When you want to draw a coordinate graph for the simple functions, reach this page. Here you will learn the complete details about graphs of simple functions quickly and with fewer efforts. Generally, a function is a process or a relation that associates with the elements. You can check the examples of various basic functions and functions which are multiples of numbers.

Common Functions

The list of common functions and their graphs are provided below.

1. Linear Function:

It is a straight line and the function which is in the form of y = mx + c is called the linear function.

2. Square Function:

The function which is in the form of f(x) = x² is known as the square function.

3. Cube Function:

The function which is in the form of y = x³ is called the cube function.

4. Square Root Function:

If the function is in the form of y = √x, then it is called the square root function.

5. Absolute Value Function:

For every function we have positive and negative value possibilities i.e f(x) = I x I is called the absolute value function.

6. Logarithmic Function:

A function that has log in it is known as the logarithmic function.

7. Reciprocal Function:

A function that is in the form of fraction and numerator value is 1 is called the reciprocal function.

8. Exponential Function:

The function that has an exponent is called the exponential function.

9. Sine Function:

A function that includes trigonometric sine is called the sine function.

10. Cosine Function:

The function which contains trigonometric cosine is known as the cosine function.

11. Tangent Function:

A function that has trigonometric tan is called the tangent function.

Steps to Draw Graph of Function

• First of all, take some function.
• Find the values of x for the corresponding values of y. That means substitute random values of x to get the y values.
• Place those values in a table.
• Plot the obtained points on a graph paper.
• If you wanted the respective function value at a particular point, then figure it out from the graph.

Solved Example Questions

1. Draw the graph of the function y = 5x. From the graph, find the value of y, when x = 0.5, 0.2.

Solution:

Given function is y = 5x

For some different values of x, the corresponding values of y are given below.

Plot the points P (0, 0), Q (1, 5), R (-1, -5) on the graph paper and join those points to form a straight line.

Reading values from the graph of sample function y = 5x

On the x-axis, take the point L at x = 0.5.

Draw LB ⊥ x-axis, meeting the graph at B.

Clearly, AL = 2.5 units.

Therefore, x = 0.5 ⇒ y = 2.5.

Take a point M at x = 0.2

Draw MA ⊥ x-axis, meeting the graph at A.

MA = 1 unit

Therefore, x = 0.2 ⇒ y = 1

2. Draw the graph of the function y = 2x². From the graph, find the value of y, when x = -1.5, 0.5.

Solution:

Given function is y = 2x²

For some different values of x, the corresponding values of y are given below.

Plot the points A (-1, 2), B (0, 0), C (1, 2), D (2, 8) on a graph and join those points.

Reading values from the graph of sample function y = 2x²

On the x-axis, take the point L at x = -1.5.

Draw LP ⊥ x-axis, meeting the graph at A.

Clearly, PL = 4.5 units.

Therefore, x = -1.5 ⇒ y = 4.5.

Take a point M at x = 0.5

Draw MB ⊥ x-axis, meeting the graph at AB

MA = 0.5 unit

Therefore, x = 0.5 ⇒ y = 0.5

3. Draw the graph of the function p(n) = 3n – 1. From the graph, find the value of p(n), when n = 2, 1.5

Solution:

Given function is p(n) = 3n – 1

For some different values of x, the corresponding values of y are given below.

Plot the points x (-1, -4), y (0, -1), z (1, 2) on the graph paper and join those points to form a straight line.

Reading values from the graph of sample function p(n) = 3n – 1

On the x-axis, take the point L at n = 2.

Draw LA ⊥ x-axis, meeting the graph at A.

Clearly, AL = 5 units.

Therefore, n = 2 ⇒ p(n) = 5

Take a point M at n = 1.5

Draw MB ⊥ x-axis, meeting the graph at B

MB = 3.5 units

Threrefore, 2 = 1.5 ⇒ p(n) = 3.5

The post Graphs of Simple Function | Basic Functions and their Graphs appeared first on Learn CBSE.

Are you preparing coordinate geometry? If yes, then the most important concept of geometry is graphing linear equations. This topic plays a vital role while preparing for the exams. So, check out the useful information about the linear equations, graphing linear equations in two variables, and solved example questions from the following sections.

Linear Equations Definition

A linear equation is a first-order equation which is having the power of the variable as one. If an equation has only one variable, then it is called the linear equations in one variable. When the linear equations have two variables, it is known as the linear equations in two variables, etc. The standard form the linear equations in two variables is ax + by + c = 0.

Here,

a, b, c are the real numbers

a, b is not equal to zero.

The solution of a linear equation in two variables is a pair of numbers, they are x, and y which satisfies the equation. The example equation is x + 3y = 12. Some of the solutions are (0, 4), (12, 0), (3, 3), (2, 6).

Properties for the graphing linear equation:

• Every linear equation has infinite solutions.
• Every point (x, y) on the line gives the solutions.
• Every point on the line satisfies the equation.
• To draw an exact line on the graph paper you can plot as many points you like, but it is necessary to plot a minimum of three points.

Therefore, every linear equation in two variables can be represented geometrically as a straight line in a coordinate plane.

How to Graph of Linear Equation in Two Variables?

Below provided are the simple steps that are helpful to you to draw a linear equation in a graph. Go through those steps and use them.

• Convert the equation in the form of y = mx + c (slope-intercept form)
• Use the trial and error method to get 3 pairs of points that satisfy the equation.
• Plot those three points on a graph.
• Join those points marked on the graph paper to get the straight line that represents the equation graphically.

Example Questions

1. Draw the graph of the linear equation y = 5x?

Solution:

Given linear equation is y = 5x

Equation is already available in the form of y = mx + c [here c = 0]

Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = 5x.

When the value of x = 0, then y = 5 × 0 = 0

When the value of x = 1, then y = 5 × 1 = 5

When the value of x = -1, then y = 5 × -1 = -5

Arrange the values of the linear equation y = 5x in the table.

Linear Equation Table is

Now, plot the points P (0, 0), Q (1, 5), R (-1, -5) on the coordinate graph

Join those points to get a straight line.

2. Draw the graph of the linear equation y = x + 2.

Solution:

Given linear equation is y = x + 2

Equation is already available in the form of y = mx + c [here c = 2, m = 1]

Now we will apply the trial and error method to find the values of ordered pairs that satisfy the given equation y = x + 2.

When the value of x = -2, then y = -2 + 2 = 0

When the value of x = -1, then y = -1 + 2 = 1

When the value of x = 0, then y = 0 + 2 = 2

When the value of x = 1, then y = 1 + 2 = 3

When the value of x = 2, then y = 2 + 2 = 4

Arrange the values of the linear equation y = x + 2 in the table.

Linear Equation Table is

Now, plot the points A (-2, 0), B (-1, 1), C (0, 2), D (1, 3), E (2, 4) on the coordinate graph

Join those points to get a straight line.

3. Draw the graph of the linear equation y = 3x − 6.

Solution:

Given linear equation is y = 3x – 6

It is in the slope-intercept form y = mx + c [ where m = 3, c = -6]

Now, apply the trial and error method to get the values of the points that satisfy the equation y = 3x – 6

When the value of x = -1, then y = 3(-1) -6 = -3 – 6 = -9

When the value of x = 0, then y = 3(0) – 6 = -6

When the value of x = 1, then y = 3(1) – 6 = 3 – 6 = -3

When the value of x = 2, then y = 3(2) – 6 = 6 – 6 = 0

Arrange the values of the linear equation y = 3x – 6 in the table.

Now, plot the points A (-1, -9), B (0, -6), C (1, -3), D (2, 0) on the coordinate graph

Join those points to get a straight line.

The post Graph of Linear Equation | Graphing Linear Equation with Two Variables appeared first on Learn CBSE.

You will no longer feel drawing 2 or more linear equations on a graph difficult anymore with our article. Get a detailed procedure to represent simultaneous equations graphically easily. Furthermore, have a look at the example questions provided below to get clarity on the topic and solve related problems.

To solve a pair of simultaneous equations graphically, we first draw two equations on the graph. Those two equations form straight lines intersecting each other at a common point. This common point gives the solution of the pair of simultaneous equations.

How to Solve Simultaneous Linear Equations Graphically?

Let us take two first-order linear equations

p₁x + q₁y + r₁ = 0

p₂x + q₂y + r₂ = 0

Draw these two lines on a coordinate graph. These lines always form a straight line on the graph. Suppose L₁ represent the graph of p₁x + q₁y + r₁ = 0 and L₂ represent the graph of p₂x + q₂y + r₂ = 0. By solving the simultaneous equations graphically we will get three possible solutions. They are as follows.

1. When the two lines L₁, L₂ interest at a single point

• Here, two lines meet at a point called p (x, y).
• So, the obtained point x coordinate, y coordinate is the unique solution of the given linear equations.
• This system is called independent.

2. When two lines L₁, L₂ are coincident

• Here, two equations represent the single line.
• Therefore, the equations have infinitely many solutions.
• This system is called dependent.

3. When two lines L₁, L₂ are parallel to each other

• The two equations have no common solutions.
• This system is called inconsistent.

Examples of Simultaneous Equations Graphically

1. Solve this system of equations by graphing: y = x + 1 and x + y = 5

Solution:

Given linear equations are y = x + 1 and x + y = 5

On a graph paper to solve simultaneous equations graphically, draw a horizontal line X’OX and a vertical line YOY’ as the x-axis and y-axis respectively.

The first equation y = x + 1 is in the slope intercept form y = mx + c [m = 1, c = 1]

Now apply the trial and error method to get the 3 pairs of points (x, y) which satisfy the equation y = x + 1.

If the value of x = -1 then y = -1 + 1 = 0

If the value of x = 0 then y = 0 + 1 = 1

If the value of x = 1 then y = 1 + 1 = 2

If the value of x = 2 then y = 2 + 1 = 3

Arrange these values of the linear equation y = x + 1 in the table

Plot the points of the equation y = x + 1; A (-1, 0), B (0, 1), C (1, 2), D (2, 3) on the graph.

Join the points A, B, C, and D to get the line equation AD.

So, the line AD represents the equation y = x + 1.

Convert the second equation x + y = 5 into the slope-intercept form.

y = 5 – x [ here, m =-1, c = 5]

Apply the trial and error method to get the points that satisfy the given equation.

If the value of x = -1 then y = 5 – (-1) = 5 + 1 = 6

If the value of x = 0 then y = 5 – 0 = 5

If the value of x = 1 then y = 5 – 1 = 4

If the value of x = 2 then y = 5 – 2 = 3

Arrange the values of the linear equation in the table.

Plot the points P (-1, 6), Q (0, 5), R (1, 4), S (2, 3) on the graph

Join the points to get a straight line of x + y = 5

We get two straight lines intersecting each other at (2, 3).

Therefore, x = 2 and y = 3 is the solution of the given system of equation.

2. Solve graphically the system if linear equation y = 2x = 4 and y + 2x = 1.

Solution:

Given two linear equations are y + 2x = 4 and y + 2x = 1

Convert the first equation y + 2x = 4 into the slope intercept form.

y = 4 – 2x [ here m = -2, c = 4]

Apply the trial and error method to get the points on the line.

If the value of x = -1 then y = 4 – 2 (-1) = 4 + 2 = 6

If the value of x = 0 then y = 4 – 2 (0) = 4 – 0 = 4

If the value of x = 1 then y = 4 – 2 (1) = 4 – 2 = 2

If the value of x = 2 then y = 4 – 2 (2) = 4 – 4 = 0

Arrange these values of the linear equation y = 4 – 2x in the table.

Now plot the points A (-1, 6), B (0, 4), C (1, 2), D (2, 0) on the graph.

Join the points to get the linear equation y + 2x = 4.

Convert the second equation y + 2x = 1 into the slope intercept form.

y = 1 – 2x [m = -2, c = 1]

Apply the trial and error method to get the points on the equation.

When the value of x = -1 then y = 1 – 2(-1) = 1 + 2 = 3

When the value of x = 0 then y = 1 – 2(0) = 1 – 0 = 1

When the value of x = 1 then y = 1 – 2(1) = 1 – 2 = -1

When the value of x = 2 then y = 1 – 2(2) = 1 – 4 = -3

Arrange these values of the equation in the table.

Plot the points P (-1, 3), Q (0, 1), R (1, -1), S (2, -3) on the graph.

Join the points to form a linear equation y + 2x = 1

We get from the graph that two straight lines are parallel to each other.

Therefore, the given system of equations has no solutions.

3. Solve this system of equations by graphing: x – 2y = 2 and 2y + 2 = x.

Solution:

Given linear equations are x – 2y = 2 and 2y + 2 = x.

Convert the first equation into y = mx + c form

y = x / 2 – 1 [ here m = 1/2, c = -1]

Apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation x – 2y = 2.

If the value of x = -1 then y = -1/2 – 1 = -3/2

If the value of x = 0 then y = -0/2 – 1 = -1

If the value of x = 1 then y = 1/2 – 1 = -1/2

Arrange these values in the table

Now plot the points of the equation A (-1, -3/2) B (0, -1), C (1, -1/2)

Join the points A, B, C to get the graph line x – 2y = 2

Convert the second equation 2y + 2 = x into the slope-intercept form.

y = ½ (x – 2) [ m = 1/2, c = -1]

Apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation 2y + 2 = x.

If the value of x = -1 then y = ½ (-1 – 2) = -3/2

If the value of x = 0 then y = ½ (0 – 2) = -1

If the value of x = 1 then y = ½ (1 – 2) = -1/2

Arrange these values in the table

Now plot the points of the equation 2y + 2 = x; A (-1, -3/2) B (0, -1), C (1, -1/2) on the graph paper.

Join the points of A, B, and C; to get the graph line AC.

Thus line AC is the graph of 2y + 2 = x.

We find that the two straight lines coincide.

Therefore, the given system of equations has an infinite number of solutions.

The post Solving Simultaneous Equations Graphically | Graphical Method Examples with Solution appeared first on Learn CBSE.

Are you searching for a tool that plots a distance vs time graph for you? No need to worry, here we are providing the easy and simple steps that are helpful to draw a graph of distance vs time. Get useful information such as the distance traveled by an object in time, distance-time graph definition, and its importance from this page. You will also get some example questions in the below sections.

Graph of Distance vs. Time Definition

Distance vs time graph shows how far a moving object traveled in a specific amount of time. It is a simple graph line that shows the displacement of an object. We will take distance along the y-axis, time along the x-axis.

Importance of the Time Distance Graph

Distance and time graph deals with the motion of bodies. You need to record the distance, time of a moving body and plot those points on a graph to get the Graph of Distance Versus Time. From this graph, you can get the uniform velocity, speed of the object.

The different types of graph lines are listed here.

1. If the graph is a linear line, then the object is moving with either fast or steady speed.

2. If the graph line is a curve, then the object speed is getting faster.

3. If the graph line is a straight line parallel to the time, then the object speed is constant.

4. If the graph line is moving upwards, then speed is steady. When the graph is moving downwards, then speed is returning to start.

Steps to Draw a Graph of Distance Vs Time

Below are the rules you need to check for while drawing a distance vs time graph. You will get an idea in detail by referring to the lines outlined below.

• Get the equation between distance and time of the object.
• Find the distance for the random values of the time and put them on a table.
• Now plot the points on a graph paper and join them to get a line.

Solved Examples of Distance Time Graph

Example 1.

A bus driver drives at a constant speed which is indicated by the speedometer and the driver measures the time taken by the bus for every kilometer. The driver notices that the bus travels 10 kilometers every 20 minutes. Draw a graph for the given details and find the distance covered when the time is 60 minutes, 75 minutes.

Solution:

Given that,

The bus covers 10 kilometers of distance in 20 minutes duration.

10 Km = 20 minutes

D = 2T

When T = 10, D = 2 * 10 = 20

When T = 20, D = 2 * 20 = 40

When T = 30, D = 2 * 30 = 60

When T = 40, D = 2 * 40 = 80

Along the x-axis: Take 1 small square = 10 minutes.

Along the y-axis: Take 1 small square = 20 km.

Plot the points A (10, 20), B (20, 40), C (30, 60), D (40, 80) on a graph.

Join those points to get a linear equation.

From the graph, we can observe that the distance covered in 60 minutes or 1 hour is 120 km.

The distance covered in 75 minutes is 150 km.

Example 2.

A person traveled 30 km away from the starting point and that took 2 hours duration.

(a) After that, he takes a rest for 1 hour.

(b) And, he moved further 30 km in 30 minutes.

(c) He traveled the remaining 60 km back to the starting point in 1 hour 30 minutes.

Solution:

Given that,

Along the x-axis: Take 1 small square = 30 minutes.

Along the y-axis: Take 1 small square = 10 km.

Now, plot the points A (2, 30), B (3, 30), C (3.5, 60), D (4, 30), E (5, 0)

Join those points to get the graph line.

Example 3.

Draw a graph of distance vs time. When the distance covered by an object is 7 times the time taken. Find the distance covered at 7.5 minutes, 8.5 minutes?

Solution:

Given that,

Distance D = 7 * Time

D = 7T

When T = 0.5, D = 7 * 0.50 = 3

When T = 1, D = 7 * 1 = 7

When T = 1.5, D = 7 * 1.5 = 10.5

When T = 2, D = 7 * 2 = 14

When T = 2.5, D = 7 * 2.5 = 17.5

Along the x-axis: Take 1 small square = 0.5 units.

Along the y-axis: Take 1 small square = 5 units.

Now plot the points A (0.5, 3), B (1, 7), C (1.5, 10.5), D (2, 14), E (2.5, 17.5) on a graph.

Join the points.

From the graph, we can observe that the distance covered at 7.5 minutes is 52.5.

The distance covered at 8.5 minutes is 59.5 units.

The post Graph of Perimeter vs. Length of the Side of a Square appeared first on Learn CBSE.

Solve the given questions on Percentage and get solutions with detailed explanations. All the Questions on Percentages are primarily based on calculating percentages along with its applications. Enhance your Conceptual Knowledge taking the help of the Solved Questions on Percentage. Identify the knowledge gap and practice the questions accordingly thereby score better grades in your exams.

1.  30 % of a number is 10 less than two-third of that number. Find the number?

Solution:

Let the number be x

From the given data (2/3)*x – 30% of x = 10

(2/3)*x – (30/100)*x = 10

x(2/3- 30/100) = 10

x((6-90)/300) = 10

x(-84/300) = 10

x = (10*300/-84)

= 3000/-84

= -250/7

2. When 50 is subtracted from a number; it reduces to 40 %. Find the three-fifth of that number?

Solution:

Let the number be x

From given data 50 subtracted from a number reduced to 40% which means

x- 50 = (40/100)*x

x- (40/100)x = 50

x(1-40/100) = 50

x(60/100) = 50

x= (50*100)/60

= 250/3

3. If 10% of x is equal to 5% of y, then 14% of x will be equal to how much % of y?

Solution:

From given data, we can write the equation as such

10% of x = 5% of y

14% of x = ? % of y

(10/100)*x = (5/100)*y

(1/10)*x = (1/20)*y

x/y = (1/20)/(1/10)

= 1/20*10/1

= 1/2

x/ y = 1/2

From this ratio, we can say that 14% of x = 7% of y

4. A shopkeeper marks the price of his goods at 20% higher than the original price. After that, he allows for a discount of 10%. What profit or loss does he get?

Solution:

Let the Cost Price of the goods be 100

Since the shopkeeper marked it 20% higher the price becomes 120.

Later the shopkeeper allows a discount of 10% i.e. 120 – 10 = 110

Since the Shopkeeper sold it for a higher price than the original price he will get a profit.

Profit % = ((Selling Price – Original Price)/Original Price)*100

= ((110-100)/100)*100

= 10%

Therefore, shopkeeper makes a profit of 10%.

5. A number is reduced by 50%. Its present value is 230. What was its original value?

Solution:

Let the number be x

The number reduced by price = x – 50/100*x

x -50/100*x = 230

x(1-50/100) = 230

x(50/100) = 230

x = (230*100)/50

= 460

Original Value is 460.

6. A student multiplied a number by 5/7 instead of 7/5, What is the percentage error in the calculation?

Solution:

Let the number be x

Ideally, the Number Multiplied by 7/5 is (7/5)x = 7x/5

Number Multiplied by 5/7 is (5/7)x = 5x/7

Error = (7x/5 – 5x/7)

= (49x -25x)/35

= 24x/35

Error % = (Error/True Value)*100

= ((24x/35)/(7x/5))*100

= (24x/35*5/7x)*100

= (120/245)*100

= 48.97%

7. Ram gets 45 % of total valid votes in an election. If the total votes were 9000, what is the number of valid votes that Ram got?

Solution:

Total number of votes cast = 9000

Ram got 45% of Votes i.e. 45/100*9000

= 45*90

= 4050

8. In a plot of 4000 sq. m., only 2500 sq. m. is allowed for construction. What percent of the plot is to be left without construction?

Solution:

Percentage of Plot allowed for construction = (2500/4000)*100

= 62.5%

Percentage of Plot not allowed for construction = (1500/4000)*100

= 37.5%

The post Problems on Percentage | Percentage – Aptitude Questions and Answers appeared first on Learn CBSE.

Real Life Problems on Percentage provided here covers different types of questions that you might come across regarding the concept Percentage. To be clear with the concept go through the step by step solutions provided for all the Problems on Percentage. Assess your preparation standards and identify the areas you are lagging and concentrate on them. Practice the Solved Examples on Percentage and enhance your conceptual knowledge.

1. Jim needs 20% to pass. If he scored 225 marks and falls short by 25 marks, what were the maximum marks he could have got?

Solution:

If Jim Scored 25 Marks he would have got 20% pass

Thus Jim required = 225+25

= 250 Marks

Let the Maximum Marks be m

20% of m = 250

20/100*m = 250

m = (250*100)/20

= 1250

2. A fruit seller had some oranges. He sells 30% oranges and still has 350 oranges. How many Oranges Originally, he had?

Solution:

Let suppose he had x oranges

x – 30% of x = 350

x(1-30/100) = 350

x = 350/(70/100)

= (350*100)/70

= 500

Therefore, fruit seller had 500 oranges originally.

3. If 10% of a = b, then b% of 10 is the same as

Solution:

10/100*a = b

a/10 = b

a = 10b ……(1)

b% of 10 = b/100*10

= 10b/100

= b/10

Substitute the value of b obtained in equation (1)

= (a/10)/10

= a/10*1/10

= a/100

= 1/100*a

= 1% of a

Therefore, b% of 10 is same as 1% of a.

4. In an election between two candidates, one got 50% of the total valid votes. If the total number of votes was 8000, the number of valid votes that the candidate got, was?

Solution:

Total Number of Votes Polled = 8000

Candidate got 50% of Total Votes

= 50% of 8000

= 50/100*8000

= 50*80

= 4000

5. Three candidates X, Y, Z contested in an election and received 1100, 7536, and 11500 votes respectively. What percentage of the total votes did the winning candidate get?

Solution:

The candidate Z with the highest votes will win

Total Votes = 1100+7536+11500

= 20136

Winning Percentage = (Votes Candidate Z got/ Total Votes)*100

= (11500/20136)*100

= 57.11%

Therefore, the Winning Percentage is 57.11%

6. The population of a town increased from 1,75,000 to 2,50,500 in a decade. What is the average percent increase of population per year is?

Solution:

Increase in Population in 10 years = 2, 50, 500- 1, 75, 000

= 75500

Average Percent Increase in Population per Decade = (75500/175000)*100

= 43.14%

Required Average per year = (43.14/10)%

= 4.314%

7. Robin scored 5 marks more than what he did in the previous examination in which he scored 25. John scored 40 marks more than she did in the previous examination in which she scored 30. Who showed less improvement?

Solution:

Robin Improvement = (5/25)*100

= 20%

John Improvement = 40/30*100

= 4000/30

= 133.33%

Therefore, Robin showed less percentage.

The post Real Life Problems on Percentage | Basic Problems on Percentage appeared first on Learn CBSE.

For those looking for Word Problems on Percentage have arrived at the right destination. Make use of the Percent Word Problems available and solve similar kinds of questions easily. Practice the Problems on Percentage with Solutions regularly and get a detailed explanation to cross-check where you went wrong. After referring to our article you can solve various problems related to the percentage quickly and efficiently.

1.  10% of the books in a public library are Science books. If there are 90,000 books in the library, find the number of Science books available?

Solution:

Total number of books = 90000

10% of books are science

therefore, 10/100*90,000

= 9000

The Total Number of Science Books available in the library are 9000.

2. A baseball pitcher won 70% of the games he pitched. If he pitched 40 ballgames, how many games did he win?

Solution:

70/100 =x /40

Cross Multiplying we have = (70*40)/100

= 28

Therefore, he won 28 games.

3. David took a math test and got 25 correct and 15 incorrect answers. What was the percentage of correct answers?

Solution:

Total Answers = 25+15

= 40

Percentage of Correct Answers = 25/40*100

= 2500/40

= 62.5%

Therefore, David got 62.5% of correct answers.

4. There are 45 Carpenters in a Crew. On a certain day, 27 were present. What Percent showed up for work?

Solution:

Percentage of Carpenters Showed up for work = (27/45)*100

= 2700/45

= 60%

Therefore, 60% of Carpenters Showed up for Work.

5. A metal bar weighs 7 Kg. 93% of the bar is Silver. How many Kgs of silver is there in the bar?

Solution:

From the given data

93% of the bar is silver

= 93% of 7kg

= 93/100*7

= 651/100

= 6.51 Kg

Therefore,  6.51 Kg of Silver is there in the bar.

6. A packet contained 540 chocolates. 55% of the Chocolates were distributed to the children. Calculate how many chocolates are distributed?

Solution:

No. of Chocolates distributed = 55% of 540

= 55/100*540

= 29700/100

= 297

297 Chocolates are distributed.

7. A man had $4,50,000 worth of property. He left 50% of his property to his son. How much did his son get?

Solution:

Total Property worth = $4, 50, 000

Property his son got = 50% of 4, 50, 000

= 50/100*4, 50, 000

= 2,25,000

Therefore, the son got $2, 25, 000.

8. In an Exam Asha secured 300 marks. If she secured 75 % marks, find the maximum marks?

Solution:

Asha secured 300 marks

Let the Maximum Marks be m

75% of m = 300

(75/100)*m = 300

m = (300*100)/75

= 400

The maximum marks in the Exam are 400.

9. An alloy contains 24 % of copper. What quantity of alloy is required to get 280 g of copper?

Solution:

Let the quantity of copper needed = m

24% of m = 280g

24/100*m = 280

m = (280*100)/24

= 1166.66 gms

The Quantity of Alloy Required to get 280g of Copper is 1166.66 gms.

10. Geetha scores 60 marks out of 80 in her Maths exam. Convert her Marks into Percent?

Solution:

Percent of Marks Geetha got = (60/80)*100

= 6000/80

= 75%

Thus, Geetha Scored 75% of Marks in her Maths Exam.

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Percentages are often used for calculations involving money. We get to see various scenarios in our day to day life where we can apply the concept of Percentage. Percentages are used in many types of problems and situations. Applications of Percentages help you solve different types of real-life percent problems. For better understanding, we have provided step by step explanation of all the Problems on Percentages.

1.  In a survey of 100 students, 60% of students liked Science, and the rest of the students liked Arts. How many number of students liked Science?

Solution:

Total Number of Students Participated in the Survey = 100

60% of Students liked Science = 60/100*100

= 60

Therefore, 60 Students Liked Science.

2. The Price of a Good increase from $18 to $20. Express the Percentage Increase of the Good?

Solution:

Increased Price = $20 – $18

= $2

Percentage Increase = (Increased Value/Original Price)*100

= ($2/$18)*100

= 100/9

= 11.11%

Thus, the Price of Good increases by 11.11%

3. Father’s Weight is 30 % more than that of his son. What Percent is Son’s Weight Less than Father’s Weight?

Solution:

Let Son’s Weight be 100 Kg

Father’s weight is 30% more than son’s i.e. 130 Kg

If Father’s Weight is 130kg then Son’s weight is 100kg

If Father’s Weight is 1 kg then Son’s Weight is 130/100

If Father’s Weight is 100kg then Son’s Weight is (100/130*100) Kg

Thus, Son’s Weight is 23.08 % less compared to his father.

4. What number is 30% of 90?

Solution:

Let the number to be m

30% of m = 90

30/100*m = 90

m = (90*100)/30

= 9000/30

= 300

Therefore, the number is 300.

5. Komali and her sister enjoyed dinner in a restaurant, and the bill was $75.50. If she wants to leave 15% of the total bill as her tip, how much should she leave?

Solution:

Total Bill = $75.50

From the given data Komali wants to leave 15% of Bill as tip = 15% of 75.50

= (15/100)*75.50

= $11.325

Thus, Komali has to leave $11.325 as a tip in the restaurant.

6. One serving of rice has 110 mg of sodium, which is 10% of the recommended daily amount. What is the recommended daily amount of sodium in total?

Solution:

Let total daily amount of sodium required = m

From given data 10% of m = 110 mg

10/100*m = 110

m = (110*100)/10

= 1100

Therefore, daily amount of sodium required in total is 1100 mg.

7. Cierra is making muffins from a mix and each muffin had 240 calories and 80 calories are fat. What Percent of Total Calories is Fat?

Solution:

Fat Calories Percent = 80/240*100

= 8000/240

= 33.33%

Thus, 33.33% Percent of Total Calories is Fat.

8. Kiara requires 30% to pass. She gets 180 marks and falls short by 20 marks. Find the maximum numbers she could have got(round to the nearest)?

Solution:

To get 30% kiara should score 180+20 = 200 Marks

Let the maximum numbers = m

30% of m  = 200

30/100*m = 200

m = (200*100)/30

= 666

Therefore, Kiara Should have got 666 Marks.

8. If the tax rate is 8% what would be sales tax if the price of the truck is $24,000?

Solution:

Tax Rate = 8%

Sales Tax = 8% of $24, 000

= 8/100*24, 000

= 1920

Sales Tax of the Truck is $1920.

SEO Title: Application of Percentage Formula | Uses of Percentages in Daily Life

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Set Theory is a branch of mathematics and is a collection of objects known as numbers or elements of the set. Set theory is a vital topic and lays stronger basics for the rest of the Mathematics. You can learn about the axioms that are essential for learning the concepts of mathematics that are built with it. For instance, Element a belongs to Set A can be denoted by a ∈ A and a ∉ A represents the element a doesn’t belong to Set A.

{ 3, 4, 5} is an Example of Set. In this article of Introduction to Set Theory, you will find Representation of Sets in different forms such as Statement Form, Roster Form, and Set Builder Form, Types of Sets, Cardinal Number of a Set, Subsets, Operations on Sets, etc.

Set Definition

Set can be defined as a collection of elements enclosed within curly brackets. In other words, we can describe the Set as a Collection of Distinct Objects or Elements. These Elements of the Set can be organized into smaller sets and they are called the Subsets. Order isn’t that important in Sets and { 1, 2, 4} is the same as { 4,2, 1}.

Examples of Sets

• Odd Numbers less than 20, i.e., 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
• Prime Factors of 15 are 3, 5
• Types of Triangles depending on Sides: Equilateral, Isosceles, Scalene
• Top two surgeons in India
• 10 Famous Engineers of the Society.

Among the Examples listed the first three are well-defined collections of elements whereas the rest aren’t.

Important Sets used in Mathematics

N: Set of all natural numbers = {1, 2, 3, 4, …..}

Q: Set of all rational numbers

R: Set of all real numbers

W: Set of all whole numbers

Z: Set of all integers = {….., -3, -2, -1, 0, 1, 2, 3, …..}

Z+: Set of all positive integers

Representation of a Set

Set can be denoted using three common forms. They are given along the following lines by taking enough examples

• Statement Form
• Roaster Form or Tabular Form
• Set Builder Form

Statement Form: In this representation, elements of the set are given with a well-defined description. You can see the following examples for an idea

Example:

Consonants of the Alphabet

Set of Natural Numbers less than 20 and more than 5.

Roaster Form or Tabular Form: In Roaster Form, elements of the set are enclosed within a pair of brackets and separated by commas.

Example:

N is a set of Natural Numbers less than 7 { 1, 2, 3, 4, 5, 6}

Set of Vowels in Alphabet = { a, e, i, o, u}

Set Builder Form: In this representation, Set is given by a Property that the members need to satisfy.

{x: x is an odd number divisible by 3 and less than 10}

{x: x is a whole number less than 5}

Size of a Set

At times, we are curious to know the number of elements in the set. This is called cardinality or size of the set. In general, the Cardinality of the Set A is given by |A| and can be either finite or infinite.

Types of Sets

Sets are classified into many kinds. Some of them Finite Set, Infinite Set,  Subset, Proper Set, Universal Set, Empty Set, Singleton Set, etc.

Finite Set: A Set containing a finite number of elements is called Finite Set. Empty Sets come under the Category of Finite Sets. If at all the Finite Set is Non-Empty then they are called Non- Empty Finite Sets.

Example: A = {x: x is the first month in a year}; Set A will have 31 elements.

Infinite Set: In Contrast to the finite set if the set has infinite elements then it is called Infinite Set.

Example: A = {x : x is an integer}; There are infinite integers. Hence, A is an infinite set.

Power Set: Power Set of A is the set that contains all the subsets of Set A. It is represented as P(A).

Example:  If set A = {-5,7,6}, then power set of A will be:

P(A)={ϕ, {-5}, {7}, {6}, {-5,7}, {7,6}, {6,-5}, {-5,7,6}}

Sub Set: If Set A contains the elements that are in Set B as well then Set A is said to be the Subset of Set B.

Example:

If set A = {-5,7,6}, then Sub Set of A will be:

P(A)={ϕ, {-5}, {7}, {6}, {-5,7}, {7,6}, {6,-5}, {-5,7,6}}

Universal Set:

This is the base for all the other sets formed. Based on the Context universal set is decided and it can be either finite or infinite. All the other Sets are Subsets of Universal Set and is given by U.

Example: Set of Natural Numbers is a Universal Set of Integers, Real Numbers.

Empty Set: 

There will be no elements in the set and is represented by the symbol ϕ or {}. The other names of Empty Set are Null Set or Void Set.

Example: S = { x | x ∈ N and 9 < x < 10 } = ∅

Singleton Set:

If a Set contains only one element then it is called a Singleton Set.

Example: A = {x : x is an odd prime number}

Operations on Sets

Consider Two different sets A and B, they are several operations that are frequently used

Union: Union Operation is given by the symbol U. Set A U B denotes the union between Sets A and B. It is read as A union B or Union of A and B. It is defined as the Set that contains all the elements belonging to either of the Sets.

Intersection: Intersection Operation is represented by the symbol ∩. Set A ∩ B is read as A Intersection B or Intersection of A and B. A ∩ B is defined in general as a set that contains all the elements that belong to both A and B.

Complement: Usually, the Complement of Set A is represented as A^c or A^‘ or ~A. The Complement of Set A contains all the elements that are not in Set A.

Power Set: The power set is the set of all possible subsets of S. It is denoted by P(S). Remember that Empty Set and the Set itself also comes under the Power Set. The Cardinality of the Power Set is 2ⁿ in which n is the number of elements of the set.

Cartesian Product: Consider A and B to be Two Sets. The Cartesian Product of the two sets is given by AxB i.e. the set containing all the ordered pairs (a, b) where a belong to Set A, b belongs to Set B.

Representation of Cartesian Product A × B = {(a, b) | a ∈ A ∧ b ∈ B}.

The cardinality of AxB is N*M where N is the cardinality of A and M is the Cardinality of B. Remember that AxB is not the same as BxA.

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Set is a well-defined collection of objects or elements. A Set is represented using the Capital Letters and the elements are enclosed within curly braces {}. Refer to the entire article to know about Representation of Set in three different ways such as Statement Form, Set Builder Form, Roster Form. For a Complete idea on this refer to the Set Theory and clear all your queries. Check out Solved Examples for all three forms explained step by step.

Representation of a Set

Sets can be represented in three different forms. Let us discuss each of them in detail by taking enough examples.

• Statement Form or Descriptive Form
• Roster or Tabular Form
• Rule or Set Builder Form

Statement Form: In this Form, well-defined descriptions are provided for the elements or members in the set. Verbal Description of Elements is given. Statement Form is also known as Descriptive Form. Elements of the set are enclosed within curly brackets {}.

Example: Set of Odd Numbers Less than 10.

In Statement Form, the elements can be expressed as {1, 3, 5, 7, 9}

Set of Students in Class V having a height above 5 ft.

Set of Numbers greater than 30 and less than 40.

Roster Form: In the Roster Form elements of the set are represented within {} and are separated by commas. In this representation order of the elements doesn’t matter but the elements must not be repeated. Roster Form is also known as Tabular Form.

Example:

1.  Set of Natural Numbers less than 10

Set of Natural Numbers Less than 10  = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Set N in Roster Form is {1, 2, 3, 4, 5, 6, 7, 8, 9}

2. Set of Natural Numbers that divide 10

Y = { 1, 2, 5, 10}

3. W is the Set of Vowels in the Word Elephant

W = {E, A}

Set Builder Form: In this form, all the elements possess a single property in order to be members of the set. In this representation of the set, the element is denoted by the symbol x or any variable followed by the symbol : or |. After this symbol write down the property possessed by the elements of the set and enclose the entire description within brackets.

Example:

1. Write the following Set in Set Builder Form = { 3, 6, 9, 12}

Set Builder Form is A = {x: x= multiples of 3,  n ∈ N and  n ≤ 15}

2. D = {x: x is an integer and – 2 < x < 11}

3. X = {m: m is a negative integer < -10}

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The collection of distinct objects is called Sets. Sets can be classified into different types of sets and some of them are explained in the below sections taking a few examples. Get Definitions of various types of sets in the following sections. To gain conceptual knowledge on this you can look at Set Theory and clear your queries regarding the concept if any. See the Example Problems on Sets and learn which one falls under what kind of set.

Empty or Null or Void Set

Any Set that doesn’t have any elements in it is called an Empty Set. It is also called as Void or Null Set. The Symbol used to denote the Empty Set is {} or φ.

Example:

1.  A = {x : 7 < x < 8, x is a natural number}

The Given Set will be a Null Set because there is No natural number between numbers 7 and 8.

2. Q = {z: z is a whole number which is not a natural number,z ≠ 0}

0 is the only whole number that is not a natural number. Since z ≠ 0 there is no other possibility thus the Set Q is {} or φ.

Singleton Set

If a Set has only one element then it is called a Singleton Set.

Example:

1.  A = {x : x is neither prime nor composite}

Set A is Singleton Set since it has only one element i.e. 1 as it is neither prime nor composite.

2. B = {y : y ∈ N and y² = 9}

Set B is Singleton Set since there is only one element whose square is 9 and it is 3.

3. B = {z : z is a even prime number}

Set B is Singleton Set as 2 is the only even prime number.

Finite Set

Any Set that is either empty or contains a finite number of elements i.e. countable elements is called a Finite Set.

Example

S = { x | x ∈ N and 50 > x > 30 } is finite since the number of elements that satisfies the condition given are countable.

B = {a, e, i, o, u} is finite since it represents the vowel letters in the English Alphabet.

Infinite Set

Contrary to the Finite Set the number of elements in the Set is infinite then it is Called an Infinite Set.

Example

A = {y: y is a point on a line} is an infinite set as there will be an infinite number of points on a line.

A = {x: x is a real number} is an infinite set because there will be an infinite number of real numbers.

Equal and Unequal Sets

Two Sets A and B are said to be equal if they have the same elements irrespective of the order of the elements in the Set. Equal Sets are denoted by A = B whereas Unequal Sets are denoted by A ≠ B.

Example:

If A = { a, e, i, o, u} B = { i, e, u, o, a} then A = B since both the Sets have Same Elements no matter what the order is.

C ={ 4, 6, 7, 8} D = { 4, 5, 7,9 } then C ≠ D as it has different elements.

A = {x, y, z} and B = {u, v, x, y, z} in this case also A = B because both the Sets have same elements.

Equivalent Set

Equivalent Sets are those which contain the same number of elements irrespective of the elements within them.

A = {1, 2, 3, 4, 5} and B = {8, 9, 10, 11, 12} are equivalent sets because both these sets have 5 elements each.

S = {1³,2³, 3³ …} and T = {y : y³ ϵ Natural number} are also equivalent sets.

Universal Set

Universal Set is the base for all other sets found. However, this depends on the context and can be either finite or infinite. All other sets are subsets of Universal Set and Universal Set is represented by U.

Example:

Real Numbers is a Universal Set for all Whole numbers, natural, odd and even rational, irrational numbers.

Power Set

Before heading into the concept of Power Sets you need to be aware of Sub Sets. Power Set is nothing but the Collection of all the Subsets. If a set has n elements the total number of subsets that can be formed are 2ⁿ.

Example:

If Set A = {-5,11,8}

The Power Set of A is P(A) = {ϕ, {-5}, {11}, {8}, {-5,11}, {11,8}, {8,-5}, {-5,11,8}}

Cardinal Number of a Set

The number of distinct elements present in the Set is called Cardinal Number of a Set. Consider a Set A and its Cardinal Number is represented as n(A).

Example:

A = Set of Letters in the Word Mathematics

The distinct letters in the given word are as such

A = { M A T H E I C S}

n(A)  = 7

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Finite and Infinite Sets are two different kinds of Sets. Word Finite means countable and infinite stands for uncountable. Learn about Definite, Indefinite Sets Definition, Properties along with solved examples. To know more about this refer to the Set Theory and know more about different types of Sets.

Finite Sets Definition

A Finite Set is a Set in which a number of elements are countable. In other words, Finite Sets are also called Countable Sets as they can be counted.

Example:

Set A has Months in a year i.e. { January, February, March, April, May, June, July, August, September, October, November, December}

Here n(A) = 12 countable elements thus it is a Finite Set.

Set B includes Vowels in English Alphabet i.e. { a, e, i, o, u}

n(B) = 5 countable elements thus it is a Finite Set.

Cardinality of Finite Sets: If a denotes the Cardinality of Finite Set A then n(A) = a. We can list out all the elements of a finite set and list them in curly braces.

Properties of Finite Set

The following finite set conditions always hold true and are always finite.

• Union of Two Finite Sets.
• The subset of Finite Set
• Power Set of a Finite Set

Examples

P = {5, 12, 14, 20, 24, 30}

Q = {5, 10, 15, 20, 25, 30}

R = {5, 20, 30}

R ⊂ P, i.e. R is a Subset of P since all the elements of R are present in P. Thus, the Subset of Finite Set is Finite.

P, Q, R are finite sets since the number of elements in it is countable.

P U Q = { 5, 10, 12, 14, 15, 20, 24, 25, 30}

The power set of 2³ = 8, the number of elements of Set P is 3. This indicates that the power set of a finite set is finite.

Non-Empty Finite Set

In this kind of Set, the number of elements in the set is quite large and only the beginning and ending of numbers are given. We can represent it as n(A) and if n(A) is a natural number then we consider it at some point as a Finite Set.

Consider M is a Set of Natural Numbers less than the number m. Thus we can say that the Cardinality of Set M is m.

Infinite Sets Definition

If a Set is not finite or has uncountable elements then the Set is considered to be an Infinite Set. Since the number of elements is not countable it is also called Uncountable Set. We can’t represent Infinite Sets in Roster Form and the elements of Infinite Set are represented using …. denoting the infinity of the set.

Example

• Set of Points on the line
• Set of all Natural Numbers
• Set of all Integers.

Cardinality of Infinite Sets: 

Cardinality of Set A is n(A) = x where x is the number of elements in the Set A. Cardinality of Infinite Set is n(A) = ∞ since the number of elements in it is unlimited.

Properties of Infinite Sets

• The Superset of an infinite set is also infinite
• The Union of two infinite sets is infinite
• The Power set of an infinite set is infinite

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Power Set is a set that includes all the Subsets along with Empty Set and the Original Set itself. To know more about Sets check out Set Theory and get a good grip on the concept. Check out Power Set Notation, Properties, How to Calculate Power Set, Power Set of Empty Set in the coming modules. For a better understanding of the Concept Power Set we even provided solved examples.

Example:

If set A = {u,v,w} is a set, then all its subsets {u}, {v}, {w}, {u,v}, {v,w}, {u,w}, {u,v,w} and {} are the elements of powerset, such as Power set of A, P(A) = {u}, {v}, {w}, {u,v}, {v,w}, {u,w}, {u,v,w} and {}

Where P(A) represents the Powerset.

Definition of Power Set

Power Set of A is defined as all the subsets within the Set A along with the Null Set and the Set itself. It is represented by P(A) and is a combination of the null set, set itself, and subsets.

How to Calculate Power Set?

If a Set has n elements then the Power Set can be obtained using the Formula 2ⁿ. It even denotes the Cardinality of a Power Set.

Example

Let us assume Set A = { x, y, z }

Number of elements: 3

Therefore, the subsets of the set are:

{ } which is the null or the empty set

{ x } { y } { z } { x, y } { y, z} { z, x } { x, y, z }

The power set P(A) = { { } , { x } { y } { z } { x, y } { y, z} { z, x } { x, y, z } }

Now, the Power Set has 2³ = 8 elements.

Power Set Notation

The number of elements of a power set is given by |A|, If A has n elements then it can be represented as |P(A)| = 2ⁿ

Properties of Power Set

• Power Set is much larger compared to the Original Set.
• The number of elements in the Power Set A is 2ⁿ where n represents the number of elements in Set A.
• Power Set of Finite Set if Finite and Countable.
• For a Set of Natural Numbers, we can do one-one mapping of the resultant set, P(S) with real numbers.
• P(S) of Set S if performed Operations like Union, Intersection, Complement denotes the Boolean Algebra.

Power Set of Empty Set

In general, Empty Set has no elements and the Power Set of Empty Set denotes the following

• A Set containing Null or Void Set.
• Empty Set is the only Subset.
• It Contains No Elements in the Set.

Solved Examples on Power Set

1. Find the Power Set of Z = {4,7,8}?

Solution:

Given Set Z= {4,7,8}

Number of Elements n = 3

2³ = 8, that shows there will be eight elements of power set of Z

Subsets of Z are {{} {4}, {7}, {8}, {4, 7} {7, 8} {4, 8} {4, 7, 8}}

2. What is the Power set of an Empty set?

Solution:

Number of Elements in Empty Set = 0

No. of Elements in Power Set = 2⁰

= 1

Thus, there is only one element in the power set and that is an empty set.

Power Set of Empty Set P(E) = 1.

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To solve problems on the union of sets you need to be aware of what is meant by a union. Get the Formulas associated with the Union of Sets. For better understanding, we have listed solved examples on finding the union of sets. After going through this article, you can get a fair idea of how to find the union of two or more sets. To know more on Operations of Sets check out the Set Theory and learn the concepts effectively.

Union of Sets

Union of Two Sets A and B is the Set of elements in Set A or B or in both. It is represented as A U B and is areas as A union B or Union of A and B.

(i) Union of Disjoint Sets

If A and B are two finite sets and if A ∩ B = ∅ then

n(A U B) = n(A) +n(B)

Union of Disjoint Sets represented using the Venn Diagram as such. A ∩ B = ∅ since no two sets have common elements.

(ii) Union of Two Sets

If A and B are two finite sets then A union B is given by

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

You can simply say that the Union of A and B is nothing but the summation of cardinal numbers of Set A and B minus Intersection of them.

In the above diagram, the three disjoint sets are A – B and B – A, A ∩ B their sum represents the A U B. Thus, we can say

n (A ∪ B) = n (A – B) + n(B – A) + n(A ∩ B)

(iii) Union of Three Sets

If A, B, C are three finite sets then n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

From the Venn diagram, it is quite clear that the Union of Three Sets is the summation of cardinal numbers of Set A, Set B, Set C, and the common elements of three sets excluding the common elements of sets taken in pairs.

Solved Examples on Union of Sets

1. A = {0, 1, 3, 5, 7, 9}, B = {2, 4, 6, 8} and C = {1, 5, 7, 9}. Find (A ∪ B) ∪ C?

Solution:

(A U B) = {0, 1, 3, 5, 7, 9} U {2, 4, 6, 8}

= { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(A ∪ B) ∪ C = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} U {1, 5, 7, 9}

= { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

2. X = {3, 4, 5, 6}, Y = {2, 5, 7} and Z = {4, 5, 6}. Verify X U Y = Y U X

Solution:

X U Y = {3, 4, 5, 6} U {2, 5, 7}

= { 2, 3, 4, 5, 6, 7}

Y U X = { 2, 5, 7} U {3, 4, 5, 6}

= { 2, 3, 4, 5, 6, 7}

Therefore X U Y = Y U X.

3. Let A = {x : x is a natural number and a factor of 12} and B = {x : x is a natural number and less than 4}. Find A ∪ B?

Solution:

A = { 1, 2, 3, 4, 6, 12}

B = { 1, 2, 4}

A U B = { 1, 2, 3, 4, 6, 12} U { 1, 2, 4}

= { 1, 2, 3, 4, 6, 12}

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Usually, the Set Intersection Operation takes only the elements that are in both sets. The intersection is nothing but the overlap of two sets. Solved Problems on Intersection of Sets helps you get an idea of how to find the intersection of two or more sets. To learn more about the Operations of Sets refer to the Set Theory and clarify all your concerns. Check out the Example Problems on Intersection of Sets explained step by step.

Intersection of Sets

The intersection of Two Sets A and B is nothing but the set of all the elements that are common to both A and B. It is denoted as (A ∩ B) and is read as A Intersection B or Intersection of A and B.

The Intersection of Sets A and B is represented as A ∩ B = {x: x ∈ A and x ∈ B}

Solved Examples on Intersection of Sets

1. If A = {1, 4, 3, 7, 9, 12} B = {1, 8, 3, 4, 7}. Find A ∩ B?

Solution:

A ∩ B = {1, 4, 3, 7, 9, 12} ∩ {1, 8, 3, 4, 7}

= {1, 3, 4, 7}

2. If set A = {2, 4, 6, 8, 9, 12}, set B = {1, 3, 5, 9, 15, 18} and set C = {2, 3, 5, 6, 7, 10}.

(i) Find Intersection of Sets A and B?

(ii) Find Intersection of Two Sets B and C?

(iii) Find Intersection of the Sets A and C?

Solution:

A ∩ B = {2, 4, 6, 8, 9, 12} ∩ {1, 3, 5, 9, 15, 18}

= {9}

B ∩ C = {1, 3, 5, 9, 15, 18} ∩ {2, 3, 5, 6, 7, 10}

={3, 5}

A ∩ C = {2, 4, 6, 8, 9, 12} ∩ {2, 3, 5, 6, 7, 10}

= {2, 6}

3. If A = {x, y, z} and B = {ф}. Find intersection of two given sets A and B?

Solution:

A ∩ B = {x, y, z} ∩ {ф}

= {}

The post Problems on Intersection of Sets appeared first on Learn CBSE.

Are you searching for a tool that plots a distance vs time graph for you? No need to worry, here we are providing the easy and simple steps that are helpful to draw a graph of distance vs time. Get useful information such as the distance traveled by an object in time, distance-time graph definition, and its importance from this page. You will also get some example questions in the below sections.

Graph of Distance vs. Time Definition

Distance vs time graph shows how far a moving object traveled in a specific amount of time. It is a simple graph line that shows the displacement of an object. We will take distance along the y-axis, time along the x-axis.

Importance of the Time Distance Graph

Distance and time graph deals with the motion of bodies. You need to record the distance, time of a moving body and plot those points on a graph to get the Graph of Distance Versus Time. From this graph, you can get the uniform velocity, speed of the object.

The different types of graph lines are listed here.

1. If the graph is a linear line, then the object is moving with either fast or steady speed.

2. If the graph line is a curve, then the object speed is getting faster.

3. If the graph line is a straight line parallel to the time, then the object speed is constant.

4. If the graph line is moving upwards, then speed is steady. When the graph is moving downwards, then speed is returning to start.

Steps to Draw a Graph of Distance Vs Time

Below are the rules you need to check for while drawing a distance vs time graph. You will get an idea in detail by referring to the lines outlined below.

• Get the equation between distance and time of the object.
• Find the distance for the random values of the time and put them on a table.
• Now plot the points on a graph paper and join them to get a line.

Solved Examples of Distance Time Graph

Example 1.

A bus driver drives at a constant speed which is indicated by the speedometer and the driver measures the time taken by the bus for every kilometer. The driver notices that the bus travels 10 kilometers every 20 minutes. Draw a graph for the given details and find the distance covered when the time is 60 minutes, 75 minutes.

Solution:

Given that,

The bus covers 10 kilometers of distance in 20 minutes duration.

10 Km = 20 minutes

D = 2T

When T = 10, D = 2 * 10 = 20

When T = 20, D = 2 * 20 = 40

When T = 30, D = 2 * 30 = 60

When T = 40, D = 2 * 40 = 80

Along the x-axis: Take 1 small square = 10 minutes.

Along the y-axis: Take 1 small square = 20 km.

Plot the points A (10, 20), B (20, 40), C (30, 60), D (40, 80) on a graph.

Join those points to get a linear equation.

From the graph, we can observe that the distance covered in 60 minutes or 1 hour is 120 km.

The distance covered in 75 minutes is 150 km.

Example 2.

A person traveled 30 km away from the starting point and that took 2 hours duration.

(a) After that, he takes a rest for 1 hour.

(b) And, he moved further 30 km in 30 minutes.

(c) He traveled the remaining 60 km back to the starting point in 1 hour 30 minutes.

Solution:

Given that,

Along the x-axis: Take 1 small square = 30 minutes.

Along the y-axis: Take 1 small square = 10 km.

Now, plot the points A (2, 30), B (3, 30), C (3.5, 60), D (4, 30), E (5, 0)

Join those points to get the graph line.

Example 3.

Draw a graph of distance vs time. When the distance covered by an object is 7 times the time taken. Find the distance covered at 7.5 minutes, 8.5 minutes?

Solution:

Given that,

Distance D = 7 * Time

D = 7T

When T = 0.5, D = 7 * 0.50 = 3

When T = 1, D = 7 * 1 = 7

When T = 1.5, D = 7 * 1.5 = 10.5

When T = 2, D = 7 * 2 = 14

When T = 2.5, D = 7 * 2.5 = 17.5

Along the x-axis: Take 1 small square = 0.5 units.

Along the y-axis: Take 1 small square = 5 units.

Now plot the points A (0.5, 3), B (1, 7), C (1.5, 10.5), D (2, 14), E (2.5, 17.5) on a graph.

Join the points.

From the graph, we can observe that the distance covered at 7.5 minutes is 52.5.

The distance covered at 8.5 minutes is 59.5 units.

The post Graph of Distance vs. Time appeared first on Learn CBSE.

Here we will learn about the graph of simple interest vs. the number of years. On a graph paper, take the time or the number of years on the x-axis, simple interest amount on the y-axis. Get the simple steps to draw a graph of simple interest vs time and solved example questions in the below sections.

How to draw a Graph of Simple Interest vs. Number of Years?

Go through the following sections to find the steps to draw a graph of simple interest vs the number of years. You have to find the relation between the simple interest amount, time period for a constant interest rate.

• Let us take the total amount and number of years to calculate the simple interest.
• Treat the constant interest rate of 1%.
• Calculate the interest amount for every year and write it as a table.
• Get the points and plot them on the graph.
• Join those points to get the required graph line.
• To read the numbers from the graph, substitute the time and find the total amount and draw on the graph.

Example Questions & Answers

Example 1.

Draw the graph between simple interest vs Number of years on amount Rs. 2000 at the rate of interest of 4% per annum?

Solution:

Given that,

Amount taken for interest p = Rs. 2000/-

Rate of interest r = 4%

Percentage of simple interest = (prt) / 100

= (2000 * 4 * t) / 100

= 80t

Consider 80t as a simple function.

By putting t = 1, 2, 3 successively, we will get the corresponding values of 80t. We get the table given below.

Along the x-axis: Take 1 small square = 1 unit.

Along the y-axis: Take 1 small square = 25 units.

Plot the points A (1, 80), B (2, 160), C (3, 240) on a graph paper.

Join these points to get the graph.

Example 2.

Simple interest on a certain sum is $ 50 per year. Then, S = 50 t, where t is the number of years.
1. Draw a graph of the above function.
2. From the graph find the value of S, when (a) t = 5 (b) t = 6?

Solution:

Given sample function is S = 50 t

Put t = 1, 2, 3, 4 successively and getting the corresponding value of S. we get the table given below.

Along the x-axis: Take 1 small square = 1 unit.

Along the y-axis: Take 1 small square = 25 units.

Plot the points A (1, 50), B (2, 100), C (3, 150), D (4, 200) on a graph.

Join the points to get a graph line.

Reading off from the graph of simple interest vs. a number of years:

(a) On the x-axis, take the point L at t = 5.

Draw LP ⊥ x-axis, meeting the graph at P.

Clearly, PL = 250 units.

Therefore, t = 5 ⇒ S = 250.

(b) On the x-axis, take the point M at t = 6

Draw MQ ⊥ x-axis, meeting the graph at Q.

So, MQ = 300 units.

Therefore, t = 6 ⇒ S = 300.

Example 3.

Draw the graph between simple interest vs Number of years on amount Rs. 1500 at the rate of interest of 2% per annum. From the graph find the value of Simple interest, when (a) number of years = 5 years 6 months (b) t = 8 years 9 months?

Solution:

Given that,

Amount taken for interest p = Rs. 1500

Rate of interest r = 2%

Percentage of simple interest = (ptr) / 100

= (1500 * 2 * t) / 100

= 30 t

Let us take S = 30t as the function.

Find the value of simple interest, when the number of years is 1, 2, 3. Write them on a table.

Along the x-axis: Take 1 small square = 1 unit.

Along the y-axis: Take 1 small square = 25 units.

Plot the points A (1, 30), B (2, 60), C (3, 90), D (4, 120) on the coordinate graph.

Join the points ABCD to get the required graph line.

Reading off from the graph of simple interest vs. a number of years:

(a) On the x-axis, take the point L at t = 5 years 6 months.

Draw LP ⊥ x-axis, meeting the graph at P.

Clearly, PL = 165 units.

Therefore, t = 5 ⇒ S = 165.

(b) On the x-axis, take the point M at t = 8 years 9 months

Draw MQ ⊥ x-axis, meeting the graph at Q.

So, MQ = 262.5 units.

Therefore, t = 6 ⇒ S = 262.5.

The post Graph of Simple Interest vs. Number of Years appeared first on Learn CBSE.

This page defines the relationship between the square side length and square area via a coordinate graph. Take the square area, square side as the coordinates of a point. And plot those points on the graph and read the unknown values from the graph easily. You can get the solved example questions on how to draw a graph of area vs side of a square in the following sections.

Relation Between Square Side Length & Area

The square area is defined as the product of the length of each side with itself. Its formula is given as side_length².

So, the relationship between the square side and the area is a square graph.

Square Area A = side² = s².

Solved Example Questions

Example 1.

Draw a graph of area vs side of a square. From the graph, find the value of the area, when the side length of the square = 3.

Solution:

Square Area A = side² = s².

For different values of s, we get the corresponding value of A.

When s = 0, A = 0² = 0

When s = 1, A = 1² = 1

When s = 2, A = 2² = 4

Thus, we have the points O (0, 0), A (1, 1), B (2, 4).

Plot these points on a graph paper and join them successively to obtain the required graph given below.

Reading off from the graph of area vs. side of a square:

On the x-axis, take the point L at s = 3.

Draw LP ⊥ x-axis, meeting the given graph at P.

Clearly, PL = 9 units.

Therefore, s = 3 ⇒ A = 9.

Thus, when s = 3 units, then A = 9 sq. units

Example 2.

Draw a graph for the following.

Side of the square (in cm) 1, 2, 3, 4 and Area (in cm) 1, 4, 9, 16 Is it a linear graph?

Solution:

Area of the square A = side² = s².

Draw these square side, area on a table.

Take the side of the square on the x-axis, area on the y-axis.

Plot the points P (1, 1), Q (2, 4), R (3, 9), s (4, 16) on the graph paper.

From the graph, we can say that square area vs side does not form a linear graph. It forms a square graph.

Example 3.

(a). Consider the relation between the area and the side of a square, given by A = s². Draw a graph of the above function.

(b). From the graph, find the value of A, when s = 2.5, 3.5.

Solution:

Given that,

Square Area A = s².

For different values of s, we get the corresponding value of A.

s = 0 ⇒ A = 0² = 0

s = 0.5 ⇒ A = 0.5² = 0.25

s = 1 ⇒ A = 1² = 1

s = 1.5 ⇒ A = 1.5² = 2.25

Thus, we get the points O (0, 0), A (0.5, 0.25), B (1, 1), C 1.5, 2.25)

Plot these points on a graph paper and join them successively to obtain the required graph given below.

(b). Reading off from the graph of area vs. side of a square:

On the x-axis, take the point L at s = 2.5.

Draw LP ⊥ x-axis, meeting the given graph at P.

Clearly, PL = 6.25 cm².

Therefore, s = 2.5 ⇒ A = 6.25.

Thus, when s = 2.5 cm, then A = 6.25 cm²

On the x-axis, take the point M at s = 3.5.

Draw MQ ⊥ x-axis, meeting the given graph at Q.

Clearly, MQ = 12.25 cm².

Therefore, s = 3.5 ⇒ A = 12.25 cm²

Thus, when s = 3.5 cm, then A = 12.25 cm²

The post Graph of Area vs. Side of a Square appeared first on Learn CBSE.

Wanna solve challenging Time and Work Questions? If yes, here is the best practice material for you. We are providing Practice Test on Time and Work and quiz along with solutions. You have to practice so many questions to improve your time management skills. With the help of the below-mentioned practice question and answers, you will come to know the various types of questions asked in the exam. Evaluate your performance with the help of the questions given below.

Practice Test on Time and Work

1.  X alone can do a piece of work in 8 hours while Y alone can do it in 12 hours. Both X and Y together can complete the work in how many days?

A. 21/4 Hours

B. 10 Hours

C. 5 Hours

D. 24/5 Hours

Solution: D (24 /5 Hours)

2. Worker X takes 8 hours to do a job. Worker Y takes 10 hours to do the same job. How long it will take when both X and Y are working together but independently, on the same job?

A. 40/9

B. 60/3

C. 44/2

D. 55/4

Solution: A (40/9)

3. X and Y together can complete a piece of work in 4 days. If X alone can complete the same work in 12 days. In how many days can Y alone can complete that work?

A. 4 Days

B. 6 Days

C. 12 Days

D. 24 Days

Solution: B (6 Days)

4. X is twice as good as a workman as Y and together they can finish a piece of work in 18 days. How many days will X alone will take to complete the work?

A. 27 Days

B. 24 Days

C. 15 Days

D. 20 Days

Solution: A (27 Days)

5. 45 men can finish a job in 16 days. 30 more men joined working, 6 days after they initiated working. How many days more will they now take to finish the remaining work?

A. 15 Days

B. 10 Days

C. 6 Days

D. 8 Days

Solution: C (6 Days)

Quiz on Time and Work Problems

6. X, Y, and Z can complete a piece of work in 24,6 and 12days respectively. In how many days, they will finish the complete work?

A. 4 Days

B. 8 Days

C. 10 Days

D. 15 Days

Solution: A (4 Days)

7. 3 boys and 3 men can complete a piece of job in 10 days while 2 boys and 2 men can complete the same piece of job in 8 days. How many days can 2 men and 1 boy finish the work?

A. 60/4

B. 25/2

C. 20/2

D. 15/3

Solution: B (25/2)

8. X and Y undertake to do a piece of work for Rs.600, X alone can do it in 6 days while Y alone can do it in 8 days. By taking the help of Z, they complete it in 3 days. Find each member’s share?

A. Rs.300, Rs.225, Rs.75

B. Rs.500, Rs.100, Rs.55

C. Rs.200, Rs.75, Rs.50

D. Rs.400, Rs.200, Rs.100

Solution: A (Rs.300, Rs.225, Rs.75)

9. X can do a piece of job in 12 days. Y is 60% more efficient than X. If Y alone works, then how many days he will take to complete the work?

A. 8.5 Days

B. 9 Days

C. 7.5 Days

D. 7 Days

Solution: C (7.5 Days)

10. X can work 5 times faster than Y and takes 60 days less compared to B to complete the work. How many days do A and B take individually to complete the work?

A. 20 Days, 60 Days

B. 15 Days, 75 Days

C. 25 Days, 80 Days

D. 15 Days, 50 Days

Solution: B (15 Days, 75 Days)

Time and Efficiency Quick Mock Tests

11. If 24 men can finish the work in 10 days, then find the number of days required to complete the same work by 30 men?

A. 12 Days

B. 15 Days

C. 8 Days

D. 20 Days

Solution: C (8 Days)

12. X can finish the work in 3 days. Y can finish the same piece of work in 6 days and Z can do the same piece of work in 7 days. How many days X, Y, and Z take to complete the job together?

A. 14/9 Days

B. 24 Days

C. 18 Days

D. 15/6 Days

Solution: A (14/9 Days)

13. P and Q can do the work in 12 days. Q and R can do the same work in 16 days, R and P can do it in 24 days. Find the number of days in which P, Q, and R can finish the work together?

A. 30/4 Days

B. 32/3Days

C. 60/4 Days

D. 70/6 Days

Solution: B (32/3 Days)

14. If 3 men can finish a job in 2 days and 4 boys can finish the same job in 6 days. Then, how many days will 8 men and 8 boys take to finish the same job?

A. 30/5 Days

B. 5/30 Days

C. 10/6 Days

D. 6/10 Days

Solution: D (6/10 Days)

15. Gita and Sita can complete a job in 20 days and 25 days respectively. Both start the job together but Gita leaves the job after a few days. After Gita leaves the job, Sita completes the work in 10 days. After how many days did Gita leave?

A. 20/3 Days

B. 60/4 Days

C. 20 Days

D. 15 Days

Solution: A (20/3 Days)

Practice Test on Time and Work – Worksheets

16. When P alone does the job, he takes 25 days more than the time taken by P and Q working together to finish the work. But Q alone takes 9 days more than the time taken by P and Q working together to finish the work. In what time P and Q together will finish this work?

A. 10 Days

B. 20 Days

C. 15 Days

D. 25 Days

Solution:  C (15 Days)

17. X can complete the job in 20 days. X and Y together can complete the job in 12 days. How long does Y take to finish that job?

A. 20 Days

B. 40 Days

C. 30 Days

D. 15 Days

Solution: C (30 Days)

18. Pipe X can fill the complete tank in 40 minutes. Pipe Y alone can fill the tank in 60 minutes. X and Y together can fill the complete tank in how many mins?

A. 120 mins

B. 60 mins

C. 40 mins

D. 80 mins

Solution: A (120 mins)

19. X can complete the work in 20 days, Y in 30 days, and C in 40 days. All of them started working together. But after 5 days, X left. After 3 more days Y also left the work. Z can complete the remaining work in how many days?

A. 35/6 Days

B. 34/3 Days

C. 45/7 Days

D. 60/8 Days

Solution: B (34/3 Days)

20. W can complete the work in 25 days, X in 30 days, Y in 50 days, Z in 75 days. W and X started working together and later Y joined after 7 days of work. After 3 more days Z replaced X. After 2 more days, W and Y left the work. In how many days did Z complete the work?

A. 8/3

B. 9/2

C. 6/2

D. 10/6

Solution: B (9/2)

Here are the most interesting and various questions regarding Time and Work. Get Practice Test on Time and Work here and score better marks. Stay tuned to our site to get the latest updates on Mock Tests, Preparation Tips, Study Materials, Practice Papers, etc.

The post Objective Time and Work Questions and Answers | Practice Tests & Quiz appeared first on Learn CBSE.

Are you preparing for competitive exams? If yes, then you must check this article to know Time & Work shortcuts and Tricks. Understand Time and Work Quantitative Aptitude Questions and Formula. Check shortcuts mentioned here and practice them accurately to solve those questions in a limited time.

Time & Work Quick Tricks

Solving aptitude questions in a limited time is really a great blessing for the candidates preparing for competitive exams. It can only be possible through regular practice. We are providing a few solved example questions, easy steps, and shortcuts to prepare for the exam. Aptitude is actually easy and interesting if you know shortcuts and steps to solve the problem. You have to follow a few tips and tricks to solve Time and Work Questions.

This section of aptitude is tricky when compared with other sections and you need to practice more. Also, make your basics strong so that you can solve many problems in less time. Time problems deal with the efficiency of an individual or group or the time taken to complete the work in time. Work is the effort put in to accomplish the task or produce a deliverable.

The relation between Efficiency and Time Formulas

1. If X and Y can together complete a workpiece in A days, X and Y in B days, and Z and X in C days, then
   a) X, Y, and Z working together will complete the work in (2ABC/AB+BC+CA) days.
   b) X alone will complete the work in (2ABC/AB+BC – CA) days.
   c) Y alone will complete the work in (2ABC/CA+AB – BC) days.
   d) Z alone will complete the work in (2ABC/CA+BC – AB) days.
2. If X and Y working together can complete a piece of work in A days and Y is k times efficient than X, then the time taken by X working alone to finish the piece of work is (k+1)A and Y working alone to finish the work is (k+1/k)A
3. Suppose X can finish a job in A days and Y is k times efficient than X, then the time taken by both X and Y working together to finish the work is A/(1+k).
4. If X does 1/nth of a job in an hour, then to finish the full job X will take n*a hours
5. If 1/n of a job is done by A in one day, then X will take n days to finish the full work.
6. Suppose X, Y, and Z, while working alone, can finish work in A, B, and C days respectively, then they will work together to finish the work in ABC/(AB+BC+CA) days.
7. If X can do a piece of work in A days and Y can do the same work in B days, then both of them together will finish the same work in AB/(A+B) days.

Formulas on Time and Work

There are various approaches to solve efficiency and work problems. We are going to discuss those approaches with you here. Before starting the practice know all the approaches and formulae and be perfect in that.

1st Approach: The Chocolate Method

To make calculations more instinctive, the job can be supposed as chocolates to be consumed instead of units of work to be finished. We consider the LCM of all the “total no of days” given in the question. This is done with the goal that the work will be multiple of “number of days” and hence calculating efficiency will be simpler.

Example:

Question: If X can do work in 9 days and Y can do the same work in 18 days, in how many days can they finish it working together?

Solution:

With the above-mentioned chocolate method, you can solve this in the following steps.

• Assume work as chocolates
• Hence, the total no of chocolates to be consumed =18 units (LCM of 9 and 18)
• This means X can eat 18 chocolates in 9 days.
• X consumes chocolates in one day = 2 units.
• Y can eat 18 chocolates in 18 days.
• Y consumes chocolates in one day = 1 unit per day
• Chocolates consumed by X and Y in one day together = 3 units
• Time taken by both of them to consume 18 chocolates = 18/3 = 6 days

2nd Approach: Per Day’s Work

In the event that X can finish the work in ‘a’ days and Y can finish a similar work in ‘b’ days, when they cooperate, the time taken to finish the work is given as following.

X can finish the work in ‘a’ days. So in one day, he will do 1/a of the work. Y can finish the work in ‘b’ days. So in one day, he will do 1/b of the work. Complete work done by both in one day = (1/a) + (1/b). Thus, the total time needed to fulfill the work = (ab)/(a +b) days.

Question: If X finishes the entire work in 10 days, Y finishes the work in 12 days. In how many days can X, Y day finish the work when worked together?

Solution: 60/11 days

• Since X finishes the entire work in 10 days, X completes 1/10th of the work in 1 day.
• Since Y finishes the entire work in 12 days, Y completes 1/12th of the work in 1 day.
• Working together they can finish the work =1/10+1/12 = 11/60 of the work in 1 day. Hence total days taken by both of them to finish the work = 60/11 days.

3rd Approach: LCM Method

In this method, we expect the total sum of work to be finished as a finite divisible and dependent on it, we continue with the calculation. To make the calculation easier, suppose the total sum of work to be finished as the LCM of time taken by various people to finish a similar work.

In the event that X can finish the work in ‘a’ days and Y can finish a similar work in ‘b’ days when they cooperate, the time taken to finish the work is given as following.

Question: If X finishes the entire work in 10 days, Y finishes the work in 12 days. In how many days can X, Y day finish the work when worked together?

Solution:

Let the amount of piece of work be 60 units (LCM of 10 and 12). Since X works 60 units in 10 days, so he works 6 units every day. Since Y works 60 units in 12 days, he works 5 units every day. Working together, they do 6 + 5 = 11 units each day. Hence to complete 60 units of work, both together will take 60/11 days.

Work Equivalence

In the man-days concept questions, we assume that all men work with the same efficiency unless it is given in the question. The relationship between the number of people working(N), the total number of hours worked per day(H), the total number of days worked(D) and quantity of work done(W) for different cases is as follows:

N(1) x D(1) x H(1)/W(1) = N(2) x D(2) x H(2)/W(2)

For the above equations, the relationship between variables is as follows:

• The total number of people working and work done are directly proportional
• The amount of work done and the number of days worked is directly proportional
• The total number of people working and the number of days worked is inversely proportional.

We hope that now you got the complete idea of Time and Work. Follow all the strategies and tricks mentioned above to improve your problem-solving skills. If you need any further clarification, you can ask us through the contact us page or in the comment box mentioned below. Bookmark our site to get all instant and latest updates.

The post Solved Time and Work Problems, Tricks | Time and Work Questions and Answers appeared first on Learn CBSE.