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:

1. Rearrange the data in ascending order:
   10, 11, 15, 15, 15, 15, 16, 16, 16, 17, 17, 17, 17, 18, 18, 19

2. Create a frequency table:

   Expense Amount	Frequency
   10	1
   11	1
   15	4
   16	3
   17	4
   18	2
   19	1

3. 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:

1. 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

2. Create a frequency table:

   Height (cm)	Frequency
   130	4
   131	6
   132	5
   133	4
   134	2
   135	5

3. 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

1. 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

2. Frequency Table:

   Value	Frequency
   2	1
   3	4
   4	12
   5	9
   6	2
   7	1
   8	1

3. 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

1. 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

2. 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

3. 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.

Solution:

1. The data is already arranged in a frequency table.

2. 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.

Solution:

1. Identify the modal class: The class with the highest frequency (75) is 18-20.

2. 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 \( = L + (\frac{f_1-f_0}{2f_1-f_0-f_2}) \times h \)
   Mode \( = 18 + (\frac{75-45}{2 \times 75-45-38}) \times 2 \)
   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.

Solution:

1. Identify the modal class: The class interval with the highest frequency (22) is 20-25.

2. 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

3. Apply the mode formula:

   Mode \( = L + (\frac{f_1-f_0}{2f_1-f_0-f_2}) \times h \)
   Mode \( = 20 + (\frac{22-16}{2 \times 22-16-11}) \times 5 \)
   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 20-25 mark range.

Q7: Let us find the mode of the frequency distribution table given below.

Solution:

1. Identify the modal class: The class interval with the highest frequency (28) is 15-20.

2. 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

3. Apply the mode formula:

   Mode \( = L + (\frac{f_1-f_0}{2f_1-f_0-f_2}) \times h \)
   Mode \( = 15 + (\frac{28-18}{2 \times 28-18-17}) \times 5 \)
   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.

Solution:

First we will change the table into exclusive limits:

Identify the modal class: The class interval with the highest frequency (32) is 75-84.

Organize the values:

Apply the mode formula:

Mode \( = L + (\frac{f_1-f_0}{2f_1-f_0-f_2}) \times h \)
Mode \( = 74.5 + (\frac{32-19}{2 \times 32-19-12}) \times 9 \)
Mode \( = 74.5 + (\frac{13}{33}) \times 9 \)
Mode ≈ 78.44

Answer: The mode of this frequency distribution is approximately 78.44.

Q9-A) 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

Q9-B) 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.

Q9-C) 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)

Let us find the difference between the upper-class limit in the median class and the lower-class limit of the modal class of the above frequency distribution table.

Solution:

1. Median Class:

   • Total frequency = 77

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

   • The median class is 125-145.

   • Upper class limit of median class = 145.

2. Modal Class:

   • The modal class is the class with the highest frequency, which is 125-145.

   • Lower class limit of modal class = 125.

3. Difference: 145 – 125 = 20

Answer: The difference between the upper-class limit of the median class and the lower-class limit of the modal class is 20.

Q10 (ii) The following frequency distribution shows the time taken to complete a 100-meter hurdle race of 150 athletes:

Let us find the difference between the upper-class limit of the modal class and the lower class limit of the modal class.

Solution:

1. Modal Class: The class with the highest frequency (71) is 14.4-14.6.

2. Upper Class Limit of Modal Class: 14.6

3. Lower Class Limit of Modal Class: 14.4

4. Difference: 14.6 – 14.4 = 0.2

Answer: The difference between the upper-class limit and lower-class 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 ​\( u_i =\frac{x_i-25}{10} \)​, ​\( \sum f_i u_i = 20 \)​ and ​\( \sum f_i= 100 \)​, let us find the value of  ​\( \overline{x} \)​.

Solution:

We know the formula for the mean using step-deviation 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 ​\( u_i = \frac{x_i-25}{10} \)​)

• 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.

Solution:

1. Convert the cumulative frequency table to a regular frequency table:

1. Identify the modal class: The class interval with the highest frequency is 30-40.

Answer: The modal class for the given frequency distribution is 30-40.