Sandeep Garg 2018 Solutions for Class 11 Commerce Economics Chapter 4 Graphic Presentation are provided here with simple stepbystep explanations. These solutions for Graphic Presentation are extremely popular among Class 11 Commerce 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 2018 Book of Class 11 Commerce Economics Chapter 4 are provided here for you for free. You will also love the adfree experience on Meritnationâ€™s Sandeep Garg 2018 Solutions. All Sandeep Garg 2018 Solutions for class Class 11 Commerce 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.
MidValues  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 midpoint to get the following class intervals.
MidValues  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 (â‚¹)  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 (midpoints)  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
MidValues  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:
Classinterval  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 â‚¹)  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 (â‚¹ 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  2008  2009  2010  2011  2012  2013  2014 
Profits (in â‚¹ 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  2009  2010  2011  2012  2013  2014  2015 
Imports (â‚¹ 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 2008 to 2014.
Year  2008  2009  2010  2011  2012  2013  2014 
Imports (â‚¹ crores)  15  22  35  45  52  55  60 
Exports (â‚¹ 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 (â‚¹)  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 2009 to 2015:
Years  2009  2010  2011  2012  2013  2014  2015 
Expenditure (â‚¹ 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.
Classinterval  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 2011 to 2105. Make a twoVariable Arithmetic Line Graph.
Year  2011  2012  2013  2014  2015 
Production (in tonnes)  20  24  30  42  55 
Sales (â‚¹ in crores)  28  32  40  57  65 
Answer:
TwoVariable Arithmetic Line Graph
Page No 7.41:
Question 21:
Present the data given in the table below in the form of histogram.
Midpoint  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 (â‚¹'000) 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 (â‚¹ 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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