RS Aggarwal Solutions Class 10 Ex 9B

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,  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

\(\Rightarrow \frac{N}{2} = 24.5 \)

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,  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,  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,  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,  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

\(\Rightarrow a+b = 15\)………….(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

\(\Rightarrow x+y = 35 \)………….(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

\(\Rightarrow f_{1}+ f_{2} = 9\)

\(\Rightarrow f_{2} = 9 – f_{1}\)……..(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,

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

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

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

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

\(\Rightarrow 60 – 10f_{1}  = 30 \)

\(\Rightarrow 10f_{1} = 30 \)

\(\Rightarrow f_{1}  = 3 \)

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


Practise This Question

Which of the following figures are similar figures?