# 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 9AExercise 9BExercise 9DExercise 9EExercise 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, xk = 26, h = 4, fk = 25, fk+1 = 22

∴ Mode, M0 = xk + {h × (f k – fk – 1)/(2fk – fk - 1 – fk+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, xk = 1500, h = 500, fk = 40, fk-1 = 24, fk+1 = 31

∴ Mode, M0 = xk + {h × (fk – fk-1)/(2fk – fk-1 – fk+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, xk = 5000, h = 5000, fk = 150, fk+1 = 100

∴ Mode, M0 = xk + {h × (fk – fk -1)/(2fk – fk-1 – fk+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, xk = 15, h = 5, fk = 24, fk-1 = 18, fk + 1 = 17

∴ Mode, M0 = xk + {h × (fk – fk -1)/(2fk – fk-1 – fk+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, xk = 85, h = 10, fk = 32, fk+1 = 6

∴ Mode, M0 = xk + {h × (fk – fk-1)/(2fk – fk-1 – fk+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, xk = 20.5, h = 5, fk = 28, fk-1 = 18, fk+1 = 20

∴ Mode, M0 = xk + {h × (fk – fk-1)/(2fk – fk-1 – fk+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.