Removal of Brackets | Expanding and Removing Brackets for Simplifying Expressions

Removal of brackets involves step by step process. We are providing a detailed procedure and various rules to follow while removing the brackets. Also, check the various methods to remove different types of brackets. Rewrite the expression by simplifying it with the help of the brackets removal function. Know the process of writing in equivalent forms by removing the brackets. Refer to the definitions of “expanding” or “removing” brackets.

How to Remove Brackets from an Equation?

There are various steps involved in bracket removal. In the further sections, we will see how the equations can be written in equivalent forms by removing the braces. This process is called “removing” or “expanding” brackets.

The actual procedure to remove the brackets is to multiply the term which is outside the brackets with the term which is inside the brackets. This is also called the law of distribution.

Steps to Brackets Removal

In Order to take off brackets, you need to follow the step by step guidelines listed below and they are as follows

Step 1:

Check for the given question or the expression whether it contains a vinculum(The horizontal line which is used in the mathematical notation for the desired purpose) or not. If there is a vinculum, then perform the operations in it. Or else go to step 2

Step 2:

Now, go for the equation and check for the innermost bracket and then perform the operation within that bracket.

Step 3:

To perform step 3, there are various rules to be followed.

Rule 1: If the equation is preceded by addition or plus sign, then remove it by writing the terms as mentioned in the equation.

Rule 2: If the equation is preceded by a negative or minus sign, then change the negative signs within it to positive or vice versa.

Rule 3: The indication of multiplication is there will be no sign between a grouping symbol and a number.

Rule 4: If there is any number before the brackets, then we multiply that number inside the brackets with the number which is outside the brackets.

Step 4:

Check for the next innermost bracket and perform the calculations or operations within it. Remove the next innermost bracket by using Step III rules. Follow this process until all the brackets are removed.

Removing Brackets by FOIL Method

Brackets can be easily removed and rules can be remembered by using the FOIL Method. Whenever multiplication is necessary outside the brackets, multiply each of the terms in the first bracket by each of the terms in the second bracket.

To avoid confusion, we apply the FOIL Method and make the calculations simple further.

In FOIL, F indicates First, O indicates Outside, I indicates Inside, L indicates Last

First: Multiply the initial terms of each bracket i.e., the first term of the first bracket with the first term in the second bracket.

Outside: Outside indicates multiplying the 2 outside terms i.e., the first term in the first bracket with the second term in the second bracket.

Inside: Multiplication of 2 inside terms inside Inside in FOIL Method i.e., the second term in the first bracket with the first term in the second bracket.

Last: Last indicates multiplication of last terms from both the brackets i.e., the second term in the first bracket with the second term in the second bracket.

Example:

(x+5)(x+10)

= (x+5)x + (x+5)10

x2+ 5x + 10x + 50

= x2+15x + 50

Removing Brackets Questions

Consider an algebraic expression which contains parentheses or round brackets ( ), square brackets [ ], curly brackets { }.

5{[4(y-4)+15] – [2(5y-3)+1]}

Step 1:

Innermost grouping symbols removal is the first step to simplify the expression. Consider (y-4), to remove the brackets we multiply the equation (y-4) by 4, here comes as 4y-16.

Step 2:

Follow the same procedure to remove parentheses in 2(5y-3), we have multiplied the equation (5y-3) by 2, here comes the final equation as 10y-6

After performing 2 steps, the result equation will be

5{[4y-16+15] – [10y-6+1]}

Step 3:

Now, the equation is further simplified and it contains only square brackets and curly brackets, we should perform all the set of operations within two brackets.

5{[4y-16+15] – [10y-6+1]}

=5{[4y-1]-[10y-5]}

Step 4:

As the brackets of two sets are removed. Therefore, the negative sign before the 2nd set implies for all the terms present in the 2nd set. Therefore, it is multiplied by -1

5{[4y-1]-[10y-5]}

=5{4y-1-10y+5}

Step 5:

As all other brackets are removed and the expression contains only curly braces. Perform all the operations that are possible within the brackets.

5{4y-1-10y+5}

=5{-6y+4}

Step 6:

To the open curly braces, apply the distributive law.

5{-6y+4}

=-30y+20

Example 2:

Simplify the equation 95 – [144 ÷ (12 x 12) – (-4) – {3 – 17 – 10}]

Solution:

Step 1:

Innermost grouping symbols removal is the first step to simplify the expression. Consider {3-17-10}, to remove the brackets we subtract the equation {3-17-10}, here comes as {3-7}

Step 2:

Follow the same procedure to remove parentheses in (12 x 12), we have multiplied the equation (12 x12), here comes the final equation as 144

After performing 2 steps, the result equation will be

95 – [144 ÷ 144 – (-4) – {3-7}]

Step 3: 

Now, the equation is further simplified and it contains only square brackets and curly brackets, we should perform all the set of operations within two brackets.

95 – [1 – (-4) – (-4)]

Step 4:

As the brackets of two sets are removed. Therefore, change the negative signs and rewrite the equation.

= 95 – [1 + 4 + 4]

Step 5:

As all other brackets are removed and the expression contains only square braces. Perform all the operations that are possible within the brackets.

= 95 – 9

Step 6:

Therefore, after subtraction the final solution is

=86

Example 3:

Simplify the equation 197 – [1/9{42 + (56 – 8 + 9)} +108]

Solution:

Step 1:

Innermost grouping symbols removal is the first step to simplify the expression. Consider {56-8+9}, to remove the brackets we subtract the equation {56-8+9}, here comes as {56-17}

= 197 – [1/9 {42 + (56 – 17)} + 108]

Step 2:

Follow the same procedure to remove parentheses in {42 + (56 – 17)}, we have subtracted the equation (56-17), here comes the final equation as {42 + 39}

After performing 2 steps, the result equation will be

197 – [1/9 {42 + 39} + 108]

Step 3:

Now, the equation is further simplified and it contains only square brackets and round brackets, we should perform all the set of operations within two brackets.

197 – [(81/9) + 108]

Step 4:

Now, simplify the equation (81/9), then the result will be 9. The final equation will be

197 – [9 + 108]

Step 5:

As all other brackets are removed and the expression contains only square braces. Perform all the operations that are possible within the brackets.

= 197 – 117

Step 6:

To get the final result, subtract 117 from 197, therefore the final result will be 80

How to Expand Brackets?

There are few different methods to expand brackets and simplify expressions. We have explained each of them in detail and even took expanding brackets examples for explaining the entire process. They are as such

1. Multiplying two bracketed terms together

If we have a situation to multiply two bracketed terms, then we multiply each term in the first bracket with each term in the second bracket. These types of multiplications lead to quadratic expressions.

Example:

(x+5)(x+10)

=(x+5)x + (x+5)10

2. Dealing with nested brackets

These are the collection of expressions nested in various sets of brackets.

Example:

Simplify -{5x-(11y-3x)-[5y-(3x-6y)]}

-[5x-(11y-3x)-[5y-3x+6y]}

-{5x-11y+3x-5y+3x-6y}

-{11x-22y}

-11x+22y

Quadrilateral- Definition, Types, Properties, Formulas, Notes

Different Geometry shapes and objects are named based on the number of sides. If an object has three sides, then it is classified as Triangle, An object with 4 sides classified as Quadrilateral, etc. Let us learn about the Quadrilateral definition, types, formula, properties, etc. in detail in this article. Every concept is explained separately on our website. Access every topic and easily get a grip on the Quadrilateral concept.

List of Quadrilateral Concepts

Find different concepts of Quadrilateral by checking out the below links. All you need is simply tap on them to have an idea of the related concept.

Quadrilateral Definition

A quadrilateral defined as a figure that has four sides or edges. Also, the quadrilateral consists of four vertices. rectangle, square, trapezoid, and kite, etc. are some of the examples of Quadrilateral.

Types of Quadrilaterals

There are various types of Quadrilaterals available. All the Quadrilaterals must have 4 sides. Also, the sum of the angles of the Quadrilateral is 360 degrees.

  1. Trapezium
  2. Kite
  3. Parallelogram
  4. Rectangle
  5. Squares
  6. Rhombus

Also, the quadrilaterals are classified differently. They are
Convex Quadrilaterals: It is defined as both diagonals of a quadrilateral are always present within a figure.
Concave Quadrilaterals: Concave Quadrilaterals one diagonals present outside of the figure.
Intersecting Quadrilaterals: The pair of non-adjacent sides intersect in Intersecting Quadrilaterals. These are also called self-intersecting or crossed quadrilaterals.

Quadrilateral Formula

Check out the below formula of a Quadrilateral.

Area of a Parallelogram = Base x Height
Area of a Square = Side x Side
Area of a Rectangle = Length x Width
Area of a Kite = 1/2 x Diagonal 1 x Diagonal 2
Area of a Rhombus = (1/2) x Diagonal 1 x Diagonal 2

Quadrilateral Properties

Know the different properties of a Quadrilateral PQRS.

  • Four sides: PQ, QR, RS, and SP
  • ∠P and ∠Q are adjacent angles
  • Four vertices: Points P, Q, R, and S.
  • PQ and QR are the adjacent sides
  • Four angles: ∠PQR, ∠QRS, ∠RSP, and ∠SPQ.
  • ∠P and ∠R are the opposite angles
  • PQ and RS are the opposite sides

Important Properties of Quadrilateral

  • Every quadrilateral consists of 4 sides, 4 angles, and 4 vertices.
  • Also, the total of interior angles = 360 degrees

Properties of a Square

  • The sides of a square are parallel to each other.
  • Also, all the sides are equal in measure.
  • The diagonals of a square perpendicular bisect each other.
  • All the interior angles of a square are at 90 degrees.

Rectangle Properties

  • The diagonals of a rectangle bisect each other.
  • The opposite sides consist of equal length in a rectangle.
  • All the interior angles of a rectangle are at 90 degrees.
  • The opposite sides are parallel to each other

Properties of a Rhombus

  • By adding two adjacent angles of a rhombus we get 180 degrees.
  • The opposite sides are parallel to each other in a rhombus.
  • The diagonals perpendicularly bisect each other
  • All four sides of a rhombus are of equal measure.
  • The opposite angles are of the same measure.

Parallelogram Properties

  • The opposite angles of a parallelogram are of equal measure.
  • The opposite side is of the same length in a parallelogram.
  • The sum of two adjacent angles of a parallelogram is equal to 180 degrees.
  • The opposite sides are parallel to each other in a Parallelogram.
  • The diagonals of a parallelogram bisect each other.

Trapezium Properties

  • The two adjacent sides are supplementary in a trapezium.
  • Only one pair of the opposite side is parallel to each other in a trapezium.
  • The diagonals of a trapezium bisect each other in the same ratio

Kite Properties

  • The large diagonal bisects the small diagonal of a kite.
  • The pair of adjacent sides have the same length in a kite.
  • Only one pair of opposite angles are of the same measure.

Notes on Quadrilateral

  • A quadrilateral is a parallelogram is 2 pairs of sides are parallel to each other.
  • Also, a quadrilateral is a trapezoid or a trapezium if two of its sides are parallel to each other.
  • A quadrilateral is a rhombus if all the sides are of equal length and the two pairs of sides are parallel to each other.
  • The quadrilateral becomes kite when 2 pairs of adjacent sides are equal to each other.
  • Quadrilateral becomes Square and Rectangle when all internal angles are right angles, all angles are right angles, and also the opposite sides of a rectangle are same.
  • Furthermore, the Quadrilateral becomes Square and Rectangle when Opposite sides of a rectangle and square are parallel. The sides of a square are of the same length.

Construct Different Types of Quadrilaterals | How to Construct Quadrilaterals of Different Types?

Do you want to know How to construct different types of quadrilaterals? Different types of quadrilaterals are developed depending on the sides, diagonals, and also angles. Have a look at a step by step explanation to construct various types of quadrilaterals. We have given different problems on the construction of quadrilaterals along with steps for better understanding. Look at them and practice all the problems given below and enhance your conceptual knowledge.

How to Construct Quadrilaterals? | Steps of Construction

You can refer to the below available various questions on constructing quadrilaterals along with a detailed explanation. For the sake of your comfort, we even jotted Steps of Construction for each and every problem so that you can solve similar kinds of questions easily.

1. Construct a parallelogram PQRS in which PQ = 7 cm, QR = 5 cm and diagonal PR = 7.8 cm.

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.
construction of quadrilateral 13

1. Draw a line segment of length 7 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 7.8 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 5 cm. Mark the point as R where the two arcs cross each other. Join the points Q and R as well as P and R.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point P as a center, draw an arc with a radius of 5 cm.
5. By taking the point R as a center, draw an arc with a radius of 7 cm.
6. Mark the point as S where the two arcs cross each other. Join the points R and S as well as P and S.

PQRS is a required parallelogram.

construction of quadrilateral 14

2. Construct a parallelogram, one of whose sides is 7.2 cm and whose diagonals are 8 cm and 8.4 cm.

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.
construction of quadrilateral 15

1. Draw a line segment of length 7.2 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 4.2 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 4 cm. Mark the point as O where the two arcs cross each other. Join the points Q and O as well as P and O.
4. By taking the point O as a center, draw an arc with the required radius.
5. Produce PO to R such that OR = PO and produce QO to S such that OS = OQ.
6. Join PS, QR, and RS.

PQRS is a required parallelogram.

3. Construct a parallelogram whose diagonals are 5.6 cm and 6.4 cm and an angle between them is 70°.

Steps of Construction:
1. Draw a line segment of length 5.6 cm and mark the ends as P and R.
2. Take the point O as a center in between P and R.
3. Next, take point O as a center and make a point by taking 70º using a protector. Draw a line XO to Y.
4. Set off OQ = 1/2 (6.4) = 3.2 cm and OS = 1/2 (6.4) =3.2 cm as shown.
5. Join PQ, QR, RS, and SP.
construction of quadrilateral 16

PQRS is a required parallelogram.

4. Construct a rectangle PQRS in which side QR = 5.2 cm and diagonal QS = 6.4 cm.

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.
construction of quadrilateral 17

1. Draw a line segment of length 5.2 cm and mark the ends as Q and R.
2. Take the point R as a center and draw a perpendicular line to QR.
3. Next, take point Q as a center and draw an arc by taking the radius 6.4 cm. Mark the point as S where the line and arc cross each other. Join the points Q and S as well as R and S.
4. By taking the point S as a center, draw an arc with the required radius of 5.2 cm.
5. Take the point Q as a center and draw a perpendicular line to QR. Mark the point as P where the point and arc cross each other. Join the points Q and P as well as P and S.

PQRS is a required rectangle.

construction of quadrilateral 18

5. Construct a square PQRS, each of whose diagonals is 5.4 cm.

Steps of Construction:
1. Draw a line segment of length 5.4 cm and mark the ends as P and R.
2. Draw the right bisector XY of PR, meeting PR at O.
3. From O set off OQ = 1/2 (5.4) = 2.7 cm along OQ and OS = 2.7 cm along OX.
4. Join PQ, QR, RS, and SP.

PQRS is a required square.
construction of quadrilateral 19

6. Construct a rhombus with a side of 4.4 cm and one of its angles equal to 67°.

Steps of Construction:
Given that a rhombus with a side of 4.4 cm and one of its angles equal to 67°.
The adjacent angle = (180° – 67°) = 113°.
1. Draw a line segment of length 4.4 cm and mark the ends as Q and R.
2. Make ∠RQX = 113° and ∠QRY = 67°.
3. Set off QP = 4.4 cm along with QX and RS = 4.4 cm along with RY.
4. Join PS.

PQRS is a required rhombus.

construction of quadrilateral 20

 

Calculating Profit Percent and Loss Percent | Profit and Loss Problems and Solutions

The profit and loss are the basic components of the income statement that summarizes the revenues, costs, and expenses subjected during a certain period. We use basic concepts in the calculation of profit percentage and loss percentage. Find Formulas for Calculating Profit % and Loss % and Problems on the Same in the forthcoming modules.

Cost Price (CP)

The price at which we purchase an item is called the cost price. In short, it is written as CP.

Example: A merchant bought an item at a cost of Rs. 150 and gain a profit of Rs. 30.

In the above example, Rs.150 is the cost price.

Selling Price (SP)

The price at which we sell an item is called the selling price. In short, it is written as SP.

Example: A shop keeper purchased a pen at Rs. 40 and he sold it at Rs.50 to a customer.

In the above example Rs. 50 is the selling price.

Profit or Gain

If the selling price of an item is more than the cost price of the same item, then it is said to be gain (or) profit i.e. S.P. > C.P.

Net profit= S.P. – C.P.

Loss

If the selling price of an item is less than the cost price of the same item, then it is said to be a loss i.e. S.P.  < C.P.

Net loss= C.P. – S.P.

NOTE:

It is important to note that the profit or loss is always calculated based on the cost price of an item.

Profit and Loss Formulas for Calculating Profit % and Loss%

Have a glance at the Profit and Loss Formulas for finding the Profit and Loss Percentage. They are along the lines

  • Net Gain = (S.P.) – (C.P.)
  • Net Loss = (C.P.) – (S.P.)
  • Gain % = (S.P. – C.P./C.P. *100)% = (gain/C.P. *100)%
  • Loss % = (C.P. – S.P./C.P. *100)%  = (loss/C.P. * 100)%
  • To find S.P. when C.P. and gain% or loss% are given :
    • P. = [(100 + Gain %) /100] * C.P.
    • P. = [(100 – Loss %) /100] * C.P.
  • To find C.P. when S.P. and gain% or loss% are given :
    • P. = [(100 + Gain %) /100] * S.P.
    • P. = [(100 – Loss %) /100] * S.P.

Profit and Loss Percent Questions and Answers

Question 1:

A man buys a book for Rs. 60 and sells it for Rs. 90. Find his gain/loss percentage?

Solution:

By seeing the question we can understand that the Selling price of the book is more than its cost price, therefore the man has profited on his total transaction.

Given data:

Cost price (C.P.) = Rs.60

Selling price (S.P.) =Rs. 90

Net profit = S.P. – C.P.

= 90 – 60

= 30

Profit % = ((Net profit)/C.P. *100)

= (30/60 *100)

= 50%

The total gain percentage is 50%.

Question 2:

If a fruit vendor purchases 9 oranges for Rs.8 and sells 8 oranges for Rs. 9. How much profit or loss percentage does he makes?

Solution:

By seeing the question we can understand that he bought 9 oranges at cost prices Rs.8 and sells 8 oranges at Rs. 9.

Given data:

Buying 9 oranges at Rs. 8

And Selling 8 oranges at Rs. 9

Here we are making quantities equal by multiplying the prices on both sides

We get,

Quantity            Price

Buying                 9 * 8                8 * 8

Selling                 8 * 9                9 * 9

By calculating,

Quantity            Price

Buying                 72                      64

Selling                 72                       81

By observing the above calculation we can see that the vendor bought 72 oranges for Rs. 64 while he sold 72 oranges at Rs. 81. This shows that the vendor is having profit in the entire transaction.

Profit percentage = ((S.P. – C.P.)/C.P. * 100)

= ((81-64)/64 * 100)

= 26.56 %

The total gain percentage for fruit vendors selling oranges is 26.56%.

 

 

Word Problems on Profit and Loss | Profit and Loss Questions with Solutions

The word problems on profit and loss are solved here to get the basic idea of how to use the formulae of profit and loss in terms of cost price and selling price. We have explained the entire concept of profit and loss and various formulae when C.P, S.P, Profit %, or Loss % is given. Solve Different Questions on Profit and Loss available here to test your grip on the fundamentals of the concept.

Profit or Gain

If the selling price of an item is more than the cost price of the same item, then it is said to be gain (or) profit i.e. S.P. > C.P.

Net profit= S.P. – C.P.

Loss

If the selling price of an item is less than the cost price of the same item, then it is said to be a loss i.e. S.P.  < C.P.

Net loss = C.P. – S.P.

Profit and Loss Word Problems with Answers

Question 1:

A laptop was brought for $ 80,000 and sold at a loss of $ 5000. Find the selling price.

Solution:

Given data:

The cost price of the laptop is $ 80,000

Loss = $ 5000

We know that,

Loss = C.P. – S.P.

$ 5000 =$ 80,000 – S.P.

S.P. = $ 80,000 – $ 5000

S.P. = $ 75,000

Therefore, the selling price of the laptop is $ 75,000.

Question 2:

Abhi sold his water purifier for $ 4000, at a loss of $ 300. Find the cost price of the water purifier.

Solution:

Given Data:

The selling price of water purifier = $ 4000

Loss = $ 300

We know that, Loss = C.P. – S.P.

From this, we can note that,

Cost price = loss + selling price

= $ 300 + $ 4000

= $ 4300

Hence, the cost price of a water purifier is $ 4300.

Question 3:

Deepika sold her gold necklace for $ 60,000 at a profit of $ 10,000. Find the cost price of the gold necklace.

Solution:

Given data

The selling price of gold necklace = $ 60,000

Gained a profit = $ 10,000

From the formula

Gain = Selling price (S.P.) – Cost price (C.P.)

We get,

Cost price (C.P.) = Selling price (S.P.) – Gain

= $ 60,000 – $ 10,000

= $ 50,000.

Hence, the cost price (C.P.) of the gold necklace is $ 50,000.

Question 4:

Karthik buys a watch for $ 6000 and sells it at a gain of 5⅓ %. For how much does he sell it?

Solution:

Given Data:

Cost price (C.P.) of watch = $ 6000

Gain = 5⅓% = 16/3 %

We know that

Gain% = ((S.P. – C.P.)/C.P. *100) %

From above,

S.P. = [{(100 + gain %) /100) * C.P.]

= $ [{(100 + 16/3)/100} * 6000]

= ${(103.33/100) * 6000]

= $ 6199.8

Hence, karthik sells his watch at an amount of $ 6199.8.

Question 5:

Siva ram bought an old bike for $ 15000 and spends $ 2000 on repairs. If he sells the bike for $ 21150, what is his gain percentage?

Solution:

Given data:

Cost price (C.P.) of bike = $ 15000

Repair cost = $ 2000

Total Cost Price = Original Price of the Bike + Repair Cost

= $ 15000+$2000

= $17000

Selling price (S.P.) of bike = $ 21150

As the selling price (S.P.) is more than the cost price (C.P.) of the bike then it is said to be in gain

Therefore, Gain = Selling price (S.P.) – Cost Price (C.P.)

= $ 21150 – $ 17000

= $ 4150.

Gain % = ((S.P. – C.P.)/C.P. *100) %

= $ (4150/17000 * 100) %

= 24. 41%

Hence, He got a 24.41% gain on his bike.

Question 6:

If the selling price of an object is doubled, the profit of the object triples. Find the profit percentage.

Solution:

Given data:

Let the cost price of the object be $ ‘a’

And selling price of the object be $ ‘b’

According to the question, the profit is tripled and selling price is doubled hence

Profit = $ 3(b – a)

Profit = S.P. – C.P.

3(b – a) = 2b –a

3b – 3a = 2b – a

Therefore, b = 2a.

Profit =$ b – a

= 2a –a

= $ a.

Profit% = ((S.P. – C.P.)/C.P. *100) %

= (a/a * 100) %

= 100%.

Question 7:

The percentage profit earned by selling an article for $ 3000 is equal to the percentage loss incurred by selling the same article for $ 2500. At what price should the article be sold to make a 20% profit?

Solution:

The above question says that the % profit earned by selling the article is equal to the % loss incurred

by the same article.

Given data:

Let the C.P. of the article be ‘P’

We know that

Profit % = ((S.P. – C.P.)/C.P. *100) %

And loss % = ((C.P. – S.P.)/C.P. *100) %

((3000 – p)/p * 100) = ((p – 2500)/p * 100)

2p = 7500

P = $ 3750.

Calculating selling price at a profit of 20%

Profit % = ((S.P. – C.P.)/C.P. *100) %

From above,

S.P. = $ [{(100 + gain %) /100) * C.P.]

= $ ((100 + 20)/100) * 3750)

= $ 4500.

Question 8.

If mangoes are bought at prices ranging from $ 300 to $ 450 are sold at prices ranging from $ 400 to $ 525, what is the greatest possible profit that might be made in selling ten mangoes?

Solution:

The question says that the mangoes are bought at a certain range and sold at a certain range. It says to find the greatest profit on selling ten mangoes.

Given data:

Cost price (C.P.) of mangoes ranging from = $ 300 – $ 450

Selling price (S.P.) of mangoes ranging from = $ 400 – $ 525

Considering,

Least cost price (C.P.) for ten mangoes = $ 300*10

= $ 3000

Greatest selling price (S.P.) of mangoes = $ 525*10

= $ 5250

Profit = S.P. – C.P.

= $ 5250 – $ 3000

= $ 2250.

Hence, the required profit obtained is $ 2250 on selling mangoes.

Question 9:

On selling 18 toys at $ 800, there is a loss equal to the cost price of 7 toys. The cost price of the toys is?

Solution:

Given data:

Let the cost price (C.P.) of the toys be ‘m’

Given, on selling 18 toys at $ 800, there is a loss equal to the cost price of 7 toys

According to the question, the equation is written as:

18m – 800 = 7m

Solving the above equation

We get m = $ 32

Therefore, the cost price of the toy is $ 32.

 

Uses of Brackets | How to Use Brackets in Math? | Brackets Rule(DMAS Rule)

Uses of Brackets are given here. Find the various uses and types of brackets. Before knowing the complete details regarding the brackets, know the fundamental operations like addition, subtraction, multiplication, and division. Know the history, rules, and precedence order of brackets. Go through the below sections to know the complete details regarding the usage of brackets.

History of Brackets

The word bracket is derived from the French word “Braguette” which the “piece around”. Anything that is written in brackets is defined as the piece of that bracket. Most brackets are used to enclose notes, references, explanations, etc. which are called crotchets as per the typographical brackets. Later, brackets were used as the group bracketed together for equal standing in some of the graded systems which are mostly used for sports brackets.

Brackets Usage

Mathematical brackets are known as symbols and parentheses which are most often used to create groups or that which clarify the order in which operations are to be done in the given algebraic expression.

Brackets symbol             Name

            ( )                         Parentheses or common brackets

            { }                        Braces or Curly brackets

            [ ]                         Brackets or square brackets or box brackets

             _                          Vinculum

The Left Part of the Bracket indicates the start of the bracket and the right part indicates the end of the bracket.

While Writing mathematical expressions having more than one bracket, parenthesis is used in the innermost part followed by braces, and these two are covered by square brackets. You need to know about the Uses of brackets to perform a  set of operations prior to the others.

Types of Brackets, Braces, Paranthesis in Math

Mathematical brackets are used for grouping. These brackets can include:

  • ( )
  • [ ]
  • { }

In the grouping of numbers, brackets come in pairs. There will be a pair of sets i.e., the opening bracket and a closing bracket. Brackets are generally used to give clarity in the order of operations.

Suppose that there is an expression: 3+5*7-2. You cannot understand which operation to perform at the beginning. Therefore, we include brackets to understand the precedence of operations. If the problem is given as (3+5)*(7-2) = 8*5 = 40.

In the above problem, the parentheses will tell you the usual order of operations and will give you visual clarity.

“( ) Brackets”

The symbols “(” “)” are known as parentheses. These are called Brackets or Round Brackets. They are called Round Brackets as they are not curly or square braces. The input of the function is enclosed in parentheses. Parenthesis means “to put beside” in Geek language. Things like additional information, asides, clarifications, citations are defined by Paranteseis. Any type of information written in parentheses can be as short as a word or number or a few sentences. If something is given in parenthesis, then that sentence must have the capability of standing on its own.

Example: The little puppy (Rocky) skipped across the garden to her mother.

“Square Brackets [ ]”

The square brackets are denoted inside the parentheses to define something to the sub-ordinate clause. Square brackets are defined by the symbol “[ ]”. The main purpose of using square brackets is writing in the conjunction or to insert a name or word for clarification.

Square brackets are needed to add clarifications.

Example: She [Rosy] hit the policeman

They are also used to add additional information

Example: Two teams in the FIFA football final match were from South America [ Argentina and Uruguay]

Missing words can also be added with the help of square brackets.

Example: It is [a] good answer.

Authorial Comments or editorial can also be added using square brackets.

Example: There are not present [my emphasis]

Square brackets are also used to modify the direct quotation.

Example: The direct quotation is “I love traveling” which can be modified as ” He love[s] traveling”.

These can also be used for nesting.

Example: (We use square brackets [in this way] inside the round brackets).

“Curly Brackets { }”

Mostly curly brackets are not used for general purposes. They are mostly used in programming or math concepts. They are mainly used to hold terms or terms and hold a list of items.

Example:

3{2+[5(3+1)+4]}

This is the example of lists

Example of programming languages:

$value=0;

do{

$value++;

if ($value >=20)

{

Print(“Value is equal to or greater than 10. Ending.”);

exit;

}

}

until ($value >=200);

Curly brackets can also be used in sets in mathematics.

Grouping of numbers resembles Sets.

Example: {1,2,4,6,9}

Difference between ( ) and [ ] in math

Square brackets [ ] are used for commands whereas parenthesis or round brackets represent braces and to describe some special words.

Multiple Level of Grouping

If you planning for an equation, then grouping within other grouping is a little bit confusing. To avoid that confusion, we use various brackets and group the numbers with order precedence. If we are using the brackets, then we can define the level of operations within the equation.

Example:

2 + {1+[2+3*(5+4)]}

In the above example, first, we go for the inner calculations.

Therefore, first, we have to go for the grouping of (5+4) = 9

Then, we go for the multiplication of 3 and 9 i.e., 18

For the next order, we go for the addition of 3 with 18, the result will be 21.

The next precedence will be the addition of 1 and 21, the result will be 22.

Then, we go for the last grouping of 2 and 22, Therefore, the result of the addition is 24.

With the above example, it is clear that finding a way of solving the equation will be easy.

Questions on How to Use Brackets

Problem 1:

Simplify the expression : [(3*2) + (4*5)]/(7-3)

Solution:

Start solving the problem by simplifying the equation that is inside the parentheses i.e., 4 *5 = 20

Then, solve the equation of other parentheses, i.e., 3 * 2 = 6

Then simplify the equation within the square brackets i.e., [6 + 20]= 26

Now, solve the equation outside the parenthesis i.e., 7-3 = 4

Now finish the division of the two values = 26/4 = 6.5

Therefore, the final solution is 6.5

Problem 2:

Simplify the equation, [(3+2)*4 – (5*6)]/[1-(3+4)*2]

Solution:

Start solving the problem by simplifying the equation inside the parenthesis i.e., 5 *6 = 30

Then, solve the equation (3+2)*4 = %*4 = 20

Then, go for the next equation of 3+4 = 7, 7*4 = 28

Now, go for the other equation in the brackets i.e., 5*6=30

The second part of the equation is 3 + -7, then 7*2 = 14

Now, solve the equation of 1-14 = -13

Finally, divide the values -10/-13 = 0.769

Therefore, the final solution is -0.769

Thus, the three types of brackets are helpful for solving various equations. Check all the uses and importance of using the brackets and also know the preceding order.

Construction of Quadrilaterals | How to Construct a Quadrilateral? | Steps of Construction

Construction Of Quadrilaterals is easy if you have a complete grip on the concept. A quadrilateral is a polygon with 4 sides, 4 angles, and also 4 vertices. When you add the interior angles of a quadrilateral, then you can get 360 degrees. The quadrilateral side lengths and angles may different. Depending on the lengths and angles of the sides, you can easily know what is the name of the quadrilateral. You can easily construct a quadrilateral by considering the following criteria.

(i) 4 sides and 1 diagonal is given.
(ii) 3 sides and including 2 angles are given
(iii) 2 sides and three angles are given
(iv) 3 sided and 2 diagonals are given
(v) 4 sides and 1 angle is given

1. Construct the Quadrilateral when 4 sides and 1 diagonal is given
PQ = 5 cm
QR = 4.5 cm
RS = 3.8 cm
PS = 4.4 cm
Diagonal PQ = 6 cm

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.
construction of quadrilateral 1

1. Draw a line segment of length 5 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 6 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 4.5 cm. Mark the point as R where the two arcs cross each other. Join the points Q and R as well as P and R.
4. By taking the point P as a center, draw an arc with a radius of 4.4 cm.
5. By taking the point R as a center, draw an arc with a radius of 3.8 cm.
6. Mark the point as S where the two arcs cross each other. Join the points R and S as well as P and S.

The final result is the required quadrilateral.
construction of quadrilateral 2

2. Construct the Quadrilateral when 3 sides and including 2 angles are given
PQ = 3.8 cm
QR = 4.2 cm
PS = 5.2 cm
∠QPS = 180º
∠PQR = 80º

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.
construction of quadrilateral 6

1. Draw a line segment of length 3.8 cm and mark the ends as P and Q.
2. Take point P as a center and make a point by taking 180º using a protector.
3. Next, take point P as a center and draw an arc by taking the radius 5.2 cm. Mark the point as S where the point and arc cross each other. Join the points P and S.
4. By taking the point Q as a center, make a point by taking 80º using a protector.
5. By taking the point Q as a center, draw an arc with a radius of 4.2 cm.
6. Mark the point as R where the point and arc cross each other. Join the points Q and R.
7. Finally, join the points R and S and draw a line segment.

The final result is the required quadrilateral.

construction of quadrilateral 7

3. Construct the Quadrilateral when 3 sides and including 2 angles are given
AB = 4.7 cm
∠ABC = 120°,
BC = 4 cm,
∠BCD = 100°, and ∠BAD = 60°.

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.
construction of quadrilateral 8

1. Draw a line segment of length 4.7 cm and mark the ends as A and B.
2. Take point A as a center and make a point by taking 60º using a protector.
3. Next, take point B as a center and make a point by taking 120º using a protector. Mark the point as D where the two points are meet at a point. Join the points A and D.
4. By taking the point B as a center, draw an arc with a radius of 4 cm.
5. Mark the point as C where the point and arc cross each other. Join the points B and C.
6. Finally, join the points C and D and draw a line segment.

The final result is the required quadrilateral.

construction of quadrilateral 9

4. Construct the Quadrilateral when 3 sided and 2 diagonals are given
PQ = 4.2 cm
QR = 4 cm
PS = 3.2 cm
Diagonal PR = 5 cm
Diagonal QS = 4.6 cm

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.
construction of quadrilateral 3

1. Draw a line segment of length 4.2 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 5 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 4 cm. Mark the point as R where the two arcs cross each other. Join the points Q and R as well as P and R.
4. By taking the point P as a center, draw an arc with a radius of 3.2 cm.
5. By taking the point Q as a center, draw an arc with a radius of 4.6 cm.
6. Mark the point as S where the two arcs cross each other. Join the points R and S as well as P and S.

The final result is the required quadrilateral.
construction of quadrilateral 4

5. Construct the Quadrilateral when 4 sides and 1 angle is given.
PQ = 4 cm, QR = 3.6 cm, RS = 4.7 cm, PS = 5.2 cm and ∠B = 80°.

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.
construction of quadrilateral 11

1. Draw a line segment of length 4 cm and mark the ends as P and Q.
2. Take point Q as a center and make a point by taking 80º using a protector.
3. Next, take point Q as a center and draw an arc with a radius of 3.6 cm. Mark the point as R where the two points are meet at a point. Join the points Q and R.
4. By taking the point P as a center, draw an arc with a radius of 5.2 cm.
5. By taking the point R as a center, draw an arc with a radius of 4.7 cm.
6. Mark the point as S where the two arcs cross each other. Join the points P and S, R and S.

The final result is the required quadrilateral.

construction of quadrilateral 12

Examples on Fundamental Operations | Questions on Fundamental Operations

Examples on Fundamental Operations are here. Get step by step procedure and ways to apply the formulae to the problems. Know the various operations like addition, subtraction, division, multiplication problems. Get all the simple and easy tips to complete the problem without any confusion. Refer to all the problems in the below sections and also follow various methods to solve those problems.

Basic Mathematical Operations Examples

Before knowing the examples, first, you must know the important points of fundamental operations on integers. We are giving the complete details regarding the integer’s fundamental operations.

Important Points

  • The integers are defined by the numbers …,-4, -3,-2,-1, 0,1, 2, 3, 4…
  • 1, 2, 3, 4,… are called positive (+) integers and -1,-2,-3,… are called negative (-) integers. 0 is defined as neither a positive value nor a negative value.
  • Integer 0 is considered as a value that is less than every positive number and greater than every negative number.
  • The absolute value of an integer is considered as its numerical value of that integer without the consideration of its sign.
  • The absolute value of an integer is considered as either positive or zero which cannot be a negative number.
  • The sum of two integer numbers having the same sign is the sum of their absolute values with a positive sign.
  • The sum of two integer numbers having the opposite signs is the “difference” of their absolute values and indicates the sign of the greater absolute value.
  • To subtract an integer y from x, we change the sign of y and add, i.e., x + (-y)
  • The product of two integer numbers having the same sign is positive (+).
  • The product of two integer numbers having different signs is negative (-).
  • 2 integers, which when added give 0, are called the additive inverse of each other.
  • The additive inverse of zero is zero.

Four Fundamental Operations Problems

Problem 1:

Two lakh sixty-three thousand nine hundred fifty-three visitors visited the trade fair on Sunday, four lakh thirty-three thousand visited on Monday, and three lakh twenty thousand six hundred fifty-six visited on Tuesday. How many visitors in all visited the trade fair in these three days?

Solution:

As per. the given question,

Visitors visited the trade fair on Sunday = 2,63,953

Visitors visited the trade fair on Monday = 4,33,000

Vistors visited the trade fair on Tuesday = 3,20,656

To find the no of visitors in all visited the trade fair, we have to apply the addition fundamental operation.

Therefore, the total visitors on 3 days = 2,63,953 + 4,33,000 + 3,20,656

= 10,17,609

Thus, the final solution is 10,17,609 visitors

Problem 2:

From a book store, 12,685 books were bought for the primary section of the school library. 15,790 books were bought for the middle section; and 13,698 books for the senior section. What was the total number of books bought?

Solution:

As per the given question,

Number of books bought for Primary section = 12,685

Number of books bought for Middle Section = 15,790

Number of books bought for Senior Section = 13,698

To find, the total no of books bought, the fundamental operation of addition is to be applied

Therefore, the total number of books = 12,685+15,790+13,698 = 42,173

Thus, the final solution is 42,173 books

Problem 3:

The masons used 1,75,692 bricks for the construction of Mr. Sharma’s house, and 2,16,785 bricks for the construction of Mr. Verma’s house. For whose house the masons used more bricks and how many bricks were used?

Solution:

As per the given question,

Bricks used for Mr. Sharms’s house = 1,75,692

More bricks were used for the construction of Mr. Verma’s house.

To find the number of bricks that were used, we apply the fundamental operation of subtraction.

The number of more bricks used for Mr. Verma’s house = 2,16,785 – 1,75,692

= 41,093

Thus, the final solution is 41,093 bricks.

Problem 4:

In a month, 1,23,498 people travel from one station to another by metro. If the distance fare Rs.32 paid by each traveler, then how much money will be collected in a month?

Solution:

As per the question,

Number o0f people travel by metro = ,1,23,498

Distance fair paid by each traveler = Rs.32

To find the money, collected in a monthy, we apply the fundamental operation of multiplication.

The amount of money collected in a month = 1,23,498 x 32 = 39,51,936

Thus, the final solution is 39,51,936 Rs

Problem 5:
If a dictionary contains 3,215 pages and there are 215 words arranged on each page, then how many words are there in the whole dictionary?

Solution:

As per the given question,

Total number of pages in the dictionary = 3,215

Number of words on each page = 215

To find the number of words in the whole dictionary, we apply the fundamental operation of multiplication.

Therefore, the number of words in whole dictionary = 3,215 x 215 = 6,91,225 words

Thus, a total number of words in the dictionary =6,91,225 words.

Problem 6:

An organizer has 25,90,488 tickets to be equally sold among 358 singing concerts. How many people will be there in each singing concert?

Solution:

As per the given question,

Total number of tickets = 25,90,488

Number of singing concerts = 358

To find the number of people in each singing concert, we apply the fundamental operation of division.

Therefore, the number of people in each singing concert = 25,90,488 / 358 =7,236 people

Thus, the final solution is 7,236 people.

Problem 7:

A contractor sent 76,95,940 bricks for the construction of 70 chambers. If an equal number of bricks was required for each chamber, how many bricks were used for each chamber?

Solution:

As per the given question,

Total number of bricks = 76,95,940

Number of chambers = 70

To find the total number of bricks used for each chamber, we apply the fundamental operation of division.

Therefore, The total number of bricks used for each chamber = 76,95,940 / 70 =1,09,942 bricks.

Thus, the final solution is 1,09,942 bricks.

Problem 8:

Mohan bought a table for Rs 12,450 and a chair for Rs. 5,400. Find the total money spend by him on buying two items?

Solution:

As per the given equation,

Cost of the table = Rs 12,450

Cost of the chair = Rs 5,400

To find the total cost, we apply the fundamental operation of addition.

Therefore, the total cost of items = Rs 12,450 + 5,400 = 17,850

Thus, the total money spent by him in buying two items is Rs. 17,850

Hence, the final solution is 17,850 Rs.

Problem 9:

Rahul bought a laptop for Rs 1,02,500, video game for Rs 5,600 and furniture for Rs 72,500. How much did he spend in all?

Solution:

As per the given question,

The cost of the laptop = Rs 1,02,500

The cost of the video game = Rs 5,600

The cost of the furniture = Rs 72,500

To find the total cost, we apply the fundamental operation of addition

Therefore, total cost = 102500 + 5600 + 72500 =180600

Thus, the total amount spent in all is Rs 180600

Hence, the final solution is 180600 Rs.

Problem 10:

The cost of the toy car is Rs 7,300 and the cost of a toy scooter is Rs 5,286. Which toy is cheaper and by how much?

Solution:

As per the given question,

The cost of a toy car = Rs 7,300

The cost of the toy scooter = Rs 5,286

To find the cheapest toy, we apply fundamental operations for subtraction.

Hence, the cheaper toy = 7300-5286 =2014

Therefore, the amount of the toy that is cheap = 2014 Rs

Thus, the final solution = Rs 2014

 

Fundamental Operations in Mathematics | Four Basic Mathematical Operations for Simplifying Algebraic Terms

Wanna start your preparation from basics? Then here is the most basic and important concept. Learn all the Fundamental Operations and their usage in day to day life. Solve all the important problems involved in fundamental algebra operations. Know the definitions, rules, strategies, tips, tricks, and problems. Refer to the upcoming sections to get more information regarding mathematical expressions and operations.

Fundamental Operations – Introduction

Fundamental Operations are the basic concept to solve any type of problem. We perform the operations at one time and generally, it starts from the left towards the right. If the expression contains more than 1 fundamental operation, then you can’t perform them in the order they appear in the question. There are certain rules to follow to perform various operations when more than 1 fundamental value is available. The precedence order has to be followed to solve the Fundamental Operation. The precedence order will be shown in the below sections.

Order of Expression – DMAS Rule

The fundamental operations are expressed in an order. Generally, the order will be in the format of “DMAS” where “D” stands for “Division”, “M” stands for “Multiplication”, “A” stands for “Addition”, “S” stands for “Subtraction”. This order is considered and sequentially performed from left to right.

As there are fundamental operations in arithmetic, there are in a wat that three pairs of arithmetic operations, and for each pair of operations is the reversal of co-operation within the pairs. As addition and subtraction are reverse operations of each other. It is similar for other operations too. In regard to Algebra, everyone must understand that it is a set of rules based on fundamental operations which helps for faster calculations and an easy approach for variable problem-solving.

Basic Fundamental Operations to Simplify Mathematical Expressions

Addition

The addition is defined by the operator (+). It is the fundamental operation of the arithmetic operators. It implies the combination of distinct quantities or sets. It involves counting numbers from one to one by incrementing the values. The result value after adding all the numbers is called the “sum”. The numbers and the initial number are called addends.

The addition of natural numbers involves 2 important properties i.e., associativity and commutativity.

Addition Operation on Positive and Negative Integers

Positive (+) + Positive (+) = Positive (+)
Negative (-) + Positive (+) = Negative (-)
Positive (+) + Negative (-) = Sign of the largest Number
Negative (-) + Positive (+) = Sign of the largest number

Examples:
5+4 = 9
(-5) + (-4) = -9
(-5) + 4 = -1
4 + (-5) = -1
(-4) + 5 = -1
5 + (-4) = 1

Example Problem:

An empty water tank of 8 feet high is available. A monkey is sitting at the bottom of that tank. The monkey tried jumping to the top of the tank. While jumping it jumps 3 ft up and slides down to 2 ft. How many jumps will the monkey take to reach the top of the water tank?

Solution:

As per the given question,

Monkey’s first jump = 3 ft up and 2 ft down

= (+3)-2=1

2nd Jump of the Monkey = 3 ft up and 2 ft down = (1+3) = 4

= (+4)-2=2

Therefore, the monkey covers 2 ft after the 2nd jump

3rd Jump of the Monkey = 3 ft up and 2 ft down = (2+3) = 5

=(+5)-2 = 3

Therefore, the monkey covers 3 ft after the 3rd jump

4th Jump of the Monkey = 3 ft up and 2 ft down = (3+3) = 6

=(+6)-2 = 4

Hence, the monkey covers 4 ft after the 4th jump

5th Jump of the Monkey = 3 ft up and 2 ft down = (4+3)=7

=(+7)-2 = 5

Therefore, the monkey covers 5 ft after the 5th jump.

6th Jump of the Monkey = 3 ft up = (5+3) = 8

As the water tank is 8 feet high, the monkey will reach the top of the water tank in the 6th jump.

Subtraction

The subtraction is defined by the operator (-). Subtraction is the inverse property of addition. It means that removing one quantity from another quantity. It involves incrementing down from the given number. The result value after subtracting all the numbers is called the “difference”. The number from which other number is subtracted is called subtrahend. The subtracted number is called minuend.

Subtraction Operation on Positive and Negative Integers

Negative (-) – Positive (+) = Negative (-)
Positive (+) – Negative (-) = Positive (+)
Negative (-) – Negative (-) = Sign of the largest Number
Negative (-) + Positive (+) = Sign of the largest number

Examples:

(-5)-4 = (-5) + (-4) =- 9
5 – (-4) = 5 +4 = 9
(-5) – (-4) = (-5) + 4 = -1
(-4) – (-5) = (-4) + 5 = 1

Example Problem:

At a height of 3000 feet above sea level, a plane is flying. At some time, there is a submarine that is exactly above and floating 700 feet below sea level. Calculate the vertical distance using the concept of subtraction of integers?

Solution:

Height of the plane that is flying = 3000 feet

Depth of the submarine = -700 feet ( It is negative, as it is below the sea level).

To calculate the vertical distance, we use the subtraction of integers operation.

3000-(-700) = 3000 + 700 = 3700 feet

Therefore, the vertical distance is 3700 feet.

Multiplication

The multiplication is defined by the operator (“x” o “*”). Multiplication is the repeated property of addition. It means that a particular number is being repeatedly added to itself several times. It involves adding the number with the same number. The result value after multiplying all the numbers is called the “product”. The number to which another number is multiplied is called multiplicand. The multiplied number is called a multiplier.

Multiplication Operation on Positive and Negative Integers

Negative (-) x Positive (+) = Negative (-)
Positive (+) x Negative (-) = Negative (-)
Negative (-) x Negative (-) = Negative (-)
Positive (+) + Positive (+) = Positive (+)

Examples:

5×4 = 12

(-3) x (-4) = 12

(-3) x 4 = – 12

3 x (-4) = -12

Example Problems:

The submarine descends 40 feet per minute from sea level. Find the relation of the submarine with the sea level 5 minutes after it starts descending?

Solution:

The submarine below sea level =40 feet

As it is below sea level, it is negative. Therefore, it is -40.

After 5 minutes, the submarine is at = -40 x 5 = -200 feet

Therefore, the final solution is -200 feet.

Division

The division is defined by the operator (÷). The division is the inverse property of multiplication. It means that a particular number or quantity is split into equal or the same parts. It involves the splitting of numbers in the same proportions. The result value after dividing the numbers is called the “quotient”. The original number is called “dividend”. The number which is used for dividing into groups is called “divisor”. The final result is “quotient”.

Division Operation on Positive and Negative Integers

Negative (-) ÷ Positive (+) = Negative (-)

Positive (+) ÷ Negative (-) = Negative (-)

Negative (-) ÷ Negative (-) = Positive (+)

Positive (+) ÷ Positive (+) = Positive (+)

Examples:

20÷4 = 5

(-20)÷(-4) = 5

(-20) ÷ 4 = -5

20÷(-4) = -5

Example Problem:

In a living room, there are 120 books in total and they are placed on 6 shelves. Consider that each shelf has an equal number of books, Calculate the number of books on each shelf?

Solution:

As per the given question,

The total no of books = 120

No of seats placed on shelves = 6

To determine the number of books on each shelf = 120/6 =20

Therefore, the number of books on each shelf = 20 books.

Thus the final solution is 20 books.

Different Types of Quadrilaterals – Definition, Properties

Depending upon the length and angles, quadrilaterals are classifieds in different ways. Let us check Different Types of Quadrilaterals and their definition, properties along with their diagrams. A quadrilateral can be explained using the below properties

  • The sum of the interior angles is 360 degrees in a quadrilateral.
  • A quadrilateral consists of 4 sides and 4 vertices, and also 4 angles.
  • The sum of the interior angle from the formula of the polygon using (n – 2) × 180 where n is equal to the number of sides of the polygon.

The main types in a quadrilateral are squares and rectangles, etc., with the same angles and sides.

Various Types of Quadrilaterals

Mainly quadrilaterals are classified into six types. They are

  1. Parallelogram
  2. Rhombus
  3. Rectangle
  4. Square
  5. Trapezium
  6. Kite

Parallelogram

A quadrilateral is said to be a parallelogram when it has two pairs of parallel sides and opposite sides are parallel and equal in length. Also, the opposite angles are equal in a parallelogram. Let us take a parallelogram PQRS, then the side PQ is parallel to the side RS. Also, the side PS is parallel to a side QR.

Two diagonals are present in the parallelogram and they intersect each other at a midpoint. From the figure, PR and QS are two diagonals. Also, the diagonals are equal in length from the midpoint.

PQ ∥ RS and PS ∥ QR.

parallelogram 6

Rhombus

Rhombus is a quadrilateral when all the four sides of a quadrilateral having equal lengths. In a rhombus, opposite sides are parallel and opposite angles are equal.

Rhombus

From the above figure, PQRS is a rhombus in which PQ ∥ RS, PS ∥ QR, and PQ = QR = RS = SP.

Rectangle

A quadrilateral is considered as a rectangle when all 4 angles of it are equal and each angle is 90 degrees. Also, both pairs of opposite sides of a rectangle are parallel and have equal lengths.

rectangle

From the above figure, PQRS is a quadrilateral in which PQ ∥ RS, PS ∥ QR and ∠P = ∠Q = ∠R = ∠S = 90°.

So, PQRS is a rectangle.

Square

A square is a quadrilateral consists all the sides and angles are equal. Also, every angle of a square is 90 degrees. The pairs of opposite sides of a square are parallel to each other.

square

From the above figure, PQRS is a quadrilateral in which PQ ∥ RS, PS ∥ QR, PQ = QR = RS = SP and ∠P = ∠Q = ∠R = ∠S = 90°.

So, PQRS is a square.

Trapezium

A quadrilateral is called a trapezium when it has one pair of opposite parallel sides.

trapezium

From the above figure, PQRS is a quadrilateral in which PQ ∥ RS. So, PQRS is a trapezium. A trapezium its non-parallel sides are equal is called an isosceles trapezium.

Kite

A quadrilateral is said to be a kite that has two pairs of equal-length sides and the sides are adjacent to each other.

kite

From the above figure, PQRS is a quadrilateral. PQ = PS, QR = RS, PS ≠ QR, and PQ ≠ RS.

So, PQRS is a kite.

Important Points to Remember for Quadrilaterals

Look at some of the important points need to remember about a quadrilateral.

  • A square is a rectangle and also it becomes a rhombus.
  • The rectangle and rhombus do not become a square.
  • A parallelogram is a trapezium.
  • Square, rectangle, and rhombus are types of parallelograms.
  • A trapezium is not a parallelogram.
  • Kite is not a parallelogram.