Mathematics Part I Solutions Solutions for Class 10 Math Chapter 6 Statistics are provided here with simple step-by-step explanations. These solutions for Statistics are extremely popular among Class 10 students for Math Statistics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### 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 $\sum _{}{f}_{i}=50$ $\sum _{}{x}_{i}{f}_{i}=218$

Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{218}{50}$
= 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 xi di = xi − 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 $\underset{}{\sum {f}_{i}=420}$ $\sum _{}{f}_{i}{d}_{i}=-12000$

Required Mean = $A+\frac{\sum _{}{f}_{i}{d}_{i}}{\sum _{}{f}_{i}}$
$=550-\frac{12000}{420}$
= 550 − 28.57
​= 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 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 $\sum _{}{f}_{i}=50$ $\sum _{}{x}_{i}{f}_{i}=141$

Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{141}{50}$
= 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:

 Class (Production in Thousand rupees) Class Mark xi di = xi − 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 $\underset{}{\sum {f}_{i}=80}$ $\sum _{}{f}_{i}{d}_{i}=-175$

Required Mean = $A+\frac{\sum _{}{f}_{i}{d}_{i}}{\sum _{}{f}_{i}}$
$37.5-\frac{175}{80}$
= 37.5 − 2.19
​= 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:

 Class (Production in Thousand rupees) Class Mark xi di = xi − A ${u}_{i}=\frac{{d}_{i}}{h}$ 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 $\underset{}{\sum {f}_{i}=120}$ $\sum _{}{f}_{i}{u}_{i}=-63$

Required Mean = $A+h\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}$
$=1250-\left(\frac{63}{120}\right)500$
= 1250 − 262.5
= 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:

 Class (Weekely wages rupees) Class Mark xi di = xi − A ${u}_{i}=\frac{{d}_{i}}{h}$ 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 $\underset{}{\sum {f}_{i}=150}$ $\sum _{}{f}_{i}{u}_{i}=-65$

Required Mean = $A+h\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}$
$=3500-\left(\frac{65}{150}\right)1000$
= 3500 − 433.33
= 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) 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 = $L+\left(\frac{\frac{N}{2}-cf}{f}\right)×h$

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) 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 = $L+\left(\frac{\frac{N}{2}-cf}{f}\right)×h$
$=150+\left(\frac{\frac{250}{2}-63}{90}\right)×50$
= 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
(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 = $L+\left(\frac{\frac{N}{2}-cf}{f}\right)×h$
$=70+\left(\frac{\frac{200}{2}-44}{55}\right)×4$
= 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) 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 = $L+\left(\frac{\frac{N}{2}-cf}{f}\right)×h$
$=50+\left(\frac{\frac{105}{2}-47}{20}\right)×10$
= 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.
(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 = $L+\left(\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}\right)×h$
$=4+\left(\frac{80-70}{2×80-70-60}\right)×1$
= 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.
(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 = $L+\left(\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}\right)×h$
$=60+\left(\frac{100-70}{2×100-70-80}\right)×20$
= 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.
(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 = $L+\left(\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}\right)×h$
$=9+\left(\frac{35-20}{2×35-20-18}\right)×2$
= 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.
(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 = $L+\left(\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}\right)×h$
$=9.5+\left(\frac{50-32}{2×50-32-36}\right)×5$
= 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 $\frac{80}{200}×360=144°$ 25 - 30 60 $\frac{60}{200}×360=108°$ 30 - 35 35 $\frac{35}{200}×360=63°$ 35 - 40 25 $\frac{25}{200}×360=45°$ 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 $\frac{50}{400}×360=45°$ Marathi 70 $\frac{70}{400}×360=63°$ Science 80 $\frac{80}{400}×360=72°$ Mathematics 90 $\frac{90}{400}×360=81°$ Social science 60 $\frac{60}{400}×360=54°$ Hindi 50 $\frac{50}{400}×360=45°$ 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 $\frac{40}{360}×360=40°$ 6th 50 $\frac{50}{360}×360=50°$ 7th 75 $\frac{75}{360}×360=75°$ 8th 50 $\frac{50}{360}×360=50°$ 9th 70 $\frac{70}{360}×360=70°$ 10th 75 $\frac{75}{360}×360=75°$ Total 360

The pie digram showing the above data is given below: #### 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 $\frac{30}{100}×360=108°$ Sweet lime 15 $\frac{15}{100}×360=54°$ Apples 25 $\frac{25}{100}×360=90°$ Cheeku 20 $\frac{20}{100}×360=72°$ Oranges 10 $\frac{10}{100}×360=36°$ Total 100

The pie digram showing the above data is given below:
​ #### 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
$=\frac{72°}{360°}×10000$
​= 2000

(2)
Number of workers working in the administration
$=\frac{36°}{360°}×10000$
= 1000

(3)
Percentage of workers working in production
$=\frac{90°}{360°}×100%$
= 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

(2)
Money invested in bank​
$=\frac{1}{4}×12000$
= Rs 3000

(3)
Difference in the central angle between immovable property and mutual fund = 120º − 60º
= 60º
Therefore, required mooney
$=\frac{1}{6}×12000$
= Rs 2000

(4)
Money invested in post​
$=\frac{1}{12}×12000$
= 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 $\overline{)\mathrm{X}}=\mathrm{A}+\frac{\sum {f}_{i}{u}_{i}}{\sum {f}_{i}}×hg$ In the formula u i = . ..
 (A) $\frac{{x}_{i}+\mathrm{A}}{g}$ (B) $\left({x}_{i}-\mathrm{A}\right)$ (C) $\frac{{x}_{i}-\mathrm{A}}{g}$ (D) $\frac{\mathrm{A}-{x}_{i}}{g}$

(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 = $\frac{40}{100}×360°=144°$
Hence, the correct option is (D).

(2)
Central angle for expenditures incurred on the construction = 75°

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
$\overline{)\mathrm{X}}=\mathrm{A}+\frac{\sum {f}_{i}{u}_{i}}{\sum {f}_{i}}×hg$ .
In this formula,
${u}_{i}=\frac{{x}_{i}-\mathrm{A}}{g}$

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 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 $\sum _{}{f}_{i}=84$ $\sum _{}{x}_{i}{f}_{i}=4410$

Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

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 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 $\sum _{}{f}_{i}=100$ $\sum _{}{x}_{i}{f}_{i}=6540$

Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

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 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 $\sum _{}{f}_{i}=120$ $\sum _{}{x}_{i}{f}_{i}=510000$

Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{510000}{120}$
= Rs 4250
​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 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 $\sum _{}{f}_{i}=50$ $\sum _{}{x}_{i}{f}_{i}=3620$

Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$

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) 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 = $L+\left(\frac{\frac{N}{2}-cf}{f}\right)×h$
$=220+\left(\frac{\frac{250}{2}-100}{80}\right)×10$
= 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) 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 = $L+\left(\frac{\frac{N}{2}-cf}{f}\right)×h$
$=20+\left(\frac{\frac{400}{2}-140}{100}\right)×20$
= 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.
(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 = $L+\left(\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}\right)×h$
$=250+\left(\frac{60-10}{2×60-10-25}\right)×250$
= 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 #### 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
​​ #### 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 = $\frac{30%}{100%}×360°=108°$

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

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

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

Central angle for two whellers = $\frac{40%}{100%}×360°=144°$

(2)
Number of two wheelers = 1200

#### 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% $\frac{10}{100}×360=36°$ Traffic 50% $\frac{50}{100}×360=180°$ Aircraft take offs 9% $\frac{9}{100}×360=32.4°$ Industry 20% $\frac{20}{100}×360=72°$ Trains 11% $\frac{11}{100}×360=39.6°$ 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 =
$=\frac{81°}{360°}×1000$
= 225
Hence, 225 students like cricket.
(2)
Number of students like football =
$=\frac{63°}{360°}×1000$
= 175
Hence, 175 students like football.
(3)
Number of students like other games =
$=\frac{72°}{360°}×1000$
= 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 $\frac{50}{180}×360=100°$ Suffered from cataract 10 $\frac{10}{180}×360=20°$ Respiratory disorders 25 $\frac{25}{180}×360=50°$ Healthy 95 $\frac{95}{180}×360=190°$ Total 180

The pie digram showing the above data is given below:
​​​ #### 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 $\frac{20}{120}×360=60°$ Behada 28 $\frac{28}{120}×360=84°$ Arjun 24 $\frac{24}{120}×360=72°$ Bakul 22 $\frac{22}{120}×360=66°$ Kadunimb 26 $\frac{26}{120}×360=78°$ Total 180

The pie digram showing the above data is given below:
​​​ View NCERT Solutions for all chapters of Class 10