Maths Ex 14.2 Class 10 NCERT Solution

Maths Ex 14.2 Class 10 is about Mode of Group Data. Here are the Solutions of NCERT Textbook of Maths Class 10

Class 10 Maths Chapter 14

Chapter 14 of Mathematics NCERT Textbook is “STATISTICS” . In this chapter , we study about the Mean of grouped data, medium of grouped data and mode of the grouped data. This chapter also teaches to find the cumulative frequencies values, draw the cumulative frequencies curves , etc . This chapter has 4 exercise which have questions based on statistics. The solution of each exercise of Maths Class 10 NCERT Textbook are provided below.

Class 10 Mathematics Exercise 14.1 – Mean of Grouped Data
Class 10 Mathematics Exercise 14.2 – Mode of Grouped Data
Class 10 Mathematics Exercise 14.3 – Median of Grouped Data
Class 10 Mathematics Exercise 14.4 – Graphical Representation of Cumulative Frequency Distribution

Maths Ex 14.2 Class 10 NCERT Solutions

SubjectMathematics
Class 10
Chapter NameChapter 14 ( Statistics )
Exercise Exercise 14.2 (Mode of Grouped Data)
Total Questions6 Questions
Text BookNCERT Text Book (Updated Text Book )

Exercise 14.2 Class 10 Maths Question 1
Q1 The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Age (in Years) 5 -1515 – 2525 – 3535 – 4545 – 5555 – 65
Number of Patients6112123145

Solution :
First Calculating Mode
Maximum Frequency = 23
Modal Class = 35 – 45
Lower Limit of Modal Frequency (L) =35
Upper Limit of Modal Frequency (H) =
Size (h) = H – L = 45-35 = 10
Frequency of Modal Class (F1) = 23
Frequency of Class proceeding the modal class (F0) = 21
Frequency of Class succeeding the modal Class (F2) = 14
We know the Formula to calculate of Mode
Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)
Putting all the values in above formula of Mode,
we get
Mode = \( 35\ +\ \frac{23\ -\ 21}{2 \times\ 23\ -\ 21\ -\ 14}\ \times\ 10 \)
Mode = \( 35\ +\ \frac{2}{11}\ \times\ 10 \)
Mode = 35 + 1.8
Mode = 36.8 years

Now calculating Mean Using Step Deviation Method
let us take a = 40 and h = 10

Age (in years)Number of patients (fi)Class Marks (xi)di = xi – 40ui = (xi – 40) / 10 fiui
5-15610-30-3-18
15-251120-20-2-22
25-352130-10-1– 21
35-452340000
45-55145010114
55-6556020210
TOTAL\( \sum \) fi = 80\( \sum \)fiui = -37

Mean = \( a\ +\ \left(\frac{\sum{f_iu_i}}{\sum f_i}\right)\ \times\ h \)
Mean = \( 40\ +\ \left(\frac{-37}{80}\right)\ \times\ 10 \)
Mean = 35.37

Exercise 14.2 Class 10 Maths Question 2
Q2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components . Determine the modal lifetimes of the components.

Lifetimes(in hours)0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120
Frequency103552613829

Solution :
Maximum Frequency (F1) = 61
Modal Class = 60-80
Therefore , L = 60 and H = 80
Size (h) =H – L = 80 – 60 = 20
Frequency of Modal Class (F1) = 61
Frequency of Class proceeding the modal class (F0) = 52
Frequency of Class succeeding the modal Class (F2) = 38
Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)
Mode = \( 60\ +\ \frac{61\ -\ 52}{2\ \times\ 61\ -\ 52\ -\ 38}\ \times\ 20 \)
Mode = 60 + 5.625 = 65.625

Exercise 14.2 Class 10 Maths Question 3
Q3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure

Expenditure (in Rs)Number of Families
1000 – 150024
1500 – 200040
2000 – 250033
2500 – 3000 28
3000 – 350030
3500 – 400022
4000 – 4500 16
4500 – 50007

Solution :
Calculating Mode
Class with maximum frequency ( Modal Class ) = 1500 – 2000
L = 1500
H = 2000
Class Size (h) = H – L = 2000 – 1500 = 500
Frequency of Modal Class (F1) = 40
Frequency of Class proceeding the modal class (F0) = 24
Frequency of Class succeeding the modal Class (F2) = 33
Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)
Mode = \( 1500\ +\ \frac{40\ -\ 24}{2 \times\ 40\ -\ 24\ -\ 33}\ \times\ 500 \)
Mode = 1500 + 347.826
Mode = 1847.826

Now, Calculating Mean
let a = 2750

Expenditure (in Rs)Number of Families(fi)xidi = xi – auifiui
1000 – 1500241250-1500-3-72
1500 – 2000401750-1000-2-80
2000 – 2500332250-500-1-33
2500 – 3000 282750 = a000
3000 – 3500303250500130
3500 – 40002237501000244
4000 – 4500 1642501500348
4500 – 5000747502000428
Total\( \sum \) fi =200\( \sum \) fidi = -35

Mean = \( a\ +\ \left(\frac{\sum{f_iu_i}}{\sum f_i}\right)\ \times\ h \)
Mean = \( 2750\ +\ \left(\frac{-35}{ 200}\right)\ \times\ 500 \)
Mean = 2750 – 87.5
Mean = 2662.5

Exercise 14.2 Class 10 Maths Question 4
Q4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of Students per teacherNumber of states / U.T.
15-203
20-258
25-309
30-3510
35-403
40-450
45-500
50-552

Solution :
Calculating Mode
Class with maximum frequency ( Modal Class ) = 30 – 35
L = 30
H = 35
Class Size (h) = H – L = 35 – 30 = 5
Frequency of Modal Class (F1) = 10
Frequency of Class proceeding the modal class (F0) = 9
Frequency of Class succeeding the modal Class (F2) = 3
Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)
Mode = \( 30\ +\ \frac{10\ -\ 9}{2 \times\ 10\ -\ 9\ -\ 3}\ \times\ 5 \)
Mode = 30 + 0.625
Mode = 30.628

Now, Calculating Mean

Number of Students per teacherNumber of states / U.T.(fi)xidi = xi – auifiui
15-20317.5-15-3-9
20-25822.5-10-2-16
25-30927.5-5-1-9
30-351032.5 =a 000
35-40337.5513
40-45048.51020
45-50047.51530
50-55252.52048
TOTAL\( \sum \) fi = 35\( \sum \) fiui = -23

Mean = \( a\ +\ \left(\frac{\sum{f_iu_i}}{\sum f_i}\right)\ \times\ h \)
Mean = \( 32.5\ +\ \left(\frac{-23}{ 35}\right)\ \times\ 5 \)
Mean = 32.5 – 3.28
Mean = 29.22

Exercise 14.2 Class 10 Maths Question 5
Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.

Runs ScoredNumber of Batsmen
3000 -40004
4000-500018
5000-60009
6000-70007
7000-80006
8000-90003
9000-100001
10000-110001

Solution :
Calculating Mode
Maximum Frequency = 18
Modal Class = 4000 – 5000
Therefore , L = 4000 and H = 5000
Size (h) =H – L = 5000 – 4000 = 1000
Frequency of Modal Class (F1) = 18
Frequency of Class proceeding the modal class (F0) = 4
Frequency of Class succeeding the modal Class (F2) = 9
Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)
Mode = \( 4000\ +\ \frac{18\ -\ 4}{2 \times\ 18\ -\ 4\ -\ 9}\ \times\ 1000 \)
Mode = 4000 + 608.7
Mode = 4608.7

Exercise 14.2 Class 10 Maths Question 6
Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Number of Cars0-1010-2020-3030-4040-5050-6060-7070-80
Frequency71413122011158

Solution :
Calculating Mode
Maximum Frequency = 20
Modal Class = 40 – 50
Therefore , L = 40 and H = 50
Size (h) =H – L = 50 – 40 = 100
Frequency of Modal Class (F1) = 20
Frequency of Class proceeding the modal class (F0) = 12
Frequency of Class succeeding the modal Class (F2) = 11
Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)
Mode = \( 40\ +\ \frac{20\ -\ 12}{2 \times\ 20\ -\ 12\ -\ 11}\ \times\ 10 \)
Mode = 40 + 4.7
Mode = 44.7