Class 10 Maths 1 Statistics Practice Set 6.3 Solutions | Maharashtra Boarda

Class 10 Mathematics 1 – Statistics

Practice Set 6.3 – Questions and Answers

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In this article, you will find the complete solutions to Class 10 Mathematics 1, Chapter 6: Statistics, Practice Set 6.3, according to the Maharashtra State Board syllabus.

These solutions explain how to calculate the mode of grouped frequency distributions using the appropriate formula. All four questions are solved step by step to help students understand the method and prepare for the SSC Board examination.

Read the solutions carefully and practise the problems to improve your mathematical skills.

 

Class 10 Mathematics 1 – Statistics

Practice Set 6.3 – Questions and Answers

Formula for Mode:

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

Where:

• l = Lower limit of the modal class
• f₁ = Frequency of the modal class
• f₀ = Frequency of the preceding class
• f₂ = Frequency of the succeeding class
• h = Class width

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Question 1

The following table shows the information regarding the milk collected from farmers at a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.

Content of fat (%)	2–3	3–4	4–5	5–6	6–7
Milk collected (litres)	30	70	80	60	20

Solution:

The highest frequency is 80.

Modal class = 4–5

Here,

l = 4, f₁ = 80, f₀ = 70, f₂ = 60, h = 1

Using the formula,

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

Mode = 4 + [(80 − 70) / (2 × 80 − 70 − 60)] × 1

Mode = 4 + 10 / (160 − 130)

Mode = 4 + 10 / 30

Mode = 4 + 0.3333

Final Answer: Mode ≈ 4.33%

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Question 2

Electricity used by some families is shown in the following table. Find the mode of electricity usage.

Use of electricity (units)	0–20	20–40	40–60	60–80	80–100	100–120
Number of families	13	50	70	100	80	17

Solution:

The highest frequency is 100.

Modal class = 60–80

Here,

l = 60, f₁ = 100, f₀ = 70, f₂ = 80, h = 20

Using the formula,

Mode = 60 + [(100 − 70) / (2 × 100 − 70 − 80)] × 20

Mode = 60 + 30 / (200 − 150) × 20

Mode = 60 + (30 / 50) × 20

Mode = 60 + 12

Final Answer: Mode = 72 units

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Question 3

The grouped frequency distribution of the supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.

Milk supplied (litres)	1–3	3–5	5–7	7–9	9–11	11–13
Number of hotels	7	5	15	20	35	18

Solution:

The highest frequency is 35.

Modal class = 9–11

Here,

l = 9, f₁ = 35, f₀ = 20, f₂ = 18, h = 2

Using the formula,

Mode = 9 + [(35 − 20) / (2 × 35 − 20 − 18)] × 2

Mode = 9 + 15 / (70 − 38) × 2

Mode = 9 + (15 / 32) × 2

Mode = 9 + 30 / 32

Mode = 9 + 0.9375

Final Answer: Mode = 9.9375 litres ≈ 9.94 litres

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Question 4

The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of the ages of the patients.

Age (years)	Less than 5	5–9	10–14	15–19	20–24	25–29
Number of patients	38	32	50	36	24	20

Solution:

The highest frequency is 50.

Modal class = 10–14

Since the class intervals are inclusive, convert them into continuous intervals.

The modal class becomes 9.5–14.5.

Here,

l = 9.5, f₁ = 50, f₀ = 32, f₂ = 36, h = 5

Using the formula,

Mode = 9.5 + [(50 − 32) / (2 × 50 − 32 − 36)] × 5

Mode = 9.5 + 18 / (100 − 68) × 5

Mode = 9.5 + (18 / 32) × 5

Mode = 9.5 + 90 / 32

Mode = 9.5 + 2.8125

Final Answer: Mode = 12.3125 years ≈ 12.31 years

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Summary of Answers

Question	Final Answer
Q.1	4.33%
Q.2	72 units
Q.3	9.94 litres
Q.4	12.31 years

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