Common Core Math

A binary number is a number expressed in the base 2 numeral system which uses only two symbols 0 and 1. Each digit in the binary is called a bit or binary digit. The addition is one of the basic arithmetic operations. Binary addition is one of the binary operations. The binary addition works similarly to the base 10 decimal system. Check out more about the binary addition using 2s complement with solved example questions in the following sections of this page.

Also, Read

• Hexadecimal Addition and Subtraction
• Binary Subtraction

Binary Addition

The binary addition is similar to the decimal system, but it is a base 2 system. The binary system has two digits 0 and 1. Almost all functionalities of the computer system use the binary number system. The process of the addition operation is very familiar to the decimal system by just adjusting the base 2 of the numbers.

Before attempting the binary addition process, we should have complete knowledge of how the place works in the binary number system. Most modern digital computers and electric circuits perform binary operations by representing each bit as a voltage signal. the bit 0 in the binary system denotes the “OFF” state, bit 1 in the binary system denotes the “ON” state.

2’s Complement of a Binary Number

To get the 2’s complement of a given binary number, invert the given number and add 1 to the least significant bit (LSB) of the given result. The various uses of 2’s complement of binary numbers are signed binary number representation, to perform arithmetic operations on binary numbers.

Example:

2’s complement of 101

Invert 101 = 010

Add 1 to LSB of 010 = 010 + 1

= 011

So, 2’s complement of 101 is 011.

Binary Addition using 2’s Complement

Binary Addition using 2’s Complement is similar to the normal addition of two binary numbers. When you add two positive numbers, then the result is a positive number. When you add two negative numbers, then the addition will be a negative number. The basic rules of binary addition is 1 + 1 = 10 (1 is carry), 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1. If two numbers have different signs then follow these cases.

Case I: When the positive number has a greater magnitude.

In this case, the carry obtained is discarded and the final result is the addition of two numbers

Let us take the 5-bit numbers 1011 and -0101

2’s complement of 0101 = 1011

1 0 1 1 ⇒ 0 1 0 1 1

– 0 1 0 1 ⇒ 1 1 0 1 1 (2’s Complement)

⇒ 0 0 1 1 0 (Carry 1 is discarded)

So, 1011 – 0101 = 00110

Case II: When the negative number has a greater magnitude.

If the negative number has a greater magnitude then no carry will be generated in the sign bit. The result of an addition will be negative and the final result obtained by taking 2’s complement of the magnitude bits of the result.

Let us take two numbers as + 0100 and -0111

+ 0 1 0 0 ⇒ 0 0 1 0 0

– 0 1 1 1 ⇒ 1 1 0 0 1 (2’s complement)

⇒ 1 1 1 0 1

2’s complement of 1101 is 0011.

Hence the required sum is – 0011.

Case III: When the numbers are negative.

When two negative numbers are added a carry will be generated from the sign bit which will be discarded. 2’s complement of the magnitude bits of the operation will be the final sum.

Let us take two numbers as -0111 and -0010

– 0 1 1 1 ⇒ 1 1 0 0 1 (2’s complement)

– 0 0 1 0 ⇒ 1 1 1 1 0 (2’s complement)

⇒ 1 0 1 1 1 (Carry 1 discarded)

2’s complement of 0111 is 1001

Hence the required sum is -1001.

Solved Examples on Binary Addition using 2’s Complement

Example 1:

Find the sum of -1101 and -1110 using the 2’s complement.

Solution:

Given numbers are -1101, -1110

Find the 2’s complement of the negative numbers

So, 2’s complement of 01110 is 10010 and 01101 is 10011

Add the complemet numbers

1 0 0 1 0 + 1 0 0 1 = 1 0 0 1 0 1

End carry 1 of the sum is dicarded

2’s complement of the result 00101 is 11011

So, sum of -1101, -1110 is -11011.

Example 2:

Find 10101 – 10111 using the 2’s complement concept.

Solution:

The given numbers are 10101, -10111

2’s complement of the negative number 10111 is 01001

Add first number and 2’s complement of negative number

1 0 1 0 1 + 0 1 0 0 1 = 1 1 1 1 0

2’s complement of the result 11110 is 00010.

So, 10101 – 10111 = -00010

Example 3:

Find the addition of 00110 and 01001 using the 2’s complement.

Solution:

Given numbers are 00110, 01001

0 0 1 1 0 + 0 1 0 0 1 = 0 1 1 1 1

Hence, the sum is 0 1 1 1 1.

FAQs on Binary Addition using 2’s Complement

1. What are the Rules of Binary Addition?

The four basic rules of the binary addition are 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 10.

2. How to add two binary numbers?

Add the first column if two numbers are 1 then the result is 10. Here, 1 is the carry. Continue adding till no digit is left on the left side. The result is the sum of two binary numbers.

3. How to find 2’s complement of a binary number?

Invert the given numbers and add one to the least significant bit to get the 2’s complement of the binary number.

Time is defined as the continued progress of existence in past, present, and future. Using the unit of time you can measure the existence of events. The most commonly used units of time are second, minute and hour. Time is an interesting topic and everyone is familiar with it. Want to know about the units of time then go through the following sections. Learn Converting Units of Time such as Hours, Minutes, Seconds from one unit to another by referring through solved examples available.

SI Unit of Time

The SI Unit of Time is Second and is accurately defined as the time interval equal to 9192631770 periods of radiation. Unit second is often represented as s or sec.

Different Units of Time

Some of the common and frequently used time units are minutes, hour, day, week, month, and year. If you want to measure a long duration of time, then you can use a decade which is equal to 10 years, a century which is equal to 100 years, a millennium which is equal to 1000 years, and a mega-annum which is equal to 1,000,000 years. The popular units of time are given below.

The most commonly used Units of Time are

• Hours
• Minutes
• Seconds

1000 milliseconds	1 second
60 seconds	1 minute
60 minutes	1 hour
24 hours	1 day
7 days	1 week
28, 29, 30, or 31 days	1 month
365 or 366 days	1 year
12 months	1 year
10 years	1 decade
100 years	1 century
1000 years	1 millennium

How Time Became So Important?

The top ten reasons why time is so important are mentioned here.

• Every single thing in the universe is affected by time.
• Time is the most precious resource because you can’t get it back.
• Because of privileges, not everyone truly has the same amount of time in a day.
• No one knows how much time they have.
• The only time we actually have is the present.
• How we see time impacts happiness.
• Managing it poorly or well has a huge impact on life.
• Skills are impacted by how much time you invest.
• Relationships are made or broken by how much time you invest.
• Time is a teacher or a healer.

Also, Read

• Units of Time Conversion Chart
• Units of Measurement

Time Conversions

To convert 1 unit of time to another, you have to know the units of time. You can use the multiplication or division operations to convert the time.

• To convert minutes into seconds, you need to multiply each minute by 60 seconds.
• To convert seconds into minutes, you need to divide each second by 60 minutes.

You can use these parameters for any type of time conversion.

Solved Examples on Converting Units of Time

1. Convert the following

(i) 5 hours 30 minutes into minutes

(ii)90 minutes to seconds

Solution:

(i) 5 hours 30 minutes into minutes

We know 1 hr = 60 minutes

5 hours = 5*60 minutes

= 300 minutes

5 hrs 30 minutes = 300 minutes +30 minutes

= 330 minutes

Therefore, 5 hours 30 minutes = 330 minutes

(ii)90 minutes to seconds

We know 1 minute = 60 seconds

90 minutes = 90*60 seconds

= 5400 seconds

Therefore, 90 minutes = 5400 seconds

2. Find the total time

5 hours 40 minutes and 3 hours 20 minutes

Solution:

Firstly add the hours i.e. 5 hours +3 hours

= 8 hours

Now add the minutes individually i.e. 40 minutes +20 minutes

= 60 minutes

= 1 hour

Now, add this to the hours we got in the earlier step i.e. 8 hours +1 hour

= 9 hours

Therefore, 5 hours 40 minutes and 3 hours 20 minutes is equal to 9 hours

Frequently Asked Questions on Units of Time

1. What is the SI unit of time?

The SI unit of time is seconds.

2. What are the 3 possible units of time?

The three most used units of time are seconds, minutes, and hours. 1 minute = 60 seconds, 1 hour = 60 minutes = 3600 seconds.

3. What are the different units of a second?

The various units of a second from the smallest to the largest values are along the lines:
Decisecond (1/10th of a second), centisecond (1/100th of a second), millisecond (1/1000th of a second), microsecond (one-millionth of a second), nanosecond (one-billionth of a second), picosecond (one-trillionth of a second), femtosecond (one-quadrillionth of a second), attosecond (one-quintillionth of a second), zeptosecond (one-sextillionth of a second), yoctosecond (one-septillionth of a second), and Planck time.

4. What is the largest unit of time?

The largest unit of time is the supereon. It is the combination of eons, eras, periods, epochs, and ages.

In the metric system of measurement, the meter is the basic unit of length, a gram is the basic unit of mass and liter is the basic unit of capacity.  We can use a centimeter(cm) to measure the length. Centimeter and Millimeter are very small units to measure the length, so we use another unit called meters.

Learn completely about the Units of a Measurement- Definition, Units Conversion, Prefix for Length, Time, Weight, and Volume or Capacity. Get to know the Importance of SI Units, Solved Examples on How to Convert one unit to another, etc.

Metric System – Introduction

The French are widely credited with originating the metric system of measurement, the system is officially adopted in 1795. It was originated in the year 1799. Metric System is basically a system used for measuring distance, length, volume, weight, and temperature. The term metric system is used as another word for SI or the international system of units.  Based on three basic units we can measure almost everything in the world, those are M- Meter, used to measure the length, Kg- Kilogram, used to measure the mass, and S- Second, used to measure time.

Units of Measurement – Definition

The SI system, also called the metric system, is used around the world. SI units stand for standard International System of the units. Seven basic units in the SI  system, give proper definitions for meter, kilogram, and the second. It also specifies and defines remaining four different  units:

1. Kelvin(K)- used to measure the Temperature

2. Ampere(A)- used to measure the Electric current

3. Candela(cd)- used to measure the Luminous Intensity

4. Mole(mol)- used to measure  the Material Quantity

Also, Read:

• Units of Length Conversion Charts
• Math Conversion Chart

Units of Measurement Conversion

To convert among units in the metric system, identify the unit that you have, the unit that you want to convert to, and then count the number of units between them. Some units are connected with each other by the following relation:

1 Kilometer (km) = 1000 meter (m)

1 meter (m) = 100 centimeter (cm)

1 centimeter (cm) =  10 millimeter (mm)

Metric Units Prefix

A metric prefix is a unit prefix that precedes a basic unit of measure to indicate a multiple or submultiple of the unit. To convert from one unit to another within the metric system usually means moving a decimal point. you can convert within the metric system relatively easily by simply multiplying or dividing the number by the value of prefix.

In order to remember the proper movements of units, arrange the prefixes from the largest to the smallest.

Now, let us discuss some of the units for length, weight, volume, time.

Units of Measurement Length

The most common unit used to measure the length are as follows. Centimeters and millimeters are very small to measure the length so, we use another unit that is the meter (m). Even meter is too small when we measure the distance between two cities, we use kilometers (km).

Sets are defined as a collection of well-defined elements that do not vary from person to person. It can be represented either in set-builder form or roster form. Generally, sets can be represented using curly braces {}. The different types of sets are empty set, finite set, singleton set, infinite set, equivalent set, disjoint sets, equal sets, subsets, superset, and universal sets. Get to know more about the Laws of Algebra of Sets for a better understanding of the students.

Laws of Algebra of Sets

The operations of sets are union, intersection, and complementation. The binary operations of set union, intersection satisfy many identities. The seven fundamental laws of the algebra of sets are commutative laws, associative laws, idempotent laws, distributive laws, de morgan’s laws, and other algebra laws.

1. Commutative Laws

For any two finite sets A and B

• A U B = B U A
• A ∩ B = B ∩ A

2. Associative Laws

For any three finite sets A, B, and C

• (A U B) U C = A U (B U C)
• (A ∩ B) ∩ C = A ∩ (B ∩ C)

So, union and intersection are associative.

3. Idempotent Laws

For any finite set A

• A U A = A
• A ∩ A = A
• A ∩ A’ = ∅
• ∅’ = U
• ∅ = U’

4. Distributive Laws

For any three finite sets A, B, and C

• A U (B ∩ C) = (A U B) ∩ (A U C)
• A ∩ (B U C) = (A ∩ B) U (A ∩ C)

Thus, union and intersection are distributive over intersection and union respectively.

5. De morgan’s Laws

For any two finite sets A and B

• A – (B U C) = (A – B) ∩ (A – C)
• A – (B ∩ C) = (A – B) U (A – C)

De Morgan’s Laws can also be written as

• Law of union: (A U B)’ = A’ ∩ B’
• Law of intersection: (A ∩ B)’ = A’ U B’

6. Complement Law

For any finite set A

• A ∪ A’ = A’ ∪ A =U
• A ∩ A’ = ∅

More laws of the algebra of sets:

7. For any two finite sets A and B;

• A – B = A ∩ B’
• B – A = B ∩ A’
• A – B = A ⇔ A ∩ B = ∅
• (A – B) U B = A U B
• (A – B) ∩ B = ∅
• A ⊆ B ⇔ B’ ⊆ A’
• (A – B) U (B – A) = (A U B) – (A ∩ B)

8. For any three finite sets A, B, and C;

• A – (B ∩ C) = (A – B) U (A – C)
• A – (B U C) = (A – B) ∩ (A – C)
• A ∩ (B – C) = (A ∩ B) – (A ∩ C)
• A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

Also, Read

• Proof of De Morgan’s Law in Boolean Algebra
• Intersection of Sets
• Subsets
• Different Notations in Sets
• Subsets of a Given Set

Solved Examples on Laws of Algebra of Sets

Example 1:

If E = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3, 4, 5}, B = {2, 5, 7} show that

(i) (A U B)’ = A’ ∩ B’

(ii) (A U B) = B U A

(iii) A ∩ B = B ∩ A

(iv) (A ∩ B)’ = A’ U B’

Solution:

Given that

E = {1, 2, 3, 4, 5, 6, 7}, A = {1, 2, 3, 4, 5}, B = {2, 5, 7}

(i) (A U B)’ = A’ ∩ B’

A U B = {{1, 2, 3, 4, 5} U {2, 5, 7}}

= {1, 2, 3, 4, 5, 7}

(A U B)’ = {1, 2, 3, 4, 5, 7}’

= {6}

A’ = {6, 7}

B’ = {1, 3, 4, 6}

A’ ∩ B’ = {6, 7} ∩ {1, 3, 4, 6}

= {6}

So, L.H.S = R.H.S

(ii) (A U B) = B U A

A U B = {{1, 2, 3, 4, 5} U {2, 5, 7}}

= {1, 2, 3, 4, 5, 7}

B U A = {2, 5, 7} U {1, 2, 3, 4, 5}

= {1, 2, 3, 4, 5, 7}

So, L.H.S = R.H.S

(iii) A ∩ B = B ∩ A

L.H.S = A ∩ B

= {1, 2, 3, 4, 5} ∩ {2, 5, 7}

= {2, 5}

R.H.S = B ∩ A

= {2, 5, 7} ∩ {1, 2, 3, 4, 5}

= {2, 5}

So, L.H.S = R.H.S

(iv) (A ∩ B)’ = A’ U B’

L.H.S = (A ∩ B)’

= {{1, 2, 3, 4, 5} ∩ {2, 5, 7}}’

= {2, 5}’

= {1, 3, 4, 6, 7}

R.H.S = A’ U B’

= {1, 2, 3, 4, 5}’ U {2, 5, 7}’

= {6, 7} U {1, 3, 4, 6}

= {1, 3, 4, 6, 7}

L.H.S = R.H.S

Hence, proved.

Example 2:

If X = {a, b, c, d}, Y = {b, d, f}, Z = {a, c, e} verify that

(i) (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z)

(ii) (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z)

Solution:

Given that,

X = {a, b, c, d}, Y = {b, d, f}, Z = {a, c, e}

(i) (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z)

L.H.S = (X ∪ Y) ∪ Z

= ({a, b, c, d} U {b, d, f}) U {a, c, e}

= {a, b, c, d, f} U {a, c, e}

= {a, b, c, d, e, f}

R.H.S = X ∪ (Y ∪ Z)

= {a, b, c, d} U ({b, d, f} U {a, c, e})

= {a, b, c, d} U {a, b, c, d, e, f}

= {a, b, c, d, e, f}

So, L.H.S = R.H.S

(ii) (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z)

L.H.S = (X ∩ Y) ∩ Z

= ({a, b, c, d} ∩ {b, d, f}) ∩ {a, c, e}

= {b, d} ∩ {a, c, e}

= ∅

R.H.S = X ∩ (Y ∩ Z)

= {a, b, c, d} ∩ ({b, d, f} ∩ {a, c, e})

= {a, b, c, d} ∩ ∅

= ∅

So, L.H.S = R.H.S

Hence verified.

Example 3:

If A = {p, q, r, s}, B = {u, q, s, v} find

(i) A – B

(ii) B – A

(iii) A ∩ B

Solution:

Given that,

A = {p, q, r, s}, B = {u, q, s, v}

(i)

A – B = {p, q, r, s} – {u, q, s, v}

= {p, r}

(ii)

B – A = {u, q, s, v} – {p, q, r, s}

= {u, v}

(iii)

A ∩ B = {p, q, r, s} ∩ {u, q, s, v}

= {q, s}

FAQs on Laws of Algebra of Sets

1. What is a set? Give an example?

A set is a collection of elements or objects or numbers represented using the curly brackets {}. The example is {1, 2, 3, 5} is a set of numbers.

2. What are the five basic properties of sets?

The five basic properties of sets are commutative property, identity property, associative property, complement property, and distributive property.

3. What are the 4 operations of sets?

The 4 set operations include set union, set intersection, set difference, the complement of a set, and cartesian product.

Worksheet on Area and Perimeter of Rectangle Problems will help the students to explore their knowledge of Rectangle word Problems. Solve all the Problems to learn the formula of Area of Rectangle and Perimeter of a Rectangle. To know the definition, properties, derivation, Problems with Solutions, Formulas of Rectangle you can visit our website. We have given the complete Rectangle concept along with examples. Check out the Area and Perimeter of Rectangle Problems Worksheet and know the various strategies to solve problems in an easy and understandable way.

Also Read :

• Perimeter and Area of Rectangle
• Practice Test on Area and Perimeter of Rectangle

Perimeter and Area of a Rectangle – Definitions

A Rectangle is a quadrilateral with two equal sides and two parallel lines and four right angles. Four right angles vertices are equal to 90 degrees, it is also called an equiangular quadrilateral.

The perimeter of the rectangle is defined as the sum of all the sides of the rectangle.  Rectangle has two lengths and breadths, it is denoted by P, it is measured in units. For finding the perimeter of the rectangle we have to add the length and breadth.

Perimeter of the Rectangle, P = 2(l + b)

The area of the rectangle is defined as to calculate the length and breadth of the two- dimensional closed figure. For finding the area of the rectangle we have to multiply the length and breadth, it is denoted by A, measured in square units.

Area of the rectangle , A = l x b

Problems on Area and Perimeter of the Rectangle

1. Find the Area and Perimeter of the following rectangles whose dimensions are :

(i) length = 15 cm             breadth = 12 cm

(ii) length = 7.9 m            breadth = 6.2 m

(iii) length = 4 m              breadth = 36 cm

(iv) length = 2 m              breadth = 6 dm

Solution:

(i) Given, length = 15 cm, breadth = 12 cm

we know that, Perimeter of rectangle = 2 (length + breadth)

substitute the given values in above formula, we get

Perimeter of rectangle = 2 (15 + 12) cm

= 2 × 27 cm

= 54 cm

We know that, area of rectangle = length × breadth

Therefore, substituting the  values in above formula, we get

Area of rectangle = 15 cm x 12 cm

= (15 × 12) cm²

= 195 cm²

Therefore, Area of rectangle is 195 cm²

(ii) Given, length = 7 m, breadth = 6.2 m

we know that, Perimeter of rectangle = 2 (length + breadth)

substitute the values in the formula , we get

Perimeter of rectangle = 2 (7.9 + 6.2) m

= 2 × 14. 1 m

= 28. 2 m

We know that, area of rectangle = length × breadth

Therefore, substituting the value we get,

Area of rectangle = 7.9 m x 6.2 m

= (7.9 × 6.2) m²

= 48. 98m²

Therefore, Area of rectangle = 48. 98 m²

(iii) Given, length = 4 m

breadth = 36 cm = 36/ 100 = 0. 36 m ( cm is converted to m)

we know that, Perimeter of rectangle = 2 (length + breadth)

substitute the values in the formula, we get

Perimeter of  a rectangle  = 2 (4 + 36) m

= 2 × 40 m

The perimeter of a rectangle is 80 m

We know that, area of rectangle = length × breadth

substituting the value we get,

Area of rectangle  = 4 m× 36 m

= (4 x 36) m²

= 144 m²

Therefore, Area of a rectangle is 144 m²

(iv) Given, length = 2 m

breadth = 60 dm

1m = 10 dm

so we get, 60 dm =  6 m ( dm is converted to m)

we know that, Perimeter of rectangle = 2 (length + breadth)

substitute the values in the formula, we get

Perimeter of rectangle  = 2 (2 + 6) m

= 2 × 8 m

= 16 m

We know that, area of rectangle = length × breadth

substituting the value we get,

Area of rectangle  = 2 m× 6 m

= (2x 6) m² = 12 m²

Therefore, the Area of a rectangle is  12m².

---

2. The perimeter of the rectangle is  140 cm. If the length of the rectangle is 30 cm, find its breadth and area of the rectangle?

Solution:

Given, Perimeter of the rectangle is, 140 cm

The length of the rectangle is, 30 cm

we know that, Perimeter of the rectangle = 2(l + b)

substitute the value in the above formula, we get

140 = 2( 30 + b)

70 = 30

40 = b

Therefore, breadth = 40 cm

Now, Area of Rectangle = length x breadth

= 30 x 40 = 120  cm²

Therefore, the Area of a rectangle is 120 cm²

---

3. The area of a rectangle is 78 cm². If the breadth of the rectangle is 6 cm, find its length and perimeter?

Solution:

Given, Area of a rectangle is 78 cm²

The breadth of the rectangle is 6 cm

we know that, Area of rectangle = length x breadth

substitute the given value, we get

78 cm = length x 6 cm

78/ 6 = length

Length of the rectangle =12 cm

Now, perimeter of rectangle = 2 (l + b)

substitute the value, we get

Perimeter of rectangle = 2(12 + 6)

= 2 x 18

= 36 cm

Therefore, the perimeter of the rectangle = 36 cm

---

4. How many boxes whose length and breadth are 9 cm and 5 cm respectively are needed to cover a rectangular region whose length and breadth are 420 cm and 90 cm?

Solution:

Given,  Length of the box is 9 cm

The breadth of the box is 5 cm

Region length is 420 cm

Region breadth is 90 cm

we know the formula,

The area of a rectangle is l x b

Therefore, Area of region = l x b

substitute the value, we get

Area of region = 420 cm x 90 cm

= 37800 cm²

Again use the  area of a rectangle formula,

Area of one box is = 9 cm x 5 cm

= 45 cm²

Number of boxes = Area of region /Area of one box = 37800/45 = 840

Thus, 840 boxes are required.

---

5. If it costs $500 to fence a rectangular park of length 40 m at the rate of $25 per m², find the breadth of the park and its perimeter. Also, find the area of the field?

Solution:

Given, Cost of Rectangular park fencing is $500

Length of the  rectangular park = 20 m

Rate of fencing 1 m² = $25

Area of a rectangle = l x b

Now we find  area , therefore  Area = 500/ 25 = 20

substitute the  value  in formula, we get

20 = 20 x breadth

breadth = area / length

b = 20 / 20 = 1 m

Now finding the Perimeter,

Perimeter of a rectangle = 2 (l + b)

substituting the values ,

Perimeter  of a rectangle=  2 (20 + 1)

=  2 (21)

Therefore, the Perimeter of a rectangle =  41 m

---

6. A rectangular tile has a length equals to 20 cm and a perimeter equals 70 cm. Find its width?

Solution:

Given, Perimeter of the tile = 80 cm

Length of the tile = 20 cm

Let W be the width of the tile

we know that,

Perimeter of a rectangle = 2(length + width)

Substituting the values, we get,

The perimeter of a tile = 80 cm

Therefore, 80 = 2 (20 + Width)

80/ 2 = 20 + Width

40 = 20 + Width

40 – 20 = Width

Therefore, Width = W = 20.

---

7. Find the area of a rectangle, Perimeter of a rectangle, and diagonal of a rectangle whose length and breadth 12 cm and 16 cm respectively.

Solution:

Given, length of the rectangle = 12 cm

Breadth of the rectangle = 16 cm

we know the formulae,

Area of a rectangle = l x b

substitute the values in the above formula, we get

Area of a rectangle = 12 x 16 = 192 cm²

we know, Perimeter of a rectangle = 2 (l + b)

substitute the values, we get

Perimeter of a rectangle = 2 (12 + 16)

= 2 (192) = 384 m

Now, we finding the diagonal of a rectangle

The diagonal of a rectangle is d² = l² + b²

substitute the values, we get

d² = (12)² + (16)²

d² = (12 + 16)²

d = √(12 + 16)²

square and root both will be cancelled,

d = 12 + 16 = 28

Therefore, the Diagonal of a rectangle = 28 cm.

---

8. Find the cost of tiling a rectangular plot of land 200 m long and 120 m wide at the rate of $6 per hundred square m?

Solution:

Given,

Cost of tiling rectangular plot of land 200 m long and 120 m wide

The cost of tiling per 100 sq.m is $6

we know the area of a rectangle formula,

Area of a rectangle = length  x breadth

substituting the values in the above formula, we get

Area of a rectangle = 200 m x 120 m

= 24000 m²

Therefore, the Area of a rectangle is 24000 m²

Now, we finding the total cost of tiling

Total cost of tiling =  (6 x 24000) / 100

= 144000/100

=  $1440

Therefore, the Total cost of tiling is $1440

---

9. The length of a rectangular board is thrice its width. If the width of the board is 140 cm, find the cost of framing it at the rate of $5 for 30 cm.

Solution:

Given, the width of the board = 140 cm

length of the board is thrice

so ,length = 3 x width

length =  3 x 140 = 420 cm

30 cm rate is $5

Circumference of rectangle = 2 ( l+ b)

substitute the values in the above formula, we get

Circumference of rectangle = 2 ( 420 + 140)

= 2 x 560 = 1120 cm
Therefore, the circumference of rectangle = 1120 cm

Now, 30 cm cost is equal to rs. 5

So, 1 cm = 5/ 30

But, we want the cost of framing

So, 1120 = (5 x 1120)/ 20 =  rs. 280

Therefore, the cost of framing is Rs. 280

---

10. The Perimeter of a rectangular pool is 46 meters. If the length of the pool is 16 meters, then find its width. Here the perimeter and length of the rectangular pool are given. we have to find the width of the pool.

Solution:

Given,

The perimeter of a rectangular pool is 46 meters

The length of the pool is 16 meters

Now we find the width of the pool.

we know the formula,

Perimeter of a rectangle = 2(l + b)

substituting the values, we get

46 = 2( 16) +2( w)

46= 32 + 2w

46 – 32 = 2W

14 = 2W

W= 14/2 = 7 meters

Therefore, the width of the Pool is 7 meters.

---

11. The sides of a rectangle are in the ratio of 4: 5 and its perimeter is 90 cm. Find the dimensions of the rectangle and hence its area.

Solution:

Given, Perimeter of a rectangle is 90 cm

Length of the sides = 4 : 5

Let the common ratio be X

So the sides will be 4X and 5X

we know that,

The Sum of all sides of the rectangle is equal to the perimeter.

so, Perimeter of a rectangle = 2 ( length + breadth)

substituting the values, we get

90 = 2(l) + 2(b)

90 = 2(4X) + 2 (5X)

90 = 8X + 10 X

18 X = 90

Therefore, X = 90/18 = 5

Hence , length = 4X  and breadth = 5X

substitute the ‘X’ value, we get

length = 4(5) = 20 , Breadth = 5(5) = 25

Now we find the area of a rectangle,

Area of a rectangle = length x breadth

substitute the values in the formula, we get

Area of a rectangle = 20  cm x 25 cm

= 500 cm²

Therefore, the Area of a rectangle is 500 cm².

---

A quadrilateral can be defined as a closed geometric, two-dimensional shape having 4 straight sides. It has 4 vertices and angles. The types of quadrilaterals are parallelograms, squares, rhombus, and rectangle. The sum of all interior angles of a quadrilateral is equal to 360°. The angle is formed when two line segments meet at a common point. The angle can be measured in degrees or radians. The angles of a quadrilateral are the angles formed inside the closed shape.

Sum of Angles of a Quadrilateral Theorem & Proof

The sum of interior angles of a quadrilateral is 360 degrees.

In the quadrilateral ABCD

∠ABC, ∠ADC, ∠DCB, ∠CBA are the interior angles

AC is the diagonal of the quadrilateral

AC splits the quadrilateral into two triangles ∆ABC and ∆ADC

We know that sum of angles of a quadrilateral is 360°

So, ∠ABC + ∠ADC + ∠DCB + ∠CBA = 360°

Let’s prove that sum of all interior angles of a quadrilateral is 360 degrees.

We know that the sum of angles in a triangle is 180°

In triangle ADC

∠CAD + ∠DCA + ∠D = 180° —- (i)

In the triangle ABC

∠B + ∠BAC + ∠BCA = 180° —- (ii)

Add both the equations

∠CAD + ∠DCA + ∠D + ∠B + ∠BAC + ∠BCA = 180° + 180°

∠D + (∠CAD + ∠BAC) + (∠BCA + ∠DCA) + ∠B = 360°

We can see that ∠CAD + ∠BAC = ∠DAB, ∠BCA + ∠DCA = ∠BCD

So, ∠D + ∠DAB + ∠BCD + ∠B = 360°

∠D + ∠A + ∠C + ∠B = 360°

Therefore, the sum of angles of a quadrilateral is 360°

Quadrilateral Angles Sum Propoerty

Each quadrilateral has 4 angles. The sum of its interior angles is always 360 degrees. So, we can find the angles of the quadrilateral if we know the remaining 3 angles or 2 angles or 1 angle and 4 sides. For a square or rectangle, the value of all angles is 90 degrees.

Also, Read

• What is a Quadrilateral?
• Construct Different Types of Quadrilaterals
• Different Types of Quadrilaterals

Examples on Quadrilateral Angles

Example 1:

Find the fourth angle of the quadrilateral if three angles are 85°, 100°, 60°?

Solution:

The given three angles of a quadrilateral are 85°, 100°, 60°

We know that the sum of angles of a quadrilateral is 360°

So, ∠A + ∠B + ∠C + ∠D = 360°

85° + 100° + 60° + x° = 360

245° + x° = 360°

x° = 360° – 245°

x° = 115°

Therefore, the fourth angle of the quadrilateral is 115°.

Example 2:

Find the measure of the missing angles in a parallelogram if ∠A = 75°?

Solution:

We know that the opposite angles of a parallelogram are equal.

So, ∠C = ∠A, ∠B = ∠D

Sum of angles is 360°

∠A + ∠B + ∠C + ∠D = 360°

75° + ∠B + 75° + ∠D = 360°

150° + ∠B + ∠D = 360°

∠B + ∠D = 360° – 150°

∠B + ∠D = 210°

∠B + ∠B = 210°

2∠B = 210°

∠B = \(\frac { 210° }{ 2 } \)

∠B = 105°

So, other angles of a parallelogram are 105°, 75°, 105°.

Example 3:

The angle of a quadrilateral are (3x + 2)°, (x – 3)°, (2x + 1)°, 2(2x + 5)° respectively. Find the value of x and the measure of each angle?

Solution:

The given angles are ∠A = (3x + 2)°, ∠B = (x – 3)°, ∠C = (2x + 1)°, ∠D = 2(2x + 5)°

We know that the sum of angles of a quadrilateral is 360°

∠A + ∠B + ∠C + ∠D = 360°

(3x + 2)° + (x – 3)° + (2x + 1)° + 2(2x + 5)° = 360°

3x + 2 + x – 3 + 2x + 1 + 4x + 10 = 360

10x + 10 = 360

10x = 360 – 10

10x = 350

x = \(\frac { 350 }{ 10 } \)

x = 35

The measurement of each angle of a quadrilateral is ∠A = (3x + 2)° = (3(35) + 2) = 105 + 2 = 107°

∠B = (x – 3)° = (35 – 3) = 32°

∠C = (2x + 1)° = (2(35) + 1) = 70 + 1 = 71°

∠D = 2(2x + 5)° = 2(2(35) + 5) = 2(70 + 5) = 2(75) = 150°

Example 4:

The three angles of a closed 4 sided geometric figure are 20.87°, 53.11°, 8.57°. Find the fourth angle?

Solution:

The given angles are ∠A = 20.87°, ∠B = 53.11°, ∠C = 8.57°

We know that the sum of angles of a quadrilateral is 360°

∠A + ∠B + ∠C + ∠D = 360°

20.87° + 53.11° + 8.57° + x° = 360°

82.55° + x° = 360°

x = 360 – 82.55

x = 277.45°

Therefore, the fourth angle of the closed 3 sided geometric figure is 277.45°.

FAQs on Sum of Angles of a Quadrilateral

1. What is the sum of the internal angles of a quadrilateral?

The sum of angles of a quadrilateral is 360 degrees.

2. What are the properties of a quadrilateral?

The three different properties of a quadrilateral are it has four sides, four vertices, four angles. And it is a closed 2-dimensional geometric figure. The sum of within angles is 360 degrees.

3. How do you prove the angle sum property of a quadrilateral?

To prove the sum property of a quadrilateral, draw a diagonal to divide it into two triangles. The sum of all interior angles of a triangle is 180 degrees. thus, the sum of angles of a quadrilateral becomes 360°.

4. What is the sum of all interior angles of a pentagon?

Draw one diagonal that should divide the pentagon into one triangle, one quadrilateral. The sum of angles of a triangle is 180 degrees, the sum of angles of a quadrilateral is 360 degrees. So, the sum of all interior angles of a pentagon is 180 + 360 = 540°.

Do you need assistance in memorizing 7 table and solving the multiplication problems? Then, stay on this page. The 7 Times Table is used to find the difficult math concepts like square roots, GCF, LCM, HCF, and others. So, learning and remembering the 7 Times Table Multiplication Chart is very important. On our site, students can discover all basic information about the Math Tables like how to learn, how to read, how to write, and tips, tricks to memorize 7 Table.

Multiplication Table of Seven | 7 Times Table Chart

Multiplication Table of 7 Chart is given here in an image format for a better understanding and quick reference of students. So, you can easily download 7 Times Multiplication Chart from here and try to memorize it regularly for doing quick calculations in competitive exams. Moreover, you can also have a quick revision of the Seven Multiplication Table by taking a printed copy and pasting it on your room wall.

Importance of Learning 7 Times Table Chart

Here you will get an answer for why one should learn the multiplication chart of 7.

• Learning Multiplication Table of 7 helps you to solve mental math problems easily. This table can be quite handy while you solve real-world problems.
• This multiplication table saves your time while performing division and multiplication questions.

How to Read 7 Times Table Multiplication Chart?

Zero times seven is 0.

One time seven is 7.

Two times seven is 14.

Three times seven is 21.

Four times seven is 28.

Five times seven is 35.

Six times seven is 42.

Seven times seven is 49.

Eight times seven is 56.

Nine times seven is 63

Ten times seven is 70.

7 Times Multiplication Table up to 20

Multiplication Table of 7 is the easiest multiplication table to remember. Refer to the below-mentioned 7 Times Table Multiplication Chart up to 20 and understand how to write it mathematically. By seeing the Seven Multiplication Table, you can study and grasp the multiplication facts about the table easily. You can also improve your math skills and speed in answering the math problems in exams. Make use of 7 Times Tabular format and do fast calculations.

7	x	0	=	0
7	x	1	=	7
7	x	2	=	14
7	x	3	=	21
7	x	4	=	28
7	x	5	=	35
7	x	6	=	42
7	x	7	=	49
7	x	8	=	56
7	x	9	=	63
7	x	10	=	70
7	x	11	=	77
7	x	12	=	84
7	x	13	=	91
7	x	14	=	98
7	x	15	=	105
7	x	16	=	112
7	x	17	=	119
7	x	18	=	126
7	x	19	=	133
7	x	20	=	140

Check More Math Multiplication Tables

Multiplication Table of 0	Multiplication Table of 1	Multiplication Table of 2
Multiplication Table of 3	Multiplication Table of 4	Multiplication Table of 5
Multiplication Table of 6	Multiplication Table of 8	Multiplication Table of 9
Multiplication Table of 10	Multiplication Table of 11	Multiplication Table of 12
Multiplication Table of 13	Multiplication Table of 14	Multiplication Table of 15
Multiplication Table of 16	Multiplication Table of 17	Multiplication Table of 18
Multiplication Table of 19	Multiplication Table of 20	Multiplication Table of 21
Multiplication Table of 22	Multiplication Table of 23	Multiplication Table of 24
Multiplication Table of 25

Tips & Tricks to Memorize Multiplication Table of 7

• Seven number has infinite multiples and can be multiplied by any whole number.
• You can also get multiples of 7 by skip counting by 7.
• As 7 is a prime number it doesn’t repeat itself unit 7 x 10.

Solved Example Questions on 7 Table

Example 1:

What does 7 x 13 mean? What number is equal to?

Solution:

7 x 13 means 7 times 13

7 x 13 = 91

91 is equal to 7 x 13.

Example 2:

Varsha has 7 glasses. She puts 7 straws in each glass. How many straws are there in all?

Solution:

As Varsha puts 7 straws in each glass,

The total number of straws = 7 x 7

= 49.

Example 3:

Using the multiplication table of 7, find the value of 7 times 7 minus 4?

Solution:

Expressing the given statement in the form of mathematical expression we get

(7 x 7) – 4 = 49 – 4

= 45.

A Decimal Number consists of a whole number part and a fractional part separated by a decimal point. If you wish to learn completely about Decimal Places and How to Learn to Count Decimal Places. Know about Decimal Place Value Chart Definition, Facts, Solved Examples in the further modules. We are sure you will be familiar with the Decimal Places by the end of this article.

What is meant by Decimal Places?

A Decimal Number consists of both whole number part and decimal number part. The digits right to the decimal point are called the Decimal Part and the digits left to the decimal point are called the whole number part. The Number of Digits present in the decimal part of the given decimal number is known as Decimal Places.

Based on the position of the digit in the number it has a value named place value. For Example, the Place Value of the digit 2 in 1234.45 is 200 as 2 is in the hundreds place. However, if you interchange the digits 3 and 2 we get a new number i.e. 1324.45. In 1324.45 the place value of a digit is 20 as it is in the tens place.

How to Learn to Count Decimal Places?

The number of digits present in the decimal part of the given decimal number is nothing but the Decimal Places. Check out the below-listed examples to understand how to read and count the decimal places for the given decimal number.

For Example:

Decimal Number 5.34 has 2 decimal places.

Decimal Number 0.376 has 3 decimal places.

The Number 86.261 has 3 decimal places.

The Number 912.67 has 2 decimal places.

To better understand the Decimal Numbers you need to be aware of the Place Value.

For Example Decimal Number 51.048053 Place Value is explained clearly below.

In the above-illustrated example, 51 is the whole number part and 048053 is the decimal or fractional part.

Whole Number Part

Place of 1 is Ones and its place value is 1

Place of 5 is Tens and its place value is 50

Decimal Number Part

Place of 0 is Tenths and its place value is 0*\(\frac { 1 }{ 10 } \) = 0

Place of 4 is hundredths and its place value is 4*\(\frac { 1 }{ 100 } \) = \(\frac { 4 }{ 100 } \) = 0.04

Place of 8 is Thousandths and its place value is 8*\(\frac { 1 }{ 1000 } \) = \(\frac { 8 }{ 1000 } \) = 0.008

Place of 0 is Ten Thousandths and its place value is 0*\(\frac { 1 }{ 10,000 } \) = 0

Place of 5 is Hundred Thousandths and its place value is 5*\(\frac { 1 }{ 100,000 } \) = \(\frac { 5 }{ 100,000 } \) = 0.00005

Place of 3 is Millionths and its place value is 3*\(\frac { 1 }{ 1000000 } \) = \(\frac { 3 }{ 1000000 } \) = 0.000003

= 5*10+1*1+0*10^-1+4*10^-2+8*10^-3+0*10^-4+5*10^-5+3*10^-6

= 51.048053

FAQs on Decimal Places

1. What are Decimal Places?

Decimal Places are nothing but the number of digits next to the decimal point or in the decimal part.

2. How do you find the Decimal Places?

Firstly, count the number of digits after the decimal point and the number itself tells the Decimal Places for a particular decimal number.

3. How many decimal places are there in the Decimal Number 32.4356?

The number of Decimal Places in the Decimal Number 32.4356 is 4.

Before recalling about Decimal Fraction let us learn the fundamentals of what is a fraction. A fraction is formed up of two parts namely numerator and denominator. A Decimal Fraction is a Fraction having a denominator of 10 or multiples of 10 such as 100, 1000, 10000, …., etc. This article helps you to be well versed with Decimals Fractions such as Definitions, Facts, Operations performed on Decimal Fractions, Solved Examples.

What is a Decimal Fraction?

A Decimal Fraction is a Fraction in which the denominator is a power of 10 such as 10, 100, 1000, etc. You can write Decimal Fractions with a Decimal Point instead of a Denominator. By expressing the decimal fractions using decimal point calculations of addition, subtraction, multiplication, and division will be much simpler.

Examples:

\(\frac { 13 }{ 100 } \) = 0.13

\(\frac { 54 }{ 10 } \) = 5.4

Also, See:

• Conversion of a Decimal Fraction into a Fractional Number
• Decimals
• Converting Decimals to Fractions

Operations on Decimal Fractions

Addition and Subtraction of Decimal Fractions: Given Numbers are placed under each other so that the decimal points lie in each column and below one another. Later, the numbers are added and subtracted in a regular way.

For Example:

Add 0.0045 and 3.0423

The Sum of 0.0045 and 3.0423 is 3.0468

Multiplication of Decimal Fractions: Multiply the given numbers without considering the decimal point. After that, the decimal point is marked off to obtain the decimal places that is the sum of decimal places in the given numbers.

For Example 0.3*0.03*0.003

Multiply the numbers without considering the decimal point

3*3*3 = 27

Now find the number of decimal places to be marked by adding the decimal places in the given numbers i.e. (1+2+3) = 6 i.e. 0.000027

Dividing Decimal Fraction by a Counting Number: Divide the given number without considering the decimal point by counting the number. Later, after obtaining the quotient place as many decimal places are there in the dividend. If we were to divide 0.0028÷7 we will firstly divide 28÷7 and the quotient is 4. As there are 4 decimal places in the given number place the same in quotient obtained i.e. 0.0004

Thus, 0.0028÷7 = 0.0004

Dividing Decimal Fraction by a Decimal Fraction: Multiply both the dividend and divisor with suitable powers of 10 so that you can make them as a whole number and then proceed.

Thus, \(\frac { 0.00077 }{ 0.11 } \) = \(\frac { 0.00077*100 }{ 0.11*100 } \)

= \(\frac { 0.077 }{ 11 } \)

= 0.007

Solved Examples on Decimal Fractions

1. Convert (i) 0.60 and (ii) 4.008 into vulgar fractions?

Solution:

(i) 0.60 = \(\frac { 60 }{ 100 } \) = \(\frac { 3 }{ 5 } \)

(ii) 4.008 = \(\frac { 4008 }{ 1000 } \) = \(\frac { 501 }{ 125 } \)

In order to convert decimal into vulgar fractions firstly place 1 in the denominator and annex with as many zeros as the number of digits after the decimal point in the given number. Later, remove the decimal point and note the whole number in the numerator. Reduce it to Lowest Form. Remember Annexing Zeros to the Right of Decimal Fraction doesn’t change the value.

2. Add 35.2 + 4.098?

Solution:

Given Numbers are placed under each other so that the decimal point lies in one column. Numbers can be subtracted or added in a usual way.

3. Evaluate i) 5.3029×100

Solution:

i) 5.3029×100

To Multiply a Decimal Fraction by a Power of 10 shift the decimal point to the right as many places of decimal that is to the power of 10.

5.3029×100 = 530.29

4. Find the product

i) 2.232×0.1

Simply multiply the numbers without considering the decimal point. i.e. 2232*1= 2232

Later, count the number of decimal places in the given numbers i.e. (3+1) = 4

Now put the decimal after 4 that counts 4 digits from right thus 0.02232

5. Evaluate 0.81÷ 9?

Solution:

Divide the given number as if there is no decimal point. 81÷9 = 9

After obtaining the quotient count the number of decimal places in the given number i.e. 0.81 =2

Hence place the decimal point on the left of 0.09

Looking for ways on How to Add Mixed Fractions? If so, halt your search as we have listed all about Mixed Fractions Addition and different methods of it clearly in the later modules. Mixed Fractions are one of the types of fractions. These are also called Mixed Numbers. Go through the entire article to be well versed with the details like Adding Mixed Fractions Definition, How to Add Mixed Fractions with Same and Different Denominators, Examples, etc.

Mixed Fraction – Definition

A Mixed Fraction is a form of a fraction that has a whole number next to a fraction.

Example: 3 \(\frac { 1 }{ 5 } \) where 3 is a whole number and \(\frac { 1 }{ 5 } \) is a fraction.

How to Add Mixed Numbers?

When it comes to Adding Mixed Fractions we can have either the same or different denominators for both the fractions to be added. There are two different methods for Adding Mixed Numbers with Like, Unlike Denominators. Here is a Step by Step Procedure on How to Add Mixed Fractions. They are as such

Method 1: Adding the Whole Numbers and Fractions Separately

• In the first step add the whole numbers separately.
• In order to add fractions with the same denominator, simply add the numerators and keep the denominator unaltered. However, if you have different or unlike denominators take the LCM of them and change to Like Fractions.
• Once, you have a Common Denominator adding fractions is much simpler.
• Find the Sum of Whole Numbers and Fractions in Simplest Form.

Method 2: Convert Mixed Numbers to Improper Fractions and then Add them

• Initially, change the given Mixed Fractions to Improper Fractions.
• If Denominators of the Improper Fractions are the same simply add the numerators. If the Denominators of Improper Fractions are Unlike or Different take the LCM of Denominators and change them to like fractions.
• Add Like Fractions and express the sum to its Simplest Form.

Also, Check:

• Like and Unlike Fractions
• Equivalent Fractions
• Types of Fractions

Examples on Adding Mixed Fractions using Method 1

1. Add 3 \(\frac { 1 }{ 3 } \), 2 \(\frac { 1 }{ 4} \)?

Solution:

3 \(\frac { 1 }{ 3 } \)+ 2 \(\frac { 1 }{ 4} \)

Let us add the whole numbers and fraction parts separately i.e.

Whole Numbers Part 3+2 = 5

Fractions Part = \(\frac { 1 }{ 3 } \)+ \(\frac { 1 }{ 4} \)

Since the denominators of the fractions are not same find the LCM of the Denominators to make them like fractions

LCM(3,4) = 12

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

= \(\frac { 4 }{ 12 } \) + \(\frac { 3 }{ 12 } \)

= \(\frac { (4+3) }{ 12 } \)

= \(\frac { 7 }{ 12 } \)

Now add the like fractions and express the sum to its simplest form

= 5 \(\frac { 7 }{ 12 } \)

Therefore, 3 \(\frac { 1 }{ 3 } \), 2 \(\frac { 1 }{ 4} \) when added gives 5 \(\frac { 7 }{ 12 } \)

2. Add 5 \(\frac { 1 }{ 4} \), 2 \(\frac { 1 }{ 5} \), \(\frac { 1 }{ 6 } \)?

Solution:

5 \(\frac { 1 }{ 4} \) + 2 \(\frac { 1 }{ 5} \) + \(\frac { 1 }{ 6 } \)

Firstly, let us add the whole numbers and fraction parts separately i.e.

Whole Numbers Part (5+2+0) = 7

Fractions Part = \(\frac { 1 }{ 4} \) + \(\frac { 1 }{ 5} \) + \(\frac { 1 }{ 6 } \)

Since the Denominators of Fractions aren’t the same find the LCM of Denominators and express them as like fractions

LCM(4, 5, 6) = 60

= \(\frac { (1*15) }{ (4*15) } \) + \(\frac { (1*12) }{ (5*12)} \) + \(\frac { (1*10) }{ (6*10) } \)

= \(\frac { 15 }{ (60) } \) + \(\frac { 12 }{ 60} \) + \(\frac { 10 }{ 60 } \)

= \(\frac { (15+12+10) }{ 60 } \)

= \(\frac { 37 }{ 60 } \)

Now add the like fractions and express the sum to its simplest form

= 7 \(\frac { 37 }{ 60 } \)

Therefore, 5 \(\frac { 1 }{ 4} \), 2 \(\frac { 1 }{ 5} \), \(\frac { 1 }{ 6 } \) when added gives 7 \(\frac { 37 }{ 60 } \)

Adding Mixed Fractions Examples using Method 2

1. Add 5 \(\frac { 1 }{ 4 } \), 3 \(\frac { 1 }{ 2 } \)?

Solution:

5 \(\frac { 1 }{ 4 } \) + 3 \(\frac { 1 }{ 2 } \)

Change the given Mixed Numbers to Improper Fractions

= \(\frac { (5*4+1) }{ 4 } \) +  \(\frac { (3*2+1) }{ 2 } \)

= \(\frac { 21 }{ 4 } \) +  \(\frac { 7 }{ 2 } \)

Since the Denominators aren’t same find the LCM and express them as Like Fractions

LCM(4,2) = 2

= \(\frac { (21*1) }{ 4*1 } \) +  \(\frac { (7*2) }{ (2*2) } \)

= \(\frac { 21 }{ 4 } \) +  \(\frac { 14 }{ 4 } \)

= \(\frac { (21+14) }{ 4 } \)

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

Thus, 5 \(\frac { 1 }{ 4 } \), 3 \(\frac { 1 }{ 2 } \) added results in \(\frac { 35 }{ 4 } \)

2. Add 6 \(\frac { 1 }{ 4 } \), 7 \(\frac { 1 }{ 4 } \), 3 \(\frac { 1 }{ 4 } \)?

Solution:

6 \(\frac { 1 }{ 4 } \) + 7 \(\frac { 1 }{ 4 } \) + 3 \(\frac { 1 }{ 4 } \)

Change the given Mixed Fractions to Improper Fractions

= \(\frac { (6*4+1) }{ 4 } \) +  \(\frac { (7*4+1) }{ 4 } \) + \(\frac { (3*4+1) }{ 4 } \)

= \(\frac { 25 }{ 4 } \) + \(\frac {29 }{ 4 } \) + \(\frac { 13 }{ 4 } \)

= \(\frac { (25+29+13) }{ 4 } \)

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

Thus, 6 \(\frac { 1 }{ 4 } \), 7 \(\frac { 1 }{ 4 } \), 3 \(\frac { 1 }{ 4 } \) results in \(\frac { 67 }{ 4 } \)

FAQs on Adding Mixed Fractions

1. What is a Mixed Fraction with Example?

A Mixed Fraction is a form of a fraction that has a whole number next to a fraction. For Example 6 \(\frac { 1 }{ 4 } \) is a Mixed Fraction where 6 is a whole number and \(\frac { 1 }{ 4 } \) is the fraction part.

2. What does a Mixed Fraction look like?

Mixed Fraction is simply an improper fraction written as the sum of a whole number and a proper fraction. For instance, improper fraction \(\frac { 5 }{ 2 } \) can be written as Mixed Fraction 2 \(\frac { 1 }{ 2 } \).

3. How to add Mixed Fractions Step by Step?

Follow the simple and easy steps listed below to add Mixed Fractions and they are as such

• Convert the given Mixed Fractions to Improper Fractions
• Find the LCM of Denominators and then make them like fractions.
• Add the Like Fractions and Express the Sum to its Simplest Form.

In this article, you will learn about the Decimal and Fractional Expansion of a Decimal Number. Before Proceeding further know the definitions of Decimal, Fraction, and the Place Value Chart. A Decimal is any number in the base 10 number system and is used to separate units place from tenths place in decimal. The decimal point present in between separates the Whole Number Part and Decimal Part.

Also, Read:

• Thousandths Place in Decimals
• Place Value
• Decimal Place Value Chart

Decimal Expansion of a Number

Decimal Expansion of a Number is its representation in the base-10 system. In this System, each decimal place consists of digits 0-9 arranged such that each digit is multiplied by a power of 10 decreasing from left to right and a decimal place with 10^0 is the one’s place.

Decimal Expansion of Number 1423.25 is defined as

1423.25 = 1*10³+4*10²+2*10¹+3*10⁰+2*10^-1+5*10^-2

= 1000+400+20+3+0.2+0.05

Decimal Expansion of a Number may terminate and in such case, the number is called a regular number or finite decimal. At times, the Decimal Expansion of a Number may become periodic and in such case, it is called a repeating decimal. However, the expansion may continue infinitely without repeating and it is called an irrational number.

Fractional Expansion of Decimals

In the Expanded Form of Decimal Fractions, you will learn how to read and write the Decimal Numbers. Decimal Numbers can be written in expanded form using the Place Value Chart.

Let us understand the same by considering an example

384.264

384.264 = 3 × 100 + 8 × 10 + 4 × 1 + 2 × \(\frac { 1 }{10 } \) + 6 × \(\frac { 1 }{100 } \) + 4 × \(\frac { 1 }{1000 } \)

= 300+80+4+\(\frac { 2 }{10 } \)+\(\frac { 6 }{100 } \)+\(\frac { 4 }{1000 } \)

Solved Examples on Decimal and Fractional Expansion

1. Write the decimal and fractional expansion of 334.252?

Solution:

In Decimal Expansion

334.252 = 3*100+3*10+4*1+2*\(\frac { 1 }{10 } \)+5*\(\frac { 1 }{100 } \)+2*\(\frac { 1 }{1000 } \)

= 300+30+4+\(\frac { 2 }{10 } \)+\(\frac { 5 }{100 } \) + \(\frac { 2 }{1000 } \)

= 300+30+4+0.2+0.05+0.002

In Fractional Expansion

= 3*100+3*10+4*1+2*\(\frac { 1 }{10 } \)+5*\(\frac { 1 }{100 } \)+2*\(\frac { 1 }{1000 } \)

= 300+30+4+\(\frac { 2 }{10 } \)+\(\frac { 5 }{100 } \) + \(\frac { 2 }{1000 } \)

2. Write the decimal and fractional expansion of 543.32?

Solution:

In Decimal Expansion

543.32 = 5*100+4*10+3*1+3*\(\frac { 1 }{10 } \)+2*\(\frac { 1 }{100 } \)

= 500+40+3+\(\frac { 3 }{10 } \)+\(\frac { 2 }{100 } \)

= 500+40+3+0.3+0.02

In Fractional Expansion

= 5*100+4*10+3*1+3*\(\frac { 1 }{10 } \)+2*\(\frac { 1 }{100 } \)

= 500+40+3+\(\frac { 3 }{10 } \)+\(\frac { 2 }{100 } \)

3. Write the Decimal and Fractional Expansion of 647.345?

Solution:

In Decimal Expansion

647.345 = 6*100+4*10+7*1+3*\(\frac { 1 }{10 } \)+4*\(\frac { 1 }{100 } \)+5*\(\frac { 1 }{1000 } \)

= 600+40+7+\(\frac { 3 }{10 } \)+\(\frac { 4 }{100 } \)+\(\frac { 5 }{1000 } \)

= 600+40+7+0.3+0.04+0.005

In Fractional Expansion

647.345 = 6*100+4*10+7*1+3*\(\frac { 1 }{10 } \)+4*\(\frac { 1 }{100 } \)+5*\(\frac { 1 }{1000 } \)

= 600+40+7+\(\frac { 3 }{10 } \)+\(\frac { 4 }{100 } \)+\(\frac { 5 }{1000 } \)

FAQ’s on Decimal and Fractional Expansion

1. What is Decimal in Expanded Form?

Expanded form notation for the decimal numbers is the mathematical expression that shows the sum of the values of each digit in the number.

2. What is the Decimal Expansion of Number 164.38?

Decimal Number 164.38 can be written in expanded form by writing it as the sum of the place value of all the digits i.e. 1*100+6*10+4*1+3*\(\frac { 1 }{10 } \)+8*\(\frac { 1 }{100 } \) = 100+60+4+0.3+0.08

3. What is the Fractional Expansion of Number 94.38?

Fractional Expansion of Number 94.38 is 9*10+4*10+3*\(\frac { 1 }{10 } \)+8*\(\frac { 1 }{100 } \) which inturn results in 90+40+\(\frac { 3 }{10 } \)+\(\frac { 8 }{100 } \)