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 |

Answer:

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)

- These solutions are easy to understand and explained in a simple language.
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- It provides different techniques to solve difficult and tricky questions.
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