WBCHSE – Let us Work Out 26.4 – WB Class 10
Q1: Fifteen friends noted their daily expenses for school and other activities. Their expenses were: 15, 16, 17, 18, 17, 19, 17, 15, 15, 10, 17, 16, 15, 16, 18, 11. Find the mode of their daily expenses.
Solution:

Rearrange the data in ascending order:
10, 11, 15, 15, 15, 15, 16, 16, 16, 17, 17, 17, 17, 18, 18, 19 
Create a frequency table:
Expense Amount Frequency 10 1 11 1 15 4 16 3 17 4 18 2 19 1 
Identify the mode:
The expense amount “15” and “17” both appear the most (4 times each).
Answer: The modes of the daily expenses are Rs.15 and Rs.17.
Q2: The heights (in cm) of students in a class were recorded as follows: 131, 130, 130, 132, 131, 133, 131, 134, 131, 132, 132, 131, 133, 130, 132, 130, 133, 135, 131, 135, 131, 135, 130, 132, 135, 135, 134, 133. Determine the mode height of the students.
Solution:

Rearrange the data in ascending order:
130, 130, 130, 130, 131, 131, 131, 131, 131, 131, 132, 132, 132, 132, 132, 133, 133, 133, 133, 134, 134, 135, 135, 135, 135, 135 
Create a frequency table:
Height (cm) Frequency 130 4 131 6 132 5 133 4 134 2 135 5 
Identify the mode:
The height “131 cm” appears most frequently (6 times).
Answer: The mode height of the students is 131 cm.
Q3: Find the mode(s) for each of the following datasets:
(i) 8, 5, 4, 6, 7, 4, 4, 3, 5, 4, 5, 4, 4, 5, 5, 4, 3, 3, 5, 4, 6, 5, 4, 5, 4, 5, 4, 2, 3, 4
(ii) 15, 11, 10, 8, 15, 18, 17, 15, 10, 19, 10, 11, 10, 8, 19, 15, 10, 18, 15, 16, 3, 16, 14, 17, 2
Solution:
(i) 8, 5, 4, 6, 7, 4, 4, 3, 5, 4, 5, 4, 4, 5, 5, 4, 3, 3, 5, 4, 6, 5, 4, 5, 4, 5, 4, 2, 3, 4

Rearrange the data:
2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8 
Frequency Table:
Value Frequency 2 1 3 4 4 12 5 9 6 2 7 1 8 1 
Mode: The value “4” appears most frequently (12 times).
(ii) 15, 11, 10, 8, 15, 18, 17, 15, 10, 19, 10, 11, 10, 8, 19, 15, 10, 18, 15, 16, 3, 16, 14, 17, 2

Rearrange the data:
2, 3, 8, 8, 10, 10, 10, 10, 10, 11, 11, 14, 15, 15, 15, 15, 15, 16, 16, 17, 17, 18, 18, 19, 19 
Frequency Table:
Value Frequency 2 1 3 1 8 2 10 5 11 2 14 1 15 6 16 2 17 2 18 2 19 2 
Mode: The value “10” and “15” appear most frequently (5 and 6 times respectively).
Answer:
(i) The mode of the dataset is 4.
(ii) The modes of the dataset are 10 and 15.
Q4: A shoe store recorded the sizes of shoes sold in a day. The data is represented in the following frequency distribution table. Determine the mode(s) of shoe sizes sold.
Size (x)  2  3  4  5  6  7  8  9 
Frequency (f)  3  4  5  3  5  4  3  2 
Solution:

The data is already arranged in a frequency table.

Identify the mode:
The sizes “4” and “6” both have the highest frequency (5).
Answer: The modes of shoe sizes sold are 4 and 6.
Q5: The following frequency distribution table shows the ages of examinees who took an entrance examination. Determine the modal age group using the mode formula.
Age (year)  No. of examinees (f) 
1618  45 
1820  75 
2022  38 
2224  22 
2426  20 
Solution:

Identify the modal class: The class with the highest frequency (75) is 1820.

Apply the mode formula:
Term Value L (Lower limit of modal class) 18 f1 (Frequency of modal class) 75 f0 (Frequency preceding modal class) 45 f2 (Frequency succeeding modal class) 38 h (Class width) 2 Mode \( = L + (\frac{f_1f_0}{2f_1f_0f_2}) \times h \)
Mode \( = 18 + (\frac{7545}{2 \times 754538}) \times 2 \)
Mode \( = 18 + (\frac{30}{67}) \times 2 \)
Mode ≈ 18.89
Answer: The modal age of the examinees is approximately 18.89 years (approx) old.
Q6: Eighty students took a periodical examination. Their marks are presented in the frequency distribution table below. Find the modal mark range using the mode formula.
Marks  No. of students (f) 
05  2 
510  6 
1015  10 
1520  16 
2025  22 
2530  11 
3035  8 
3540  5 
Solution:

Identify the modal class: The class interval with the highest frequency (22) is 2025.

Organize the values:
Term Value L (Lower limit of modal class) 20 f1 (Frequency of modal class) 22 f0 (Frequency preceding modal class) 16 f2 (Frequency succeeding modal class) 11 h (Class width) 5 
Apply the mode formula:
Mode \( = L + (\frac{f_1f_0}{2f_1f_0f_2}) \times h \)
Mode \( = 20 + (\frac{2216}{2 \times 221611}) \times 5 \)
Mode \( = 20 + (\frac{6}{17}) \times 5 \)
Mode ≈ 21.76
Answer: The modal mark range for this periodical examination is approximately 21.76, which falls within the 2025 mark range.
Q7: Let us find the mode of the frequency distribution table given below.
Class Interval  Frequency (f) 
05  5 
510  12 
1015  18 
1520  28 
2025  17 
2530  12 
3035  8 
Solution:

Identify the modal class: The class interval with the highest frequency (28) is 1520.

Organize the values:
Term Value L (Lower limit of modal class) 15 f1 (Frequency of modal class) 28 f0 (Frequency preceding modal class) 18 f2 (Frequency succeeding modal class) 17 h (Class width) 5 
Apply the mode formula:
Mode \( = L + (\frac{f_1f_0}{2f_1f_0f_2}) \times h \)
Mode \( = 15 + (\frac{2818}{2 \times 281817}) \times 5 \)
Mode \( = 15 + (\frac{10}{21}) \times 5\)
Mode ≈ 17.38
Answer: The mode of this frequency distribution is approximately 17.38.
Q8: Let us find the mode of the frequency distribution table given below.
Class Interval  Frequency (f) 
4554  8 
5564  13 
6574  19 
7584  32 
8594  12 
95104  6 
Solution:
First we will change the table into exclusive limits:
Class Interval (Inclusive)  Class Interval (Exclusive)  Frequency 
4554  44.554.5  8 
5564  54.564.5  13 
6574  64.574.5  19 
7584  74.584.5  32 
8594  84.594.5  12 
95104  94.5104.5  6 
Identify the modal class: The class interval with the highest frequency (32) is 7584.
Organize the values:
Term  Value 
L (Lower limit of modal class)  74.5 
f1 (Frequency of modal class)  32 
f0 (Frequency preceding modal class)  19 
f2 (Frequency succeeding modal class)  12 
h (Class width)  9 
Apply the mode formula:
Mode \( = L + (\frac{f_1f_0}{2f_1f_0f_2}) \times h \)
Mode \( = 74.5 + (\frac{3219}{2 \times 321912}) \times 9 \)
Mode \( = 74.5 + (\frac{13}{33}) \times 9 \)
Mode ≈ 78.44
Answer: The mode of this frequency distribution is approximately 78.44.
Q9A) M.C.Q
(i) The median of a given frequency distribution is found graphically with the help of
(a) Frequency curve
(b) Frequency polygon
(c) Histogram
(d) Ogive
Answer: (d) Ogive
Explanation: An ogive (cumulative frequency graph) is used to determine the median graphically.
(ii) If the mean of numbers \( 6, 7, x, 8, y, 14 \) is \( 9 \), then
(a) x + y = 21
(b) x + y = 19
(c) x − y = 21
(d) x − y = 19
Solution:

Sum of the numbers = 6 + 7 + x + 8 + y + 14 = 35 + x + y

Mean \( = \frac{ \text{Sum}}{ \text{Number of values}} = \frac{35 + x + y}{6} = 9 \)

35 + x + y = 54

x + y = 19
Answer: (b) x+y=19
(iii) If \( 35 \) is removed from the data \( 30, 34, 35, 36, 37, 38, 39, 40 \) then the median increases by
(a) 2
(b) 1.5
(c) 1
(d) 0.5
Solution:

Original Data: 30, 34, 35, 36, 37, 38, 39, 40 (Median \(=\frac{36+37}{2} = 36.5 \))

Data after removing 35: 30, 34, 36, 37, 38, 39, 40 (Median = 37)

Increase in Median = 37 – 36.5 = 0.5
Answer: (d) 0.5
(iv) If the mode of data \( 16, 15, 17, 16, 15, x, 19, 17, 14 \) is \(15 \), then the value of \( x \) is –
(a) 15
(b) 16
(c) 17
(d) 19
Answer: (a) 15
Explanation: The mode is the most frequent value. Since 15 is the mode, x must be 15 to make it appear most often.
(v) If the median of arranging the ascending order of data \( 8, 9, 12, 17, x+2, x+4, 30, 31, 34, 39 \), is \( 15 \), then the value of \( x \) is
(a) 22
(b) 12
(c) 20
(d) 24
Solution:
The data is already nearly in ascending order.
Since the median is 15, the two middle values must average to 15.
The middle values are 17 and (x+2).
∴ \(\frac{17 + x + 2}{2} = 15 \)
⇒ 19 + x = 30
∴ x = 11
Answer: (b) 12
Q9B) True or False
(i) Value of mode of data \( 2, 3, 9, 10, 9, 3, 9 \) is \( 10\).
Answer: False
Explanation: The mode is 9, as it appears most frequently.
(ii) Median of data \(3, 14, 18, 20, 5 \) is \( 18 \).
Answer: False
Solution:

Order the data: 3, 5, 14, 18, 20

The median (middle value) is 14.
Q9C) Fill in the Blanks
(i) Mean, median, mode are the measures of __________.
Answer: Central Tendency
Explanation: These values represent the central or typical value in a dataset.
(ii) If the mean of \( x_1, x_2, x_3,…., x_n \) is \( \overline{x} \), the mean of \( ax_1, ax_2, ax_3,…, ax_n \) is __________.
Answer: \( a \overline{x} \)
Explanation: If you multiply each data point by a constant (‘a’), the mean also gets multiplied by that constant.
(iii) At the time of finding arithmetic mean, the lengths of all classes are __________.
Answer: Equal
Explanation: When calculating the mean for grouped data, we assume the data is evenly distributed within each class interval. This means the class widths should be equal.
Q10. Short Answer Type Questions
(i)
Class  6585  85105  105125  125145  145165  165185  185205 
Frequency  4  15  3  20  14  7  14 
Let us find the difference between the upperclass limit in the median class and the lowerclass limit of the modal class of the above frequency distribution table.
Solution:

Median Class:

Total frequency = 77

Median class is the class containing the \( \frac{77+1}{2} = 39th \ \ value \).

The median class is 125145.

Upper class limit of median class = 145.


Modal Class:

The modal class is the class with the highest frequency, which is 125145.

Lower class limit of modal class = 125.


Difference: 145 – 125 = 20
Answer: The difference between the upperclass limit of the median class and the lowerclass limit of the modal class is 20.
Q10 (ii) The following frequency distribution shows the time taken to complete a 100meter hurdle race of 150 athletes:
Time (seconds)  13.814  1414.2  14.214.4  14.414.6  14.614.8  14.815 
No. of athletes  2  4  5  71  48  20 
Let us find the difference between the upperclass limit of the modal class and the lower class limit of the modal class.
Solution:

Modal Class: The class with the highest frequency (71) is 14.414.6.

Upper Class Limit of Modal Class: 14.6

Lower Class Limit of Modal Class: 14.4

Difference: 14.6 – 14.4 = 0.2
Answer: The difference between the upperclass limit and lowerclass limit of the modal class is 0.2 seconds.
Q10 (iii) The mean of a frequency distribution is \( 8.1 \). If \( \sum f_i x_i= 132+5k \) and \( \sum f_i= 20 \), let us find the value of \(k \).
Solution:
We know the formula for the mean of a frequency distribution is:
Mean (x̄) = \( \frac{\sum f_i x_i}{ \sum f_i} \)
We are given:

Mean (x̄) = 8.1

\( \sum f_i x_i = 132+5k \)

\( \sum f_i = 20 \)
Substitute these values into the formula:
\( 8.1 = \frac{132 + 5k}{20} \)
Solve for k:
162 = 132 + 5k
⇒ 30 = 5k
∴ k = 6
Answer: The value of k is 6.
Q10 (iv) If \( u_i =\frac{x_i25}{10} \), \( \sum f_i u_i = 20 \) and \( \sum f_i= 100 \), let us find the value of \( \overline{x} \).
Solution:
We know the formula for the mean using stepdeviation method is:
\( \overline{x}= A + (\frac{\sum f_i u_i}{ \sum f_i}) \times h \)
Where:

A = Assumed mean

h = Class width
From the given information, we can deduce:

A = 25 (since \( u_i = \frac{x_i25}{10} \))

h = 10
Substitute the values into the formula:
\( x̄ = 25 + \frac{20}{100} \times 10 \)
⇒ x̄ = 25 + 2
∴ x̄ = 27
Answer: The value of x̄ (the mean) is 27.
Q10 (v) Let us write the modal class from the above frequency distribution table.
Marks  less than 10  less than 20  less than 30  less than 40  less than 50  less than 60 
No. of students  3  12  27  57  75  80 
Solution:

Convert the cumulative frequency table to a regular frequency table:
Marks  Frequency (f) 
010  3 
1020  9 
2030  15 
3040  30 
4050  18 
5060  5 

Identify the modal class: The class interval with the highest frequency is 3040.
Answer: The modal class for the given frequency distribution is 3040.