Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 24 Measures Of Central Tendency

Exercise 24.1

Question 1. Find the mean of the first 12 even numbers.
Answer: The first 12 even numbers are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)
n = 12

\( \Rightarrow \bar{X} = \frac{2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24}{12} \)

\( \Rightarrow \bar{X} = \frac{156}{12} \)

\( \Rightarrow \bar{X} = 13 \)

Therefore, Mean of first 12 even numbers = 13.
In simple words: We add all 12 even numbers and divide by 12 to get the average. The answer is 13.

📝 Teacher's Note: Show students the pattern - even numbers are 2, 4, 6, 8... The formula for mean is total divided by number of items. Practice with smaller sets first.

🎯 Exam Tip: Always write the formula first, then substitute values. Show all steps clearly. Write "Therefore" before the final answer.

Question 2. Find the mean of the first 10 prime numbers.
Answer: The first 10 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)
n = 10

\( \Rightarrow \bar{X} = \frac{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29}{10} \)

\( \Rightarrow \bar{X} = \frac{129}{10} \)

\( \Rightarrow \bar{X} = 12.9 \)

Therefore, Mean of first 10 prime numbers = 12.9
In simple words: Prime numbers can only be divided by 1 and themselves. We add the first 10 prime numbers and divide by 10.

📝 Teacher's Note: Remind students that prime numbers are special numbers like 2, 3, 5, 7... They cannot be divided by any other number except 1 and themselves.

🎯 Exam Tip: List all prime numbers clearly before calculating. Prime numbers are easy to forget, so write them carefully. Show the addition step.

Question 3. Find the mean of numbers from 7 to 17.
Answer: The numbers are: 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)
n = 11

\( \Rightarrow \bar{X} = \frac{7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17}{11} \)

\( \Rightarrow \bar{X} = \frac{132}{11} \)

\( \Rightarrow \bar{X} = 12 \)

Therefore, Mean of numbers from 7 to 17 = 12
In simple words: We take all whole numbers from 7 to 17. Add them up and divide by how many numbers we have (which is 11).

📝 Teacher's Note: Count carefully from 7 to 17 - there are 11 numbers, not 10. Students often make this counting mistake.

🎯 Exam Tip: When finding numbers "from A to B", count carefully. From 7 to 17 has 11 numbers, not 10. Always double-check your count.

Question 4. Find the mean of odd numbers from 5 to 20. What will be the mean if each number is multiplied by 4?
Answer: The odd numbers are: 5, 7, 9, 11, 13, 15, 17, 19

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)
n = 8

\( \Rightarrow \bar{X} = \frac{5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{8} \)

\( \Rightarrow \bar{X} = \frac{96}{8} \)

\( \Rightarrow \bar{X} = 12 \)

Therefore, Mean of odd numbers from 5 to 20 = 12

If numbers are multiplied by 4, the numbers are: 20, 28, 36, 44, 52, 60, 68, 76

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)
n = 8

\( \Rightarrow \bar{X} = \frac{20 + 28 + 36 + 44 + 52 + 60 + 68 + 76}{8} \)

\( \Rightarrow \bar{X} = \frac{384}{8} \)

\( \Rightarrow \bar{X} = 48 \)

Therefore, Mean of odd numbers from 5 to 20 when multiplied by 4 = 48
In simple words: Odd numbers are 5, 7, 9, 11... When we multiply each number by 4, the mean also gets multiplied by 4. So 12 × 4 = 48.

📝 Teacher's Note: Teach students the shortcut - when all numbers are multiplied by the same value, the mean also gets multiplied by that value. No need to recalculate everything.

🎯 Exam Tip: Remember the rule: if all data is multiplied by a number, the mean also gets multiplied by that number. This saves time in calculations.

Question 5. Find the mean of natural numbers from 32 to 46. What will be the mean if each number is diminished by 5?
Answer: The numbers are: 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)
n = 15

\( \Rightarrow \bar{X} = \frac{32+33+34+35+36+37+38+39+40+41+42+43+44+45+46}{15} \)

\( \Rightarrow \bar{X} = \frac{585}{15} \)

\( \Rightarrow \bar{X} = 39 \)

Therefore, Mean of natural numbers from 32 to 46 = 39

If numbers are diminished by 5, the numbers are: 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)
n = 15

\( \Rightarrow \bar{X} = \frac{27+28+29+30+31+32+33+34+35+36+37+38+39+40+41}{15} \)

\( \Rightarrow \bar{X} = \frac{510}{15} \)

\( \Rightarrow \bar{X} = 34 \)

Therefore, Mean of natural numbers from 32 to 46 when diminished by 5 = 34
In simple words: Natural numbers are counting numbers 1, 2, 3... When we subtract 5 from each number, the mean also decreases by 5. So 39 - 5 = 34.

📝 Teacher's Note: Explain that "diminished by" means "subtract from". When the same number is subtracted from all data, the mean also decreases by that amount.

🎯 Exam Tip: Rule for subtraction: if all data is decreased by a number, the mean also decreases by that number. This is a quick shortcut.

Question 6. If the numbers 8, 14, 20, x, 12 have mean 13, find the value of x.
Answer: Numbers are = 8, 14, 20, x, 12
Mean = 13
n = 5

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)

\( \Rightarrow 13 = \frac{8+14+20+x+12}{5} \)

\( \Rightarrow 13 = \frac{54+x}{5} \)

\( \Rightarrow 54 + x = 65 \)

\( \Rightarrow x = 11 \)

The value of x = 11
In simple words: We know the mean and all numbers except x. We use the mean formula backwards to find the missing number.

📝 Teacher's Note: This is reverse calculation. Students should multiply mean by n to get total, then subtract known numbers to find the unknown.

🎯 Exam Tip: For missing value problems, write mean = sum/n, then rearrange to find the unknown. Show each algebra step clearly.

Question 7. If the numbers 11, 14, p, 26, 10, 12, 18, 6 have mean 15, find the value of p.
Answer: Numbers are = 11, 14, p, 26, 10, 12, 18, 6
Mean = 15
n = 8

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)

\( \Rightarrow 15 = \frac{11+14+p+26+10+12+18+6}{8} \)

\( \Rightarrow 15 = \frac{97+p}{8} \)

\( \Rightarrow 97 + p = 120 \)

\( \Rightarrow p = 23 \)

The value of p = 23
In simple words: Add all known numbers (97), then use mean formula to find what p must be to make the average equal to 15.

📝 Teacher's Note: Students should add all known numbers first, then solve for the unknown. Check the answer by substituting back.

🎯 Exam Tip: Always verify your answer. Put p = 23 back in the original list and check if mean = 15. This catches calculation errors.

Question 8. The mean monthly salary of 10 people is Rs. 8,670. If a new person with salary Rs. 9000 joins, what will be the new mean monthly salary?
Answer: Mean monthly salary of 10 people = Rs. 8,670
n = 10

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)

\( Rs.8,670 = \frac{\sum x_n}{10} \)

\( \Rightarrow \sum x_n = Rs.86,700 \)

Salary of new person = Rs. 9000

\( \sum x_n = Rs.(86,700 + 9,000) \)
\( \sum x_n = Rs.95,700 \)

n = 11

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)

\( \Rightarrow \bar{X} = \frac{Rs.95,700}{11} \)

\( \Rightarrow \bar{X} = Rs.8,700 \)

The new mean monthly income = Rs. 8,700
In simple words: We first find total salary of 10 people. Then add the new person's salary and divide by 11 people to get new mean.

📝 Teacher's Note: Explain that when someone new joins, both the total and the count change. Students often forget to increase n from 10 to 11.

🎯 Exam Tip: Step 1: Find total using old mean. Step 2: Add new value to total. Step 3: Divide by new count. Don't forget to increase n by 1.

Question 9. Find the mean height of 9 students if their heights are: 142 cm, 158 cm, 152 cm, 143 cm, 139 cm, 144 cm, 146 cm, 148 cm, 151 cm.
Answer: The heights are: 142 cm, 158 cm, 152 cm, 143 cm, 139 cm, 144 cm, 146 cm, 148 cm, 151 cm
n = 9

\( \bar{X} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \)

\( \bar{X} = \frac{142 + 158 + 152 + 143 + 139 + 144 + 146 + 148 + 151}{9} \)

\( \bar{X} = \frac{1323}{9} \)

\( \bar{X} = 147cm \)

The mean height = 147 cm
In simple words: We add all 9 heights and divide by 9 to find the average height of the students.

📝 Teacher's Note: Use real examples like student heights to make mean more meaningful. Show how mean gives us the "typical" or "average" value.

🎯 Exam Tip: Always include units (cm) in your final answer. Add all values carefully - use a calculator if allowed. Show the addition step clearly.

Answer 10.

(i)

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ \overline{x} = \frac{605}{25} \]
\[ \overline{x} = 24.2 \] Mean = 24.2

(ii)

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ \overline{x} = \frac{1910}{50} \]
\[ \overline{x} = 38.2 \] Mean = 38.2

(iii)

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ \overline{x} = \frac{630}{40} \]
\[ \overline{x} = 15.75 \] Mean = 15.75

(iv)

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ \overline{x} = \frac{1980}{40} \]
\[ \overline{x} = 49.5 \] Mean = 49.5

(v)

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ \overline{x} = \frac{3900}{50} \]
\[ \overline{x} = 78 \] Mean = 78

(vi)

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ \overline{x} = \frac{1610}{60} \]
\[ \overline{x} = 26.83 \] Mean = 26.83

(vii)

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ \overline{x} = \frac{13020}{100} \]
\[ \overline{x} = 130.2 \] Mean = 130.2

Answer 11.

\[ \sum f_i = x_1 + x_2 + \ldots + x_n \]
50 = 7 + x + 15 + y + 10
\[ \implies x + y + 32 = 50 \]
\[ \implies x + y = 18 \ldots \ldots \ldots (i) \]
\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
25.8 = \( \frac{860 + 15x + 35y}{50} \)
\[ \implies 15x + 35y + 860 = 1290 \]
\[ \implies 15x + 35y = 430 \]
\[ \implies 3x + 7y = 86 \ldots \ldots \ldots (ii) \] Multiplying (i) by 3 and subtracting from (ii) 4y = 32
\[ \implies y = 8 \] Putting value of y in (i) x + 8 = 18 x = 10 Therefore, x = 10 and y = 8

📝 Teacher's Note: Show students how to set up two equations with two unknowns. Make them understand that the sum of all frequencies must equal the total given. This is a useful skill for solving frequency problems.

🎯 Exam Tip: Always write down the formula for mean first. Then substitute the given values carefully. Set up equations for sum of frequencies and mean formula. Solve step by step to avoid calculation mistakes.

Answer 12.

\[ \overline{x} = A + \frac{\sum f_i d}{\sum f_i} \]
\[ \overline{x} = 25 + \frac{80}{60} \]
\[ \overline{x} = 25 + 1.33 \]
\[ \overline{x} = 26.33 \] Mean = 26.33

📝 Teacher's Note: Explain that assumed mean method makes calculations easier when we have large numbers. Choose the middle value as A to get smaller deviations. This reduces calculation errors.

🎯 Exam Tip: Always state clearly which value you are taking as assumed mean (A). Show the deviation formula d=x-A. Calculate deviations carefully - check plus and minus signs. Write the final formula and substitute values step by step.

Answer 13.

\[ \overline{x} = A + \frac{\sum f_i d}{\sum f_i} \]
\[ \overline{x} = 35.5 + \frac{0}{80} \]
\[ \overline{x} = 35.5 + 0 \]
\[ \overline{x} = 35.5 \] Mean = 35.5

📝 Teacher's Note: Point out to students that when the sum of deviations is zero, the assumed mean is exactly equal to the actual mean. This is a special case that makes calculation very easy.

🎯 Exam Tip: When you get \( \sum f_i d = 0 \), the mean equals the assumed mean. Double-check your deviation calculations to make sure this zero is correct. This saves you from making calculation errors in the final step.

Exercise 24.2

Answer 14.
Answer:

Given: A=125 and \( h_i = 10 \)
Formula: \( \bar{x} = A + h \times \frac{\sum f_i u}{\sum f_i} \)
Step 1: \( \bar{x} = 125 + 10 \times \frac{-3}{100} \)
Step 2: \( \bar{x} = 125 - 0.3 \)
Step 3: \( \bar{x} = 124.70 \)
Therefore, Mean = 124.70

📝 Teacher's Note: In step method, we choose A as the middle class mark. This makes calculations easy. The u values become small numbers like -2, -1, 0, 1, 2.

🎯 Exam Tip: Always write the formula first. Show each step clearly. Check your answer - it should be close to A value (125 in this case).

Answer 15.
Answer:

Given: A=70 and \( h_i = 20 \)
Formula: \( \bar{x} = A + h \times \frac{\sum f_i u}{\sum f_i} \)
Step 1: \( \bar{x} = 70 + 20 \times \frac{62}{300} \)
Step 2: \( \bar{x} = 70 + 4.13 \)
Step 3: \( \bar{x} = 74.13 \)
Therefore, Mean = 74.13

📝 Teacher's Note: When class interval is 20, we use h=20 in the formula. Pick A as the class with highest frequency. This reduces calculation errors.

🎯 Exam Tip: Write all values in the table first. Double check your \( \sum f_i u \) calculation. One mistake here will give wrong final answer.

Answer 1.
Answer: Arranging the given data in descending order: 45 kg, 45 kg, 44 kg, 43 kg, 42 kg, 41 kg, 40 kg, 39 kg, 37 kg, 36 kg, 36 kg. The middle term is 41 kg which is 6th term. Therefore, Median of weights = 41 kg.
In simple words: Put all weights from biggest to smallest. The middle number is the median. Here we have 11 weights, so 6th number is the middle.

📝 Teacher's Note: For odd number of values (like 11), the median is the middle value. Count from both ends to find the middle position.

🎯 Exam Tip: Always arrange data in order first. Count total values to find middle position. For 11 values, median is the 6th value.

Answer 2.
Answer: Arranging the given data in descending order: 98, 96, 91, 88, 86, 84, 79, 75, 72, 68. The middle terms are 86 and 84 which are 5th and 6th terms. \( \text{median} = \frac{86 + 84}{2} = \frac{170}{2} = 85 \). Therefore, Median of marks = 85.
In simple words: We have 10 marks. Since 10 is even, we take the average of 5th and 6th marks. That gives us the median.

📝 Teacher's Note: For even number of values (like 10), median is the average of two middle values. The two middle positions are n/2 and (n/2 + 1).

🎯 Exam Tip: For even number of values, always find average of two middle numbers. Show the calculation clearly - add them and divide by 2.

Answer 3.
Answer: The first 15 whole numbers are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. Arranging the given data in descending order: 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0. The middle term is 7 which is 8th term. Therefore, Median of numbers = 7.
In simple words: First 15 whole numbers start from 0 and go till 14. The middle number out of 15 numbers is the 8th number, which is 7.

📝 Teacher's Note: Whole numbers start from 0, not 1. Count carefully: 0, 1, 2... up to 14 gives us 15 numbers total.

🎯 Exam Tip: Remember first 15 whole numbers are 0 to 14, not 1 to 15. For 15 values, median is the 8th value.

Answer 4.
Answer: The prime numbers between 20 and 50 are: 23, 29, 31, 37, 41, 43, 47. Arranging the given data in descending order: 47, 43, 41, 37, 31, 29, 23. The middle term is 37 which is 4th term. Therefore, Median of numbers = 37.
In simple words: Prime numbers can only be divided by 1 and themselves. Between 20 and 50, we get 7 such numbers. The middle one is 37.

📝 Teacher's Note: Prime numbers are numbers that have exactly two factors: 1 and the number itself. Examples: 23, 29, 31, 37, 41, 43, 47.

🎯 Exam Tip: List all prime numbers first, then arrange in order. For 7 values, median is the 4th value. Write the definition of prime if asked.

Answer 5.
Answer:

(i) Given data: 1, 8, 15, 5, 9, 4, 19, 6, 18
Arranging the given data in descending order: 19, 18, 15, 11, 9, 8, 6, 5, 4
The middle term is 9 which is 5th term
Therefore, Median of numbers = 9. (ii) Given data: 25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Arranging the given data in descending order: 35, 34, 32, 31, 29, 26, 25, 23, 22, 20
The middle terms are 29 and 26 which are 5th and 6th terms
\( \text{median} = \frac{29 + 26}{2} = \frac{55}{2} = 27.5 \)
Therefore, Median of numbers = 27.5 (iii) Given data: 3x, x+5, x+7, x+9, x+11, x+13
Arranging the given data in descending order: x+13, x+11, x+9, x+7, x+5, 3x
The middle terms are x+9 and x+7 which are 3rd and 4th terms
\( \text{median} = \frac{x+9+x+7}{2} = \frac{2x+16}{2} = x+8 \)
Therefore, Median = x+8. In simple words: For the first set (9 numbers), median is the 5th number. For the second set (10 numbers), median is average of 5th and 6th numbers. For algebra, we follow the same rules.

📝 Teacher's Note: Show students that median rules work the same for numbers and algebra. Count positions carefully in each case.

🎯 Exam Tip: For odd numbers of values, median is middle value. For even numbers, median is average of two middle values. This works for both numbers and algebra.

Answer 6.
Answer:

(i) Weight data:

No. of terms = 100
The mean of 50th and 51st term is the median
50th and 51st terms lay under 40 and 40
\( \text{median} = \frac{40 + 40}{2} = 40 \)
Hence, Median = 40 (ii) Salary data:

No. of terms = 75
\( \text{median} = \frac{75 + 1}{2} = 38\text{th term} \)
38th term lies under 4500
Hence, Median = 4500 In simple words: Make a cumulative frequency table. Find which value contains the middle position. For 100 students, middle position is between 50th and 51st. For 75 people, middle position is 38th.

📝 Teacher's Note: Cumulative frequency helps find median quickly. Add frequencies one by one until you reach the middle position.

🎯 Exam Tip: Always make cumulative frequency column. For n values, median position is n/2 (if even) or (n+1)/2 (if odd). Find which class contains this position.

Answer 7.
Answer:

No. of terms = 50
The mean of 25th and 26th term is the median
25th and 26th terms lay under 140 and 140
\( \text{median} = \frac{140 + 140}{2} = 140 \)
Hence, Median = 140
Upper Quartile (Q₃): \( \frac{n \times 3}{4} = \frac{50 \times 3}{4} = 37.5\text{th term} = 141 \)
Lower Quartile (Q₁): \( \frac{n}{4} = \frac{50}{4} = 12.5\text{th term} = 139 \)
Upper Quartile = 141 and Lower Quartile = 139 In simple words: Median is the middle value. Q₁ is the value at 1/4 position. Q₃ is the value at 3/4 position. These divide data into four equal parts.

📝 Teacher's Note: Quartiles divide data into 4 parts. Q₁ is at n/4 position, Q₂ (median) at n/2, Q₃ at 3n/4 position.

🎯 Exam Tip: Find positions first: n/4, n/2, 3n/4. Then use cumulative frequency to find which values are at these positions. Show each calculation step.

Answer 8.
Answer:

No. of terms = 50 In simple words: This table shows shoe sizes and how many people wear each size. We can find median and quartiles using cumulative frequency method.

📝 Teacher's Note: This is frequency distribution of discrete data. Each shoe size is a separate value, not a range. Median will be average of 25th and 26th values.

🎯 Exam Tip: For 50 values, median is average of 25th and 26th positions. Look at cumulative frequency to find which shoe sizes are at these positions.

Question (i). From the given frequency distribution table, calculate the lower quartile, upper quartile, interquartile range and semi-interquartile range.

Answer:
Step 1: Create cumulative frequency table

No. of terms = 70
Step 2: Calculate Lower Quartile (Q₁)
Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{50}{4} \) = 12.5th term = 7
Step 3: Calculate Upper Quartile (Q₃)
Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{50 \times 3}{4} \) = 37.5th term = 9
Step 4: Calculate Interquartile range
Interquartile range = Q₃ - Q₁ = 9 - 7 = 2
Step 5: Calculate Semi-interquartile range
Semi-interquartile range = \( \frac{Q₃ - Q₁}{2} = \frac{9 - 7}{2} \) = 1
Hence, Lower quartile = 7, upper quartile = 9, interquartile range = 2, semi-interquartile range = 1
In simple words: We first made a table showing how many students got marks up to each point. Then we found the positions where 25% and 75% of students fall. The difference between these positions gives us the range.

📝 Teacher's Note: Show students how to find the quarter positions first. Then locate which marks these positions represent. Make a clear cumulative frequency table step by step.

🎯 Exam Tip: Always write "No. of terms = n" first. Then calculate n/4 and 3n/4 positions clearly. Show all formula steps to get full marks.

Question (ii). From the given frequency distribution table, calculate the lower quartile, upper quartile, interquartile range and semi-interquartile range.

Answer:
No. of terms = 70
Step 1: Calculate Lower Quartile (Q₁)
Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{70}{4} \) = 17.5th term = 30
Step 2: Calculate Upper Quartile (Q₃)
Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{70 \times 3}{4} \) = 52.5th term = 45
Step 3: Calculate ranges
Interquartile range = Q₃ - Q₁ = 45 - 30 = 15
Semi-interquartile range = \( \frac{Q₃ - Q₁}{2} = \frac{45 - 30}{2} \) = 7.5
Hence, Lower quartile = 30, upper quartile = 45, interquartile range = 15, semi-interquartile range = 7.5
In simple words: We found where the 17.5th student and 52.5th student would be positioned. Their marks tell us the quartile values. The difference shows how spread out the middle 50% of students are.

📝 Teacher's Note: Help students see that 17.5th position means we look at the cumulative frequency to find which mark group contains this position. Same for 52.5th position.

🎯 Exam Tip: Write the formulas first: Q₁ = n/4, Q₃ = 3n/4. Then find which mark values these positions correspond to. Always include units in your final answer.

Question (iii). From the given frequency distribution table, calculate the lower quartile, upper quartile, interquartile range and semi-interquartile range.

Answer:
Step 1: Create cumulative frequency table

No. of terms = 20
Step 2: Calculate quartiles
Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{20}{4} \) = 5th term = 12
Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{20 \times 3}{4} \) = 15th term = 16
Step 3: Calculate ranges
Interquartile range = Q₃ - Q₁ = 16 - 12 = 4
Semi-interquartile range = \( \frac{Q₃ - Q₁}{2} = \frac{16 - 12}{2} \) = 2
Hence, Lower quartile = 12, upper quartile = 16, interquartile range = 4, semi-interquartile range = 2
In simple words: We found where the 5th and 15th values fall in our ordered list. The 5th value is 12 and 15th value is 16. The gap between them shows how spread the middle data is.

📝 Teacher's Note: With smaller data sets, students can actually count the positions. Show them how the 5th term in cumulative frequency gives the quartile value directly.

🎯 Exam Tip: For discrete data like this, the quartile positions give exact variate values. No interpolation needed. Count carefully using cumulative frequency.

Answer 9. Draw an ogive for the following data and read median, lower quartile and upper quartile from the graph.

Answer:
Step 1: Construct cumulative frequency table

Step 2: Draw the ogive
Take a graph paper and draw both the axes.
On the x-axis, take a scale of 1cm=10 to represent the class intervals.
On the y-axis, take a scale of 1cm=10 to represent the frequency.
Now, plot the points (10,4), (20,16), (30,37), (40,55), (50,70), (60,77), (70,80)
Join them by a smooth curve to get the ogive.

[Diagram: An ogive curve showing cumulative frequency plotted against class intervals, with points marked for quartile calculations]

Step 3: Read values from graph
No. of terms = n = 80
∴ Median = \( \frac{40 + 41}{2} \) = 40.5th term
Through mark of 40.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.
The value of B is the median which is 32.
Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{80}{4} \) = 20th term
Through mark of 20 on y-axis draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular to x-axis which meets it at Q.
The value of Q is the lower quartile which is 23.
Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{80 \times 3}{4} \) = 60th term
Through mark of 60 on y-axis draw a line parallel to x-axis which meets the curve at R. From R, draw a perpendicular to x-axis which meets it at S.
The value of S is the upper quartile which is 43.5.
In simple words: We made a curved graph showing how many students got marks below each level. Then we drew horizontal lines at the 25%, 50%, and 75% positions to read the quartile values.

📝 Teacher's Note: Show students how to plot the upper boundaries of class intervals on x-axis. The ogive should be a smooth curve, not straight lines joining the points.

🎯 Exam Tip: Always label your axes clearly with scales. Draw neat horizontal and vertical lines to read values. Write the final answers: Median = 32, Q₁ = 23, Q₃ = 43.5.

Question (i). Draw an ogive for the following data and read median, lower quartile and upper quartile from the graph.

Answer:
Step 1: Create cumulative frequency table

Step 2: Draw the ogive
Take a graph paper and draw both the axes.
On the x-axis, take a scale of 1cm=20 to represent the marks.
On the y-axis, take a scale of 1cm=10 to represent the number of boys.
Now, plot the points (40,10), (50,22), (60,36), (70,48), (80,57), (90,64), (100,70)
Join them by a smooth curve to get the ogive.

[Diagram: An ogive curve for marks distribution showing cumulative frequency against marks, with quartile lines marked]

Step 3: Read values from graph
No. of terms = 70
∴ Median = \( \frac{35 + 36}{2} \) = 35.5th term
Through mark of 35.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.
The value of B is the median which is 60.
Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{70}{4} \) = 17.5th term
Through mark of 17.5 on y-axis draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular to x-axis which meets it at Q.
The value of Q is the lower quartile which is 47.5.
Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{70 \times 3}{4} \) = 52.5th term
Through mark of 52.5 on y-axis draw a line parallel to x-axis which meets the curve at R. From R, draw a perpendicular to x-axis which meets it at S.
The value of S is the upper quartile which is 74.5.
In simple words: We plotted how many boys scored below each mark level. Then we found where the 25%, 50%, and 75% lines cross the curve to get our quartile values.

📝 Teacher's Note: Help students understand that we plot at the upper boundary of each class interval. The ogive shows "less than" cumulative frequency.

🎯 Exam Tip: Mark the scale clearly. Read values carefully from where your construction lines meet the x-axis. Write final answers with proper labels.

Question (iii). Draw an ogive for the following data and read median, lower quartile and upper quartile from the graph.

Answer:
Given data is a less than cumulative data, so draw the ogive as it is.

Take a graph paper and draw both the axes.
On the x-axis, take a scale of 1cm=10 to represent marks less than.
On the y-axis, take a scale of 1cm=20 to represent the number of students.
Now, plot the points (10,5), (20,15), (30,30), (40,54), (50,72), (60,86), (70,94), (80,100).
Join them by a smooth curve to get the ogive.

[Diagram: An ogive curve for "less than" cumulative frequency data showing a smooth ascending curve with quartile construction lines]

No. of terms = 100
∴ Median = \( \frac{50 + 51}{2} \) = 50.5th term = 50th term
Through mark of 50 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.
The value of B is the median which is 40.
Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{100}{4} \) = 25th term
Through mark of 25 on y-axis draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular to x-axis which meets it at Q.
The value of Q is the lower quartile which is 28.
Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{100 \times 3}{4} \) = 75th term
Through mark of 75 on y-axis draw a line parallel to x-axis which meets the curve at R. From R, draw a perpendicular to x-axis which meets it at S.
The value of S is the upper quartile which is 52.
In simple words: This data already shows how many students scored less than each mark. So we directly plot the curve and read where the quarter positions fall on the marks scale.

📝 Teacher's Note: Point out that "less than" data is already in cumulative form. No need to add frequencies. Students can directly plot and read values.

🎯 Exam Tip: For "less than" data, plot directly. Mark construction lines clearly. Write final answers: Median = 40, Q₁ = 28, Q₃ = 52.

[Diagram: This shows a cumulative frequency curve (ogive) with marks on x-axis (0 to 90) and cumulative frequency on y-axis. The curve shows quartile positions marked as points.]

No. of terms = 100

∴ Median = \( \frac{50 + 51}{2} = 50.5^{th} \) term

Through mark of 50.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

The value of B is the median which is 38.

Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{100}{4} = 25^{th} \) term

Through mark of 25 on y-axis draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular to x-axis which meets it at Q.

The value of Q is the lower quartile which is 28.

Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{100 \times 3}{4} = 75^{th} \) term

Through mark of 75 on y-axis draw a line parallel to x-axis which meets the curve at R. From R, draw a perpendicular to x-axis which meets it at S.

The value of S is the upper quartile which is 51.

(iv)

Given data is cumulative data, so draw the ogive as it is.

Take a graph paper and draw both the axes.

On the x-axis, take a scale of 1cm=10 to represent the Age (in yrs) under.

On the y-axis, take a scale of 1cm=10 to represent the no. of males.

Now, plot the points (10,6), (20,10), (30,25), (40,32), (50,43), (60,50).

Join them by a smooth curve to get the ogive.

[Diagram: This shows an ogive curve plotted on a graph with Age (in years) under on x-axis and No. of students on y-axis. Points are marked and connected with a smooth curve.]

No. of terms = 50

∴ Median = \( \frac{25 + 26}{2} = 25.5^{th} \) term

Through mark of 25.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

The value of B is the median which is 30.

Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{50}{4} = 12.5^{th} \) term

Through mark of 12.5 on y-axis draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular to x-axis which meets it at Q.

The value of Q is the lower quartile which is 22.

Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{50 \times 3}{4} = 37.5^{th} \) term

Through mark of 37.5 on y-axis draw a line parallel to x-axis which meets the curve at R. From R, draw a perpendicular to x-axis which meets it at S.

The value of S is the upper quartile which is 44.

(v)

Given data is cumulative data, so draw the ogive as it is.

Take a graph paper and draw both the axes.

On the x-axis, take a scale of 1cm=10 to represent the marks (more than).

On the y-axis, take a scale of 1cm=10 to represent the no. of students.

Now, plot the points (0,80), (10,80), (20,78), (30,70), (40,60), (50,48), (60,34), (70,22), (80,13), (90,6).

Join them by a smooth curve to get the ogive.

[Diagram: This shows a decreasing ogive curve plotted on a graph with Marks (more than) on x-axis and No. of students on y-axis. The curve shows a downward trend from left to right.]

[Diagram: This shows another cumulative frequency curve (ogive) with marks more than on x-axis (0 to 100) and cumulative frequency on y-axis. The curve shows quartile positions marked as points.]

No. of terms = 80

∴ Median = \( \frac{40 + 41}{2} = 40.5^{th} \) term

Through mark of 40.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

The value of B is the median which is 55.

Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{80}{4} = 20^{th} \) term

Through mark of 20 on y-axis draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular to x-axis which meets it at Q.

The value of Q is the lower quartile which is 71.

Upper Quartile (Q₃) = \( \frac{n \times 3}{4} = \frac{80 \times 3}{4} = 60^{th} \) term

Through mark of 60 on y-axis draw a line parallel to x-axis which meets the curve at R. From R, draw a perpendicular to x-axis which meets it at S.

The value of S is the upper quartile which is 40.

Answer 10.
We construct cumulative frequency table of the given distribution:

Take a graph paper and draw both the axes.

On the x-axis, take a scale of 1cm=20 to represent the marks.

On the y-axis, take a scale of 1cm=50 to represent the no. of students.

Now, plot the points (10,5), (20,15), (30,26), (40,46), (50,73), (60,111), (70,151), (80,180), (90,194), (100,200).

Join them by a smooth curve to get the ogive.

[Diagram: This shows an ogive curve plotted on a graph paper with marks on x-axis and number of students on y-axis. The curve shows an S-shaped pattern with quartile positions marked.]

(i) No. of terms = 200

∴ Median = \( \frac{100 + 101}{2} = 100.5^{th} \) term

Through mark of 100.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

The value of B is the median which is 58.5.

(ii) Lower Quartile (Q₁) = \( \frac{n}{4} = \frac{200}{4} = 50^{th} \) term

Through mark of 50 on y-axis draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular to x-axis which meets it at Q.

The value of Q is the lower quartile which is 41.

(iii) From marks = 80 draw a line parallel to y-axis and meet the curve at R. From R, Draw a perpendicular on y-axis which meets it at S. The difference of the value obtained when subtracted from 200 gives the number of students who scored more than 80%.

📝 Teacher's Note: An ogive is a graph that shows cumulative frequency. The median is the middle value. For 200 students, the median is the average of 100th and 101st values.

🎯 Exam Tip: Always draw neat lines parallel to axes. Mark the points clearly. Write the final answers with proper labels.

Answer 11.
Answer: who scored more than 80%.
\( \Rightarrow 200 - 180 = 20 \)
20 students scored more than 80%

We construct cumulative frequency table of the given distribution:

Take a graph paper and draw both the axes.

On the x-axis, take a scale of 1 cm = 10 to represent the marks.

On the y-axis, take a scale of 1 cm = 50 to represent the no. of students.

Now, plot the points (19.5,7), (29.5,18), (39.5,38), (49.5,84), (59.5,141), (69.5,178), (79.5,193), (89.5,200).

Join them by a smooth curve to get the ogive.

[Diagram: This shows a cumulative frequency graph (ogive) with marks percentage on x-axis and number of students on y-axis. The curve starts from bottom left and rises steeply to the right.]

(i) No. of terms = 200
\( \therefore \text{Median} = \frac{100 + 101}{2} = 100.5^{\text{th}} \text{term} \)

Through mark of 100.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

The value of B is the median which is 52.

(ii) From marks % = 35 draw a line parallel to y-axis and meet the curve at R. From R, Draw a perpendicular on y-axis which meets it at S. The difference of the value obtained when subtracted from 200 gives the number of students who scored more than 35%.

\( \Rightarrow 200 - 23 = 172 \)

172 students scored more than 35%

📝 Teacher's Note: The ogive curve shows cumulative frequency. To find how many scored MORE than a value, subtract from total. To find median, go to the middle value position.

🎯 Exam Tip: Always draw neat lines and mark points clearly on graph. Write "Therefore median = " clearly. Show your subtraction step for "more than" questions.

Exercise 24.3

Answer 1.
Answer:
(i) 6, 7, 1, 8, 6, 5, 9, 4, 6, 7, 1, 3, 2, 6, 7, 8
Mode = 6 because it occurs 4 times.

(ii) 21, 22, 28, 23, 24, 21, 26, 22, 29, 27, 21, 21, 25, 24, 23
Mode = 21 because it occurs 4 times.

(iii) 3, 4, 5, 7, 6, 3, 5, 4, 3, 5, 6, 4, 7, 5, 4, 5, 4, 3, 4, 5, 7, 6, 5, 6, 6, 7
Mode = 5 because it occurs 7 times.

(iv) 15, 17, 16, 17, 10, 12, 14, 15, 19, 12, 16, 15, 16
Mode = 16 because it occurs 4 times.

(v) 20, 20, 30, 30, 30, 30, 35, 40, 40, 45, 45, 45, 50, 55, 60, 60, 60, 65, 70, 70, 70
Mode = 30 because it occurs 4 times.

In simple words: Mode is the value that appears most often in a data set. Count how many times each number appears and pick the one with highest count.

📝 Teacher's Note: Ask students to make tally marks next to each number. This helps them count frequencies easily. Mode can have more than one value if two numbers appear the same maximum times.

🎯 Exam Tip: Always write "Mode = [value] because it occurs [number] times." This shows you understand what mode means and gets full marks.

Answer 2.
Answer:

Mode = 23 because it occurs maximum number of times i.e. 35

Mode = 100 because it occurs maximum number of times i.e. 18

Mode = 5 because it occurs maximum number of times i.e. 10

In simple words: For frequency tables, look at the frequency column. The value with the highest frequency is the mode.

📝 Teacher's Note: Teach students to scan the frequency column first to find the highest number. Then look at which value has that frequency. This avoids confusion.

🎯 Exam Tip: Write "Mode = [value] because it has maximum frequency = [number]." Always mention both the mode value and its frequency for full marks.

Answer 3.
Answer:

[Diagram: This shows a histogram with marks on x-axis (0 to 80) and number of students on y-axis (0 to 30). The highest bar is at 30-40 marks range with frequency 24.]

(a) Take 1 cm = 1 unit and plot marks on x-axis and no. of students on y-axis.

(b) Draw a bar graph for the given data.

(c) From the histogram it is clear that class 30-40 has highest frequency i.e. 24

(d) Join the ends of the corresponding frequencies which meet at P and drop a perpendicular on the x-axis from P to Q. Q is the mode.
Therefore, Mode = 35

[Diagram: This shows a histogram with I.Q. Score on x-axis (80 to 200) and number of students on y-axis (0 to 18). The highest bar is at 120-140 score range with frequency 16.]

(a) Take 1 cm = 1 unit and plot I. Q. Score on x-axis and no. of students on y-axis.

(b) Draw a bar graph for the given data.

(c) From the histogram it is clear that class 120-140 has highest frequency i.e. 16

(d) Join the ends of the corresponding frequencies which meet at P and drop a perpendicular on the x-axis from P to Q. Q is the mode.
Therefore, Mode = 134

In simple words: For grouped data, find the class with highest frequency. Then use the histogram method or formula to find the exact mode value within that class.

📝 Teacher's Note: Show students how to draw lines from the top corners of the modal class to meet at a point. Drop a line from that point to x-axis to get the mode.

🎯 Exam Tip: First identify the modal class (highest frequency). Then clearly show the construction lines on your histogram. Write the final mode value clearly.

Question (a). Take 1cm = 1 unit and plot mangoes on x-axis and no. of trees on y-axis.
Answer: This requires drawing a histogram using the given data from the table. The x-axis should show mangoes per tree (19.5-29.5, 29.5-39.5, 39.5-49.5, 49.5-59.5) and y-axis should show number of trees (20, 14, 6, 4). Each bar should be 1cm wide representing 10 unit intervals.
In simple words: Draw bars on graph paper. Each bar shows how many trees give that many mangoes. The tallest bar is for 19.5-29.5 mangoes.

[Diagram: This shows a histogram with 4 bars. The first bar (19.5-29.5) is tallest with height 20, second bar (29.5-39.5) has height 14, third bar (39.5-49.5) has height 6, and fourth bar (49.5-59.5) has height 4.]

📝 Teacher's Note: Show students how to draw equal width bars. The height of each bar = frequency. Make sure students label both axes clearly.

🎯 Exam Tip: Always label the axes. Write "Mangoes per tree" on x-axis and "Number of trees" on y-axis. Use the scale given in the question.

Question (b). Draw a bar graph for the given data.
Answer: A bar graph uses separate bars (not touching) to show the data. Draw 4 separate bars with gaps between them. Bar heights should be 20, 14, 6, and 4 respectively for the four classes.
In simple words: Bar graph has separate bars with spaces. Histogram has bars touching each other. Both show the same data but look different.

📝 Teacher's Note: Explain the difference clearly - histogram bars touch, bar graph bars have gaps. Students often get confused between these two.

🎯 Exam Tip: If question asks for "bar graph", draw separate bars with gaps. If it asks for "histogram", draw touching bars.

Question (c). From the histogram it is clear that class 19.5-29.5 has highest frequency i.e. 20
Answer: Yes, this is correct. Looking at the histogram, the class 19.5-29.5 has the tallest bar with frequency 20. This means most trees (20 trees) produce between 19.5 to 29.5 mangoes.
In simple words: The tallest bar in the graph shows which group has the most trees. Here, 20 trees give 19.5-29.5 mangoes each.

📝 Teacher's Note: Teach students to read histograms by finding the tallest bar. The tallest bar always shows the highest frequency or modal class.

🎯 Exam Tip: Write "highest frequency" and give the exact number. Also mention which class interval it belongs to.

Question (d). Join the ends of the corresponding frequencies which meet at P and drop a perpendicular on the x-axis from P to Q. Q is the mode. Therefore, Mode = 23.5
Answer: This method finds the mode using the histogram. We draw lines from the top corners of the modal class bar to the adjacent bars. Where these lines meet (point P), we drop a line straight down to the x-axis. This point Q gives us the mode value of 23.5.
In simple words: We draw two slanting lines on the tallest bar. Where they cross, we draw a straight line down. This gives us the mode (most common value).

📝 Teacher's Note: Practice this graphical method step by step. Students must draw neat lines and measure carefully to get the right answer.

🎯 Exam Tip: Draw clear lines and mark point P and Q clearly. Write "Mode = " with the exact value you get from the graph.

Question (a). Take 1cm = 1 unit and plot marks % on x-axis and no. of students on y-axis.
Answer: This requires drawing a histogram for the student marks data. The x-axis shows marks percentage ranges and y-axis shows number of students. The scale is 1cm = 1 unit, so each class interval of 10 marks will be 1cm wide.
In simple words: Draw bars showing how many students got marks in each range. The tallest bar shows most students got 69.5-79.5 marks.

[Diagram: This shows a histogram with 7 bars representing marks distribution. The tallest bar is for 69.5-79.5 marks with height 114 students.]

📝 Teacher's Note: This is similar to the previous histogram but with different data. Students should follow the same steps - equal width bars with heights equal to frequency.

🎯 Exam Tip: Label axes as "Marks %" and "Number of students". Draw bars touching each other since it's a histogram.

Question (b). Draw a bar graph for the given data.
Answer: Draw 7 separate bars with gaps between them. Bar heights should be 14, 26, 40, 92, 114, 78, and 36 respectively for the seven mark ranges.
In simple words: Same data as histogram but bars have spaces between them. This makes it easy to read each value separately.

📝 Teacher's Note: Remind students again about the difference. Bar graphs show discrete data, histograms show continuous data like class intervals.

🎯 Exam Tip: Keep equal gaps between bars. Make sure bar heights match the frequency values exactly.

Question (c). From the histogram it is clear that class 69.5-79.5 has highest frequency i.e. 114
Answer: Yes, this is correct. The class 69.5-79.5 has the highest frequency of 114 students. This means most students scored between 69.5% to 79.5%.
In simple words: 114 students got marks between 69.5% and 79.5%. This is more than any other group, so this bar is tallest.

📝 Teacher's Note: Point out that this shows most students performed well (above 69%). This is good performance data.

🎯 Exam Tip: Always identify the modal class correctly. Write the class interval and its frequency clearly.

Question (d). Join the ends of the corresponding frequencies which meet at P and drop a perpendicular on the x-axis from P to Q. Q is the mode. Therefore, Mode = 73
Answer: Using the graphical method on the histogram, we draw lines from the top of the modal class bar to adjacent bars. These lines meet at point P. From P, we drop a perpendicular to the x-axis at point Q. The mode value at Q is 73.
In simple words: We use the same line-drawing method as before. The lines cross at P, and the point below P on the x-axis gives us mode = 73.

📝 Teacher's Note: This shows most students scored around 73%. The graphical method gives a more precise value than just saying "69.5-79.5".

🎯 Exam Tip: Draw the construction lines clearly. Mark points P and Q. Write the final answer as "Mode = 73" with proper units if needed.