Conversion of Seconds into Hours Definitions, Formula, Examples | How to Convert Sec to Hr?

Conversion Of Seconds Into Hours

Time is written in terms of Hours, Minutes, Seconds. Time can be converted from one unit to another with the help of conversion formulas. Among all Time conversions, Conversion of Seconds into Hours is the trickest one and highly seek help by students to solve their homework or assignments. To aid your preparation and homework, we have curated the details about seconds to hours conversion like definitions, conversion formula, the procedure to convert sec to hr, and some solved examples on second to hour conversion.

Go ahead and dig deep regarding Conversion of Seconds(sec) into Hours(hr) also look at the Math Conversion Chart to learn more about length, mass, capacity conversions along Time Conversions.

Definitions of Seconds & Hours

Seconds(sec)

A Second is a Unit of Time and its symbol is ‘s or sec’. As per the SI definition, the definition of second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the unperturbed ground state of the caesium 133 atoms at zero kelvins. According to the conversion base, 1 sec = 0.000277778 hr.

Hour(hr)

An Hour is a Unit of Time measurement, determined as 1/24th of a day or 60 Minutes. The Hour is the SI unit of time taken for the Metric System. The abbreviation of Hours is ‘hr’. As per the conversion base, 1 hr = 60 Minutes or 3600 Seconds.

Formula for Converting Second to Hour

Have a glance at the provided box and learn the mathematical formula which is used for converting seconds to hours:

Hours = Seconds ÷ 3,600

Second to Hour Conversion Table | Sec to Hr Conversion Chart

Here is the list of seconds converted to hours. Check out the conversion chart of sec to hrs and get some idea of How much is particular second in Hours.

SecondsHoursHours, Minutes & Seconds
10.0002780 hr 0 min 1 sec
1000.0277780 hr 1 min 40 sec
2000.0555560 hr 3 min 20 sec
3000.0833330 hr 5 min
4000.1111110 hr 6 min 40 sec
5000.1388890 hr 8 min 20 sec
6000.1666670 hr 10 min
7000.1944440 hr 11 min 40 sec
8000.2222220 hr 13 min 20 sec
9000.250 hr 15 min
1,0000.2777780 hr 16 min 40 sec
2,0000.5555560 hr 33 min 20 sec
3,0000.8333330 hr 50 min
3,60011 hr
4,0001.11111 hr 6 min 40 sec
5,0001.38891 hr 23 min 20 sec
6,0001.66671 hr 40 min
7,0001.94441 hr 56 min 40 sec
8,0002.22222 hr 13 min 20 sec
9,0002.52 hr 30 min
10,0002.77782 hr 46 min 40 sec
20,0005.55565 hr 33 min 20 sec
30,0008.33338 hr 20 min
40,00011.1111 hr 6 min 40 sec
50,00013.8913 hr 53 min 20 sec
60,00016.6716 hr 40 min
70,00019.4419 hr 26 min 40 sec
80,00022.2222 hr 13 min 20 sec
86,4002424 hr
90,0002525 hr
100,00027.7827 hr 46 min 40 sec

Conversion of Seconds into Hours Process | How to Convert Seconds(sec) to Hours(hr)?

One of the easiest methods to convert seconds to hours is to divide the number of seconds by 3,600. In order to explain the reason for this conversion, first set up the conversion tables where first convert the number of seconds to minutes, and then the number of minutes to hours. These can be helpful to convert the sec to hr easily.

If not, you may also make use of the simple conversion formula to convert seconds to hours. The formula for second to hour unit of time conversion is given here.

We know 1 second is equal to 1/3600 hour

Hence, the Seconds to Hours Conversion Formula is given as

Hours = Seconds ÷ 3,600

Once, you have the formula for conversion between sec to hr just apply the given inputs in the formula and find out the hours within less time. For more learning knowledge, please go with the Conversion Of Seconds Into Hours solved examples explained by us and learn completely how to convert Sec to Hr easily within no time.

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Worked-out Examples on Conversion of Seconds(sec) into Hours(hr)

1. Convert 1500 seconds into hours?

Solution:

First, we start converting seconds into minutes and then minutes into hours.

As we know the formula for conversion of sec into min, so apply it in step 1:

1500 seconds = (1500 ÷ 60) minutes (1 second = 1/60 minute)

= 25 minutes

Now, convert the minutes into hours by using the conversion formula:

= (25 ÷ 60) hours (1 minute = 1/60 hour)

= 0.416667 hours or 0 hr 25 min.

Therefore, 1500 seconds(sec) = 0 hr 25 min.

2. Convert 76 Seconds into Hours using conversion formula?

Solution:

We have 76 seconds and the conversion formula for Seconds(sec) to Hours(hr) conversion is as follows:

Hours = Seconds ÷ 3600

Now, apply the given time values in the formula and covert the sec to hr,

Hours = 76 ÷ 3600

= 0.021111 hr or 0 hr 1 min 16 sec

Hence, 76 seconds in hours is 0.021111 hr or 0 hr 1 min 16 sec. 

FAQs on Converting Seconds to Hours

1. What is the formula for converting Seconds to Hours?

The conversion formula for seconds to hours is Hours = Seconds ÷ 3600

2. How much is 15 seconds in hours? 

15 sec = 0.004167 hr or 15 sec = 0 hr 0 min 15 sec.

3. How to convert Sec to Hr?

As there are 3600 seconds in an hour to convert from Seconds to hours divide the time value by 3600.

How to Find the Percent of a Given Number? | Definition & Word Problems on Percentage

How To Find The Percent Of A Given Number

Percentages topic is widely utilized by the people in various fields like shopping for deals, buying things at veggie or fruit markets, etc. Also, it is commonly used in accounting and finance scenarios like Profits, Interest Rates, Sales, and Taxation. Moreover, the percentage is helpful for grading the student’s annual marks. So, finding percentages can be tricky but an easy mathematical process. If you have to calculate the ratio or portion of a quantity then you need help with percentages. Hence, check out this article properly and learn what is percentage, how to find the percent of a given number or quantity along with worked-out examples.

What is Percentage?

A percentage is a number or ratio as a fraction of 100. In other words, the word percent indicated one part in a hundred. Always, the number of a percentage is represented by a percent symbol (%) or simply “percent”. Here is the percentage illustration:

5 %, 10 %, 33 \(\frac { 1 }{ 6} \) %, 75 %

For example, 60 percent (or 60%) means 60 out of 100.

However, the percentage is the outcome when a particular number is multiplied by a percent. So, learn how to calculate the percentages for a given number or quantity in the below modules with solved word problems.

How to Find the Percent of a Given Number?

To calculate the % of a given number so easily, please follow the below steps:

  • Take the number, say x.
  • Let the percent as p%.
  • To find the formula is P% of x
  • Now, write these as a proportion as \(\frac { P }{ 100 } \) = \(\frac { ? }{ x } \)
  • Finally, do cross multiplication and calculate the value of the “?” mark.

Solved Examples on Percentages

1. What is 3 ⅓ of 60 km?

Solution :

Given expression is 3 ⅓ of 60 km

\(\frac { 3 ⅓ }{ 100 } \) = \(\frac { x }{ 60 } \)

Now, Convert mixed fraction to improper fraction

\(\frac { 10 }{ 3 } \)/100 = \(\frac { x }{ 60 } \)

\(\frac { 10 }{ 3 } \) x100 = \(\frac { x }{ 60 } \)

\(\frac { 10 }{ 300 } \) = \(\frac { x }{ 60 } \)

Cross multiply and find the x value

300x = 10 . 60

300x = 600

x = \(\frac { 600 }{ 300 } \)

x = 2 km.

2. Find 41% of 400.

Solution:

Given is 41% of 400

Now find the % of a given number

41% of 400 = 41 × \(\frac { 1 }{ 100 } \) × 400

= \(\frac { 41 }{ 100 } \) × 400

= \(\frac { 41 × 400 }{ 100 } \)

= \(\frac { 16400 }{ 100 } \) (Finally divide 16400 by 100 and get the result)

= 164.

3. What is the sum of the money of which 5 % of $750?

Solution:

Let the required sum of money be $m.

5 % of $m = $750

⇒ \(\frac { 5 }{ 100 } \) × m = 750

⇒ m = \(\frac { 750 × 100 }{ 5 } \)

⇒ m = 15000

Hence, sum of the money = $15000.

4. Find 17% of $4500?

Solution:

Given expression is 17% of $4500

Now, convert 17% into decimal form

Then write it as 0.17 x $4500

Multiply  17 × $4,500 = $76,500

Finally, keep the decimal point

Therefore the result for 17% of $4500 is $765.00

5. The price of a TV was reduced by 40% to $500. What was the original price?

Solution:

To find the original price,

First, determine the percentage of the actual price by subtracting 40% from 100.

Later, Product the final price by 100 ie., 500 x 100 = 50000.

Now, divide the result by the percentage computed in step 1 above.

Then, \(\frac { 50000 }{ 60 } \) = $ 833.33

The actual price of a TV is $ 833.33.

Ascending Order or Descending Order – Definition, Symbol, Ordering, Facts, and Examples

Ascending Order or Descending Order

Ascending order is arranging or placing numbers from smallest to largest. In ascending order, the numbers are in increasing order. For example, 1, 2, 3, etc are in ascending order. The reverse process of ascending order is called descending order. The decreasing order is arranging or placing numbers from largest to smallest. Examples for decreasing order are 3, 2, 1. The ascending order is represented by the less than symbol ‘<‘ where descending order is represented by the greater than symbol ‘>‘.

Ascending Order – Definition & Symbol

Ascending order is the arrangement of numbers from the lowest to the highest. In the case of Ascending Order, the Smallest Number will be on the top of the list when sorted. To represent the order of numbers we use the symbol “<”

Examples of Ascending Order

  • For numbers or amounts, the ascending order is 5, 8, 11, 18, 23, 31.
  • For words and letters, the ascending order is A, B, C, D, E……Y, Z.
  • Also, for dates, the ascending order will be from the oldest dates to recent dates.

Problems on Ascending Order

1. Arrange the below numbers in an Ascending Order

(i) 2, 14, 3, 59, 46
(ii) 25, 8, 97, 47, 3
(iii) 5, 6, 82, 31, 24
(iv) 6, 7, 35, 14, 4
(v) 24, 8, 15, 94, 119

Solution:
(i) Given numbers are 2, 14, 3, 59, 46.
Compare the values and write down the smallest number.
Write down the smallest number first, and then compare with all the remaining numbers with the same number of digits.
2, 3, 14, 46, 59.

The Ascending Order of the numbers is 2, 3, 14, 46, 59.

(ii) Given numbers are 25, 8, 97, 47, 3.
Compare the values and write down the smallest number.
Write down the smallest number first, and then compare with all the remaining numbers with the same number of digits.
3, 8, 25, 47, 97.

The Ascending Order of the numbers is 3, 8, 25, 47, 97.

(iii) Given numbers are 5, 6, 82, 31, 24.
Compare the values and write down the smallest number.
Write down the smallest number first, and then compare with all the remaining numbers with the same number of digits.
5, 6, 24, 31, 82.

The Ascending Order of the numbers is 5, 6, 24, 31, 82.

(iv) Given numbers are 6, 7, 35, 14, 4.
Compare the values and write down the smallest number.
Write down the smallest number first, and then compare with all the remaining numbers with the same number of digits.
4, 6, 7, 14, 35.

The Ascending Order of the numbers is 4, 6, 7, 14, 35.

(v) Given numbers are 24, 8, 15, 94, 119.
Compare the values and write down the smallest number.
Write down the smallest number first, and then compare with all the remaining numbers with the same number of digits.
8, 15, 24, 94, 119.

The Ascending Order of the numbers is 8, 15, 24, 94, 119.

Descending Order – Definition & Symbol

Descending Order is the arrangement of numbers from the highest to the lowest. Descending Order is the Contradictory of Ascending Order. Here Numbers are arranged from bigger to smaller. The number that is largest among the sorted list will fall at the top of the Descending Order List and the Smallest Number will be at the Last. It is denoted by the symbol  ‘>’.

Examples of Descending Order

  • For numbers or amounts, the descending order is 31, 23, 18, 11, 8, 5.
  • For words and letters, the descending order is Z, Y, X, W……B, A.
  • Also, for dates, the descending order will be from the recent dates to the oldest dates.

Problems on Descending Order

1. Arrange the below numbers in an Descending Order

(i) 2, 14, 3, 59, 46
(ii) 25, 8, 97, 47, 3
(iii) 5, 6, 82, 31, 24
(iv) 6, 7, 35, 14, 4
(v) 24, 8, 15, 94, 119

Solution:
(i) Given numbers are 2, 14, 3, 59, 46.
Compare the values and write down the highest number.
Write down the highest number first, and then compare with all the remaining numbers with the same number of digits.
59, 46, 14, 3, 2.

The Descending Order of the numbers is 59, 46, 14, 3, 2.

(ii) Given numbers are 25, 8, 97, 47, 3.
Compare the values and write down the highest number.
Write down the highest number first, and then compare with all the remaining numbers with the same number of digits.
97, 47, 25, 8, 3.

The Descending Order of the numbers is 97, 47, 25, 8, 3.

(iii) Given numbers are 5, 6, 82, 31, 24.
Compare the values and write down the highest number.
Write down the highest number first, and then compare with all the remaining numbers with the same number of digits.
82, 31, 24, 6, 5.

The Descending Order of the numbers is 82, 31, 24, 6, 5.

(iv) Given numbers are 6, 7, 35, 14, 4.
Compare the values and write down the highest number.
Write down the highest number first, and then compare with all the remaining numbers with the same number of digits.
35, 14, 7, 6, 4.

The Descending Order of the numbers is 35, 14, 7, 6, 4.

(v) Given numbers are 24, 8, 15, 94, 119.
Compare the values and write down the highest number.
Write down the highest number first, and then compare with all the remaining numbers with the same number of digits.
119, 94, 24, 15, 8.

The Descending Order of the numbers is 119, 94, 24, 15, 8.

FAQs on Ascending or Descending Order

1. What does Ascending Order mean?

When the Numbers are arranged in increasing order i.e. from smallest to largest then they are said to be in Ascending Order.

2. What does Descending Order mean?

Descending Order or Decreasing Order is the way of arranging numbers from biggest to smallest.

3. What are the signs used to represent Ascending Order and Descending Order?

Ascending order is denoted by the ‘<‘ (less than) symbol, whereas descending order is denoted using the ‘>’ (greater than) symbol.

Regular and Irregular Polygon Definition, Formula, Examples | Difference Between Regular and Irregular Polygons

Regular and Irregular Polygon

Learn Regular and Irregular Polygon Concepts and properties along with examples. Know the difference between Regular Polygons and Irregular Polygons and find out the given shape is a regular or irregular polygon. Check out the step-by-step process to find the given shape is a polygon or not? Also, we have given clear details about Regular and Irregular Polygons. Go through the Solved Examples on Regular and Irregular Polygons and learn how to solve problems on them quickly and easily.

Regular Polygon – Definition

A polygon that consists of equal sides with equal length and also by having equal angles called a regular polygon.

Examples of a Regular Polygon

Let us have a look at the different Examples of a Regular Polygon below.

Equilateral Triangle:
An equilateral triangle is a triangle in which all three sides have the same length and all angles are equal.
In the below figure of an equilateral triangle PQR there are three sides i.e., PQ, QR, and RP are equal and there are three angles i.e., ∠PQR, ∠QRP, and ∠RPQ are equal.
Therefore, an equilateral triangle is a regular polygon.

equilateral triangle

Square:
A Square is a quadrilateral that has four equal sides and four equal angles.
In the below figure of a square PQRS there are four sides i.e., PQ, QR, RS and SP are equal and there are four angles i.e., ∠PQR, ∠QRS, ∠RSP, and ∠SPQ are equal.

Therefore, a square is a regular polygon.

square 2

Regular Pentagon:
A polygon is regular when all angles are equal and all sides are equal.
In the below figure of a regular pentagon PQRST there are five sides i.e., PQ, QR, RS, ST, and TP are equal and there are five angles i.e., ∠PQR, ∠QRS, ∠RST, ∠STP, and ∠TPQ are equal.

Therefore, a regular pentagon is a regular polygon.
regular pentagon

Irregular Polygon – Definition

A polygon that consists of unequal sides with unequal lengths and also having unequal angles called an irregular polygon.

Examples of Irregular Polygon

Let us have a look at the different Examples of a Regular Polygon below.

Scalene Triangle:
A scalene triangle is a triangle that has all three sides are in different lengths, and all three angles are of different measures.
In the adjoining figure of a scalene triangle PQR there are three sides i.e., PQ, QR, and RS are unequal and there are three angles i.e., ∠PQR, ∠RSP, and ∠SPQ are unequal.

Therefore, a scalene triangle is an irregular polygon.
Scalene triangle

Rectangle:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. Each angle is at right angles. The opposite sides of the rectangle are equal in length which makes it different from a square.
In the adjoining figure of a rectangle PQRS there are four sides i.e., PQ, QR, RS, and SP where the opposite sides are equal i.e., PQ = RS, and QR = PS. So, all the sides are not equal to each other.
Similarly, among the four angles i.e., ∠PQR, ∠QRS, ∠RSP, and ∠SPQ where the opposite angles are equal i.e., ∠PQR = ∠RSP and ∠QRS = ∠SPQ. So, all the angles are not equal to each other.

Therefore, a rectangle is an irregular polygon.
rectangle PQRS

Irregular Hexagon:
An irregular hexagon is a six-sided shape whose sides are not equal.
In the adjoining figure of an irregular hexagon PQRSTU there are six sides i.e., PQ, QR, RS, ST, TU, and UP are equal and there are six angles i.e., ∠PQR, ∠QRS, ∠RST, ∠STU, ∠TUP, and ∠UPQ are equal.

Therefore, an irregular hexagon is an irregular polygon.
irregular hexagon

Rounding Decimals to the Nearest Whole Number Definition, Rules, Examples | How to Round to the Nearest Whole Number?

Rounding Decimals to the Nearest Whole Number

Rounding Decimals to the Nearest Whole Number guide is available here. Check the steps to find out the process of rounding the decimals worksheet. Make use of number lines and visual models to round decimal numbers. Know the definition, rules, and methods to round the decimal values. Follow the below sections to gather the information regarding rounding decimal values to the nearest whole numbers. Also, check the solved problems and solution procedure.

Rounding Decimals to the Nearest Whole Number – Definition

The rounding of decimal values is a process to estimate a particular number in context. The rounding of decimals refers to the gaining of accuracy to a certain degree. We can easily round the decimals to the nearest whole numbers i.e., tenths or hundreds. With the help of rounding decimals to the nearest whole numbers, you can easily estimate the solution quickly and easily. It is also used to get the average score of the pupil in the class. There are various procedures to round off the whole numbers.

Basic Rules of Rounding to Whole Numbers

While rounding to the nearest whole numbers, you have to follow a few basic rules.

  • Identify the value in the place you are rounding to. The smaller value you consider at the place value, the most accurate will be the final result.
  • Look at the next smallest number in place value is the number to the right of the place value which you are rounding to. For suppose, if you are looking to round to the nearest ten, then you would look at the one’s place.
  • If the number in the succeeding smallest place value is less than five (0, 1, 2, 3, 4 or 5), you have to leave the digit you want to round as a whole number to as in. Any digits after the tens place value (include the smallest place value you looked at) drop-off it or become zeros which are located after the decimal point. It is called rounding down.
  • If the number in the succeeding smallest place value is greater than or equal to five (5, 6, 7, 8, or 9), you have to increase the digit value you are rounding to +1 (one). As seen in the previous step, the remaining digits before the decimal point become zero, and any values after the decimal point are dropped. It is called rounding up.

Also, Read: Round off to Nearest 100

How to Round to the Nearest Whole Number?

There are a few steps to round numbers to the nearest whole number.

Step 1: First, look at the number that you want to round.

Step 2: As you have to round the number to the nearest whole number, we mark the number in the one’s place.

Step 3: Now look at the digit which is at right to the decimal point, i.e., tenth’s place

Step 4: (i) If the number in the tenth column has 0,1,2,3,4, we have to round down the digit at the ones place to the nearest whole number.

(ii) If the number in the tenths columns has 5,6,7,8 or 9, we have to round up the digit at one’s place to the nearest whole number.

Step 5: You have to remove all the numbers after the decimal point. The desired answer will be the left-out number.

Rounding to Nearest Tenths

We have to follow the given steps in rounding numbers to the nearest tenths.

Step 1: First, look at the number that you want to round.

Step 2: As you have to round the number to the nearest tenths, we mark the number in the tenths place.

Step 3: Now look at the digit which is at right to the tenth’s point, i.e., hundredth’s place

Step 4: (i) If the number in the hundredth’s column has 0,1,2,3,4, we have to round down the digit at the tenths place to the nearest tenths.

(ii) If the number in the hundredth’s column has 5,6,7,8 or 9, we have to round up the digit at the tenth’s place to the nearest tenths.

Step 5: You have to remove all the numbers after the tenths column. The desired answer will be the left out number.

Rounding to Nearest Hundredths

We have to follow the given steps in rounding numbers to the nearest hundredths

Step 1: First, look at the number that you want to round.

Step 2: As you have to round the number to the nearest hundredths, we mark the number in the hundredth’s place.

Step 3: Now look at the digit which is at right to the hundredth’s point, i.e., thousandth’s place

Step 4: (i) If the number in the thousandth’s column has 0,1,2,3,4, we have to round down the digit at the hundredth’s place to the nearest hundredth’s.

(ii) If the number in the thousandth’s column has 5,6,7,8 or 9, we have to round up the digit at the hundredth’s place to the nearest hundredth’s.

Step 5: You have to remove all the numbers after the hundredth’s column. The desired answer will be the left out number.

How to Round Decimals Using a Calculator?

Follow the below procedure to round decimals to the whole numbers.

Step 1: Enter the decimal or the number in the given input field.

Step 2: Find the button “round” and click on it to get the final result.

Step 3: You will notice the value that is displayed in the output field.

Examples on Rounding Numbers to Nearest Whole Number

Problem 1:

Consider the number 46.8. Round the digit to the nearest whole number?

Solution:

As the given number = 46.8

The number in the tenths place = 8 which is greater than 5

As the value is greater than 5, then the whole part of the number increases with 1 and the number to the right of the decimal value means the tenth’s place becomes zero.

Hence, the final value of the rounded-up = 47

Problem 2:

Consider another number 45.379. Round the digit to the nearest whole number?

Solution:

As the given number is 45.379

The ones digit in the number = 5 and tenths digit is 3

Since the value in the tenths digit is 3, then the one’s digit will remain unchanged, i.e., 45.379

Rewrite the digit by dropping the decimal point and all the digits after it.

Therefore, 45 is the final answer.

Problem 3:

Round the following digits to the nearest tenths: 0.437

Solution:

As the given number is 0.437

The hundredth’s digit is 3 and therefore the tenth digit will remain unchanged

As the hundredth’s digit is less than 5, we have to drop off the remaining digits.

The final answer is 0.4

Read a Watch or a Clock | How to Read A Time Shown in Analogue Clock or Digital Clock?

how to read a watch or a clock

Read a Watch or Clock: Reading a clock or watch is an ability to master easily in a short time and with less effort. There are various types of clocks like Analogue clock, Digital clock, electronic word clock, musical clock, etc. Analogue clocks are distributed over a circle and move the hour hand and minute hand individually. This helps people to read a time from a clock or watch.

Whereas in the digital type of clock, you can easily read the hour and minutes. Well, reading a watch or clock can be confused when it comes to Roman numerals and military time. With a bit of hard work and smart work, you guys can read a clock or watch with ease. So, practice more by viewing the below modules regarding How to Read a Watch or a Clock.

How to Read a Watch or a Clock?

There are two methods to read a watch or clock. The first method is reading an Analogue clock and the second method is reading a Digital clock. Let’s start the process of the first method and learn the concept to read a watch or clock easily.

Method 1: Reading an Analogue Clock

  • In step 1, we will discuss learning how a clock is divided. A clock is split into 12 sections. At the top, you will see the number ’12’. On the right-hand side, you will see a ‘1’ followed by 2, 3, 4, 5, ..so on up to 11 in a clockwise direction.
  • Here the numbers labelling each section are the hours.
  • The sections between the two numbers are divided into 5 divisions/segments.
  • Above we have seen the numbers arranged in the clock.
  • In a clock, there are 2 hands. A little hand indicated the hours and a big hand indicates minutes. For instance, if the little hand is showing ‘1’, then we have to read the time like this 1 o’clock.
  • Now, use the big hand and read the minutes. To read the minutes, first, take the indicated number and multiply it by 5 to get the minutes.
    • If the big hand is pointing to “3,” you’ll know that it is 15 minutes past the hour.
    • If the big hand is between “1” and “2,” note what dash it is pointing to. For example, if it is on the 3rd dash after the “1,” it is 8 minutes past the hour. (1 x 5 + the number of dashes).
  • After learning how to read hour and minutes, you can tell time easily. For instance,
    • If the little hand is pointing to “2” and the big hand is pointing to “12,” it is “two o’clock.”
    • If the little hand is pointing to “3” and the big hand is pointing to “2,” it is “3 hours 10 minutes” or “ten minutes past three.”
  • To tell AM or PM, first, we have to know the time of day. From midnight to noon the next day, the time is in AM. From noon to midnight, the time is in PM.

Method 2: Reading a Digital Clock

  • The digital clock is the opposite of the analogue clock. It displays only two numbers separated by a colon.
  • The first number on a digital clock indicates the hours. If the first number reads “7”, then it is read as 7 o’clock hour.
  • Now, come to the second number on a digital clock or watch, which is found after the colon, tells the minutes into the hour. For instance, if it displays 12, then it is read as 12 minutes into the hour.
  • Finally, put them together to read the exact time. For example, if the clock displays ‘7:12’ then it means 7 hours 12 minutes, or seven-twelve, or twelve past seven.
  • Also, to identify whether its AM or PM, some digital clocks show AM or PM directly on the clock.

Solved Examples on How to Tell Time in English?

1. Read the time shown in the given clock?

reading analogue clock or watch to tell time

Solution:

In the given figure, the hour hand is showing the number 10 then 10 hours and the minute hand is showing the number 12 then 0 minutes. So, we can read it as a 10 o’clock.

2. Read the time shown in the below clock image?

how to read a clock or watch example image

Solution:

From the given clock image, the hour hand is indicating between 4 and 5. It is almost near to 5 but for now, we will read as a 4 hour or 4 o’clock.

The minute hand is indicating 7 then 7 x 5 is 40. So, we read as 40 minutes.

By joining both together, we read it as 4 hours 40 minutes or 40 minutes past 4. 

Cardinal Number of a Set Definition, Examples | How to find the Cardinal Number of a Set?

Cardinal Number of a Set

Cardinal Numbers are Numbers that are used for counting something. They are also called the Cardinals. Cardinals are meant by how many of anything is existing in a group. In other words, cardinal numbers are a collection of ordinal numbers. Learn about the Cardinal Number of a Set Definition, Solved Examples explained in detail in the further modules.

Cardinal Number of a Set – Definition

The Number of Distinct Elements present in a finite set is called the Cardinal Number of a Set. Usually, we define the size of a set using cardinality. The Cardinal Number of a Set A is denoted as n(A) where A is any set and n(A) represents the number of members in Set A.

Consider a set of even numbers less than 15.

Set A = {2, 4, 6, 8, 10, 12, 14}

As Set A has 7 elements, the Cardinal Number of the Set is n(A) = 7

Note:

(i) Cardinal number of an infinite set is not defined.

(ii) Cardinal number of the empty set is 0 since it has no element.

Also, Check:

How to find the Cardinal Number of a Set?

1. Find the cardinal number of the following set

E = { x : x < 0, x ∈ N }

Solution:

x<0 means negative integers and they don’t fall under Natural Numbers.

Therefore, the above set will not have any elements

Cardinal Number of Set E is n(E) =0

2. Find the Cardinal Number of the Following Set

Q = { x : – 4 ≤ x ≤ 3, x ∈ Z }

Solution:

Given Q = { x : – 4 ≤ x ≤ 3, x ∈ Z }

x={-4, -3, -2, -1, 0, 1, 2, 3 }

Number of Elements in the above set is 8

Therefore, Cardinal Number of Set Q is n(Q) = 8

3. Find the Cardinal Number of the Set

A = { x : x is even prime number }

Solution:

Among all the prime numbers 2 is the only even prime number and the set has only one element

A ={2}

Cardinal Number of a Set n(A) = 1

4. Set D = {3, 4, 4, 5, 6, 7, 8, 8, 9}

Solution:

We know the cardinal number of a set is nothing but the number of distinct elements in the set

Cardinal Number of Set D is n(D) = 7

5. Find the Cardinal Number of a Set X = {letters in the word APPLE}

Solution:

Set X = {letters in the word APPLE}

We know the cardinal number of a set is nothing but the number of distinct elements in the set

x = {A, P, L, E}

Cardinal Number of Set n(X) = 4

6. Find the Cardinal Number of a Set

P = {x | x ∈ N and x2 <25}

Solution:

Given P = {x | x ∈ N and x2 <25}

Then P = {1, 2, 3, 4}

Cardinal Number of Set P is 4 and is denoted by n(P) = 4

Perimeter of a Triangle Definition, Formula & Examples | How to Find the Perimeter of a Triangle?

Perimeter of a Triangle Definition, Formula & Examples

Polygons are the most important topic among all math concepts. One of the simplest polygons is Triangle and the easiest way to work with polygons is by calculating their perimeter. The term perimeter is a path that encloses an area. It completely refers total length of the edges or sides of a given polygon or a two-dimensional figure with angles. So, make sure you move down the page till to an end and learn the perimeter of a triangle definition, formula, and how to calculate the triangle perimeter easily & quickly.

What is the Perimeter of a Triangle?

The definition of Perimeter of a Triangle is the sum of the lengths of the side of the Triangle. It denotes as,

Perimeter = Sum of the three sides

In real-life problems, a perimeter of a triangle can be useful in building a fence around the triangular parcel, tying up a triangular box with ribbon, or estimating the lace required for binding a triangular pennant, etc.

Always, the result of the triangle perimeter should be represented in units. If the side lengths of the triangle are measured in centimeters, then the final result needs to be in centimeters.

Formula to Calculate Perimeter of Triangle

The basic formula is surprisingly uncomplicated. Simply add up the lengths of all of the triangle sides and you get the perimeter value of the given triangle. In the case of the triangle, if the sides are a,b,c then the perimeter of a triangle formula is P = a + b + c.

How to Find the Perimeter of a Triangle?

Between Area and Perimeter of a Triangle calculation, finding the perimeter of the triangle is the easiest one and it has three ways to calculate the triangle perimeter. All the three ways used to find the triangle perimeter are mentioned here for your sake of knowledge and understanding the concept efficiently. The ways to find the perimeter of a triangle are as follows:

  1. The first & simple way is when side lengths are given, then we have to add them together to get the perimeter of the given triangle.
  2. If we have two sides and then solve for a missing side using the Pythagorean theorem.
  3. In case, we have the side-angle-side information in the given question, then we can solve for the missing side with the help of the Law of Cosines.

For a better explanation of the concept, we have listed out some worked-out examples of calculating the perimeter of a triangle below. Have a look at the solved examples and understand the concept behind solving the perimeter of a polygon ie., a Triangle.

Solved Examples on Finding Perimeter of a Triangle

1. Find the perimeter of the triangle where the three sides of the triangle are 20 cm, 34 cm, 15 cm?

Solution:

Given Sides of the triangle are a = 20 cm, b = 34 c, c = 15 cm

Now, use the Perimeter of Triangle Formula and find the result,

Perimeter = (a + b + c) 

= 20 + 34 + 15 = 69 cm.

2. Find the missing side whose perimeter is 40 cm and two sides of the triangles are 15 cm?

Solution:

Given,

a = 15 cm
b = 15 cm
P = 40 cm

Find c, Let’s assume c=x

Perimeter of the triangle P = a + b + c

40 cm = 15 cm + 15 cm + x

40 cm = 30 cm + x

x = 40 cm – 30 cm

x = 10 cm

Therefore, the length of the third side of the triangle is 10 cm.

FAQs on Calculating Triangle Perimeter

1. What are the types of Triangles and their perimeter formulas?

There are 4 types of triangles. They are listed below with their perimeter formulas:

  1. Equilateral triangle: Perimeter (P) = 3 x l
  2. Right triangle: Perimeter (P) = a + b + c
  3. Isosceles triangle: Perimeter (P) = 2 x l + b
  4. Scalene triangle: Perimeter (P) = b + p + h where h2 = b2 + p2

2. What is the formula of the perimeter of a triangle?

If a triangle has three sides a, b and c, then, the formula for the perimeter of a triangle is Perimeter, P = a + b +c.

3. How do you calculate the perimeter of a triangle with known sides?

If we knew the sides of a triangle, then finding the perimeter is so simple, just apply the perimeter of a triangle formula ie., P = a + b +c. and substitute all the sides and add them together.

Units for Measuring Length | Standard Metric Units of Length & Its Measurements, Definitions & Examples

Units for measuring length

In real-life mathematical tasks, we always do measuring the lengths in various situations. To learn about what is measuring length and what are the units for measuring length, this guide will help you a lot. Hence, gain knowledge about measuring length by understanding various units to measure the length from this page. Also, you may grasp some important information about units of measurements like temperature, time, mass, volume, etc. from our provided Math Conversion Chart.

What is Measuring Length?

Measuring of Length of any object includes in our daily life, we always measure the length of the cloth for dresses, wall-length for wallpapers, and many more other tasks.

You all want to know what is measuring length? Measuring length is a measurement of an object in a unit of length by using measuring tools such as scale or ruler.

For instance, the length of a pen can be measured in inches with the help of a ruler.

What is Length?

Length is determined as “Distance between two points” OR “The maximum extended dimension of an object”. Some of the tools used to measure length are Vernier caliper and Tapes.

Also Check: Units of Mass and Weight Conversion Chart 

What is a Unit of Length?

A unit of length refers any arbitrarily chosen and accepted reference standard for measurement of length. In modern use, the most common units are the metric units, which are utilized across the world. In the United States, the U.S. Customary units are also used mostly. For some purposes in the UK & other countries, British Imperial units are used. The metric system is classified into SI and non-SI units.

Units for Measuring Length

According to the metric system, the standard unit of length is a meter (m). Based upon the measuring of length, the meter can be converted into various units like millimeters (mm), centimeter (cm), and kilometer (km). In accordance with the length conversion charts, the different units of lengths and their equivalents are tabulated below:

Units of Length
10 millimeters (mm)=1 centimeter (cm)
10 centimeters=1 decimeter (dm)
10 centimeters=100 millimeters
10 decimeters=1 meter (m)
10 decimeters=1000 millimeters
10 meters=1 dekameter (dam)
10 dekameters=1 hectometer (hm)
10 dekameters=100 meters
10 hectometers=1 kilometer (km)
10 hectometers=1000 meters

See Related Articles:

Units of Length Definitions & Examples

Length describes how long a thing is from one end to the other.

Metric UnitUS Standard Unit (English or Customary Unit)
Meter (m): Used to measure big lengths.

Example: Length of a classroom.

Kilometer (km): Used to measure very long-distance or lengths.

Example: Distance between two places.

Centimeter (cm): Used to measure small lengths.

Example: Length of a pencil.

Millimeter (mm): Used to measure very short lengths or thicknesses.

Example: Length of a pencil tip.

Yard (yd): It is bigger than a foot.

Example: Length of a football field.

Inch (in): Used to measure the length of small objects.

Example: Length of a bed.

Mile (mi): Used to measure long distances.

Example: Distance between two places.

Foot (ft): Used to measure short distances and heights.

Example: Heights of buildings.

FAQs on Unit of Measuring Length

1. What is the standard unit of measuring length?

‘Meter(m)’ is the standard unit of measuring length

2. Which is the standard tool to measure the length?

The standard tool among all to measure the length of any object is a ‘Ruler’. Just by placing the ruler beside the object and measure the object from start to end with the help of readings given on the ruler or scale.

3. What are the basic units of measurements?

There are seven SI base units and they are as follows:

  • Length – meter (m)
  • Mass – kilogram (kg)
  • Temperature – kelvin (K)
  • Time – second (s)
  • Electric current – ampere (A)
  • Luminous intensity – candela (cd)
  • Amount of substance – mole (mole)

Logarithm Rules Or Log Rules | Expanding Laws of Logarithm | Formulas & Examples

Logarithm Rules Or Log Rules

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: loga (MN) = loga M + loga N
  • Power Rule Law: loga Mn = n loga M
  • Quotient Rule Law: loga (M/N) = loga M – loga N
  • Change of Base Rule Law: loga M = logb M × loga 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.

logb(x ∙ y) = logb(x) + logb(y)

Proof of Log Product Rule Law: 

loga(MN) = loga M + loga N

Let loga M = x ⇒ a sup>x = M

and loga N= y ⇒ ay = N

Now ax ∙ ay = MN or, ax+y = MN

Therefore from definition, we have,

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

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

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

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

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

logb(xy) = y ∙ logb(x)

Proof of Log Power Rule Law:

logaMn = n loga M

Let loga Mn = x ⇒ ax = Mn

and loga M = y ⇒ ay = M

Now, ax = Mn = (ay)n = any

Therefore, x = ny or, loga Mn = n loga 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.

loga(x / y) = loga(x) – loga(y)

Proof of Log Quotient Rule Formula:

Let M = ax and N = ay, then it follows that loga(M) = x and loga(N) = y,

We can now prove the quotient rule as follows:

loga (M/N) = loga (ax/ay)

= loga(ax-y)

= x – y [Put the values of x and y]

= loga M – loga N

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

= loga M + loga N + loga P – (loga R + loga S + loga 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.

logb M = loga M/loga b

Proof of Change of base Rule Law:

loga M = logb M × loga b

Assume loga M = x ⇒ ax = M,

logb M = y ⇒ by = M,

and loga b = z ⇒ az = b.

Now, ax= M = by – (az)y = ayz

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

Corollary:

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

loga a = logb a × loga b or, logb a × loga b = 1 [since, loga a = 1]

or, logb a = 1/loga 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,

logb M = loga M/loga 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) loga 1 = 0

(ii) loga a = 1

(iii) a loga M = M

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

(v) loga (M/N) = loga M – loga N

(vi) loga Mn = n loga M

(vii) loga M = logb M × loga b

(viii) logb a × loga b = 1

(ix) logb a = 1/loga b

(x) logb M = loga M/loga 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.

Convert Exponentials and Logarithms | Relationship between Exponentials and Logarithms

Convert Exponentials and Logarithms

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 = by for y

For x>0, b>0 b≠ 1, y = logb x is equivalent to by = 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.

by = 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 by = x

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

by = 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 3x = 27 in Logarithmic Form?

Solution:

3x = 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  6y = 98 in Logarithmic Form?

Solution:

Given Equation is 6y = 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

Coin Toss Probability Formula, Solved Examples | Possible Events that Occur when a Coin is Tossed

Coin Toss Probability

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

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

 

 

Trigonometrical Ratios of (180° – θ) | Solved Problems on Trigonometric Functions of 180 Minus Theta | Relation Between All Six Quadrants

Trigonometrical Ratios of (180° - θ)

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.

ASTC Formula

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:

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 | Multiplying Fractions Worksheets with Answers

Worksheet on Multiplication of Fractions

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?

Solution:

As given in the question,

Gita has marbles = 8

She gave to younger brother = \(\frac { 1 }{ 4 } \) of 8

No of marbles her brother got = \(\frac { 1*8 }{ 4 } \)

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

= 2

Therefore, her brother got 2 marbles

Thus, the final solution is 2 marbles


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?

Solution:

As given in the question,

Mr. Gupta puts = 3\(\frac { 1 }{ 4 } \) litres of petrol

He used = \(\frac { 1 }{ 3 } \) of 3\(\frac { 1 }{ 4 } \)

\(\frac { 1 }{ 3 } \) * \(\frac { 13 }{ 4 } \)

\(\frac { 1*13 }{ 3*4 } \)

3\(\frac { 13 }{ 12 } \)

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

Therefore, Mr. Gupta used 1\(\frac { 1 }{ 12 } \) litres of petrol.

Thus, the final solution is 1\(\frac { 1 }{ 12 } \) litres


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?

Solution:

As given in the question,

Neha spends = \(\frac { 3 }{ 5 } \) hours/day

Hours she spends in a week = \(\frac { 3 }{ 5 } \) * 7

\(\frac { 3*7 }{ 5 } \) \(\frac { 21 }{ 5 } \)

4\(\frac { 1 }{ 5 } \) hours

Therefore, Neha spends 4\(\frac { 1 }{ 5 } \) hours in morning exercises.

Thus, the solution is 4\(\frac { 1 }{ 5 } \) hours


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?

Solution:

As given in the question,

Weight of one plum cake = \(\frac { 3 }{ 4 } \) kg

No of cakes = 5

Amount of cake he bought = \(\frac { 3 }{ 4 } \) * 5

\(\frac { 3*5 }{ 4 } \) \(\frac { 15 }{ 4 } \)

Therefore, \(\frac { 15 }{ 4 } \) kilograms of cake, Mr. Ramesh buy

Thus, the final solution is \(\frac { 15 }{ 4 } \)


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?

Solution:

As given in the question,

The cost of one kilogram apple = 25\(\frac { 1 }{ 2 } \)

The cost of 1\(\frac { 1 }{ 2 } \) kilogram apples = 25\(\frac { 1 }{ 2 } \) * 1\(\frac { 1 }{ 2 } \)

\(\frac { 51 }{ 2 } \) * \(\frac { 3 }{ 2 } \)

\(\frac { 153 }{ 4 } \)

Therefore, \(\frac { 153 }{ 4 } \) is the price for 1\(\frac { 1 }{ 2 } \) kilogram apples.

Thus, the final solution is \(\frac { 153 }{ 4 } \)


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?

Solution:

As given in the question,

Thickness of book = 1\(\frac { 1 }{ 4 } \)

No of books = 16

The thickness of a pile of 16 books = 1\(\frac { 1 }{ 4 } \) * 16

\(\frac { 5 }{ 4 } \) * 16

\(\frac { 80 }{ 4 } \)

= 20

Therefore, the thickness of a pile of 16 books is 20cm.

Thus, the final solution is 20cm.


Problem 7:

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

Solution:

As given in the question,

Cost of 1 litre milk = 16\(\frac { 3 }{ 4 } \)

Cost of \(\frac { 32 }{ 5 } \) litre milk = 16\(\frac { 3 }{ 4 } \)

= \(\frac { 67 }{ 4 } \)

The cost of 6\(\frac { 2 }{ 5 } \) is \(\frac { 67 }{ 4 } \) * \(\frac { 32 }{ 5 } \)

= \(\frac { 67*8 }{ 5 } \)

= \(\frac { 536 }{ 5 } \)

= 107\(\frac { 1 }{ 5 } \)

Therefore, the cost of 6\(\frac { 2 }{ 5 } \) litre of milk is 107\(\frac { 1 }{ 5 } \)

Thus, the final solution is 107\(\frac { 1 }{ 5 } \)


Problem 8:

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

Solution:

As given in the question,

Length of the rectangular park = 41\(\frac { 2 }{ 3 } \)

Breadth of the rectangular park = 18\(\frac { 3 }{ 5 } \)

Area of rectangle = length * breadth

= \(\frac { 125 }{ 3 } \) * \(\frac { 93 }{ 5 } \)

= 775 meters

Therefore, the area of the rectangular park is 775 meters

Thus, the final solution is 775 meters.


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

Solution:

As given in the question,

The cost of milk = 236\(\frac { 1 }{ 4 } \)

No of litres = 5\(\frac { 2 }{ 5 } \)

Cost of \(\frac { 27 }{ 5 } \) litres of milk = \(\frac { 945 }{ 4 } \)

Cost of 1 litre of milk = \(\frac { 945%4 }{ 27%5 } \)

= \(\frac { 945 }{ 4 } \) * \(\frac { 5 }{ 27 } \)

= \(\frac { 35*5 }{ 4 } \)

= \(\frac { 175 }{ 4 } \)

= 43\(\frac { 3 }{ 4 } \)

Therefore, the cost per litre = 43\(\frac { 3 }{ 4 } \)

Thus, the final solution is 43\(\frac { 3 }{ 4 } \)


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?

Solution:

As given in the question,

Amount of sugar Margaret has = 2\(\frac { 1 }{ 3 } \)

No of cups she is using to make cookies = 1\(\frac { 1 }{ 4 } \)

Amount of sugar will she have left = 2\(\frac { 1 }{ 3 } \) – 1\(\frac { 1 }{ 4 } \)

= \(\frac { 7 }{ 3 } \) – \(\frac { 5 }{ 4 } \)

= \(\frac { 7 }{ 3 } \) * \(\frac { 4 }{ 4 } \) – \(\frac { 5 }{ 4 } \) * \(\frac { 3 }{ 3 } \)

= \(\frac { 28 }{ 12 } \) – \(\frac { 15 }{ 12 } \)

= \(\frac { 28-15 }{ 12 } \)

= \(\frac { 13 }{ 12 } \)

Therefore, \(\frac { 13 }{ 12 } \) cups of sugar will she have left.

Thus, the final solution is \(\frac { 13 }{ 12 } \)


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?

Solution:

As given in the question,

Amount of money Russell spent on movies = \(\frac { 1 }{ 3 } \)

Amount of money Russell spent on new video game = \(\frac { 1 }{ 2 } \)

Total amount of money he spent of his allowance = \(\frac { 1 }{ 3 } \) + \(\frac { 1 }{ 2 } \)

= \(\frac { 1 }{ 3 } \) * \(\frac { 2 }{ 2 } \) + \(\frac { 1 }{ 2 } \) * \(\frac { 3 }{ 3 } \)

= \(\frac { 2 }{ 6 } \) + \(\frac { 3 }{ 6 } \)

= \(\frac { 2+3 }{ 6 } \)

= \(\frac { 5 }{ 6 } \)

Therefore, the total amount of money spent = \(\frac { 5 }{ 6 } \)

Thus, the final solution is \(\frac { 5 }{ 6 } \)


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?

Solution:

As given in the question,

Acres of land Sam has = 3\(\frac { 1 }{ 8 } \)

Acres of land he planted potatoes = 2\(\frac { 2 }{ 3 } \)

Amount of land left to plant the corn = 3\(\frac { 1 }{ 8 } \) – 2\(\frac { 2 }{ 3 } \)

= \(\frac { 25 }{ 8 } \) – \(\frac { 8 }{ 3 } \)

= \(\frac { 25 }{ 8 } \) * \(\frac { 3 }{ 3 } \) – \(\frac { 8 }{ 3 } \) * \(\frac { 8 }{ 8 } \)

= \(\frac { 75 }{ 24 } \) – \(\frac { 64 }{ 24 } \)

= \(\frac { 75-64 }{ 24 } \)

= \(\frac { 11 }{ 24 } \)

Therefore, \(\frac { 11 }{ 24 } \) acres of land left to plant the corn.

Thus, the final solution is \(\frac { 11 }{ 24 } \)


Problem 13:

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

Solution:

As given in the question,

Amount of spaghetti = 80 pounds

No of families = 7

Amount of spaghetti each family receive = \(\frac { 80 }{ 7 } \)

Therefore, \(\frac { 80 }{ 7 } \) pounds of spaghetti each family will 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?

Solution:

As given in the question,

No of people = 11

Length of the rope = 70 feet

Length of rope each person will receive = \(\frac { 70 }{ 11 } \)

Therefore, each person will receive \(\frac { 70 }{ 11 } \) feet of rope.

Thus, the final solution is \(\frac { 70 }{ 11 } \).


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?

Solution:

As given in the question,

Fraction of pizza left = \(\frac { 1 }{ 4 } \)

No of people who share the pizza = 2

Amount of pizza each will get = \(\frac { 1%4 }{ 2 } \)

= \(\frac { 1 }{ 4 } \) * \(\frac { 2 }{ 1 } \)

= \(\frac { 2 }{ 4 } \)

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

Therefore, each gets \(\frac { 1 }{ 2 } \) of pizza.

Thus, the final solution is \(\frac { 1 }{ 2 } \)


Worksheet on Division of Fractions | Dividing Fractions Worksheet with Answers

Worksheet on Division of Fractions

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?

Solution:

As given in the question,

Amount of chocolate = \(\frac { 2 }{ 21 } \)

No of bags = Total lb/lbs per bag

\(\frac { 2 }{ 3 } \) % \(\frac { 2 }{ 21 } \)

= \(\frac { 2 }{ 3 } \) % \(\frac { 2 }{ 21 } \)

= \(\frac { 2 }{ 3 } \) * \(\frac { 21 }{ 2 } \)

= \(\frac { 2 }{ 3 } \) + \(\frac { 2 }{ 21 } \)

= \(\frac { 2 }{ 3 } \) * \(\frac { 21 }{ 21 } \) + \(\frac { 2 }{ 21 } \) * \(\frac { 3 }{ 3 } \)

= \(\frac { 42 }{ 63 } \) + \(\frac { 6 }{ 63 } \)

= \(\frac { 48 }{ 63 } \)


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?

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 get = \(\frac { 1%5 }{ 3 } \)

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

= \(\frac { 1 }{ 15 } \)

Therefore, \(\frac { 1 }{ 15 } \) fraction of original bag each dog gets

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


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?

Solution:

As given in the question,

Fraction 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

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

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

Therefore, \(\frac { 1 }{ 12 } \) gallons of salt water

Thus, the final solution is \(\frac { 1 }{ 12 } \)


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?

Solution:

As given in the question,

Length of the board = 4ft

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

No of pieces David will get = 4 % \(\frac { 1 }{ 4 } \)

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

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

= 16 pieces

Therefore, Devin split the board into 16 pieces.


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?

Solution:

As given in the question,

No of hours Jasmine left to finish math problems = \(\frac { 1 }{ 4 } \)

No of problems = 4

The time she have to spend on each problem = \(\frac { 1 }{ 4 } \) % 4

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

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

Therefore, \(\frac { 1 }{ 16 } \) fraction of time she spend on each problem

Thus, the final solution is \(\frac { 1 }{ 16 } \)


Problem 6:

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

Solution:

As given in the question,

The fraction of homework problems = \(\frac { 1 }{ 3 } \)

The fraction that is equivalent to wht Sandra finished = \(\frac { 1 }{ 3 } \) * \(\frac { 2 }{ 2 } \)

= \(\frac { 2 }{ 6 } \)

= \(\frac { 1 }{ 3 } \) * \(\frac { 3 }{ 3 } \)

= \(\frac { 3 }{ 9 } \)

Therefore, the fractions \(\frac { 2 }{ 6 } \) and \(\frac { 3 }{ 9 } \) are equivalent to Sandra.

Thus, the final solution is \(\frac { 2 }{ 6 } \) and \(\frac { 3 }{ 9 } \)


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?

Solution:

As given in the question,

No of cups of sugar = 9

The fraction of cake he needs = \(\frac { 3 }{ 4 } \)

No of cups = 9 % \(\frac { 3 }{ 4 } \)

= \(\frac { 9 }{ 1 } \) * \(\frac { 4 }{ 3 } \)

= \(\frac { 36 }{ 3 } \)

= 12

Therefore, Berlin needs 12 cups of sugar.

Thus, the final solution is 12 cups.


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?

Solution:

As given in the question,

Amount of yards of ribbon per project = 3\(\frac { 1 }{ 2 } \)

No of yards = 28

The greatest number of projects = 28 % 3\(\frac { 1 }{ 2 } \)

= 28 % \(\frac { 7 }{ 2 } \)

= \(\frac { 28 }{ 1 } \) * \(\frac { 2 }{ 7 } \)

= \(\frac { 56 }{ 7 } \)

= 8

Therefore, Mahathi can complete 8 projects with this ribbon.

Thus, the final solution is 8 projects


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?

Solution:

As given in the question,

No of cookies = 40

No of cookies of chocolate chip = \(\frac { 3 }{ 5 } \)

No of cookies of peanut butter = \(\frac { 1 }{ 4 } \)

No of chocolate chips = 40 * \(\frac { 3 }{ 5 } \)

= \(\frac { 120 }{ 5 } \)

= 24

No of peanut cookies = 40 * \(\frac { 1 }{ 4 } \)

= \(\frac { 40 }{ 4 } \)

= 10

Therefore, a total of 10 chocolate and peanut cookies are 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?

Solution:

As given in the question,

No of hours Gillian take to build a dog house = 6\(\frac { 1 }{ 2 } \)

No of hours she worked each night = 1\(\frac { 3 }{ 4 } \)

No of nights he took = 6\(\frac { 1 }{ 2 } \) % 1\(\frac { 3 }{ 4 } \)

= \(\frac { 13 }{ 2 } \) % \(\frac { 7 }{ 4 } \)

= \(\frac { 13 }{ 2 } \) * \(\frac { 4 }{ 7 } \)

= \(\frac { 52 }{ 14 } \)

= 3\(\frac { 10 }{ 14 } \)

Therefore, 3\(\frac { 10 }{ 14 } \) nights it took to build a dog house.

Thus, the final solution is 3\(\frac { 10 }{ 14 } \)


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?

Solution: 

As given in the question,

No of students in class = 40

No of boys = \(\frac { 2 }{ 5 } \)

No of boys present = \(\frac { 40 }{ 1 } \) * \(\frac { 2 }{ 5 } \)

= \(\frac { 80 }{ 5 } \)

= 16 boys

Therefore, the total no of girls = 40 – 16

= 24 girls

Therefore, there are 24 girls present in the class

Thus, the final solution is 24 girls.


Problem 12:

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

Solution: 

As given in the question,

Amount of Php, John had = 960

Amount of money he spent on food = \(\frac { 1 }{ 3 } \)

Amount of money left out after spending on food = \(\frac { 960 }{ 1 } \) * \(\frac { 1 }{ 3 } \)

= \(\frac { 960 }{ 3 } \)

= 320

Total amount of money left = 960 – 320 = 640

Therefore, John has 640 Php left with him.

Thus, the final solution is Php 640.


Problem 13:

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

Solution: 

As given in the question,

No of apples for sale = 480

Fraction of apples he sold = \(\frac { 3 }{ 5 } \)

No of apples left = \(\frac { 480 }{ 1 } \) * \(\frac { 3 }{ 5 } \)

= \(\frac { 1440 }{ 5 } \)

= 288

No of apples left after selling = 480 – 288

= 192

Therefore, there were 192 apples left after selling

Thus, the final solution is 192 apples.


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?

Solution: 

As given in the question,

The fraction of cubic yard Rose shoveled = \(\frac { 1 }{ 3 } \)

The fraction of cubic yard Mary shoveled = \(\frac { 1 }{ 2 } \)

The amount of soil they put into the garden together = \(\frac { 1 }{ 3 } \) + \(\frac { 1 }{ 2 } \)

Multiply the first fraction with \(\frac { 2 }{ 2 } \) and second fraction with \(\frac { 3 }{ 3 } \)

= \(\frac { 2 }{ 6 } \) + \(\frac { 3 }{ 6 } \)

= \(\frac { 5 }{ 6 } \)

Therefore, \(\frac { 5 }{ 6 } \) fraction of soil they put into the garden bed altogether

Thus, the solution is \(\frac { 5 }{ 6 } \)


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?

Solution: 

As given in the question,

Amount of chocolate = \(\frac { 4 }{ 5 } \)

Otto would have chocolate = \(\frac { 1 }{ 3 } \)

Amount of chocolate Carl’s have = \(\frac { 4 }{ 5 } \) – \(\frac { 1 }{ 3 } \)

Multiply the first equation with \(\frac { 3 }{ 3 } \) and the second equation with \(\frac { 5 }{ 5 } \)

= \(\frac { 12 }{ 15 } \) – \(\frac { 5 }{ 15 } \)

= \(\frac { 7 }{ 15 } \)

Therefore, Carl’s have \(\frac { 7 }{ 15 } \) fraction of chocolate

Thus, the final solution is \(\frac { 7 }{ 15 } \).