NCERT Solutions Maths Class 10 Exercise 14.1

# NCERT Solutions Maths Class 10 Exercise 14.1

## Maths Class 10 Exercise 14.1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

 Number of plants 0 – 2 2 – 4 4 – 6 6 – 8 8 – 10 10 – 12 12 – 14 Number of houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

Solution: Since the number of plants and the number of houses are small in their values, so we use direct method.

Mean = 8.1

Hence, the mean number of plants per house is 8.1.

2. Consider the following distribution of daily wages of 50 workers of a factory.

 Daily wages (in Rs.) 100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 Number of workers 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

From given data, let assumed mean (a) = 150

Width of the class (h) = 20

= 150 + 20(0.24)

= 150 – 4.8

= 145.2

Hence, the mean daily wages of the workers of the factory is Rs. 145.20.

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency (f).

 Daily pocket allowance (in Rs.) 11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25 Number of children 7 6 9 13 f 5 4

Solution:

From given data, let assumed mean (a) = 18

⇒ 2f = 40

f = 40/2

f = 20

Hence, the missing frequency is 20.

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows:

 Number of heart beats per minute 65 – 68 68 – 71 71 – 74 74 – 77 77 – 80 80 – 83 83 – 86 Number of women 2 4 3 8 7 4 2

Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:

From given data, let assumed mean (a) = 75.5

Width of the class (h) = 3

= 75.5 + 3(0.13)

= 75.5 + 0.39

= 75.89

= 75.9    (approx.)

Hence, the mean heart beats per minute for these women is 75.9.

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

 Number of mangoes 50 – 52 53 – 55 56 – 58 59 – 61 62 – 64 Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution: Since the number of mangoes and the number of boxes are large numerically, so we use step-deviation method.

From given data, let assumed mean (a) = 57

Width of the class (h) = 3

= 25/40

= 0.0625   (approx.)

= 57 + 3(0.0625)

= 57 + 0.1875

= 57.1875

= 57.19      (approx.)

Hence, the mean number of mangoes kept in a packing box is 57.19.

6. The table below shows the daily expenditure on food of 25 households in a locality.

 Daily expenditure (in Rs.) 100 – 150 150 – 200 200 – 250 250 – 300 300 – 350 Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Solution:

From given data, let assumed mean (a) = 225

Width of the class (h) = 50

= 7/25

= –0.28

= 225 + 50(– 0.28)

= 225 – 14

= 211

Hence, the mean daily expenditure on food is Rs. 211.

7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

 Concentration of SO2 (in ppm) Frequency 0.00 – 0.04 4 0.04 – 0.08 9 0.08 – 0.12 9 0.12 – 0.16 2 0.16 – 0.20 4 0.20 – 0.24 2

Find the mean concentration of SO2 in the air.

Solution: From given data, let assumed mean (a) = 0.10

Width of the class (h) = 0.04

= 0.10 + 0.04(– 0.033)

= 0.10 – 0.00132

= 0.09868

= 0.099     (approx.)

Hence, the mean concentration of SO2 in the air is 0.099 ppm.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

 Number of days 0 – 6 6 – 10 10 – 14 14 – 20 20 – 28 28 – 38 38 – 40 Number of students 11 10 7 4 4 3 1

Solution:

From given data, let assumed mean (a) = 17

= 17 + (–181)/40

= 17 – 4.52

= 12.48

Hence, the mean number of days a student was absent is 12.48 days.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

 Literacy rate (in %) 45 – 55 55 – 65 65 – 75 75 – 85 85 – 95 Number of cities 3 10 11 8 3

Solution:

From given data, let assumed mean (a) = 70

Width of the class (h) = 10

= –2/35

= –0.057

= 70 + 10(– 0.057)

= 70 – 0.57

= 69.43

Hence, the mean literacy rate is 69.43%.