| | ||||||||||
Q) What is graph?
A) Show/hide Answer
Q) What is pie chart?
A) Show/hide Answer
Q) What is line graph?
A) Show/hide Answer
Q) What is Linear Graphs?
A) Show/hide Answer
Q) What is Location of a point?
A) Show/hide Answer
Q) The point where the x-axis and y-axis intersect is called the?
A) Show/hide Answer
Q) The ordinate of a point is its distance from the
A) Show/hide Answer
Q) How to Construct a Bar Graph?
A) Show/hide Answer
Q) 150 students of class VI have popular school subjects as given below:
| subject | Number Of Students |
| French | 40 |
| English | 55 |
| Maths | 55 |
| Geography | 19 |
| Science | 20 |
A) Show/hide Answer
Q) Find the mode of the following data? 14, 8, 4, 8, 1,9, 9, 41, 6, 10, 12, 0.
A) Show/hide Answer
Q) The weights in kg of 10 students are given below: 29, 43, 36, 48, 46, 41, 33,74, 44,73 Find the mode of this data. Is there more than 1 mode? If yes, why?
A) Show/hide Answer
Q) Find the mean of first 10 prime numbers?
A) Show/hide Answer
Q) Find the mean of first twenty whole numbers?
A) Show/hide Answer
Q) Find the mean of the following data? 16, 8, 8, 8, 7, 8, 9, 21, 9, 10, 12, 8.
A) Show/hide Answer
Q) The equation y = mx is always passing through the origin?
A) Show/hide Answer
Q) The equation representing y-axis is?
A) Show/hide Answer
Q) The point (1,-3) lies in the __________ quadrant.
A) Show/hide Answer
Q) Equation of y-axis is x = 0. The standard form of this equation is x + 0.y =?
A) Show/hide Answer
Q) y = _____ is a graph of straight line parallel to x-axis and standard form of this equation is 0.x + y = b.
A) Show/hide Answer
Q) The runs scored in a cricket match by 11 players is as follows: 7, 42, 131, 51, 202, 81, 7, 16, 68, 11, 28 Find the mean, mode, median of this data?
A) Show/hide Answer
Q) Find the median of the following data? 37, 39,39, 70, 31, 98, 38.
A) Show/hide Answer
Q) The following observations are arranged in ascending order. The median of the data is 25 find the value of x. 17, x, 54, x +8 , 35, 39, 40?
A) Show/hide Answer
Q) The mean of 9, 31, 6, 24, x and 13 is 66. Find the value of the observation x?
A) Show/hide Answer
Q) The mean of 6, 8, x + 7, 10, 9x - 6, and 2 is 11. Find the value of x and also the value of the observation in the data?
A) Show/hide Answer
Q) The following observations are arranged in ascending order. The median of the data is 45 find the value of x? 6, x, 24, x +8, 39, 39, 40.
A) Show/hide Answer
Q) x = a is a graph of straight line parallel to y-axis and standard form of this equation is x + 0.y =?
A) Show/hide Answer
Q) x = a is a graph of straight line parallel to y-axis and standard form of this equation is x + 0.y =?
A) Show/hide Answer
Q) In the below image What is the different of expenditures minimum?
A) Show/hide Answer
Q) In below figure,How much measure of central angle of playing?
A) Show/hide Answer
Q) In below figure,What is the Number of vehicles in Time in hours of 8-9?
A) Show/hide Answer
- Bar Graphs
- Pie Chart & Pie graph
- A Line Graph
- Worksheet on Line Graph
- Linear Graphs
- Location of a point
- Coordinates
- Frequency Distribution
- 1. Graphical presentation of data is easier to understand.
- 2. (i) A bar graph is used to show comparison among categories.
(ii) A pie graph is used to compare parts of a whole.
(iii) A Histogram is a bar graph that shows data in intervals. - 3. A line graph displays data that changes continuously over periods of time.
- 4. A line graph which is a whole unbroken line is called a linear graph.
- 5. For fixing a point on the graph sheet we need, x-coordinate and y-coordinate.
- 6. The relation between dependent variable and independent variable is shown through a graph.
- It consists of rectangular bars of equal width.
- The space between the two consecutive bars must be the same.
- Bars can be marked both vertically and horizontally but normally we use vertical bars.
- The height of bar represents the frequency of the corresponding observation.
For example, let us observe the following data of the bar graph.
The following data gives the information of the number of children involved in different activities.
| Activities | Dance | Music | Art | Cricket | Football |
| No. Of Children | 30 | 40 | 25 | 20 | 53 |
How to Construct a Bar Graph?
- On a graph, draw two lines perpendicular to each other, intersecting at 0.
- The horizontal line is x-axis and vertical line is y-axis.
- Along the horizontal axis, choose the uniform width of bars and uniform gap between the bars and write the names of the data items whose values are to be marked.
- Along the vertical axis, choose a suitable scale in order to determine the heights of the bars for the given values. (Frequency is taken along y-axis).
- Calculate the heights of the bars according to the scale chosen and draw the bars.
Bar graph gives the information of the number of children involved in different activities.
1. The percentage of total income spent under various heads by a family is given below.
| Different Heads | Food | Clothing | Health | Education | House Rent | Miscellaneous |
| % Age of Total Number |
40% | 10% | 10% | 15% | 20% | 5% |
Represent the above data in the form of bar graph.
2. 150 students of class VI have popular school subjects as given below:
| Subject | French | English | Maths | Geography | Science |
| Number Of Students | 30 | 20 | 26 | 38 | 34 |
Draw the column graph/bar graph representing the above data.
Solution:
Take the subjects along x-axis, and the number of students along y-axis
Bar graph gives the information of favourite subjects of 150 students.
3. The vehicular traffic at a busy road crossing in a particular place was recorded on a particular day from 6am to 2 pm and the data was rounded off to the nearest tens.
| Time in hours | 6-7 | 7-8 | 8-9 | 9-10 | 10-11 | 11-12 | 12-1 | 1-2 |
| Number of vehicles | 100 | 450 | 1250 | 1050 | 750 | 600 | 550 | 200 |
Bar graph gives the information of number of vehicles passing through the crossing during different intervals of time.
It is a circular graph which is used to represent data. In this :
- Various observations of the data are represented by the sectors of the circle.
- The total angle formed at the centre is 360°.
- The whole circle represents the sum of the values of all the components.
- The angle at the centre corresponding to the particular observation component is given by
Value of the component × 360°
`bar ( Total Value )`
- If the values of observation/components are expressed in percentage, then the centre angle corresponding to particular observation/component is given by
Percentage Value of component × 360°
`bar ( 100 )`
Steps of construction of pie chart for a given data:
- Find the central angle for each component using the formula given on the previous page.
- Draw a circle of any radius.
- Draw a horizontal radius
- Starting with the horizontal radius, draw radii, making central angles corresponding to the values of respective components.
- Repeat the process for all the components of the given data.
- These radii divide the whole circle into various sectors.
- Now, shade the sectors with different colours to denote various components.
- Thus, we obtain the required pie chart.
- Calculate the central angle for each component, given by
- Draw a circle of convenient radius.
- Within this circle, draw a horizontal radius.
- Starting with the horizontal radius, draw radii making central angles corresponding to the values of the respective components, till all the components are exhausted. These radii divide the whole circle into various sectors.
- Shade each sector with different design.
- Draw a circle of any convenient radius.
- Draw a horizontal radius of this circle.
- Draw sectors starting from the horizontal radius with central angles of 105 degree, 65 degree, 70 degree, 80 degree and 40 degree respectively.
- Shade the sectors differently using different colors and label them.
- Draw a circle of any convenient radius.
- Draw a horizontal radius of this circle.
- Draw sectors starting from the horizontal radius with central angles of 60 degree, 90 degree, 120 degree , 40 degree and 50 degree respectively.
- Shade the sectors differently using different colors and label them.
- Draw a circle of any convenient radius.
- Draw a horizontal radius of this circle.
- Draw sectors starting from the horizontal radius with central angles of 125 degree, 100 degree, 60 degree, 45 degree and 30 degree respectively.
- Shade the sectors differently using different colors and label them.
- Draw a circle of any convenient radius.
- Draw a horizontal radius of the circle.
- Draw sectors starting from the horizontal radius with central angles of 54 degree, 72 degree, 36 degree, 90 degree and 108 degree respectively.
- Shade the sectors differently using different colors and label them.
Solved example on construction of pie chart/pie graph:
1. The following table shows the numbers of hours spent by a child on different events on a working day.
Represent the adjoining information on a pie chart
| Activity | No. Of Hours |
|---|---|
| School | 6 |
| Sleep | 8 |
| Playing | 2 |
| Study | 4 |
| T.V. | 4 |
| Others | 3 |
The central angles for various observations can be calculated as:
| Activity | No. Of Hours | Measure Of Central Angle |
|---|---|---|
| School | 6 | ((6/24)×360)°=90° |
| Sleep | 8 | ((8/24)×360)°=120° |
| Playing | 2 | ((2/24)×360)°=30° |
| Study | 4 | ((4/24)×360)°=60° |
| T.V. | 4 | ((1/24)×360)°=15° |
| Others | 3 | ((3/24)×360)°=45° |
Now, we shall represent these angles within the circle as different sectors. Then we now make the pie chart:
2. The favourite flavours of ice-cream for the children in a locality are given in percentage as follow. Draw the pie chart to represent the given information
| Flavours | % of students prefer the flavours |
|---|---|
| Vanilla | 25% |
| Strawberry | 15% |
| Chocolate | 10% |
| Kesar-Pista | 30% |
| Mango Zap | 20% |
The central angles for various observations can be calculated as:
| Flavours | % of students prefer the flavours | Measure Of Central Angles |
|---|---|---|
| Vanilla | 25% | ((25/100)×360)°=90° |
| Strawberry | 15% | ((15/100)×360)°=54° |
| Chocolate | 10% | ((10/100)×360)°=36° |
| Kesar-Pista | 30% | ((30/100)×360)°=108° |
| Mango Zap | 20% | ((20/100)×360)°=72° |
Now, we shall represent these angles within a circle to obtain the required pie graph.
In a pie chart, the various observations or components are represented by the sectors of a circle and the whole circle represents the sum of the values of all components.
The central angle for a component is given by:
Central angle for a component = (Value of the component/Sum of the values of all components × 360°)
Construction of making a pie chart or graph from the given data.
Steps of pie graphs Construction:
Central angle of component = (Value of the component/total value × 360°)
This will be the required pie chart for the given data.
1. Mr. Bin's with a yearly salary of $ 10800 plans his budget for a year as given below:
| Item | Food | Education | Rent | Savings | Miscellaneous |
|---|---|---|---|---|---|
| Amount(in dollar) | 3150 | 1950 | 2100 | 2400 | 1200 |
Represent the above data by a pie graph.
Solution:
Total amount earned by Mr. Bin in a year = $ 10800.
Central angle of component = (Value of the component/total value × 360°)
Calculation of central angles
| Item | Amount (in $) | Central Angle |
|---|---|---|
| Food | 3150 | ((3150/10800)×360)°=105° |
| Education | 1950 | ((1950/10800)×360)°=65° |
| Rent | 2100 | ((2100/10800)×360)°=70° |
| Savings | 2400 | ((2400/10800)×360)°=80° |
| Miscellaneous | 1200 | ((1200/10800)×360)°=40° |
Construction of making pie chart
Steps of construction:
Thus, we obtain the required pie chart, as shown in the given figure.
2. The data on the mode of transport used by 720 students are given below:
| Mode Of Transport | Bus | Cycle | Train | Car | Scooter |
|---|---|---|---|---|---|
| No. Of Students | 120 | 180 | 240 | 80 | 100 |
Represent the above data by a pie chart.
Solution:
Total number of students = 720.
Central angle for a mode of transport
= (number of students using that mode/total number of students × 360°)
Calculation of central angles
| Mode Of Transport | No of Students | Central Angle |
|---|---|---|
| Bus | 120 | ((120 ÷720) × 360)° = 60° |
| Cycle | 180 | ((180 ÷720) × 360)° = 90° |
| Train | 240 | ((240 ÷720) × 360)° = 120° |
| Car | 80 | ((80 ÷720) × 360)° = 40° |
| Scooter | 100 | ((100 ÷720) × 360)° = 50° |
Construction for creating pie chart
Steps of construction:
Thus, we obtain the required pie chart, as shown in the given figure.
3. There are 216 workers in a factory as per list given below:
| Cadre | Labourer | Mechanic | Fitter | Supervisor | Clerk |
|---|---|---|---|---|---|
| No. Of Workers | 75 | 60 | 36 | 27 | 18 |
Represent the above data by a pie chart.
Solution:
Total number of workers = 216.
Central angle for a cadre = (number of workers in that cadre/ total number of workers × 360°)
Calculation of central angles
| Cadre | No of Workers | Central Angle |
|---|---|---|
| Labourer | 75 | ((75 ÷ 216) × 360)° = 125° |
| Mechanic | 60 | ((60 ÷ 216) × 360)° = 100° |
| Fitter | 36 | ((36 ÷216) × 360)° = 60° |
| Supervisor | 27 | ((27 ÷ 216) × 360)° = 45° |
| Clerk | 18 | ((18 ÷ 216) × 360)° = 30° |
Construction to make a pie graph
Steps of construction:
Thus, we obtain the required pie chart, as shown in the given figure.
4. The following table shows the expenditure in percentage incurred on the construction of a house in a city:
| Item | Brick | Cement | Steel | Labour | Miscellaneous |
|---|---|---|---|---|---|
| Expenditure (in percentage) | 15% | 20% | 16% | 25% | 30% |
Represent the above data by a pie chart.
Solution:
Total percentage = 100.
Central angle for a component
= (value of the component/100 × 360°)
Calculation of central angles
| Item | Expenditure (in percentage) | Central Angle |
|---|---|---|
| Brick | 15% | ((15 ÷ 100) × 360)° = 54° |
| Cement | 20% | ((20 ÷ 100) × 360)° = 72° |
| Steel | 10% | ((10 ÷ 100) × 360)° = 36° |
| Labour | 25% | ((25 ÷ 100) × 360)° = 90° |
| Miscellaneous | 30% | ((30 ÷ 100) × 360)° = 108° |
Construction for creating pie chart
Steps of construction:
Thus, we obtain the required pie chart, shown in the adjoining figure.
The data which changes over a period of time can be displayed through a line graph.
In line graph:
- Points are plotted on the graph related to two variables
- Points are joined by the line segments.
- Linear equations have infinitely many solutions.
- Every point (h, k) on the line AB gives the solution x = h and y = k.
- Every point which lies on AB satisfies the equation of AB.
- To draw an exact line on the graph paper you can plot as many points you like, but it is necessary to plot minimum three points.
- Convert the given equation in the form of y = mx + b (slope intercept form).
- Apply trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation.
- Plot these points on the graph paper.
- Join the points marked on the graph paper to get a straight line which represent the given equation graphically.
- Linear equation in two variables has infinitely many solutions.
- A graph of linear equation is always a straight line.
- Every point on the straight line is the solution of the linear equation.
- Equation of y-axis is x = 0. The standard form of this equation is x + 0.y = 0.
- Equation of x-axis is y = 0. The standard form of this equation is 0.x + y = 0.
- x = a is a graph of straight line parallel to y-axis and standard form of this equation is x + 0.y = a
- y = b is a graph of straight line parallel to x-axis and standard form of this equation is 0.x + y = b.
- The equation y = mx is always passing through the origin (0, 0).
Steps of construction of line graph:
On a graph, draw two lines perpendicular to each other intersecting at O. The horizontal line is x-axis and vertical line is y-axis. Mark points at equal intervals along x-axis and write the names of the data items whose values are to be marked.
Along the y-axis, choose an appropriate scale considering the given values.
Now, make the points
Join each point with the successive point using a ruler. Thus, a line graph is obtained
Solve Examples on Line Graph:
1. The following table gives the information of the sum scored by
McKay in 10 matches. Represent this information using line graph.
| Match | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Runs Scored | 80 | 30 | 50 | 90 | 40 | 60 | 75 | 20 | 15 | 90 |
2. Mobile phones sold by a shop in a certain week are as follows:
| Days | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday |
| No of Mobile Phones sold | 20 | 16 | 12 | 24 | 4 | 10 |
Represent the above data by a line graph.
1. The following are the runs scored by a team in the first 5 overs:
| Match | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Runs Scored | 6 | 4 | 2 | 0 | 10 | 20 | 15 | 6 | 18 | 12 |
Draw the line graph for the above data.
2. Draw the line graph showing the following information. The table shows the colours favoured by a group of people.
| Colours | Yellow | Pink | Blue | Green | Orange |
| No. Of People | 16 | 20 | 30 | 26 | 34 |
3. Can there be time-temperature graphs as follows? Give reason.
4. The following tables give the information about a patient's body temperature recorded in the hospital every hour.
| Time | 9 am | 10 am | 11 am | 12 noon | 1 pm | 2 pm | 3 pm |
| Temperature | 35°C | 36°C | 39°C | 38°C | 36.5°C | 36.5°C | 37°C |
Represent this information on a line graph.
Answers for the worksheet on line graph are given below to check the exact graph and the answers of the above question.
3. (a) and (c) possible (b) not possible - time being constant
Properties for graphing linear equation:
Method to draw the graph of linear equation in two variables:
Note:
Learn the steps for graphing linear equation in two variables:
1. Draw the graph of the linear equation y = 2x.
Solution:
The given linear equation y = 2x is already in the form of y = mx + b [here b = 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 = 2x.
When the value of x = 0, then y = 2 × 0 = 0
When the value of x = 1, then y = 2 × 1 = 2
When the value of x = 3, then y = 2 × 3 = 6
When the value of x = -1, then y = 2 × -1 = -2
When the value of x = -2, then y = 2 × -2 = -4
Arrange the values of the linear equation y = 2x in the table.
| X | 1 | 2 | 3 | 4 |
| Y | 4 | 3 | 2 | 1 |
Now, plot the points P (0, 0), Q (1, 2), R (2, 4), S (3, 6), T (-1, -2), U (-2, -4) on the graph paper.
Join the points of P, Q, R, S, T and U.
We get a straight line passing through origin. This straight line is the graph of the equation y = 2x.
2. Draw the graph of the equation 4x - y = 3.
Solution:
The given linear equation 4x - y = 3.
Now convert the given equation in the form of y = mx + b
4x - y = 3
⇒ 4x - 4x - y = - 4x + 3
⇒ - y = - 4x + 3
⇒ y = 4x - 3
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = 4x - 3.
When the value of x = 0, then y = (4 × 0) - 3 = - 3
When the value of x = 1, then y = (4 × 1) - 3 = 1
When the value of x = 2, then y = (4 × 2) - 3 = 5
Arrange these value of the linear equation y = 4x - 3 in the table.
| X | 0 | 1 | 2 |
| Y | -3 | 1 | 5 |
Now, plot the point P (0, -3), Q (1, 1), R(2,5) on the graph.
Join the points of P, Q and R.
We get a straight line passing through origin. This straight line is the graph of linear equation 4x - y = 3.
The fundamental geometrical concepts depend on three basic concepts - point, line and plane. The terms cannot be precisely defined. However, the meanings of these terms are explained through examples.
Point:
- It is the mark of position and has an exact location.
- It has no length, breadth or thickness.
- It is denoted by a dot made by the tip of a sharp pencil.
- It is denoted by capital letter.
- In the given figure P, Q, R represents different points.
Line:
- It is a straight path which can be extended indefinitely in both the directions.
- It is shown by two arrowheads in opposite directions.
- It does not have any fixed length.
- It has no endpoints.
- It is denoted as AB↔ or BA↔ and is read as line AB or line in the BA.
- It can never be measured.
- Infinite number of points lie on the line.
- Sometimes it is also denoted by small letters of the English alphabet.
Ray:
- It is a straight path which can be extended indefinitely in one direction only and the other end is fixed.
- It has no fixed length.
- It has one endpoint called the initial point.
- It cannot be measured.
- It is denoted as OA→ and is read as ray OA.
- A number of rays can be drawn from an initial point O.
- Ray OA and ray OB are different because they are extended in different directions.
- Infinite points lie on the ray.
Line Segment:
- It is a straight path which has a definite length.
- It has two endpoints.
- It is a part of the line.
- It is denoted as AB or BA.
- It is read as line segment AB or line segment BA.
- The distance between A and B is called the length of AB.
- Infinite number of points lies on a line segment.
- Two line segments are said to be equal if they have the same length.
Plane:
A smooth, flat surface gives us an idea of a plane. The surface of the table, wall, blackboard, etc., is smooth and flat. It extends endlessly in all the directions. It has no length, breadth or thickness. Here, we have shown a portion of a particular plane. We can draw certain figures like square, rectangle, triangle, and circle on the plane. Hence, these figures are also called plane figures.
Incidence Properties of Lines in a Plane:
- An infinite number of many lines can be drawn to pass through a given point in a plane.
Through a given point in a plane, infinitely many lines can be drawn to pass through. - Two distinct points in a plane determine a unique line.
One and only one line can be drawn to pass through two given points, i.e., two distinct points in a plane. This line lies wholly in the plane. - Infinite number of points lie on the line in a plane.
- Two lines in a plane either intersect at a point or they are parallel to each other.
Collinear Points:
Two or more points which lie on the same line in a plane are called collinear points.
- The line is called the line of collinearity.
- Two points are always collinear.
- In the adjoining figure.......
Points A, B, C are collinear lying on line
Points X, Y, Z are not collinear because all the three points do not lie on a line.
Hence, they are called non-collinear points.
Similarly, here points M, N, O, P, Q are collinear points and A, B, C are non-collinear points.
Note:
Two points are always collinear.
Find the Coordinates of a Point:
In the adjoining figure, for locating the coordinates of a point draw XOX' and YOY' are co-ordinate axes.
To locate the position of point P, we draw a perpendicular from P on X'OX, i.e., PT ⊥ XOX' So, the co-ordinate of point P are (OT, PT).
Example to find the coordinates of a point:
1. In the adjoining figure, XOX' and YOY' are the co-ordinate axes. Find out the coordinates of point A, B, C and D.
Solution:
To locate the position of point A, draw AQ ⊥ X'OX.
Then the co-ordinate of point A are (OQ, QA) i.e., A (5, 2). These points lie in the I quadrant.
To locate the position of point B, draw BP ⊥ X'OX.
Then the co-ordinate of point B are (OP, PB) i.e., B (-3, 4). These points lie in the II quadrant.
To locate the position of point C, draw CS ⊥ X'OX.
Then the co-ordinate of point C are (OS, SC), i.e., C (-4, -2). These points lie in the III quadrant.
To locate the position of point D, draw DR ⊥ X'OX.
Then the co-ordinate of point D are (OR, RD) i.e., D (3, -2). These points lie in the IV quadrant.
2. In the adjoining figure, XOX' and YOY' are the co-ordinate axes. Find out the coordinates of point P, Q, R, S, T and U. Also write the abscissa and ordinate in each case.
Solution:
To locate the position of point Q:
Point Q is the I quadrant where abscissa and ordinate both are positive.
Perpendicular distance of Q from y-axis is 4 units.
So, x-co-ordinate of Q is 4.
Perpendicular distance of Q from x-axis is 3 units.
So, y-co-ordinate of Q is 3.
Therefore, co-ordinate of Q are (4, 3).
To locate the position of point P:
Point P is the II quadrant where abscissa is negative and ordinate is positive.
Perpendicular distance of P from y-axis is 2 units.
So, x-co-ordinate of P is -2
Perpendicular distance of P from x-axis is 5 units.
So, y-co-ordinate of P is 5
Therefore, co-ordinate of P are (-2, 5)
To locate the position of point S:
Point S is the III quadrant where abscissa and ordinate both are negative.
Perpendicular distance of S from y-axis is 4 units.
So, x-co-ordinate of S is -4.
Perpendicular distance of S from x-axis is 1 unit.
So, y-co-ordinate of S is -1.
Therefore, co-ordinate of S are (-4, -1)
To locate the position of point R:
Point R is the IV quadrant where abscissa is positive and ordinate is negative.
Perpendicular distance of R from y-axis is 2 units.
So, x-co-ordinate of R is 2
Perpendicular distance of R from x-axis is 4 units.
So, y-co-ordinate of R is -4
Therefore, co-ordinate of R are (2, -4)
To locate the position of point T:
Point T is in the positive x-axis. We know, that the co-ordinate of a point on x-axis are of the form (x, 0)
Perpendicular distance of T from y-axis is 2 units.
So, x-co-ordinate of T is 2
Perpendicular distance of T from x-axis is 0 unit.
So, y-co-ordinate of T is 0
Therefore, co-ordinate of T are (2, 0)
To locate the position of point U:
Point U is in the negative y-axis. We know, that the co-ordinate of a point on y-axis are of the form (0, y)
Perpendicular distance of U from y-axis is 0 units.
So, x-co-ordinate of U is 0
Perpendicular distance of U from x-axis is 4 units.
So, y-co-ordinate of U is -4
Therefore, co-ordinate of U are (0, -4)
In worksheet on frequency distribution the questions are based on arranging data in ascending order or descending order and constructing the frequency distribution table.
1. Arrange the following data in ascending order.
(a) 7, 2, 10, 14, 0, 6, 15, 24, 8, 3
(b) 4.6, 8.1, 2.0, 3.5, 0.7, 9.3, 1.4, 0.8
2. Arrange the following data in descending order.
(a) 14, 2, 0, 10, 6, 1, 22, 13, 28, 4, 8, 16
(b) 1.2, 3.5, 0.1, 0.3, 2.4, 8.6, 5.0, 3.7, 0.7, 0.9
3. Construct the frequency table for each of the following.
(a) 4, 3, 6, 5, 2, 4, 3, 3, 6, 4, 2, 3, 2, 2, 3, 3, 4, 5, 6, 4, 2, 3, 4
(b) 6, 7, 5, 4, 5, 6, 6, 8, 7, 9, 6, 5, 6, 7, 7, 8, 9, 4, 6, 7, 6, 5
4. The marks obtained out of 25 by 30 students of a class in the examination are given below.
20, 6, 23, 19, 9, 14, 15, 3, 1, 12, 10, 20, 13, 3, 17, 10, 11, 6, 21, 9, 6, 10, 9, 4, 5, 1, 5, 11, 7, 24
Represent the above data as a grouped data taking the class interval 0 - 5
5. Complete the table given below.
(a)
(b)
6. Weekly pocket expenses (in $) of 30 students of class VIII are 37, 41, 39, 34, 71, 26, 56, 61, 58, 79, 83, 72, 64, 39, 75, 39, 37, 59, 57, 37, 53, 38, 49, 45, 70, 82, 44, 37, 79, 76.
Construct the grouped frequency table with the class interval of equal width such as 30 - 35. Also, find the range of the weekly pocket expenses.
7. Pulse rate (per minute) of 25 persons were recorded as
61, 75, 71, 72, 70, 65, 77, 72, 67, 80, 77, 62, 71, 74, 79, 67, 80, 77, 62, 71, 74, 61, 70, 80, 72, 59, 78, 71, 72.
Construct a frequency table expressing the data in the inclusive form taking the class interval 61-65 of equal width. Now, convert this data again into the exclusive form in the separate table.
8. The frequency distribution of weights (in kg) of 40 persons is given below.
(a) What is the lower limit of fourth class interval?
(b) What is the class size of each class interval?
(c ) Which class interval has the highest frequency?
(d) Find the class marks of all the class intervals?
9. Construct the frequency distribution table for the data on heights (cm) of 20 boys using the class intervals 130 - 135, 135 - 140 and so on. The heights of the boys in cm are: 140, 138, 133, 148, 160, 153, 131, 146, 134, 136, 149, 141, 155, 149, 165, 142, 144, 147, 138, 139. Also, find the range of heights of the boys.
10. Construct a frequency distribution table for the following weights (in gm) of 30 oranges using the equal class intervals, one of them is 40-45 (45 not included). The weights are: 31, 41, 46, 33, 44, 51, 56, 63, 71, 71, 62, 63, 54, 53, 51, 43, 36, 38, 54, 56, 66, 71, 74, 75, 46, 47, 59, 60, 61, 63.
(a) What is the class mark of the class intervals 50-55?
(b) What is the range of the above weights?
(c) How many class intervals are there?
(d) Which class interval has the lowest frequency?
Answers for worksheet on frequency distribution are given below to check the exact answers of the above questions on presentation data.
Answers:
1. (a) 0, 2, 3, 6, 7, 8, 10, 14, 15, 24
(b) 0.7, 0.8, 1.4, 2.0, 3.5, 4.6, 8.1, 9.3
2. (a) 28, 22, 16, 14, 13, 10, 8, 6, 4, 2, 0, 1
(b) 8.6, 5.0, 3.7, 3.5, 2.4, 1.2, 0.9, 0.7, 0.3, 0.1
3. (a)
(b)
4.
5.
(b)
6.
Range = $ 57
7. Inclusive form
Exclusive form
8. (a) 45
(b) 5
(c) 40 - 45
(d) 32.5, 37.5, 42.5, 47.5, 52.5
9.
Range = 34 cm
10.
(a) 52.5
(b) 44 gm
(c) 10
(d) 65 - 70, 75 - 80
1. The following graph shows the temperature of a patient in a hospital, recorded
every hour.
(a) What was the patient's temperature at 1 p.m. ?
(b) When was the patient's temperature 38.5° C?
(c) The patient's temperature was the same two times during the period given.
What were these two times?
(d) What was the temperature at 1.30 p.m.? How did you arrive at your answer?
(e) During which periods did the patients' temperature showed an upward trend?
2. The following line graph shows the yearly sales figures for a manufacturing company.
(a) What were the sales in (i) 2002 (ii) 2006?
(b) What were the sales in (i) 2003 (ii) 2005?
(c) Compute the difference between the sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared
to its previous year?
3. Use the tables below to draw linear graphs.
(a) The number of days a hill side city received snow in different years.
| Year | 2003 | 2004 | 2005 | 2006 |
| Days | 8 | 10 | 5 | 12 |
(b) Population (in thousands) of men and women in a village in different years.
| Year | 2003 | 2004 | 2005 | 2006 | 2007 |
| Number of Men | 12 | 12.5 | 13 | 13.2 | 13.5 |
| Number of Women | 11.3 | 11.9 | 13 | 13.6 | 12.8 |
1. Plot the following points on a graph sheet. Verify if they lie on a line
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5)
2. Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at
which this line meets the x-axis and y-axis.
3. Write the coordinates of the vertices
of each of these adjoining figures.
4. State whether True or False. Correct
that are false.
(i) A point whose x coordinate is zero
and y-coordinate is non-zero will
lie on the y-axis.
(ii) A point whose y coordinate is zero
and x-coordinate is 5 will lie on
y-axis.
(iii) The coordinates of the origin
are (0, 0).