Maharashtra Board Class 10 Math Part – 1 Solution Chapter 6 Problem Set 6 – Statistics
Balbharati Maharashtra Board Class 10 Math Part – 1 Solution Chapter 6: Statistics. Marathi or English Medium Students of Class 10 get here Statistics full Exercise Solution.
Std | Maharashtra Class 10 |
Subject | Math Part 1 Solution |
Chapter | Statistics |
Problem Set | 6 |
Miscellaneous Problems – 6
(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°
Solution:
Given, blood group O – 40% in pie diagram
∴ Central angle = 40/100 × 360
= 144° (D)
(2) Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure ₹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
Solution:
Given, sector angle for expenditure of ₹45000 is 75°
let, total expenditure be x.
∴ Central angle = 45000/x × 360
Or, 75 = 45000 × 360/3000x
Or, x = 45000 × 360/75
= ₹216000 (A)
(3) Cumulative frequencies in a grouped frequency table are useful to find . . .
(A) Mean
(B) Median
(C) Mode
(D) All of these
Solution:
Cumulative frequencies are useful to find the media of the group. (B)
(4) The formula to find mean from a grouped frequency table is X = A + Σfiui/Σfi× hg in the formula ui = …..
(A) xi + A/g
(B) (xi – A)
(C) xi – A/g
(D) A – xi/g
Solution:
The formula for ui = xi – A/g (C)
(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
Solution:
The median is in the group 16 – 18 (C)
(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)
Solution:
Coordinates for class 4-16 is (5, 8) (C)
(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 |
Solution:
Class (income) × 1000 | Class mark xi | Frequency fi | xifi |
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 |
Σfi = 84 | Σfixi = 4410 |
∴ Mean X = Σfixi/Σfi = 4410/84 = 52.5
∴ The mean of their income is ₹52500
(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 |
Solution:
Class Loan ×₹1000 | Class mark xi | Frequency fi | fixi |
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 |
Σfi = 100 | Σfixi = 6540 |
∴ Mean X = Σfixi/Σfi = 6540/100 = 654
∴ The mean of loans —– = ₹65400
(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 |
Solution:
Class weekly wages in ₹ | Class mark xi | Frequency fi | fixi |
0 – 2000 | 1000 | 15 | 15000 |
2000 – 4000 | 3000 | 35 | 105000 |
4000 – 6000 | 5000 | 50 | 250000 |
6000 – 8000 | 7000 | 20 | 140000 |
Σfi = 120 | Σfixi = 510000 |
∴ Mean, X = Σfixi/Σfi = 510000/120 = 4250
∴ Mean of weekly wages = ₹4250
(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 (Thousand rupees) | 50 – 60 | 60 – 70 | 70 – 80 | 80 – 90 | 90 – 100 |
No. of families | 7 | 13 | 20 | 6 | 4 |
Solution:
Class, amount of air ×₹1000 | Class mark xi | Frequency fi | xifi |
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 |
Σfi = 50 | Σfixi = 3620 |
∴ Mean X = Σfixi/Σfi = 3620/50 = 72.4
∴ Mean of amount of aid given = ₹72.4 × 1000
= ₹72400
(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 distances.
Distance (km) | 200 – 210 | 210 – 220 | 220 – 230 | 230 – 240 | 240 – 250 |
No. of buses | 40 | 60 | 80 | 50 | 20 |
Solution:
Given,
Class distance (km) | Frequency (f) No. of buses | Cumulative frequency (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 |
∴ N/2 – 250/2 = 125,
Lower limit of median class L = 220
h = 230 – 220 = 10, cf = 100, f = 80
∴ Median = L + [N/2 – cf/f] × h
= 220 + [125-100/80] × 10
= 220 + 25/1
= 220 + 3.125
= 223.125
∴ The median distance covered = 223.125 km.
(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 |
Solution:
Class, price in ₹ | Frequency (f) No. of articles | Cumulative frequency (cf) |
0 – 20 | 140 | 140 ->cf |
20 – 40 | 100 -> f | 240 |
40 – 60 | 80 | 320 |
60 – 80 | 60 | 380 |
80 – 100 | 20 | 400 = N |
∴ The median class is 20 – 40
∴ L = 20, h = 60 – 40 = 20,
f = 100, cf = 140, N/2 = 400/2 = 200
∴ Median = L + [N/2 – cf/f] × h
= 20 + [200-140/100] × 20
= 20 + 60/5
= 20 + 12
= 32
∴ ₹32 is the median price:
(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 |
Solution:
From the given frequency table,
Maximum number of customer prefers 250 – 500 weigh sweets so this is our model class.
∴ f1 = 60, f0 = 10, f2 = 25
L = 250, h = 500 – 250 = 250
∴ Mode = L + [f1 – f0/2f1 – f0 – f2] × h
= 250 + [60 – 10/2×60 – 10 – 25] × 250
= 250 + 50/85 × 250
= 250 + 147.06
= 397.06
∴ The mode of demand of sweets is 397.06 gm.
(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 |
Solution:
(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 |
Solution:
(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 |
Solution:
The following table represents the histogram.
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 |
The following table represents the frequency polygon.
Class, time required for experiment (in min) | Class mark | Frequency No. of students | Coordinates |
20 – 22 | 21 | 8 | (21, 8) |
22 – 24 | 23 | 16 | (23, 16) |
24 – 26 | 25 | 22 | (25, 28) |
26 – 28 | 27 | 18 | (27, 18) |
28 – 30 | 29 | 14 | (29, 14) |
30 – 32 | 31 | 12 | (31, 12) |
(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 |
Solution:
The given, frequency table is not continuous so we have to make it continuous.
Difference in the consecutive classes = 25 – 24 = 1
∴ In order to make it continuous. We have to subtract and add 1/2 = 0.5 to each lower and upper limits of each respectively.
The following table represents the frequency polygon.
Class Age of donors (In yrs) | Continuous clean | Class Mark | Frequency No. of blood donors | Coordinates |
20 – 24 | 14.5 – 24.5 | 22 | 38 | (22, 38) |
25 – 24 | 24.5 – 29.5 | 27 | 46 | (27, 46) |
30 – 34 | 29.5 – 34.5 | 32 | 35 | (32, 35) |
35 – 39 | 34.5 – 39.5 | 37 | 24 | (37, 24) |
40 – 44 | 39.5 – 44.5 | 42 | 15 | (42, 15) |
45 – 49 | 44.5 – 49.5 | 47 | 12 | (47, 12) |
(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 |
Solution:
The following table represents frequency polygon.
Class avg. rainfall (cm) | Class mark | Frequency No. of towns | Coordinates |
0 – 20 | 10 | 14 | (10, 14) |
20 – 40 | 30 | 12 | (30, 12) |
40 – 60 | 50 | 36 | (50, 36) |
60 – 80 | 70 | 48 | (70, 41) |
80 – 100 | 90 | 40 | (90, 40) |
(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.
Solution:
From pie diagram,
it is given that,
Sector of cars = 30%
Sector of tempos = 12%
Sector of Auto = 8%
Sector of Auto = 10%
Sector of two wheelers = 40%
No. of two wheelers = 12000.
Central angle for class = 30/100 × 360
= 108°
Central angle for tempos = 12/100 × 360
= 43.2°
Central angle for buses = 8/100 × 360
= 28.8°
Central angle for autos = 10/100 × 360
= 36°
Central angle for two wheelers = 40/100 × 360
= 144°
Now, no. of two wheelers is 1200
∴ Let, total no. of vehicles be x
∴ No. of two wheelers = central angle/360° × total vehicle.
Or, 1200 = 144/360 × x
Or, x = 1200 × 360/144 = 3000
∴ Total no. of vehicles use 3000.
(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% |
Solution:
From the given frequency table let us find the central angle in the following table.
Class | Percentage of sector | Central angle measurement |
Construction | 10% | 10/100 × 360 = 36° |
Traffic | 50% | 50/100 × 360 = 180° |
Aircraft Take offs | 9% | 9/100 × 360 = 32.4° |
Industry | 20% | 20/100 × 360 = 72° |
Trains | 11% | 11/100 × 360 = 39.6° |
The above pie diagram represents the adjacent table.
(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?
Solution:
From the pie diagram the following is given,
Central angle for cricket = 81°
Central angle for football = 63°
Central angle for hockey = 54°
Central angle for other = 72°
Central angle for khokho = 45°
Central angle for kabbaddi = 45°
Total no. of students = 1000
No. of cricket students = 81/360 × 1000
= 225 students
No. of football students = 63/360 × 1000
= 175 students
No. of students in other sports = 72/360 × 1000
= 200 students
(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.
Solution:
Total no. of women = 180
No. of women short of haemoglobin = 50
No. of women with cataract = 10
No. of women with respiratory disorder = 25
No. of women healthy = 180 – (50 + 10 + 25)
= 180 – 85
= 95
The following table represents the health of women as pie diagram.
Class | No. of women | Measurement of central angle |
Short of haemoglobin | 50 | 50/180 × 360 = 100° |
Cataract | 10 | 10/180 × 360 = 20° |
Respiratory Disorder | 25 | 25/180 × 360 = 50° |
Healthy | 95 | 95/180 × 360 = 190° |
The above pie diagram represents the adjacent – table.
(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 |
Solution:
From the given, frequency table let as find the central angle in the following table.
Total no. of trees = 120
Tree name | No. of trees | Measurement of central angle |
Karanj | 20 | 20/120 × 360 = 60° |
Behada | 28 | 28/120 × 360 = 84° |
Arjun | 24 | 24/120 × 360 = 72° |
Bakul | 22 | 22/120 × 360 = 66° |
Kadunimb | 26 | 26/120 × 360 = 78° |
The above pie diagram represents the adjacent table.
Here is your solution of Maharashtra Board Class 10 Math Part – 1 Chapter 6 Statistics Problem Set 6
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