Sunshine Math Answers

Introduction to sunshine math answers:-

SUNSHINE MATH, this course is planned to improve your child’s voyage from first to last in mathematics. It is a math fortification course offered by Somerset Academy. Mission of Sunshine Maths is to put forward each and every students the chance to defy themselves in all category of math conception by giving them additional learning activities.

Sunshine Math Word Problems with Answers:-


Prob 1 :

Ram can cut a piece of wood in to three pieces in 20 minutes. How long it takes for him to cut the piece of wood in to 6 pieces.

Sol :

Given:-

Ram can cut a piece of wood in to three pieces in 20 minutes.

Find in how many minutes ram cut 6 pieces of wood.

3 pieces = 20 minutes. ------------> (1)

6 pieces = ? Minutes

6 is two times of 3.

So he takes double the time he took for cutting the piece in to three pieces.

The time taken to cut the wood in to three pieces is 20 minutes.

Double the times of 20 minutes is 40 minutes.

The time taken to cut the wood in to six pieces is 40 minutes.

Ans : 40 minutes



More examples:-Sunshine Math Equation Problems with Answers:-


Prob 2 :

When I was at the park I saw some girls and cats. Counting heads and legs I got 32 and 104 respectively. Find the total number of girls and cats in the park.

Sol :

Given there are totally 32 heads and 104 legs respectively.

The equation for head can be written as x + y = 32.

Let x – girls, y – cats.

The equation for legs can be written as 2x + 4y = 104.

By solving the above two equation we get

y = 20 and x = 12 .

Ans : The number of girls in the park is 12 and number of cats in the park is 20.

Prob 3 :

23 + 21 + 5+ 3 + 56 = 159.

Find the given expression correct or not.

Sol :

The given expression is 23 + 21 + 5+ 3 + 56 = 159.

The sum of numbers in right hand side is 106.

Ans : So the given expression is algebraically wrong .

Prob 4 :

13 + 41 + 51+ 32 + 34 = 171.

Find the given expression correct or not.

Sol:-

The given expression is 13 + 41 + 51+ 32 + 34 = 171.

The sum of numbers in right hand side is171.

Ans : So the given expression is algebraically correct.

Prob 5 :

3 + 1 + 5+ 32 - 34 = 11.

Find the given expression correct or not.

Sol :

The given expression is 3 + 1 + 5+ 32 - 34 = 11.

The sum of numbers in right hand side is 7.

Ans :So the given expression is algebraically wrong.

Prime Numbers 1-100

Introduction to prime numbers 1-100:

Prime numbers are the numbers it can be divided only by 1 and the number itself.Prime numbers cannot have any thing or factors except one and the number by itself. The +ve(positive) integers are divided into two such as prime numbers and composite integers.Totally there is  25 prime numbers between 1 and 100. So remaining are the composite numbers.

The numbers are like 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97 are 25 prime numbers between 1 to 100.That twenty five prime numbers are in between 1-100 These numbers cannot be divide by any other number except 1 and the number itself.And the same is called prime numbers.


The prime numbers from 1-100


In the prime number composite numbers have thing or factors  compare than other is one and itself Except the 25 prime numbers between 1 to 100 all the other numbers are composite numbers.In that,the list of  composite numbers 4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30,32,33,34,35,36,38,39,40,42,44,45,46,48,49,50,51,52,54,55,56,57,58,60,

62,63,64,65,66,68,69,70,72,74,75,76,77,78,80,81,82,84,85,86,87,88,90,91,92,93,94,95,96,98,99,100.......

These were the composite numbers between 1 to 100.


Problems of prime numbers between 1-100


solving prime numbers 1-100:

Here we mentioned the prime numbers from 1 to 100 only

So many prime numbers are there.

Let us see some examples,

Example 1:

To determine the given number is prime or not :one hundred forty-three ?

The factors of 143 are 1, 11, 13,143

So here we have factors for 143 other than 1 and 143

So this is not a prime number

The remaining prime number compare than other numbers are called composite number

So 143 is a composite number

Example 2:

To determine the one hundred thirty-one is prime or not?

Solution

The factors of 131 is 1, 131

So here we don’t have any other factor except 1 and the number 131

So the number 131 is a prime number

Example 3

To determine the prime numbers between thirty to thirty-five?

Solution

30    - Factors  - 1,2,3,5,6,10,15,30 – Not a prime number

31    – Factors – 1,31 – Prime number

32    – Factors – 1,2,,4,8,16,32 –Not a prime number

33     - Factors  - 1,3,11,33  - Not a prime number

34     - Factors  - 1,2,,17,34 – Not a prime number

35     - Factors – 1,5,7,35 –Not a prime number

So from 30 to 35 we have only one prime number

31 is only prime between 30 to 35

Definition of Conclusion

Introduction on definition of conclusion:

The term definition of conclusion in maths is used to define us about the problem that we solve and when we produce the final result at the end then that stage of processes is called as conclusion. In this chapter let us discuss about the term definition of conclusion in detail with suitable examples and explanations.

Representation of Definition of Conclusion:

The final result that we produce or the result that we produce at the end of the sum is called as conclusion.

Example Problems Based on Definition of Conclusion:

Example 1: Based on definition of conclusion

Solve the given problem by giving a conclusion by finding the value of A if A=33/2?

Solution:

Given: `A=33/2`

To produce an conclusion by finding the value of A

Step 1: The term division to found to be calculated in the given problem

Step 2: The conclusion or result will be given after finding the value of `A`

`A= 33/2`

`A= 16.5`

Hence, we conclude the problem by finding the value of `A=16.5`

Example 2: Based on definition of conclusion

Solve the given problem by giving a conclusion by finding the value of X if X=150/9?

Solution:

Given:` X=150/9`

To produce an conclusion or result

Step 1: The term division to found to be calculated in the given problem

Step 2: The conclusion or result will be given after finding the value of `X`

`X=150/9`

`X=16.66`

Hence, we conclude the problem by finding the value of `X =16.66`



Exercise Problems Based on Definition of Conclusion:

Solve the given problem by giving a conclusion by finding the value of Z if Z=330/5?
Answer: We conclude the problem by finding the value of Z=66

Solve the given problem by giving a conclusion by finding the value of Y if Y=88/2?
Answer: We conclude the problem by finding the value of Y=44

Summation Math

Introduction for summation math:

Summation math is assumed to be the sequence form of addition. Summation is the process of combine a sequence of numbers by means of addition. The summation can be written as n1 + n2 + n3 + n4 + n5 + ……… nn. Summation math is a continuous function for one variable in closed interval. Summation  of an infinite series of value is not constantly possible. In this topic we will discuss the some examples  problems.

Summation Math Rules:

Summation of infinite numbers is called summation,

`sum_(n=1)^oo`  =  1 + 2 + 3 + 4 + 5 + 6 + ...... + `oo`

Summation Rules:

Summation of a number

`sum_(n=1)^oo` a = a + a + a + ...... a (infinite times) = a * (`oo` )

Summation of square of infinite number.

`sum_(n =1)^oo`  i2 = 12 + 22 + 32 + 42 + 52 + ….. n2 = `(n (n + 1) (2n + 1))/(6)`
Summation of cube of infinite number.

`sum_(n=1)^oo` i3 = 13 + 23 + 33 + 43 + 53 + ….. n3 = `(n^2 (n + 1)^2/4) `

`sum_(i=1)^n`  si = s1 + s2 + s3 + … + sn

(n times) = sn, where s is constant.

`sum_(i=1)^n` i = 1 + 2 + 3 + … + n = `(n (n +1)) /(2) `

Example for Summation Math:

Example 1: Determine the value of the summation math `sum_(n=1)^4` ` (3 + 4n)`

Solution:

`sum_(n=1)^oo`  `= 1 + 2 + 3 + 4 + 5 + 6 + ...... + n`

`sum_(n=1)^5` `(3 + 4n) = (3 + 4(1)) + (3 + 4(2)) + (3 + 4(3)) + (3 + 4(4)) `

`= (3 + 4) + (3 + 8) + (3 + 12) + (3 + 16) `

`= 7 + 11 + 15 + 19 = 52`

Answer: 52

Example 2: Determine the value of the summation math `sum_(n=1)^6` `(n + 1)^2`

Solution:

Here, `(n + 1)^2`

`(n + 1)^2 = n^2 + 2n +1`

so, we wirte the summation as `sum_(n=1)^6` ` (n^2 + 2n+ 1)`

the summation, we get

`sum_(n=1)^6` `n^2` + `sum_(n=1)^6` `2n` +  `sum_(n=1)^6`

`= {12 + 22 + 32 + 42 + 52 + 6 ^2} + {(2 * (1)) + (2 * (2)) + (2 * (3)) + + (2 * (4)) + + (2 * (5)) + + (2 * (6))} + (1 * (6))`

= `((6 (6 + 1) ((2 * (6)) + 1))/(6)) + 42 + 6` 


`=91 + 42 + 6`

Answer: 139

Simple Solution Math Book

Introduction for simple solution mathematics book:

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.  The mathematician Benjamin Peirce called mathematics "the science that draws necessary conclusions". Arithmetic operations and numbers are mainly used in mathematics. Let us see some simple solution math book.  (Source: Wikipedia)



Example Problems for Simple Math Book:

Let us see some simple math book here:

Problem 1:

Solve the equation 2x - 24 = 78

Solution:

Given equation 2x - 24= 78

Add 24 on both the sides of the equation, we get

2x = 102

Divide the obtained equation by 2 on both the sides, we get

x = `102 / 2` = 51

Therefore the final answer for this equation is x =51

Problem 2:

Solve:

Given function `(x / 4)` = 14

Solution:

Given, `(x / 4)` = 14

Multiply the above equation by 4 on both the sides, we get

x = 14 * 4

x = 56

Therefore the final answer is x = 56

Problem 3:

Divide 75 by 5

Solution:

Given, 75 divided by 5

It can be written as,

75 ÷ 5

Divide the value of 75 by 5, we get

75 ÷ 5 = 15

Therefore the final answer for the division is 15

Problem 4:

Add this two equations x - y2 + 4 and x2 + 2x + y2 - 22

Solution:

The given equations are, x - y2 + 4 and x2 + 2x + y2 - 22

Now we have add the two equations, we get

= x - y2 + 4 + x2 + 2x + y2 - 22

= x2 + 3x - 18

Therefore the final equation will be x2 + 3x - 18

Problem 5:

Subtract the two equations (x - 2y + 18) and (3x + 8y + 60)

Solution:

Given equations are (x - 2y + 18) and (3x + 8y + 60)

Subtract the two equations, we get

= (x - 2y + 18) - (3x + 8y + 60)

Expand the above equation, we get

= x - 2y + 18 - 3x - 8y - 60

= - 2x - 10y - 42

Divide the obtained equation by - 2, we get

= x + 5y + 21

Therefore the final answer is x + 5y + 21



Practice Problems for Simple Math Book:

Problem 1:

Solve x + 24 = 3x - 38

Solution:

Therefore x = 31

Problem 2:

Solve x2 - 144 = 0

Solution:

Therefore x = ± 12

Problem 3:

x + y = 23, find the value of x at y = 2

Solution:

Therefore x = 21

Long Math Problem

Introduction on long math problem

This article is about long math problem. Long math problem will contain a lot of step in solving the problem to obtain the final solution. In some of the long math problem we use different formula to obtain the solution. There are many long math problem in many different math chapters. Each topics of mathematics have different solving properties theorem methods and formulas. There are many websites and online tutors to help with long math problem. Many stiudents favorite and best website to solve long math problem is tutor vista. Below we can see one long math problem.

Long Math Problem

Long math problem does not means that the given problem must be long. It also means the steps taken to solve some problems lead to long math problem. Below we solved one math problem which have a long steps to obtain the entire solution.
1. Solve the equations: p + 2q + 3r = 14, 3p + q + 2r = 11, 2p + 3q + r = 11.

Solution: Let the given equations be identified as follows:

p + 2q + 3r = 14------- (1)

3p + q + 2r = 11------ (2)

2p + 3q + r = 11------ (3)

Consider the equations (1) and (3)

(1) ? p + 2q + 3r = 14

(3) × 3 ? 6p + 9q + 3r = 33 subtracting

–5p – 7q = –19

5p + 7q = 19 (4)

Consider the equations (2) and (3)

(2) ? 3p + q + 2r = 11

(3) × 2 ? 4p + 6q + 2r = 22 subtracting

–p – 5q = –11

p + 5q = 11 (5)

Consider the equations (4) and (5)

(4) ? 5p + 7q = 19

(5) × 5 ? 5p + 25q = 55 subtracting

–18q = –36; ?q = 2

Substitute q = 2 in (5) we get

p + 5(2) = 11; p + 10 = 11; ?p= 1

Substitute p = 1, q = 2 in (3) we get

2(1) + 3(2) + r = 11; 2 + 6 + r = 11 ? r = 3

The solution is p = 1, q = 2, r = 3.


Long Math Problem

Similar to the above problem there are more number of math problems to solve with long steps. In some problems the given math problem will be long and the solution will be a shorter one. In some cases the solution and the problem both will be a longer one. Here are some problem similar to the solved problem above for your practice. The solution also given below and you have to try the steps similar to the problem solved above.

1. Solve: 3x – 3y + 4z = 14; –9x – 6y + 2z = 1; 6x + 3y + z = 5

Answer: The solution is x = 1, y = –1, z = 2

2. Solve: a + b = 3, b + c = –5, c + a = 2.

Answer: The solution is a = 5, b = –2, c = –3.

Multiply Symbol

Introduction of multiply symbol:

Multiplication or Times is fundamental operation on whole numbers. The fundamental operations on whole numbers are addition, subtraction, multiplication, and division. The time is the process of repeated addition of a number. Times or multiply denoted by the symbol × or *. Elementary school students use the symbol × for multiplication (multiply) or times. Leibniz uses the cap symbol `nn`  for multiplication or times. This symbol is used to indicate intersection in set.

Multiplication Example

Multiplication or times indicates by placing the quantities to be multiplied side by side (juxtaposition). Multiplication or times is a short form of addition of the same number several times. The important of multiplication or times process is to place the digits value of the factors in the proper columns. That is, units number must be placed in the units column, tens in tens column, and hundreds in hundreds column. Notice that it is not essential to write the zero in the case of 15 tens (150) since the 1 and 5 are written in the proper columns.

Ex 1:   Multiply 283 by 101

Sol:    283 × 101 means 101 times 283

Or 100 times 283 + one time 283

Or 28300 + 283

Or 28583

Therefore 283 × 101 = 28583

Ex 2:  John earns four times, of what his brother earns. If his brother earns $1800 in a month, how much does John earn?

Sol :  John brother earn $1800

John earn three times more than his brother that is $1800 times 4 (Here the times indicate the symbol ×)

Therefore $1800 × 4

John earn $7200

Ex 3: Multiply 650 × 99

Sol :  650 × 99 means 99 times 650

Or 100 times 650 – one time 650 (Here the times indicate the symbol ×)

Or 65000 – 650

Or 64350

Therefore 650 × 99 = 64350

Practice Problem

1. Kim earns 5 times, of what his brother earns. If his brother earns $1500 in a month, how much does Kim earn?

Ans :  $7500

2. Multiply 350 × 470

Ans : 164500

3. In a town have four post offices. In each post office there are six workers. How many workers do the post offices have in total?

Ans : 24

4. Four children are playing cricket. They all brought 7 balls. How many balls do they have totaled?

Ans : 28