RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9B Class 10 Maths
Chapter Name  RS Aggarwal Chapter 9 Mean, Mode and Median 
Book Name  RS Aggarwal Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 9B Solutions
1. In a hospital, the ages of diabetic patient were recorded as follows. Find the median age.
Age (in years) 
0  15 
15  30 
30  45 
45  60 
60  75 
No. of patients 
5 
20 
40 
50 
25 
Solution
We prepare the cumulative frequency table, as shown below:
Age (in years) 
Number of patients (f_{i}) 
Cumulative Frequency (cf) 
0  15 
5 
5 
15  30 
20 
25 
30  45 
40 
65 
45  60 
50 
115 
60 – 75 
25 
140 
Total 
N = Î£ f_{i }= 140 

Now, N = 140
⇒ 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.
∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + ((N/2 – cf)/f) × h
= 45 + (140/2 – 65)/50 × 15
= 45 + (70 – 65)/50 × 15
= 45 + 1.5
= 46.5
Hence, the median age is 46.5 years.
2. Compute mean 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 
Solution
Class 
Frequency (f) 
Cumulative Frequency (cf) 
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 = Î£f = 50 
Now, N = 50
⇒ N/2 = 25
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35  42
Thus, the median class is 35  42.
∴ l = 35, h = 7, f = 16, cf = c.f. of preceding class=25 and N/2 = 25.
Now,
Median = l + ((N/2 – cf)/f) × h
= 35 + 7 × (25 – 25)/16
= 35 + 0
= 35.
Hence, the median age is 46.5 years
3. The following table shows the daily wages of workers in a factory:
Daily wages in (₹) 
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.
Solution
Class 
Frequency (f) 
Cumulative Frequency (cf) 
0 – 100 
40 
40 
100  200 
32 
72 
200  300 
48 
120 
300  400 
22 
142 
400  500 
8 
150 

N = Î£f = 150 

Now, N = 150
⇒ N/2 = 75
The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 – 300.
Thus, the median class is 200 – 300.
∴ l = 200, h = 100, f = 48, cf = c.f. of preceding class=72 and N/2 = 75.
Now,
Median, M = l + {h × ((N/2 – cf)/f)} × h
= 200 + {100 × (75 – 72)/48}
= 200 + 6.25
= 206.25
Hence, the median daily wage income of the workers is RS 206.25.
4. Calculate the median from the following frequency distribution table:
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 
Solution
Class 
Frequency (f) 
Cumulative Frequency (cf) 
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 = Î£f = 49 
Now, N = 49
⇒ N/2 = 24.5
The cumulative frequency just greater than 24.5 is 26 is 26 and the corresponding class is 15 – 20.
Thus, the median class is 15 – 20.
∴ l = 15, h = 5, f = 15, cf = c.f. of preceding class=11 and N/2 = 24.5.
Now,
Median, M = l + {h × ((N/2 – cf))/f)}
= 15 + {5 × (24.5 – 11)/15)}
= 15 + 4.5
= 19.5
Hence, the median = 19.5.
5. Given below is the number of units of electricity consumed in a week in a certain locality:
Class 
65  85 
85  105 
105  125 
125  145 
145  165 
165  185 
185  200 
Frequency 
4 
5 
13 
20 
14 
7 
4 
Calculate the median.
Solution
Class 
Frequency (f) 
Cumulative Frequency (cf) 
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 = Î£f = 67 

Now, N = 67
⇒ N/2 = 33.5
The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125 – 145.
Thus, the median class is 125  145.
∴ l = 125, h = 20, f = 20,
cf = c.f. of preceding class =22 and N/2 = 33.5.
Now,
Median, M = l + {h
Median, M = l + {h × ((N/2 – cf))/f)}
= 125 + {20 × (33.5 – 22)/20)}
= 125 + 11.5
= 136.5
Hence, the median = 136.5.
6. Calculate the median from the following data:
Height (in cm) 
135  140 
140  145 
145  150 
150  160 
155  160 
160  165 
165  170 
170  175 
Frequency 
6 
10 
18 
22 
20 
15 
6 
3 
Solution
Class 
Frequency (f) 
Cumulative Frequency (cf) 
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 = Î£f = 100 

Now, N = 100
⇒ N/2 = 50.
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 – 155.
Thus, the median class is 150 – 155.
∴ l = 150, h = 5, f = 22, cf = c.f. of preceding class=34 = N/2 = 50.
Now,
Median, M = l + {h × ((N/2 – cf)/f)}
= 150 + {5 × ((50 – 34)/22)}
= 150 + 3.64
= 153.64
Hence, the median = 153.64
7. Calculate the missing frequency from the following distribution, it being given that the median of distribution is 24.
Class 
0  10 
10  20 
20  30 
30  40 
40  50 
Frequency 
5 
25 
? 
18 
7 
Solution
Class 
Frequency (f_{i}) 
Cumulative Frequency (cf) 
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
∴ Median class=20 – 30
Let the unknown frequency be x.
Here, l = 20, n/2 = (x + 55)/2, c.f. of the preceding class=c.f = 30, f = x, h = 10
Now,
Median, M = {l + (n/2 – cf)/f} × h
⇒ 24 = 20 + ((x + 55)/2 – 30)/x × 10
⇒ 24 = 20 + ((x + 55 – 60)/2)/x × 10
⇒ 24 = 20 + (x – 5)/2x × 10
⇒ 24 = 20 + (5x – 25)/x
⇒ 24 = (20x + 5x – 25)/x
⇒ 24x = 25x – 25
⇒ x =  25
⇒ x = 25
Hence, the unknown frequency is 25.
8. 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 
12 
a 
12 
15 
b 
6 
6 
4 
Solution
Class 
Frequency (f) 
Cumulative Frequency (cf) 
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 = Î£f_{i} = 70 
Let a and b be the missing frequencies of class intervals 5 – 10 and 20 – 25 respectively.
Then, 55 + a + b = 70 ⇒ a + b = 15 ...(1)
Median is 16, which lies in 15 – 20. So, the median class is 15 – 20.
∴ l = 15, h = 5, N = 70, f = 15 and cf = 24 + a
Now,
Median, M = l + (N/2 – cf/f)) × h
⇒ 16 = 15 + (70/2 – (24 + a)/15) × 5
⇒ 16 = 15 + (35 – 24 – a)/3
⇒ 16 = 15 + (11 – a)/3
⇒ 16 – 15 = (11 – a)/3
⇒ 1 × 3 = 11 – a
⇒ a = 11 – 3
⇒ a = 8
∴ b = 15 – a [From (1)]
⇒ b = 15 – 8
⇒ b = 7
Hence, a = 8, and b = 7.
9. In the following data the median of the runs scored by 60 top batsmen of the world in oneday 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 
Number of batsman 
5 
x 
y 
12 
6 
2 
Solution
We prepare the cumulative frequency table, as shown below:
Runs scored 
Number of batsman (fi) 
Cumulative Frequency (cf) 
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 = Î£f_{i} = 60 
Let x and y be the missing frequencies of class intervals 3500 – 4500 respectively. Then,
25 + x + y = 60 ⇒ x + y = 35 ...(1)
Median is 5000, which lies in 4500 – 5500. So, the median class is 4500 – 5500.
∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x
Now,
Median, M = l + ((N/2 – cf)/f) × h
⇒ 5000 = 4500 + (60/2 – (5 + x)/y) × 1000
⇒ 5000 – 4500 = (30 – 5 – x)/y × 1000
⇒ 500 = (25 – x)/y × 1000
⇒ y = 50 – 2x
⇒ 35 – x = 50 – 2x [From (1)]
⇒ 2x – x = 50 – 35
⇒ x = 15
∴ y = 35 – x [From (1)]
⇒ y = 35 – 15
⇒ y = 20
Hence, x = 15 and y = 20.
10. If the median of the following frequency distribution is 32.5, find the values of f_{1}and f_{2}.
Class 
0  10 
10  20 
20 – 30 
30  40 
40  50 
50 – 60 
60  70 
Total 
Frequency 
f_{i} 
5 
9 
12 
f_{2} 
3 
2 
40 
Solution
Class 
Frequency (f) 
Cumulative Frequency (cf) 
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 = Î£f = 40 

Now, f_{1} + f_{2} + 31 = 40
⇒ f_{1} + f_{2} = 9
⇒ f_{2} = 9 – f_{1}
The median is 32.5 which lies in 30 – 40.
Hence, Median class=30  40
Hence, l = 30, N/2 = 40/2 = 20, f = 12 and cf = 14 + f_{1}
Now, Median = 32.5.
⇒ l = ((N/2 – cf)/f) × h = 32.5
⇒ 30 + {(20 – 14 + f_{1})/12) × 10 = 32.5
⇒ (6 – f_{1})/12 × 10 = 2.5
⇒ (60 – 10f_{1})/12 = 2.5
⇒ 60 – 10f_{1} = 30
⇒ 10f_{1} = 30
⇒ f_{1} = 3
From equation (i), we have:
f_{2} = 9 – 3
⇒ f_{2} = 6
11. Calculate the median for the following data:
Class 
19  25 
26  32 
33  39 
40  46 
47  53 
54  60 
Frequency 
35 
96 
68 
102 
35 
4 
Solution
First, we will convert the data into exclusive form.
Class 
Frequency (f) 
Cumulative frequency (cf) 
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 
435 
340 
N = Î£f = 340 
Now, N = 340
⇒ N/2 = 70
The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 – 39.5.
Thus, the median class is 32.5 – 39.5
∴ l = 32.5, h = 7, f = 68, cf = c.f. of preceding class=131 and N/2 = 170.
∴ Median, M = l + {h × (N/2 – cf/f))}
= 32.5 + {7 × ((170 – 131)/68)}
= 32.5 + 4.01
= 36.51
Hence, the median = 36.51
12. Find the median wages for the following frequency distribution:
Wages per day (in ₹) 
61  70 
71  80 
81  90 
91  100 
101  110 
111 – 120 
No. of women workers 
5 
15 
20 
30 
20 
8 
Solution
Class 
Frequency (f) 
Cumulative Frequency (cf) 
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 = Î£f = 98 
Now, N = 98
⇒ N/2 = 49
The cumulative frequency just greater than 49 is 70 and the corresponding class is 90.5 – 100.5.
Thus, the median class is 90.5 – 100.5.
Now, l = 90.5, h = 10, f = 30, cf = c.f of preceding class=40 and N/2 = 49.
∴ Median, M = l + {h × (N/2 – cf/f))}
= 90.5 + {10 × ((49 – 40)/30)}
= 90.5 + 3
= 93.5
Hence, median wages = Rs 93.50.
13. Find the median from the following data:
Class 
1  5 
6  10 
11  15 
16  20 
21  25 
26  30 
31  35 
35  40 
40  45 
Frequency 
7 
10 
16 
32 
24 
16 
11 
5 
2 
Solution
Converting into exclusive form we get:
Class 
Frequency (f) 
Cumulative Frequency (cf) 
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 – 33.5 
11 
116 
35.5 – 40.5 
5 
121 
40.5 – 45.5 
2 
123 
N = Î£f = 123 
Now, N = 123
⇒ N/2 = 61.5
The cumulative frequency just greater than 61.5 is 65 and the corresponding class is 15.5 – 20.5.
Thus, the median class is 15.5 – 20.5
∴ l = 15.5, h = 5, f = 32, cf = c.f. of preceding class=33 and N/2 = 61.5.
∴ Median, M = l + {h × (N/2 – cf/f))}
= 15.5 + {5 × ((61.5 – 33)/32)}
= 15.5 + 4.45
= 19.95
Hence, median = 19.95.
14. Find the median from the following data:
Marks 
No. of students 
Below 10 
12 
Below 20 
32 
Below 30 
57 
Below 40 
80 
Below 50 
92 
Below 60 
116 
Below 70 
164 
Below 80 
200 
Solution
Class 
Cumulative frequency (cf) 
Frequency (f) 
0  10 
12 
12 
10  20 
32 
20 
20  30 
57 
25 
30  40 
80 
23 
40  50 
92 
12 
50  60 
116 
24 
60 – 70 
164 
48 
70  80 
200 
36 
N = Î£f = 200 
Now, N = 200
⇒ N/2 = 100
The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 – 60.
Thus, the median class is 50 – 60.
∴ l = 50, h = 10, f = 24,
cf = c.f. pf preceding class=92 and N/2 = 100.
∴ Median, M = l + {h × (N/2 – cf/f))}
= 50 + {10 × (100 – 92)/24)}
= 50 + 3.33
= 53.33
Hence, median = 53.33.