What is GCF in Math

Introduction to gcf in math:

In mathematics, Gcf is used to simplify the common fractions and carrying out basic operations. The greatest common divisor (gcd), as well-known as the greatest common factor (gcf), greatest common denominator, or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divide the numbers without a remainder. In mathematics “Greatest Common Factor” short form gcf. GCF of those numbers is largest factor which commonly divides the specified set of numbers (two or more).


Examples problem for gcf in math:


1. Find the gcf of 8, 10, 6, 14 and 16.

Solution:

Given numbers are 8, 10, 6, 14 and 16.

The Greatest Common Factor (GCF) of the numbers 8, 10, 6, 14 and 16 is 2.

2 is the greatest number that divides common into all of them.

Gcf of this problem is 2.

2. Find the gcf of 150, 250,300 and 350.

Solution:

Given numbers are 150, 250, 300 and 350.

The Greatest Common Factor (GCF) of the numbers 150, 250, 300 and 350 is 50.

50 is the greatest number that divides common into all of them.

Gcf of this problem is 50.

3. Find the gcf of 25, 55, 17, 16 and 50.

Solution:

Given numbers are 25, 55, 17, 16 and 50.

The Greatest Common Factor (GCF) of the numbers 25, 55, 17, 16 and 50 is 1.

1 is the greatest number that divides common into all of them.

Gcf of this problem is 1.

4. Find the gcf of 15, 45, 18, 27 and 60.

Solution:

Given numbers are 15, 45, 18, 27 and 60.

The Greatest Common Factor (GCF) of the numbers 15, 45, 18, 27 and 60 is 3.

3 is the greatest number that divides common into all of them.

Gcf of this problem is 3.

5. Find the gcf of 10, 30, 50, 70 and 80.

Solution:

Given numbers are 10, 30, 50, 70 and 80.

The Greatest Common Factor (GCF) of the numbers 10, 30, 50, 70 and 80 is 10.

10 is the greatest number that divides common into all of them.

Gcf of this problem is 10.

6. Find the Greatest Common Factor (gcf) of 14, 24 and 42.

lowest Factors of 14: 1,2,7,14
lowest Factors of  24 : 1, 2,4, 6,8,12,24
lowest Factors of  42  : 1, 2, 3, 6, 7, 14, 21, 42

Common Factors: 1, 2.
1, 2 divides 14, 24, 42 therefore they are common factors.

Answers for gcf = 2

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Practice problems for gcf in math:


1. Find the gcf of 20, 52, 10, 16 and 50.

Answers for gcf = 2

2. Find the gcf of 25, 55, 15, 35 and 50.

Answers for gcf = 5

3. Find the gcf of 6, 12, 15, 21 and 30.

Answers for gcf = 3

Help Kids in Math

Introduction to help kids in math:
Mathematics is a group of terms which is used to declare various functions. In general mathematics is always based on the number system formats .
The following areas are covered in mathematics,they are integers, fractions, number types,algebra problems, geometry problems, measurement of shapes, trigonometry and calculus an etc.. Here, in this article will help kids to learn math.

Math topics with examples to help kids:


Here are some basic math topics to help kids.

Number system:

In number system we have many numbers to describe,they are

Positive number:

Positive number is a number which identified by the symbol  '+' . Generally the number starts with 1,2,3...

Example: 4+ 6 = 10   ( the result is also a positive number)

Negative number:

Negative number is a number which identified by the symbol ' - '. These numbers are starting with  -1 , -2 , -3 ....

Example : -4 + -6 = -10.

Fractions:

Fraction is a whole number which consist of two parts, they are numerator ( top part )  and the denominator ( the bottom part)

normally the numerator is greater than the denominator.

Example: 10 / 2.

Arithmetic operations:

The following operations are called arithmetic operations, they are , Addition, subtraction, multiplication and division.

Examples:

Addition:

Its a normal addition, just add the given values and place the larger number symbol

10 + 20 = 30

Subtraction:

Its a normal subtraction, just subtract the smaller number from the larger number and place the larger number symbol at the result.

10 - 20 = -10 .

Multiplication:

multiplication is a process of product the given values.

5 * 4 = 20 ( positive * positive =positive )

5 * (-4) =  -20 ) ( positive * negative = negative )

- 5 * 4 = -20 ( negative * positive = negative )

-5 * -4 20 ( negative * negative =positive )

Division:

Division properties also same like multiplication.

Example: 10 / 2  = 5.


Example math problems to help kids:


Let we see some algebra problems to help kids:

Solve the problem using order of operation method.

5( 2+3) - 10 +3( 5-2)

Solution:

step1 : As per the order of operations rule, we need to take the parentheses values

= 5 ( 2+3) - 10 + 3 (5-2)

= 5(5) - 10 +3 (3)

step2 : Multiply the values

= 25 - 10 +9

step3: add the values

= 25 +9 -10

= 34 - 10

step4 : subtract the values

= 24.

Math problem 2 ) 3x  - 5  =  25 where x = 10, check whether the x value satisfy the equation

solution: apply the x value in the above equation

3x - 5 = 25

3 ( 10 ) - 5  = 25

30 -5 =25

25 = 25 .

Answer : The x value is true and its satisfy the equation.

Solve : 3x + 6 = 12 -3x, find x value.

Solution:

step1: group the terms,we get

3x +3x = 12 - 6

6x = 6

step2:  divide by 6 on both sides,we get

6x / 6 = 6/6

x =1 .

Math problem3 )  a+ 2a+3a+4a+5a+6a -84 = 0 find the value of a.

Solution:

step1 : add all the coefficients of a ,we get

= a + 2a+3a+4a+5a+6a -84=0

21a - 84 = 0

step2: add 84 on both sides, we get

21a - 84 +84 = 0+ 84

21a = 84

step3: divide by 21 on both sides,we get

21a /21 = 84 / 21

a = 4

Math Grade Six Probability

Introduction math grade six probability:

Generally probability is defined as the ratio of the number do ways of an event occur to the total number of possible outcomes, probability is used in the area of statistics, finance, gambling and science.

Probability formula for math grade six probabilities

The probability of event P (A)  =no of possible events n (a) /the total number of the events n(s)


Example problem 1 - math grade six probability


Suppose a bag contains the number up to 1 to 10, find the probability of sleeting a prime number?

Solution:

Bag contains the number from 1 to 10 here we have to know which the prime number between the numbers 1 to 10 is

So the prime number is 2.3,5,7

So the probability p (prime) =4/10 or 2/5

Example problem 2 - math grade six probability

A quiz competition is conducted by the royal club there are 25 members are attended in the competition, in that 7 members are selected in the competition find the probability of the selected members

Solution:

There are 25 members are attended in the quiz competition (total number), 7 members are selected so the probability of the past person is 7/25

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Example problem 3 - math grade six probability

Find the probability of the red, blue and green color ball?

Solution:

From the given diagram w e can calculate the total number of balls.
So total number of balls=number of red color ball + blue color ball + green color ball

Total number of red color ball=5

Total number of blue color ball=4

Total number of green color ball=2

So the total number of red, blue and yellow color ball is

Probability of taking red color ball=5/11

Probability of taking blue color ball=4/11

Probability of taking green color ball=2/11

Example problem 4 - math grade six probability:

Find the probability of blue, yellow and pink color?

Solution:

Here the rectangle is divided into 8 parts, so the total number of divided part is 8

Here pink color is 2

Number of part colored with blue color is 2

Number of part colored with yellow color is 4

So the probability of pink color 2/8

Probability of blue color 2/8

Probability of blue color 4/8

Range in Math Terms

Introduction to range in math terms:

In math terms range is a difference between the maximum and minimum value in the set of numbers or elements. In math terms, group of numbers or elements is called set. A set can have finite number of elements. There are two steps to calculate the range the range of set of numbers.

Step 1: Arrange the numbers in ascending order by size.

Step 2: Subtract minimum value by maximum value.

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Range in Math Terms - Examples


Example 1: Calculate the range of set of numbers: {20, 30, 32, 51, 56, 64, 33, 35, 61, 27}

Solution:

Arrange the set of numbers in ascending order

{20, 27, 30, 32, 33, 35, 51, 56, 61, 63}

Range = Maximum value – Minimum value

= 63 – 20 = 43

Therefore range of set of numbers is 43.

Example 2: Calculate the range of set of numbers: {38, 44, 49, 57, 23, 54, 124, 158, 264, 142}

Solution:

Arrange the set of numbers in ascending order

{23, 38, 44, 49, 54, 57, 124, 142, 158, 264}

Range = Maximum value – Minimum value

= 264 – 23 = 241

Therefore range is 241.

Example 3: Ten student’s weight is as follows {37, 47, 87, 76, 99, 48, 75, 62, 57, 62}. Calculate the range.

Solution:

Arrange the set of numbers in ascending order

{37, 47, 48, 57, 62, 62, 75, 76, 87, 99}

Range = Maximum value – Minimum value

= 99 – 37 = 62

Therefore range of ten student’s height is 62.

Example 4: Ten student’s heights as follows {135, 122, 162, 172, 173, 142, 167, 144, 137, 160}.  Calculate the range of set of numbers:

Solution:

Arrange the set of numbers in ascending order

{122, 135, 137, 142, 144, 160, 162, 167, 172, 173}

Range = Maximum value – Minimum value

= 173 – 122 = 51

Therefore range of ten student’s height is 51.

Example 5: Calculate the range of set of numbers: {9.52, 7.25, 5.65, 10.47, 9.52, 12.60, 8.29, 5.20, 7.68, 4.34}

Solution:

Arrange the set of numbers in ascending order

{4.34, 5.2, 5.65, 7.25, 7.68, 8.29, 9.52, 10.52, 11.47, 12.6}

Range = Maximum value – Minimum value

= 12.6 – 4.34 = 8.26

Therefore range of set of numbers is 8.26.

Example 6: Twelve students’ marks in math as follows {37, 45, 100, 99, 75, 83, 67, 97, 39, 44, 48, 59}. Calculate the range of set of numbers:

Solution:

Arrange the set of numbers in ascending order

{37, 39, 44, 45, 48, 59, 67, 75, 83, 97, 99, 100}

Range = Maximum value – Minimum value

= 100 – 37 = 63

Therefore range of set of numbers is 63.

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Range in Math Terms - Practice


Problem 1: Ten students’ marks in math as follows {33, 55, 77, 51, 79, 48, 61, 76, 99, 27}. Calculate the range of marks.

Answer: 72

Problem 2: Calculate the range of set of numbers: {29, 29, 27, 29, 24, 54, 26, 42, 27, 95}

Answer: 71

Problem 3: Calculate the range of set of numbers: {3.45, 5.27, 8.25, 4.47, 8.52, 23.6, 14.9, 34.3, 27.87, 4.04}

Answer: 30.85

Distance and Rate Word Problems Math

Introduction to distance and rate word problem:

In mathematics education, the term word problem is often used to refer to any mathematical exercise where significant background information on the problem is presented as text rather than in mathematical notation. As word problems often involve a narrative of some sort, they are occasionally also referred to as story problems.

Here, we are going to see about word problems on distance and rate measurement in math. A few examples of word problems on distance and rate is given below which helps you for learning distance and rate word problems in math.

(Source: Wikipedia)

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Example of word problems on distance and rate:


Word problem 1:

Rohit left office and drove his car at the rate of 55 miles per hour for 3 hours. He stopped his car for few minutes and then he drove it for another 2 hours at the rate of 50 miles per hour to reach his home. Calculate how many miles did Rohit drive?

Solution:

The total distance travelled by Rohit is given by

Distance, D = (55 * 3) + ( 50 * 2)

= 165 + 100

= 265

So, total distance travelled by Rohit is 265 miles.

Word problem 2:

Nick travelled from place A to place B by bike in 4 hours. Chan travelled the same distance in 3 hours at a rate 30 miles per hour greater than Nick's. Calculate the distance between the two places.

Solution:

Let X be the rate at which Nick travelled between the two places.

Therefore, X + 30 be the rate at which Chan travelled between the two places.

So, the distance travelled by Nick is given by

Distance, D = Rate * Time = 4X ....................... (1)

Distance travelled by Chan is given by

Distance, D = Rate * Time = 3(X + 30) ............... (2)

From equation (1), we get

X = D/4

Substitute X = D/4 in equation (2).

D = 3(X + 30)

D = 3(D/4 + 30)

4D = 3(D + 120)

4D = 3D + 360

D = 360

Therefore, the distance between the two places is 360 miles.

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Homework problems:


1) Mick left from city A and drove his car at the rate of 70 miles per hour for 4 hours. He stopped his car for few minutes and then he drove it for another 4 hours at the rate of 60 miles per hour to reach city B. Calculate how many miles did Mick drive?

2) Morkel travelled from City A to City B by car in 3 hours. Joseph travelled the same distance in 2 hours at a rate 20 miles per hour greater than Morkel's. Calculate the distance between the two cities.

Solutions:

1) Total distance travelled by Mick is 520 miles.

2) Distance between the two cities 120 miles.

Math Word Problems Grade 9

Introduction to math word problem grade 9:

In mathematics, especially in the area of abstract algebra known as combinatorial group theory, the word problem for a recursively presented  group G is the algorithmic problem of deciding whether two words represent the same element. Although it is common to speak of the word problem for the group G strictly speaking it is a presentation of the group that does or does not have solvable word problem. (Source: Wikipedia)

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Example problems for math word problems grade 9:


Math word problems grade 9 – Example: 1

In the box 12 cards are placed as 1 to 12, mixed up thoroughly and  then a card is drawn at random from the box. if it is known that the number on the card is more than 3, find the probability that it is an even number.

Solution:

Let S be the sample space. then,

S = {1, 2, 3, 4, ......10, 11, 12}

Let A = Event of getting a card having a number more than 3.

And B = Event of getting a card having an even number.

Then, A = {4,5,6,7,8,9,10,11,12} and B = {2,4,6,8,10,12}

Therefore A B ={4,6,8,10,12}.

Therefore P(A) = `(n(A))/(n(S))` = `9/12` = `3/4`

P(B) = `(n(B))/(n(S))` = `6/12` = `1/12`

and P(A`nn` B) =`(P(AnnB))/(n(S))` = `5/12`

Suppose A has already occured and then B occurs.

Then, We have to find P(`B/A` )

Therefore P(`B/A` ) = P(A`nn` B) = `(P(AnnB))/(P(A))` = `(5/12)/(3/4)` = `(5/12 * 4/3)` = `5/9`

Math word problems grade 9 – Example: 2

The odds in favor of occurrence of an event 5:13. Find the odds against the occurrence of the event. Find the probability that it will occur.

Solution:

Odds in favor = `5/13`

Therefore Odds against = `13/5`

Number of favorable outcomes = 5

Number of unfavorable outcomes = 5 + 13

Required probability = `5/18`

Math word problems grade 9 – Example: 3

One card is drawn from a well shuffled deck of cards. Find the probability that the card is red ace.

Solution:

The number of possible outcomes n(s) = 52

Let A: getting a red ace card.

There are 2 red ace cards in the deck of the  cards n(A)  = 2.

P(ace) = P(A) = `(n(A))/(n(S))`

= `2/52`

= `1/26`

Math word problems grade 9 – Example: 4

A pair of dice is thrown. Find the probability of getting a doublet.

Solution:

When two dice are thrown, there are 36 sample points  n(S) = 36

A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

n(A) = 6

P(a doublet) = `(n(A))/(n(S))`

= `6/36`

= `1/6`

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Practice problems for math word problems grade 9:


1. A, B and C shoot to hit a target. If A hits the target 4 times in 5 trials; B hits it 3 times in 4 trials and C hits it 2 times in 3 trials, what is the probability that the target is hit by at least 2 persons?

[Answer: The required probability is 5/6]

2.A card is drawn from a well-shuffled deck of 52 cards and without replacing this card, a second card is drawn. Find the probability that the first card is a club and the second card is spade.

[Answer: `13/204` ]

Grade 10 Math Elimination

Introduction to Grade 10 math elimination:

Elimination method is used to solve if we have two unknown variables. In grade 10 math elimination we are going to solve word problems. In grade 10 math elimination problems we have to use the multiplication operation to eliminate the variables. After eliminating we will get the equation with one variable. Using that we have to find the x and y values. Here we will see an example problem for grade 10 math elimination method.

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Example problems grade 10 math elimination methods:


Example:

Solve the following equation using elimination method.

5x + y = 10

2x + 3y = - 9

Solution:

Given equations are,

5x + y = 10    ………………… (1)

2x + 3y = - 9     ...……………....  (2)

Here both variables having different co – efficient. So multiply the first equation with 3 for eliminating y.

15x + 3 y = 30   …………………. (3)

2x + 3 y = - 9    ………………… (4)

Now subtract 3 and 4

We get,

13x = 39

Now we have to divide by 13 on both side of the equation,

So x = 3

Now plug the value in any of the equation.

5(3) + y = 10

15 + y = 10

So y = - 5

In this the variables with values given in the equation directly. Now we will see an example to make an equation and then we have to solve.

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Example 2:


Alex and Jerry is have the total age is 30. Three years ago, the sum of twice Alex's age and three times Tom's age was 60 years. Find both of their ages using elimination method.

Solution:

Sum of Alex and Jerry’s age is 30

x + y = 30 ……………….. (1)

Three years ago gives x -3 and y – 3

2 (x – 3) + 3 (y – 3) = 60

2x – 6 + 3y – 9 = 60

2x + 3y - 15 = 60

2x + 3y = 75    ……………………. (2)

Multiply (1) with 2

2x + 2y = 60   .......................... (3)

2x + 3y = 75    ......................... (4)

Now subtract the equation

We get,

-y = - 15

So y = 15 (Jerry’s age)

From this Alex’s age x + 15 = 30 than x = 15

So Alex age = 15 and Jerry’s age = 15

These are the some example problems for grade 10 math elimination.