in a given expression.

For Example a + b = b + a

and a b = b a.

Restate 3 x 4 x X.

Now in this question what we have to do is shift the numbers and the variables to make different possible combinations. So the different possible combinations could be 4 x 3 x X, 4 x X x 3, 3 x X x 4, X x 3 x 4, and X x 4 × 3.

Now we solve a problem to see what the properties say :

Solve 3 a – 5 b + 7 a.

Commutative property – 3 a + 7 a – 5 b.

Associative property - ( 3 a + 7 a) – 5 b.

Distributive property – a ( 3 +7 ) - 5 b. 

In next post we will talk on Percentage in Grade VI. For more information on class 12 ICSE syllabus, you can visit our website

Similar Triangles in Grade VI

Hello friends, in today's session we are going to learn about the similar triangles and how to solve problems based on proportional reasoning (some portion taken from ICSE class 9 syllabus).
Similar triangles are those triangles which have same shape but different size. If the size of the triangles also become equal then they are no more called as similar but are called as Congruent triangles.
The same definition is true even for other Polygons. The similarity between two figures is represented by using a symbol ‘<’, which says “ is similar to”.
Just like to find congruency we use different methods in a same manner to find similarity we have different methods which are very much similar to the congruency.

1. AAA( Angle Angle Angle) – If all the three angles of one triangle are equal to the three angles of the other triangle then they can be called similar.
   
2. SSS( Side Side Side) – If all the three sides of one triangle are equal to the three sides of the other triangle, then the triangles are said to be similar.
   
3. SAS( Side Angle Side) – If two sides and the included angle of one triangle are in the same ratio with the other triangle then also they are similar.
   

If we take two triangles ∆ABC and ∆DEF, then we can find that they are similar or not using the above three properties. If they are similar then they can be written as ∆ABC < ∆DEF.
Now how to use proportional reasoning in similarity and solve problems.
Let's understand this with the help of an example.
Which score among this is better?
50 runs on 10 balls or 40 from 10 balls.
In this case we can easily say that 50 runs from 10 balls is better. As the ratio for the first one is 50/10 = 5 and the ratio for the second one is 40/10 = 4. So 5 is greater than 4 so the first one is better.
For solving problems based on proportional reasoning we have to follow some simple steps.

1. describe the given ratio in words,
   
2. convert them into same unit of measurement.
   
3. put them in the ratio form, and write the missing one as x.
   
4. find the value of x by cross multiplication or by any other method.
   
5. Convert in the appropriate unit of measurement.
   

For Example -

1. ∆ EFG is similar to ∆ XYZ. The sides of ∆ EFG measure 5, 6, and 14. Two sides of ∆ XYZ measure 15 and 18. The third side measures
   

Now to solve this problem we will use proportional reasoning.
So will be 5/15 = 6/18 = 14/x.
1/3 = 14/x.
x = 14 x 3.
x = 42.
So this is the length of the third side of the triangle.

2. if ∆ ABC is similar to ∆ MNO. AB/MN = BC/NO =?
   

As the all the sides of the triangle are in a similar ratio. So the ratio will be AB/MN = BC/NO = AC/MO.
So our answer is AC/MO.

In upcoming posts we will discuss about Math Blog on Grade VI. Visit our website for information on rational expressions

Circles in Grade VI

Hello friends, in today' session we are going to learn about a very interesting topic of Geometry (brief definition taken from syllabus for class 9th ICSE), Circles. So let's just start by understanding its definition.
A circle is defined as a simple shape which consists of those points in a plane which are at a given distance from a given point. The given point is called as the center and the given distance is called the radius. The radius could also be understood as the distance from the center to the circumference of the circle.
Another more simple definition of circle could be the curve formed by a point that moves so that its distance from a given point is always constant.
The terminology of the circle includes the following -
Diameter – it is the length of the line segment whose end points lie on the circle and it passes through the center. The diameter of the circle divides it into two equal halves.
Radius – the half of the diameter of the circle is called the radius.
Chord – the line segment whose end points lie on the circle.
Tangent – it is a straight line which touches the outside of the circle at a single point.
Arc – it is any part of the circumference of the circle.
Sector – it is a region bounded by the two radii of the circle.
Segment- it is the region bounded by the chord and the arc of a circle.
The radius of the circle in general is represented as r and the diameter as d.
The circumference of the circle is given by the formula C = 2 π r.
and the area is given by A = π r ².
Now you have got the basic knowledge on what circle is and how to find its area and circumference.
So let's now move towards the properties of the circle. The properties will help you a lot in solving the problems on circles.

1. 
   
   1. Circles with equal radii are always congruent.
   2. Circles which have a different radii are similar.
   3. The central angle which intercepts an arc is double of any inscribed angle that intercepts the same arc.
   4. If the radius is perpendicular to the chord then it will bisect the chord i.e. divide it into two equal halves.
   5. If the distance of the chord is equal from the center then the chords are also equal in length.
   6. The tangent of the circle is always at right angle to the radius at the point of contact.
   7. Two tangents drawn to the circle from the exterior are always equal in length.
   8. The angle subtended at the center of the circle by its circumference is equal to 360⁰.
   9. Circumference of two circles of different radii is proportional to their corresponding radii.
   10. Arcs formed by the same circle are proportional to their corresponding radii.
   11. Radii of the circle is always constant.
   12. If the chords of a circle are equal in length then they have equal circumference.
   13. The diameter of the circle is its longest chord.
   14. If the radii of two circles are equal then the circles are equal.
   15. If the circles are equal then their circumference is also equal.
       
       In next post we will talk on Similar Triangles in Grade VI. For more information on math word problems, you can visit our website

Fractions in Grade Vi

Welcome guys! Let us start with the point where we had left in previous article and that is Number system of grade VI Algebra (also covered in 8th grade math). In our last article we have discussed about the factorization technique to sort out the problems of rational numbers and today we are going to follow the Number system content whose next topic is Fractions and variety of arithmetic operations that are applied on its queries to be sorted out.
Let us start with the demonstration of Fraction first. Fraction of any number means a part of something and presented by using two terms that are numerator and denominator. These terms are related by division operator to form the fraction, for example:
2/3; here 2 is numerator and 3 is denominator. Students are well aware with the general presentation of the fraction because it was included in grade 5 math syllabus as well, so here we are going to be specific with the operations that are entrenched in this particular grade VI math content.
So now move towards the general operations that are required to learn for solving fraction queries, firstly start with Least Common Multiple. When any fraction requires the calculation of Least Common Multiple that means student has to convert the unlike terms in the denominator of the fractions into their corresponding like terms and this can be done in following ways:
1. Like general multiplication the product of two denominator value provides a common multiple that can be explained by following example:
 Common multiple of 10 and 30 is 10 x30 = 300.
But 300 may not be the least common multiple of them.
So to find LCM of 10 and 30 first find multiples of each as
Multiples of 10 are: 10, 20, 30 …
Multiples of 30 are:  30, 60, 90 …
so, the least common multiple of 10 and 30 is 30.

2. Another way to find LCM is by removing the frequent occurrence of common factors. We have 5 as a factor common to both 10 and 30. We remove the repeat occurrence of 5 from the product to get the LCM.
Product = 10 x15 = ( 2 x 5) x (2 x 3 x 5) = 300
Least common multiple = ( 2 x 5) x ( 3) = 30
Its reverse procedure includes the use of prime factors as: prime factor of LCM value contains the prime factor of the numbers.
30 = 5 x 2 x 3 = (5 x 2) x 3 = 10 x 3 and 30 is a multiple of 10
30 = 5 x 2 x 3 = (5 x 3) x 2 = 15 x 2 and 30 is a multiple of 15
3. The above phenomenon can be applied by using short division technique that is executed as:
5 | 10, 15 (factor out 5)
       2,   3

Since, there are no other common factors to the quotients in the bottom so multiply the common factor to the remaining quotients to evaluate the LCM of the given numbers.

LCM = 5 x 2 x 3 = 30
Let us take one more example to understand the procedure more easily:

Determine LCM of 42 and 63.
7 | 42, 63 (factor out 7 from the numbers)
3 |   6,  9 (here 3 is common factor)
       2,  3 (no other common factor)
 Now as told in above part, multiply the common factors with the quotient value to get the LCM as:
LCM = 7 x 3 x 2 x 3 = 126


4. When there are more than two numbers then find all the prime factors first that are common to at-least two of those numbers to find the LCM.
Example: calculate LCM of 9, 14 and 21.
Use the short division technique
3 | 9, 14, 21 ( 3 is a prime factor for all these values)
7 | 3, 14, 7  ( here we are taking 7 because it is prime to 7 and 14)
3, 2, 1 (here no prime factor can be manipulated in parallel with condition mentioned above)
So find the LCM by following the similar procedure of multiplying quotient with common factors
LCM = 3 x 7 x 3 x 2 x 1 = 126
Now check whether LCM contains the prime factors of the number or not;.
126 = 3 x 7 x 3 x 2
= (3 x 3) x 2 x 7 = 9 x 14
= (7 x 2) x 3 x 3 = 14 x 9
= (3 x 7) x 2 x 3 = 21 x 6

This is how to evaluate least common multiple of the fraction so let us move towards the next operation. Addition and subtraction like general operations evaluation on fraction is also included in grade 5 syllabus and here also we are not going to explore that operation because you guys are already aware of it, so let's start with Multiplication and division operation of fractions.

Whenever you perform multiplication or division that time the overall operation is a mixed form of division and multiplication, still there are some ways to deal out with this kind of fraction queries as explained below with proper example:

1. Multiplication of fractions may be converted to mixed operations involving multiplication and division with natural numbers.
3/4 x 5/8 = (3 ÷ 4) x (5 ÷ 8) = 3 ÷ 4 x 5 ÷ 8
In mixed operations that involve multiplication and division, we may divide the product of dividends by the product of divisors.
3 ÷ 4 x 5 ÷ 8 = (3 x 5) ÷ (4 x 8)
Similarly, in multiplication of fractions, we may directly “divide” the product of numerators by the product of denominators.
3/4 X 5/8 =  (3 x 5)/ (4 x 8) = 15/32

2. Similar to mixed operations, we may cancel out factors that are common to  numerators and denominators.
 Example: Multiply 8/27 by 15/16
                 8/27 X 15/16 = 1/27 X 15/2 (factor out 8)

                  Now factor out 3
                = 1/9 X 5/2
                = 5/18



3. If there is “no denominator,” then the number is considered a “numerator.”
(a) Three fifths of a class of 35 is girls. How many girls are in that class?
3/5 of 35 = 3/5 X 35
Factor out 5
= 3/1 X 7/1
= 21

4. To multiply mixed numbers, convert them to improper fractions first.
Example: what is 1(1/2) of 2(1/2)?
2(1/2) X 1(1/2) = 5/2 X 3/2
                         = 15/4
                        = 3 (3/4)

5. The reciprocal of a number is obtained by switching numerator and denominator. The product of a number and its reciprocal is always 1.
 (a) The reciprocal of 2/3 is 3/2, and their product is 1.
      2/3 X 3/2 = (cancel out the common terms) = 1/1 = 1

 (b) The reciprocal of 2 is 1/2 because the denominator of whole number is 1.
      2 X 1/2 = 2/1 X ½ = 1/1 = 1

6. When situation arises where mixed numbers are involved with the division operator that time they have to be converted first into normal fraction form.
Example: divide 6(2/5) by 2(2/15)
                           6(2/5) ÷ 2(2/15) = 32/5 ÷ 32/15
                          Factor out 5 cancel common terms
                           1/1 X 3/1 = 3/1 = 3

Some points which students should memorize while solving fraction problems:

1.    “Like’ fractions are added by doing the simple addition of numerators. Subtraction is also performed in similar way by just replacing addition operation with subtraction but denominator remains always same.
2.    Additional step included when unlike fractions are added or subtracted that time first convert them to like fractions.
3.    To convert unlike fractions into like fractions, we first calculate the LCM (least common multiple) of all the unlike denominators of the fraction and then we will calculate the equivalent fractions for unlike fractions with LCM (as the new denominator of the fraction).
4.    To multiply fractions, just simply multiply numerators together to have product of the numerators and multiply the denominators to get the denominator’s product. To divide fraction by a fraction, reciprocal of the second fraction is taken and then multiplied with the first one.
5.    In general practice, a fraction in the final answer, always expressed in its lowest terms and this is done by eliminating all the common factors out of the numerator and the denominator.
6.    A “division” notation is not the only notation possible to express fractions, students also can prefer “Decimal notation”.

For more tips and tricks to sort out fraction problems students can use the online math tutoring service, where proficient math tutors are always available to sort out students’ queries.
This is all about fraction that we have explored in this session with its frequent arithmetic operations and we will be continuing with rest of the topics in Number System unit of grade VI Algebra in our successive article, so keep following and practicing with the various worksheets provided on the math website.

In upcoming posts we will discuss about Circles in Grade VI. Visit our website for information on class 9 ICSE syllabus

Factors in Grade VI

Math article for VI grade Algebra (Number system:  Rational numbers, Factors/multiples, GCF)

Hey guys! Welcome to another descriptive session of grade VI math. In previous article we have explored some initial topics of Algebra but in this article we are going to explore another unit of grade VI math content and that is Number system. Most of the students get relaxed when this unit is included in their syllabus because of simple arithmetic operations execution is the only clause they need to handle in this unit. But to reach to that level student has to learn various kinds of number presentation on which actual arithmetic operations are implied. Today in this article we will discuss all about rational numbers and its related operations like multiple factors and GCF (greatest common factor).

Let us start with introduction of Rational numbers, all the fraction form presentation of the numbers is known as Rational numbers but with a condition that the number in denominator is not equal to zero. For example “k/l” and “m/n” are rational numbers for all numbers but with condition “l and n is not equal to zero”. Now let’s talk about general arithmetic operations on rational numbers:
1.    Addition of rational numbers: One common thing to be remembered while doing addition, multiplication or division with Rational numbers is that these numbers are following Associative law, Distributive law and commutative law. On the basis of above two statements, the addition procedure of rational expressions includes following three assumptions:
1.    If two rational numbers are x and y so their simple addition will results (x + y) and it should satisfy associative and commutative law.
2.      (x + 0) = x; x is a rational number.
3.     x + x* = 0, let us prove this;
Suppose a and b are two fractions then from statement (1)
a + b = (general addition)
Subtract 3^rd statement equation from first one
a* + b* = (a + b)*; e.g. 3* + 8* = 11*
a + b  = (a− b) if a > b, e.g., 7 + 4' = (7 − 4)' = 3.
a + b' = (b− a)' if a< b, e.g., 2 + 8' = (8 − 2)' = 6'.


2.    Multiplying rational numbers: Almost same approach is taken for implying this arithmetic operation on rational numbers, similar kind of assumptions for this are as follows:
1.  If  two rational numbers are x and y then there is a way to multiply
them that results to another rational number as xy, that, if x and y are infraction form, xy is the usual product of them and also satisfies the associative, commutative and distributive law.

2.  1 · x = x; when x is a rational number.
3.  0 · x = 0
If x and y both are non-zero fractions then the following can be proved:

          (−x)y = −(xy)
          x(−y) = −(xy)
(−x)(−y) = xy


4.  Dividing rational numbers: The process of division of rational numbers is as same as that of dividing fractions. But before exploring it in detail a required theorem is needed to be explained here:
theorem states that: “if x, y are two rational numbers with y != 0. Then there is one and only one rational number’ z’ such that product of z and y equals to x; x = zy”.
So if we write x = zy
Then it can be rewritten as x/y = z
Therefore ‘z’ is called the division of x by y.
Form of x/y is also known as Quotient of x and y, where y is not equal to zero.

5.  Comparing rational numbers: this process is executed by the actual relation of numbers in respect of number line, for example if student requires to explain a<b then it will be as that ‘a’ is to the left of ‘b’ on the number line. There are three mutual exclusive conditions that arise when the comparison of rational expression is done, these conditions are: a = b or a <b or a> b.
This phenomenon is also called trichotomy law.
If a, b and c are three rational numbers then their possible inequalities are as follows:

(i) For any a, b, a < b is equivalent to:−a > −b.
(ii) For any a, b, c, a < b is equivalent to:  a + c < b + c.
(iii) For any a, b, a < b is equivalent to:  b − a > 0.
(iv)  For any a, b, c, if c > 0, then a < b is equivalent to:ac <   bc.
(v) For any a, b, c, if c < 0, then a < b is equivalent to:  ac > bc.
(vi)  For any a, a > 0 is equivalent to:  1/a > 0


This is all about Rational numbers and general arithmetic operations implementation on them. Now let us move towards the next topic of Number system and that is Factors. Factors are two numbers that, when they are multiplied together then it results a new number called product. Every integer number except 1 has at least two factors and Composite numbers may have more than two factors as their solution (also read on how to factor polynomials).
One more important term exhibits when factor is explained and that term is multiples. Multiples are the whole integer numbers of any particular value that can be easily divisible with all of them or we can say multiples are the result of two integer numbers from which one is always the same( whose multiples are to be find).
For example:
If we need to find multiples of 3 then;
3 X 1 , 3 X 2 ……. 3 X n results the set of its multiples as
3, 6, 9 , 12…….

Now when any number is decomposed into its multiples then that form of presentation is called its factors. For example 15 can be divided into   3 and 5, similarly 12 can be divided into 6 and 2, and 6 can be further divided in to 2 and 3; therefore the final factors of 12 are 2, 2, and 3.
A number can have different sets of factors, for example 12 can also be factorized as 3 and 4, and then 4 can be further decomposed in 2 and 2. In this case the final factors of 12 are: 3, 2, and 2. While finding factors of any number, student can use the following clues to make the task easier.

·         Any even number must include 2 in its factors
·         Any integer number ending with numeral 5 has a factor of 5.
·         Any number that is more than 0 and ends with numeral 0 will have 2 and 5 in its factors list.
Now let us talk about the last term of today’s session and that is Greatest Common factor, it is pretty understood that two or more numbers may have similar factors those are termed as common factors but the largest factor of them, as far as numeral value concern, is called greatest common factor. In general two methods are preferred to find out the GCF of numbers:
First method: it is known as listing factors and include following steps;
a.    First List all the factors of each number.
b.    Then identify their common factors.
c.    Greatest of the common factors is the GCF of the numbers.

Second method (Use prime factors) : this method based on the calculation of prime factors therefore includes following steps:
a.    Fine prime factors of each number
b.    Then identify their common prime factors.
c.    Then product of these common prime factors is the GCF of the numbers.


Let us take an example to explain it practically:
Q. Determine the GCF of 15 and 18.
Solution: by using Listing Factors:
Factors of 15 are: 1, 3, 5, 15
Factors of 18 are: 1, 2, 3, 6, 9 , 18
Common factor of 15 and 18 are 1 and 3, greater of them is 3 so it is said to be GCF of 15 and 18.

That’s all for today, in this article we have explained the major terms of Number system unit and remaining of them will be included in its successive article. Content of grade VI math is simple but still it has some depth so proper attention to each topic is the key to pass out this class with extreme knowledge of important math topics.

At any stage if students like you feels any sort of difficulty to sort out the math query then they can prefer Online math tutoring websites. This math tutoring provider has all the relevant data of every math topic, which is categorized according to math grades for your ease. Internet is the friendliest environment in present time that’s why math tutoring services has chosen this platform for spreading their service and it is working for them, in past few years this moderate educational scheme has gained much positive feedback from its users because of it is 24 x 7 hours assistance given by expert online math tutors.

In next post we will talk on Fractions in Grade Vi. For more information on ICSE class 8 syllabus, you can visit our website

Ratio and Proportion in VI Grade

Hello friends! Welcome back to another important session for enhancing your knowledge of sixth grade math of Tamilnadu education board. In previous article we had discussed the whole syllabus of grade 6 math which student needs to learn throughout this course and from today we are going to start explaining each and every topic of this syllabus in total demonstrative way that will help students to solve the relative problems in their exams. So, topic for the day is Ratios and proportions along with some percent problems. This content is included in your first unit of grade VI math as Algebra.
Let’s start with the short introduction of ratios, nothing much needed to be explained about this because students are getting through this topic since their previous classes. Still; Ratio is a kind of presentation that includes division or ‘:’ between two numbers as:
A:B = A/B Ratio is also called the fraction , where upper part of division is numerator and lower one is denominator. A ratio is a pair of numbers that compares two quantities or describes a rate. In general Ratio can be represented in three manners as:
When explained in words then as: 2 to 3
By using a colon: 2:3
Similarly it can be represented as a fraction: 2/3

There are some conditions where equal ratios are used; equal ratios make same comparisons and to determine the equal ratios. Both terms of ratios are multiplied or divided by same number. Let us take an example to find equal ratios of any given one:

If given ratio is 10/12 then

Divide both terms by 2 as
10 ÷ 2/ 12 ÷ 2 = 5/6
 Now by multiplying 2 in both terms
10 X 2/ 12 X 2 = 20/24
Various more can be determined in similar ways.

Now let us take one more example and see the word problems of ratios:
Example: if the ratios of boys to girls is 4 to 3 where there are total 35 students in the class then how many boys are there in class?

Solution; suppose boys as B and girls as G so given is that
4:3 = B:G
Also given in question that B + G = 35
So if we take the ratio of boys to total students then it is as:
4/7 = B / 35
Now implement cross multiplication
B = 4/7 X 35
B = 4 X 5
B = 20
So, there are 20 boys among 35 students in the class.

This is how any ratio problem is solved but as done in the above problem, every time the use of proportion and cross multiplication is needed to implement while solving the query. So let’s elaborate about these terms in detail. Starting with proportions:

A proportion is a representation of two equal ratios. Any proportion shows that numbers in different ratios. Those numbers related by proportion are compared to each other in same manner.

For example, 2/3 = 10 /15
Now it can be said as “2 is to 3 as 10 is to 15”.
In general to present proportion, a double colon symbol is used as;
2/3 = 10 /15 also can be represent as 2:3 :: 10:15

While solving proportions, the use of cross multiplication is often used because it is equal. Every proportion includes two terms as means and extremes. In proportion 2/3 = 10 /15, 3 and 10 are means and 2, 15 are extremes.

So when cross multiplication is done the next equation includes:
Product of means = product of extremes
3 X 10 = 2 X 15
This phenomenon is really useful when in any proportion in which one of the terms is unknown. Let us take an example of solving proportion:

A is a product that is preferred by 9 out of 10 people., if this statement is true then how many people in  numbers out of 250 should prefer A?

Solution:  suppose ‘n’ is the number of peoples who prefers A among 250 people. Then set up the appropriate proportion as:

n / 250 = 9 /10
use the cross multiplication
Product of means = product of extremes
250 X 9 = n X 10
10 n = 2250
n = 2250/10
n= 225


So, 225 out of 250 peoples should prefer product A.

There is a useful application existing for proportions and that is Unit Pricing. The Unit price of any product is the price of per unit measure. To calculate the unit price students need to implement the proportion as follows

Total price paid/ quantity bought in units = unit price/1 unit
Because the denominator of LHS is one the proportion remains as:
Unit price = (Total price paid)/ (quantity bought in units)

Let us take an example of calculating unit pricing:
Q. which is the better but, 5 bars of soap in Rs. 229 or 4 bars for Rs.189?

Solution: here the unit is 1 bar of soap that can be evaluated as
229/5 = 45.8
For the second option
189/4= 47.2

It is clear that the first option is better to buy.

This is all about Ratios and proportions, now it’s time to elaborate next topic that is Percent. Queries of percent evaluation are surely related with ratios and proportions because like ratios, percent form also include fraction representation.

Percent, eventually in mathematics means each hundred like several of percents means ratio of that percent with hundred. A symbol % is used to represent percentage of any number that means 1/100. So if we say 100 % then


100 % =100/100 = 1

While going through the Algebra of grade VI, students will eventually learn that decimals, fractions and percentage can be converted into each other quite easily. Let us see some of the points related to this conversion:
1.    For writing a decimal as a percent number, just multiply the given decimal number by 100 and align a percent sign with it.
2.    For writing a percent as a decimal: divide the percent by 100 and remove the percentage symbol as well.
3.    To write a fraction as a percent: it is executed in two steps, firstly convert the fraction into decimal form and then use the above method to convert the resulted decimal into fraction.
4.    Last one is to write a percent as a fraction: remove the percent sign and add hundred into its denominator, now convert it into its lower fraction form.

There are 3 forms of percent evaluation problems: finding percent of a number, finding the number when percent is known and the third one is to calculate the percent of one number to another. All these three types of problems can be solved by using proportion.
Let us take an example and see how proportion principle is used to sort out various percent related problems:

 What number is 70% of 250?

One way to sort out these kinds of problems is to first convert the percent into fraction and then eventually use the proportion structure to evaluate the unknown variable.
70 % = 70/100 (fraction form of percent)
Now to set up a proportion, students must find the ratio equal to the above one. So the proportion is
70/100 = n /250
Convert the fraction into its lower form
7/10 = n /250
Now imply the general rule of solving proportion that is

Product of means = product of extremes
10 X n = 7 X 250
n = 7 X 250 / 10
 n= 7 X 25
 n = 175

So, 70 % of 250 is 175 that means 175 to 250 is the same ratio as 70 to 100.

This is all about Ratios, proportions and percent problems of grade 6 algebra problems. In this article most of the terms related to these topics  are explained but still to learn more in better manner students can really use online math tutoring websites, where proficient math tutors are available to sort out your mathematical queries and explain various fundamentals of it. The key thing is that every time when students access the online math education website for having the assistance then tutors are instantly available to help him through remote connection. Now to develop a friendly communication so that students can ask their problems more frequently with tutors. Options like live online chat, video conference, and worksheets solving sessions are available.
Students can schedule their regular math learning session with online tutors so that they can be regular with their math lessons to enhance their mathematical skills. This all facility is available for 24 x 7 hours that really generates flexibility in your study schedule. Everything is categorized according to the math grades so nothing gets messed up when you visit the online math websites.

In upcoming posts we will discuss about Factors in Grade VI. Visit our website for information on statistics help

Syllabus of VI Grade

Mathematics, a subject which is never going to stop following you, so its better that students gets friendly with it rather then moving away. Today we are going to put spotlight on the grade VI syllabus of maths and explain the basic things which students need to execute while perusing this standard. Although grade VI is an earlier and easy standard but still it has its own importance mostly in math subject. The higher complexity problems which are introduced in upper classes can only be resolved when all the basic fundamentals are very strongly prepared by students and 6th grade math have that prior content which helps students to learn. Most of the mathematics rules of following math branches: Algebra, Number system, Geometry, and Statistics are included in this grade These six branches of mathematics cover the whole grade VI syllabus that is categorized in six units. These units are Algebra, Geometry, Measurement, Number system, Probability and statistics, and Mathematical reasoning. Every unit of this syllabus (which is also covered in syllabus for class 8th ICSE) starts with proper introductory topics which puts the use of that particular branch of mathematics in front of the student.

Let us take every unit one by one and see what actually students going to learn in that, first of all students are introduced with Algebra. Hint of algebraic equations formation and what exactly algebra is given to students in previous classes. So here they need to do evaluation in parallel with formation of algebraic equations. Ease for students that even in VIth standard, Algebra problems evaluation does not consider more than two or three steps. In this unit students will start their grade VIth problems related to Ratios, Proportion, and percentage. Individually students of this standard are very well aware of these terms but here they get to know what kind of scenario occurs when all of these terms are combined or used together to form a different kind of math query. Other major topics that are included in this Algebra introduction unit are Describing relationships, Generating formulas, Equations for linear relationships, Evaluating algebraic expressions,
Multi step problems, Geometric quantities/ expressions, Evaluating formulas, Solving two-step equations. When students come to the end of this unit, at that time they have pretty much information of Algebra standard formulas because all of them are evaluated by students while dealing with the Generating formulas segment of first unit.

Now just put a step forward and enter into second unit conversation which is Geometry. Geometry deals with linear shapes formation or their analysis related fundamentals. Nothing new is to be done here but this unit is focused on circles analysis and plane co-ordinate system. But students have to start with various angle types and their relationship fundamentals along with triangles theory. To make geometrical concepts clearer, grade VI syllabus includes proportional reasoning of similar triangles. Geometry unit of grade VI math is not on the heavier side, it just shows its presence in this syllabus by introducing concepts of circles and triangles with their corresponding properties. Although having small content these units provide much required information of geometrical standard theorems.

Very soon the popularization of geometry ends and students are introduced with third unit that is Measurement. Nothing new is there for students, like previous classes this unit includes exact estimation of any query and Unit conversion. This Unit conversion is included in this class as well because one thing in math that can be seen on both sides as advantage or disadvantage of mathematics, “a problem can be represented or asked in several ways and this task is executed by using Unit conversion. So once students get aware of all the relevant units of particular value then its easy to recognize its various forms. Apart from this, Measurement part of the grade VI math includes several kinds of problems session where some kinds of measurement is needed to be done.

After completion of third unit, students reach into the mid of grade VI math syllabus where from level of problems gets increased with variety of topics to be learnt. Instantly students have to face the headache of Number system and operations that is termed as fourth unit of this grade syllabus. Let us start with the major content list which needs to be studied throughout this unit. Topics included in this unit are Rational numbers, Factors/multiples, GCF, Prime factorization, Addition and subtraction, Operations on fractions, decimals, integers, exponents, Commutative, associative, distributive property, Ratios and rates, Identity and inverse properties, Estimation of solutions, Order of operations, Percentages, and Proportions. Most of the topics are followed from 5th grade math syllabus but this time there evaluation is a bit tough. There are some standard math expression decomposition principles like commutative property and distributive property, some more like them are included in this part which actually avails some sort of ease while dealing with any complex form query. This unit includes the most calculative math queries in respect to other units of grade VI math syllabus.

After dealing with the above units students are introduced with most mystifying mathematical branch that is Statistics and the headache grows up because of Probability problems introduction in this syllabus. The VI math only includes the topics with which students gets pretty much aware of probability problems and its properties. According to the mind level of students of grade VI, syllabus is smartly configured by including such topics which makes students knowledgeable and pretty informed about probability and various other statistics problems rather then forcing them to simplify those problems concurrently. The topics which are representing the Statistics part in grade VI syllabus are Constructing sample spaces, Selecting a sample, Sampling errors, biased sample, Probability, complement, Collect/organize/graph data, predictions, Measures of central tendency/dispersion, Possible outcomes, Estimating probability, Representing probability, Independent, dependent, and equally likely events. To solve queries of probability like terms the mind of students should be alert and able to solve various puzzles like situation, that's why an extra unit is invoked in this syllabus which helps students to do some mind exercise. This final unit of VIth grade math syllabus is Mathematical reasoning.

Let us explore the content and use of every topic of Mathematical reasoning. Students need to go through from topics like Mathematics in daily life, Steps in problem solving, Problem solving strategies, Tools to solve problems, Informal and mathematical language Make generalizations, and Justify solutions. The important factor of this unit is that it makes student that much enable so he can check his evaluated answer by own, whether it is right or not.

Syllabus of grade VI math is not pretty vast but still it includes various predefined formulas and principles that clear the doubts of students related to any query. One more advantage of including the description of standard formulas in this grade is that whenever student implies those formulas to solve any other query in upper classes that time he will be very sure of the formula evaluation process because he already knows that “how this formula is being defined earlier.”

Everything gets easier to be learnt when there is somebody available to assist you in that, Following this statement Online math tutoring are working in present time through the Global Internet platform. Impressive assistance is given by online math tutors. As we know competition results better performance and this is also a reason why various Online math educational websites adds moderated features and suitable options for students to have much reliable and extended math learning sessions. These online math tutoring websites prefer the E-learning concept which implements one-on-one session between student and an online tutor to provide proper attention to every student. For making this service accessible by various students at a time, these service managers maintain the heavy bench strength of expert and proficient online math tutors. Student can open his regular account on any online educational website for having scheduled regular math lessons otherwise random access is always an option for them.

As far as grade VI math is concerned, the ratio of explanation of math fundamentals is much larger than the simplification of problems. That’s why online math tutors use various video aids to explain the facts of maths in front of students. Options like live online chatting and video conference adds much more interactivity between students and tutor throughout the session. So if all the students put proper attention to this class contents then for sure they can build pretty strong math basics for themselves and virtual presence of an online math tutor is always with them to clarify any of their doubt instantly. In the above session we have explained the contents of sixth grade math with the required steps which should be initiated by students to survive with ease in this class and being already prepared to face the next class where some more newer terms with a bit higher difficulty level are waiting for them.

In next post we will talk on Ratio and Proportion in VI Grade. For more information on chemistry help, you can visit our website