Chapter 7 from Textbook Mathematics Part 1 Solutions for Class 9 MATHS

Question 1:

The following table shows the number of Buses and Trucks in nearest lakh units. Draw percentage bar-diagram. (Approximate the percentages to the nearest integer)

Year	No of trucks	No of buses
2005-2006 2007-2008 2008-2009 2009-2010	47 56 60 63	9 13 16 18

Answer:

The following table shows the number of Buses and Trucks in nearest lakh units.

Year	No of trucks	Percentage of trucks	No of buses	Percentage of buses
2005-2006	47	$= \frac{47}{(47 + 9)} \times 100 = \frac{4700}{56} = 83.92 \approx 84$	9	$= \frac{9}{(47 + 9)} \times 100 = \frac{900}{56} = 16.07 \approx 16$
2007-2008	56	$= \frac{56}{(56 + 13)} \times 100 = \frac{5600}{69} = 81.15 \approx 81$	13	$= \frac{13}{(56 + 13)} \times 100 = \frac{1300}{69} = 18.84 \approx 19$
2008-2009	60	$= \frac{60}{(60 + 16)} \times 100 = \frac{6000}{76} = 78.94 \approx 79$	16	$= \frac{16}{(60 + 16)} \times 100 = \frac{1600}{76} = 21.05 \approx 21$
2009-2010	63	$= \frac{63}{(63 + 18)} \times 100 = \frac{6300}{81} = 77.77 \approx 78$	18	$= \frac{18}{(63 + 18)} \times 100 = \frac{1800}{81} = 22.22 \approx 22$

The percentage bar-diagram of the above table is as follows:

Question 2:

In the table given below, the information is given about roads. Using this draw sub-divided and percentage bar-diagram (Approximate the percentages to the nearest integer)

Year	Permanent Roads ( Lakh km.)	Temporary Roads ( Lakh km.)
2000-2001 2001-2002 2003-2004 2007-2008	14 15 17 20	10 11 13 19

Answer:

In the table given below, the information is given about roads.

Year	Permanent Roads ( Lakh km.)	Percentage of permanent roads	Temporary Roads ( Lakh km.)	Percentage of temporary roads
2000-2001	14	$= \frac{14}{(14 + 10)} \times 100 = \frac{1400}{24} = 58.33 \approx 58$	10	$= \frac{10}{(14 + 10)} \times 100 = \frac{1000}{24} = 41.66 \approx 42$
2001-2002	15	$= \frac{15}{(15 + 11)} \times 100 = \frac{1500}{26} = 57.69 \approx 58$	11	$= \frac{11}{(15 + 11)} \times 100 = \frac{1100}{26} = 42.30 \approx 42$
2003-2004	17	$= \frac{17}{(17 + 13)} \times 100 = \frac{1700}{30} = 56.66 \approx 57$	13	$= \frac{13}{(17 + 13)} \times 100 = \frac{1300}{30} = 43.33 \approx 43$
2007-2008	20	$= \frac{20}{(20 + 19)} \times 100 = \frac{2000}{39} = 51.28 \approx 51$	19	$= \frac{19}{(20 + 19)} \times 100 = \frac{1900}{39} = 48.71 \approx 49$

The sub-divided and percentage bar-diagram of the above table is as follows:

Question 1:

Classify following information as primary or secondary data.

(i) Information of attendance of every student collected by visiting every class in a school.

(ii) The information of heights of students was gathered from school records and sent to the head office, as it was to be sent urgently.

(iii) In the village Nandpur, the information collected from every house regarding students not attending school.

(iv) For science project, information of trees gathered by visiting a forest.,

Answer:

(i) Since, the information of attendance of every student of every class is already collected by each class.

So, the information of attendance of every student collected by visiting every class in a school is a secondary data.

(ii) Since, the information of heights of students was gathered from school records which is already taken.

So, the information of heights of students was gathered from school records and sent to the head office, as it was to be sent urgently is a secondary data.

(iii) Since, the information is collected from every house.

So, the information collected from every house in the village Nandpur regarding students not attending school is a primary data.

(iv) Since, information of trees gathered by visiting a forest.

So, the information of trees gathered by visiting a forest for science project is a primary data.

Question 1:

For class interval 20-25 write the lower class limit and the upper class limit.

Answer:

For class interval 20-25.

The lower class limit is 20 and the upper class limit is 25.

Question 2:

Find the class-mark of the class 35-40.

Answer:

Since, $\frac{35 + 40}{2} = 37.5$

So, the class-mark of the class 35-40 is 37.5

Question 3:

If class mark is 10 and class width is 6 then find the class.

Answer:

We have, class mark = 10 and class width = 6.

Since, $\frac{6}{2} = 3$

The lower class limit = $10 - 3 = 7$ and the upper class limit = $10 + 3 = 13$ .

So, the class is 7-13.

Question 4:

Complete the following table.

Classes (age)	Tally marks	Frequency (No. of students)
12-13		$1234$
13-14		$1234$
14-15		$1234$
15-16		$1234$
		N=∑f = 35

Answer:

Classes (age)	Tally marks	Frequency (No. of students)
12-13		$5$
13-14		$14$
14-15	\|\|	$12$
15-16		$4$
		N=∑f = 35

Question 5:

In a ‘tree plantation’ project of a certain school there are 45 students of 'Harit Sena.' The record of trees planted by each student is given below :

3, 5, 7, 6, 4, 3, 5, 4, 3, 5, 4, 7, 5 , 3, 6, 6, 5, 3, 4, 5, 7, 3, 5, 6, 4, 4, 3, 5, 6, 6, 4, 3, 5, 7, 3, 4, 5, 7, 6, 4, 3, 5, 4, 4, 7.

Prepare a frequency distribution table of the data.

Answer:

The record of trees planted by each student is given below:

3, 5, 7, 6, 4, 3, 5, 4, 3, 5, 4, 7, 5 , 3, 6, 6, 5, 3, 4, 5, 7, 3, 5, 6, 4, 4, 3, 5, 6, 6, 4, 3, 5, 7, 3, 4, 5, 7, 6, 4, 3, 5, 4, 4, 7.

The frequency distribution table of the given data is as follows:

Number of trees planted	Tally marks	Number of students / Frequency
3	$\| \| \| \|$ $\| \| \| \|$	10
4	$\| \| \| \|$ $\| \| \| \|$ $\|$	11
5	$\| \| \| \|$ $\| \| \| \|$ $\|$	11
6	$\| \| \| \|$ $\|$ $\|$	7
7	$\| \| \| \|$ $\|$	6
	Total	45

Question 6:

The value of π upto 50 decimal places is given below :

3.14159265358979323846264338327950288419716939937510 From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.

Answer:

The value of π upto 50 decimal places is given below :

3.14159265358979323846264338327950288419716939937510

The ungrouped frequency distribution table of digits appearing after the decimal point is as follows:

Digits	Tally marks	Frequency
0	$\| \|$	2
1	$\| \| \| \|$	5
2	$\| \| \| \|$	5
3	$\| \| \| \|$ $\| \| \| \|$	9
4	$\| \| \| \|$	4
5	$\| \| \| \|$	5
6	$\| \| \| \|$	4
7	$\| \| \|$	3
8	$\| \| \| \|$	5
9	$\| \| \| \|$ $\| \| \|$	8
	Total	50

Question 7:

In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.

(i)

Class width	Frequency
5	3
15	9
25	15
35	13

(ii)

Class width	Frequency
22	6
24	7
26	13
28	4

Answer:

In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.

(i)

Class-mark	Frequency
5	3
15	9
25	15
35	13

We have, class size =

15 - 5 = 10

Since, the lower limit of class-mark 5

= 5 - \frac{10}{2} = 0

and the upper limit

= 5 + \frac{10}{2} = 10

.

So, the exclusive frequency table is given by:

Class	Frequency
0-10	3
10-20	9
20-30	15
30-40	13

Also, the inclusive frequency table is given by:

Class	Frequency
0-10	3
11-20	9
21-30	15
31-40	13

(ii)

Class-mark	Frequency
22	6
24	7
26	13
28	4

We have, class size =

24 - 22 = 2

Since, the lower limit of class-mark 22

= 22 - \frac{2}{2} = 21

and the upper limit

= 22 + \frac{2}{2} = 23

.

So, the exclusive frequency table is given by:

Class	Frequency
21-23	6
23-25	7
25-27	13
27-29	4

Also, the inclusive frequency table is given by:

Class	Frequency
20-22.5	6
23-25.5	7
25-27.5	13
27-29.5	4

Disclaimer: The exclusive frequency table can vary from student to student. The "Class width" is replaved by "Class-marks" according to the question.

Question 8:

In a school, 46 students of 9th standard, were told to measure the lengths of the pencilsin their compass-boxes in centimeters. The data collected was as follows.

16, 15, 7, 4.5, 8.5, 5.5, 5, 6.5, 6, 10, 12, 13, 4.5, 4.9, 16, 11, 9.2, 7.3, 11.4, 12.7, 13.9, 16, 5.5, 9.9, 8.4, 11.4, 13.1, 15, 4.8, 10, 7.5, 8.5, 6.5, 7.2, 4.5, 5.7, 16, 5.7, 6.9, 8.9, 9.2, 10.2, 12.3, 13.7, 14.5, 10

By taking inclusive classes 0-5, 5-10, 10-15.... prepare a grouped frequency distribution

table.

Answer:

Given:

In a school, 46 students of 9th standard, were told to measure the lengths of the pencilsin their compass-boxes in centimeters. The data collected was as follows.

16, 15, 7, 4.5, 8.5, 5.5, 5, 6.5, 6, 10, 12, 13, 4.5, 4.9, 16, 11, 9.2, 7.3, 11.4, 12.7, 13.9, 16, 5.5, 9.9, 8.4, 11.4, 13.1, 15, 4.8, 10, 7.5, 8.5, 6.5, 7.2, 4.5, 5.7, 16, 5.7, 6.9, 8.9, 9.2, 10.2, 12.3, 13.7, 14.5, 10

The grouped frequency distribution table for the given data is as follows:

Class Interval	Tally Marks	Frequency
0-5	$\| \| \| \|$	5
5-10	$\| \| \| \|$ $\| \| \| \|$ $\| \| \| \|$ $\| \| \| \|$	20
10-15	$\| \| \| \|$ $\| \| \| \|$ $\| \| \| \|$	15
15-20	$\| \| \| \|$ $\|$	6
	Total	46

Question 9:

In a village, the milk was collected from 50 milkmen at a collection center in litres as given below :

27, 75, 5, 99, 70, 12, 15, 20, 30, 35, 45, 80, 77, 90, 92, 72, 4, 33, 22, 15, 20, 28, 29, 14, 16, 20, 72, 81, 85, 10, 16, 9, 25, 23, 26, 46, 55, 56, 66, 67, 51, 57, 44, 43, 6, 65, 42, 36, 7, 35.

By taking suitable classes, prepare grouped frequency distribution table.

Answer:

Given:

In a village, the milk was collected from 50 milkmen at a collection center in litres as given below :

27, 75, 5, 99, 70, 12, 15, 20, 30, 35, 45, 80, 77, 90, 92, 72, 4, 33, 22, 15, 20, 28, 29, 14, 16, 20, 72, 81, 85, 10, 16, 9, 25, 23, 26, 46, 55, 56, 66, 67, 51, 57, 44, 43, 6, 65, 42, 36, 7, 35.

The grouped frequency distribution table for the above data is as follows:

Class interval	Tally marks	Frequency
0-10	$\| \| \| \|$	5
10-20	$\| \| \| \|$ $\| \|$	7
20-30	$\| \| \| \|$ $\| \| \| \|$	10
30-40	$\| \| \| \|$	5
40-50	$\| \| \| \|$	5
50-60	$\| \| \| \|$	4
60-70	$\| \| \|$	3
70-80	$\| \| \| \|$	5
80-90	$\| \| \|$	3
90-100	$\| \| \|$	3
	Total	50

Question 10:

38 people donated to an organisation working for differently abled persons. The amount in rupees were as follows :

101, 500, 401, 201, 301, 160, 210, 125, 175, 190, 450, 151, 101, 351, 251, 451, 151, 260, 360, 410, 150, 125, 161, 195, 351, 170, 225, 260, 290, 310, 360, 425, 420, 100, 105, 170, 250, 100.

(i) By taking classes 100-149, 150-199, 200-249... prepare grouped frequency distribution table.

(ii) From the table, find the number of people who donated rupees 350 or more.

Answer:

Given:

38 people donated to an organisation working for differently abled persons. The amount in rupees were as follows:

101, 500, 401, 201, 301, 160, 210, 125, 175, 190, 450, 151, 101, 351, 251, 451, 151, 260, 360, 410, 150, 125, 161, 195, 351, 170, 225, 260, 290, 310, 360, 425, 420, 100, 105, 170, 250, 100.

(i) The grouped frequency distribution table of the given data is as follows:

Class interval	Tally marks	Frequency
100-149	$\| \| \| \| \| \| \|$	7
150-199	$\| \| \| \| \| \| \| \| \| \|$	10
200-249	$\| \| \|$	3
250-299	$\| \| \| \| \|$	5
300-349	$\| \|$	2
350-399	$\| \| \| \|$	4
400-449	$\| \| \| \|$	4
450-500	$\| \| \|$	3
	Total	38

(ii) As, 4 + 4 + 3 = 11

So, the number of people who donated rupees 350 or more is 11.

Question 1:

Complete the following cumulative freuqency table :

Class (Height in cm)	Frequency (No. of students)	Less than type frequency
150-153	05	05
153-156	07	05 + $1234$ = $1234$
156-159	15	$1234$ + 15 = $1234$
159-162	10	$1234$ + $1234$ = 37
162-165	05	37 + 5 = 42
165-168	03	$1234$ + $1234$ = 45
	Total N = 45

Answer:

The cumulative frequency table:

Class (Height in cm)	Frequency (No. of students)	Less than type frequency
150-153	05	05
153-156	07	05 + $07$ = $12$
156-159	15	$12$ + 15 = $27$
159-162	10	$27$ + $10$ = 37
162-165	05	37 + 5 = 42
165-168	03	$42$ + $03$ = 45
	Total N = 45

Question 2:

Complete the following Cumulative Frequency Table :

Class (Monthly income in Rs.)	Frequency (No. of individuals)	More than or equal to type cumulative frequency
1000-5000	45	...........
5000-10000	19	..........
10000-15000	16	..........
15000-20000	02	.........
20000-25000	05	.........
	Total N = 87

Answer:

The Cumulative Frequency Table:

Class (Monthly income in Rs.)	Frequency (No. of individuals)	More than or equal to type cumulative frequency
1000-5000	45	45 + 42 = 87
5000-10000	19	19 + 23 = 42
10000-15000	16	16 + 07 = 23
15000-20000	02	02 + 05 = 07
20000-25000	05	05
	Total N = 87

Question 3:

The data is given for 62 students in a certain class regarding their mathematics marks out of 100. Take the classes 0-10, 10-20.. and prepare frequency distribution table and cumulative frequency table more than or equal to type.

55, 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10, 75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64, 42, 58, 31, 82, 27, 11, 78, 97, 07, 22, 2 36, 35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17, 77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45, 47, 49.

From the prepared table, answer the following questions :

(i) How many students obtained marks 40 or above 40 ?

(ii) How many students obtained marks 90 or above 90 ?

(iii) How many students obtained marks 60 or above 60 ?

(iv) What is the cumulative frequency of equal to or more than type of the class 0-10?

Answer:

Given:

The data is given for 62 students in a certain class regarding their mathematics marks out of 100.

55, 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10, 75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64, 42, 58, 31, 82, 27, 11, 78, 97, 07, 22, 2 36, 35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17, 77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45, 47, 49.

The frequency distribution table for the given data is as follows:

Class (Marks obtained)	Tally marks	Frequency (No. of students)
0-10	$\| \| \| \|$	4
10-20	$\| \| \|$	3
20-30	$\| \| \| \| \| \| \|$	8
30-40	$\| \| \| \| \| \| \| \|$	9
40-50	$\| \| \| \| \| \| \| \| \| \| \|$	13
50-60	$\| \| \| \| \|$	6
60-70	$\| \| \| \|$	5
70-80	$\| \| \| \| \|$	6
80-90	$\| \| \| \|$	5
90-100	$\| \| \|$	3
		Total N = 62

The cumulative frequency table more than or equal to type for the given data is as follows:
â€‹

Class (Marks obtained)	Frequency (No. of students)	More than or equal to type cumulative frequency
0-10	4	62
10-20	3	58
20-30	8	55
30-40	9	47
40-50	13	38
50-60	6	25
60-70	5	19
70-80	6	14
80-90	5	8
90-100	3	3
	Total N = 62

(i) The number of students who obtained marks 40 or above 40 is 38.

(ii) The number of students who obtained marks 90 or above 90 is 3.

(iii) The number of students who obtained marks 60 or above 60 is 19.

(iv) The cumulative frequency of equal to or more than type of the class 0-10 is 62.

Question 4:

Using the data in example (3) above, prepare less than type cumulative frequency table and answer the following questions.

(i) How many students obtained less than 40 marks ?

(ii) How many students obtained less than 10 marks ?

(iii) How many students obtained less than 60 marks ?

(iv) Find the cumulative frequency of the class 50-60.

Answer:

Given:

The data is given for 62 students in a certain class regarding their mathematics marks out of 100.

55, 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10, 75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64, 42, 58, 31, 82, 27, 11, 78, 97, 07, 22, 2 36, 35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17, 77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45, 47, 49.

The frequency distribution table for the given data is as follows:

Class (Marks obtained)	Tally marks	Frequency (No. of students)
0-10	$\| \| \| \|$	4
10-20	$\| \| \|$	3
20-30	$\| \| \| \| \| \| \|$	8
30-40	$\| \| \| \| \| \| \| \|$	9
40-50	$\| \| \| \| \| \| \| \| \| \| \|$	13
50-60	$\| \| \| \| \|$	6
60-70	$\| \| \| \|$	5
70-80	$\| \| \| \| \|$	6
80-90	$\| \| \| \|$	5
90-100	$\| \| \|$	3
		Total N = 62

The cumulative frequency table of less than type for the given data is as follows:
â€‹

Class (Marks obtained)	Frequency (No. of students)	Less than type cumulative frequency
0-10	4	4
10-20	3	7
20-30	8	15
30-40	9	24
40-50	13	37
50-60	6	43
60-70	5	48
70-80	6	54
80-90	5	59
90-100	3	62
	Total N = 62

(i) The number of students who obtained less than 40 marks is 24.

(ii) The number of students who obtained less than 10 marks is 4.

(iii) The number of students who students who obtained less than 60 marks is 43.

(iv) The cumulative frequency of the class 50-60 is 43.

Question 1:

Yield of soyabean per acre in quintal in Mukund's field for 7 years was 10, 7, 5, 3, 9, 6, 9. Find the mean of yield per acre.

Answer:

$Mean = \frac{10 + 7 + 5 + 3 + 9 + 6 + 9}{7} = \frac{49}{7} = 7$

So, the mean of yield per acre is 7.

Question 2:

Find the median of the observations, 59, 75, 68, 70, 74, 75, 80.

Answer:

The observations in ascending order: 59, 68, 70, 74, 75, 75, 80.

Since, the number of observation, n = 7, which is odd.

$\Rightarrow Median = (\frac{n + 1}{2}) th = (\frac{7 + 1}{2}) th = (\frac{8}{2}) th = 4 th Observation = 74$

So, the median of the observations is 74.

Question 3:

The marks (out of 100) obtained by 7 students in Mathematics' examination are given below. Find the mode for these marks.

99, 100, 95, 100, 100, 60, 90

Answer:

Given:

The marks (out of 100) obtained by 7 students in Mathematics' examination are given below:

99, 100, 95, 100, 100, 60, 90

Since, 100 repeats for maximum times.

So, the mode for these marks is 100.

Question 4:

The monthly salaries in rupees of 30 workers in a factory are given below.

5000, 7000, 3000, 4000, 4000, 3000, 3000, 3000, 8000, 4000,

4000, 9000, 3000, 5000, 5000, 4000, 4000, 3000, 5000, 5000,

6000, 8000, 3000, 3000, 6000, 7000, 7000, 6000, 6000, 4000

From the above data find the mean of monthly salary.

Answer:

Given:

The monthly salaries in rupees of 30 workers in a factory are given below:

5000, 7000, 3000, 4000, 4000, 3000, 3000, 3000, 8000, 4000,

4000, 9000, 3000, 5000, 5000, 4000, 4000, 3000, 5000, 5000,

6000, 8000, 3000, 3000, 6000, 7000, 7000, 6000, 6000, 4000

The frequency table of the above data is as follows:

Salaries (Rs.) (x_i)	Number of workers (f_i)	f_ix_i
3000	8	24000
4000	7	28000
5000	5	25000
6000	4	24000
7000	3	21000
8000	2	16000
9000	1	9000
	Total N = 30	Total = 147000

∵ Mean = \frac{147000}{30} = 4900

So, the mean of monthly salary is Rs 49000.

Question 5:

In a basket there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows:
60, 70, 90, 95, 50, 65, 70, 80, 85, 95.

Find the median of the weights of tomatoes.

Answer:

Given:

The weight of 10 tomatoes in a basket in grams in ascending order is as follows:

50, 60, 65, 70, 70, 80, 85, 90, 95, 95

Since, the number of observations is 10, which is even.

So, median = \frac{1}{2} [(\frac{10}{2}) th observation + (\frac{10}{2} + 1) th observation] = \frac{1}{2} [5 th observation + (5 + 1) th observation] = \frac{1}{2} [5 th observation + 6 th observation] = \frac{1}{2} [70 + 80] = \frac{150}{2} = 75

Find the median of the weights of tomatoes is 75 grams.

Question 6:

A hockey player has scored following number of goals in 9 matches.

5, 4, 0, 2, 2, 4, 4, 3, 3

Find the mean, median and mode of the data.

Answer:

Given:

The scores of the hockey player in 9 matches in ascending order:

0, 2, 2, 3, 3, 4, 4, 4, 5

Mean = \frac{5 + 4 + 0 + 2 + 2 + 4 + 4 + 3 + 3}{9} = \frac{27}{9} = 3

Median = (\frac{9 + 1}{2}) th = (\frac{10}{2}) th = 5 th observation = 3

â€‹Mode = The observation with the maximum occurrence = 4

Question 7:

The calculated mean of 50 observations was 80. It was later discovered that observation 19 was recorded by mistake as 91. What was the correct mean?

Answer:

It was later discovered that observation 19 was recorded by mistake as 91. What was the correct mean?

We have,

The calculated mean of 50 observations was 80.

$\Rightarrow$ The total incorrect sum of the observations = $80 \times 50 = 4000$

$\Rightarrow$ The total correct sum of the observations $= 4000 - 19 + 91 = 4072$

So, the correct mean $= \frac{4072}{50} = 81.44$

Question 8:

Following 10 observations are arranged in ascending order as follows.

2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20

If the median of the data is 11, find the value of x

Answer:

Given:

The 10 observations are arranged in ascending order as follows:

2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20

Since, the median of the data is 11.

As, the number of observations = 10, which is an even number.

\Rightarrow Median = 11 \Rightarrow \frac{1}{2} [(\frac{10}{2}) th observation + (\frac{10}{2} + 1) th observation] = 11 \Rightarrow \frac{1}{2} [5 th observation + 6 th observation] = 11 \Rightarrow \frac{1}{2} [x + 1 + x + 3] = 11 \Rightarrow 2 x + 4 = 2 \times 11 \Rightarrow 2 x = 22 - 4 \Rightarrow x = \frac{18}{2} = 9

So, the value of x is 9.

Question 9:

The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observation is 25. Find the 18th observation.

Answer:

We have,

The mean of 35 observations is 20.
Let x be the sum of 35 observations

$\Rightarrow$ mean = $\frac{sum of observations}{number of observations}$
$\Rightarrow 20 = \frac{x}{35} \Rightarrow x = 700$

Let y be the sum of first 18 observations.
Mean of first 18 observations is 15.
$\frac{y}{18} = 15 \Rightarrow y = 270$
And z be the sum of last 18 observations
Mean of last 18 observations is 25.
$\frac{z}{18} = 25 \Rightarrow z = 450$
let the 18th term be a.
18th term is common in first 18 and last 18 observations
$x + a = y + z 700 + a = 270 + 450 a = 720 - 700 a = 20$
Hence, the 18th observation is 20.

Question 10:

The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.

Answer:

We have,

The mean of 5 observations is 50.

⇒ The total sum of 5 observations = 50 × 5 = 250

Also, the mean of 4 observations is 45.

⇒ The total sum of 4 observations = 45 × 4 = 180

So, the observation which was removed = 250 − 180 = 70.

Question 11:

There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.

Answer:

We have,

Number of students in a class = 40, and

Number of boys in a class = 14,

Mean of marks obtained by boys = 33, and

Mean of marks obtained by girls = 35

Number of girls in the class = $40 - 15 = 25$

Sum of marks obtained by boys = $33 \times 15 = 495$ , and

Sum of marks obtained by girls = $35 \times 25 = 875$

So, the mean of all students in the class = $\frac{495 + 875}{40} = \frac{1370}{40} = 34.25$ .

Question 12:

The weights of 10 students (in kg) are given below :

40, 35, 42, 43, 37, 35, 37, 37, 42, 37.

Find the mode of the data

Answer:

Given:

The weights of 10 students (in kg) are given below:

40, 35, 42, 43, 37, 35, 37, 37, 42, 37.

Since, the observation that occurs maximum = 37

So, the mode of the data is 37.

Question 13:

In the following table, the information is given about the number of families and the siblings in the families less than 14 years of age. Find the mode of the data.

No. of siblings	1	2	3	4
Families	15	25	5	5

Answer:

The number of families and the siblings in the families less than 14 years of age is given in the following information table:

No. of siblings	1	2	3	4
Families	15	25	5	5

Since, the number of siblings that occurs the most = 2.

So, the mode of the data is 2.

Question 14:

Find the mode of the following data.

Marks	35	36	37	38	39	40
No of students	09	07	09	04	04	02

Answer:

We have,

Marks	35	36	37	38	39	40
No of students	09	07	09	04	04	02

Since, the marks that occurs the most are 35 and 37.

So, the mode of the given data are 35 and 37.

Question 1:

Write the correct alternative answer for each of the following questions.

(i) Which of the following data is not primary ?

(A) By visiting a certain class, gathering information about attendence of students.

(B) By actual visit to homes, to find number of family members.

(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

(D) Review the cleanliness status of canals by actually visitiing them.

(ii) What is the upper class limit for the class 25-35 ?

(A) 25

(B) 35

(C) 60

(D) 30

(iii) What is the class-mark of class 25-35 ?

(A) 25

(B) 35

(C) 60

(D) 30

(iv) If the classes are 0-10, 10-20, 20-30... then in which class should the observation 10 be included ?

(A) 0-10 (B) 10-20 (C) 0-10 and 10-20 in these 2 classes (D) 20-30

(v) If

\bar{x}

is the mean of

x_{1,} x_{2} . . . . . . . . . . . . . x_{n}

and

\bar{y}

is the mean of

y_{1}, y_{2} . . . . . . . . . . . . . . . . y_{n}

and

\bar{z}

is the mean

x_{1}, x_{2} . . . . . . . . . . . . . . x_{n}, y_{1}, y_{2}, . . . . . . . . . . . . . . . . . . . . . y_{n}

then z = ?

(A)

\frac{\bar{x} + \bar{y}}{2}

( B)

\bar{x} + \bar{y}

(C)

\frac{\bar{x} + \bar{y}}{n}

(D)

\frac{\bar{x} + \bar{y}}{2 n}

(vi) The mean of five numbers is 80, out of which mean of 4 numbers is 46, find the 5th number :

(A) 4

(B) 20

(C) 434

(D) 66

(vii) Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.

(A) 40.6

(B) 40.4

(C) 40.3

(D) 40.7

(viii) What is the the mode of 19, 19, 15, 20, 25, 15, 20, 15?

(A) 15,

(B) 20

(C) 19
(D) 25

(ix) What is the median of 7, 10, 7, 5, 9, 10 ?

(A) 7

(B) 9

(C) 8

(D) 10

(x) From following table, what is the cumulative frequency of less than type for the class 30-40 ?

Class	0-10	10-20	20-30	30-40	40-50
Frequency	7	3	12	13	2

(A) 13

(B) 15

(C) 35

(D) 22

Answer:

(i)

(A) By visiting a certain class, gathering information about attendence of students.

Since, the information related to the attendence has been already recorded.

So, it is not primary.

(B) By actual visit to homes, to find number of family members.

Since, the information is collected by visiting.

So, it is a primary data.

(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

Since, the information is collected by each member.

So, it is a primary data.

(D) Review the cleanliness status of canals by actually visitiing them.

Since, this information is gathered by visiting.

So, it is a primary data.

Hence, the correct alternative answer is (A).

(ii) Since, the upper class limit for the class 25-35 is 35.

Hence, the correct alternative answer is (B).

(iii) Since, the class-mark of class 25-35

\frac{25 + 35}{2} = \frac{60}{2} = 30

Hence, the correct alternative answer is (D).

(iv) If the classes are 0-10, 10-20, 20-30..,. then the class in which the observation 10 should be included 10-20.

Hence, the correct alternative answer is (B).

(v) If

\bar{x}

is the mean of

x_{1,} x_{2} . . . . . . . . . . . . . x_{n}

and

\bar{y}

is the mean of

y_{1}, y_{2} . . . . . . . . . . . . . . . . y_{n}

and

\bar{z}

is the mean

x_{1}, x_{2} . . . . . . . . . . . . . . x_{n}, y_{1}, y_{2}, . . . . . . . . . . . . . . . . . . . . . y_{n}

,

Then z =

\frac{n \bar{x} + n \bar{y}}{2 n} = \frac{n (x + y)}{2 n} = \frac{x + y}{2}

Hence, the correct alternative answer is (A).

(vi) Since, the mean of five numbers is 80, out of which mean of 4 numbers is 46

So, the 5th number

= Sum of 5 observations - Sum of 4 observations = (80 \times 5) - (46 \times 4) = 400 - 184 = 216

.

Disclaimer: The options provided are all incorrect.

(vii) We have,

Mean of 100 observations is 40.

Total sum of observations =

100 \times 40 = 4000

If the 9th observation is 30 and is replaced by 70 keeping all other observations same.

New, total sum of observations =

4000 - 30 + 70 = 4040

So, the new mean =

\frac{4040}{100} = 40.4

Hence, the correct alternative answer is (B).

(viii) Since, the observation 15 is occuring the most in the data: 19, 19, 15, 20, 25, 15, 20, 15

So, the mode of the given data is 15.

Hence, the correct alternative answer is (A).

(ix) Since, the ascending order of the given data is as: 5, 7, 7, 9, 10, 10.

And, the number of observations is 6, which is even.

So, the median of the given data = \frac{1}{2} [(\frac{6}{2}) th observation + (\frac{6}{2} + 1) th observation] = \frac{1}{2} [3 rd observation + 4 th observation] = \frac{1}{2} [7 + 9] = \frac{16}{2} = 8

Hence, the correct alternative answer is (C).

(x) We have,

Class	0-10	10-20	20-30	30-40	40-50
Frequency	7	3	12	13	2

So, the cumulative frequency of less than type for the class 30-40 =

7 + 3 + 12 + 13 = 35

Hence, the correct alternative answer is (C).

Question 2:

The mean salary of 20 workers is Rs.10,250. If the salary of office superintendent is added, the mean will increase by Rs.750. Find the salary of the office superintendent.

Answer:

We have,

The mean salary of 20 workers is Rs.10,250.

The sum of salary of 20 workers = $10250 \times 20 = Rs . 2, 05, 000$

Since, the mean will increase by Rs.750 if the salary of office superintendent is added

$\Rightarrow$ The mean salary of 21 workers including office superintendent = $10250 + 750 = Rs . 11, 000$ $10250 + 750 = Rs . 11, 000$

$\Rightarrow$ The sum of salary of 21 workers including office superintendent = $11000 \times 21 = Rs . 2, 31, 000$

So, the salary of the office superintendent = $231000 - 205000 = Rs . 26, 000$

Question 3:

The mean of nine numbers is 77. If one more number is added to it then the mean increases by 5. Find the number added in the data.

Answer:

We have, the mean of nine numbers is 77.

The sum of nine numbers = $77 \times 9 = 693$ .

The new mean when one more number is added to it = 77 + 5 = 82.

The sum of ten numbers = $82 \times 10 = 820$ .

So, the number added in the data = $820 - 693 = 127$ .

Question 4:

The monthly maximum temperature of a city is given in degree celcius in the following data. By taking suitable classes, prepare the grouped frequency distribution table

29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5, 29.0, 29.5, 29.9, 33.2, 30.2

From the table, answer the following questions.

(i) For how many days the maximum temperature was less than 34^oC ?

(ii) For how many days the maximum temperature was 34^oC or more than 34^oC ?

Answer:

Given:

The monthly maximum temperature of a city is given in degree celcius in the following data:

29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5, 29.0, 29.5, 29.9, 33.2, 30.2

The grouped frequency distribution table of the given data is as follows:

Class (Temperature)	Tally marks	Frequency
28-29	\|\|	2
29-30	$\| \| \| \|$ \|	6
30-31	$\| \| \| \|$	5
31-32	\|\|\|	3
32-33	$\| \| \| \|$	5
33-34	\|\|\|	3
34-35	\|\|\|	3
35-36	\|\|	2
36-37	\|	1
		Total N = 30

(i) The number of days the maximum temperature was less than 34^oC = 2 + 6 + 5 + 3 + 5 + 3 = 24.

(ii) The number of days the maximum temperature was 34^oC or more than 34^oC = 3 + 2 + 1 = 6.

Question 5:

If the mean of the following data is 20.2, then find the value of p.

$x_{i}$	10	15	20	25	30
$f_{i}$	6	8	p	10	6

Answer:

We have,

$x_{i}$	$f_{i}$	$f_{i}$ $x_{i}$
10	6	60
15	8	120
20	p	20p
25	10	250
30	6	180
	Total N = (30 + p)	Total sum = (610 + 20p)

Since, mean = 20.2

\Rightarrow \frac{(610 + 20 p)}{(30 + p)} = 20.2 \Rightarrow 610 + 20 p = 20.2 (30 + p) \Rightarrow 610 + 20 p = 606 + 20.2 p \Rightarrow 20.2 p - 20 p = 610 - 606 \Rightarrow 0.2 p = 4 \Rightarrow p = \frac{4}{0.2} ∴ p = 20

Question 6:

There are 68 students of 9th standard from model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.

70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68, 80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43, 36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55, 56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78, 80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37, 45, 42, 70, 37, 45, 66, 56, 47

By taking classes 30-40, 40-50, .... prepare the less than type cumulative frequency table

Using the table, answer the following questions :

(i) How many students, have scored marks less than 80 ?

(ii) How many students have scored marks less than 40 ?

(iii) How many students have scored marks less than 60 ?

Answer:

Given:

The marks scored by 68 students of 9th standard from model Highschool, Nandpur out of 80, in written exam of mathematics is as follows:

70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68, 80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43, 36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55, 56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78, 80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37, 45, 42, 70, 37, 45, 66, 56, 47

Class (Marks)	Frequency	Less than type cumulative frequency
30-40	14	14
40-50	20	34
50-60	11	45
60-70	12	57
70-80	9	66
80-90	2	68
	total N = 68

(i) The number of students, who have scored marks less than 80 is 66.

(ii) The number of students who have scored marks less than 40 is 14.

(iii) The number of students who have scored marks less than 60 = 45.

Question 7:

By using data in example (6), and taking classes 30-40, 40-50... prepare equal to or more than type cumulative frequency table and answer the following questions based on it.

(i) How many students have scored marks 70 or more than 70 ?

(ii) How many students have scored marks 30 or more than 30 ?

Answer:

Given:

The marks scored by 68 students of 9th standard from model Highschool, Nandpur out of 80, in written exam of mathematics is as follows:

70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68, 80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43, 36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55, 56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78, 80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37, 45, 42, 70, 37, 45, 66, 56, 47

The equal to or more than type cumulative frequency of the above data:

Class (Marks)	Frequency	Equal to or more than type cumulative frequency
30-40	14	68
40-50	20	54
50-60	11	34
60-70	12	23
70-80	9	11
80-90	2	2
	total N = 68

(i) The number of students who have scored marks 70 or more than 70 is 11.

(ii) The number of students who have scored marks 30 or more than 30 is 68.

Question 8:

There are 10 observations arranged in ascending order as given below.

45, 47 ,50 ,52, x, x+2, 60,62,63,74. The median of these observations is 53.

Find the value of x. Also find the mean and the mode of the data.

Answer:

We have,

The 10 observations are arranged in ascending order as given below.

45, 47, 50, 52, x, x + 2, 60 ,62 ,63 ,74.

Since, the number of observation is 10, which is an even number and,

The median of the given observations is 53.

\Rightarrow Median = 53 \Rightarrow \frac{1}{2} [(\frac{10}{2}) th observation + (\frac{10}{2} + 1) th observation] = 53 \Rightarrow \frac{1}{2} [5 th observation + 6 th observation] = 53 \Rightarrow \frac{1}{2} [x + x + 2] = 53 \Rightarrow 2 x + 2 = 53 \times 2 \Rightarrow 2 x = 106 - 2 \Rightarrow x = \frac{104}{2} ∴ x = 52

Mean of the given data = \frac{45 + 47 + 50 + 52 + 52 + 54 + 60 + 62 + 63 + 74}{10} = \frac{559}{10} = 55.9

Since, the observation that is occurring the maximum is 52.

So, the mode of the data is 52.

Digits	Tally marks	Frequency
0	$\| \|$	2
1	$\| \| \| \|$	5
2	$\| \| \| \|$	5
3	$\| \| \| \|$ $\| \| \| \|$	9
4	$\| \| \| \|$	4
5	$\| \| \| \|$	5
6	$\| \| \| \|$	4
7	$\| \| \|$	3
8	$\| \| \| \|$	5
9	$\| \| \| \|$ $\| \| \|$	8
	Total	50

Mathematics Part i Solutions Solutions for Class 9 Maths Chapter 7 - Statistics

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