# 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 9AExercise 9CExercise 9DExercise 9EExercise 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.