Solutions to Practice Sets 6.1 to 6.6 and Miscellaneous Problems 6, with worked tables for mean, median and mode, and drawn histograms, frequency polygons and pie diagrams. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 1 (Algebra).
1The 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 0-2, 2-4, 4-6, 6-8, 8-10 hrs; students 7, 18, 12, 10, 3)
Time (hrs)
Class mark xi
Frequency fi
xi fi
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
Total
N = 50
Σxifi = 218
X = ΣxifiΣfi = 21850 = 4.36
Mean time spent on studies = 4.36 hours
2The toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by ‘assumed mean’ method. (Toll ₹ 300-400, …, 700-800; vehicles 80, 110, 120, 70, 40)
Toll (₹)
Class mark xi
di = xi − 550
fi
fidi
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
Total
N = 420
Σfidi = −12000
Assumed mean A = 550
d = ΣfidiΣfi = −12000420 = −28.5714
X = A + d = 550 + (−28.5714) = 521.43
Mean toll = ₹ 521.43
3A milk centre sold milk to 50 customers. Find the mean of the milk sold by direct method. (Milk 1-2, …, 5-6 litre; customers 17, 13, 10, 7, 3)
Milk sold (litre)
Class mark xi
Frequency fi
xi fi
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
Total
N = 50
Σxifi = 141
X = ΣxifiΣfi = 14150 = 2.82
Mean milk sold = 2.82 litre
4A frequency distribution table for the production of oranges of some farm owners is given. Find the mean production of oranges by ‘assumed mean’ method. (Production 25-30, …, 45-50 thousand rupees; farm owners 20, 25, 15, 10, 10)
Production (thousand ₹)
Class mark xi
di = xi − 37.5
fi
fidi
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
Total
N = 80
Σfidi = −175
Assumed mean A = 37.5
d = ΣfidiΣfi = −17580 = −2.1875
X = A + d = 37.5 + (−2.1875) = 35.31
Mean production = ₹ 35.31 thousand
5Funds collected by 120 workers for drought affected people are given. Find the mean of the funds by ‘step deviation’ method. (Fund ₹ 0-500, …, 2000-2500; workers 35, 28, 32, 15, 10)
Fund (₹)
xi
di = xi − 1250
ui = di500
fi
fiui
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
Total
N = 120
Σfiui = −63
A = 1250, g = 500
u = ΣfiuiΣfi = −63120 = −0.525
X = A + u × g = 1250 + (−0.525) × 500 = 987.5
Mean fund = ₹ 987.5
6Weekly wages of 150 workers of a company are given. Find the mean of the weekly wages by ‘step deviation’ method. (Wages ₹ 1000-2000, …, 4000-5000; workers 25, 45, 50, 30)
Weekly wages (₹)
xi
di = xi − 3500
ui = di1000
fi
fiui
1000 − 2000
1500
−2000
−2
25
−50
2000 − 3000
2500
−1000
−1
45
−45
3000 − 4000
3500 = A
0
0
50
0
4000 − 5000
4500
1000
1
30
30
Total
N = 150
Σfiui = −65
A = 3500, g = 1000
u = ΣfiuiΣfi = −65150 = −0.4333
X = A + u × g = 3500 + (−0.4333) × 1000 = 3066.67
Mean weekly wage = ₹ 3066.67
The mean is ₹ 3066.67 (to two decimals). Any assumed mean gives the same answer.
Practice Set 6.2
1The 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. (Hours 8-10, 10-12, 12-14, 14-16; workers 150, 500, 300, 50)
Daily no. of hours
Frequency fi
Cumulative frequency (less than) cf
8 − 10
150
150 → cf
10 − 12
500 → f
650
12 − 14
300
950
14 − 16
50
1000
N = 1000 ∴ N2 = 500
The cumulative frequency just greater than or equal to 500 is 650, so the median class is 10 − 12.
L = 10, cf = 150, f = 500, h = 2
Median = L + [N2 − cff] × h
= 10 + [500 − 150500] × 2
= 10 + 350 × 2500 = 10 + 1.4
= 11.4
Median = 11.4 hours
2The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data. (Mangoes 50-100, …, 250-300; trees 33, 30, 90, 80, 17)
No. of mangoes
Frequency fi
Cumulative frequency (less than) cf
50 − 100
33
33
100 − 150
30
63 → cf
150 − 200
90 → f
153
200 − 250
80
233
250 − 300
17
250
N = 250 ∴ N2 = 125
The cumulative frequency just greater than or equal to 125 is 153, so the median class is 150 − 200.
L = 150, cf = 63, f = 90, h = 50
Median = L + [N2 − cff] × h
= 150 + [125 − 6390] × 50
= 150 + 62 × 5090 = 150 + 34.4444
= 184.44
Median = 184.44 mangoes (approximately 184 mangoes)
3The table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data. (Speed 60-64, 64-69, 70-74, 75-79, 79-84, 84-89 km/hr; vehicles 10, 34, 55, 85, 10, 6)
The classes are printed as 60-64, 64-69, 70-74, 75-79, 79-84, 84-89, which overlap. Following the pattern, they are read as 60-64, 65-69, 70-74, 75-79, 80-84, 85-89 and made continuous by subtracting and adding 0.5.
Speed (km/hr)
Continuous class
Frequency fi
Cumulative frequency (less than) cf
60 − 64
59.5 − 64.5
10
10
65 − 69
64.5 − 69.5
34
44
70 − 74
69.5 − 74.5
55
99 → cf
75 − 79
74.5 − 79.5
85 → f
184
80 − 84
79.5 − 84.5
10
194
85 − 89
84.5 − 89.5
6
200
N = 200 ∴ N2 = 100
The cumulative frequency just greater than or equal to 100 is 184, so the median class is 74.5 − 79.5.
L = 74.5, cf = 99, f = 85, h = 5
Median = L + [N2 − cff] × h
= 74.5 + [100 − 9985] × 5
= 74.5 + 1 × 585 = 74.5 + 0.0588
= 74.56
Median speed = 74.56 km/hr
4The production of electric bulbs in different factories is shown. Find the median of the productions. (Bulbs 30-40, …, 90-100 thousand; factories 12, 35, 20, 15, 8, 7, 8)
Bulbs (thousands)
Frequency fi
Cumulative frequency (less than) cf
30 − 40
12
12
40 − 50
35
47 → cf
50 − 60
20 → f
67
60 − 70
15
82
70 − 80
8
90
80 − 90
7
97
90 − 100
8
105
N = 105 ∴ N2 = 52.5
The cumulative frequency just greater than or equal to 52.5 is 67, so the median class is 50 − 60.
L = 50, cf = 47, f = 20, h = 10
Median = L + [N2 − cff] × h
= 50 + [52.5 − 4720] × 10
= 50 + 5.5 × 1020 = 50 + 2.75
= 52.75
Median production = 52.75 thousand bulbs
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Practice Set 6.3
1The table shows the milk collected from farmers and the content of fat in the milk. Find the mode of fat content. (Fat 2-3, …, 6-7 %; milk 30, 70, 80, 60, 20 litre)
Fat content (%)
Frequency
2 − 3
30
3 − 4
70 → f0
4 − 5
80 → f1
5 − 6
60 → f2
6 − 7
20
The maximum frequency is 80, so the modal class is 4 − 5.
L = 4, h = 1, f1 = 80, f0 = 70, f2 = 60
Mode = L + [f1 − f02f1 − f0 − f2] × h
= 4 + [80 − 702(80) − 70 − 60] × 1
= 4 + 10 × 130 = 4 + 0.3333
= 4.33
Mode of fat content = 4.33 %
2Electricity used by some families is shown. Find the mode for use of electricity. (Units 0-20, …, 100-120; families 13, 50, 70, 100, 80, 17)
Use of electricity (unit)
Frequency
0 − 20
13
20 − 40
50
40 − 60
70 → f0
60 − 80
100 → f1
80 − 100
80 → f2
100 − 120
17
The maximum frequency is 100, so the modal class is 60 − 80.
L = 60, h = 20, f1 = 100, f0 = 70, f2 = 80
Mode = L + [f1 − f02f1 − f0 − f2] × h
= 60 + [100 − 702(100) − 70 − 80] × 20
= 60 + 30 × 2050 = 60 + 12
= 72
Mode = 72 units
3Grouped frequency distribution of supply of milk to hotels is given. Find the mode of the supply of milk. (Milk 1-3, …, 11-13 litre; hotels 7, 5, 15, 20, 35, 18)
Milk (litre)
Frequency
1 − 3
7
3 − 5
5
5 − 7
15
7 − 9
20 → f0
9 − 11
35 → f1
11 − 13
18 → f2
The maximum frequency is 35, so the modal class is 9 − 11.
L = 9, h = 2, f1 = 35, f0 = 20, f2 = 18
Mode = L + [f1 − f02f1 − f0 − f2] × h
= 9 + [35 − 202(35) − 20 − 18] × 2
= 9 + 15 × 232 = 9 + 0.9375
= 9.94
Mode of supply of milk = 9.94 litre
4The frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients. (Age: less than 5, 5-9, 10-14, 15-19, 20-24, 25-29; patients 38, 32, 50, 36, 24, 20)
Age (years)
Continuous class
Frequency
Less than 5
0 − 4.5
38
5 − 9
4.5 − 9.5
32 → f0
10 − 14
9.5 − 14.5
50 → f1
15 − 19
14.5 − 19.5
36 → f2
20 − 24
19.5 − 24.5
24
25 − 29
24.5 − 29.5
20
The maximum frequency is 50, so the modal class is 9.5 − 14.5.
L = 9.5, h = 5, f1 = 50, f0 = 32, f2 = 36
Mode = L + [f1 − f02f1 − f0 − f2] × h
= 9.5 + [50 − 322(50) − 32 − 36] × 5
= 9.5 + 18 × 532 = 9.5 + 2.8125
= 12.31
Mode of ages = 12.31 years
The classes are not continuous, so 0.5 is subtracted from each lower limit and added to each upper limit. Only the modal class 9.5-14.5 and its neighbours are used in the formula.
Practice Set 6.4
1Draw a histogram of the following data. (Height 135-140, 140-145, 145-150, 150-155 cm; students 4, 12, 16, 8)
Histogram. Scale — X-axis: 1 cm = 5 cm; Y-axis: 1 cm = 4 students
2The table shows the yield of jowar per acre. Show the data by histogram. (Yield 2-3, 4-5, 6-7, 8-9, 10-11 quintal; farmers 30, 50, 55, 40, 20)
The classes are not continuous, so they are made continuous:
1.5-3.5, 3.5-5.5, 5.5-7.5, 7.5-9.5, 9.5-11.5
Histogram. Scale — X-axis: 1 cm = 2 quintal; Y-axis: 1 cm = 10 farmers
3The investment made by 210 families is shown. Present it in the form of a histogram. (Investment 10-15, …, 30-35 thousand ₹; families 30, 50, 60, 55, 15)
Histogram. Scale — X-axis: 1 cm = ₹ 5 thousand; Y-axis: 1 cm = 10 families
4Time allotted for the preparation of an examination by some students is shown. Draw a histogram. (Time 60-80, …, 140-160 minutes; students 14, 20, 24, 22, 16)
Histogram. Scale — X-axis: 1 cm = 20 minutes; Y-axis: 1 cm = 4 students
Practice Set 6.5
1Observe the frequency polygon (points (25, 0), (35, 20), (45, 40), (55, 50), (65, 60), (75, 35), (85, 15), (95, 0); X-axis: marks, Y-axis: students) and answer the questions.
Frequency polygon of fig 6.6 (redrawn)
(1) Which class has the maximum number of students?
The highest point is (65, 60). Class mark 65 belongs to the class 60-70.
Class 60-70 (60 students)
(2) Write the classes having zero frequency.
The points (25, 0) and (95, 0) have frequency 0.
Classes 20-30 and 90-100
(3) What is the class mark of the class having frequency of 50 students?
The point with frequency 50 is (55, 50).
Class mark 55
(4) Write the lower and upper class limits of the class whose class mark is 85.
Class mark 85 is the mid-point of 80-90.
Lower limit 80, upper limit 90
(5) How many students are in the class 80-90?
Class mark of 80-90 is 85; the point is (85, 15).
15 students
2Show the following data by a frequency polygon. (Electricity bill ₹ 0-200, …, 800-1000; families 240, 300, 450, 350, 160)
Class
Class mark
Frequency
Coordinates
0-200
100
240
(100, 240)
200-400
300
300
(300, 300)
400-600
500
450
(500, 450)
600-800
700
350
(700, 350)
800-1000
900
160
(900, 160)
Frequency polygon
To close the polygon, a class of zero frequency is assumed before the first class and after the last class. Their class marks are −100 and 1100, so the end points are (−100, 0) and (1100, 0). These are only for closing the figure.
3The table shows the classification of percentages of marks of students. Draw a frequency polygon. (Percentage 30-40, …, 90-100; students 7, 33, 45, 65, 47, 18, 5)
Class
Class mark
Frequency
Coordinates
30-40
35
7
(35, 7)
40-50
45
33
(45, 33)
50-60
55
45
(55, 45)
60-70
65
65
(65, 65)
70-80
75
47
(75, 47)
80-90
85
18
(85, 18)
90-100
95
5
(95, 5)
Frequency polygon; end points (25, 0) and (105, 0)
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Practice Set 6.6
1The age group and number of persons who donated blood in a camp is given. Draw a pie diagram. (20-25: 80, 25-30: 60, 30-35: 35, 35-40: 25)
Component
No. of persons
Measure of central angle
20-25 yrs
80
80200 × 360° = 144°
25-30 yrs
60
60200 × 360° = 108°
30-35 yrs
35
35200 × 360° = 63°
35-40 yrs
25
25200 × 360° = 45°
Total
200
360°
Pie diagram (sectors drawn clockwise from the top)
2The marks obtained by a student in different subjects are shown. Draw a pie diagram. (English 50, Marathi 70, Science 80, Mathematics 90, Social science 60, Hindi 50)
Component
Marks
Measure of central angle
English
50
50400 × 360° = 45°
Marathi
70
70400 × 360° = 63°
Science
80
80400 × 360° = 72°
Mathematics
90
90400 × 360° = 81°
Social science
60
60400 × 360° = 54°
Hindi
50
50400 × 360° = 45°
Total
400
360°
Pie diagram (sectors drawn clockwise from the top)
3In a tree plantation programme, the number of trees planted by students of different classes is given. Draw a pie diagram. (5th 40, 6th 50, 7th 75, 8th 50, 9th 70, 10th 75)
Component
No. of trees
Measure of central angle
5th
40
40360 × 360° = 40°
6th
50
50360 × 360° = 50°
7th
75
75360 × 360° = 75°
8th
50
50360 × 360° = 50°
9th
70
70360 × 360° = 70°
10th
75
75360 × 360° = 75°
Total
360
360°
Pie diagram (sectors drawn clockwise from the top)
The total is 360 trees, so the central angle of each sector equals the number of trees in degrees.
4The percentages of demands for different fruits with a fruit vendor are given. Show the information by a pie diagram. (Mango 30, Sweet lime 15, Apples 25, Cheeku 20, Oranges 10)
Component
Percentage of demand
Measure of central angle
Mango
30
30100 × 360° = 108°
Sweet lime
15
15100 × 360° = 54°
Apples
25
25100 × 360° = 90°
Cheeku
20
20100 × 360° = 72°
Oranges
10
10100 × 360° = 36°
Total
100
360°
Pie diagram (sectors drawn clockwise from the top)
5The pie diagram (fig 6.13) shows the proportions of different workers in a town: Production 90°, Hotels 72°, Construction 72°, Food 48°, Agriculture 24°, Electricity 18°, Administration 36°. Answer the questions.
(1) If the total workers is 10,000, how many are in the field of construction?
Workers in construction = 72360 × 10,000 = 2000
2000 workers
(2) How many workers are working in the administration?
Workers in administration = 36360 × 10,000 = 1000
1000 workers
(3) What is the percentage of workers in production?
Percentage = 90360 × 100 = 25%
25%
6The annual investments of a family are shown in the pie diagram (fig 6.14): Immovable property 120°, Shares 60°, Post 30°, Deposit in bank 90°, Mutual fund 60°. Answer the questions.
(1) If the investment in shares is ₹ 2000, find the total investment.
60° represents ₹ 2000.
Total investment = 2000 × 36060 = ₹ 12,000
₹ 12,000
(2) How much amount is deposited in bank?
90360 × 12,000 = ₹ 3000
₹ 3000
(3) How much more money is invested in immovable property than in mutual fund?
Immovable property = 120360 × 12,000 = ₹ 4000
Mutual fund = 60360 × 12,000 = ₹ 2000
Difference = 4000 − 2000 = ₹ 2000
₹ 2000 more
(4) How much amount is invested in post?
30360 × 12,000 = ₹ 1000
₹ 1000
Miscellaneous Problems 6
1Find the correct answer from the alternatives given.
(1) The persons of O-blood group are 40%. What should be the measure of angle for them in a pie diagram? (A) 114° (B) 140° (C) 104° (D) 144°
40100 × 360° = 144°
(D) 144°
(2) The expenditure ₹ 45,000 on cement was shown by a sector of central angle 75°. What was the total expenditure? (A) 2,16,000 (B) 3,60,000 (C) 4,50,000 (D) 7,50,000
Total = 45,000 × 36075 = ₹ 2,16,000
(A) 2,16,000
(3) Cumulative frequencies in a grouped frequency table are useful to find … (A) Mean (B) Median (C) Mode (D) All of these
The median formula uses cf, the cumulative frequency of the class before the median class.
(B) Median
(4) In the formula X = A + (ΣfiuiΣfi) × g, ui = … (A) (xi + A)/g (B) (xi − A) (C) (xi − A)/g (D) (A − xi)/g
ui = dig = xi − Ag
(C) (xᵢ − A)/g
(5) Distance covered per litre (km) 12-14, 14-16, 16-18, 18-20; cars 11, 12, 20, 7. The median is in the group … (A) 12-14 (B) 14-16 (C) 16-18 (D) 18-20
N = 50, N/2 = 25
Cumulative frequencies: 11, 23, 43, 50
The 25th value lies in the class with cf 43.
(C) 16-18
(6) No. of trees planted 1-3, 4-6, 7-9, 10-12; students 7, 8, 6, 4. In a frequency polygon the coordinates of the point for the class 4-6 are … (A) (4, 8) (B) (3, 5) (C) (5, 8) (D) (8, 4)
Class mark of 4-6 = (4 + 6) ÷ 2 = 5; frequency = 8
(C) (5, 8)
2The table shows the income of farmers in a grape season. Find the mean of their income. (Income 20-30, …, 70-80 thousand ₹; farmers 10, 11, 15, 16, 18, 14)
Income (thousand ₹)
Class mark xi
Frequency fi
xi fi
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
Total
N = 84
Σxifi = 4410
X = ΣxifiΣfi = 441084 = 52.5
Mean income = ₹ 52.5 thousand
3The loans sanctioned by a bank for construction of farm ponds are shown. Find the mean of the loans. (Loan 40-50, …, 80-90 thousand ₹; farm ponds 13, 20, 24, 36, 7)
Loan (thousand ₹)
Class mark xi
Frequency fi
xi fi
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
Total
N = 100
Σxifi = 6540
X = ΣxifiΣfi = 6540100 = 65.4
Mean loan = ₹ 65.4 thousand
4The weekly wages of 120 workers in a factory are shown. Find the mean of the weekly wages. (Wages ₹ 0-2000, 2000-4000, 4000-6000, 6000-8000; workers 15, 35, 50, 20)
Weekly wages (₹)
xi
di = xi − 5000
ui = di2000
fi
fiui
0 − 2000
1000
−4000
−2
15
−30
2000 − 4000
3000
−2000
−1
35
−35
4000 − 6000
5000 = A
0
0
50
0
6000 − 8000
7000
2000
1
20
20
Total
N = 120
Σfiui = −45
A = 5000, g = 2000
u = ΣfiuiΣfi = −45120 = −0.375
X = A + u × g = 5000 + (−0.375) × 2000 = 4250
Mean weekly wage = ₹ 4250
5The amount of aid given to 50 flood affected families is shown. Find the mean of the amount of aid. (Aid 50-60, …, 90-100 thousand ₹; families 7, 13, 20, 6, 4)
Aid (thousand ₹)
Class mark xi
Frequency fi
xi fi
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
Total
N = 50
Σxifi = 3620
X = ΣxifiΣfi = 362050 = 72.4
Mean aid = ₹ 72.4 thousand
6The distances covered by 250 public transport buses in a day are shown. Find the median of the distances. (Distance 200-210, …, 240-250 km; buses 40, 60, 80, 50, 20)
Distance (km)
Frequency fi
Cumulative frequency (less than) cf
200 − 210
40
40
210 − 220
60
100 → cf
220 − 230
80 → f
180
230 − 240
50
230
240 − 250
20
250
N = 250 ∴ N2 = 125
The cumulative frequency just greater than or equal to 125 is 180, so the median class is 220 − 230.
L = 220, cf = 100, f = 80, h = 10
Median = L + [N2 − cff] × h
= 220 + [125 − 10080] × 10
= 220 + 25 × 1080 = 220 + 3.125
= 223.13
Median distance = 223.13 km
7The prices of different articles and demand for them are shown. Find the median of the prices. (Price: less than 20, 20-40, 40-60, 60-80, 80-100 ₹; articles 140, 100, 80, 60, 20)
Price (₹)
Continuous class
Frequency fi
Cumulative frequency (less than) cf
Less than 20
0 − 20
140
140 → cf
20 − 40
20 − 40
100 → f
240
40 − 60
40 − 60
80
320
60 − 80
60 − 80
60
380
80 − 100
80 − 100
20
400
N = 400 ∴ N2 = 200
The cumulative frequency just greater than or equal to 200 is 240, so the median class is 20 − 40.
L = 20, cf = 140, f = 100, h = 20
Median = L + [N2 − cff] × h
= 20 + [200 − 140100] × 20
= 20 + 60 × 20100 = 20 + 12
= 32
Median price = ₹ 32
The class ‘less than 20’ is taken as 0-20.
8The demand for a sweet and the number of customers is shown. Find the mode of demand of sweet. (Weight 0-250, …, 1000-1250 gram; customers 10, 60, 25, 20, 15)
Weight of sweet (gram)
Frequency
0 − 250
10 → f0
250 − 500
60 → f1
500 − 750
25 → f2
750 − 1000
20
1000 − 1250
15
The maximum frequency is 60, so the modal class is 250 − 500.
L = 250, h = 250, f1 = 60, f0 = 10, f2 = 25
Mode = L + [f1 − f02f1 − f0 − f2] × h
= 250 + [60 − 102(60) − 10 − 25] × 250
= 250 + 50 × 25085 = 250 + 147.059
= 397.06
Mode of demand = 397.06 gram
9Draw a histogram for the following frequency distribution. (Electricity 50-70, …, 150-170 units; families 150, 400, 460, 540, 600, 350)
Histogram. Scale — X-axis: 1 cm = 20 units; Y-axis: 1 cm = 100 families
10In a handloom factory the number of workers and the days required to weave a saree are given. Present the information by a frequency polygon. (Days 8-10, …, 18-20; workers 5, 16, 30, 40, 35, 14)
Frequency polygon; end points (7, 0) and (21, 0)
11The time required for students to do a science experiment is shown. Show the information by a histogram and also by a frequency polygon. (Time 20-22, …, 30-32 minutes; students 8, 16, 22, 18, 14, 12)
Histogram
Frequency polygon drawn by joining the mid-points of the tops of the rectangles; end points (19, 0) and (33, 0)
12Draw a frequency polygon for the following grouped frequency distribution. (Age of donor 20-24, 25-29, …, 45-49 yrs; donors 38, 46, 35, 24, 15, 12)
Class marks: 22, 27, 32, 37, 42, 47 (the class marks do not change when the classes are made continuous).
Frequency polygon; end points (17, 0) and (52, 0)
13The average rainfall in 150 towns is shown. Show the information by a frequency polygon. (Rainfall 0-20, …, 80-100 cm; towns 14, 12, 36, 48, 40)
Frequency polygon; end points (−10, 0) and (110, 0)
The end point (−10, 0) comes from the imaginary class −20 to 0 that is used only to close the polygon.
14The pie diagram (fig 6.15) shows the percentages of vehicles passing a signal: Two wheelers 40%, Cars 30%, Tempos 12%, Autorickshaws 10%, Buses 8%. (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.
(1) Central angles
Vehicle
Percentage
Central angle
Two wheelers
40
40100 × 360° = 144°
Cars
30
108°
Tempos
12
43.2°
Autorickshaws
10
36°
Buses
8
28.8°
Total
100
360°
(2) Number of all vehicles
40% of all vehicles = 1200
All vehicles = 1200 × 10040 = 3000
3000 vehicles
15The following table shows causes of noise pollution. Show it by a pie diagram. (Construction 10%, Traffic 50%, Aircraft take offs 9%, Industry 20%, Trains 11%)
Component
Percentage
Measure of central angle
Construction
10
10100 × 360° = 36°
Traffic
50
50100 × 360° = 180°
Aircraft take offs
9
9100 × 360° = 32.4°
Industry
20
20100 × 360° = 72°
Trains
11
11100 × 360° = 39.6°
Total
100
360°
Pie diagram (sectors drawn clockwise from the top)
16The pie diagram (fig 6.16) shows the games students like: Cricket 81°, Football 63°, Hockey 54°, Other 72°, Kho-kho 45°, Kabaddi 45°. If the total number of students is 1000, find how many students like (1) cricket (2) football (3) other games.
(1) Cricket
81360 × 1000 = 225
225 students
(2) Football
63360 × 1000 = 175
175 students
(3) Other games
72360 × 1000 = 200
200 students
17Medical check up of 180 women was conducted in a health centre. 50 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.
Healthy women = 180 − (50 + 10 + 25) = 95
Component
No. of women
Measure of central angle
Short of haemoglobin
50
50180 × 360° = 100°
Cataract
10
10180 × 360° = 20°
Respiratory disorders
25
25180 × 360° = 50°
Healthy
95
95180 × 360° = 190°
Total
180
360°
Pie diagram (sectors drawn clockwise from the top)
18On an environment day, students planted 120 trees. Show the information by a pie diagram. (Karanj 20, Behada 28, Arjun 24, Bakul 22, Kadunimb 26)
Component
No. of trees
Measure of central angle
Karanj
20
20120 × 360° = 60°
Behada
28
28120 × 360° = 84°
Arjun
24
24120 × 360° = 72°
Bakul
22
22120 × 360° = 66°
Kadunimb
26
26120 × 360° = 78°
Total
120
360°
Pie diagram (sectors drawn clockwise from the top)
