Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 24 Measures Of Central Tendency Mean Median Quartiles Mode

Exercise 24A

Question 1. Find the mean of the following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6
Answer:
(i)
\( \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} \)
Here n = 5
\( \bar{x} = \frac{6 + 9 + 11 + 12 + 7}{5} = \frac{45}{5} = 9 \)

(ii)
\( \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} \)
Here n = 8
\( \bar{x} = \frac{11 + 14 + 23 + 26 + 10 + 12 + 18 + 6}{8} = \frac{120}{8} = 15 \)
In simple words: To find the mean, add all numbers and divide by how many numbers you have. Mean tells us the average value of all numbers.

📝 Teacher's Note: Show students that mean is like sharing equally. If 5 students share 45 chocolates, each gets 9 chocolates on average.

🎯 Exam Tip: Always count the numbers carefully. Write the formula first, then put the values and calculate step by step.

Question 2. Marks obtained (in mathematics) by 9 student are given below: 60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) find the arithmetic mean
(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.
Answer:
(a) Here n = 9
\( \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} \)
\( \bar{x} = \frac{60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56}{9} = \frac{531}{9} = 59 \)

(b)
If marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63
In simple words: The average marks is 59. When we add 4 to each student's marks, the average also increases by 4.

📝 Teacher's Note: When we add the same number to all values, the mean also increases by that same number. This is a very useful property.

🎯 Exam Tip: For part (b), just add the increase to the original mean. Don't calculate everything again. This saves time in exams.

Question 3. Find the mean of the natural numbers from 3 to 12.
Answer:
Numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Here n = 10
\( \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} \)
\( \bar{x} = \frac{3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12}{10} = \frac{75}{10} = 7.5 \)
In simple words: Natural numbers from 3 to 12 means all whole numbers from 3 to 12. The average of these 10 numbers is 7.5.

📝 Teacher's Note: List all numbers first to avoid missing any. Students often forget to include both end numbers (3 and 12).

🎯 Exam Tip: Write all numbers clearly before adding. Check that you have the right count of numbers.

Question 4. (a) Find the mean of 7, 11, 6, 5, and 6
(b) If each number given in (a) is diminished by 2, find the new value of mean.
Answer:
(a) The mean of 7, 11, 6, 5 and 6
\( \bar{x} = \frac{7 + 11 + 6 + 5 + 6}{5} = \frac{35}{5} = 7 \)

(b)
If we subtract 2 from each number, then the mean will be 7-2 = 5
In simple words: The average is 7. When we subtract 2 from each number, the new average becomes 5. The average also decreases by 2.

📝 Teacher's Note: This shows another property of mean. When we subtract the same number from all values, the mean also decreases by that number.

🎯 Exam Tip: For part (b), just subtract 2 from the original mean. Don't recalculate everything - this property saves time.

Question 5. If the mean of 6, 4, 7, 'a' and 10 is 8. Find the value of 'a'
Answer:
No. of terms = 5
Mean = 8
Sum of numbers = 8 × 5 = 40 ...(i)
But, sum of numbers = 6+4+7+a+10 = 27+a ...(ii)
From (i) and (ii)
27+a = 40
a = 13
In simple words: We know the average is 8 for 5 numbers. So the total sum must be 8×5 = 40. We add the known numbers and find what 'a' must be.

📝 Teacher's Note: Teach students this trick: Mean × Number of values = Total sum. Then find the missing value by subtraction.

🎯 Exam Tip: Always write "Sum = Mean × Number of terms" first. This is the key formula for finding missing values.

Question 6. The mean of the number 6, 'y', 7, 'x' and 14 is 8. Express 'y' in terms of 'x'.
Answer:
No. of terms = 5 and mean = 8
Sum of numbers = 5 × 8 = 40 ...(i)
but sum of numbers = 6+y+7+x+14 = 27+y+x ...(ii)
from (i) and (ii)
27 + y + x = 40
x + y = 13
y = 13 – x
In simple words: We have two unknown values x and y. Using the given mean, we find that x + y = 13. So y = 13 - x.

📝 Teacher's Note: This shows how to handle two unknowns. We can express one in terms of the other using the mean condition.

🎯 Exam Tip: Write the equation clearly: Sum = Mean × Number of terms. Then solve for the relationship between unknowns.

Question 7. The ages of 40 students are given in the following table:

Find the arithmetic mean.
Answer:

\( \bar{x} = \frac{\sum f_i x_i}{\sum f} = \frac{614}{40} = 15.35 \)
In simple words: When data is given in frequency table, we multiply each age by how many students have that age, then add all products and divide by total students.

📝 Teacher's Note: Make sure students understand that frequency means "how many times". Multiply each value by its frequency before adding.

🎯 Exam Tip: Always make the f×x column. Add all frequencies to check total is correct. Use the formula: Mean = Σfx/Σf.

Question 8. If 69.5 is the mean of 72, 70, 'x', 62, 50, 71, 90, 64, 58 and 82, find the value of 'x'.
Answer:
No. of terms = 10
Mean = 69.5
Sum of the numbers = 69.5 × 10 = 695 ...(i)
But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82
= 619 + x ...(ii)
from (i) and (ii)
619 + x = 695
x = 76
In simple words: We know the average and count of numbers. So total sum must be 69.5×10 = 695. We add all known numbers and find what x must be.

📝 Teacher's Note: Students should add the known numbers carefully. A common mistake is addition errors when many numbers are involved.

🎯 Exam Tip: Double-check your addition of known numbers. Write Sum = Mean × n first, then solve for the unknown.

Question 9. The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of 'f'.

Answer:

Mean = 60.95

\( \frac{1718 + 60f}{28 + f} = 60.95 \)

\( \implies \) 1718 + 60f = 60.95(28 + f)

\( \implies \) 1718 + 60f = 1706.6 + 60.95f

\( \implies \) (60.95 – 60)f = 1718.0 – 1706.6

\( \implies \) 0.95f = 11.4

\( \therefore f = \frac{11.4}{0.95} = 12 \)
In simple words: We have a missing frequency 'f'. Using the given mean, we set up an equation and solve for f. The answer is f = 12.

📝 Teacher's Note: This is harder because frequency is unknown. Set up the mean formula carefully and solve the equation step by step.

🎯 Exam Tip: Write the mean formula with unknowns first. Cross multiply carefully and solve the linear equation. Check your answer by substituting back.

Question 10. From the data given below, calculate the mean wage, correct to the nearest rupee.

(i) If the number of workers in each category is doubled, what would be the new mean wage?
(ii) If the wages per day in each category are increased by 60%; what is the new mean wage?
(iii) If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%, what would be the new mean wage?
Answer:

\( \bar{x} = \frac{\sum fx}{\sum f} = \frac{3360}{42} = 80 \)

(i) Mean remains the same if the number of workers in each category is doubled.
Mean = 80

(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%
New mean = \( 80 \times \frac{160}{100} = 128 \)

(iii) No change in the mean if the number of workers is doubled but if wages per worker is reduced by 40%, then
New mean = \( 80 \times \frac{60}{100} = 48 \)
In simple words: The original mean wage is Rs. 80. Doubling workers doesn't change mean. Increasing wages by 60% makes mean Rs. 128. Reducing wages by 40% makes mean Rs. 48.

📝 Teacher's Note: Show students that multiplying all frequencies by same number doesn't change the mean. But changing the values directly affects the mean.

🎯 Exam Tip: Remember: changing frequencies (number of items) doesn't change mean. Changing values changes the mean by the same percentage.

Question 11. The contents of 100 match boxes were checked to determine the number of matches they contained.

(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches.
Answer:

(i) \( \overline{x} = \frac{\sum fx}{\sum f} = \frac{3813}{100} = 38.13 \)
(ii) In the second case,
New mean = 39 matches
Total contents = 39 × 100 = 3900
But total number of matches already given = 3813
Number of new matches to be added = 3900 - 3813 = 87
In simple words: First we find the average by adding all matches and dividing by total boxes. Then to get 39 average, we need 3900 total matches, so we add 87 more matches.

📝 Teacher's Note: Use a table to organize data. Show students how to multiply each value by its frequency. This makes the calculation clear and prevents errors.

🎯 Exam Tip: Always write the mean formula first. Show all working steps clearly. Round to the correct number of decimal places as asked in the question.

Question 12. If the mean of the following distribution is 3, find the value of p.

Answer:

Now, Mean = \( \frac{\sum fx}{\sum f} \)
\( \implies 3 = \frac{6p + 87}{33} \)
\( \implies 99 = 6p + 87 \)
\( \implies 6p = 12 \)
\( \implies p = 2 \)
In simple words: We make a table with fx column. The mean formula gives us an equation with p. We solve this equation to find p = 2.

📝 Teacher's Note: Show students how to handle algebraic expressions in frequency tables. Make sure they understand that (p + 4) × 6 = 6p + 24.

🎯 Exam Tip: Set up the mean equation carefully. Write each step clearly when solving for p. Check your answer by substituting back.

Question 13. In the following table, \( \sum f = 200 \) and mean = 73. Find the missing frequencies f₁ and f₂.

Answer:

Given, \( \sum f = 200 \)
\( \implies 86 + f₁ + f₂ = 200 \)
\( \implies f₁ + f₂ = 114 \) ...(i)
Mean = \( \frac{\sum fx}{\sum f} \)
\( \implies 73 = \frac{7000 + 50f₁ + 100f₂}{200} \)
\( \implies 7000 + 50f₁ + 100f₂ = 14600 \)
\( \implies 50f₁ + 100f₂ = 7600 \)
\( \implies f₁ + 2f₂ = 152 \) ...(ii)
Subtracting (i) from (ii), we get
f₂ = 38
\( \implies f₁ = 114 - 38 = 76 \)
Hence, f₁ = 76 and f₂ = 38
In simple words: We make two equations - one from total frequency and one from mean. We solve these two equations together to find the missing frequencies.

📝 Teacher's Note: This is a system of two equations problem. Show students how to form equations from given conditions. Practice more examples with missing frequencies.

🎯 Exam Tip: Always write both equations clearly. Label them (i) and (ii). Show the substitution method step by step. Check your answer by adding frequencies.

Question 14. Find the arithmetic mean (correct to the nearest whole-number) by using step-deviation method.

Answer: Let the assumed mean A = 30

∴ Mean = \( A + \frac{\sum ft}{\sum f} \times i \)
= \( 30 + \frac{(-598)}{384} \times 5 \)
= \( 30 - \frac{299 \times 5}{192} \)
= \( 30 - \frac{1495}{192} \)
= \( \frac{5760 - 1495}{192} \)
= \( \frac{4265}{192} \)
= 22.21
= 22
In simple words: Step-deviation method makes calculation easier by choosing a middle value and using equal steps. We get the same answer as direct method but with smaller numbers.

📝 Teacher's Note: Choose the assumed mean from middle values with highest frequency. Class interval (i) is the common difference between consecutive x values. Here i = 5.

🎯 Exam Tip: Write the step-deviation formula clearly. Choose A wisely to make calculations simple. Round the final answer as asked in the question.

Question 15. Find the mean (correct to one place of decimal) by using short-cut method.

Answer: Let the assumed mean A = 45

∴ Mean = \( A + \frac{\sum fd}{\sum f} \)
= \( 45 + \frac{(-88)}{200} \)
= \( 45 - \frac{11}{25} \)
= \( \frac{1125 - 11}{25} \)
= \( \frac{1114}{25} \)
= 44.6
In simple words: Short-cut method uses an assumed mean to make calculations easier. We find deviations from this assumed mean, then adjust the final answer.

📝 Teacher's Note: Choose assumed mean as a value from the data that makes calculations simple. Usually pick a middle value with high frequency.

🎯 Exam Tip: Write the short-cut formula clearly. Show all calculation steps. Give the final answer to the correct decimal places as asked.

Exercise 24B

Question 1. The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.

Answer: The solution shows grouped data where we need to find the class midpoint first. For age group 16-18, midpoint = (16+18)/2 = 17. This problem requires finding midpoints of each class interval and then calculating the mean using the midpoint method.
In simple words: For grouped data, we use the middle value of each group as the representative value. Then we calculate mean like a normal frequency table.

📝 Teacher's Note: Explain that for grouped data, we cannot use exact values. We use class midpoints as representative values. Show how to find midpoint: (lower limit + upper limit) ÷ 2.

🎯 Exam Tip: Always find midpoints first for grouped data. Make a proper table with midpoints, frequencies, and products. Show all working clearly.

Question 2. The following table gives the weekly wages of workers in a factory.

Calculate the mean by using:
(i) Direct Method
(ii) Short - Cut Method

Answer:

(i) Direct Method

\[ \overline{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{5520}{80} = 69 \]

(ii) Short - cut method

\[ \overline{x} = A + \frac{\sum f_id_i}{\sum f_i} = 72.5 + \left(\frac{-280}{80}\right) = 72.5 - 3.5 = 69 \]

In simple words: We found the average wage using two methods. Both gave us Rs. 69 as the mean wage. This is the middle value that best represents all workers' wages.

📝 Teacher's Note: Show students that both methods give the same answer. The direct method adds all wages and divides. The short-cut method uses a middle value (A) as a base to make calculations easier.

🎯 Exam Tip: Always show the formula first, then substitute values. In short-cut method, choose A from the middle of the data to get smaller numbers.

Question 3. The following are the marks obtained by 70 boys in a class test:

Calculate the mean by:
(i) Short - cut method
(ii) Step - deviation method

Answer:

(i) Short - cut method

\[ \overline{x} = A + \frac{\sum f_id_i}{\sum f_i} = 65 + \left(\frac{-270}{70}\right) = 65 - 3.86 = 61.14 \]

(ii) Step - deviation method

Here A = 65 and h = 10

\[ \overline{x} = A + h \times \frac{\sum f_iu_i}{\sum f_i} = 65 + 10 \times \left(\frac{-27}{70}\right) = 65 - 3.86 = 61.14 \]

In simple words: The average marks of all 70 boys is 61.14. This means most students scored around 61 marks out of 100. Both methods give the same answer.

📝 Teacher's Note: Step-deviation method uses h (class interval width) to make calculations even simpler. Here h = 10, so we divide deviations by 10 to get smaller numbers.

🎯 Exam Tip: In step-deviation method, always write h value clearly. Remember to multiply by h in the final formula. Both methods must give the same answer.

Question 4. Find mean by step - deviation method:

Answer:

Here A = 87.50 and h = 7

\[ \overline{x} = A + h \times \frac{\sum f_iu_i}{\sum f_i} = 87.5 + 7 \times \frac{29}{150} = 87.5 + 1.35 = 88.85 \]

In simple words: The mean value is 88.85. This is the average of all the data values. We used step-deviation method to make the calculation easier.

📝 Teacher's Note: Choose A as the mid-value of the class with highest frequency. Here 84-91 has 38 frequency, so A = 87.50. This reduces calculation work.

🎯 Exam Tip: Write h value clearly at the start. Check that the sum of \(u_i\) values is small - if it's too big, you chose wrong A value.

Question 5. The mean of the following frequency distribution is \(21\frac{1}{7}\). Find the value of 'f':

Answer:
Given: Mean = \(21\frac{1}{7} = \frac{148}{7}\)

Using mean formula:
\[ \overline{x} = \frac{\sum f_ix_i}{\sum f_i} \]
\[ \frac{148}{7} = \frac{1235 + 35f}{63 + f} \]

Cross multiplying:
\[ 148(63 + f) = 7(1235 + 35f) \]
\[ 9324 + 148f = 8645 + 245f \]
\[ 9324 - 8645 = 245f - 148f \]
\[ 679 = 97f \]
\[ f = \frac{679}{97} = 7 \]

Therefore, f = 7

In simple words: We used the mean formula and put in the given mean value. Then we solved the equation to find f = 7. This makes the data complete.

📝 Teacher's Note: This is a reverse problem - we know the mean and need to find missing frequency. Set up the mean formula as an equation and solve for the unknown value.

🎯 Exam Tip: Convert mixed fraction to improper fraction first. Cross multiply carefully and solve step by step. Always verify your answer by substituting back.

Question 6. Using step-deviation method, calculate the mean marks of the following distribution.

Answer:
Given data:

Step 1: Using the mean formula
\( \overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1235 + 35f}{63 + f} \)
Step 2: Given that mean \( \overline{x} = 21\frac{1}{7} = \frac{148}{7} \)
Step 3: Setting up the equation
\( \frac{148}{7} = \frac{1235 + 35f}{63 + f} \)
Step 4: Cross multiplying
\( 148(63 + f) = 7(1235 + 35f) \)
\( 9324 + 148f = 8645 + 245f \)
Step 5: Solving for f
\( 245f - 148f = 9324 - 8645 \)
\( 97f = 679 \)
\( f = 7 \)
Therefore, f = 7
In simple words: We used the mean formula to find the missing frequency. When the mean is given, we can make an equation and solve it. The unknown frequency f turns out to be 7.

📝 Teacher's Note: Show students how to set up the equation step by step. A common mistake is forgetting to include f in both the numerator and denominator when calculating totals.

🎯 Exam Tip: Always write "Given mean =" and "Required: f =". Cross multiply carefully and check your arithmetic. Show each step clearly for full marks.

Question 7. Using the information given in the adjoining histogram, calculate the mean.

[Diagram: This histogram shows frequency on y-axis (0-30) and class intervals on x-axis (15-25, 25-35, 35-45, 45-55, 55-65). The bars show frequencies of approximately 10, 20, 25, 15, and 5 respectively.]

Answer:
Step 1: Let the assumed mean A = 72.5

Step 2: Using step deviation method
Mean = \( A + \frac{\sum f_i d_i}{\sum f_i} = 72.5 + \left(\frac{-280}{80}\right) = 72.5 - 3.5 = 69 \)
Therefore, mean = 69
In simple words: We picked a middle value as our assumed mean. Then we found how far each class is from this assumed mean. Finally, we adjusted our assumed mean to get the real mean.

📝 Teacher's Note: Explain that step deviation method makes calculations easier when dealing with large numbers. Choose the assumed mean from the middle of the data to minimize calculations.

🎯 Exam Tip: Always show the table with columns for fi, xi, di, and fidi. Write the final formula clearly. Double-check the sign of your final answer.

Question 8. If the mean of the following observations is 54, find the value of 'p'.

Answer:
Step 1: Creating the frequency table

Step 2: Using mean formula
\( \overline{x} = \frac{\sum fx}{\sum f} = \frac{2370 + 30p}{39 + p} \) ...(i)
Step 3: Given mean = 54
From (i): \( \frac{2370 + 30p}{39 + p} = 54 \)
Step 4: Cross multiplying
\( 2370 + 30p = 54(39 + p) \)
\( 2370 + 30p = 2106 + 54p \)
Step 5: Solving for p
\( 54p - 30p = 2370 - 2106 \)
\( 24p = 264 \)
\( p = 11 \)
Therefore, p = 11
In simple words: We made a table with the mid-values and used the mean formula. When we put the given mean value in the formula, we got an equation which we solved to find p.

📝 Teacher's Note: Remind students to always include the unknown frequency 'p' in both the total frequency and the total fx. Show them how to set up the equation step by step.

🎯 Exam Tip: Write "Given: Mean = 54" clearly. Show the cross multiplication step. Always verify your answer by substituting back into the original equation.

Question 9. The mean of the following distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f₁ and f₂.

Answer:
Step 1: Setting up the frequency table

Step 2: From given conditions
\( \sum f = 30 + f₁ + f₂ \) and \( \sum fx = 2060 + 30f₁ + 70f₂ \) ...(i)
\( \sum f = 50 \); mean = 62.8 ...(ii)
Step 3: Finding f₁ + f₂
From (i) and (ii): \( 30 + f₁ + f₂ = 50 \)
\( f₁ + f₂ = 20 \) ...(iii)
Step 4: Using mean formula
mean = \( \frac{2060 + 30f₁ + 70f₂}{50} \)
\( 62.8 = \frac{2060 + 30f₁ + 70f₂}{50} \)
\( 2060 + 30f₁ + 70f₂ = 62.8 × 50 \)
\( 2060 + 30f₁ + 70f₂ = 3140 \)
\( 30f₁ + 70f₂ = 1080 \)
\( 3f₁ + 7f₂ = 108 \) ...(iv)
Step 5: Solving equations (iii) and (iv)
From (iii): \( f₁ = 20 - f₂ \)
Substituting in (iv): \( 3(20 - f₂) + 7f₂ = 108 \)
\( 60 - 3f₂ + 7f₂ = 108 \)
\( 4f₂ = 48 \)
\( f₂ = 12 \)
Step 6: Finding f₁
From (iii): \( f₁ = 20 - 12 = 8 \)
Therefore, f₁ = 8 and f₂ = 12
In simple words: We used two given facts - total frequency is 50 and mean is 62.8. This gave us two equations with two unknowns. We solved these equations to find the missing frequencies.

📝 Teacher's Note: Show students how to form two separate equations from the given conditions. Emphasize that we need two conditions to find two unknown values.

🎯 Exam Tip: Always write both conditions clearly: "Sum of frequencies = 50" and "Mean = 62.8". Show the substitution method step by step. Verify by checking if f₁ + f₂ = 20.

Question 10. Calculate the mean of the distribution, given below, using the short cut method:

Answer:
Step 1: Setting up the table with assumed mean A = 45.5

Step 2: Using short cut method formula
Mean = \( A + \frac{\sum fd}{\sum f} \)
= \( 45.5 + \frac{70}{50} \)
= \( 45.5 + 1.4 \)
= \( 46.9 \)
Therefore, mean = 46.9
In simple words: We picked a middle value (45.5) as our assumed mean. We found the difference of each class from this value. Then we adjusted our assumed mean using these differences to get the real mean.

📝 Teacher's Note: Explain why we choose the assumed mean from the middle of the data. Show students that the class with d = 0 is our assumed mean class. This method saves calculation time.

🎯 Exam Tip: Choose A (assumed mean) as a mid-value from the data to make calculations easier. Always show the table with f, x, d, and fd columns. Write the formula clearly before substituting values.

Question 11. Calculate the mean of the following distribution:

Answer:
Step 1: Setting up the frequency table

Step 2: Using direct method
Mean = \( \frac{\sum fx}{\sum f} = \frac{3600}{100} = 36 \)
Therefore, mean = 36
In simple words: We found the middle value of each class interval. Then we multiplied each middle value by its frequency. Finally, we divided the total by the total frequency to get the mean.

📝 Teacher's Note: Show students how to find the mid-value: add the two limits and divide by 2. For example, for 0-10, mid-value = (0+10)/2 = 5. This is the direct method for finding mean.

🎯 Exam Tip: Always make a proper table with columns for class interval, frequency, mid-value, and fx. Add up all fx values and all f values carefully. Show the division step clearly for full marks.

Exercise 24C

Question 1. A student got the following marks in 9 questions of a question paper. 3, 5, 7, 3, 8, 0, 1, 4 and 6. Find the median of these marks.
Answer:
Step 1: Arranging the given data in descending order:
8, 7, 6, 5, 4, 3, 3, 1, 0
Step 2: The middle term is 4 which is the 5th term.
Median = 4
In simple words: We arrange all marks from highest to lowest. The middle number is the median. Since there are 9 marks, the 5th mark is in the middle.

📝 Teacher's Note: Show students how to count positions. With 9 numbers, the middle is position 5. Always arrange data first before finding the median.

🎯 Exam Tip: Always write the arranged data clearly. Then count to find the middle position. For odd numbers, the median is exactly the middle term.

Question 2. The weights (in kg) of 10 students of a class are given below: 21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24. Find the median of their weights.
Answer:
Step 1: Arranging the given data in descending order:
28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5
Step 2: The middle terms are 24 and 24, 5th and 6th terms
Step 3: \( \text{median} = \frac{24 + 24}{2} = \frac{48}{2} = 24 \)
In simple words: When there are 10 numbers (even), we take the average of the 5th and 6th numbers after arranging them. Here both middle numbers are 24, so the median is 24.

📝 Teacher's Note: For even number of values, the median is the average of two middle values. Students often forget to divide by 2.

🎯 Exam Tip: Write "even number of terms" and clearly show which two middle terms you are averaging. Always divide their sum by 2.

Question 3. The marks obtained by 19 students of a class are given below: 27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28. Find: (i) median (ii) lower quartile (iii) upper quartile (iv) interquartile range
Answer:
Step 1: Arranging in ascending order:
22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 31, 32, 32, 33, 35, 35, 36, 36, 37

(i) Median:
Middle term is 10th term i.e. 29
Median = 29

(ii) Lower quartile:
\( Q_1 = \frac{n + 1}{4} \text{th term} \)
\( Q_1 = \frac{19 + 1}{4} \text{th term} \)
\( Q_1 = 5\text{th term} = 26 \)

(iii) Upper quartile:
\( Q_3 = \frac{3(n + 1)}{4} \text{th term} \)
\( Q_3 = \frac{3(19 + 1)}{4} \text{th term} \)
\( Q_3 = 15\text{th term} = 35 \)

(iv) Interquartile range:
Interquartile range = \( Q_3 - Q_1 = 35 - 26 = 9 \)
In simple words: Median is the middle value. Lower quartile is the middle of the lower half. Upper quartile is the middle of the upper half. Interquartile range shows the spread of the middle 50% data.

📝 Teacher's Note: Teach students to find positions using formulas. For 19 terms: median is 10th, Q1 is 5th, Q3 is 15th term.

🎯 Exam Tip: Always use the formulas for quartile positions. Show your calculations clearly. Write the final answers with labels.

Question 4. From the following data, find: (i) Median (ii) Upper quartile (iii) Inter-quartile range 25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
Answer:
Step 1: Arrange in ascending order:
0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95

(i) Median:
Median is the mean of 8th and 9th term
\( = \frac{40 + 45}{2} = \frac{85}{2} = 42.5 \)

(ii) Upper quartile:
\( Q_3 = \frac{3(n)}{4} \text{th term} \)
\( Q_3 = \frac{3 \times 16}{4} \text{th term} = 12\text{th term} = 65 \)

(iii) Interquartile range:
\( Q_1 = \frac{16}{4} \text{th term} = 18; Q_3 = 65 \)
Interquartile range = \( Q_3 - Q_1 = 65 - 18 = 47 \)
In simple words: With 16 numbers, the median is the average of 8th and 9th positions. The upper quartile is at the 12th position. The range between quartiles shows how spread out the middle half of data is.

📝 Teacher's Note: For 16 terms (even number), median is average of 8th and 9th terms. Q1 is 4th term, Q3 is 12th term.

🎯 Exam Tip: Count carefully to find the right positions. Show all steps for median calculation when you have even number of terms.

Question 5. The ages of 37 students in a class are given in the following table:

Find the median.
Answer:

Number of terms = 37
\( \text{Median} = \frac{37 + 1}{2} \text{th term} = 19\text{th term} \)
Median = 14

In simple words: We make a cumulative frequency table. The 19th student (middle position) has age 14. So the median age is 14 years.

📝 Teacher's Note: Make students understand cumulative frequency. It shows "how many students up to this age". The median position is where we cross the middle.

🎯 Exam Tip: Always make the cumulative frequency column. Find the median position first, then see which value contains that position.

Question 6. The weight of 60 boys are given in the following distribution table:

Find: (i) median (ii) lower quartile (iii) upper quartile (iv) interquartile range
Answer:

Number of terms = 60

(i) median:
median = the mean of the 30th and the 31st terms
\( \therefore \text{median} = \frac{39 + 39}{2} = \frac{78}{2} = 39 \)

(ii) lower quartile (Q₁):
\( = \frac{60}{4} \text{th term} = 15\text{th term} = 38 \)

(iii) upper quartile (Q₃):
\( = \frac{3 \times 60}{4} \text{th term} = 45\text{th term} = 40 \)

(iv) Interquartile range:
= Q₃ - Q₁ = 40 - 38 = 2

In simple words: With 60 boys, the median is the average of 30th and 31st positions (both are 39kg). Q₁ is at 15th position (38kg) and Q₃ is at 45th position (40kg).

📝 Teacher's Note: For 60 terms, median positions are 30th and 31st. Use cumulative frequency to find which weight each position falls in.

🎯 Exam Tip: Make the cumulative frequency table first. Then find the exact positions for median, Q₁, and Q₃. Check which values contain these positions.

Question 7. Estimate the median for the given data by drawing an ogive:

Answer:
To find the median using an ogive (cumulative frequency curve), we need to:

Step 1: Make cumulative frequency table
Step 2: Draw the ogive curve
Step 3: Find N/2 = 50/2 = 25th value
Step 4: From the graph, the median value corresponds to the 25th position

In simple words: An ogive is a curved graph that shows cumulative frequency. We find the middle position (25th) and read the corresponding value from the graph to get the median.

[Diagram: The ogive would be a smooth curve plotted with class boundaries on x-axis and cumulative frequencies on y-axis, rising from left to right.]

📝 Teacher's Note: Students should practice drawing smooth curves. The median is found by drawing a horizontal line from N/2 on y-axis to the curve, then down to x-axis.

🎯 Exam Tip: Always label your axes clearly. Draw a neat smooth curve. Show the construction lines to find the median value from the graph.

Question 8. By drawing an ogive, estimate the median for the following frequency distribution:

Answer:

Number of terms = 55
\( \text{Median} = \frac{55 + 1}{2} \text{ term} = 28^{th} \text{ term} \)

[Diagram: This shows an ogive (cumulative frequency curve) with Weight (kg) on x-axis and Cumulative frequency on y-axis. A horizontal line from 28 on y-axis meets the curve at point A. A vertical line from A meets x-axis at point B showing the median value.]

Through mark of 28 on the 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 x-axis at B. The value of B is the median which is 18.4 kg
In simple words: We make a step-by-step frequency table. Then we draw a smooth curve. The middle value (28th position) gives us the median weight of 18.4 kg.

📝 Teacher's Note: Show students how to find the middle position first. Then use the graph to read the median value. Make them practice reading values from the curve.

🎯 Exam Tip: Always make the cumulative frequency table first. Mark the middle position clearly. Draw neat lines on the graph to show your work.

Question 9. From the following cumulative frequency table, find:
(i) median
(ii) lower quartile
(iii) upper quartile

Answer:

Number of terms = 80
(i) Median: \( \text{Median} = \frac{80}{2} = 40^{th} \text{ term} \)

[Diagram: This shows an ogive with Marks on x-axis and Cumulative frequency on y-axis. A horizontal line from 40 on y-axis meets the curve at A. A vertical line from A meets x-axis at B showing the median value.]

Through 40th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B. Value of B is the median = 40
(ii) Lower quartile (Q₁): \( Q_1 = \frac{80}{4} = 20^{th} \text{ term} = 18 \)
(iii) Upper Quartile (Q₃): \( Q_3 = \frac{3 \times 80}{4} = 60^{th} \text{ term} = 66 \)
In simple words: We find the middle value (median), the 1/4 position value (lower quartile), and the 3/4 position value (upper quartile) using the curve. These divide the data into four equal parts.

📝 Teacher's Note: Explain that quartiles divide data into four equal parts. Show the positions: Q₁ at n/4, median at n/2, Q₃ at 3n/4. Use the graph to read values.

🎯 Exam Tip: Calculate the positions first (n/4, n/2, 3n/4). Then use the ogive to find the values at these positions. Show your working clearly.

Question 10. In a school, 100 pupils have heights as tabulated below:

Find the median height by drawing an ogive.
Answer:

Number of terms = 100
\( \text{Median} = \frac{100}{2} = 50^{th} \text{ term} \)

[Diagram: This shows an ogive with Height on x-axis and Cumulative frequency on y-axis. A horizontal line from 50 on y-axis meets the curve at A. A vertical line from A meets x-axis at B showing the median height.]

Through 50th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B. Value of B is the median = 148
Median height = 148cm
In simple words: We make a cumulative frequency table and draw a smooth curve. The middle student (50th position) has a height of 148 cm. This is the median height.

📝 Teacher's Note: Explain that median is the middle value when data is arranged in order. With 100 students, the median is at the 50th position. Use the graph to find this value.

🎯 Exam Tip: Always find n/2 first (here 100/2 = 50). Draw horizontal line from 50 on y-axis to curve, then drop vertical line to x-axis to read median value.

Question 11. Attempt this question on a graph paper. The table shows the distribution of marks gained by a group of 400 students in an examination.

Using a scale of 2 cm to represent 10 marks and 2 cm to represent 50 students, plot these points and draw a smooth curve through the points. Estimate from the graph:
(i) the median marks
(ii) the quartile marks
Answer:

Number of terms = 400
(i) Median: \( \text{Median} = \frac{400}{2} = 200^{th} \text{ term} \)

[Diagram: This shows an ogive with Marks on x-axis and Cumulative frequency on y-axis. A horizontal line from 200 on y-axis meets the curve at A. A vertical line from A meets x-axis at B showing the median value.]

Through 200th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B. Value of B is the median = 62
(ii) Lower Quartile: \( Q_1 = \frac{400}{4} = 100^{th} \text{ term} = 49 \)
Upper Quartile: \( Q_3 = \frac{3 \times 400}{4} = 300^{th} \text{ term} = 74 \)
In simple words: We plot the given points on graph paper and draw a smooth curve. Using this curve, we find the median (middle value) and quartiles (quarter positions) by reading from the graph.

📝 Teacher's Note: Make sure students use the correct scale given in the question. Plot points carefully and draw a smooth curve through all points. Practice reading values from the curve.

🎯 Exam Tip: Use the exact scale given (2 cm = 10 marks, 2 cm = 50 students). Calculate positions first: n/4, n/2, 3n/4. Show your line drawing on the graph clearly.

Question 12.
Attempt this question on graph paper.

(i) Construct the 'less than' cumulative frequency curve for the above data. Using 2 cm = 10 years on one axis and 2 cm = 10 casualties on the other.
(ii) From your graph determine:
(a) the median
(b) the lower quartile
Answer:
Cumulative Frequency Table:

Number of terms = 83

[Diagram: This shows a cumulative frequency curve graph with age on x-axis (0 to 80) and cumulative frequency on y-axis (0 to 90). The curve shows increasing trend with points marked at each age group.]

(i) Median calculation:
Median = \( \frac{83}{2} = 41.5^{th} \) term
Through 41.5th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.
Value of B is the median = 43

(ii) Lower Quartile calculation:
Q₁ = \( \frac{83}{4} = 20.75^{th} \) term = 29

Upper Quartile = \( 83 \times \frac{3}{4} = 62.25^{th} \) term = 53
In simple words: We made a table to add up casualties. Then we drew a smooth curve on graph paper. The median is the middle value when all data is arranged in order.

📝 Teacher's Note: Show students how to make the cumulative frequency table first. Then plot points carefully on graph paper. The curve should be smooth, not straight lines between points.

🎯 Exam Tip: Always write the cumulative frequency table first. Label your graph axes clearly. Show all working for median and quartile calculations to get full marks.

Exercise 24D

Question 1.
Find the mode of the following data:
(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6
(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8
Answer:
(i) Mode = 7
Since 7 occurs 4 times
(ii) Mode = 11
Since it occurs 4 times
In simple words: Mode is the number that appears most often in the list. Count how many times each number appears and pick the one with highest count.

📝 Teacher's Note: Ask students to make tally marks for each number. This makes it easy to see which number appears most. Some data sets may have no mode or more than one mode.

🎯 Exam Tip: Write "Mode = [number]" clearly. Then write "because it occurs [count] times" to show your reasoning. This gets you full marks.

Question 2.
The following table shows the frequency distribution of heights of 50 boys:

Find the mode of heights.
Answer: Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.
In simple words: Look at the frequency row. Find the highest number (18). The height above that highest frequency is the mode (122 cm).

📝 Teacher's Note: In a frequency table, students should look for the highest frequency first. The value corresponding to that frequency is the mode. Make this clear with colored pens.

🎯 Exam Tip: Always identify the highest frequency first. Then write the corresponding value as the mode. Include the unit (cm, kg etc.) in your answer.

Question 3.
Find the mode of following data, using a histogram:

Answer:

[Diagram: This shows a histogram with class intervals on x-axis (0-10, 10-20, 20-30, 30-40, 40-50) and frequency on y-axis (0 to 25). The tallest bar is at 20-30 class with frequency 20.]

Mode is in 20-30, because in this class there are 20 frequencies.
In simple words: In the histogram, the tallest bar shows the modal class. The class 20-30 has the tallest bar, so mode lies in this range.

📝 Teacher's Note: Explain that for grouped data, we find the modal class (not exact mode). The modal class is the one with highest frequency. The tallest bar in histogram shows this.

🎯 Exam Tip: For grouped data, write "Modal class is [range]" not just "Mode is [number]". Always identify the class with maximum frequency first.

Question 4.
The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:

Answer:

[Diagram: This shows a histogram with expenditure classes on x-axis (20-25, 25-30, 30-35, 35-40, 40-45, 45-50) and number of students on y-axis (0 to 25). The tallest bar is at 30-35 class with frequency 23.]

Mode is in 30-35 because it has the maximum frequency.
In simple words: Most boys (23 students) spent money in the range Rs 30-35. This is the modal class because it has the highest number of students.

📝 Teacher's Note: Point out that 23 is the highest frequency in the table. The expenditure range corresponding to this frequency is the modal class. Draw the histogram to make it visual.

🎯 Exam Tip: Scan the frequency column to find the maximum value first. Write "Modal class is [range]" and mention it has maximum frequency. Include currency units.

Question 5.
Find the median and mode for the set of numbers:
2, 2, 3, 5, 5, 5, 6, 8 and 9
Answer:
Median = \( \frac{9 + 1}{2} = 5^{th} \) term which is 5.
Mode = 5 because it occurs maximum number of times.
In simple words: Median is the middle value when numbers are arranged in order. Mode is the number that appears most often (5 appears three times).

📝 Teacher's Note: The data is already arranged in ascending order. For median with odd number of terms, use the middle term formula. For mode, count frequency of each number.

🎯 Exam Tip: Check if data is arranged in order first. Write the position formula for median, then find the actual value. For mode, state which number repeats most.

Question 6.
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
(i) What are his modal marks?
(ii) What are his median marks?
(iii) What are his total marks?
(iv) What are his mean marks?
Answer:
Arranging the given data in ascending order:
7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19

(i) Mode = 16 as it occurs maximum number of times.

(ii) Median = \( \frac{11 + 1}{2} = 6^{th} \) term = 15

(iii) Total marks = 7+10+12+12+14+15+16+16+16+17+19 = 154

(iv) Mean = \( \frac{x_1 + x_2 + x_3 +...+ x_n}{n} = \frac{154}{11} = 14 \)
In simple words: Mode is 16 (appears 3 times). Median is the middle value (15). Mean is total marks divided by number of tests.

📝 Teacher's Note: First arrange data in order. This makes finding median and mode easier. Show students how to count frequency for mode and use position formula for median.

🎯 Exam Tip: Always arrange data in ascending order first. Show your working for each part clearly. Include proper units and double-check your arithmetic for mean calculation.

Question 7.
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7 and 8.
Answer:
(i) Mean = \( \frac{0 + 0 + 2 + 2 + 3 + 3 + 3 + 4 + 5 + 5 + 5 + 5 + 6 + 6 + 7 + 8}{16} = \frac{64}{16} = 4 \)

(ii) Median = mean of 8th and 9th term
= \( \frac{4 + 5}{2} = \frac{9}{2} = 4.5 \)

(iii) Mode = 5 as it occurs maximum number of times.
In simple words: Mean is the average of all marks. Median is the middle value (average of 8th and 9th terms). Mode is 5 because it appears 4 times.

📝 Teacher's Note: For even number of terms, median is average of two middle terms. The data is already arranged in order. Count carefully to find which number appears most often.

🎯 Exam Tip: For 16 terms, median is average of 8th and 9th terms. Show the calculation clearly. State that mode is the value with highest frequency (4 times).

Question 8.
At a shooting competition the score of a competitor were as given below:

(i) What was his modal score?
(ii) What was his median score?
(iii) What was his total score?
(iv) What was his mean score?
In simple words: This table shows how many times the competitor scored each number. We need to find the most common score, middle score, total and average score.

📝 Teacher's Note: This is a frequency distribution table. Students need to understand that "No. of shots" means how many times each score occurred. The modal score has the highest frequency.

🎯 Exam Tip: For frequency tables, find modal score by identifying highest frequency. For median and mean, you may need to list all individual scores first, then calculate.

Exercise 24E

Question 1. The following distribution represents the height of 160 students of a school.

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:
i. The median height.
ii. The interquartile range.
iii. The number of students whose height is above 172 cm.
Answer: Step 1: Create cumulative frequency table.

Step 2: Find the answers from the ogive graph.

[Diagram: This shows an ogive graph with height on x-axis and cumulative frequency on y-axis. Points are plotted at the upper class boundaries and connected with a smooth curve.]

(i) Median height:
Median = \( \frac{160}{2} = 80^{th} \) term
From the graph, median height = 157.5 cm (ii) Interquartile range:
Lower quartile (Q₁) = \( \frac{160}{4} = 40^{th} \) term = 152 cm
Upper quartile (Q₃) = \( \frac{3 \times 160}{4} = 120^{th} \) term = 164 cm
Interquartile range = Q₃ - Q₁ = 164 - 152 = 12 cm (iii) Students above 172 cm height:
From the graph, students up to 172 cm = 144
Students above 172 cm = 160 - 144 = 16 students
In simple words: We draw a curved line using cumulative frequencies. Then we read values from the graph by drawing horizontal and vertical lines at specific points.

📝 Teacher's Note: Show students how to plot points at upper class boundaries. Make sure they understand that ogive is a smooth curve, not straight lines between points.

🎯 Exam Tip: Always mark the median point clearly on the graph. Write "from graph" when reading values. Show your working for quartile calculations step by step.

Question 2. The following table gives the weekly wages of workers in a factory:

Calculate: (i) the mean, (ii) the modal class, (iii) the number of workers getting weekly wages below Rs.80 and (iv) the number of workers getting Rs.65 or more but less than Rs.85 as weekly wages.
Answer: Step 1: Create frequency table with class marks and fx.

(i) Mean:
Mean = \( \frac{\sum fx}{\sum f} = \frac{5520}{80} = 69 \) (ii) Modal class:
Modal class = 55-60 as it has maximum frequency of 20. (iii) Number of workers getting wages below Rs.80:
Workers below Rs.80 = 60 (iv) Number of workers getting Rs.65 or more but less than Rs.85:
Workers from 65-70, 70-75, 75-80 = 10 + 9 + 6 = 25
Also workers from 80-85 = 12
Total = 25 + 12 = 37 workers
In simple words: We find the average wage by adding all fx values and dividing by total workers. Modal class is the group with most workers.

📝 Teacher's Note: Teach students that class mark is the middle value of each class. For 50-55, class mark = (50+55)/2 = 52.5. This helps in calculating mean.

🎯 Exam Tip: Always show the formula for mean clearly. Write "Modal class" not just "Mode". Count cumulative frequency carefully for range questions.

Question 3. Draw an ogive for the data given below and from the graph determine: (i) the median marks (ii) the number of students who obtained more than 75% marks

Answer: Step 1: Create cumulative frequency table.

Total number of students = 120

[Diagram: This shows an ogive graph with marks on x-axis (0-100) and cumulative frequency on y-axis (0-120). Points are plotted and connected with a smooth curve.]

(i) Median marks:
Median position = \( \frac{120}{2} = 60^{th} \) student
From the graph, median marks = approximately 45 marks (ii) Students who got more than 75% marks:
75% means 75 marks and above
From the graph, students up to 75 marks = approximately 110
Students above 75 marks = 120 - 110 = 10 students
In simple words: We draw the ogive curve and find the middle student's marks for median. For 75% marks, we see how many students are below 75 and subtract from total.

📝 Teacher's Note: Remind students that for continuous data, we use class boundaries (like 9.5, 19.5) instead of class limits (like 9, 19) when drawing ogive.

🎯 Exam Tip: Mark the median point clearly on your graph. Show the horizontal and vertical lines used to read the median value. Always check your answer makes sense.

Question 4. The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m-1 and median q. Find p and q.
Answer:
Step 1: Find the mean m of the first set of numbers.
Mean of 1, 7, 5, 3, 4 and 4 = \( \frac{24}{6} = 4 \)
\( \implies \) m = 4

Step 2: Find p using the mean condition.
The numbers 3, 2, 4, 2, 3, 3 and p have mean m-1 = 4-1 = 3
Therefore, \( \frac{17 + p}{7} = 3 \)
\( \implies \) 17 + p = 21
\( \implies \) p = 4

Step 3: Find the median q by arranging in order.
Arranging in ascending order: 2, 2, 3, 3, 3, 4, 4
Median = 4th term = 3
Therefore, q = 3

Final Answer: p = 4, q = 3
In simple words: We first found m by adding all numbers and dividing by count. Then we used the mean formula to find p. Finally, we arranged all numbers in order to find the middle value (median).

📝 Teacher's Note: Show students how to arrange numbers in order first. The median is always the middle number when arranged from smallest to biggest. For 7 numbers, the 4th number is the median.

🎯 Exam Tip: Always write "arranging in ascending order" and show the ordered list. Then clearly mark which position is the median. This gets you full marks.

Question 5. The marks of 200 students in a test were recorded as follows:

Construct the cumulative frequency table. Draw the ogive and use it to find:
(i) the median and
(ii) the number of students who score more than 35% marks.
Answer:
Cumulative Frequency Table:

[Diagram: This shows an ogive (cumulative frequency curve) plotted on a graph with marks on x-axis and cumulative frequency on y-axis. The curve rises from bottom-left to top-right showing the cumulative distribution.]

(i) Finding the median:
\( \text{Median} = \frac{200}{2} = 100^{th} \text{ term} \)
From the ogive, at cumulative frequency 100, the corresponding mark is 52.5
Median = 52.5 marks

(ii) Finding students with more than 35% marks:
35% of total marks = \( \frac{35}{100} \times 100 = 35 \) marks
From the ogive, at mark 35, cumulative frequency = 28
Number of students scoring more than 35% = 200 - 28 = 172
172 students scored more than 35% marks
In simple words: We made a table showing how many students got marks up to each level. Then we drew a curve. The median is the mark of the middle student (100th out of 200). For 35% question, we found how many got less than 35 marks, then subtracted from 200.

📝 Teacher's Note: Teach students that cumulative frequency means "add up all the frequencies so far". The ogive is like climbing stairs - it only goes up, never down. Use the graph to read values directly.

🎯 Exam Tip: Always draw lines on the ogive to show your reading. Mark the points clearly. Write "From the graph" to show you used the ogive method. This gets you method marks.

Question 6. The marks of 20 students in a test were as follows: 2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20. Calculate: (i) the mean (ii) the median (iii) the mode
Answer:
Given data: 2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20
Number of terms = 20

(i) Mean calculation:
\( \sum x = 2+6+8+9+10+11+11+12+13+13+14+14+15+15+15+16+16+18+19+20 = 257 \)
\( \text{Mean} = \frac{\sum x}{n} = \frac{257}{20} = 12.85 \)

(ii) Median calculation:
\( \text{Median} = \frac{10^{th} \text{ term} + 11^{th} \text{ term}}{2} = \frac{13 + 14}{2} = \frac{27}{2} = 13.5 \)

(iii) Mode calculation:
Mode = 15 (as it appears 3 times, which is the maximum frequency)

Final Answer: Mean = 12.85, Median = 13.5, Mode = 15
In simple words: Mean is the average - add all numbers and divide by count. Median is the middle value when arranged in order. Mode is the number that appears most often (15 appears 3 times).

📝 Teacher's Note: Show students that data is already arranged in order, so finding median is easy. For 20 numbers, median is average of 10th and 11th values. Mode needs counting - which number repeats most?

🎯 Exam Tip: Always write "arranging in ascending order" even if data is already sorted. Show the sum calculation clearly for mean. Circle the mode value and write its frequency. This shows complete working.

Question 7. The marks obtained by 120 students in a mathematics test is given below:

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:
(i) the median
(ii) the number of students who obtained more than 75% in test.
(iii) the number of students who did not pass in the test if the pass percentage was 40.
(iv) the lower quartile
Answer:
Cumulative Frequency Table:

[Diagram: This shows an ogive (cumulative frequency curve) with marks on x-axis (0 to 110) and cumulative frequency on y-axis (0 to 140). The curve shows the cumulative distribution rising smoothly from bottom-left to top-right.]

(i) Median:
\( \text{Median} = \frac{120 + 1}{2} = 60.5^{th} \text{ term} \)
From the ogive, the median = 43 marks

(ii) Students with more than 75% marks:
From the ogive, at 75 marks, cumulative frequency = 110
Number of students with more than 75% = 120 - 110 = 10

(iii) Students who did not pass (less than 40% marks):
From the ogive, at 40 marks, cumulative frequency = 52
Number of students who did not pass = 52

(iv) Lower quartile:
\( Q_1 = 120 \times \frac{1}{4} = 30^{th} \text{ term} = 30 \)
In simple words: We made a cumulative table and drew a smooth curve. The median is the mark of the middle student (60th out of 120). We read values from the graph by drawing lines to find answers for each part.

📝 Teacher's Note: Teach students to draw neat perpendicular lines on the ogive. Start from the required frequency on y-axis, go right to the curve, then down to read the marks. Always label your lines clearly.

🎯 Exam Tip: Use graph paper and choose good scale (like 1 cm = 10 marks). Draw smooth curve through all points. Show all construction lines clearly. Write "From the ogive" before each answer to get full marks.

Question 8. Using a graph paper, draw an ogive for the following distribution which shows a record of the width in kilograms of 200 students.

Use your ogive to estimate the following:
(i) The percentage of students weighing 55 kg or more
(ii) The weight above which the heaviest 30% of the student fall
(iii) The number of students who are
(a) underweight
(b) overweight,
if 55-70 kg is considered as standard weight.

Answer:

[Diagram: An ogive curve is drawn on graph paper with weight on x-axis (0 to 90) and cumulative frequency on y-axis (0 to 220). The curve shows points at (45,5), (50,22), (55,44), (60,89), (65,140), (70,171), (75,191), (80,200) connected by a smooth S-shaped curve.]

(i) Number of students weighing more than 55 kg = 200-44 = 156
Therefore, percentage of students weighing 55 kg or more = \( \frac{156}{200} \times 100 = 78\% \)

(ii) 30% of students = \( \frac{30 \times 200}{100} = 60 \)
Heaviest 60 students in weight = 9 + 21 + 30 = 60
weight = 65 kg (from table)

(iii)
(a) underweight students when 55-70 kg is standard = 46 (approx) from graph
(b) overweight students when 55-70 kg is standard = 200- 55-70 = 154 (approx) from graph

📝 Teacher's Note: First make a cumulative frequency table. Then draw the ogive by plotting points. Use the graph to read values carefully. Show students how to read from both axes.

🎯 Exam Tip: Always make the C.f. table first. Plot points correctly. Use a smooth curve. Read values from graph carefully. Show all working steps for percentage calculations.

Question 9. The distribution, given below, shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Answer:

Number of terms = 25

(i) Mean = \( \frac{171}{25} = 6.84 \)

(ii) Median = \( \frac{25+1}{2} \)th term = 13th term = 7

(iii) Mode = 6 as it has maximum frequencies i.e. 6

In simple words: Mean is the average of all marks. Median is the middle value when arranged in order. Mode is the mark that appears most often (9 students got 6 marks).

📝 Teacher's Note: Make the table with fx column. For median in ungrouped data, find the middle position. Mode is the value with highest frequency. Students often mix up median position formula.

🎯 Exam Tip: Write the formula first. Calculate step by step. For median, find position first, then find the value at that position. Mode is easy - just look for highest frequency.

Question 10. The mean of the following distribution is 52 and the frequency of class interval 30-40 is 'f'. Find f.

Answer:

Mean = \( \frac{\sum fx}{\sum f} = \frac{1940 + 35f}{36 + f} \)......(i)

But, mean = 52.....(ii)
From (i) and (ii)

\( \frac{1940 + 35f}{36 + f} = 52 \)
1940 + 35f = 1872 + 52f
1940 + 35f = 1872 + 52f
17f = 68
f = 4

In simple words: We use the mean formula with unknown f. Then we solve the equation to find f. The missing frequency is 4.

📝 Teacher's Note: Make the table carefully with mid-values. Set up the mean equation. Cross multiply and solve for f. Students often make calculation errors, so check the arithmetic twice.

🎯 Exam Tip: Write Mean formula first. Make the fx column with f as unknown. Cross multiply carefully. Always check your answer by substituting back into the original equation.

Question 11. The monthly income of a group of 320 employees in a company is given below:

Draw an ogive of the given distribution on a graph paper taking 2 cm = Rs 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine:
(i) the median wage.
(ii) number of employees whose income is below Rs 8500.
(iii) If salary of a senior employee is above Rs 11,500, find the number of senior employees in the company.
(iv) the upper quartile.

Answer:

Number of employees = 320

[Diagram: An ogive curve is drawn on graph paper with monthly income (thousands) on x-axis (0 to 14) and number of employees on y-axis (0 to 350). Points are plotted at (7,20), (8,65), (9,130), (10,225), (11,285), (12,315), (13,320) and connected with a smooth curve. Horizontal and vertical lines are drawn to read values.]

(i) Median = \( \frac{320}{2} = 160 \)th term

From the graph, median wage = Rs 9,200 (approximately)

(ii) Number of employees with income below Rs 8,500 = 95 (from graph)

(iii) Number of senior employees (salary above Rs 11,500) = 320 - 310 = 10 (from graph)

(iv) Upper quartile = \( \frac{3 \times 320}{4} = 240 \)th term = Rs 10,400 (from graph)

In simple words: We draw a smooth curve using cumulative frequency. Then we read values from the graph. Median is the middle value. Quartiles divide data into four equal parts.

📝 Teacher's Note: First make cumulative frequency table. Plot points carefully with correct scale. Draw smooth curve. Show students how to read values by drawing horizontal and vertical lines on graph.

🎯 Exam Tip: Use the given scale correctly. Plot all points first, then draw smooth curve. For median find n/2 position. For quartiles find 3n/4 position. Read values carefully from graph.

Question 12. A mathematics aptitude test of 50 students was recorded as follows:

Draw a histogram for the above data using a graph paper and locate the mode.

Answer:

[Diagram: This shows a histogram with marks on x-axis (50-60, 60-70, 70-80, 80-90, 90-100) and number of students on y-axis. The bars show heights of 4, 8, 14, 19, and 5 respectively. Lines are drawn to find the mode.]

(i) Draw the histogram

(ii) In the highest rectangle which represents modal class draw two lines AC and BD intersecting at P.

(iii) From P, draw a perpendicular to x-axis meeting at Q.

(iv) Value of Q is the mode = 82 (approx)

In simple words: We draw bars for each group. The tallest bar (80-90 marks with 19 students) is the modal class. We draw special lines to find the exact mode value inside this bar.

📝 Teacher's Note: Show students how to draw the histogram first. Then explain that the modal class is where the tallest bar is. The special lines help us find the exact mode point.

🎯 Exam Tip: Always label your axes clearly. Draw the histogram neatly. Show the construction lines for finding mode. Write "Mode = 82 marks" as your final answer.

Question 13. Marks obtained by 200 students in an examination are given below:

Draw an ogive of the given distribution on a graph paper taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph:

(i) the median wage.

(ii) number of students who failed if minimum marks required to pass is 40

(iii) if scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.

Answer:

First, we create a cumulative frequency table:

Number of students = 200

[Diagram: This shows an ogive (cumulative frequency curve) with marks on x-axis and cumulative frequency on y-axis. The curve rises smoothly from 0 to 200 students.]

(i) Median = \( \frac{200}{2} = 100^{th} \) term

Through mark 100, draw a parallel line to x-axis which meets the curve at A. From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 57 marks (approx)

(ii) The number of students who failed (if minimum marks required to pass is 40) = 46 (approx from the graph)

(iii) The number of students who secured grade one in the examination = 200 - 188 = 12 (approx from the graph)

In simple words: An ogive is a curved line that shows how many students scored up to each mark. We use it to find the middle score (median) and count students in different groups.

📝 Teacher's Note: Help students understand that cumulative frequency means "total up to this point." Start with small numbers and build up. Show how to read values from the graph by drawing lines.

🎯 Exam Tip: Always make the cumulative frequency table first. Draw the ogive carefully with smooth curves. Show all construction lines clearly. Write answers with "approx" when reading from graph.

Question 14. The marks obtained by 40 students in a short assessment is given below, where a and b are two missing data.

If mean of the distribution is 7.2, find a and b.

Answer:
Given: Mean = 7.2, Total students = 40
Step 1: Use the mean formula.
Mean = \( \frac{\sum f_i x_i}{\sum f_i} \)
Step 2: Set up the equation.
\( 7.2 = \frac{6 \times 5 + a \times 6 + 16 \times 7 + 13 \times 8 + b \times 9}{6 + a + 16 + 13 + b} \)
\( 7.2 = \frac{246 + 6a + 9b}{35 + a + b} \)
\( \implies \) \( 1.2a - 1.8b = -6 \).......(i)
Step 3: Use total number of students.
Total number of students = \( 6 + a + 16 + 13 + b \)
\( \implies \) \( 40 = 35 + a + b \)
\( \implies \) \( a + b = 5 \).......(ii)
Step 4: Solve the equations.
Multiply equation (ii) by 1.8 and add it to equation (i)
1.8a + 1.8b = 9
1.2a - 1.8b = -6
3a = 3
\( \implies \) a = 1
Step 5: Find b.
Substituting a = 1 in equation (ii) we get,
1 + b = 5
\( \implies \) b = 4
Therefore: a = 1, b = 4
In simple words: We used two facts - the total students is 40 and the average mark is 7.2. This gave us two equations to solve for the missing numbers a and b.

📝 Teacher's Note: Remind students that mean formula needs both numerator (sum of marks × frequency) and denominator (total frequency). Always check that total students adds up correctly.

🎯 Exam Tip: Write "Given", set up equations clearly, and show each step. Check your answer by substituting back - the mean should come out as 7.2 and total should be 40.

Question 15. Find the mode and the median of the following frequency distribution.

Answer:
Finding Mode:
Since the frequency for x = 14 is maximum (9).
So Mode = 14.
Finding Median:
First, create cumulative frequency table:

N = 29
Median = \( \left(\frac{n+1}{2}\right)^{th} \) term
= \( \left(\frac{30}{2}\right)^{th} \) term
= \( 15^{th} \) term
= frequency of the \( 15^{th} \) term
According to the table it can be observed that the value of x from the \( 13^{th} \) term to the \( 17^{th} \) term is 13.
So the median = 13.
Mode = 14, Median = 13
In simple words: Mode is the value that appears most often (14 appears 9 times). Median is the middle value when arranged in order - the 15th value out of 29 is 13.

📝 Teacher's Note: For mode, just find the highest frequency. For median in discrete data, count to the middle position using cumulative frequency. Show students how to read the table step by step.

🎯 Exam Tip: For mode, clearly state "highest frequency" in your working. For median, always make the cumulative frequency table first. Show the calculation for finding the middle term position.

Question 16. The median of the observations 11, 12, 14, (x - 2) (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Answer:
Given:
Data in ascending order: 11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47
Total number of observations = n = 9 (odd)
Median = 24
Step 1: Find the median term.
\( \implies \) Median = \( \left(\frac{n+1}{2}\right)^{th} \) term = \( \left(\frac{9+1}{2}\right)^{th} \) term = 5th term
Step 2: Set up the equation.
Given, median = 24
\( \implies \) 5th term = 24

Question 17. The number 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x.
Answer:
Data in ascending order: 6, 8, 10, 12, 13 and x
Total number of observations = n = 6 (even)

Step 1: Find the mean.
Mean = \( \frac{\sum x}{n} = \frac{6 + 8 + 10 + 12 + 13 + x}{6} = \frac{49 + x}{6} \)

Step 2: Find the median.
For even number of observations:
Median = \( \frac{\left(\frac{n}{2}\right)^{th} \text{ term} + \left(\frac{n}{2} + 1\right)^{th} \text{ term}}{2} \)
Median = \( \frac{3^{rd} \text{ term} + 4^{th} \text{ term}}{2} = \frac{10 + 12}{2} = \frac{22}{2} = 11 \)

Step 3: Use the given condition.
Given: Mean = Median
\( \frac{49 + x}{6} = 11 \)
\( \implies \) 49 + x = 66
\( \implies \) x = 17

In simple words: We found the middle value (median) first. Then we made the average equal to this middle value. This gave us x = 17.

📝 Teacher's Note: Show students that median stays the same because x comes after 13 in order. Only the mean changes when x changes. This helps them see the difference between mean and median.

🎯 Exam Tip: Always write "Given: Mean = Median" and show the median calculation clearly. The median formula for even numbers is very important to remember.

Question 18. (Use a graph paper for this question). The daily pocket expenses of 200 students in a school are given below: Draw a histogram representing the above distribution and estimate the mode from the graph.
Answer:

[Diagram: A histogram showing rectangular bars for each class interval. The highest bar is at 20-25 with frequency 50. Two diagonal lines AC and BD are drawn from the top corners of this highest rectangle, intersecting at point E. A perpendicular line from E meets the x-axis at L = 21.5]

The modal class is 20-25 (highest frequency = 50).
Using the graphical method:
In the highest rectangle which represents modal class, draw two lines AC and BD intersecting at E.
From E, draw a perpendicular to x-axis meeting at L.
Value of L is the mode. Hence, mode = 21.5

In simple words: We draw the tallest bar on the graph. Then we draw two diagonal lines that cross inside this bar. Where they cross, we drop a line down to find the mode.

📝 Teacher's Note: Use real graph paper and show students how to draw the diagonal lines carefully. The crossing point gives the exact mode value. This method is more accurate than just using the middle of the class.

🎯 Exam Tip: Always identify the modal class first (highest frequency). Draw the diagram neatly and mark all points clearly. Write "mode = 21.5" as your final answer.

Question 19. The marks obtained by 100 students in a mathematics test are given below: Draw an ogive for the given distribution on a graph sheet. Use a scale of 2 cm = 10 units on both the axes. Use the ogive to estimate: (i) Median (ii) Lower quartile (iii) Number of students who obtained more than 85% marks in the test. (iv) Number of students failed, if the pass percentage was 35.
Answer:

[Diagram: An ogive (cumulative frequency curve) showing a smooth S-shaped curve rising from (0,0) to (100,100). Horizontal lines are drawn at y = 25, 50, and 95 to find quartiles and other values.]

From the ogive:
(i) Median = \( \left(\frac{N}{2}\right)^{th} \) term = \( \left(\frac{100}{2}\right)^{th} \) term = 50th term = 45

(ii) Lower quartile = \( \left(\frac{N}{4}\right)^{th} \) term = \( \left(\frac{100}{4}\right)^{th} \) term = 25th term = 32

(iii) Number of students who obtained more than 85% marks = 100 - 94 = 6

(iv) Number of students who failed = 29

In simple words: We plot points showing how many students scored up to each mark. Then we draw a smooth curve. We can read any value by drawing horizontal and vertical lines on this graph.

📝 Teacher's Note: Make sure students understand that ogive shows "less than" values. For median, we go to N/2 = 50 on y-axis and read the x-value. Practice drawing horizontal and vertical lines on graph paper.

🎯 Exam Tip: Always make a cumulative frequency table first. Plot points carefully and draw a smooth curve. For any quartile, use the formula position = (n×quartile)/4 and read from the graph.

Question 20. The mean of following numbers is 68. Find the value of 'x'. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence, estimate the median.
Answer:
Step 1: Use the mean formula.
Mean = \( \frac{\text{Sum of all observations}}{\text{Total number of observations}} \)

\( 68 = \frac{45 + 52 + 60 + x + 69 + 70 + 26 + 81 + 94}{9} \)

\( 68 = \frac{497 + x}{9} \)
\( \implies \) 612 = 497 + x
\( \implies \) x = 612 - 497
\( \implies \) x = 115

Step 2: Find the median.
Data in ascending order: 26, 45, 52, 60, 69, 70, 81, 94, 115

Since the number of observations is odd, the median is the \( \left(\frac{n + 1}{2}\right)^{th} \) observation.
\( \implies \) Median = \( \left(\frac{9 + 1}{2}\right)^{th} \) observation = 5th observation

Hence, the median is 69.

In simple words: We used the given mean to find x. Then we arranged all numbers in order and picked the middle one (5th number out of 9).

📝 Teacher's Note: Show students to always arrange data in order before finding median. For odd numbers, the middle position is easy to find: (n+1)/2. Make them count carefully to avoid mistakes.

🎯 Exam Tip: First find x using the mean formula. Then arrange ALL numbers including x in ascending order. For 9 numbers, median is the 5th number. Show your working clearly.

Question 21. The marks of 10 students of a class in an examination arranged in ascending order is as follows: 13, 35, 43, 46, x, x + 4, 55, 61, 71, 80. If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Answer:
Here the number of observations n = 10, which is even.

Step 1: Find the median formula for even numbers.
Median = \( \frac{\left(\frac{n}{2}\right)^{th} \text{ term} + \left(\frac{n}{2} + 1\right)^{th} \text{ term}}{2} \)

\( 48 = \frac{\left(\frac{10}{2}\right)^{th} \text{ term} + \left(\frac{10}{2} + 1\right)^{th} \text{ term}}{2} \)

\( 48 = \frac{5^{th} \text{ term} + 6^{th} \text{ term}}{2} \)

\( 48 = \frac{x + (x + 4)}{2} \)

Step 2: Solve for x.
\( 96 = 2x + 4 \)
\( 92 = 2x \)
\( x = 46 \)

Step 3: Find the complete data and mode.
So, the given data is: 13, 35, 43, 46, 46, 50, 55, 61, 71, 80
In the given data, 46 occurs most frequently.
\( \therefore \) Mode = 46

In simple words: We found that the 5th and 6th numbers are x and x+4. Their average is 48. This gave us x = 46. The number 46 appears twice, so it is the mode.

📝 Teacher's Note: Make students see that x = 46 and x+4 = 50 are the middle two values. Mode is the value that repeats most. Here 46 appears twice while others appear once.

🎯 Exam Tip: For even numbers, median is average of two middle terms. Write the complete data after finding x. Mode is the most frequent value - count how many times each number appears.

Question 22. The daily wages of 80 workers in a project are given below. Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs. 50 on x-axis and 2 cm = 10 workers on y-axis). Use your ogive to estimate: i. the median wages of the workers. ii. The lower quartile wage of workers. iii. the number of workers who earn more than Rs. 625 daily.
Answer:

[Diagram: An ogive showing cumulative frequency curve for the wages distribution. The curve starts at (400,0) and rises smoothly to (750,80). Horizontal lines are drawn at y = 20, 40, and specific points to find quartiles.]

From the ogive:
i. Median wages = \( \left(\frac{N}{2}\right)^{th} \) observation = \( \left(\frac{80}{2}\right)^{th} \) = 40th observation ≈ Rs. 580

ii. Lower quartile = \( \left(\frac{N}{4}\right)^{th} \) observation = \( \left(\frac{80}{4}\right)^{th} \) = 20th observation ≈ Rs. 540

iii. Number of workers earning more than Rs. 625: From the graph, about 62 workers earn up to Rs. 625. So workers earning more than Rs. 625 = 80 - 62 = 18

In simple words: We draw a curve showing how many workers earn up to each wage level. Then we read values by drawing lines from the required positions on the y-axis.

📝 Teacher's Note: Show students how to make cumulative frequency table first. For 80 workers, median is at position 40, lower quartile at position 20. Draw horizontal lines from these y-values to read x-values.

🎯 Exam Tip: Always use the given scale (2 cm = Rs. 50). Mark points carefully on graph paper. For "more than Rs. 625", find cumulative frequency up to 625, then subtract from total 80.

Question 23. The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to:
(i) Frame a frequency distribution table.
(ii) To calculate mean.
(iii) To determine the Modal class.

[Diagram: A histogram showing marks on X-axis (0 to 50) and number of students on Y-axis (0 to 11). The bars show: 0-10 marks has 2 students, 10-20 marks has 5 students, 20-30 marks has 8 students, 30-40 marks has 4 students, and 40-50 marks has 6 students.]

Answer:

(i) The frequency distribution table is as follows:

(ii) To calculate mean:

Mean = \( \frac{\Sigma fx}{\Sigma f} = \frac{695}{25} = 27.8 \)

(iii) Here the maximum frequency is 8 which is corresponding to class 20-30.
Hence, the modal class is 20-30.

In simple words: We read the heights of bars from the histogram to make the table. For mean, we find the middle point of each class and multiply by frequency. Modal class is the one with the highest bar (most students).

📝 Teacher's Note: Show students how to read bar heights from histogram. The middle point of 0-10 is 5, of 10-20 is 15, and so on. This helps them understand class marks easily.

🎯 Exam Tip: Always make the frequency table first. Write "Modal class = 20-30" clearly. For mean, show the fx column and write the formula. Check that total frequency matches the given number of students.