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Page No 138:

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
xi
Frequency
(Number of students)
fi
Class mark × Frequency
xifi
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
    fi=50 xifi=218

Mean = xififi
=21850
= 4.36 hours
Hence, the mean time spent by students for their studies is 4.36 hours.

Page No 138:

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
xi
di = x− A Frequency
(Number of vehicles
fi
Frequency × deviation
fi × di
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
      fi=420 fidi=-12000

Required Mean = A+fidifi
=550-12000420
= 550 − 28.57
​= Rs 521.43  

Hence, the mean of toll is Rs 521.43.

Page No 138:

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
xi
Frequency
(Number of customers)
fi
Class mark × Frequency
xifi
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
    fi=50 xifi=141

Mean = xififi
=14150
= 2.82 litres
Hence, the mean of the milk sold is 2.82 litres.

Page No 138:

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:

Class
(Production in
Thousand rupees)
Class Mark
xi
di = x− A Frequency
(Number of farm owners)
fi
Frequency × deviation
fi × di
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
      fi=80 fidi=-175

Required Mean = A+fidifi
37.5-17580
= 37.5 − 2.19
​= 35.31 thousand rupees
= Rs 35310

Hence, the mean production of oranges is Rs 35310.

Page No 138:

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:

Class
(Production in
Thousand rupees)
Class Mark
xi
di = x− A ui=dih Frequency
(Number of farm owners)
fi
Frequency × deviation
fi × ui
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
        fi=120 fiui=-63

Required Mean = A+hfiuifi
=1250-63120500
= 1250 − 262.5
= Rs 987.5
Hence, the mean of the funds is Rs 987.5.

Page No 138:

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:

Class
(Weekely wages rupees)
Class Mark
xi
di = x− A ui=dih Frequency
(Number of workers)
fi
Frequency × deviation
fi × ui
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
        fi=150 fiui=-65

Required Mean = A+hfiuifi
=3500-651501000
= 3500 − 433.33
= Rs 3066.67
Hence, the mean of the weekly wages is Rs 3066.67.



Page No 145:

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)
fi
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
(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+N2-cff×h
=10+10002-150500×2=10+1.4=11.4 hours
Hence, the median of the number of hours they work is 11.4 hours.

Page No 145:

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)
fi
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
(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+N2-cff×h
=150+2502-6390×50
= 150 + 34.44
= 184.44 mangoes
= 184 mangoes
Hence, the median of data is 184 mangoes.

Page No 145:

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
(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+N2-cff×h
=70+2002-4455×4
= 70 + 4.09
= 74.09
=  75 vehicles
Hence, the median of data is 75 vehicles.



Page No 146:

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)
fi
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
(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+N2-cff×h
=50+1052-4720×10
= 50 + 2.75
= 52.75 thousand lamps
= 52750 lamps 
Hence, the median of the productions is 52750 lamps.



Page No 149:

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.
(the lower limit of modal class) =  4
f1 (frequency of the modal class) = 80 
fo (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+f1-f02f1-f0-f2×h
=4+80-702×80-70-60×1
= 4 + 0.33
= 4.33
Hence, the modal fat content is 4.33 litres.

Page No 149:

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.
(the lower limit of modal class) =  60
f1 (frequency of the modal class) = 100
fo (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+f1-f02f1-f0-f2×h
=60+100-702×100-70-80×20
= 60 + 12
= 72
Hence, the modal electricity is 72 units.

Page No 149:

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.
(the lower limit of modal class) =  9
f1 (frequency of the modal class) = 35 
fo (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+f1-f02f1-f0-f2×h
=9+35-202×35-20-18×2
= 9 + 0.94
= 9.94
Hence, the the mode of the supply of milk is 9.94 litres.

Page No 149:

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.
(the lower limit of modal class) =  9.5
f1 (frequency of the modal class) = 50 
fo (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+f1-f02f1-f0-f2×h
=9.5+50-322×50-32-36×5
= 9.5 + 2.81
= 12.31
Hence, the modal age is 12.31 years.



Page No 153:

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

Page No 153:

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:

 

Page No 153:

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

Page No 153:

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



Page No 157:

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.

Page No 157:

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

Page No 157:

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



Page No 163:

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 80200×360=144°
25 - 30 60 60200×360=108°
30 - 35 35 35200×360=63°
35 - 40 25 25200×360=45°
Total 200  

The pie digram showing the above data is given below:

Page No 163:

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 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  

The pie digram showing the above data is given below:
 



Page No 164:

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 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  

The pie digram showing the above data is given below:

Page No 164:

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 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  

The pie digram showing the above data is given below:
​ 

Page No 164:

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=central angle for construction360°×10000
=72°360°×10000
​= 2000

(2)
Number of workers working in the administration=central angle for administration360°×10000
=36°360°×10000
= 1000

(3)
Percentage of workers working in production=central angle for production360°×100%
=90°360°×100%
= 25%

Page No 164:

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
60°360°×Total investment=2000Total investment=360°60°×2000=Rs 12000

(2)
Money invested in bank​=90°360°×Total investment
=14×12000
= Rs 3000

(3)
Difference in the central angle between immovable property and mutual fund = 120º − 60º
= 60º
Therefore, required mooney=60°360°×Total investment
=16×12000
= Rs 2000

(4)
Money invested in post​=30°360°×Total investment
=112×12000
= Rs 1000

Page No 164:

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+fiuifi×hg In the formula u i = . ..
(A) xi+Ag (B) xi-A (C) xi-Ag (D) A-xig

(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 = 40100×360°=144°
Hence, the correct option is (D).

(2)
Central angle for expenditures incurred on the construction = 75°
 Expenditures on constructionTotal expenditures×360°=75°Total expenditures=45000×360°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+fiuifi×hg .
In this formula, 
ui=xi-Ag

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).


 



Page No 165:

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
xi
Frequency
(Number of farmers)
fi
Class mark × Frequency
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
    fi=84 xifi=4410

Mean = xififi
=441084=52.5 thousand rupees=52500
Hence, the mean of the income is Rs 52500.

Page No 165:

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
xi
Frequency
(Number of farm ponds)
fi
Class mark × Frequency
xifi
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
    fi=100 xifi=6540

Mean = xififi
=6540100=65.4 thousand rupees=65400
Hence, the mean of the loans is Rs 65400.



Page No 166:

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
xi
Frequency
(Number of workers)
fi
Class mark × Frequency
xifi
0 - 2000 1000 15 15000
2000 - 4000 3000 35 105000
4000 - 6000 5000 50 250000
6000 - 8000 7000 20 140000
    fi=120 xifi=510000

Mean = xififi
=510000120
= Rs 4250
​Hence, the mean of the weekly wages is Rs 4250.

Page No 166:

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
xi
Frequency
(Number of families)
fi
Class mark × Frequency
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

 
  fi=50 xifi=3620

Mean = xififi
=362050=72.4 thousand rupees=Rs 72400
Hence, the mean of the amount of acid is Rs 72400.

Page No 166:

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)
fi
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
(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+N2-cff×h
=220+2502-10080×10
= 220 + 3.13
= 223.13 km
Hence, the median of the distances is 223.13 km.

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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)
fi
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
(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+N2-cff×h
=20+4002-140100×20
= 20 + 12
= Rs 32 
Hence, the  median of the prices is Rs 32.

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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.
(the lower limit of modal class) =  250
f1 (frequency of the modal class) = 60 
fo (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+f1-f02f1-f0-f2×h
=250+60-102×60-10-25×250
= 250 + 147.06
= 397.06
Hence, the modal demand of sweet is 397.06 grams.

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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 
 



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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

Page No 167:

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

Page No 167:

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

Page No 167:

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
​​

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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 = 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 auto rikshaw = 10%100%×360°=36°

Central angle for two whellers = 40%100%×360°=144°

(2)
Number of two wheelers = 1200
 40%100%×Total number of vehicles=1200Total number of vehicles=100%40%×1200=3000

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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% 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%  

The pie digram showing the above data is given below:



Page No 168:

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 = Central angle for cricket360°×Total number of students
=81°360°×1000
= 225 
Hence, 225 students like cricket.
(2)
Number of students like football = Central angle for football360°×Total number of students
=63°360°×1000
= 175 
Hence, 175 students like football.
(3)
Number of students like other games = Central angle for other games360°×Total number of students
=72°360°×1000
= 200
Hence, 200 students like other games.

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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 50180×360=100°
Suffered from cataract 10 10180×360=20°
Respiratory disorders 25 25180×360=50°
Healthy 95 95180×360=190°
Total 180  

The pie digram showing the above data is given below:
​​​

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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 20120×360=60°
Behada 28 28120×360=84°
Arjun 24 24120×360=72°
Bakul 22 22120×360=66°
Kadunimb 26 26120×360=78°
Total 180  

The pie digram showing the above data is given below:
​​​



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