# RS Aggarwal Solutions Class 10 Ex 9C

## RS Aggarwal Class 10 Ex 9C Chapter 9

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

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:

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

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.

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

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

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

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:

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

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:

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.

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.

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>

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

Ammonia gas dissolves in water to give ammonium ion.In this reaction water acts as: