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

RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9C 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 9B Exercise 9D Exercise 9E Exercise 9F
Related Study	NCERT Solutions for Class 10 Maths

Exercise 9C Solutions

1. Find the mode of the following distribution:

Marks	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60
Frequency	12	35	45	25	13

Solution

Here, the maximum class frequency is 45, and the class corresponding to this frequency is 30 – 40. So, the modal class is 30 – 40.

Now,

Modal class=30 – 40, lower limit (l) of modal class=30, class size (h) = 10,

Frequency (f₁) of the modal class=45,

Frequency (f₀) of class preceding the modal class=35,

frequency (f₂) of class succeeding the modal class=25

Now, let us substitute these values in the formula:

Mode = l + (f₁ – f₀)/(2f₁ – f₀ – f₂) × h

= 30 + (45 – 35)/(90 – 35 – 45) × 10

= 30 + (10/30) × 10

= 30 + 3.33

= 33.33

Hence, the mode is 33.33.

2. Compute the mode of the following data:

Class	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100
Frequency	25	16	28	20	5

Solution

Here, the maximum class frequency is 28, and the class corresponding to this frequency is 40 – 60. So, the modal class is 40 – 60.

Now,

Modal class = 40 – 60, lower limit (l) of modal class=40, Class size (h) = 20,

frequency (f₁) of the modal class = 28,

frequency (f₀) of class preceding the modal class = 16,

frequency (f₂) of class succeeding the modal class = 20

Now, let us substitute these values in the formula:

Mode = l + (f₁ – f₀)/(2f₁ – f₀ – f₂) × h

= 40 + (28 – 16)/(56 – 16 – 20) × 20

= 40 + (12/20) × 20

= 40 + 12

= 52.

Hence, the mode is 52.

3. Heights of students of class X are given in the flowing frequency distribution

Height (in cm)	150 - 155	155 - 160	160 - 165	165 - 170	170 - 175
Number of students	15	8	20	12	5

Find the modal height.

Also, find the mean height. compared and intercept the two measures of central tendency.

Solution

Here, the maximum class frequency is 20, and the class corresponding to this frequency is 160 – 165. So, the modal class is 160 – 165.

Now,

Modal class=160 – 165, lower limit (l) of modal class=160, class size (h) = 5,

frequency (f₁) of the modal class=20,

frequency (f₀) of class preceding the modal class=8

frequency (f₂) of class succeeding the modal class=12

Now, let us substitute these values in the formula:

Mode = l + (f₁ – f₀)/(2f₁ – f₀ – f₂) × h

= 160 + (20 – 8)/(40 – 8 – 12) × 5

= 160 + (12/20) × 5

= 160 + 3

= 163

Hence, the mode is 163.

It represents that the height of maximum number of students is 163 cm.

Now, to find the mean let us put the data in the table given below:

Height (in cm)	Number of students (f1)	Class mark (xi)	fi xi
150 - 155	15		2287.5
155 - 160	8		1260
160 - 165	20		3250
165 - 170	12		2010
170 - 175	5		862.5
Total	Σfi = 60		Σfi xi = 9670

= 9760/60

= 161.17

Thus, mean of the given data is 161.17.

It represents that on an average, the height of a student is 161.17 cm.

4. Find the mode of the following distribution: 

Class interval	10 - 14	14 - 18	18 - 22	22 - 26	26 - 30	30 - 34	34 - 38	38 - 42
Frequency	8	6	11	20	25	22	10	4

Solution

As the class 26 – 30 has the maximum frequency, it is the modal class.

Now, x_k = 26, h = 4, f_k = 25, f_k+1 = 22

∴ Mode, M₀ = x_k + {h × (f _k – f_{k – 1})/(2f_k – f_{k - 1} – f_k+1)}

= 26 + {4 × (25 – 20)/(2 × 25 – 20 – 22)}

= 26 + {4 × 5/8}

= (26 + 2.5)

= 28.5

5. Given below is the distribution of total household expenditure of 200 manual workers in a city:

Expenditure (in Rs)	Number of manual workers
1000 - 1500	24
1500 - 2000	40
2000  - 2500	31
2500 - 3000	28
3000 – 3500	32
3500 – 4000	23
4000 – 4500	17
4500 - 5000	5

Find the average expenditure done by maximum number of manual workers.

Solution

As the class 1500 – 2000 has the maximum frequency, it is the modal class.

Now, x_k = 1500, h = 500, f_k = 40, f_k-1 = 24, f_k+1 = 31

∴ Mode, M₀ = x_k + {h × (f_k – f_k-1)/(2f_k – f_k-1 – f_k+1)}

= 1500 + {500 × (40 – 24)/(2 × 40 – 24 – 31)}

= 1500 + {500 × 16/25}

= (1500 + 320)

= 1820

Hence, mode = Rs 1820

6. Calculate the mode from the following data:

Monthly salary (in Rs)	No of employees
0 - 5000	90
5000 - 10000	150
10000 - 15000	100
15000 - 20000	80
20000 - 25000	70
25000 - 30000	10

Solution

As the class 5000 – 10000 has the maximum frequency, it is the modal class.

Now, x_k = 5000, h = 5000, f_k = 150, f_k+1 = 100

∴ Mode, M₀ = x_k + {h × (f_k – f_{k -1})/(2f_k – f_k-1 – f_k+1)}

= 5000 + {5000 × (150 – 90)/(2 × 150 – 90 -100)}

= 5000 + {5000 × 60/110}

= (5000 + 2727.27)

= 7727.27

Hence, mode = Rs 7727.27

7. Compute the mode from the following data:

Age (in years)	0 - 5	5 - 10	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35
No. of patients	6	11	18	24	17	13	5

Solution

As the class 15 – 20 has the maximum frequency, it is the modal class

Now, x_k = 15, h = 5, f_k = 24, f_k-1 = 18, f_{k + 1} = 17

∴ Mode, M₀ = x_k + {h × (f_k – f_{k -1})/(2f_k – f_k-1 – f_k+1)}

= 15 + {5 × (24 – 18)/(2 × 24 – 18 – 17)}

= 15 + {5 × 6/13}

= (15 + 2.3)

= 17.3

Hence, mode = 17.3 years

8. Compute the mode from the following series:

Size	45 - 55	55 - 65	65 - 75	75 - 85	85 - 95	95 - 105	105 - 115
Frequency	7	12	17	30	32	6	10

Solution

As the class 85 – 95 has the maximum frequency, it is modal class.

Now, x_k = 85, h = 10, f_k = 32, f_k+1 = 6

∴ Mode, M₀ = x_k + {h × (f_k – f_k-1)/(2f_k – f_k-1 – f_k+1)}

= 85 + {10 × (32 – 30)/(2 × 32 – 30 – 6)}

= 85 + {10 × 2/28}

= (85 + 0.71)

= 85.71

Hence, mode = 85.71

9. Compute the mode from the following data:

Class interval	1 - 5	6 - 10	11 - 15	16 - 20	21 - 25	26 - 30	31 - 35	36 - 40	41 - 45	46 - 50
Frequency	3	8	13	18	28	20	13	8	6	4

Solution

Clearly, we have to find the mode of the data. The given data is an inclusive series. So, we will convert it to an exclusive form as given below:

Class Interval	0.5 – 5.5	5.5 – 10.5	10.5 – 15.5	15.5 - 20.5	20.5 – 25.5	25.5 – 30.5	30.5 – 35.5	35.5 – 40.5	40.5 – 45.5	45.5 – 50.5
Frequency	3	8	13	18	28	20	13	8	6	4

As the class 20.5 – 25.5 has the maximum frequency, it is modal class.

Now, x_k = 20.5, h = 5, f_k = 28, f_k-1 = 18, f_k+1 = 20

∴ Mode, M₀ = x_k + {h × (f_k – f_k-1)/(2f_k – f_k-1 – f_k+1)}

= 20.5 + {5 × (28 – 18)/(2 × 28 – 18 – 20)}

= 20.5 + {5 × 10/18}

= (20.5 + 2.78)

= 23.78

Hence, mode = 23.28.

10. The age wise participation of students in the annual function of a school is shown in the following distribution.

Age (in years)	5 - 7	7 - 9	9 - 11	11 - 13	13 - 15	15 - 17	17 - 19
Number of students	x	15	18	30	50	48	x

Find the missing frequencies when the sum of frequencies is 181. Also find the mode of the data.

Solution

It is given that the sum of frequencies is 181.

∴ x + 15 + 18 + 30 + 50 + 48 + x = 181

⇒ 2x + 161 = 181

⇒ 2x = 181 – 161

⇒ 2x = 20

⇒ x = 10

Thus, x = 10

Here, the maximum class frequency is 50, and the class corresponding to this frequency is 13 – 15. So, the modal class is 13 – 15.

Now,

Modal class=13 – 15, lower limit (l) of modal Class = 13, class size (h) = 2,

frequency (f₁) of the modal class=50,

frequency (f₀) of class preceding the modal class=30,

frequency (f₂) of class succeeding of modal class=48

Now, let us substitute these values in the formula:

Mode = l + (f₁ – f₀)/(2f₁ – f₀ – f₂) × h

= 13 + (50 – 30)/(100 – 30 – 48) × 2

= 13 + (20/22) × 2

= 13 + 1.82

= 14.82

Hence, the mode is 14.82.