# TR Jain & Vk Ohri (2018) Solutions for Class 11 Commerce Economics Chapter 4 - Organisation Of Data

TR Jain & Vk Ohri (2018) Solutions for Class 11 Commerce Economics Chapter 4 Organisation Of Data are provided here with simple step-by-step explanations. These solutions for Organisation Of Data are extremely popular among class 11 Commerce students for Economics Organisation Of Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the TR Jain & Vk Ohri (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 Meritnationâ€™s TR Jain & Vk Ohri (2018) Solutions. All TR Jain & Vk Ohri (2018) Solutions for class 11 Commerce Economics are prepared by experts and are 100% accurate.

#### Page No 55:

#### Question 1:

Identify 10 important objects that impact your environment. Classify them, as living and non-living. Is it a quantitative classification? If not, give reasons.

#### Answer:

Objects that impact our environment are:

(i) Factories

(ii) Animals

(iii) Human beings

(iv) Automobiles

(v) Fuel

(vi) Coal

(vii) Soil

(viii) Fertilisers

(ix) Trees

(x) Plastic

The above objects can be classified as:

a)** Living**- Animals, human beings, trees

b)

**- Factories, automobiles, soil, fuel, coal, fertilisers, plastic**

*Non-living*No, it is not a quantitative classification as we cannot classify the data on the basis of numerical values into different classes or groups. This is because we cannot judge in numbers the impact of objects on environment.

#### Page No 55:

#### Question 2:

Identify the following variables as discrete and continuous: volume, height, weight, temperature, snowfall, population, crop-production, no. of scooters on the ring road of Delhi, no. of houses demolished by the MCD.

#### Answer:

Discrete Variable |
Continuous Variable |
||

1. | Population | 1. | Volume |

2. | Crop-production | 2. | Height |

3. | No. of scooters on the ring road of Delhi | 3. | Weight |

4. | No. of houses demolished by the M.C.D | 4. | Temperature |

5. | Snowfall |

#### Page No 58:

#### Question 1:

Following data relate to the pocket expenses (rupees) of 10 students of Class XI. Arrange them in the ascending and descending orders:

50, | 20, | 30, | 15, | 45, | 35, | 40, | 25, | 20, | 43. |

#### Answer:

Data Arranged in Ascending Order |
Data Arranged in Descending Order |

15 20 20 25 30 35 40 43 45 50 |
50 45 43 40 35 30 25 20 20 15 |

#### Page No 58:

#### Question 2:

In a sample investigation, 20 persons are found to have the following money in their pockets as office expenses. Arrange the data in the ascending and descending orders:

114, | 108, | 100, | 98, | 101, | 109, | 117, | 119, | 121, | 136 |

131, | 136, | 143, | 156, | 169, | 182, | 195, | 207, | 219, | 255. |

#### Answer:

Data Arranged in Ascending Order |
Data Arranged in Descending Order |

98 100 101 108 109 114 117 119 121 131 136 136 143 156 169 182 195 207 219 255 |
255 219 207 195 182 169 156 143 136 136 131 121 119 117 114 109 108 101 100 98 |

#### Page No 58:

#### Question 3:

Collect data of weekly expenditure on food by your family. Arrange the data in an order. Also, convert the data into monthly expenditure. Find the total number of observations.

#### Answer:

**Data of weekly expenditure on food**

Week |
Expenditure |

1^{st} week |
1000 |

2^{nd} week |
2000 |

3^{rd} week |
2100 |

4^{th} week |
2200 |

5^{th} week |
2500 |

6^{th} week |
2700 |

7^{th} week |
2800 |

8^{th} week |
3000 |

**Ascending and Descending order of data**

Ascending order of data |
Descending order of data |

1000 | 3000 |

2000 | 2800 |

2100 | 2700 |

2200 | 2500 |

2500 | 2200 |

2700 | 2100 |

2800 | 2000 |

3000 | 1000 |

**Monthly Expenditure**

Month |
Expenditure |

1^{st} Month |
1000 + 2000 + 2100 + 2200 = Rs 7300 |

2^{nd} Month |
2500 + 2700 + 2800 + 3000 = Rs 11,000 |

Total number of observations = 8

#### Page No 79:

#### Question 1:

In an examination, 25 students secured the following marks:

23 | 28 | 30 | 32 | 35 | 35 | 36 | 40 | 41 | 43 | 44 | 45 | 45 |

48 | 49 | 52 | 53 | 54 | 56 | 56 | 58 | 61 | 62 | 65 | 68 |

20−29, 30−39, 40−49, 50−59, and 60−69.

(ii) Arrange the data with cumulative frequencies.

#### Answer:

i) In the form of a frequency distribution, the given data can be arranged as follows.

ii) With cumulative frequencies the given data can be arranged as follows.

METHOD-1 |
METHOD-2 |
||

Marks |
No. of Students |
Marks |
No. of students |

Less than 29Less than 39Less than 49Less than 59Less than 69 |
0 + 2 = 2 2 + 5 = 7 7 + 8 = 15 15 + 6 = 21 21 + 4 = 25 |
More than 20More than 30More than 40More than 50More than 60 |
25 25 − 2 = 23 23 − 5 = 18 18 − 8 = 10 10 − 6 = 4 |

#### Page No 79:

#### Question 2:

The following data is of the age of 25 students of Class XI.

Arrange these data in the form of a frequency distribution.

15 | 16 | 16 | 17 | 18 | 18 | 17 | 15 | 15 | 16 | 16 | 17 | 15 |

16 | 15 | 16 | 16 | 18 | 15 | 17 | 17 | 18 | 10 | 16 | 15 |

#### Answer:

#### Page No 79:

#### Question 3:

Students of Class XI obtained following marks in Economics. Classify the data in the form of individual series, discrete series, continuous series and cumulative frequency series.

15 | 15 | 18 | 16 | 20 | 21 | 25 | 25 | 15 | 16 | 18 | 22 | 24 | 25 | 20 |

18 | 22 | 24 | 24 | 25 | 25 | 23 | 20 | 15 | 16 | 17 | 19 | 18 | 22 | 22 |

#### Answer:

* Individual series *is simply the arrangement of the given data in ascending (or, descending order)

15 | 15 | 15 | 15 | 16 | 16 | 16 | 17 | 18 | 18 | 18 | 18 | 19 | 20 | 20 |

20 | 21 | 22 | 22 | 22 | 22 | 23 | 24 | 24 | 24 | 25 | 25 | 25 | 25 | 25 |

* Discrete series*:

*:*

**Continuous series**

**Cumulative frequency series**Method1 |
Method2 |
||

Marks |
No. of Students |
Marks |
No. of students |

Less than 15Less than 19Less than 23Less than 27 |
0 9 + 4 = 13 13 + 9 = 22 22 + 8 = 30 |
More than 12More than 16More than 20More than 24 |
30 30 − 4 = 26 26 − 9 = 17 17 − 9 = 8 |

#### Page No 79:

#### Question 4:

Arrange the following data in the form of an exclusive frequency distribution, using 5−10 as the initial class interval:

12 | 36 | 40 | 30 | 28 | 20 | 19 | 10 | 10 | 19 | 27 | 15 | 26 | 10 |

19 | 7 | 45 | 33 | 26 | 37 | 5 | 20 | 11 | 17 | 37 | 30 | 20 |

#### Answer:

#### Page No 79:

#### Question 5:

Weight of 20 students is given in kilograms. Using class interval of 5, make a frequency distribution.

30 | 45 | 26 | 25 | 42 | 33 | 15 | 35 | 45 | 45 |

45 | 39 | 42 | 40 | 18 | 35 | 41 | 20 | 36 | 48 |

#### Answer:

In the given data, the lowest value is 15, while the highest value is 48. Accordingly, we take (15 - 20) as the initial class interval.

Weight (kg) |
Tally Bars |
No. of Students |

15−20 | 2 | |

20−25 | 1 | |

25−30 | 2 | |

30−35 | 2 | |

35−40 | 4 | |

40−45 | 4 | |

45−50 | 5 | |

Total | $\sum f=20$ |

#### Page No 80:

#### Question 6:

Convert the following data in a simple frequency distribution:

5 students obtained less than 3 marks |

12 students obtained less than 6 marks |

25 students obtained less than 9 marks |

33 students obtained less than 12 marks |

#### Answer:

The given information can be summarised as follows.

Marks |
Cumulative Frequency( c.f.) |

Less than 3Less than 6Less than 9Less than 12 |
5 12 25 33 |

Marks |
Frequency( f) |

0 − 3 3 − 6 6 − 9 9 − 12 |
5 7 (= 12 − 5) 13 (= 25 − 12) 8 (= 33 − 25) |

$\Sigma f=33$ |

#### Page No 80:

#### Question 7:

In the following statement, take the number of letters in a word as items and number of items a word (of the same size) repeats itself as frequencies. Prepare a discrete series.

"Success in the examination confers no absolute right to appointment unless government is satisfied after such an enquiry as may be considered necessary that the candidate is suitable in all respect for appointment."

#### Answer:

In the given question, the size of the items refers to the number of alphabets in the word. Accordingly, in the given statement there are eight items with size equal to 2, namely, in, no, to, is, an, as, be, is, in. Similarly, there are 5 items with size equal to three, namely, the, may, the, all, for. In the similar manner, we can find the frequency corresponding to the various items.

The frequency distribution table is prepared in the following manner.

Size of Item |
Tally Bars |
Frequency |

2 | 9 | |

3 | 5 | |

4 | 2 | |

5 | 2 | |

6 | 1 | |

7 | 3 | |

8 | 3 | |

9 | 3 | |

10 | 2 | |

11 | 3 | |

Total | $\sum f=33$ |

#### Page No 80:

#### Question 8:

An economic survey revealed that 30 families in a town incur following expenditure in a day (rupees).

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 | 38 |

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

(ii) How many families spend more than 29 rupees a day?

#### Answer:

i)

ii) Number of families spending more than Rs 29 per day = 4+2 = 6

Thus, there are 6 families that spend more than Rs 29 per day.

Percentage of families spending more than Rs 29

$=\frac{\mathrm{Number}\mathrm{of}\mathrm{families}\mathrm{spending}\mathrm{more}\mathrm{than}\mathrm{Rs}29}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{families}}\times 100\phantom{\rule{0ex}{0ex}}=\frac{6}{30}\times 100=20\%\phantom{\rule{0ex}{0ex}}\mathrm{Thus},20\%\mathrm{of}\mathrm{the}\mathrm{families}\mathrm{spend}\mathrm{more}\mathrm{than}\mathrm{Rs}29\mathrm{per}\mathrm{day}.$

#### Page No 80:

#### Question 9:

From the following data related to the weight of college students in kg, prepare a frequency distribution with a class interval of 10 exclusive and inclusive basis:

40 92 49 52 69 |
70 72 42 50 60 |
63 65 43 48 54 |
53 53 47 65 82 |
85 79 50 42 55 |

#### Answer:

Exclusive Method |
Inclusive Method |
||

Weight( in kg) |
No. of Students(f) |
Weight(in kg) |
No. of Students(f) |

40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 90 − 100 |
7 7 5 3 2 1 |
40 − 50 51 − 61 62 − 72 73 − 83 84 − 94 95 − 105 |
9 6 6 2 2 0 |

∑f = 25 |
∑f = 25 |

#### Page No 81:

#### Question 10:

Construct the simple frequency distribution from the following data:

Mid-value | 5 | 15 | 25 | 35 | 45 | 55 |

Frequency | 2 | 8 | 15 | 12 | 7 | 6 |

#### Answer:

The class interval can be calculated from the mid-points using the following adjustment 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 = 5 – 5 = 0

Upper limit of first class = 5 + 5 = 10

Thus, the first class interval is (0 − 10). Similarly, we can calculate the remaining class intervals.

Mid-value |
Class-interval |
Frequency( f) |

5 15 25 35 45 55 |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 |
2 8 15 12 7 6 |

∑f = 50 |

#### Page No 81:

#### Question 11:

Classify the following data by taking class interval such that their mid-values are 17, 22, 27, 32 and so on:

30 30 36 33 |
42 27 22 41 |
30 42 30 21 |
54 36 31 16 |
40 28 19 17 |
48 28 48 36 |
14 37 16 37 |
17 54 42 41 |
51 44 32 46 |
42 31 21 47 |
25 36 22 52 |
41 40 40 53 |

#### Answer:

The class interval can be calculated from the mid-points using the following adjustment 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.

$\mathrm{Value}\mathrm{of}\mathrm{adjustment}=\frac{17-12}{2}=2.5$

Thus, we add and subtract 2.5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 12 – 2.5 = 9.5

Upper limit of first class = 12 + 2.5 = 14.5

Thus, the first class interval is (9.5 − 14.5). Similarly, we can calculate the remaining class intervals.

Mid-value |
Class-interval |
Frequency( f) |

12 17 22 27 32 37 42 47 52 |
9.5 − 14.5 14.5 − 19.5 19.5 − 24.5 24.5 − 29.5 29.5 − 34.5 34.5 − 39.5 39.5 − 44.5 44.5 − 49.5 49.5 − 54.5 |
1 5 4 4 8 6 11 4 5 |

∑f = 48 |

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