Chapter 6 from Textbook Mathematics Part 1 Solutions for Class 10 MATHS

Question 1:

The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.

Time (hrs.)	0 - 2	2 - 4	4 - 6	6 - 8	8 - 10
No. of students	7	18	12	10	3

Answer:

Class (Time in hours)	Class Mark x_i	Frequency (Number of students) f_i	Class mark × Frequency x_if_i
0 - 2	1	7	7
2 - 4	3	18	54
4 - 6	5	12	60
6 - 8	7	10	70
8 - 10	9	3	27
		$\sum_{} f_{i} = 50$	$\sum_{} x_{i} f_{i} = 218$

Mean =

\frac{\sum_{} x_{i} f_{i}}{\sum_{} f_{i}}

â€‹

= \frac{218}{50}

= 4.36 hours
Hence, the mean time spent by students for their studies is 4.36 hours.

Question 2:

In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by 'assumed mean' method.

Toll (Rupees)	300 - 400	400 - 500	500 - 600	600 - 700	700 - 800
No. of vehicles	80	110	120	70	40

Answer:

Class (Toll in rupees)	Class Mark x_i	d_i = x_i− A	Frequency (Number of vehicles f_i	Frequency × deviation f_i× d_i
300 - 400	350	−200	80	−16000
400 - 500	450	−100	110	−11000
500 - 600	550 = A	0	120	0
600 - 700	650	100	70	7000
700 - 800	750	200	40	8000
			$\underset{}{\sum f_{i} = 420}$	$\sum_{} f_{i} d_{i} = - 12000$

Required Mean =

A + \frac{\sum_{} f_{i} d_{i}}{\sum_{} f_{i}}

= 550 - \frac{12000}{420}

= 550 − 28.57
â€‹= Rs 521.43
Hence, the mean of toll is Rs 521.43.

Question 3:

A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.

Milk Sold (Litre)	1 - 2	2 - 3	3 - 4	4 - 5	5 - 6
No. of Customers	17	13	10	7	3

Answer:

Class (Milk sold in litres)	Class Mark x_i	Frequency (Number of customers) f_i	Class mark × Frequency x_if_i
1 - 2	1.5	17	25.5
2 - 3	2.5	13	32.5
3 - 4	3.5	10	35
4 - 5	4.5	7	31.5
5 - 6	5.5	3	16.5
		$\sum_{} f_{i} = 50$	$\sum_{} x_{i} f_{i} = 141$

Mean =

\frac{\sum_{} x_{i} f_{i}}{\sum_{} f_{i}}

â€‹

= \frac{141}{50}

= 2.82 litres
Hence, the mean of the milk sold is 2.82 litres.

Question 4:

A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by 'assumed mean' method.

Production (Thousand rupees)	25 - 30	30 - 35	35 - 40	40 - 45	45 - 50
No. of Customers	20	25	15	10	10

Answer:

â€‹

Class (Production in Thousand rupees)	Class Mark x_i	d_i = x_i− A	Frequency (Number of farm owners) f_i	Frequency × deviation f_i× d_i
25 - 30	27.5	−10	20	−200
30 - 35	32.5	−5	25	−125
35- 40	37.5= A	0	15	0
40 - 45	42.5	5	10	50
45 - 50	47.5	10	10	100
			$\underset{}{\sum f_{i} = 80}$	$\sum_{} f_{i} d_{i} = - 175$

Required Mean =

A + \frac{\sum_{} f_{i} d_{i}}{\sum_{} f_{i}}

37.5 - \frac{175}{80}

= 37.5 − 2.19
â€‹= 35.31 thousand rupees
= Rs 35310
Hence, the mean production of oranges is Rs 35310.

Question 5:

A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of the funds by 'step deviation' method.

Fund (Rupees)	0 - 500	500 - 1000	1000 - 1500	1500 - 2000	2000 - 2500
No. of workers	35	28	32	15	30

Answer:

â€‹

Class (Production in Thousand rupees)	Class Mark x_i	d_i = x_i− A	$u_{i} = \frac{d_{i}}{h}$	Frequency (Number of farm owners) f_i	Frequency × deviation f_i× u_i
0 - 500	250	−1000	−2	35	−70
500 - 1000	750	−500	−1	28	−28
1000 - 1500	1250 = A	0	0	32	0
1500 - 2000	1750	500	1	15	15
2000 - 2500	2250	1000	2	10	20
				$\underset{}{\sum f_{i} = 120}$	$\sum_{} f_{i} u_{i} = - 63$

Required Mean =

A + h \frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}}

= 1250 - (\frac{63}{120}) 500

= 1250 − 262.5
â€‹= Rs 987.5
Hence, the mean of the funds is Rs 987.5.

Question 6:

The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by 'step deviation' method.

Weekly wages (Rupees)	1000 - 2000	2000 - 3000	3000 - 4000	4000 - 5000
No. of workers	25	45	50	30

Answer:

â€‹

Class (Weekely wages rupees)	Class Mark x_i	d_i = x_i− A	$u_{i} = \frac{d_{i}}{h}$	Frequency (Number of workers) f_i	Frequency × deviation f_i× u_i
1000 - 2000	1500	−2000	−2	25	−50
2000 - 3000	2500	−100	−1	45	−45
3000 - 4000	3500 = A	0	0	50	0
4000 - 5000	4500	1000	1	30	30
				$\underset{}{\sum f_{i} = 150}$	$\sum_{} f_{i} u_{i} = - 65$

Required Mean =

A + h \frac{\sum_{} f_{i} u_{i}}{\sum_{} f_{i}}

= 3500 - (\frac{65}{150}) 1000

= 3500 − 433.33
â€‹= Rs 3066.67
Hence, the mean of the weekly wages is Rs 3066.67.

Question 1:

The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.

Daily No. of hours	8 - 10	10 - 12	12 - 14	14 - 16
Number of workers	150	500	300	50

Answer:

Class (Number of working hours)	Frequency (Number of workers) f_i	Cumulaive frequency less than the upper limit
8 - 10	150	150
10 - 12 (Median Class)	500	650
12 - 14	300	950
14 - 16	50	1000
	$N = 1000$

From the above table, we get
L (Lower class limit of the median class) = 10
N (Sum of frequencies) = 1000
h (Class interval of the median class) = 2
f (Frequency of the median class) = 500
cf (Cumulative frequency of the class preceding the median class) = 150
Now, Median =

L + (\frac{\frac{N}{2} - c f}{f}) \times h

â€‹

= 10 + (\frac{\frac{1000}{2} - 150}{500}) \times 2 = 10 + 1.4 = 11.4 hours

Hence, the median of the number of hours they work is 11.4 hours.

Question 2:

The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.

No. of Mangoes	50 - 100	100 - 150	150 - 200	200 - 250	250 - 300
No. of trees	33	30	90	80	17

Answer:

Class (Number of working hours)	Frequency (Number of workers) f_i	Cumulaive frequency less than the upper limit
50 - 100	33	33
100 - 150	30	63
150 - 200 (Median Class)	90	153
200 - 250	80	233
250 - 300	17	250
	N = 250

From the above table, we get
L (Lower class limit of the median class) = 150
N (Sum of frequencies) = 250
h (Class interval of the median class) = 50
f (Frequency of the median class) = 90
cf (Cumulative frequency of the class preceding the median class) = 63
Now, Median =

L + (\frac{\frac{N}{2} - c f}{f}) \times h

â€‹

= 150 + (\frac{\frac{250}{2} - 63}{90}) \times 50

= 150 + 34.44
= 184.44 mangoes
= 184 mangoes
Hence, the median of data is 184 mangoes.

Question 3:

The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.

Average Speed of Vehicles(Km/hr)	60 - 64	64 - 69	70 - 74	75 - 79	79 - 84	84 - 89
No. of vehicles	10	34	55	85	10	6

Answer:

Class (Number of working hours)	Frequency (Number of workers)	Cumulaive frequency less than the upper limit
60 - 64	10	10
64 - 69	34	44
70 - 74 (Median Class)	55	99
75 - 79	85	184
79 - 84	10	194
84 - 89	6	200
	N = 200

From the above table, we get
L (Lower class limit of the median class) = 70
N (Sum of frequencies) = 200
h (Class interval of the median class) = 4
f (Frequency of the median class) = 55
cf (Cumulative frequency of the class preceding the median class) = 44
Now, Median =

L + (\frac{\frac{N}{2} - c f}{f}) \times h

â€‹

= 70 + (\frac{\frac{200}{2} - 44}{55}) \times 4

= 70 + 4.09
= 74.09
= 75 vehicles
Hence, the median of data is 75 vehicles.

Question 4:

The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.

No. of bulbs produced (Thousands)	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
No. of factories	12	35	20	15	8	7	8

Answer:

Class (Number of bulbs produced in thousands)	Frequency (Number of factories) f_i	Cumulaive frequency less than the upper limit
30 - 40	12	12
40 - 50	35	47
50 - 60 (Median Class)	20	67
60 - 70	15	82
70 - 80	8	90
80 - 90	7	97
90 - 100	8	105
	N = 105

From the above table, we get
L (Lower class limit of the median class) = 50
N (Sum of frequencies) = 105
h (Class interval of the median class) = 10
f (Frequency of the median class) = 20
cf (Cumulative frequency of the class preceding the median class) = 47
Now, Median =

L + (\frac{\frac{N}{2} - c f}{f}) \times h

â€‹

= 50 + (\frac{\frac{105}{2} - 47}{20}) \times 10

= 50 + 2.75
= 52.75 thousand lamps
= 52750 lamps
Hence, the median of the productions is 52750 lamps.

Question 1:

The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.

Content of fat (%)	2 - 3	3 - 4	4 - 5	5 - 6	6 - 7
Milk collected (Litre)	30	70	80	60	20

Answer:

The maximum class frequency is 80.
The class corresponding to this frequency is 4 - 5.
So, the modal class is 4 - 5.
L (the lower limit of modal class) = 4
f₁ (frequency of the modal class) = 80
f_o (frequency of the class preceding the modal class) = 70
f₂ (frequency of the class succeeding the modal class) = 60
h (class size) = 1
Mode = $L + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$
$= 4 + (\frac{80 - 70}{2 \times 80 - 70 - 60}) \times 1$
= 4 + 0.33
= 4.33
Hence, the modal fat content is 4.33 litres.

Question 2:

Electricity used by some families is shown in the following table. Find the mode for use of electricity.

Use of electricity (Unit)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
No. of families	13	50	70	100	80	17

Answer:

The maximum class frequency is 100.
The class corresponding to this frequency is 60 - 80.
So, the modal class is 60 - 80.
L (the lower limit of modal class) = 60
f₁ (frequency of the modal class) = 100
f_o (frequency of the class preceding the modal class) = 70
f₂ (frequency of the class succeeding the modal class) = 80
h (class size) = 20
Mode = $L + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$
$= 60 + (\frac{100 - 70}{2 \times 100 - 70 - 80}) \times 20$
= 60 + 12
= 72
Hence, the modal electricity is 72 units.

Question 3:

Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.

Milk (Litre)	1 - 3	3 - 5	5 - 7	7 - 9	9 - 11	11 - 13
No. of hotels	7	5	15	20	35	18

Answer:

The maximum class frequency is 35.
The class corresponding to this frequency is 9 - 11.
So, the modal class is 9 - 11.
L (the lower limit of modal class) = 9
f₁ (frequency of the modal class) = 35
f_o (frequency of the class preceding the modal class) = 20
f₂ (frequency of the class succeeding the modal class) = 18
h (class size) = 2
Mode = $L + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$
$= 9 + (\frac{35 - 20}{2 \times 35 - 20 - 18}) \times 2$
= 9 + 0.94
= 9.94
Hence, the the mode of the supply of milk is 9.94 litres.

Question 4:

Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.

Age (years)	Less than 5	5 - 9	10 - 14	15 - 19	20 - 24	25 - 29
No. of patients	38	32	50	36	24	20

Answer:

Converting the given table into continuous class, we get

Age (years)	0.5 - 4.5	4.5 - 9.5	9.5 - 14.5	14.5 - 19.5	19.5 - 24.5	24.5 - 29.5
No. of patients	38	32	50	36	24	20

The maximum class frequency is 50.
The class corresponding to this frequency is 9.5 - 14.5.
So, the modal class is 9.5 - 14.5.
L (the lower limit of modal class) = 9.5
f₁ (frequency of the modal class) = 50
f_o (frequency of the class preceding the modal class) = 32
f₂ (frequency of the class succeeding the modal class) = 36
h (class size) = 5
Mode =

L + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h

= 9.5 + (\frac{50 - 32}{2 \times 50 - 32 - 36}) \times 5

= 9.5 + 2.81
= 12.31
Hence, the modal age is 12.31 years.

Question 1:

Draw a histogram of the following data.

Height of student (cm)	135 - 140	140 - 145	145 - 150	150 - 155
No. of students	4	12	16	8

Answer:

The histogram for the given data is

Question 2:

The table below shows the yield of jowar per acre. Show the data by histogram.

Yield per acre (quintal)	2 - 3	4 - 5	6 - 7	8 - 9	10 - 11
No. of farmers	30	50	55	40	20

Answer:

The given classes are not continuous. Converting into the continuous classes, we get

Yield per acre (quintal)	Class (continuous)	No. of farmers
2 - 3	1.5 - 3.5	30
4 - 5	3.5 - 5.5	50
6 - 7	5.5 - 7.5	55
8 - 9	7.5 - 9.5	40
10 - 11	9.5 - 11.5	20

The histogram for the above data is given below:

Question 3:

In the following table, the investment made by 210 families is shown. Present it in the form of a histogram.

Investment (Thousand Rupees)	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35
No. of families	30	50	60	55	15

Answer:

The histogram for the given data is

Question 4:

Time alloted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.

Time (minutes)	60 - 80	80 - 100	100 - 120	120 - 140	140 - 160
No. of students	14	20	24	22	16

Answer:

The histogram for the given data is

Question 1:

Observe the following frequency polygon and write the answers of the questions below it.

(1) Which class has the maximum number of students ?
(2) Write the classes having zero frequency.
(3) What is the class-mark of the class, having frequency of 50 students ?
(4) Write the lower and upper class limits of the class whose class mark is 85.
(5) How many students are in the class 80-90?

Answer:

(1) The class 60 - 70 has the maximum number of students i.e., 60
(2) The classes 20 - 30 and 90 - 10 have zero frequency.
(3) The class-mark of the class having frequency of 50 students is 55.
(4) The lower and upper class limits of the class having class mark of 85 are 80 and 90.
(5) Number of students in the class 80-90 are 15.

Question 2:

Show the following data by a frequency polygon.

Electricity bill (Rs)	0 - 200	200 - 400	400 - 600	600 - 800	800 - 1000
Families	240	300	450	350	160

Answer:

Consider the following table

Class (Electricity bill in Rupees)	Class Mark	Frequency (Number of families)
0 - 200	100	240
200 - 400	300	300
400 - 600	500	450
600 - 800	700	350
800 - 1000	900	160

The frequency polygon using the class mark and frequancy given in the above table as

Question 3:

The following table shows the classification of percentages of marks of students and the number of students. Draw a frequency polygon from the table.

Result (Percentage)	30 - 40	40 - 50	50 - 60	60 -70	70 - 80	80 - 90	90 - 100
No. of students	7	33	45	65	47	18	5

Answer:

Consider the following table

Class (Result in %)	Class Mark	Frequency (Number of students)
30 - 40	35	7
40 - 50	45	33
50 - 60	55	45
60 - 70	65	65
70 - 80	75	47
80 - 90	85	18
90 - 100	95	5

The frequency polygon using the class mark and frequancy given in the above table as

Question 1:

The age group and number of persons, who donated blood in a blood donation camp is given below. Draw a pie diagram from it.

Age group(Yrs)	20 - 25	25 - 30	30 - 35	35 -40
No. of persons	80	60	35	25

Answer:

The measures of central angles are given in the table.

Age group(Yrs)	No. of persons	Central Angle
20 - 25	80	$\frac{80}{200} \times 360 = 144 °$
25 - 30	60	$\frac{60}{200} \times 360 = 108 °$
30 - 35	35	$\frac{35}{200} \times 360 = 63 °$
35 - 40	25	$\frac{25}{200} \times 360 = 45 °$
Total	200

The pie digram showing the above data is given below:

Question 2:

The marks obtained by a student in different subjects are shown. Draw a pie diagram showing the information.

Subject	English	Marathi	Science	Mathematics	Social science	Hindi
Marks	50	70	80	90	60	50

Answer:

The measures of central angles are given in the table.

Subjects	Marks	Central Angle
English	50	$\frac{50}{400} \times 360 = 45 °$
Marathi	70	$\frac{70}{400} \times 360 = 63 °$
Science	80	$\frac{80}{400} \times 360 = 72 °$
Mathematics	90	$\frac{90}{400} \times 360 = 81 °$
Social science	60	$\frac{60}{400} \times 360 = 54 °$
Hindi	50	$\frac{50}{400} \times 360 = 45 °$
Total	400

The pie digram showing the above data is given below:

Question 3:

In a tree plantation programme, the number of trees planted by students of different classes is given in the following table. Draw a pie diagram showing the information.

Standard	5 th	6th	7 th	8 th	9 th	10 th
No. of trees	40	50	75	50	70	75

Answer:

The measures of central angles are given in the table.

Age group(Yrs)	No. of persons	Central Angle
5th	40	$\frac{40}{360} \times 360 = 40 °$
6th	50	$\frac{50}{360} \times 360 = 50 °$
7th	75	$\frac{75}{360} \times 360 = 75 °$
8th	50	$\frac{50}{360} \times 360 = 50 °$
9th	70	$\frac{70}{360} \times 360 = 70 °$
10th	75	$\frac{75}{360} \times 360 = 75 °$
Total	360

The pie digram showing the above data is given below:
â€‹

Question 4:

The following table shows the percentages of demands for different fruits registered with a fruit vendor. Show the information by a pie diagram.

Fruits	Mango	Sweet lime	Apples	Cheeku	Oranges
Percentages of demand	30	15	25	20	10

Answer:

The measures of central angles are given in the table.

Fruits	Percentages of demand	Central Angle
Mango	30	$\frac{30}{100} \times 360 = 108 °$
Sweet lime	15	$\frac{15}{100} \times 360 = 54 °$
Apples	25	$\frac{25}{100} \times 360 = 90 °$
Cheeku	20	$\frac{20}{100} \times 360 = 72 °$
Oranges	10	$\frac{10}{100} \times 360 = 36 °$
Total	100

The pie digram showing the above data is given below:
â€‹

Question 5:

The pie diagram in figure shows the proportions of different workers in a town. Answer the following questions with its help.
(1) If the total workers is 10,000; how many of them are in the field of construction ?
(2) How many workers are working in the administration ?
(3) What is the percentage of workers in production ?

Answer:

(1)
Number of workers working in the field of construction $= \frac{central angle for construction}{360 °} \times 10000$
$= \frac{72 °}{360 °} \times 10000$
â€‹= 2000

(2)
Number of workers working in the administration $= \frac{central angle for administration}{360 °} \times 10000$
$= \frac{36 °}{360 °} \times 10000$ â€‹
= 1000
â€‹
(3)
Percentage of workers working in production $= \frac{central angle for production}{360 °} \times 100 %$
$= \frac{90 °}{360 °} \times 100 %$ â€‹
= 25%

Question 6:

The annual investments of a family are shown in the adjacent pie diagram. Answer the following questions based on it.
(1) If the investment in shares is Rs 2000/, find the total investment.
(2) How much amount is deposited in bank ?
(3) How much more money is invested in immovable property than in mutual fund ?
(4) How much amount is invested in post ?

Answer:

(1)
Money invested in shares = Rs 2000
$\Rightarrow \frac{60 °}{360 °} \times Total investment = 2000 \Rightarrow Total investment = \frac{360 °}{60 °} \times 2000 = Rs 12000$

(2)
Money invested in bankâ€‹ $= \frac{90 °}{360 °} \times Total investment$
â€‹ $= \frac{1}{4} \times 12000$
= Rs 3000

(3)
Difference in the central angle between immovable property and mutual fund = 120^º − 60^º
= 60^º
Therefore, required mooney $= \frac{60 °}{360 °} \times Total investment$
â€‹ $= \frac{1}{6} \times 12000$
= Rs 2000
â€‹
(4)
Money invested in postâ€‹ $= \frac{30 °}{360 °} \times Total investment$
$= \frac{1}{12} \times 12000$
= Rs 1000

Question 1:

Find the correct answer from the alternatives given.
(1) The persons of O– blood group are 40%. The classification of persons based on blood groups is to be shown by a pie diagram. What should be the measures of angle for the persons of O– blood group ?

(A) 114°)

(B) 140°

(C) 104°

(D) 144°

(2) Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure Rs 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction ?

(A) 2,16,000

(B) 3,60,000

(C) 4,50,000

(D) 7,50,000

(3) Cumulative frequencies in a grouped frequency table are useful to find . . .

(A) Mean

(B) Median

(C) Mode

(D) All of these

(4) The formula to find mean from a grouped frequency table is

X = A + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h g

In the formula u i = . ..

(A)

\frac{x_{i} + A}{g}

(B)

(x_{i} - A)

(C)

\frac{x_{i} - A}{g}

(D)

\frac{A - x_{i}}{g}

(5)

Distance Covered per litre (km)	12 - 14	14 - 16	16 - 18	18 - 20
No. of cars	11	12	20	7

The median of the distances covered per litre shown in the above data is in the group . . . . . .

(A) 12 - 14

(B) 14 - 16

(C) 16 - 18

(D) 18 - 20

(6)

No. of trees planted by each student	1 - 3	4 - 6	7 - 9	10 - 12
No. of students	7	8	6	4

The above data is to be shown by a frequency polygon. The coordinates of the points to show number of students in the class 4-6 are . . . .

(A) (4, 8)

(B) (3, 5)

(C) (5, 8)

(D) (8, 4)

Answer:

(1)
Central angle for O– blood group persons = $\frac{40}{100} \times 360 ° = 144 °$
Hence, the correct option is (D).
(2)
Central angle for expenditures incurred on the construction = 75°
$\Rightarrow \frac{Expenditures on construction}{Total expenditures} \times 360 ° = 75 ° \Rightarrow Total expenditures = 45000 \times \frac{360 °}{75 °} = Rs 216000$
Hence, the correct option is (A).
(3)
Cumulative frequencies in a grouped frequency table are used to find the median of the data.
Hence, the correct option is (B).
(4)
To find mean of a grouped frequency table using $X = A + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h g$ .
In this formula, $u_{i} = \frac{x_{i} - A}{g}$

Hence, the correct option is (C).
(5)
Since, the class 16 - 18 has the highest frequency. Thus, the median of the distances covered per litre will be lie in this class.
Hence, the correct option is (C).
(6)
The coordinates of the points to show number of students in the class 4-6 are (5, 8).
Hence, the correct option is (C).

Question 2:

The following table shows the income of farmers in a grape season. Find the mean of their income.

Income (Thousand Rupees)	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Farmers	10	11	15	16	18	14

Answer:

Class (Income in thousand rupees)	Class Mark x_i	Frequency (Number of farmers) f_i	Class mark × Frequency x_if_i
20 - 30	25	10	250
30 - 40	35	11	385
40 - 50	45	15	675
50 - 60	55	16	880
60 - 70	65	18	1170
70 - 80	75	14	1050
		$\sum_{} f_{i} = 84$	$\sum_{} x_{i} f_{i} = 4410$

Mean =

\frac{\sum_{} x_{i} f_{i}}{\sum_{} f_{i}}

â€‹

= \frac{4410}{84} = 52.5 thousand rupees = 52500

Hence, the mean of the income is Rs 52500.

Question 3:

The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.

Loan (Thousand Rupees)	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90
No. of farm ponds	13	20	24	36	7

Answer:

Class (Loan in thousand rupees)	Class Mark x_i	Frequency (Number of farm ponds) f_i	Class mark × Frequency x_if_i
40 - 50	45	13	585
50 - 60	55	20	1100
60 - 70	65	24	1560
70 - 80	75	36	2700
80 - 90	85	7	595
		$\sum_{} f_{i} = 100$	$\sum_{} x_{i} f_{i} = 6540$

Mean =

\frac{\sum_{} x_{i} f_{i}}{\sum_{} f_{i}}

â€‹

= \frac{6540}{100} = 65.4 thousand rupees = 65400

Hence, the mean of the loans is Rs 65400.

Question 4:

The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.

Weekly wages (Rupees)	0 - 2000	2000 - 4000	4000 - 6000	6000 - 8000
No. of workers	15	35	50	20

Answer:

Class (Weekly wages in thousand rupees)	Class Mark x_i	Frequency (Number of workers) f_i	Class mark × Frequency x_if_i
0 - 2000	1000	15	15000
2000 - 4000	3000	35	105000
4000 - 6000	5000	50	250000
6000 - 8000	7000	20	140000
		$\sum_{} f_{i} = 120$	$\sum_{} x_{i} f_{i} = 510000$

Mean =

\frac{\sum_{} x_{i} f_{i}}{\sum_{} f_{i}}

â€‹

= \frac{510000}{120}

= Rs 4250
â€‹Hence, the mean of the weekly wages is Rs 4250.

Question 5:

The following frequency distribution table shows the amount of aid given to 50 flood affected families. Find the mean of the amount of aid.

Amount of aid (Thosand rupees)	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
No. of families	7	13	20	6	4

Answer:

Class (Amount of aid in thousand rupees)	Class Mark x_i	Frequency (Number of families) f_i	Class mark × Frequency x_if_i
50 - 60	55	7	385
60 - 70	65	13	845
70 - 80	75	20	1500
80 - 90	85	6	510
90 - 100	95	4	380
		$\sum_{} f_{i} = 50$	$\sum_{} x_{i} f_{i} = 3620$

Mean =

\frac{\sum_{} x_{i} f_{i}}{\sum_{} f_{i}}

â€‹

= \frac{3620}{50} = 72.4 thousand rupees = Rs 72400

Hence, the mean of the amount of acid is Rs 72400.

Question 6:

The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distance.

Distance (km)	200 - 210	210 - 220	220 - 230	230 - 240	240 - 250
No. of buses	40	60	80	50	20

Answer:

Class (Distance in Kms)	Frequency (Number of buses) f_i	Cumulaive frequency less than the upper limit
200 - 210	40	40
210 - 220	60	100
220 - 230 (Median Class)	80	180
230 - 240	50	230
240 - 250	20	250
	N = 250

From the above table, we get
L (Lower class limit of the median class) = 220
N (Sum of frequencies) = 250
h (Class interval of the median class) = 10
f (Frequency of the median class) = 80
cf (Cumulative frequency of the class preceding the median class) = 100
Now, Median =

L + (\frac{\frac{N}{2} - c f}{f}) \times h

â€‹

= 220 + (\frac{\frac{250}{2} - 100}{80}) \times 10

= 220 + 3.13
= 223.13 km
Hence, the median of the distances is 223.13 km.

Question 7:

The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.

Price (Rupees)	20 less than	20 - 40	40 - 60	60 - 80	80 - 100
No. of articles	140	100	80	60	20

Answer:

Class (Prices in Rupees)	Frequency (Number of articles) f_i	Cumulaive frequency less than the upper limit
0 - 20	140	140
20 - 40 (Median Class)	100	240
40 - 60	80	320
60 - 80	60	380
80 - 100	20	400
	N = 400

From the above table, we get
L (Lower class limit of the median class) = 20
N (Sum of frequencies) = 400
h (Class interval of the median class) = 20
f (Frequency of the median class) = 100
cf (Cumulative frequency of the class preceding the median class) = 140
Now, Median =

L + (\frac{\frac{N}{2} - c f}{f}) \times h

â€‹

= 20 + (\frac{\frac{400}{2} - 140}{100}) \times 20

= 20 + 12
= Rs 32
Hence, the median of the prices is Rs 32.

Question 8:

The following frequency table shows the demand for a sweet and the number of customers. Find the mode of demand of sweet.

Weight of sweet (gram)	0 - 250	250 - 500	500 - 750	750 - 1000	1000 - 1250
No. of customers	10	60	25	20	15

Answer:

The maximum class frequency is 60.
The class corresponding to this frequency is 250 - 500.
So, the modal class is 250 - 500.
L (the lower limit of modal class) = 250
f₁ (frequency of the modal class) = 60
f_o (frequency of the class preceding the modal class) = 10
f₂ (frequency of the class succeeding the modal class) = 25
h (class size) = 250
Mode = $L + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$
$= 250 + (\frac{60 - 10}{2 \times 60 - 10 - 25}) \times 250$
= 250 + 147.06
= 397.06
Hence, the modal demand of sweet is 397.06 grams.

Question 9:

Draw a histogram for the following frequency distribution.

Use of electricity (Unit)	50 - 70	70 - 90	90 - 110	110 - 130	130 - 150	150 - 170
No. of families	150	400	460	540	600	350

Answer:

The histogram for the given data is

Question 10:

In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.

No. of days	8 - 10	10 - 12	12 - 14	14 - 16	16 - 18	18 - 20
No. of workers	5	16	30	40	35	14

Answer:

Consider the following table

Class (Number of days)	Class Mark	Frequency (Number of workers)
8 - 10	9	5
10 - 12	11	16
12 - 14	13	30
14 - 16	15	40
16 - 18	17	35
18 - 20	19	14

The frequency polygon using the class mark and frequancy given in the above table as

â€‹

Question 11:

The time required for students to do a science experiment and the number of students is shown in the following grouped frequency distribution table. Show the information by a histogram and also by a frequency polygon.

Time required for experiment (minutes)	20 - 22	22 - 24	24 - 26	26 - 28	28 - 30	30 - 32
No. of students	8	16	22	18	14	12

Answer:

Consider the following table

Class (Time required for experiment in minutes)	Class Mark	Frequency (Number of students)
20 - 22	21	8
22 - 24	23	16
24 - 26	25	22
26 - 28	27	18
28 - 30	29	14
30 - 32	31	12

The frequency polygon using the class mark and frequancy given in the above table as

Question 12:

Draw a frequency polygon for the following grouped frequency distribution table.

Age of the donor (Yrs.)	20 - 24	25 - 29	30 - 34	35 - 39	40 - 44	45 - 49
No. of blood doners	38	46	35	24	15	12

Answer:

Consider the following table

Class (Age of doctors in years)	Continuous Class	Class Mark	Frequency (Number of students)
20 - 24	19.5 - 24.5	22	38
25 - 29	24.5 - 29.5	27	46
30 - 34	29.5 - 34.5	32	35
35 - 39	34.5 - 39.5	37	24
40 - 44	39.5 - 44.5	42	15
45 - 49	44.5 - 49.5	47	12

The frequency polygon using the class mark and frequancy given in the above table as

Question 13:

The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.

Average rainfall (cm)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100
No. of towns	14	12	36	48	40

Answer:

Consider the following table

Class (Average rainfall in cm)	Class Mark	Frequency (Number of towns)
0 - 20	10	14
20 - 40	30	12
40 - 60	50	36
60 - 80	70	48
80 - 100	90	40

The frequency polygon using the class mark and frequancy given in the above table as
â€‹â€‹

Question 14:

Observe the adjacent pie diagram. It shows the percentages of number of vehicles passing a signal in a town between 8 am and 10 am
(1) Find the central angle for each type of vehicle.
(2) If the number of two-wheelers is 1200, find the number of all vehicles.

Answer:

(1)
Central angle for cars = $\frac{30 %}{100 %} \times 360 ° = 108 °$

Central angle for tempos = $\frac{12 %}{100 %} \times 360 ° = 43.2 °$

Central angle for buses = $\frac{8 %}{100 %} \times 360 ° = 28.8 °$

Central angle for auto rikshaw = $\frac{10 %}{100 %} \times 360 ° = 36 °$

Central angle for two whellers = $\frac{40 %}{100 %} \times 360 ° = 144 °$
â€‹
(2)
Number of two wheelers = 1200
$\Rightarrow \frac{40 %}{100 %} \times Total number of vehicles = 1200 \Rightarrow Total number of vehicles = \frac{100 %}{40 %} \times 1200 = 3000$

Question 15:

The following table shows causes of noise pollution. Show it by a pie diagram.

Construction	Traffic	Aircraft take offs	Industry	Trains
10%	50%	9%	20%	11%

Answer:

The measures of central angles are given in the table.

Cause of noise pollution	Percentage	Central Angle
Construction	10%	$\frac{10}{100} \times 360 = 36 °$
Traffic	50%	$\frac{50}{100} \times 360 = 180 °$
Aircraft take offs	9%	$\frac{9}{100} \times 360 = 32.4 °$
Industry	20%	$\frac{20}{100} \times 360 = 72 °$
Trains	11%	$\frac{11}{100} \times 360 = 39.6 °$
Total	100%

The pie digram showing the above data is given below:

Question 16:

A survey of students was made to know which game they like. The data obtained in the survey is presented in the adjacent pie diagram. If the total number of students are 1000,

(1) How many students like cricket ?
(2) How many students like football ?
(3) How many students prefer other games ?

Answer:

(1)
Number of students like cricket = $\frac{Central angle for cricket}{360 °} \times Total number of students$
$= \frac{81 °}{360 °} \times 1000$
= 225
Hence, 225 students like cricket.
(2)
Number of students like football = $\frac{Central angle for football}{360 °} \times Total number of students$
$= \frac{63 °}{360 °} \times 1000$
= 175
Hence, 175 students like football.
(3)
Number of students like other games = $\frac{Central angle for other games}{360 °} \times Total number of students$
$= \frac{72 °}{360 °} \times 1000$
= 200
Hence, 200 students like other games.

Question 17:

Medical check up of 180 women was conducted in a health centre in a village. 50 of them were short of haemoglobin, 10 suffered from cataract and 25 had respiratory disorders. The remaining women were healthy. Show the information by a pie diagram.

Answer:

The measures of central angles are given in the table.

Health condition	Number of women	Central Angle
Short of haemoglobin	50	$\frac{50}{180} \times 360 = 100 °$
Suffered from cataract	10	$\frac{10}{180} \times 360 = 20 °$
Respiratory disorders	25	$\frac{25}{180} \times 360 = 50 °$
Healthy	95	$\frac{95}{180} \times 360 = 190 °$
Total	180

The pie digram showing the above data is given below:
â€‹â€‹â€‹

Question 18:

On an environment day, students in a school planted 120 trees under plantation project. The information regarding the project is shown in the following table. Show it by a pie diagram.

Tree name	Karanj	Behada	Arjun	Bakul	Kadunimb
No. of trees	20	28	24	22	26

Answer:

The measures of central angles are given in the table.

Tree Name	Number of tress	Central Angle
Karanj	20	$\frac{20}{120} \times 360 = 60 °$
Behada	28	$\frac{28}{120} \times 360 = 84 °$
Arjun	24	$\frac{24}{120} \times 360 = 72 °$
Bakul	22	$\frac{22}{120} \times 360 = 66 °$
Kadunimb	26	$\frac{26}{120} \times 360 = 78 °$
Total	180

The pie digram showing the above data is given below:
â€‹â€‹â€‹

Mathematics Part i Solutions Solutions for Class 10 Maths Chapter 6 - Statistics

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