RS Aggarwal Class 10 Solutions Chapter 9 Mean, Mode and Median Exercise 9B

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	Exercise 9A Exercise 9C Exercise 9D Exercise 9E Exercise 9F
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 (fi)	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 = Σ fi = 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 = Σf = 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 = Σf = 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 = Σf = 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 = Σf = 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 = Σf = 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 (fi)	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 = Σfi = 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 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
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 = Σfi = 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 f₁and f₂.

Class	0 - 10	10 - 20	20 – 30	30 - 40	40 - 50	50 – 60	60 - 70	Total
Frequency	fi	5	9	12	f2	3	2	40

 Solution 

Class	Frequency (f)	Cumulative Frequency (cf)
0 – 10	f1	f1
10 - 20	5	f1 + 5
20 - 30	9	f1 + 14
30 – 40	12	f1 + 26
40 - 50	f2	f1 + f2 + 26
50 - 60	3	f1 + f2 + 29
60 - 70	2	f1 + f2 + 31
	N = Σf = 40

Now, f₁ + f₂ + 31 = 40

⇒ f₁ + f₂ = 9

⇒ f₂ = 9 – f₁

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 + f₁

Now, Median = 32.5.

⇒ l = ((N/2 – cf)/f) × h = 32.5

⇒ 30 + {(20 – 14 + f₁)/12) × 10 = 32.5

⇒ (6 – f₁)/12 × 10 = 2.5

⇒ (60 – 10f₁)/12 = 2.5

⇒ 60 – 10f₁ = 30

⇒ 10f₁ = 30

⇒ f₁ = 3

From equation (i), we have:

f₂ = 9 – 3

⇒ f₂ = 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 = Σf = 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 = Σf = 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 = Σf = 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 = Σf = 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.