RS Aggarwal Solutions Class 10 Ex 9C

Question-1: Find the mode of the following frequency distribution:

Marks 10-20 20-30 30-40 40-50 50-60
Frequency 12 35 45 25 13

Answer:

As the class 50-60 has maximum frequency, so it is the modal class:

Therefore, \(x_{k}=50,h=10,f_{k}=20\:and\:f_{k-1}=13\:and\:f_{k+1}=6\)

\(Mode\:M_{o}=x_{k}+\left [ h\times \frac{\left ( f_{k}-f_{k-1} \right )}{\left ( 2f_{k}-f_{k-1} -f_{k+1}\right )} \right ]\)

\(=50+\left [ 10\times \frac{\left ( 20-13 \right )}{\left ( 2\times 20-13 -6\right )} \right ]\)

= 50 +(70/21) = (50+3.33) = 53.33

Hence, mode = 53.33

Question-2:

Answer:

As the class 40 – 60 has maximum frequency, so it is the modal class:

Therefore, \(x_{k}=40,h=10,f_{k}=62\:and\:f_{k-1}=47\:and\:f_{k+1}=37\)

\(Mode\:M_{o}=x_{k}+\left [ h\times \frac{\left ( f_{k}-f_{k-1} \right )}{\left ( 2f_{k}-f_{k-1} -f_{k+1}\right )} \right ]\)

\(=40+\left [ 10\times \frac{\left ( 62-47 \right )}{\left ( 2\times 62-47 -37\right )} \right ]\)

= 40 + (150/40) = (40+3.75) = 43.75

Hence, mode = 43.75

Question-3: Find the mode of the following distribution:

Class interval 10-14 14-18 18-22 22-26 26-30 30-34 24-38 38-42
Frequency 8 6 11 20 25 22 10 4

Answer:

As the class 26 – 30 has maximum frequency, so it is the modal class:

Therefore, \(x_{k}=26,h=4,f_{k}=25\:and\:f_{k-1}=20\:and\:f_{k+1}=22\)

\(Mode\:M_{o}=x_{k}+\left [ h\times \frac{\left ( f_{k}-f_{k-1} \right )}{\left ( 2f_{k}-f_{k-1} -f_{k+1}\right )} \right ]\)

\(=26+\left [4\times \frac{\left ( 25-20 \right )}{\left ( 2\times 25-20-22\right )} \right ]\)

= 26 + (5/2) = (26+2.5) = 28.5

ence, mode = 28.5

Question-4: Given below is the distribution of total household expenditure of 200 manual workers in a city.

Expenditure (in rs.) No. of manual workers
1000-1500 24
1500-2000 40
2000-2500 31
2500-3000 28
3000-3500 32
3500-4000 23
4000-4500 17
4500-5000 5

Find the expenditure done by a maximum number of manual workers.

Answer:

As the class 1500 – 2000 has maximum frequency, so it is the modal class:

Therefore, \(x_{k}=1500,h=500,f_{k}=40\:and\:f_{k-1}=24\:and\:f_{k+1}=31\)

\(Mode\:M_{o}=x_{k}+\left [ h\times \frac{\left ( f_{k}-f_{k-1} \right )}{\left ( 2f_{k}-f_{k-1} -f_{k+1}\right )} \right ]\)

\(=1500+\left [500\times \frac{\left (40-24 \right )}{\left ( 2\times 40-24-31\right )} \right ]\)

= 1500 + (500 x 16/25)

= (1500+320) = Rs 1820

Hence, the average expenditure done by maximum number of workers = Rs. 1820

Question-5: Calculate the mode from the following data:

Monthly salary (in rs.) No. of employees
0-5000 90
5000-10000 150
10000-15000 100
15000-20000 80
20000-25000 70
25000-30000 10

Answer:

As the class 5000 – 10000 has maximum frequency, so it is the modal class:

Therefore, \(x_{k}=5000,h=5000,f_{k}=150\:and\:f_{k-1}=90\:and\:f_{k+1}=100\)

\(Mode\:M_{o}=x_{k}+\left [ h\times \frac{\left ( f_{k}-f_{k-1} \right )}{\left ( 2f_{k}-f_{k-1} -f_{k+1}\right )} \right ]\)

\(=5000+\left [5000\times \frac{\left (150-90 \right )}{\left ( 2\times 300-90-100\right )} \right ]\)

= 5000 + 2727.27

= Rs.7727.27

Hence, mode = Rs.7727.27

Question-6: Compute the mode from the following data:

Age (in years) 0-5 5-10 10-15 15-20 20-25 25-30 30-35
Number of patients 6 11 18 24 17 13 5

Answer:

As the class 15-20 has maximum frequency, so it is the modal class:

Therefore, \(x_{k}=15,h=5,f_{k}=24\:and\:f_{k-1}=18\:and\:f_{k+1}=17\)

\(Mode\:M_{o}=x_{k}+\left [ h\times \frac{\left ( f_{k}-f_{k-1} \right )}{\left ( 2f_{k}-f_{k-1} -f_{k+1}\right )} \right ]\)

\(=15+\left [5\times \frac{\left (24-18 \right )}{\left ( 2\times 48-18-17\right )} \right ]\)

= (15+2.30) = 17.3 years

Hence, mode = 17.3 years

Question-7: Compute the mode from the following series:

Size 45-55 55-65 65-75 75-85 85-95 95-105 105-115
Frequency 7 12 17 30 32 6 10

Answer:

As the class 85-95 has maximum frequency, so it is the modal class:

Therefore, \(x_{k}=85,h=10,f_{k}=32\:and\:f_{k-1}=30\:and\:f_{k+1}=6\)

\(Mode\:M_{o}=x_{k}+\left [ h\times \frac{\left ( f_{k}-f_{k-1} \right )}{\left ( 2f_{k}-f_{k-1} -f_{k+1}\right )} \right ]\)

\(=85+\left [10\times \frac{\left ( 32-30 \right )}{\left ( 2\times 64-30-6\right )} \right ]\)

= 85 + (5/7) = (85+0.71) = 85.71

Hence, mode = 85.71

Question-8:

Answer:

The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get

Class Frequency
0.5-5.5 3
5.5-10.5 8
10.5-15.5 13
15.5-20.5 18
20.5-25.5 28
25.5-30.5 20
30.5-35.5 13
35.5-40.5 8
40.5-45.5 6
45.5-50.5 3

As the class 20.5-25.5 has maximum frequency, so it is the modal class:

Therefore, \(x_{k}=20.5,h=5,f_{k}=28\:and\:f_{k-1}=18\:and\:f_{k+1}=20\)

\(Mode\:M_{o}=x_{k}+\left [ h\times \frac{\left ( f_{k}-f_{k-1} \right )}{\left ( 2f_{k}-f_{k-1} -f_{k+1}\right )} \right ]\)

\(=20.5+\left [5\times \frac{\left (28-18 \right )}{\left ( 2\times 56-18-20\right )} \right ]\)

= 20.5 + [5 x 10 /18]

= 20.5 + 2.78 = 23.28

Hence, mode = 23.28

EXERCISE – 9D

Question-1: Find the mean, mode and median of the following frequency distribution:

Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 4 4 7 10 12 8 5

Answer:

Let assumed mean be 35, h = 10, now we have

Class \(Frequency\:f_{i}\) \(Mid value\:x_{i}\) \(u_{i}=\left ( \frac{x_{i}-A}{h} \right )\) C.F \(f_{i}u_{i}\)
0-10 5 5 -3 5 -15
10-20 10 15 -2 15 -20
20-30 18 25 -1 33 -18
30-40 30 35-A 0 63 0
40-50 20 45 1 83 20
50-60 12 55 2 95 24
60-70 5 65 3 100 15
N= 100 \(\sum \left ( f_{i}u_{i} \right )=6\)

(i) Mean

= \(\bar{x}=A+h\left ( \frac{\sum f_{i}u_{i}}{N} \right )\)

= 35 + 10 x (6/100) = 35 + 0.6 = 35.6

(ii) N = 100, N/2 = 50

Cumulative frequency just after 50 is 63

Median class is 30-40

I = 30, h = 10, N = 100, c = 33, f = 30

Therefore, \(Median\:M_{e}=I+h\left ( \frac{\frac{N}{2}-c}{f} \right )=30+10\left ( \frac{50-33}{30} \right )\)

= 30 + 10 (17/30) = 30 + 5.67 = 35.67

(iii) Mode = 3 x median – 2 x mean

= 3 x 35.67 – 2 x 35.6 = 107.01 – 71.2

= 35.81

Thus, Mean = 35.6, Median = 35.67 and Mode = 35.81

Question-2:

Answer:

Let us assume mean A be 8.5. Class interval h = 3

Class \(Frequency\:f_{i}\) \(Mid value\:x_{i}\) \(u_{i}=\left ( \frac{x_{i}-A}{h} \right )\) \(f_{i}u_{i}\) C.F
1-4 6 2.5 -2 -12 6
4-7 30 5.5 -1 -30 36
7-10 40 8.5 = A 0 0 76
10-13 16 11.5 1 16 92
13-16 4 14.5 2 8 96
16-19 4 17.5 3 12 100
N=100 \(\sum \left ( f_{i}u_{i} \right )=6\)

N = total frequency = 100

(i) Mean

= \(\bar{x}=A+h\left ( \frac{\sum f_{i}u_{i}}{N} \right )\)

= 8.5+3 x -6/100 = 8.5-18/100

= 8.5-0.18 = 8.32

(ii) N/2 = 50, Cumulative frequency just after 50 is 76

Median class is 7-10

I=7, h=3, N=100, f =40,c=36

Therefore, \(Median\:M_{e}=I+h\left ( \frac{\frac{N}{2}-c}{f} \right )=7+3\left ( \frac{50-36}{40} \right )\)

= 7+3 x 14/40 = 7+21/20 = 7+1.05 = 8.05

(iii) Mode = 3 x median – 2 x mean

= 3 x 8.05 – 2 x 8.32 = 24.15 – 16.64

= 7.51

Thus, mean = 8.32, median = 8.05, mode = 7.51

Question-3: A survey regarding the heights (in cm) of 50 girls of a class was conducted and the following data was obtained:

Class 120-130 130-140 140-150 150-160 160-170 Total
Frequency 2 8 12 20 8 30

Find the mean, median and mode of the above data.

Answer:

Let the assumed mean A be 145. Class interval h = 10.

Class \(Frequency\:f_{i}\) \(Mid value\:x_{i}\) \(u_{i}=\left ( \frac{x_{i}-A}{h} \right )\) \(f_{i}u_{i}\) C.F
120-130 2 125 -2 -4 2
130-140 8 135 -1 -4 10
140-150 12 145 = A 0 0 22
150-160 20 155 1 20 42
160-170 8 165 2 16 50
N=50 \(\sum \left ( f_{i}u_{i} \right )=24\)

(i) Mean

= \(\bar{x}=A+h\left ( \frac{\sum f_{i}u_{i}}{N} \right )\)

= 145+10 x (24/50)

=145+4.8 = 149.8

(ii) N = 50, N/2 = 25

Cumulative frequency just after 25 is 42

Median class is 150-160

I = 150, h = 10, N = 50, c = 22, f = 20

Therefore, \(Median\:M_{e}=I+h\left ( \frac{\frac{N}{2}-c}{f} \right )=150+10\left ( \frac{25-22}{20} \right )\)

= 30 + 10 x 3/20 = 150 + 1.5 = 151.5

(iii) Mode = 3 x median – 2 x mean

= 3 x 151.5 – 2 x 149.8 = 454.5 – 299.6

= 154.9

Thus, mean = 149.8, median = 151.5, mode = 154.9

Question-4: The following table gives the daily income of 50 workers of a factory:

Daily income (in rs.) 100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10

Find the mean, median and mode of the above data.

Answer:

Let assumed mean A = 150 and h = 20

Class \(Frequency\:f_{i}\) \(Mid value\:x_{i}\) \(u_{i}=\left ( \frac{x_{i}-A}{h} \right )\) \(f_{i}u_{i}\) C.F
100-120 12 110 -2 -24 12
120-140 14 130 -1 -14 26
140-160 8 150 = A 0 0 34
160-180 6 170 1 6 40
180-200 10 190 2 20 50
N = 50 \(\sum \left ( f_{i}u_{i} \right )=-12\)

(i) Mean

= \(\bar{x}=A+h\left ( \frac{\sum f_{i}u_{i}}{N} \right )\)

= 150-24/5 = 150-4.8 = 145.2

(ii) N = 50, N/2 = 25

Cumulative frequency just after 25 is 26

Median class is 120-140

I = 120, h = 10, N = 50, c = 12, f = 14

Therefore, \(Median\:M_{e}=I+h\left ( \frac{\frac{N}{2}-c}{f} \right )=120+20\left ( \frac{25-12}{14} \right )\)

= 120+20 x 13/14 = 120+130/7 = 120+18.6 = 138.6

(iii) Mode = 3 x median – 2 x mean

= 3 x 138.6 – 2 x 145.2

= 415.8 – 190.4 = 125.4

Thus, mean = 145.2, median = 138.6, mode = 125.4

Question-5: The table below shows the daily expenditure on food of 30 households in a locality:

Daily expenditure (in rs.) Number of households
100-150 6
150-200 7
200-250 12
250-300 3
300-350 2

Find the mean, median and mode of the above data/p>

Answer:

Let assumed mean A = 225 and h = 50

Class \(Frequency\:f_{i}\) \(Mid value\:x_{i}\) \(u_{i}=\left ( \frac{x_{i}-A}{h} \right )\) \(f_{i}u_{i}\) C.F
100-150 6 125 2 12 6
150-200 7 175 1 7 13
200-250 12 225 0 0 25
250-300 3 275 1 3 28
300-350 2 325 2 4 30
N = 30 \(\sum \left ( f_{i}u_{i} \right )=-12\)

(i) Mean

= \(\bar{x}=A+h\left ( \frac{\sum f_{i}u_{i}}{N} \right )\)

= 225 – 20 = 205

(ii) N = 30, N/2 = 15

Cumulative frequency just after 15 is 25

Corresponding frequency interval is 200 – 250

Median class is 200 – 250

Cumulative frequency c just before this class = 13

So I = 200, f = 12, N/2 = 15, h = 50, c = 13

Therefore, \(Median\:M_{e}=I+h\left ( \frac{\frac{N}{2}-c}{f} \right )=200+50\left ( \frac{15-13}{12} \right )\)

= 200+50 x 2/12 = 200+25/3 = 200+8.33 = 208.33

Hence, Mean = 205 and Median = 208.33


Practise This Question

Carbon has a valency of four. It is capable of forming single bonds with four other carbon atoms or atoms of some other monovalent element, etc. What is this property of carbon known as?