Sandeep Garg (2017) Solutions for Class 11 Science Economics Chapter 4 Graphic Presentation are provided here with simple step-by-step explanations. These solutions for Graphic Presentation are extremely popular among class 11 Science students for Economics Graphic Presentation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Sandeep Garg (2017) Book of class 11 Science Economics Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Sandeep Garg (2017) Solutions. All Sandeep Garg (2017) Solutions for class 11 Science Economics are prepared by experts and are 100% accurate.
Page No 7.39:
Question 1:
Draw a Line frequency graph of the following data:
Marks | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
Frequency | 3 | 7 | 9 | 11 | 12 | 14 | 15 |
Answer:
The given data can be represented with the help of a line frequency graph as follows:
Page No 7.39:
Question 2:
Represent the following data by a histogram:
Marks in Statistics | 0−10 | 10−20 | 40−60 | 60−80 | 80−100 |
No. of Students | 12 | 28 | 60 | 48 | 30 |
Answer:
The given data can be represented with the help of a histogram.
Page No 7.39:
Question 3:
The following data relates to the marks in Economics of 70 students. Depict it through Histogram.
Marks | 0−10 | 10−20 | 20−50 | 50−70 | 70−80 |
No. of Students | 7 | 10 | 24 | 18 | 11 |
Answer:
Marks | No. of student | Adjusted/ Histogram |
0 − 10 10 − 20 20 − 50 50 − 70 70 − 80 |
7 10 24 18 11 |
7 10 24 ÷ 3 = 8 18 ÷ 2 = 9 11 |
Page No 7.39:
Question 4:
Represent the following frequency distribution by a histogram.
Mid-Values | 2.5 | 7.5 | 12.5 | 17.5 | 22.5 |
Frequency | 5 | 10 | 30 | 15 | 6 |
Answer:
Here,
The difference between the two mid points is 5 therefore half of the difference, i.e. 2.5 will be added and subtracted from each mid-point to get the following class intervals.
Mid-Values | Class Interval | Frequency |
2.5 7.5 12.5 17.5 22.5 |
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 |
5 10 30 15 6 |
Page No 7.39:
Question 5:
The frequency distribution of marks obtained by 60 students of a class in a college is given below:
Marks | 30−34 | 35−39 | 40−44 | 45−49 | 50−54 | 55−59 | 60−64 |
No. of Students | 3 | 5 | 12 | 18 | 14 | 6 | 2 |
Answer:
Converting the inclusive class intervals into exclusive class intervals, as given below:
Marks | No. of students |
29.5 − 34.5 34.5 − 39.5 39.5 − 44.5 44.5 − 49.5 49.5 − 54.5 54.5 − 59.5 59.5 − 64.5 |
3 5 12 18 14 6 2 |
Page No 7.39:
Question 6:
Draw a frequency polygon for the following data:
Items | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Frequency | 4 | 6 | 10 | 25 | 22 | 18 | 12 |
Answer:
Frequency polygon under discrete series can be represented as:
Page No 7.39:
Question 7:
Present the following data in the form of frequency polygon, using histogram.
Daily wages (â¹) | 60−80 | 80−100 | 100−120 | 120−140 | 140−160 | 160−180 | 180−200 |
No. of Workers | 3 | 5 | 10 | 15 | 7 | 4 | 2 |
Answer:
For drawing a frequency polygon, we simply join the top mid points of the rectangles of the histogram using a straight line.
Page No 7.40:
Question 8:
The frequency distribution of marks obtained by students in a class test is given below:
Marks (mid-points) | 45 | 55 | 65 | 75 | 85 |
No. of Students | 5 | 9 | 12 | 8 | 2 |
Answer:
The given data can be represented with the help of a frequency polygon.
Page No 7.40:
Question 9:
You are given the following marks secured by 45 students in an examination:
Marks | 20−29 | 30−39 | 40−49 | 50−59 | 60−69 | 70−79 |
No. of Students | 4 | 10 | 12 | 9 | 7 | 3 |
Answer:
Marks | No. of Students |
19.5 − 29.5 29.5 − 39.5 39.5 − 49.5 49.5 − 59.5 59.5 − 69.5 69.5 − 79.5 |
4 10 12 9 7 3 |
Page No 7.40:
Question 10:
Depict the following frequency distribution with the help of frequency polygon
Mid-Values | 5 | 15 | 25 | 35 | 45 | 55 | 65 | 75 |
Frequency | 4 | 10 | 16 | 22 | 25 | 12 | 7 | 2 |
Answer:
Frequency Polygon (without histogram) under continuos series:
Page No 7.40:
Question 11:
Make a frequency curve of the following data:
Class-interval | 20−40 | 40−60 | 60−80 | 80−100 | 100−120 | 120−140 | 140−160 |
Frequency | 3 | 7 | 11 | 15 | 13 | 6 | 2 |
Answer:
Class Interval | Frequency |
20 − 40
40 − 60 60 − 80 80 − 100 100 − 120
120 − 140 140 − 160
|
3 7 11 15 13 6 2 |
Page No 7.40:
Question 12:
From the following information, construct, less than and more than ogive.
Daily wages (in â¹) | 60−90 | 90−120 | 120−150 | 150−180 | 180−210 |
No. of Workers | 11 | 14 | 25 | 12 | 8 |
Answer:
For constructing less than ogive, first the given frequency distribution must be converted into less than cumulative frequency distribution as follows.
Daily Wages | No. of workers |
Less than 90 Less than 120 Less than 150 Less than 180 Less than 210 |
11 25 50 62 70 |
We now plot the cumulative frequencies against the upper limit of class intervals. The curve obtained on joining the points so plotted is known as less than ogive.
For constructing more than ogive, first the given frequency distribution must be converted into more than cumulative frequency distribution as follows.
Daily Wages | No. of workers |
More than 60 More than 90 More than 120 More than 150 More than 180 |
70 70 − 11 = 59 59 − 14 = 45 45 − 25 = 20 20 − 12 = 8 |
We now plot the cumulative frequencies against the lower limit of class intervals. The curve obtained on joining the points so plotted is known as more than ogive.
Page No 7.40:
Question 13:
The table given below shows the amount of sales of 100 companies:
Sales (â¹ in crores) | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
No. of Companies | 7 | 12 | 15 | 30 | 22 | 14 |
Answer:
(i) For constructing less than ogive, first the given frequency distribution must be converted into less than cumulative frequency distribution as follows.
Sales | No. of Companies |
Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 |
7 19 34 64 86 100 |
We now plot the cumulative frequencies against the upper limit of class intervals. The curve obtained on joining the points so plotted is known as less than ogive.
(ii) For constructing more than ogive, first the given frequency distribution must be converted into more than cumulative frequency distribution as follows.
Sales | No. of companies |
More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 |
100 100 − 7 = 93 93 − 12 = 81 81 − 15 = 66 66 − 30 = 36 36 − 22 = 14 |
We now plot the cumulative frequencies against the lower limit of class intervals. The curve obtained on joining the points so plotted is known as more than ogive.
(iii) 'Less than' and 'More than' ogive
Page No 7.40:
Question 14:
Represent the following data relating to annual profits of a company with the help of suitable graph.
Year | 2013 | 2014 | 2015 | 2016 | 2017 | 2018 | 2019 |
Profits (in â¹ Crores) | 25 | 37 | 45 | 35 | 50 | 54 | 60 |
Answer:
Page No 7.40:
Question 15:
The following data relates to rubbers imports of a company for seven years. Present the information using a time series graph:
Year | 2013 | 2014 | 2015 | 2016 | 2017 | 2018 | 2019 |
Imports (â¹ Crores) | 125 | 175 | 150 | 125 | 200 | 150 | 175 |
Answer:
Time Series Graph
Page No 7.41:
Question 16:
Prepare a graph to represent following data of imports and exports of a commodity from 2013 to 2019.
Year | 2013 | 2014 | 2015 | 2016 | 2017 | 2018 | 2019 |
Imports (â¹ crores) | 15 | 22 | 35 | 45 | 52 | 55 | 60 |
Exports (â¹ crores) | 20 | 28 | 42 | 60 | 65 | 70 | 75 |
Answer:
Page No 7.41:
Question 17:
The data relating to daily pocket money allowance of the students of class XI of a school is given below:
Pocket Money (â¹) | 0−5 | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 |
No. of Students | 7 | 12 | 18 | 25 | 10 | 3 |
Answer:
Pocket Money | No. of Students |
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 |
7 12 18 25 10 3 |
Page No 7.41:
Question 18:
The following tale gives details of expenditure incurred by a company from 2013 to 2019:
Years | 2013 | 2014 | 2015 | 2016 | 2017 | 2018 | 2019 |
Expenditure (â¹ in Lakhs) | 12 | 18 | 22 | 15 | 25 | 35 | 42 |
Answer:
Page No 7.41:
Question 19:
Make histogram and frequency polygon from the following distribution.
Class-interval | 0−20 | 20−30 | 30−40 | 40−60 | 60−100 |
Frequency | 10 | 4 | 6 | 14 | 16 |
Answer:
Class Interval | Frequency | Adjusted Frequency for Histogram |
0 − 20 20 − 30 30 − 40 40 − 60 60 − 100 |
10 4 6 14 16 |
10 ÷ 2 = 5 4 6 14 ÷ 2 = 7 16 ÷ 4 = 4 |
Page No 7.41:
Question 20:
The following table gives data on the production and sales of factory for 5 years between 2015 to 2019. Make a two-Variable Arithmetic Line Graph.
Year | 2015 | 2016 | 2017 | 2018 | 2019 |
Production (in tonnes) | 20 | 24 | 30 | 42 | 55 |
Sales (â¹ in crores) | 28 | 32 | 40 | 57 | 65 |
Answer:
Two-Variable Arithmetic Line Graph
Page No 7.41:
Question 21:
Present the data given in the table below in the form of histogram.
Mid-point | 15 | 25 | 35 | 45 | 55 | 65 | 75 |
Frequency | 5 | 12 | 20 | 18 | 16 | 25 | 22 |
Answer:
Mid point | Class interval | Frequency |
15 25 35 45 55 65 75 |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
5 12 20 18 16 25 22 |
Page No 7.41:
Question 22:
In a certain colony, 40 households were selected. The data on monthly income (â¹'000) is given below:
20 | 12 | 35 | 55 | 40 | 14 | 35 | 8 | 18 | 11 |
60 | 11 | 35 | 50 | 45 | 20 | 16 | 7 | 15 | 70 |
18 | 65 | 62 | 16 | 21 | 19 | 24 | 11 | 19 | 35 |
13 | 25 | 43 | 15 | 30 | 40 | 25 | 55 | 22 | 35 |
(ii) Draw a histogram and a frequency polygon form the frequency distribution.
Answer:
Class Interval | Tally | Frequency |
0 − 10 | 2 | |
10 − 20 | 14 | |
20 − 30 | 7 | |
30 − 40 | 6 | |
40 − 50 | 4 | |
50 − 60 | 3 | |
60 −70 | 3 | |
70 − 80 | 1 | |
40 |
Page No 7.41:
Question 23:
Draw a 'Less than' and 'More than' ogive from the following distribution:
Profits (â¹ in Lakhs) | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
No. of Companies | 4 | 7 | 10 | 20 | 17 | 2 |
Answer:
For constructing less than ogive, first the given frequency distribution must be converted into less than cumulative frequency distribution as follows.
Less than | No. of Companies |
Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 |
4 4 + 7 = 11 11 + 10 = 21 21 + 20 = 41 41 + 17 = 58 58 + 2 = 60 |
We now plot the cumulative frequencies against the upper limit of class intervals. The curve obtained on joining the points so plotted is known as less than ogive.
For constructing more than ogive, first the given frequency distribution must be converted into more than cumulative frequency distribution as follows.
More than | No. of companies |
More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 |
60 60 − 4 = 56 56 − 7 = 49 49 − 10 = 39 39 − 20 = 19 19 − 17 = 2 |
We now plot the cumulative frequencies against the lower limit of class intervals. The curve obtained on joining the points so plotted is known as more than ogive.
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