# RS Aggarwal Class 10 Solutions Chapter 9 - Mean Median Mode Ex 9B(9.2)

## RS Aggarwal Class 10 Chapter 9 - Mean Median Mode Ex 9B(9.2) Solutions Free PDF

The RS Aggarwal Class 10 solution of Chapter 9 we provided here are in accordance with the CBSE syllabus and are designed by our subject experts with easy and effective explanations of the solutions given in RS Aggarwal Class 10 maths textbook.Â  It helps the students in solving all the exercise questions in a step-by-step manner and develop a strong foundation of the important mathematical concepts. It also improves their question solving-skills.

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## Download PDF of RS Aggarwal Class 10 Solutions Chapter 9â€“Â Mean Median Mode Ex 9B (9.2)

Q1: In a hospital, the ages of diabetic patients were recorded as follows

 Age (in years) 0-15 15-30 30-45 45-60 60-75 Number of patients 5 20 40 50 25

Find the median age.

Sol:

We prepare the cumulative frequency table, as shown:

 Age (in years) Number of patients Cumulative frequency 0-15 5 5 15-30 20 25 30-45 40 65 45-60 50 115 60-75 25 140 Total $N= \Sigma f_{i}= 140$

Now, N= 140

$\Rightarrow \frac{N}{2} = 70$

The cumulative frequency just greater than 70 is 115 and the corresponding class is 45-60.

Thus the median class is 45-60.

We know,

$Median = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

And l= 45, h= 15, f= 50, N= 140 and cf= 65

$Median = 45+\left ( \frac{\frac{140}{2}-65}{50} \right )\times 15$

$Median = 45+\left ( \frac{70-65}{50} \right )\times 15$

$Median = 45 + 1.5$

$Median = 46.5$

Hence the median age is 46.5 years.

Q2: Compute the median from the following data :

 Marks 0-7 7-14 14-21 21-28 28-35 35-42 42-49 Number of students 3 4 7 11 0 16 9

Sol:

We prepare the cumulative frequency table, as shown:

 Marks Number of students (f) Cumulative frequency 0-7 3 3 7-14 4 7 14-21 7 14 21-28 11 25 28-35 0 25 35-42 16 41 42-49 9 50 $N= \Sigma f = 50$

N= 50

$\Rightarrow \frac{N}{2} = 25$

The cumulative frequency greater than 25 is 41 and the corresponding class is 35-42.

Thus the median class is 35-42.

Therefore, l=35, h = 7, f = 16, Â c.f. of preceding class = 25 and $\frac{N}{2} = 25$

$Median = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$Median = 35 + \left ( \frac{25-25}{16} \right )\times 7$

$Median = 35+ 0$

$Median = 35$

Q3: The following tables shows the daily wages of workers in a factory:

 Daily wages (in Rs.) 0-100 100-200 200-300 300-400 400-500 Number of workers 40 32 48 22 8

Find the median daily wage income of the workers.

Sol

We prepare the cumulative frequency table, as shown:

 Daily wages Number of workers Cumulative frequency 0-100 40 40 100-200 32 72 200-300 48 120 300-400 22 142 400-500 8 150 $N= \Sigma f = 150$

N= 150

$\Rightarrow \frac{N}{2} = 75$

The cumulative frequency greater than 75 is 120 and the corresponding class is 200-300.

Thus the median class is 200-300.

Therefore, l=200, h = 100, f = 48, c.f. of preceding class = 75 and $\frac{N}{2} = 75$

$Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$Median = 200 + \left ( \frac{75-72}{48} \right )\times 100$

$Median = 200+ 6.25$

$Median = 206.25$

Hence, the median daily wage income of the worker is Rs. 206.25

Q4: Calculate the median from the following frequency distribution:

 Class 5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 Frequency 5 6 15 10 5 4 2 2

Sol:

We prepare the cumulative frequency table, as shown:

 Class Frequency (f) Cumulative frequency 5-10 5 5 10-15 6 11 15-20 15 26 20-25 10 36 25-30 5 41 30-35 4 45 35-40 2 47 40-45 2 49 $N= \Sigma f = 49$

N= 49

$\Rightarrow \frac{N}{2} = 24.5$</amp-mathml

The cumulative frequency greater than 24.5 is 26 and the corresponding class is 15-20.

Thus the median class is 15-20.

Therefore, l= 15, h = 5, f = 15, Â c.f. of preceding class = 11 and $\frac{N}{2} = 24.5$

$Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$Median = 15 + \left ( \frac{24.5 – 11}{15} \right )\times 5$

$Median = 15 + 4.5$

$Median = 19.5$

Hence, median is 19.5.

Q5: Given below is the number of units of electricity consumed in a week in a certain locality :

 Consumption (in units) 65-85 85-105 105-125 125-145 145-165 165-185 185-205 Number of consumption 4 5 13 20 14 7 4

Sol:

We prepare the cumulative frequency table, as shown:

 Class Frequency (f) Cumulative frequency 65-85 4 4 85-105 5 9 105-125 13 22 125-145 20 42 145-165 14 56 165-185 7 63 185-205 4 67 $N= \Sigma f = 67$

N= 67

$\Rightarrow \frac{N}{2} = 33.5$

The cumulative frequency greater than 33.5 is 42 and the corresponding class is 125-145.

Thus the median class is 125-145.

Therefore, l= 125, h = 20, f = 20, Â c.f. of preceding class = 22 and $\frac{N}{2} = 33.5$

$Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$Median = 125 + \left ( \frac{33.5 – 22}{20} \right )\times 20$

$Median = 125 + 11.5$

$Median = 136.5$

Hence, median is 136.5 .

Q6: Calculate the median from the following data

 Height 135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175 No. of boys 6 10 18 22 20 15 6 3

Sol:

We prepare the cumulative frequency table, as shown/p>

 Class Frequency (f) Cumulative frequency 135-140 6 6 140-145 10 16 145-150 18 34 150-155 22 56 155-160 20 76 160-165 15 91 165-170 6 97 170-175 3 100 $N= \Sigma f = 100$

N= 100

$\Rightarrow \frac{N}{2} = 50$

The cumulative frequency greater than 50 is 56 and the corresponding class is 150-155.

Thus the median class is 150-155

Therefore, l= 150, h = 5, f = 22, Â c.f. of preceding class = 34 and $\frac{N}{2} = 50$

$Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$Median = 150 + \left ( \frac{50 – 34}{22} \right )\times 5$

$Median = 150 + 3.64$

$Median = 153.64$

Hence, median is 153.64.

Q7: Calculate the missing frequency from the distribution, it being given that the median of the distribution is 24.

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 25 ? 18 7

Sol:

We prepare the cumulative frequency table, as shown:

 Class Frequency (f) Cumulative frequency 0-10 5 5 10-20 25 30 20-30 x x+30 30-40 18 x+48 40-50 7 x+55

Median is 24 which lies in 20-30

Therefore, Median Class = 20-30

Let the unknown frequency be x.

l= 20, h = 10, f = x, Â c.f. of preceding class = 30 and $\frac{N}{2} = \frac{x+55}{2}$

$Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$24 = 20 +\frac{\frac{x+55}{2}-30}{x}\times 10$

$\Rightarrow 24 = 20 +\frac{\frac{x+55-60}{2}}{x}\times 10$

$\Rightarrow 24 = 20 +\frac{x-5}{2x}\times 10$

$\Rightarrow 24 – 20= \frac{x-5}{x}\times 5$

$\Rightarrow 4x = 5x – 25$

$\Rightarrow x = 25$

Hence, the unknown frequency is 25.

Q8: The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.

 Class 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 Frequency 120 a 12 15 b 6 6 4

Sol:

We prepare the cumulative frequency table, as shown:

 Class Frequency$f_{i}$ Cumulative frequency 0-5 12 12 5-10 a 12+a 10-15 12 24+a 15-20 15 39+a 20-25 b 39+a+b 25-30 6 45+a+b 30-35 6 51+a+b 35-40 4 55+a+b Total $N= \Sigma f_{i} = 70$

Let a,b be the missing frequencies of class intervals 5-10 and 20-25 respectively. Then,

55 + a + b = 70

$\Rightarrow a+b = 15$â€¦â€¦â€¦â€¦.(i)

Median is 16, which lies in15-20. So the median class is 15-20.

Therefore, l= 15, h= 5, N= 70, f= 15, and cf = 24+a

Now,

$Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$16 = 15 +\frac{\frac{70}{2}-(24 + a)}{15}\times 5$

$\Rightarrow 16 = 15 + \frac{35 – 24 – a}{3}$

$\Rightarrow 16 – 15 = \frac{11-a} {3}$

$\Rightarrow 1 \times 3 = 11 – a$

$\Rightarrow a = 11 – 3$

$\Rightarrow a = 8$

Therefore, $b = 15 – a$( From (i) )

$\Rightarrow b = 15 – 8$

$\Rightarrow b = 7$

Hence, a= 8 and b = 7.

Q9: In the following data the median of the runs scored by top 60 batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.

 Runs scored 2500-3500 3500-4500 4500-5500 5500-6500 6500-7500 7500-8500 No. of batsmen 5 x y 12 6 2

Sol:

We prepare the cumulative frequency table, as shown:

 Runs scored No. of batsmen Cumulative frequency 2500-3500 5 5 3500-4500 x 5+x 4500-5500 y 5+x+y 5500-6500 12 17+x+y 6500-7500 6 23+x+y 7500-8500 2 25+x+y Total $N= \Sigma f_{i} = 60$

Let x,y be the missing frequencies of class intervals 3500-4500 and 4500-5500 respectively.

Then,

25 + x + y = 60

$\Rightarrow x+y = 35$â€¦â€¦â€¦â€¦.(i)

Median is 5000, which lies in 4500-5500.

So the median class is 4500-5500.

Therefore, l= 4500, h= 1000, N= 60, f= y, and cf = 5+x

Now,

$Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$5000 = 4500 +\frac{\frac{60}{2} – (5 + x)}{y}\times 1000$

$\Rightarrow 5000 – 4500 = \frac{30 – 5 – x}{y} \times 1000$

$\Rightarrow 500 = \frac{25-x} {y} \times 1000$

$\Rightarrow y = 50 – 2x$

$\Rightarrow 35 – x = 50 – 2x$ ( From (i) )

$\Rightarrow 2x – x = 50 – 35$

$\Rightarrow x = 15$

Therefore, $y = 35 – x$

$\Rightarrow y = 35 – 15$

$\Rightarrow y = 20$

Hence, x= 15 and y = 20.

Q10: If the median of the following frequency distribution is 32.5, find the values of $f_{1} \; and\; f_{2}$.

 Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total Frequency $f_{1}$ 5 9 12 $f_{2}$ 3 2 40

Sol:

We prepare the cumulative frequency table, as shown below:

 Class Frequency Cumulative frequency 0-10 $f_{1}$ $f_{1}$ 10-20 5 $f_{1}$+5 20-30 9 $f_{1}$+14 30-40 12 $f_{1}$+26 40-50 $f_{2}$ $f_{1}$+$f_{2}$+26 50-60 3 $f_{1}$+$f_{2}$+29 60-70 2 $f_{1}$+$f_{2}$+31 $N= \Sigma f = 40$

Now, $f_{1}$+$f_{2}$+31 = 40

$\Rightarrow f_{1}+ f_{2} = 9$

$\Rightarrow f_{2} = 9 – f_{1}$â€¦â€¦..(i)

The median is 32.5 which lies in 30 – 40.

Hence, median class = 30 – 40

Therefore, l= 30,$\frac{N}{2} = 20$, f= 12, and cf = $14+f_{1}$

Now,

$Median = 32.5 \; Also, \; Median (M) = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h$

$32.5 = 30 + \frac{20 – (14 + f_{1})}{12} \times 10$

$\Rightarrow 32.5 = 30 + \frac{6 – f_{1}}{12} \times 10$

$\Rightarrow 2.5 = \frac{ 6 – f_{1} } {12} \times 10$

$\Rightarrow 60 – 10f_{1} Â = 30$

$\Rightarrow 10f_{1} = 30$

$\Rightarrow f_{1} Â = 3$

From eq. (i), we have

$f_{2}= 9-3$

$\Rightarrow f_{2}= 6$

Q11:Calculate the median for the following data:

 Age( in years) 19-25 26-32 33-39 40-46 47-53 54-60 Frequency 35 96 68 102 35 4

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

 Class Frequency $f_{i}$ C.F 18.5-25.5 35 35 25.5-32.5 96 131 32.5-39.5 68 199 39.5-46.5 102 301 46.5-53.5 35 336 53.5-60.5 4 340 $N=\sum f_{i}=340$

N =340

$\Rightarrow \frac{N}{2} =170$

Now,

The cumulative frequency just greater than 170 is 199 and corresponding class is 32.5-39.5.

Thus, the median class is 32.5-39.5.

I = 32.5, h = 7, f = 68, c.f. = cf preceding class is 131 and N/2 = 170

Median = $I+\left [ h\times \frac{\left ( \frac{N}{2}-c \right )}{f} \right ]= 32.5+\left [ 7\times \frac{\left (170-131 \right )}{22} \right ]$

= 32.5 + 4 = 36.5

Hence median is 36.5 years.

Q12: Find the median wages for the following frequency distribution:

 Wages per day ( in Rs.) 61-70 71-80 81-90 91-100 101-110 111-120 No. of workers 5 15 20 30 20 8

Sol:

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get/p>

 Wages per day(in Rs.) Frequency $f_{i}$ C.F 60.5-70.5 5 5 70.5-80.5 15 20 80.5-90.5 20 40 90.5-100.5 30 70 100.5-110.5 20 90 110.5-120.5 8 98 $N=\sum f_{i}=98$

N =98 => N/2 = 49

Now,

The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5-100.5.

Thus, the median class is 90.5-100.5

I = 90.5, h = 10, f = 30, c = C.F. preceding median class = 40 and N/2 = 49

Median = $I+\left [ h\times \frac{\left ( \frac{N}{2}-c \right )}{f} \right ]= 90.5+\left [ 10\times \frac{\left (49-40 \right )}{30} \right ]$

= 90.5+3 = Rs. 93.5

Hence, median is Rs. 93.5

Q13: Find the median from the following data:

 Class 1-5 6-10 11-15 16-20 21-25 26-30 31-35 36-40 41-45 Frequency 7 10 16 32 24 16 11 5 2

Sol:

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

 Marks Frequency $f_{i}$ C.F 0.5-5.5 7 7 5.5-10.5 10 17 10.5-15.5 16 33 15.5-20.5 32 65 20.5-25.5 24 89 25.5-30.5 16 105 30.5-35.5 11 116 35.5-40.5 5 121 40.5-45.5 2 123 $N=\sum f_{i}=123$

N =123

=> N/2 = 61.5

Now,

The cumulative frequency just greater than 61.5 is 65 and corresponding class is 15.5-20.5.

Thus, the median class is 15.5-20.5

I = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33 and N/2 = 61.5

Median = $I+\left [ h\times \frac{\left ( \frac{N}{2}-c \right )}{f} \right ]= 15.5+\left [ 5\times \frac{\left (61.5-33 \right )}{32} \right ]$

= 15.5+4.45 = 19.95

Hence, median = 19.95

Q14: Find the median from the following data:

 Marks Frequency $f_{i}$ Below 10 12 Below 20 32 Below 30 57 Below 40 80 Below 50 92 Below 60 116 Below 70 164 Below 80 200

Sol:

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

 Marks Frequency $f_{i}$ C.F 0-10 12 12 10-20 20 32 20-30 25 57 30-40 23 80 40-50 12 92 50-60 24 116 60-70 24 116 70-80 36 200 $N=\sum f_{i}=200$

N =200

=> N/2 = 100

Now,

The cumulative frequency just greater than 100 is 116 and corresponding class is 50-60.

Thus, the median class is 50-60

I = 50, h = 10, f = 24, c = C.F. preceding median class = 92 and N/2 = 100

Median = $I+\left [ h\times \frac{\left ( \frac{N}{2}-c \right )}{f} \right ]= 50+\left [ 10\times \frac{\left (100-92 \right )}{24} \right ]$

$=50+\left [ 10\times \frac{84}{24} \right ]$

= 50 + 3.33 = 53.33

Hence, Median = 53.33

### Key Features of RS Aggarwal Class 10 Solutions Chapter 9 â€“Â  Mean Median Mode Ex 9B (9.2)

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