Mathematics Part I Solutions Solutions for Class 10 Math Chapter 6 Statistics are provided here with simple stepbystep 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 adfree 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.
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 x_{i} 
Frequency (Number of students) f_{i} 
Class mark × Frequency x_{i}f_{i} 
0  2  1  7  7 
2  4  3  18  54 
4  6  5  12  60 
6  8  7  10  70 
8  10  9  3  27 
$\sum _{}{f}_{i}=50$  $\sum _{}{x}_{i}{f}_{i}=218$ 
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{218}{50}$
= 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 x_{i} 
d_{i} = x_{i }− A  Frequency (Number of vehicles f_{i} 
Frequency × deviation f_{i }× d_{i} 
300  400  350  −200  80  −16000 
400  500  450  −100  110  −11000 
500  600  550 = A  0  120  0 
600  700  650  100  70  7000 
700  800  750  200  40  8000 
$\underset{}{\sum {f}_{i}=420}$  $\sum _{}{f}_{i}{d}_{i}=12000$ 
Required Mean = $A+\frac{\sum _{}{f}_{i}{d}_{i}}{\sum _{}{f}_{i}}$
$=550\frac{12000}{420}$
= 550 − 28.57
= 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 x_{i} 
Frequency (Number of customers) f_{i} 
Class mark × Frequency x_{i}f_{i} 
1  2  1.5  17  25.5 
2  3  2.5  13  32.5 
3  4  3.5  10  35 
4  5  4.5  7  31.5 
5  6  5.5  3  16.5 
$\sum _{}{f}_{i}=50$  $\sum _{}{x}_{i}{f}_{i}=141$ 
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{141}{50}$
= 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 x_{i} 
d_{i} = x_{i }− A  Frequency (Number of farm owners) f_{i} 
Frequency × deviation f_{i }× d_{i} 
25  30  27.5  −10  20  −200 
30  35  32.5  −5  25  −125 
35 40  37.5= A  0  15  0 
40  45  42.5  5  10  50 
45  50  47.5  10  10  100 
$\underset{}{\sum {f}_{i}=80}$  $\sum _{}{f}_{i}{d}_{i}=175$ 
Required Mean = $A+\frac{\sum _{}{f}_{i}{d}_{i}}{\sum _{}{f}_{i}}$
$37.5\frac{175}{80}$
= 37.5 − 2.19
= 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 x_{i} 
d_{i} = x_{i }− A  ${u}_{i}=\frac{{{\displaystyle d}}_{i}}{h}$  Frequency (Number of farm owners) f_{i} 
Frequency × deviation f_{i }× u_{i} 
0  500  250  −1000  −2  35  −70 
500  1000  750  −500  −1  28  −28 
1000  1500  1250 = A  0  0  32  0 
1500  2000  1750  500  1  15  15 
2000  2500  2250  1000  2  10  20 
$\underset{}{\sum {f}_{i}=120}$  $\sum _{}{f}_{i}{u}_{i}=63$ 
Required Mean = $A+h\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}$
$=1250\left(\frac{63}{120}\right)500$
= 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 x_{i} 
d_{i} = x_{i }− A  ${u}_{i}=\frac{{{\displaystyle d}}_{i}}{h}$  Frequency (Number of workers) f_{i} 
Frequency × deviation f_{i }× u_{i} 
1000  2000  1500  −2000  −2  25  −50 
2000  3000  2500  −100  −1  45  −45 
3000  4000  3500 = A  0  0  50  0 
4000  5000  4500  1000  1  30  30 
$\underset{}{\sum {f}_{i}=150}$  $\sum _{}{f}_{i}{u}_{i}=65$ 
Required Mean = $A+h\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}$
$=3500\left(\frac{65}{150}\right)1000$
= 3500 − 433.33
= Rs 3066.67
Hence, the mean of the weekly wages is Rs 3066.67.
Page No 145:
Question 1:
Daily No. of hours  8  10  10  12  12  14  14  16 
Number of workers  150  500  300  50 
Answer:
Class
(Number of working hours) 
Frequency (Number of workers) f_{i} 
Cumulaive frequency less than the upper limit 
8  10  150  150 
10  12 (Median Class) 
500  650 
12  14  300  950 
14  16  50  1000 
$N=1000$ 
From the above table, we get
L (Lower class limit of the median class) = 10
N (Sum of frequencies) = 1000
h (Class interval of the median class) = 2
f (Frequency of the median class) = 500
cf (Cumulative frequency of the class preceding the median class) = 150
Now, Median = $L+\left(\frac{{\displaystyle \frac{N}{2}cf}}{f}\right)\times h$
$=10+\left(\frac{{\displaystyle \frac{1000}{2}150}}{500}\right)\times 2\phantom{\rule{0ex}{0ex}}=10+1.4\phantom{\rule{0ex}{0ex}}=11.4\mathrm{hours}$
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) f_{i} 
Cumulaive frequency less than the upper limit 
50  100  33  33 
100  150  30  63 
150  200 (Median Class) 
90  153 
200  250  80  233 
250  300  17  250 
N = 250 
From the above table, we get
L (Lower class limit of the median class) = 150
N (Sum of frequencies) = 250
h (Class interval of the median class) = 50
f (Frequency of the median class) = 90
cf (Cumulative frequency of the class preceding the median class) = 63
Now, Median = $L+\left(\frac{{\displaystyle \frac{N}{2}cf}}{f}\right)\times h$
$=150+\left(\frac{{\displaystyle \frac{250}{2}63}}{90}\right)\times 50$
= 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 MumbaiPune express way. Find the median of the data.
Average Speed of
Vehicles(Km/hr) 
60  64  64  69  70  74  75  79  79  84  84  89 
No. of vehicles  10  34  55  85  10  6 
Answer:
Class
(Number of working hours) 
Frequency (Number of workers) 
Cumulaive frequency less than the upper limit 
60  64  10  10 
64  69  34  44 
70  74 (Median Class) 
55  99 
75  79  85  184 
79  84  10  194 
84  89  6  200 
N = 200 
From the above table, we get
L (Lower class limit of the median class) = 70
N (Sum of frequencies) = 200
h (Class interval of the median class) = 4
f (Frequency of the median class) = 55
cf (Cumulative frequency of the class preceding the median class) = 44
Now, Median = $L+\left(\frac{{\displaystyle \frac{N}{2}cf}}{f}\right)\times h$
$=70+\left(\frac{{\displaystyle \frac{200}{2}44}}{55}\right)\times 4$
= 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) f_{i} 
Cumulaive frequency less than the upper limit 
30  40  12  12 
40  50  35  47 
50  60 (Median Class) 
20  67 
60  70  15  82 
70  80  8  90 
80  90  7  97 
90  100  8  105 
N = 105 
From the above table, we get
L (Lower class limit of the median class) = 50
N (Sum of frequencies) = 105
h (Class interval of the median class) = 10
f (Frequency of the median class) = 20
cf (Cumulative frequency of the class preceding the median class) = 47
Now, Median = $L+\left(\frac{{\displaystyle \frac{N}{2}cf}}{f}\right)\times h$
$=50+\left(\frac{{\displaystyle \frac{105}{2}47}}{20}\right)\times 10$
= 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.
L (the lower limit of modal class) = 4
f_{1} (frequency of the modal class) = 80
f_{o} (frequency of the class preceding the modal class) = 70
f_{2} (frequency of the class succeeding the modal class) = 60
h (class size) = 1
Mode = $L+\left(\frac{{f}_{1}{f}_{0}}{2{f}_{1}{f}_{0}{f}_{2}}\right)\times h$
$=4+\left(\frac{8070}{2\times 807060}\right)\times 1$
= 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.
L (the lower limit of modal class) = 60
f_{1} (frequency of the modal class) = 100
f_{o} (frequency of the class preceding the modal class) = 70
f_{2} (frequency of the class succeeding the modal class) = 80
h (class size) = 20
Mode = $L+\left(\frac{{f}_{1}{f}_{0}}{2{f}_{1}{f}_{0}{f}_{2}}\right)\times h$
$=60+\left(\frac{10070}{2\times 1007080}\right)\times 20$
= 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.
L (the lower limit of modal class) = 9
f_{1} (frequency of the modal class) = 35
f_{o} (frequency of the class preceding the modal class) = 20
f_{2} (frequency of the class succeeding the modal class) = 18
h (class size) = 2
Mode = $L+\left(\frac{{f}_{1}{f}_{0}}{2{f}_{1}{f}_{0}{f}_{2}}\right)\times h$
$=9+\left(\frac{3520}{2\times 352018}\right)\times 2$
= 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.
L (the lower limit of modal class) = 9.5
f_{1} (frequency of the modal class) = 50
f_{o} (frequency of the class preceding the modal class) = 32
f_{2} (frequency of the class succeeding the modal class) = 36
h (class size) = 5
Mode = $L+\left(\frac{{f}_{1}{f}_{0}}{2{f}_{1}{f}_{0}{f}_{2}}\right)\times h$
$=9.5+\left(\frac{5032}{2\times 503236}\right)\times 5$
= 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 classmark 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 8090?
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 classmark 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 8090 are 15.
Page No 157:
Question 2:
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:
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:
Age group(Yrs)  20  25  25  30  30  35  35 40 
No. of persons  80  60  35  25 
Answer:
The measures of central angles are given in the table.
Age group(Yrs)  No. of persons  Central Angle 
20  25  80  $\frac{80}{200}\times 360=144\xb0$ 
25  30  60  $\frac{60}{200}\times 360=108\xb0$ 
30  35  35  $\frac{35}{200}\times 360=63\xb0$ 
35  40  25  $\frac{25}{200}\times 360=45\xb0$ 
Total  200 
The pie digram showing the above data is given below:
Page No 163:
Question 2:
Subject  English  Marathi  Science  Mathematics  Social science  Hindi 
Marks  50  70  80  90  60  50 
Answer:
The measures of central angles are given in the table.
Subjects  Marks  Central Angle 
English  50  $\frac{50}{400}\times 360=45\xb0$ 
Marathi  70  $\frac{70}{400}\times 360=63\xb0$ 
Science  80  $\frac{80}{400}\times 360=72\xb0$ 
Mathematics  90  $\frac{90}{400}\times 360=81\xb0$ 
Social science  60  $\frac{60}{400}\times 360=54\xb0$ 
Hindi  50  $\frac{50}{400}\times 360=45\xb0$ 
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  $\frac{40}{360}\times 360=40\xb0$ 
6th  50  $\frac{50}{360}\times 360=50\xb0$ 
7th  75  $\frac{75}{360}\times 360=75\xb0$ 
8th  50  $\frac{50}{360}\times 360=50\xb0$ 
9th  70  $\frac{70}{360}\times 360=70\xb0$ 
10th  75  $\frac{75}{360}\times 360=75\xb0$ 
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  $\frac{30}{100}\times 360=108\xb0$ 
Sweet lime  15  $\frac{15}{100}\times 360=54\xb0$ 
Apples  25  $\frac{25}{100}\times 360=90\xb0$ 
Cheeku  20  $\frac{20}{100}\times 360=72\xb0$ 
Oranges  10  $\frac{10}{100}\times 360=36\xb0$ 
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$=\frac{\mathrm{central}\mathrm{angle}\mathrm{for}\mathrm{construction}}{360\xb0}\times 10000$
$=\frac{72\xb0}{360\xb0}\times 10000$
= 2000
(2)
Number of workers working in the administration$=\frac{\mathrm{central}\mathrm{angle}\mathrm{for}\mathrm{administration}}{360\xb0}\times 10000$
$=\frac{36\xb0}{360\xb0}\times 10000$
= 1000
(3)
Percentage of workers working in production$=\frac{\mathrm{central}\mathrm{angle}\mathrm{for}\mathrm{production}}{360\xb0}\times 100\%$
$=\frac{90\xb0}{360\xb0}\times 100\%$
= 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
$\Rightarrow \frac{60\xb0}{360\xb0}\times \mathrm{Total}\mathrm{investment}=2000\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Total}\mathrm{investment}=\frac{360\xb0}{60\xb0}\times 2000\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}12000$
(2)
Money invested in bank$=\frac{90\xb0}{360\xb0}\times \mathrm{Total}\mathrm{investment}$
$=\frac{1}{4}\times 12000$
= Rs 3000
(3)
Difference in the central angle between immovable property and mutual fund = 120^{º} − 60^{º}
= 60^{º}
Therefore, required mooney$=\frac{60\xb0}{360\xb0}\times \mathrm{Total}\mathrm{investment}$
$=\frac{1}{6}\times 12000$
= Rs 2000
(4)
Money invested in post$=\frac{30\xb0}{360\xb0}\times \mathrm{Total}\mathrm{investment}$
$=\frac{1}{12}\times 12000$
= 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° 
(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 
(A) $\frac{{x}_{i}+\mathrm{A}}{g}$  (B) $\left({x}_{i}\mathrm{A}\right)$  (C) $\frac{{x}_{i}\mathrm{A}}{g}$  (D) $\frac{{\displaystyle \mathrm{A}{x}_{i}}}{{\displaystyle g}}$ 
(5)
Distance Covered per litre (km)  12  14  14  16  16  18  18  20 
No. of cars  11  12  20  7 
(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 
(A) (4, 8)  (B) (3, 5)  (C) (5, 8)  (D) (8, 4) 
Answer:
(1)
Central angle for O– blood group persons = $\frac{40}{100}\times 360\xb0=144\xb0$
Hence, the correct option is (D).
(2)
Central angle for expenditures incurred on the construction = 75°
$\Rightarrow \frac{\mathrm{Expenditures}\mathrm{on}\mathrm{construction}}{\mathrm{Total}\mathrm{expenditures}}\times 360\xb0=75\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Total}\mathrm{expenditures}=45000\times \frac{360\xb0}{75\xb0}\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}216000$
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 $\overline{)\mathrm{X}}=\mathrm{A}+\frac{\sum {f}_{i}{u}_{i}}{\sum {f}_{i}}\times hg$ .
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 46 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 x_{i} 
Frequency (Number of farmers) f_{i} 
Class mark × Frequency x_{i}f_{i} 
20  30  25  10  250 
30  40  35  11  385 
40  50  45  15  675 
50  60  55  16  880 
60  70  65  18  1170 
70  80  75  14  1050 
$\sum _{}{f}_{i}=84$  $\sum _{}{x}_{i}{f}_{i}=4410$ 
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{4410}{84}\phantom{\rule{0ex}{0ex}}=52.5\mathrm{thousand}\mathrm{rupees}\phantom{\rule{0ex}{0ex}}=52500$
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 x_{i} 
Frequency (Number of farm ponds) f_{i} 
Class mark × Frequency x_{i}f_{i} 
40  50  45  13  585 
50  60  55  20  1100 
60  70  65  24  1560 
70  80  75  36  2700 
80  90  85  7  595 
$\sum _{}{f}_{i}=100$  $\sum _{}{x}_{i}{f}_{i}=6540$ 
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{6540}{100}\phantom{\rule{0ex}{0ex}}=65.4\mathrm{thousand}\mathrm{rupees}\phantom{\rule{0ex}{0ex}}=65400$
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 x_{i} 
Frequency (Number of workers) f_{i} 
Class mark × Frequency x_{i}f_{i} 
0  2000  1000  15  15000 
2000  4000  3000  35  105000 
4000  6000  5000  50  250000 
6000  8000  7000  20  140000 
$\sum _{}{f}_{i}=120$  $\sum _{}{x}_{i}{f}_{i}=510000$ 
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{510000}{120}$
= 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 x_{i} 
Frequency (Number of families) f_{i} 
Class mark × Frequency x_{i}f_{i} 
50  60  55  7  385 
60  70  65  13  845 
70  80  75  20  1500 
80  90  85  6  510 
90  100  95  4  380 
$\sum _{}{f}_{i}=50$  $\sum _{}{x}_{i}{f}_{i}=3620$ 
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
$=\frac{3620}{50}\phantom{\rule{0ex}{0ex}}=72.4\mathrm{thousand}\mathrm{rupees}\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}72400$
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) f_{i} 
Cumulaive frequency less than the upper limit 
200  210  40  40 
210  220  60  100 
220  230 (Median Class) 
80  180 
230  240  50  230 
240  250  20  250 
N = 250 
From the above table, we get
L (Lower class limit of the median class) = 220
N (Sum of frequencies) = 250
h (Class interval of the median class) = 10
f (Frequency of the median class) = 80
cf (Cumulative frequency of the class preceding the median class) = 100
Now, Median = $L+\left(\frac{{\displaystyle \frac{N}{2}cf}}{f}\right)\times h$
$=220+\left(\frac{{\displaystyle \frac{250}{2}100}}{80}\right)\times 10$
= 220 + 3.13
= 223.13 km
Hence, the median of the distances is 223.13 km.
Page No 166:
Question 7:
The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.
Price (Rupees)

20 less than  20  40  40  60  60  80  80  100 
No. of articles  140  100  80  60  20 
Answer:
Class
(Prices in Rupees) 
Frequency (Number of articles) f_{i} 
Cumulaive frequency less than the upper limit 
0  20  140  140 
20  40 (Median Class) 
100  240 
40  60  80  320 
60  80  60  380 
80  100  20  400 
N = 400 
From the above table, we get
L (Lower class limit of the median class) = 20
N (Sum of frequencies) = 400
h (Class interval of the median class) = 20
f (Frequency of the median class) = 100
cf (Cumulative frequency of the class preceding the median class) = 140
Now, Median = $L+\left(\frac{{\displaystyle \frac{N}{2}cf}}{f}\right)\times h$
$=20+\left(\frac{{\displaystyle \frac{400}{2}140}}{100}\right)\times 20$
= 20 + 12
= Rs 32
Hence, the median of the prices is Rs 32.
Page No 166:
Question 8:
The following frequency table shows the demand for a sweet and the number of customers. Find the mode of demand of sweet.
Weight of sweet (gram)

0  250  250  500  500  750  750  1000  1000  1250 
No. of customers  10  60  25  20  15 
Answer:
The maximum class frequency is 60.
The class corresponding to this frequency is 250  500.
So, the modal class is 250  500.
L (the lower limit of modal class) = 250
f_{1} (frequency of the modal class) = 60
f_{o} (frequency of the class preceding the modal class) = 10
f_{2} (frequency of the class succeeding the modal class) = 25
h (class size) = 250
Mode = $L+\left(\frac{{f}_{1}{f}_{0}}{2{f}_{1}{f}_{0}{f}_{2}}\right)\times h$
$=250+\left(\frac{6010}{2\times 601025}\right)\times 250$
= 250 + 147.06
= 397.06
Hence, the modal demand of sweet is 397.06 grams.
Page No 166:
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
Page No 167:
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
Page No 167:
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 twowheelers is 1200, find the number of all vehicles.
Answer:
(1)
Central angle for cars = $\frac{30\%}{100\%}\times 360\xb0=108\xb0$
Central angle for tempos = $\frac{12\%}{100\%}\times 360\xb0=43.2\xb0$
Central angle for buses = $\frac{8\%}{100\%}\times 360\xb0=28.8\xb0$
Central angle for auto rikshaw = $\frac{10\%}{100\%}\times 360\xb0=36\xb0$
Central angle for two whellers = $\frac{40\%}{100\%}\times 360\xb0=144\xb0$
(2)
Number of two wheelers = 1200
$\Rightarrow \frac{40\%}{100\%}\times \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{vehicles}=1200\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{vehicles}=\frac{100\%}{40\%}\times 1200\phantom{\rule{0ex}{0ex}}=3000$
Page No 167:
Question 15:
The following table shows causes of noise pollution. Show it by a pie diagram.
Construction

Traffic  Aircraft take offs  Industry  Trains 
10%  50%  9%  20%  11% 
Answer:
The measures of central angles are given in the table.
Cause of noise pollution  Percentage  Central Angle 
Construction  10%  $\frac{10}{100}\times 360=36\xb0$ 
Traffic  50%  $\frac{50}{100}\times 360=180\xb0$ 
Aircraft take offs  9%  $\frac{9}{100}\times 360=32.4\xb0$ 
Industry  20%  $\frac{20}{100}\times 360=72\xb0$ 
Trains  11%  $\frac{11}{100}\times 360=39.6\xb0$ 
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 = $\frac{\mathrm{Central}\mathrm{angle}\mathrm{for}\mathrm{cricket}}{360\xb0}\times \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}$
$=\frac{81\xb0}{360\xb0}\times 1000$
= 225
Hence, 225 students like cricket.
(2)
Number of students like football = $\frac{\mathrm{Central}\mathrm{angle}\mathrm{for}\mathrm{football}}{360\xb0}\times \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}$
$=\frac{63\xb0}{360\xb0}\times 1000$
= 175
Hence, 175 students like football.
(3)
Number of students like other games = $\frac{\mathrm{Central}\mathrm{angle}\mathrm{for}\mathrm{other}\mathrm{games}}{360\xb0}\times \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}$
$=\frac{72\xb0}{360\xb0}\times 1000$
= 200
Hence, 200 students like other games.
Page No 168:
Question 17:
Medical check up of 180 women was conducted in a health centre in a village. 50 of them were short of haemoglobin, 10 suffered from cataract and 25 had respiratory disorders. The remaining women were healthy. Show the information by a pie diagram.
Answer:
The measures of central angles are given in the table.
Health condition  Number of women  Central Angle 
Short of haemoglobin  50  $\frac{50}{180}\times 360=100\xb0$ 
Suffered from cataract  10  $\frac{10}{180}\times 360=20\xb0$ 
Respiratory disorders  25  $\frac{25}{180}\times 360=50\xb0$ 
Healthy  95  $\frac{95}{180}\times 360=190\xb0$ 
Total  180 
The pie digram showing the above data is given below:
Page No 168:
Question 18:
On an environment day, students in a school planted 120 trees under plantation project. The information regarding the project is shown in the following table. Show it by a pie diagram.
Tree name

Karanj  Behada  Arjun  Bakul  Kadunimb 
No. of trees  20  28  24  22  26 
Answer:
The measures of central angles are given in the table.
Tree Name  Number of tress  Central Angle 
Karanj  20  $\frac{20}{120}\times 360=60\xb0$ 
Behada  28  $\frac{28}{120}\times 360=84\xb0$ 
Arjun  24  $\frac{24}{120}\times 360=72\xb0$ 
Bakul  22  $\frac{22}{120}\times 360=66\xb0$ 
Kadunimb  26  $\frac{26}{120}\times 360=78\xb0$ 
Total  180 
The pie digram showing the above data is given below:
View NCERT Solutions for all chapters of Class 10