The DAV Maths Class 7 Solutions and **DAV Class 7 Maths Chapter 12 Worksheet 1** Solutions of Data Handling offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 12 WS 1 Solutions

Question 1.

The attendance in a school for a week is as follows:

425, 430, 400, 408, 410, 423

Calculate the mean of the attendance.

Answer:

Mean = \(\frac{\text { Sum of all observations }}{\text { Number of observations }}=\frac{425+430+400+408+410+423}{6}\)

\(\frac{2496}{6}\) = 416

Hence the mean attendance = 416.

Question 2.

The heights of 10 boys were measured in cm and the results were as follows: 143, 132, 149, 148, 151, 146, 135, 128, 139, 150

(i) What is the height of the tallest boy?

(ii) What is the height of the shortest boy?

(iii) What is the range of the data?

(iv) Find the mean height.

(v) Flow many boys are there whose heights are less than the mean height?

Answer:

The given data are 143, 132, 149, 148, 151, 146, 135, 128, 139, 150

Height of the tallest boy = 151

Height of the shortest boy = 128

Range = 151 – 128 = 23

Mean height = \(\frac{\text { Sum of all data }}{\text { Number of data }}\)

= \(\frac{143+132+149+148+151+146+135+128+139+150}{10}=\frac{1421}{10}\) = 142.1

(v) Number of boys whose heights are more than the mean heights:

i.e. 143,146,148,149,150,151 i.e., 6.

Question 3.

A group of students was given a special test. The test was completed by various students in the following duration of time (in minutes) 17, 19, 20, 22, 24, 24, 28, 30, 30, 36.

(i) Find the mean time taken by the students to complete the test.

(ii) Find the new mean, if another student who took 14 minutes is also included.

Answer:

(i) Mean = \(\frac{\text { Sum of all observations }}{\text { Number of observations }}\)

= \(\frac{17+19+20+22+24+24+28+30+30+36}{10}=\frac{250}{10}\)

= 25 minutes

(ii) If a student who took 14 minutes for completing the test is included

∴ Total sum of 11 students = 250 + 14 = 264

∴ New Mean = \(\frac{264}{11}\) = 24

Question 4.

The enrolment of a school during seven consecutive years was as follows: 1580, 2020, 2580, 3200, 3550, 3710, 4010.

(i) Find the mean enrolment of the school for this period.

(ii) Find the new mean, if the enrolment of the current year, that is, 4270 is also included.

Answer:

The given observations are;

1580, 2020, 2580, 3200, 3550, 3710, 4010.

(i) Mean enrolment = \(=\frac{\text { Sum of all observations }}{\text { Number of observations }}\)

= \(\frac{1580+2020+2580+3200+3550+3710+4010}{7}=\frac{20650}{7}\)

= 2950

(ii) If the enrolment for the new year is also added then total sum of 8 years = 20650 + 4270 = 24920

New Mean = \(\frac{24920}{8}\) = 3115

Question 5.

In a factory the total number of workers are 25. Out of them, the average monthly salary of 15 workers is ₹ 8500 whereas the other 10 workers have an average salary oft 9800. Find the monthly average salary of all the 25 workers.

Answer:

Average monthly salary of 15 workers = ₹ 8500

Sum of all the monthly salaries of 15 workers = 15 × 8500 = 1127,500

Average salary of 10 workers = ₹ 9800

Sum of all the monthly salaries of 10 workers = 10 × 9800 = ₹ 98000

Now, sum of all the monthly salaries of 25 workers = 1127,500 + ₹ 98000 = 1225,500

Monthly average salary of all the 25 workers = \(\frac{225,500}{25}\) = 19020

Hence, the monthly salary of all the 25 workers is 19020.

Question 6.

The average salary of 19 workers in a factory is ₹ 12,000 per month. If the salary of the manager is ₹ 42,000 per month, find the average monthly salary paid to all the employees.

Answer:

Average salary of 19 workers = 112,000

∴ Sum of their salary amount = 19 × 12,000 = 1228,000

If the salary of the manager i.e. t 42000 is also included, then the sum of salary amount of 20 employees = 1228,000 +142,000 = ₹ 270,000

∴ Average monthly salary = \(\frac{270,000}{20}\) = ₹ 13,500

Hence, the averge monthly salary paid to all the employees is 113,500.

Question 7.

The mean of 9 observations was found to be 35. Later on, it was discovered that an observation 81 was misread as 18. Find the correct mean of the observations.

Answer:

The mean of 9 observations was 35

∴ Sum = 9 × 35 = 315

Now 81 was misread as 18

∴ Correct sum is 315 + (81 – 18) = 315 + 63 = 378

∴ Correct Mean = \(\frac{378}{9}\) = 42.

Question 8.

The mean of 5 numbers is 27. If one number is excluded, then the mean is 25, find the excluded number.

Answer:

Mean of 5 numbers = 27

Their sum = 5 × 27 = 135

1 number is excluded

∴ Total number of observations = 5 – 1 = 4

Mean = 25

Their sum = 25 × 4 = 100

New excluded number = 135 – 100 = 35.

Question 9.

The mean of 8 numbers is 27. If one more number is included, then the mean is 26. Find the included number.

Answer:

The mean of 8 observations = 27

Their sum = 8 × 27 = 216

If 1 observation is added then the total number of observation = 8 + 1 = 9 Mean = 26

Their sum = 9 × 26 = 234

New observation = 234 – 216 = 18.

Question 10.

In a certain hospital, the mean birth rate of a week is 35. If the mean birth rate from Monday to Thursday was 32 and that of Thursday to Sunday was 36, find the number of births on Thursday.

Answer:

Mean birth rate of a week (7 days) = 35

Their sum = 7 × 35 = 245

Mean birth rate of 4 days (from Monday to Thursday) = 32

∴ Their sum = 4 × 32 = 128

Mean birth rate for 4 days (from Thursday to Sunday) = 36

∴ Their sum = 4 × 36 = 144

Sum of births for 8 days = 128 + 144 = 272

Number of births on Thursday = 272 – 245 = 27.

### DAV Class 8 Maths Chapter 12 Worksheet 1 Notes

- Data is a collection of numbers gathered to give some information.
- The data collected and put without any specific arrangement is called raw data.

e.g., 5, 25, 20, 8, 12, 35, 100, 85, 65 - Data needs to be organised to make it easy to understand and interpret.
- The raw data are arranged either in ascending order (increasing order) or descending order (decreasing order).

e.g. Raw data: 6, 18, 9, 20, 40, 35, 15, 14, 16, 45 - Ascending order: 6, 9, 14, 15, 16, 18, 20, 35, 40, 45
- Descending order: 45, 40, 35, 20, 18, 16, 15, 14, 9, 6

Range:

It is the difference of the highest and the lowest observation.

Range = Highest Observation – Lowest Observation.

Mean:

Mean = Sum of all observations / Number of observations

Median:

Median is the middle value of the observations arranged in ascending or descending order.

For even number, Median = Sum of two middle most observations / 2

For odd number, Median = Middle most observation

Mode :

The value which occurs the most is called mode.

Bar graphs are pictorial representation of data using bars of uniform width and equal spacing between them.

Example 1:

The weights of 8 students in kg are given below.

35, 36, 40, 32, 25, 30, 45, 27

(i) Find their mean weight.

(ii) What is the range?

(iii) What is the number of students who have their weights more than the mean weight?

Solution:

(i) Mean = \(\frac{35+36+40+32+25+30+45+27}{8}\)

= \(\frac{270}{8}\)

(ii) Maximum weight = 45 kg

Minimum weight = 25 kg

∴ Range = Max. weight – Min.

weight = 45 kg – 25 kg = 20 kg

(iii) Weights more than the mean weight are 35, 36, 40 and 45 = 4 students.

Example 2:

The mean of 20 observations is 55.6. If one observation 43 is excluded, find the mean of the remaining 9 observations.

Solution:

Mean of 10 observations = 55.6

Sum of the 10 observations = 55.6 × 10 = 556

Observation 43 is excluded

∴ Sum of 9 observations = 556 – 43 = 513

∴ Mean of the remaining 9 observations = \(\frac{513}{9}\) = 57

Hence the required mean = 57.

Example 3:

Find the median of the following data 235, 248, 250, 220, 225, 128, 136, 155.

Solution:

Arranging the given data in ascending order 120, 125, 128, 135, 136, 148, 150, 155.

n = 8 (even)

∴ Median numbers are 135 and 136

∴ Median = \(\frac{135+136}{2}\) = 135.5

Hence the required median 135.5.

Example 4:

Find the mode of the following observations 6, 8, 10, 12, 5, 4, 8, 3, 8, 4, 6, 8, 15, 8.

Solution:

Arranging the observations in increasing order

3, 4, 4, 5, 6, 6, 8, 8, 8, 8, 8, 10, 12, 15

Here we see that 8 occurs most i.e. 5 times

Hence the mode is 8.