Nm Shah 2018 Solutions for Class 11 Commerce 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 Commerce students for Economics Graphic Presentation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Nm Shah 2018 Book of Class 11 Commerce Economics Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Nm Shah 2018 Solutions. All Nm Shah 2018 Solutions for class Class 11 Commerce Economics are prepared by experts and are 100% accurate.

#### Question 1:

The frequency distribution of marks obtained by students in a class test is given below:

 Marks : 0-10 10-20 20-30 30-40 40-50 No. of Students : 3 10 14 10 3
Draw a histogram to represent the frequency distribution  of marks . Comment on the shape of the histogram.

Histogram is a joining rectangular diagram with equal class interval of size 10.

#### Question 2:

Present the data given in the table below in the form of a Histogram :

 Mid-points : 115 125 135 145 155 165 175 185 195 Frequency : 6 25 48 72 116 60 38 22 3

The two-dimensional diagrams that depict the frequency distribution of a continuous series by the means of rectangles are called histograms.
In order to construct a histogram, we first require the class intervals corresponding to the various mid-points, which is calculated using the following formula.

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.

Thus, we add and subtract 5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 115 – 5 = 110

Upper limit of first class = 115 + 5 = 120.

Thus, the first class interval is (110-120). Similarly, we can calculate the remaining class intervals.

 Mid-points Class Interval Frequency 115 125 135 145 155 165 175 185 195 110 − 120 120 − 130 130 − 140 140 − 150 150 − 160 160 − 170 170 − 180 180 − 190 190 − 200 6 25 48 72 116 60 38 22 3

#### Question 3:

Make a frequency Polygon and Histogram  using the given data:

 Marks obtained : 10-20 20-30 30-40 40-50 50-60 60-70 No. of Students : 5 12 15 22 14 4

#### Question 4:

Draw Histogram from the following data:

 Marks obtained : 10-20 20-30 30-40 40-50 50-70 70-100 No. of Students : 6 10 15 10 6 3

The data is given in the form of unequal class interval. So, we will first make appropriate adjustment in the frequencies to make the class intervals equal.  The general formula for the adjustment of the frequency is as follows.

 Marks No. of Students Adjusted Frequency 10−20 6 − 20−30 10 − 30−40 15 − 40−50 10 − 50−70 6 $\frac{10×6}{20}=3$ 70−100 3 $\frac{10×3}{30}=1$

#### Question 5:

In a certain colony a sample of 40 households  was selected . The data on daily income for this sample are given as follows:

 200 120 350 550 400 140 350 85 180 110 110 600 350 500 450 200 170 90 170 800 190 700 630 170 210 185 250 120 180 350 110 250 430 140 200 400 200 400 210 300
(a) Construct a Histogram and a frequency Polygon.
(b) Show that the area under the polygon is equal to the area under the histogram.

(a)
 Income Tally Marks Frequency Area for each class 0 − 100 2 100 × 2 = 200 100 − 200 14 100 × 14 = 1400 200 − 300 8 100 × 8 = 800 300 − 400 5 100 × 5 = 500 400 − 500 5 100 × 5 = 500 500 − 600 2 100 × 2 = 200 600 − 700 2 100 × 2 = 200 700 − 800 1 100 × 1 = 100 800 − 900 1 100 × 1 = 100 40 Total area =4000

(b) The area under a histogram and under a frequency polygon is the same (i.e equal to 4000) because of the fact that we extend the first class interval to the left by half the size of class interval as the starting point of the frequency polygon. Similarly, the last class interval is extended to the right by the same amount as the end point of the frequency polygon. This ensures that the area that was excluded while joining the mid-points is included in the frequency polygon such that the area under the frequency polygon and the area of histogram is the same.

#### Question 6:

Present the data given in the table below in Histogram:

 Marks : 25-29 30-34 35-39 40-44 45-49 50-54 55-59 Frequency : 4 5 23 31 10 8 5

Before proceeding to construct histogram, we first need to convert the given inclusive series into an exclusive series using the following formula.

The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class.

Therefore, we add 0.5 to the upper limit and subtract 0.5 from the lower limit of each class.

 Marks Frequency 24.5 − 29.5 29.5 − 34.5 34.5 − 39.5 39.5 − 44.5 44.5 − 49.5 49.5 − 54.5 54.5 − 59.5 4 5 23 31 10 8 5

#### Question 7:

A survey showed that the average daily expenditures (in rupees) of 30 households in a city were:
11, 12, 14, 16, 16, 17, 18, 18, 20, 20, 20, 21, 21, 22, 22, 23, 23, 24, 25, 25, 26, 27, 28, 28, 31, 32, 32, 33, 36, 36.
(a) Prepare a frequency distribution using class intervals:

 10-14, 15-19, 20-24, 25-29, 30-34 and 35-39.

(b) What percent of the households spend more than â‚¹ 29 each day?
(c) Draw a frequency histogram for the above data.

(a)

 Class Interval Tally Marks Frequency 10 − 14 15 − 19 20 − 24 25 − 29 30 − 34 35 − 39 3 5 10 6 4 2 30

(b) Households that spend more than 29 each day$=\frac{6}{30}×100=20%$

(c) To construct histogram, we first need to convert the given inclusive series into an exclusive series using the following formula.

The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class.

Therefore, we add 0.5 to the upper limit and subtract 0.5 from the lower limit of each class.

 Class Interval Frequency 9.5 − 14.5 14.5 − 19.5 19.5 − 24.5 24.5 − 29.5 29.5 − 34.5 34.5 − 39.5 3 5 10 6 4 2 30

#### Question 8:

Draw Histogram from a given data relating to monthly pocket money allowance of the students of class XII in a school:

 Size of classes( in â‚¹ ) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 No of students 5 10 15 20 25 15 10 5

#### Question 9:

Draw a Histogram and Frequency Polygon of the following information.

 Wages in Rupees : 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115 No. of workers : 9 12 15 11 20 20 11 2

#### Question 10:

Draw ogive$—$(a) less than, and (b) more than of the folowing data:

 Weekly wages of No of workers : 100-105 105-110 110-115 115-120 120-125 125-130 130-135 Workers (â‚¹) : 200 210 230 320 350 520 410

In order to construct the ogives, we first need to calculate the less than and the more than cumulative frequencies as as follows.

 Weekly Wages Cumulative Frequency Less than 105 Less than 110 Less than 115 Less than 120 Less than 125 Less than 130 Less than 135 200 200 + 210 = 410 410 + 230 = 640 640 + 320 = 960 960 + 350 = 1410 1410 + 520 = 1930 1930 + 410 = 2240

 Weekly Wages Cumulative Frequency More than 100 More than 105 More than 110 More than 115 More than 120 More than 125 More than 130 2240 2240$-$200 = 2040 2040$-$210 = 1830 1830$-$230 = 1600 1600$-$320 = 1280 1280$-$350 = 930 930$-$520 = 410

#### Question 11:

Prepare a less than ogive from the following data:

 Class : 0-6 6-12 12-18 18-24 24-30 30-36 Frequency : 4 8 15 20 12 6

In order to prepare a less than ogive, we first need to calculate the less than cumulative frequency distribution as follows:

 Class Cumulative Frequency Less than 6 Less than 12 Less than 18 Less than 24 Less than 30 Less than 36 4 4 + 8 = 12 12 + 15 = 27 27 + 20 = 47 47 + 12 = 59 59 + 6 = 65

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.

#### Question 12:

From the following frequency distribution prepare a 'less than ogive'.

 Capital (â‚¹ in lakhs) : 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 No. of companies : 2 3 7 11 15 7 2 3

For constructing a less than ogive, we convert the frequency distribution into a less than cumulative frequency distribution as follows:

 Capital Cumulative Frequency Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 2 2 + 3 = 5 5 + 7 = 12 12 + 11 = 23 23 + 15 = 38 38 + 7 = 45 45 + 2 = 47 47 + 3 = 50

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.

#### Question 13:

Arrange the following information on a time-series graph:

 Year : 2009-10 2010-11 2011-12 2012-13 2013-14 2014-15 2015-16 NDP(â‚¹ in '000 crores) : 35 36 37 40 41 44 44

Time Series Graph

#### Question 14:

Plot the following data of annual profits of a firm on a time series graph.

 Year : 2012 2013 2014 2015 2016 2017 Profits(â‚¹ in thousand) : 60 72 75 65 80 95

Time Series Graph

#### Question 15:

Prepare a suitable graph from the following data relating to export and import.

 Year 2010-11 2011-12 2012-13 2013-14 2014-15 2015-16 2016-17 Exports (â‚¹ in crores) 1505 2265 2070 1805 1632 1527 1845 Imports( â‚¹ in crores) 1005 1145 1980 1335 1547 1435 1740

The given data can be presented in the form of a time series graph as follows:

Here, the smallest value is 1,005 which is far higher than zero. Therefore, in this case we use a false base line starting from 1,000.

#### Question 16:

Present graphically the following sales of Delhi branch of USHA FANS.

 Year : 2008 2009 2010 2011 2012 2013 2014 2015 2016 Sales(â‚¹ in '000) : 13 15 12 19 25 31 29 27 35

The given data can be presented in the form of a time series graph as follows:

#### Question 17:

Prepare a graph showing total cost and total production of a scooter manufacturing company.

 Year : 2012 2013 2014 2015 2016 Production (in units) : 8500 9990 11700 13300 15600 Total cost (â‚¹ in lakh) : 24 29 34 45 49