Common Core Math

In this logarithm rules or log rules guide, students and teachers will learn the presented common laws of logarithms, also called ‘log rules’. Mainly, there are four log rules that are helpful in expanding logarithms, condensing logarithms, and solving logarithmic equations. Along with this, you will also find the proofs of these four log rules and additional laws of logarithms for a better understanding of the basic logarithm concept. Whenever you get confused during homework help please check out the basic logarithm rules or log rules prevailing here in this article.

Logarithm Rules Or Log Rules

There are four following math logarithm formulas:

• Product Rule Law: log_a (MN) = log_a M + log_a N
• Power Rule Law: log_a Mⁿ = n log_a M
• Quotient Rule Law: log_a (M/N) = log_a M – log_a N
• Change of Base Rule Law: log_a M = log_b M × log_a b

Also Check: Convert Exponentials and Logarithms

Descriptions of Logarithm Rules

Here, we have discussed four log rules along with proofs to grasp the concepts easily and become pro in calculating the logarithm problems. Let’s start with proof 1:

1. Logarithm Product Rule:

The logarithm of the multiplication of x and y is the sum of the logarithm of x and the logarithm of y.

log_b(x ∙ y) = log_b(x) + log_b(y)

Proof of Log Product Rule Law: 

log_a(MN) = log_a M + log_a N

Let log_a M = x ⇒ a sup>x = M

and log_a N= y ⇒ ay = N

Now a^x ∙ a^y = MN or, a^x+y = MN

Therefore from definition, we have,

log_a (MN) = x + y = log_a M + log_a N [putting the values of x and y]

Corollary: The law is true for more than two positive factors i.e.,

log_a (MNP) = log_a M + log_a N + log_a P

since, log_a (MNP) = log_a (MN) + log_a P = log_a M+ log_a N+ log_a P

Therefore in general, log_a (MNP ……. )= log_a M + log_a N + log_a P + …….

So, the product logarithm of two or more positive factors to any positive base other than 1 is equal to the sum of the logarithms of the factors to the same base.

Example: Calculate log₁₀(8 ∙ 4)?

The given expression matches the logarithm product rule.

So apply the log rule and get the result,

log₁₀(8 ∙ 4) = log₁₀(8) + log₁₀(4)

2. Logarithm Power Rule:

The logarithm of x raised to the power of y is y times the logarithm of x.

log_b(x^y) = y ∙ log_b(x)

Proof of Log Power Rule Law:

log_aMⁿ = n log_a M

Let log_a Mⁿ = x ⇒ a^x = Mⁿ

and log_a M = y ⇒ a^y = M

Now, a^x = Mⁿ = (a^y)n = a^ny

Therefore, x = ny or, log_a Mⁿ = n log_a M [putting the values of x and y].

Example: Find log₁₀(2⁹)?

Given log₁₀(2⁹) is in logarithm power rule. So, apply the log rule and calculate the output:

Hence, log₁₀(2⁹) = 9∙ log₁₀(2).

3. Logarithm Quotient Rule Formula

The logarithm of the ratio of two numbers is the logarithm of the numerator minus the logarithm of the denominator.

log_a(x / y) = log_a(x) – log_a(y)

Proof of Log Quotient Rule Formula:

Let M = a^x and N = a^y, then it follows that log_a(M) = x and log_a(N) = y,

We can now prove the quotient rule as follows:

log_a (M/N) = log_a (a^x/a^y)

= log_a(a^x-y)

= x – y [Put the values of x and y]

= log_a M – log_a N

Corollary: log_a [(M × N × P)/(R × S × T)] = log_a (M × N × P) – log_a (R × S × T)

= log_a M + log_a N + log_a P – (log_a R + log_a S + log_a T)

Example: Calculate log₁₀(10 / 5)

Now apply the log quotient rule and get the result,

Therefore, log₁₀(10 / 5) = log₁₀(10) – log₁₀(5).

4. Logarithm Base Change Rule:

The logarithm of M for base b is equal to the base a log of M divided by the base a log of b.

log_b M = log_a M/log_a b

Proof of Change of base Rule Law:

log_a M = log_b M × log_a b

Assume log_a M = x ⇒ a^x = M,

log_b M = y ⇒ b^y = M,

and log_a b = z ⇒ a^z = b.

Now, a^x= M = b^y – (a^z)y = a^yz

Therefore x = yz or, log_a M = log_b M × log_a b [putting the values of x, y, and z].

Corollary:

(i) Putting M = a on both sides of the change of base rule formula [log_a M = log_b M × log_a b] we get,

log_a a = log_b a × log_a b or, log_b a × log_a b = 1 [since, log_a a = 1]

or, log_b a = 1/log_a b

In other words, the logarithm of a positive number a with respect to a positive base b (≠ 1) is equal to the reciprocal of logarithm of b with respect to the base a.

(ii) On the basis of the log change of base rule formula we get,

log_b M = log_a M/log_a b

In other terms, the logarithm of a positive number M in respect of a positive base b (≠ 1) is equal to the quotient of the logarithm of the number M and the logarithm of the number b both with respect to any positive base a (≠1).

List of Some Other Logarithm Rules or Log Rules:

If M > 0, N > 0, a > 0, b > 0 and a ≠ 1, b ≠ 1 and n is any real number, then

(i) log_a 1 = 0

(ii) log_a a = 1

(iii) a ^{log_a M} = M

(iv) log_a (MN) = log_a M + log_a N

(v) log_a (M/N) = log_a M – log_a N

(vi) log_a Mⁿ = n log_a M

(vii) log_a M = log_b M × log_a b

(viii) log_b a × log_a b = 1

(ix) log_b a = 1/log_a b

(x) log_b M = log_a M/log_a b

Solved Examples of How to Apply the Log Rules or Logarithm Rules

1. Evaluate the expression ie., log₂ 4 + log₂ 8 by using log rules.

Solution:

Given expression is log₂ 4 + log₂ 8

First, express 4 and 8 as exponential numbers with a base of 2. Next, apply the logarithm power rule formula followed by the identity rule. Once you finished that, add the resulting values to find the final answer.

log₂ 4 + log₂ 8
= log₂ 2² + log₂2³ [apply power rule]
= 2 log₂ 2 + 3 log₂ 2 [apply identity rule]
= 2(1) + 3(1)
= 2+3
= 5
Hence, the answer for the given expression log₂ 4 + log₂ 8 is 5.

2. Evaluate the expression with Log Rules: log₃ 162 – log₃ 6

Solution:

log₃ 162 – log₃ 6

Now, we can’t express the 162 as an exponential number with base 3. Don’t worry, we have another way of solving the expression.

It is possible by applying the log rules in the reverse process. Yes, we can also apply the logarithm rules in reverse if not solved in a direct manner.

Remember that the log expression can be stated as one or a single logarithm number via using the backward Quotient Rule Law. Sounds different right, but so easy to calculate.

Take the given expression, log₃ 162 – log₃ 6
= log₃ (162/6)
= log₃ (27)
= log₃ (3³)
= 3 log₃ 3
= 3(1)
= 3

By applying the rules in reverse, we get the result as 3 for the given expression log₃ 162 – log₃ 6.

Hence, log₃ 162 – log₃ 6 = 3.

Are you looking for ways on how to convert from Exponential Form to Logarithmic Form? Then, don’t panic as we will discuss how to change Exponential Form to Logarithmic Form or Vice Versa. Get to know the Definitions of Exponential and Logarithmic Forms. Find Solved Examples on Converting between Exponential and Logarithmic Forms and learn the entire procedure.

Logarithmic Form – Definition

Logarithmic Functions are inverse of Exponential Functions. It tells us how many times we need to multiply a number to get another number. To give us the ability to solve the problem x = b^y for y

For x>0, b>0 b≠ 1, y = log_b x is equivalent to b^y = x

Example: When asked how many times we’ll need to multiply 2 in order to get 32, the answer is the logarithm 5.

Exponential Form – Definition

Exponents are when a number is raised to a certain power that tells you how many times to repeat the multiplication of a number by itself.

b^y = x

How to Convert from Exponential Form to Logarithmic Form?

To convert from exponential form to logarithmic form, identify the base of the exponential equation
and move the base to the other side of the equal to sign, and add the word “log”. Do not change anything
but the base, the other numbers or variables will not change sides.

Consider the equation b^y = x

The equation y = log_b x is said to be the Logarithmic Form

b^y = x is said to be Exponential Form

Two Equations are different ways of writing the same thing.

Solved Examples on Converting Between Exponential Form to Logarithmic Form

1. Convert the 10³ = 1000 Exponential Form to Logarithmic Form?

Solution:

10³ = 1000

log₁₀1000 = 3

In this example, the base is 10 and the base moved from the left side of the exponential equation to the right side of the logarithmic equation, and the word “log” was added.

2. Write the Exponential Equation 3^x = 27 in Logarithmic Form?

Solution:

3^x = 27

In this example, the base is 3 and the base moved from the left side of the exponential equation to the right side of the logarithmic equation, and the word “log” was added.

x = log₃27

= log₃3³

= 3log₃3

= 3.1

= 3

3. Write the Exponential Equation  6^y = 98 in Logarithmic Form?

Solution:

Given Equation is 6^y = 98

In this example, the base is 6 and the base moved from the left side of the exponential equation to the right side of the logarithmic equation, and the word “log” was added.

y = log₆ 98

Have you ever seen Tossing a Coin before Commencement of a Cricket Match? This is usually done in Matches and the Captain who predicts the Toss Correctly can choose on what his team could defend. It is the most common application of the Coin Toss Experiment. Tossing a Coin is quite useful as the Probability of obtaining Heads is as likely as Tail. There are only two outcomes when you flip a coin i.e. Head(H) and Tail(T).

However, if you Toss 2, 3, 4, or more coins than that at the same time the Probability is Different. Let us learn about the Coin Toss Probability Formula in detail in the later sections. You can check out Solved Examples on Tossing a Coin and their Probabilities here.

Tossing a Coin Probability

When Tossed a Coin you will have only two possible outcomes i.e. Head or Tail. However, you will not know which outcome you will get among Heads or Tails. Tossing a Coin is a Random Experiment and you do know the set of Outcomes but not the exact outcomes.

General Formula to Determine the Probability = \(\frac { No.\; of\; Favorable\; Outcomes }{  Total\; Number\; of\; Possible\; Outcomes } \)

On Tossing, we do have only two possible outcomes

Probability of getting Head = \(\frac { No.\;of \;Outcomes\; to\; get\; Head }{  Total\; Number\; of\; Possible\; Outcomes } \)

= \(\frac { 1 }{ 2 } \)

Probability of getting Tail = \(\frac { No.\; of\; Outcomes\; to\; get\; Tail }{  Total\; Number\; of\; Possible\; Outcomes } \)

= \(\frac { 1 }{ 2 } \)

How to Predict Heads or Tails?

• If a coin is fair or unbiased, i.e. no outcome is particularly preferred, then it is difficult to predict heads or tails. Both the outcomes are equally likely to show up.
• If a coin is unfair or biased, i.e. an outcome is preferred, then we can predict the outcome by choosing the side that has a higher probability.
  • If the probability of a head showing up is greater than 1/2, then we can predict the next outcome as a head.
  • If the probability of a tail showing up is greater than 1/2, then we can predict the next outcome as a tail.

Solved Examples on Coin Toss Probability

1. On tossing a coin twice, what is the probability of getting only one Head?

Solution:

On tossing a coin twice, the possible outcomes are {HH, TT, HT, TH}

Therefore, the total number of outcomes is 4

Getting only one Head includes {HT, TH}

Thus, the number of favorable outcomes is 2

Hence, the probability of getting exactly one head =  Probability of Favorable Outcomes/Total Number of Outcomes

= \(\frac { 1 }{ 2 } \)

2. Three fair coins are tossed simultaneously. What is the probability of getting at least three tails?

Solution:

When 3 coins are tossed, the possible outcomes are {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Thus, the total number of possible outcomes = 8
Getting at least 3 tails include the outcomes = {TTT}

No. of Favorable Outcomes = 1

Probability of getting at least three tails include = Probability of Favorable Outcomes/Total Number of Outcomes

= \(\frac { 1 }{ 8 } \)

Do you want to learn about the Relation among All Trigonometric Ratios of (180° – θ)? Then, halt your search here as we have explained the Relationship between Trigonometric Functions of 180 Minus Theta in detail along with their proofs. Find Solved Examples on finding the Trigonometric Ratios with Step by Step Explanation making it easy for you to solve related problems in no time. Know a Simple Formula to memorize the Trigonometric Functions.

How to Determine the Trigonometric Ratios of (180° – θ)?

Before diving deep into the article to understand how trigonometric ratios of (180° – θ) are determined you need to understand the ASTC Formula.

The ASTC Formula can be easily remembered by considering the Phrases provided below

“All Silver Tea Cups” or ” All Students Take Calculus”

You will better understand the concept by having a glance at the below picture. From the picture, it is clearly evident that 180 degrees minus theta fall under the 2nd Quandrant. In the 2nd Quadrant, only sin and cosecant are Positive.

sin(180° – θ)=sinθ
cos(180° – θ)=−cosθ
tan(180° – θ)=−tanθ
cosec (180° – θ)=cosec θ
sec(180° – θ)=−secθ
cot(180° – θ)=−cotθ

Read More Articles:

• Trigonometrical Ratios of 90 Degree Plus Theta
• Trigonometrical Ratios Table
• Worksheet on Trigonometric Identities

Evaluate Trigonometric Functions of (180° – θ)

1. Evaluate Sin(180° – θ)?

Solution:

Sin(180° – θ) = Sin (90° + 90° – θ)

= Sin [90° + (90° – θ)]

= Cos (90° – θ), [since Sin (90° + θ) = Cos θ]

= Sin θ[since Cos (90° – θ) = Sin θ]

Therefore, Sin (180° – θ) = Sin θ

2. Evaluate Cos(180° – θ)?

Cos (180° – θ) = Cos (90° + 90° – θ)

= Cos [90° + (90° – θ)]

= – Sin (90° – θ), [since Cos (90° + θ) = -Sin θ]

= -Cos θ [since sin (90° – θ) = cos θ]

Therefore, Cos (180° – θ) = – Cos θ

3. Evaluate Tan (180° – θ)?

Solution:

Tan (180° – θ) = Tan (90° + 90° – θ)

= Tan [90° + (90° – θ)] [since Tan (90° + θ) = -Cot θ]

= – Cot (90° – θ)

= Tan θ [since Cot (90° – θ) = Tan θ]

Therefore, Tan (180° – θ) = – Tan θ

4. Evaluate Csc (180° – θ)?

Solution:

Csc (180° – θ) = \(\frac { 1 }{ Sin(180° – θ) } \)

= \(\frac { 1 }{ Sin θ } \) [Since Sin(180° – θ) = Sin θ]

Therefore, Csc (180° – θ) = \(\frac { 1 }{ Sin θ } \)

5. Evaluate Sec (180° – θ)?

Solution:

Sec (180° – θ) = \(\frac { 1 }{ Cos(180° – θ) } \)

= = \(\frac { 1 }{ -Cos θ } \) [Since Cos(180° – θ) = -Cos θ]

= -Sec θ

Therefore, Sec (180° – θ) = -Sec θ

6. Evaluate Cot (180° – θ)?

Solution:

Cot (180° – θ) = \(\frac { 1 }{ Tan(180° – θ) } \)

= \(\frac { 1 }{ -Tan θ } \) [Since Tan(180° – θ) = -Tan θ]

= -Cot θ

Solved Examples on Trigonometric Ratios

1. Find the Value of Cos 150°?

Solution:

Cos 150° = Cos(180° – 30°)

= – Cos 30°

= – \(\frac { √3 }{ 2 } \)

2. Find the value of Cot 135°?

Solution:

Cot 135° = Cot(180° – 45°)

= -Cot 45°

= -1

Worksheet on Multiplication of Fractions is here. Get important questions involved in Fractions Multiplication. Refer to the various rules, methods, formulae of Fractions Multiplication. Follow questions regarding proper, improper, mixed fractions and how they work when multiplying them. Solved examples in Multiplying Fractions Worksheet will help you to get a piece of detailed information and also helps you to score better marks in the exam.

Problem 1:

Gita has eight marbles. She gave \(\frac { 1 }{ 4 } \) of them to her younger brother. How many marbles did her brother get?

Problem 2:

Mr. Gupta puts 3\(\frac { 1 }{ 4 } \) litres of petrol in his car. If he uses \(\frac { 1 }{ 3 } \) of it. How many litres of petrol did he use?

Problem 3:

Neha spends 3\(\frac { 3 }{ 5 } \) hours a day in morning exercises. How many hours does she spend in morning exercises in one week?

Problem 4:

One plum cake weighs \(\frac { 3 }{ 4 } \) kg. If Mr. Ramesh buys five such cakes, how many kilograms of cakes did he buy?

Problem 5:

The cost of one kilogram apples is Rupees 25\(\frac { 1 }{ 2 } \). What is the cost of 1\(\frac { 1 }{ 2 } \) kilogram apples?

Problem 6:

The thickness of the Mathematics book of Class – IV is 1\(\frac { 1 }{ 4 } \) cm. What will be the thickness of a pile of 16 such books?

Problem 7:

Milk is sold at Rs 16\(\frac { 3 }{ 4 } \) per litre, find the cost of 6\(\frac { 2 }{ 5 } \) litre of milk?

Problem 8:

Find the area of the rectangular park which is 41\(\frac { 2 }{ 3 } \) long and 18\(\frac { 3 }{ 5 } \) m board?

Problem 9.
If the cost of 5\(\frac { 2 }{ 5 } \) litres of milk is 236\(\frac { 1 }{ 4 } \), find the cost per litre?

Problem 10:

Margaret has 2\(\frac { 1 }{ 3 } \) cups of sugar. She is using 1\(\frac { 1 }{ 4 } \) cups to make cookies. How much sugar will she have left?

Problem 11:

Russell spent \(\frac { 1 }{ 3 } \) of his allowance to go to movies. He wants to buy a new video game that will cost him another \(\frac { 1 }{ 2 } \) of his allowance. If he buys the game. How much of his allowance will he have spent?

Problem 12:

Sam has 3\(\frac { 1 }{ 8 } \) acres of land. He planted potatoes on 2\(\frac { 2 }{ 3 } \) of the land. He wants to plant corn. How much land does he have left to plant the corn?

Problem 13:

I bought 80 pounds of spaghetti to share equally between 7 families. About how many pounds of spaghetti will each family receive?

Problem 14:

I have 70 feet of rope that I need to share equally between 11 people. About how many feet of rope will each person receive?

Problem 15:

Mom made pizza. There is only \(\frac { 1 }{ 4 } \) of the pizza left. Mom and her friend Sally are going to share the remaining pizza evenly. How much of pizza will they each get?

Place value is nothing but the position or place of a digit in the decimal number. In a number, every digit has someplace. The positions of digits in number starts from one’s place. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand’s, and so on. Generally, a number is formed by grouping digits together.

Place, Place Value and Face Value Definitions

In a number, every digit has a fixed position known as the digit place. And each digit has a value depending on its place called the place value of the digit. Face value of a digit for any place in the value of the digit itself.

Place Value of a digit = (face value of the digit) x (value of the place)

Place Value Table

Place value chart is helpful to ensure that the digits are in the correct places. Place value tells us how much each digit stands for. The place value table is mentioned here.

Properties of Place Value

• The place value of every one-digit number is the same as the face value.

The place value and face value of 1, 2, 3, 4, 5, 6, 7, 8, and 9 are 1, 2, 3, 4, 5, 6, 7, 8, and 9. The place value of 0 is always 0. It may present at any place in the number, its value is always zero.

Example:

(i) In the numbers 105, 270, 1025 the place of value 0 is 0.

(ii) The place of 1 in 251 is 1, 7 in 8567 is 7.

• In a two-digit number, the place value of the ten-place digit is equal to 10 times of the digit.

Example:

(i) The place value of 6 in 67 is 6 x 10 = 60

(ii) The place value of 2 in 526 is 2 x 10 = 20

• In the number 567, the digit 7 is at one’s place, the digit 6 is at ten’s place and the digit 5 is at the hundred’s place.

So, the place value of 7 is 7, 6 is 6 x 10 = 60, and 5 is 5 x 100 = 500

Thus, for the place value of a digit, the digit is multiplied by the place value of 1 it has to be that place.

Example:

In the number 286,

The place value of 6 is 6 x 1 = 6

The place value of 8 is 8 x 10 = 80

The place value of 2 is 2 x 100 = 200

• Now it is the general law that the digit possesses its place value as the product of the digit and place value of one to be at that position.

Examples:

(i) In the number 3578,

The place value of 8 is 8 x 1 = 8 because 8 is at the unit’s place.

The place value of 7 is 7 x 10 = 70 because 7 is at ten’s place.

The place value of 5 is 5 x 100 = 500 because 5 is at hundred’s place

The place value of 3 is 3 x 1000 = 3000 because 3 is at thousand’s place.

(ii) In the number 58762, the place value of each digit is as follows

2 is 2 x 1 = 2

6 is 6 x 10 = 60

7 is 7 x 100 = 700

8 is 8 x 1000 = 8000

5 is 5 x 10000 = 50000

(iii) Find the place value of digits 30589

Also, check out

• Decimal Place Value Chart
• Place Value Chart

Example Questions on Place Value

Example 1:

Write the place value of the given numbers.

(i) 5 in 38956

(ii) 2 in 2587

(iii) 6 in 6845321

Solution:

(i) The given number is 38956

The place value of 5 in 38956 is 5 x 10 = 50. Because 5 is at tens place.

(ii) The given number is 2587

The place value of 2 in 2587 is 2 x 1000 = 2000. Because 2 is at thousands place.

(iii) The given number is 6845321

The place value of 6 in 6845321 is 6 x 1000000 = 60,00,000. Because 6 is at ten lakhs place.

Example 2:

Write the place value of a highlighted digit in the given numbers

(i) 2589

(ii) 67525

(iii) 2515963

Solution:

(i) The given number is 2589

The place value of highlighted 8 in 2589 is 8 x 10 = 80 as it is located at the tens position.

(ii) The given number is 67525

The place value of highlighted digit 5 in 67525 is 5 x 100 = 500 as it is located at the hundred’s position.

(iii) The given number is 2515963

The place value of highlighted digit 1 in 2515963 is 1 x 10000 = 10000 as it is located at the ten thousand’s position.

Example 3:

Circle the following.

(i) Digit at the hundreds place in 5289.

(ii) Digit at the lakhs place in 2563891

(iii) Digit at the units place in 5280

Solution:

(i) In the number 5289,

9 is at units place, 8 is at tens place, 2 is at hundreds place and 5 is at thousand’s place

So, Digit at the hundreds place in 5289 is 2.

(ii) In the number 2563891,

1 is at units place, 9 is at tens place, 8 is at hundreds place, 3 is at thousand’s place, 6 is at ten thousand’s place, 5 is at lakhs place, 2 is at ten lakhs place.

Therefore, Digit at the lakhs place in 2563891 is 5.

(iii) In the number 5280,

0 is at the unit’s place, 8 is at the ten’s place, 2 is at the hundred’s place and 5 is at the thousand’s place.

Hence, the digit at the units place in 5280 is 0.

Example 4:

Find the place value of 5 in the given numbers.

(i) 6,00,521

(ii) 5,23,168

(iii) 2,05,387

Solution:

(i) The given number is 6,00,521

The place value of 5 is 5 x 100 = 500.

(ii) The given number is 5,23,168

The place value of 5 is 5 x 10,000 = 50,000.

(iii) The given number is 2,05,387

The place value of 5 is 5 x 1000 = 5000.

FAQs on Place Value

1. Write the differences between place value and face value?

Place value means the position of a particular digit in the number but face value represents the exact value of a digit in that number. For example in the number 2556, the place value of 2 is thousands but the face value is 2.

2. Define place value with an example?

The place value is the position of a digit in a number. The place values of digits are represented as ones, tens, hundreds, thousands, ten thousand, and so on. The example is the place value of 8 in 589 is 8 tens i.e 80.

Access Worksheet on Division of Fractions here to get acquainted with various problems on Division of Fractions. Refer to Fractions Division’s step-by-step procedure to solve the problems. Follow the study material and guidance to solve various questions and answers. Know the important methods, rules, and formulae of Dividing Fractions. Try to solve the Dividing Fractions Word Problems Worksheet available on your own to test your knowledge of the concept. We have provided detailed solutions to all the Dividing Fractions Problems available.

Dividing Fractions Word Problems Worksheets with Answers

Problem 1:

Jiya thought it would be nice to include \(\frac { 2 }{ 21 } \) of a pound of chocolate in each of the holiday gift bags she made for her friends and family. How many holiday gift bags could jiya make with \(\frac { 2 }{ 3 } \) of a pound of chocolate?

Problem 2:

Lithi has \(\frac { 1 }{ 5 } \) of a bag of dog food left. She is splitting it between her 3 dogs evenly. What fraction of the original bag does each dog get?

Problem 3:

John has \(\frac { 1 }{ 4 } \) of a gallon of saltwater that he is using for an experiment. He needs to evenly separate the saltwater into 3 separate beakers. How much salt water will be in each beaker?

Problem 4:

David has a board that measures 4ft in length. The board is going to be cut into \(\frac { 1 }{ 4 } \) ft pieces. How many pieces will David split the board into?

Problem 5:

Jasmine has \(\frac { 1 }{ 4 } \) hour left to finish 4 math problems on the test. How much time does she have to spend on each problem?

Problem 6:

Sandya finished \(\frac { 1 }{ 3 } \) of her homework problems. Which fraction is equivalent to what Sandya finished?

Problem 7:

Berlin has 9 cups of sugar. If this is \(\frac { 3 }{ 4 } \) of the number he needs to make a cake, how many cups does he need?

Problem 8:

Mahathi is working on projects that require 3\(\frac { 1 }{ 2 } \) yards of ribbon per project. Mahathi has 28 yards of ribbon. What is the greatest number of projects that Mahathi can complete with this ribbon?

Problem 9:

Arjun made 40 cookies for the upcoming bake sale. He made \(\frac { 3 }{ 5 } \) of the cookies of chocolate chip and \(\frac { 1 }{ 4 } \) of them peanut butter. How many chocolate chip and peanut butter cookies are there in together?

Problem 10:

Gillian took 6\(\frac { 1 }{ 2 } \) hours to build a dog house. If she worked for 1\(\frac { 3 }{ 4 } \) hours each night. How many nights did it take?

Problem 11:

A class consists of 40 students. Of them \(\frac { 2 }{ 5 } \) of the fraction are boys. Find the number of girls present in the class?

Problem 12:

John had Php 960. He spent \(\frac { 1 }{ 3 } \) of his money on food. How many were left to him?

Problem 13:

Efren had 480 apples for sale. He sold \(\frac { 3 }{ 5 } \) of them. How many were left?

Problem 14:

Rose and Mary were filling the class-raised garden bed with soil. Rose shoveled in \(\frac { 1 }{ 3 } \) of a cubic yard, and Mary shoveled in \(\frac { 1 }{ 2 } \) of a cubic yard. How much of the soil did they put into the garden bed altogether?

Problem 15:

Carl’s invited Otto over to his house. Carl’s liked to share with his guests, so he got out his chocolate stash from last Halloween. He still had \(\frac { 4 }{ 5 } \) of a pound of chocolate. Carl’s asked Otto how much chocolate he would like. Otto said that he would like \(\frac { 1 }{ 3 } \) of a pound of chocolate. Carl’s obliged. How much chocolate does Carl’s have left?

Worksheet on Fractions is here. Know the step-by-step procedure to solve fraction problems. Refer to addition, subtraction, multiplication, and division of fractions to know the different model problems. Follow different terminology and rules to solve fraction problems. Various methods, rules, formulae involved in fractions operations are here. Go through the below sections to know more about the fractions.

Solved Problems on Simplifying Fractions

Problem 1:

Sameer has read three-fifth of the 75 pages of his book. How many more pages he needs to read to complete the book?

Problem 2:

Ravi had Rs.675. He gave \(\frac { 13 }{ 15 } \) of the amount to Sam. Sam spent \(\frac { 9 }{ 15 } \) of the amount given to him. How much money is Sam left with?

Problem 3:

Jagadish spends \(\frac { 3 }{ 5 } \) of his income on rent and \(\frac { 1 }{ 4 } \) of remainder amount on grocery. He saves the remaining amount.

a) What fraction of income does he save?

b) If he saves Rs.1029 per month, what is his monthly income?

Problem 4:

A baby rabbit was born \(\frac { 3 }{ 4 } \) of a month early. When it was born, its weight was \(\frac { 7 }{ 8 } \) kilogram. The weight at birth was \(\frac { 9 }{ 10 } \) kilogram less than the average weight of a newborn rabbit. What is the average weight of a newborn rabbit?

Problem 5:

In the zoo, \(\frac { 3 }{ 8 } \) of the staff are male. \(\frac { 7 }{ 12 } \) of the staff works part-time at the zoo. What is the fraction of the staff that is female?

Problem 6:

The lions in the zoo are fed three times a day. During the morning feeding, \(\frac { 2 }{ 15 } \) tons of meat is fed. During the afternoon feeding, the weight of the meat will be \(\frac { 1 }{ 15 } \) ton meat fed during the morning. If the total weight of the meat in a day is \(\frac { 1 }{ 2 } \) ton, how much is fed during the feeding at night?

Problem 7:

The monkey nursery is open two times a day: \(\frac { 2 }{ 3 } \) hours at noon and \(\frac { 5 }{ 12 } \) hours in the afternoon. How much time is the monkey nursery open everyday?

Problem 8:

An octopus weighed \(\frac { 5 }{ 6 } \) kilogram. After two weeks, its weight was increased by \(\frac { 3 }{ 10 } \) kilogram. But afterwards, it lost \(\frac { 1 }{ 5 } \) kilogram in weight as it was sick. What was the current weight?

Problem 9:

Two types of fish can be found in a small tank that is 5\(\frac { 1 }{ 7 } \) feet long. A blue fish is \(\frac { 2 }{ 15 } \) feet long and an orange fish is \(\frac { 7 }{ 10 } \) feet long. How much longer is the orange fish?

Problem 10:

Ila read 25 pages of a book containing 100 pages. Lalitha read \(\frac { 2 }{ 5 } \) of the same book. Who read less?

Problem 11:

Javed was given \(\frac { 5 }{ 7 } \) of a basket of apples. What fraction of apples are left in the basket?

Problem 12:

Reena was given 1\(\frac { 1 }{ 2 } \) piece of cake and Meena was given 1\(\frac { 1 }{ 3 } \) piece of cake. Find total amount was given to both of them?

Problem 13:

Bailey’s restaurant bought 6\(\frac { 1 }{ 3 } \) pounds of onions. The restaurant bought 8 times as much potatoes as onion. How many pounds of potatoes did the restaurant buy?

Problem 14:

Betsy has \(\frac { 2 }{ 5 } \) of a cup of powdered sugar. He sprinkles \(\frac { 3 }{ 4 } \) of the powdered sugar onto a plate of brownies. How much sugar does Betsy sprinkle on the brownies?

Problem 15:

At Goshen High School, \(\frac { 1 }{ 2 } \) of the students play a sport. Of the students who play a sport, \(\frac { 1 }{ 2 } \) play football. What fraction of the students play football?

Get detailed information regarding the division of fractions here. Check all the important terms, formulae, and the usage of fractions division. Refer step by step procedure to solve the fraction division problems. Know the various methods to solve proper, improper, mixed fractions, etc. Follow the complete guide to know the day-to-day usage of fractions division problems. Go through the below sections to follow the importance and problems of division on fractions.

Division of Fractions – Introduction

Dividing the fractions also involves various methods and also a lengthy process. First of all, you have to know the difference between proper, improper, and mixed fractions. Once you come to know the difference, then you can easily analyze the method to solve them. As we have seen, for the multiplication, we have to make the fractions equivalent.

In the same way, the division of fractions requires the equivalent fractions to solve them. First, make sure you make the fractions equivalent and then follow all the steps in dividing the fractions. Generally, fraction division in the direct method requires more effort. Hence, we are providing an alternative method here.

How to Divide Fractions?

Division of fractions can be done by multiplying the fractions by writing the reciprocal of one of fraction numbers or by reversing one of two fraction integers. Reciprocal of the fraction number or reverse of the fraction number means, if the fraction is \(\frac { a }{ b } \), then \(\frac { b }{ a } \) is its reciprocal. Interchanging the position of numerator and denominator is nothing but reversing the fraction. Fractions division can be classified into 3 ways.

• Fractions division by a fraction
• Fractions division by a mixed fraction
• Fractions division by a whole number

Fractions Division by a Fraction

Fractions division can be done in three steps. For division, we convert into the multiplication of fractions and obtain the desired result. The three steps are as follows.

1. Convert the second fraction into its reciprocal and multiply it with the first fraction.
2.  Multiply the denominator and numerator of both the fractions.
3. Next, simplify the fraction numbers.

For example, suppose that \(\frac { a }{ b } \) is divided by the fraction \(\frac { c }{ d } \). We can solve the division as:

• \(\frac { a }{ b } \) ÷ \(\frac { c }{ d } \) = \(\frac { a }{ b } \) * \(\frac { d }{ c } \)
• \(\frac { a }{ b } \) ÷\(\frac { c }{ d } \) = \(\frac { a*d }{ b*c } \)
• \(\frac { a }{ b } \) ÷ \(\frac { c }{ d } \) = \(\frac { ad }{ bc } \)

As given above, we can divide the fractions in three ways. Further simplification has to be done to get the exact result.

Fractions Division by a Whole Number

Fractions Division by a whole number is the easiest process. Follow the below steps to divide a whole number.

1. The whole number is nothing but the real numbers which include zero and all positive integers. All the whole numbers can be written as the fraction values if we give the denominator value as 1.
2. Find the reciprocal of the given number
3. Now, multiply the value of the fraction by a given fraction.
4. Then, simplify the equation to get its lowest terms.

Example:

Divide the fraction \(\frac { 6 }{ 5 } \) by 10?

Solution:

As given in the question,

The equation is \(\frac { 6 }{ 5 } \) by 10.

Step 1: Conversion into the whole number

As 10 is the whole number, we convert it into fractional value i.e., we write it as \(\frac { 10 }{ 1 } \)

Step 2: Find the reciprocal of the fraction

To find the reciprocal, we get \(\frac { 1 }{ 10 } \)

Step 3: Multiply both the fractions

Multiply the fractions, \(\frac { 6 }{ 5 } \) x \(\frac { 1 }{ 10 } \)

Step 4: Now, multiply the numertors and denominators of the fractions i.e., \(\frac { 6×1 }{ 15×10 } \)

Step 5: Simplify the equation

The resutant is  \(\frac { 3 }{ 25 } \)

Fractions Division by a Mixed Fraction

Fractions Division by a mixed fraction is almost similar to fractions division by fraction. The following are the steps to divide mixed fractions.

1. Conversion of mixed fraction into an improper fraction
2. Now, find the reciprocal of the improper fraction
3. Multiply the resultant fraction by the given fraction value.
4. Simplify the fractions.

Example:

Divide the fraction \(\frac { 2 }{ 5 } \) by 3\(\frac { 1 }{ 2 } \)?

Solution:

As given in the question,

The equation is \(\frac { 2 }{ 5 } \) by 3\(\frac { 1 }{ 2 } \)

Step 1: Conversion of mixed fraction into improper fraction

Convert, 3\(\frac { 1 }{ 2 } \), we get the result as 7/2

Step 2: Now, convert the reciprocal of the improper fraction, we get \(\frac { 2 }{ 7 } \)

Step 3: Multiply \(\frac { 2 }{ 5 } \) and \(\frac { 2 }{ 7 } \)

Step 4: Multiply the numertors and denominators of the fractions i.e., \(\frac { 2×2 }{ 5×7 } \)

Step 5: Simplify the fraction

the result value is \(\frac { 4 }{ 35 } \)

Fractions Division of Decimal Values

In the above sections, we have seen how to divide the fractions into three steps. Now, we see how to divide decimals with examples.

To convert the decimal values into natural numbers, we multiply both the numerator and denominator by 10.

Example:

Divide \(\frac { 0.5 }{ 0.2 } \)

Solution:

As given in the question,

The equation is \(\frac { 0.5 }{ 0.2 } \)

To divide the decimal values, convert the values into natural numbers by multiplying both numerator and denominator with 10.

Therefore, \(\frac { 0.5×10 }{ 0.2×10 } \)

We get the solution as, \(\frac { 5 }{ 2 } \) = 2.5

Also, we use the method of dividing fractions to solve the problem.

We can write 0.5 as \(\frac { 5 }{ 10 } \) and 0.2 as \(\frac { 2 }{ 10 } \)

Therefore, to find the solution of \(\frac { 5÷10 }{ 2÷10 } \)

Take the reciprocal of the second fraction i.e., \(\frac { 10 }{ 2 } \)

= \(\frac { 5*10 }{ 10*2 } \)

= \(\frac { 50 }{ 20 } \)

= \(\frac { 5 }{ 2 } \)

= 2.5

Dividing Fractions Examples

Problem 1: Lucy has \(\frac { 1 }{ 5 } \) of a bag of dog food left. She is splitting it between her 3 dogs evenly. What fraction of the original bag does each dog get?

Solution:

As given in the question,

Amount of dog food left = \(\frac { 1 }{ 5 } \)

No of dogs = 3

The fraction of the original bag each dog gets = \(\frac { 1/5 }{ 3 } \)

= \(\frac { 1 }{ 5 } \) x \(\frac { 1 }{ 3 } \)

= \(\frac { 1 }{ 15 } \) of the bag each dog gets.

Thus the final solution is \(\frac { 1 }{ 15 } \)

Problem 2: AJ has \(\frac { 1 }{ 4 } \) of a gallon of saltwater that he is using for an experiment. He needs to evenly separate the saltwater into separate beakers. How much salt water will be in each beaker?

Solution:

As given in the question,

Amount of salt water he is using for an experiment = \(\frac { 1 }{ 4 } \)

No of beakers = 3

Amount of salt water in each beaker = \(\frac { 1/4 }{ 3 } \)

= \(\frac { 1 }{ 4 } \) x \(\frac { 1 }{ 3 } \)

= \(\frac { 1 }{ 12 } \)

Therefore, \(\frac { 1 }{ 12 } \) gallons of salt water will be in each beaker.

Problem 3: Devin has a board that measures 4 ft in length. The board is going to be cut into \(\frac { 1 }{ 4 } \) ft pieces. How many pieces will Devin split the board into?

Solution:

As given in the question,

Length of the board = 4 ft

No of pieces the board is going to be cut = \(\frac { 1 }{ 4 } \)

No of pieces Devin split the board = 4 ÷ \(\frac { 1 }{ 4 } \)

= \(\frac { 4 }{ 1 } \) x \(\frac { 4 }{ 1 } \)

= \(\frac { 16 }{ 1 } \)

Hence, the final solution is 16 pieces.

Therefore, Devin will split the board into 16 pieces.

Have you ever wondered why certain numbers divide evenly into a number while others will not? Divisibility Rules or Tests will help you determine whether the given number actually divides the other number without actually dividing. Maths may not be easy for some of us and to help you we have listed the guidelines on whether a Number is Divisible by 6 or not. Check out Solved Examples for testing whether a number is divisible by 6 or not.

Divisibility Rule of 6 – Definition

A number is divisible by 6 if it is divisible by both 2 and 3. To know whether a number is divisible by 6 one needs to know both the Divisibility Rules of 2 and 3 in advance.

How to Check whether a Number is Divisible by 6 or not?

Go through the below Divisibility Guidelines of 6 to know whether a number is divisible by 6 or not. To know if a number is divisible by 6 it has to meet two conditions in prior and they are along the lines

• It has to be divisible by 2 i.e. the last digit in the number should be even.
• It has to be divisible by  3 i.e. the sum of digits must be a multiple of 3.

If both the conditions are met then the given number is also a multiple of 6 and divisible by 6.

Solved Examples on Divisibility Tests for 6

Find whether the following numbers are divisible by 6 or not

1. 48

Solution:

Given Number is 48

To check whether it is divisible by 6 or not it has to meet two conditions

48 is divisible by 2 since the units digit of the number i.e. 8 is divisible by 2

48 is divisible by 3 since the sum of digits is a multiple of 3 i.e. 4+8 = 12 is divisible by 3

Therefore, 48 is divisible by both 2 and 3

Thus, 48 is divisible by 6.

2. 258

Solution:

Given Number is 258

258 is divisible by 2 since the units digit of the number i.e. 6 is divisible by 2

258 is divisible by 3 since the sum of digits is divisible by 3 i.e. 2+5+8 = 15 is divisible by 3

Since 258 is divisible by both 2, 3 it is also divisible by 6

3. 154

Solution:

Given Number is 154

154 is divisible by 2 since the units digit of the number i.e. 4 is divisible by 2

154 is not divisible by 3 as the sum of the digits is not divisible by 3 i.e. 1+5+4 = 10

10 is not divisible by 3

Since the given number is not divisible by 3 the given number is not divisible by 6.

4. 243

Solution:

Given Number is 243

243 is not divisible by 2 since the units digit of the number is not divisible by 2 i.e. 3 is not divisible by 2

243 is divisible by 3 since the sum of digits is a multiple of 3 i.e. 2+4+3 = 9 and 9 is divisible by 3

Since the given number is not divisible by 2 it is not divisible by 6

Therefore, 243 is not divisible by 6.

Complex Number is a combination of both Real and Imaginary Numbers. In other words, Complex Numbers are defined as the numbers that are in the form of x+iy where x, y are real numbers and i =√-1.

z = x+iy here x is the real part of the Complex Number and is denoted by Re Z and y is called the Imaginary Part and is denoted as Im Z. In the later sections, you will find What is a Complex Number and Properties of Complex Numbers. We tried explaining each and every Property of Complex Number in detail with Proofs.

What are Complex Numbers?

If x, y ∈ R, then ordered pair (x, y) = x + iy is called a complex number. It is denoted by z. Where x is the real part and is denoted as Re(z) and y is the imaginary part of the complex number and represented as Im(z).

(i) If Re(z) = x = 0, then the number z is a purely imaginary number

(ii) If Im(z) = y = 0 then the number z is a purely real number.

Properties of Complex Numbers

1. If x, y are two real numbers and x+iy =0 then x = 0 and y = 0

Proof:

Since, x + iy = 0 = 0 + i0, thus by the definition of equality of two complex numbers we can say that, x = 0 and y = 0.

2. If x, y, p, q are real and x + iy = p + iq then x = p and y = q

Proof:

Given x + iy = p + iq

rearranging the equation we get x − p = -i(y − q)

⇒ (x − p)² = i² (y − q)²

⇒ (x − p)² + (y − q)² = 0 (We know i² = -1)……..(1)

Since x, y, p, q are real, and (x − p)² and (y − q)² are both non-negative. Equation (1) is satisfied if each square is separately zero.

Thus we can write the equation as follows

(x − p)² = 0 or x = p and (y − q)² = 0 or y = q.

3. Similar to real numbers, the set of complex numbers also satisfy the commutative, associative, and distributive laws

Proof:

If z₁, z₂ and z₃ be three complex numbers then,

z₁ + z₂ = z₂ + z₁ (commutative law of addition) and z₁. z₂ = z₂. z₁ (commutative law of multiplication)

(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃) (associative law of addition) and (z₁. z₂) z3 = z₁ (z₂. z₃) (associative law of multiplication)

z₁(z₂ + z₃) = z₁ z₂ + z₁ z₃ (distributive law)

4. Sum and Product of Two Conjugate Complex Quantities are both Real.

Proof:

Consider z = x + iy is a complex number where x, y are real.

Then, the conjugate of z is = x − iy.

Now, z + \(\overline {z}\)= x + iy + x − iy = 2x, is real.

and z. \(\overline {z}\) = (x + iy)(x − iy) = x² − i²y² = x² + y² is also real.

5. For two complex quantities z₁ and z₂,  |z₁+ z₂| ≤ |z₁ | + |z₂ |

Proof:

Let z₁ = r₁(cosθ₁ + isinθ₁ ) and z₂ = r₂(cosθ₂ + isinθ₂ ).

Hence |z₁ | = r₁ and |z₂ | = r₂

Now

z₁ + z₂ = r₁(cosθ₁isinθ₁) + r₂(cosθ₂ + isinθ₂)

= (r₁(cosθ₁+ r₂cosθ₂ )+ i(r₁sinθ₁+ r₂sinθ₂)

Hence |z₁+ z₂ | = √(r₁cosθ₁+ r₂cosθ₂)₂ + (r₁sinθ₁+ r₂sinθ₂)₂

= √r₁2(cos₂θ₁+ sin2₁) + r₂2(cos2θ₂+ sin2θ₂) + 2r₁r₂ (cosθ₁ cosθ₂+ sinθ₁ sinθ₂)

= √r1₂ + r2₂ + 2r₁r₂cos (θ₁– θ₂)

Now, |cos(θ₁– θ₂)| ≤ 1

Hence |z₁+ z₂| ≤ √r1₂ + r2₂ + 2r₁r₂ or |z₁+ z₂ | ≤ |z₁| + |z₂ |

6. If the sum of two complex numbers is real and the product of two complex numbers is also real then the complex numbers are conjugate to each other.

Proof:

Let us consider z₁ = a + ib and z₂ = c + id are two complex quantities (a, b, c, d and real and b ≠ 0, d ≠0).

As per the property,

z₁ + z₂ = a+ ib + c + id = (a + c) + i(b + d) is real.

Therefore, b + d = 0

⇒ d = -b

and,

z₁.z₂ = (a + ib)(c + id) = (a + ib)(c +id) = (ac – bd) + i(ad + bc) is real.

Therefore, ad + bc = 0

⇒ -ab + bc = 0, (Since, d = -b)

⇒ b(c – a) = 0

⇒ c = a (Since, b ≠ 0)

Hence, z₂ = c + id = a + i(-b) = a – ib

Thus, we can say that z₁ and z₂ are conjugate to each other.

Are you looking for help regarding the Units of Time Conversion i.e. from Hours to Minutes? You have come the right way and we will guide you all throughout on How to Convert from Hours to Minutes. Check out the basic definitions of hours, minutes, hrs to min conversion table, the procedure for hours to mins conversion, etc. Have a glance at the Solved Examples on Hours to Minutes Conversions and understand how to solve related problems easily. You can also check out our Math Conversion Chart to learn more about length, mass, capacity conversions along with Time Conversions.

Hours & Minutes Definitions

Hour(hr)

An Hour is a Unit of Time Conventionally defined as 1/24th of a day or 60 Minutes. The Hour is the SI unit of time accepted for the Metric System. Hours can be abbreviated as hr. As per the conversion base, 1 hr = 60 Minutes or 3600 Seconds.

Minutes(min)

A Minute is a period of time conventionally equal to 1/60 of an hour or 60 seconds. The Minute is the SI unit of time accepted for the Metric System. Minutes can be abbreviated as min. According to the Conversion Base, 1 Min = 60 Seconds.

Hours to Minutes Conversion Chart | Hrs to Mins Conversion Table

Below is the list of hours converted to minutes. Refer to them and have an idea of How many minutes are there in a Particular Hour.

How to Convert Hours to Minutes?

To convert hours measurement to minutes multiply time by the conversion ratio. Use the simple formula listed below to change between Hours to Minutes easily as a part of your conversions.

We know 1 hour = 60 minutes

Time in Minutes is equal to hours multiplied by 60.

Therefore, the Hours to Minutes Conversion Formula is given as

Now that you know the formula to Convert Between Hours to Minutes simply substitute the given hrs value and calculate the minutes in no time. To understand more about hr to min Conversions check out a few worked out examples provided by us in the below modules and know how to convert Hours to Minutes easily and quickly.

Read Similar Articles

• Conversion of Hours into Seconds
• Conversion of Minutes into Seconds
• Conversion of Minutes into Hours

Solved Examples on Conversion of Hours into Minutes

1. Convert 12 hrs to min?

Solution:

We know 1 hr = 60 minutes

The formula to convert from hrs to mins is given as follows

min = hours *60

min = 12*60

= 720

Therefore, 12 hrs converted to mins is 720 mins.

2. Convert 6 hrs 30 mins to minutes?

Solution:

Given 6hrs 30 mins

We know 1 hr = 60 minutes

6hrs 30 minutes = 6*60+30

= 360+30

= 390 minutes

Therefore, 6 hrs 30 mins converted to mins is 390 minutes.

3. Convert 8 hrs 42 mins to mins?

Solution:

We know 1 hr = 60 mins

8hrs 42 mins = 8*60+42

=480+42

= 512 mins

Therefore, 8 hrs 42 mins to 512 mins.

FAQs on Converting Hours to Minutes

1. What is the formula for converting hours to minutes?

The formula for converting hours to minutes is Minutes = Hours*60

2. How many minutes are there in 1 hour?

There are 60 minutes in an hour.

3. How to convert hr to mins?

Since there are 60 mins in an hour to convert from hrs to mins multiply with 60.

Worried about how to find the Arc Length? Don’t Panic as we will guide you completely in this and help you with the definition of Arc Length, Formula and Proof for Length of the Arc explained in detail. Check out the step by step explanation provided for determining the Arc Length. Also, refer to the solved examples for calculating the length of the arc and solve related problems in no time.

In general, Arc Length is nothing but the distance along the curved line making the arc. Remember that Arc Length is greater than the Straight Line Distance present between the Endpoints. Know about Denotations in the Arc Length Formula in the later sections.

Prove that the radian measure of any angle at the centre of a circle is equal to the ratio of the arc subtending that angle at the centre to the radius of the circle?

Let, us consider XOY to be a given angle. Now, with Centre O and any radius \(\overline {O L} \) draw a circle. Consider the circle drawn intersects \(\overline {O X} \) and \(\overline {O Y} \) at Points L and M respectively.

From the figure, it is clear that Arc LM subtends ∠LOM at Centre O. Now, take an Arc LN of length equal to the radius of the circle and join \(\overline {O N} \).
From definition, ∠LON = 1 radian.

We know the ratio of two arcs in a circle is equal to the ratio of the angles subtended by the arcs at the center of the circle,

Therefore, ∠LOM/∠LON = Arc LM/Arc LN
or, ∠LOM/1 radian = Arc LM/Radius \(\overline {O L} \)
or, ∠LOM = Arc LM/Radius OL × 1 radian = Arc LM/ Radius \(\overline {O L} \) radian.
Thus, circular measures of ∠LOM is Arc LM/Radius \(\overline {O L} \)
Let us consider θ be the circular measure of ∠LOM, Arc LM = s and Radius of the circle = \(\overline {O L} \) = r then,
θ = s/r, [i.e. theta equals s over r]

or, s = r θ, [i.e. s r theta formula]

Denotations in the Arc Length Formula

• s is the Arc Length
• r is the Radius of the Circle
• θ is the Central Angle of the Arc

Formulas to Measure the Arc Length

Solved Examples on Arc Length using the Formula s = rθ

1. Calculate the Length of the Arc if the radius of the circle is 9 cm and θ = 45°?

Solution:

We know the formula to Calculate the Arc Length s = rθ

Substituting the given input values r = 9 cm, θ =45° we get

Arc Length = 9*45°

= 9*0.785398(since 45° in radians =0.785398)

= 7.06 cm

Therefore, Arc Length is 7.06 cm

2. What would be the length of the arc formed by 65° of a circle having a diameter of 14 cm?

Solution:

From the given data

θ =65°

Diameter of the Circle = 14 cm

Radius of the Circle = \(\frac { d }{ 2 } \)

= \(\frac { 14 }{ 2 } \)

= 7 cm

We know the formula to calculate the Arc Length s = rθ

= 7*65°

= 7*1.13446

= 7.94 cm

Therefore, Arc Length is 7.94cm

Temperature is written in terms of Celsius, Fahrenheit, Kelvin. Temperature can be converted from one unit to another using certain formulas. Of all the Temperature Conversions Celsius to Fahrenheit Conversion is the most commonly used. To help you understand we have provided the formula for °C to °F Conversion, the procedure to convert from Celsius to Fahrenheit in the later modules.

Both Celsius and Fahrenheit have different freezing and boiling points of water and follow the varied unit differences between each scale. Check out Solved Examples on Celsius to Fahrenheit Temperature Conversion for a better understanding of the concept.

Celsius to Fahrenheit Conversion Formula

Conversion from Celsius to Fahrenheit can be mathematically expressed using the formula

F = \(\frac { 9 }{ 5 } \)*C+32

Where F is the Temperature in Fahrenheit

C is the Temperature in Celsius

How to Convert Celsius to Fahrenheit(°C to °F)?

Check out the quick and easiest way to perform Celsius to Fahrenheit Conversion. They are along the lines

• To convert from Celsius to Fahrenheit firstly multiply the value of the celsius degrees by 9.
• Later divide the product with 5.
• Add 32 to the result obtained in the earlier step.
• That’s the required Fahrenheit Value and place °F symbol next to it.

Celsius to Fahrenheit Conversion Table

Relation between °C and °F

The Relation between Celsius and Fahrenheit is direct. Celsius is directly proportional to Fahrenheit and it implies

• If the Temperature in Celsius increases then the Fahrenheit Equivalent Temperature also increases.
• If the Temperature in Celsius decreases then the Fahrenheit Equivalent Temperature also decreases.

Both the Scales of Temperature have their own measurements. In Fahrenheit Scale 32 °F is the freezing point of water and 212 °F is the boiling point of water. However, in Celsius Scale 0 °C is the Freezing Point of water and 100 °C is the boiling point of water.

Solved Examples on Celsius to Fahrenheit Conversion

1. Convert 40°C to °F?

Solution:

We know the formula to convert Celsius to Fahrenheit is

\(F=(\frac{9}{5}\times 1)+32\)

\(F= (\frac{9}{5}\times 40)+32\)

=104°F

Therefore 40°C converted to Fahrenheit is 104°F

2. What is 60°C in Fahrenheit?

Solution:

We know the formula to convert Celsius to Fahrenheit is

\(F=(\frac{9}{5}\times 1)+32\)

\(F=(\frac{9}{5}\times 60)+32\)

= 140°F

Therefore, 60°C in Fahrenheit is 140°F

FAQs on Celsius to Fahrenheit Conversion

1. What is the formula to convert from Celsius(°C) to Fahrenheit(°F)?

The Formula to Convert from Celsius to Fahrenheit is \(F=(\frac{9}{5}\times 1)+32\)

2. What is C equal to in F?

1 Degree in Centigrade is Equal to 33.8 Fahrenheit.

3. What are the different Temperature Conversions?

The three major temperature conversions are

• Conversion Between Celsius and Kelvin
• Conversion Between Fahrenheit and Kelvin
• Conversion Between Celsius and Fahrenheit