ML Aggarwal Class 9 Maths Solutions Chapter 19 Coordinate Geometry

 

Exercise 19.1

 

Question 1. Find the mean of 8, 6, 10, 12, 1, 3, 4, 4.
Answer: Use the formula: Mean = Sum of all observations ÷ No. of observations. Adding the values: 8 + 6 + 10 + 12 + 1 + 3 + 4 + 4 = 48. Divide by 8: Mean = 48 ÷ 8 = 6.
In simple words: Add all the numbers together, then divide the total by how many numbers you have.

Exam Tip: Always count how many observations are present before dividing - missing one number in the count is a common mistake.

 

Question 2. 5 people were asked about the time in a week they spend in doing social work in their community. They replied 10, 7, 13, 20 and 15 hours, respectively. Find the mean time in a week devoted by them for social work.
Answer: Use the formula: Mean = Total hours ÷ No. of people. Add the hours: 10 + 7 + 13 + 20 + 15 = 65. Divide by 5 people: Mean = 65 ÷ 5 = 13 hours per week.
In simple words: Find the total time spent, then split it equally among all 5 people.

Exam Tip: The mean represents what each person would spend if they all spent equal time - it's the "fair share" value.

 

Question 3. The enrollment of a school during six consecutive years was as follows: 1620, 2060, 2540, 3250, 3500, 3710. Find the mean enrollment.
Answer: Apply the formula: Mean = Sum of enrollments ÷ No. of years. Add all values: 1620 + 2060 + 2540 + 3250 + 3500 + 3710 = 16680. Divide by 6: Mean = 16680 ÷ 6 = 2780 students.
In simple words: Add all the student numbers across 6 years, then divide by 6 to find the average enrollment per year.

Exam Tip: Check that you've included all six years - this type of problem often has multiple data points to add.

 

Question 4. Find the mean of the first twelve natural numbers.
Answer: The first twelve natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Sum = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78. Mean = 78 ÷ 12 = 6.5.
In simple words: For natural numbers in a row, the mean lands exactly halfway between the first and last number.

Exam Tip: For consecutive numbers, the mean is always the average of the first and last values.

 

Question 5(i). Find the mean of the first six prime numbers.
Answer: The first six prime numbers are 2, 3, 5, 7, 11, 13. Add them: 2 + 3 + 5 + 7 + 11 + 13 = 41. Mean = 41 ÷ 6 = \( \frac{41}{6} \) or 6\( \frac{5}{6} \).
In simple words: List the first six primes, add them, and divide by 6.

Exam Tip: Remember that 2 is a prime number and is the only even prime.

 

Question 5(ii). Find the mean of the first seven odd prime numbers.
Answer: The first seven odd prime numbers are 3, 5, 7, 11, 13, 17, 19. Add them: 3 + 5 + 7 + 11 + 13 + 17 + 19 = 75. Mean = 75 ÷ 7 = \( 10\frac{5}{7} \) or approximately 10.71.
In simple words: All primes except 2 are odd - add these seven numbers and divide by 7.

Exam Tip: When the mean is not a whole number, express it as a fraction or mixed number unless decimals are specifically requested.

 

Question 6(i). The marks (out of 100) obtained by a group of students in a Mathematics test are 81, 72, 90, 90, 85, 86, 70, 93 and 71. Find the mean marks obtained by the group of students.
Answer: Add all marks: 81 + 72 + 90 + 90 + 85 + 86 + 70 + 93 + 71 = 738. There are 9 students. Mean = 738 ÷ 9 = 82 marks.
In simple words: The average score across all students is 82 marks.

Exam Tip: Count the number of marks carefully - one missing or extra value changes the result significantly.

 

Question 6(ii). The mean of the age of three students Vijay, Rahul and Rakhi is 15 years. If their ages are in the ratio 4 : 5 : 6 respectively, then find their ages.
Answer: Let their ages be 4x, 5x, and 6x. Mean age = (4x + 5x + 6x) ÷ 3 = 15x ÷ 3 = 5x. Since mean = 15, we have 5x = 15, so x = 3. Therefore: Vijay's age = 4(3) = 12 years, Rahul's age = 5(3) = 15 years, Rakhi's age = 6(3) = 18 years.
In simple words: Use the ratio to write each age with x, then use the mean to find x, and multiply back.

Exam Tip: Always verify by checking that the three ages average to 15 years: (12 + 15 + 18) ÷ 3 = 15 ✓

 

Question 7. The mean of 5 numbers is 20. If one number is excluded, mean of the remaining numbers becomes 23. Find the excluded number.
Answer: If the mean of 5 numbers is 20, then sum of all 5 = 20 × 5 = 100. After one is removed, 4 numbers remain with mean 23, so sum of remaining 4 = 23 × 4 = 92. The excluded number = 100 - 92 = 8.
In simple words: Find the total of 5 numbers, then subtract the total of 4 numbers to find the one that was left out.

Exam Tip: This backward approach - finding the sum first, then the missing value - is key when you know means but need to find individual values.

 

Question 8. The mean of 25 observations is 27. If one observation is included, the mean still remains 27. Find the included observation.
Answer: The mean of 25 observations is 27, so sum = 27 × 25 = 675. When one observation is added, there are 26 total, and the mean is still 27, so the new sum = 27 × 26 = 702. The included observation = 702 - 675 = 27.
In simple words: Find both sums - before and after adding the new number - then subtract to find what was added.

Exam Tip: When the mean stays the same after adding a number, that number must equal the mean itself.

 

Question 9. The mean of 5 observations is 15. If the mean of first three observations is 14 and that of the last three is 17, find the third observation.
Answer: Sum of all 5 observations = 15 × 5 = 75. Sum of first 3 = 14 × 3 = 42. Sum of last 3 = 17 × 3 = 51. The third observation appears in both groups. Using the relationship: Third observation = Sum of first 3 + Sum of last 3 - Total sum = 42 + 51 - 75 = 18.
In simple words: The third number is counted in both the first group and the last group, so add both sums and subtract the total.

Exam Tip: Identify which observation(s) overlap between groups - this is crucial for solving overlapping-set mean problems.

 

Question 10. The mean of 8 variates is 10.5. If seven of them are 3, 15, 7, 19, 2, 17 and 8, then find the 8th variate.
Answer: Sum of 8 variates = 10.5 × 8 = 84. Sum of 7 known variates = 3 + 15 + 7 + 19 + 2 + 17 + 8 = 71. The 8th variate = 84 - 71 = 13.
In simple words: Find the total sum needed, then subtract what you already have to find the missing value.

Exam Tip: Double-check your addition of the 7 known values - a single arithmetic error will give the wrong answer.

 

Question 11. The mean weight of 8 students is 45.5 kg. Two more students having weights 41.7 kg and 53.3 kg join the group. What is the new mean weight?
Answer: Total weight of 8 students = 45.5 × 8 = 364 kg. After 2 more join: Total weight = 364 + 41.7 + 53.3 = 459 kg. Total students = 10. New mean weight = 459 ÷ 10 = 45.9 kg.
In simple words: Add the new weights to the old total, then divide by the new count of students.

Exam Tip: Always update both the total sum AND the count when members are added or removed from a group.

 

Question 12. Mean of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. Find the correct mean of the observations.
Answer: Incorrect sum = 35 × 9 = 315. The error was reading 81 as 18, a difference of 63. Correct sum = 315 - 18 + 81 = 378. Correct mean = 378 ÷ 9 = 42.
In simple words: Start with the wrong total, remove the wrong value, add the right value, then recalculate the mean.

Exam Tip: When correcting an error, subtract the wrong value and add the correct one to get the accurate sum.

 

Question 13. A student scored the following marks in 11 questions of a question paper: 7, 3, 4, 1, 5, 8, 2, 2, 5, 7, 6. Find the median marks.
Answer: Arrange in ascending order: 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8. There are 11 observations (odd number). The median is the middle value at position (11 + 1) ÷ 2 = 6th position. The 6th term is 5.
In simple words: Sort all numbers from smallest to largest, then pick the one in the middle.

Exam Tip: For an odd number of observations, the median is always a value that actually appears in the data.

 

Question 14. Calculate the mean and the median of the numbers: 2, 3, 4, 3, 0, 5, 1, 1, 3, 2.
Answer: Mean: Sum = 0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5 = 24. Mean = 24 ÷ 10 = 2.4. Median: Arrange in order: 0, 1, 1, 2, 2, 3, 3, 3, 4, 5. There are 10 values (even number). Median = (5th value + 6th value) ÷ 2 = (2 + 3) ÷ 2 = 2.5.
In simple words: For the mean, add and divide. For the median with an even count, find the two middle numbers and split the difference between them.

Exam Tip: When there's an even number of values, the median might not be a value in your list - it's the average of the two middle values.

 

Question 15. A group of students was given a special test in Mathematics. The test was completed by the various students in the following time in (minutes): 24, 30, 28, 17, 22, 36, 30, 19, 32, 18, 20, 24. Find the mean time and median time taken by the students to complete the test.
Answer: Mean: Sum of times = 24 + 30 + 28 + 17 + 22 + 36 + 30 + 19 + 32 + 18 + 20 + 24 = 300. Mean = 300 ÷ 12 = 25 minutes. Median: Arrange in order: 17, 18, 19, 20, 22, 24, 24, 28, 30, 30, 32, 36. There are 12 values. Median = (6th value + 7th value) ÷ 2 = (24 + 24) ÷ 2 = 24 minutes.
In simple words: Add all times and divide by 12 for the mean. Find the average of the two middle times for the median.

Exam Tip: Always sort the data before finding the median - this is the most common mistake in median calculations.

 

Question 16. In a Science test given to a group of students, the marks scored by them (out of 100) are 41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52. Find the mean and median of this data.
Answer: Mean: Add all marks: 41 + 39 + 52 + 48 + 54 + 62 + 46 + 52 + 40 + 96 + 42 + 40 + 98 + 60 + 52 = 822. Mean = 822 ÷ 15 = 54.8. Median: Arrange in ascending order: 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98. There are 15 observations (odd). Median is at position (15 + 1) ÷ 2 = 8th position = 52.
In simple words: Add all marks and divide by 15 for mean. For median with 15 scores, the 8th score (when sorted) is the middle value.

Exam Tip: Notice that the mean (54.8) and median (52) differ here due to outliers like 96 and 98 - the median is less affected by extreme values.

 

Question 17. The points scored by a Kabaddi team in a series of matches are as follows: 7, 17, 2, 5, 27, 15, 8, 14, 10, 48, 10, 7, 24, 8, 28, 18. Find the mean and the median of the points scored by the Kabaddi team.
Answer: Mean: Sum of points = 7 + 17 + 2 + 5 + 27 + 15 + 8 + 14 + 10 + 48 + 10 + 7 + 24 + 8 + 28 + 18 = 248. Mean = 248 ÷ 16 = 15.5 points. Median: Arrange in order: 2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48. There are 16 matches (even). Median = (8th value + 9th value) ÷ 2 = (10 + 14) ÷ 2 = 12.
In simple words: Add all 16 point totals and divide by 16. The median is the average of the two middle scores when sorted.

Exam Tip: With 16 values, the two middle positions are the 8th and 9th - carefully count to find these positions after sorting.

 

Question 18. The following observations have been arranged in ascending order. If the median of the data is 47.5, find the value of x. 17, 21, 23, 29, 39, 40, x, 50, 51, 54, 59, 67, 91, 93.
Answer: There are 14 observations (even count). The median is the average of the 7th and 8th values. The 7th value is x and the 8th value is 50. So, median = (x + 50) ÷ 2 = 47.5. Multiply both sides by 2: x + 50 = 95. Therefore, x = 45.
In simple words: Use the median formula to write an equation with x, then solve for x by multiplying and subtracting.

Exam Tip: Always identify which positions are "middle" for an even number of observations - they determine which values go into the median formula.

 

Question 14(ii). The class marks of a distribution are 9.5, 16.5, 23.5, 30.5, 37.5 and 44.5. Determine the class size and the class limits of the third class.
Answer: The class size is found by taking the gap between any two consecutive class marks. Here, the gap is 16.5 - 9.5 = 7. The third class mark in the list is 23.5. To find the lower limit, subtract half the class size: 23.5 - (7/2) = 23.5 - 3.5 = 20. To find the upper limit, add half the class size: 23.5 + (7/2) = 23.5 + 3.5 = 27. Therefore, class size = 7, lower limit = 20 and upper limit = 27.
In simple words: The gap between class marks is 7. The middle of class 3 is 23.5. So the class goes from 20 to 27.

Exam Tip: Always find class size first by subtracting consecutive class marks. Then use the formula: lower limit = class mark - (class size/2) and upper limit = class mark + (class size/2).

 

Exercise 19.3

 

Question 1. The area under wheat cultivation last year in the following states, correct to the nearest lacs hectares was:
Represent the above information by a bar graph.
Answer: To create a bar graph, follow these steps:

1. Position states along the x-axis.

2. Use 1 cm on the y-axis to represent 20 lac hectares.

3. Draw rectangles based on the data given in the table.

The bar graph is shown in the adjoining figure.
In simple words: A bar graph uses tall or short bars to show how much wheat each state grew. Taller bars mean more wheat, shorter bars mean less wheat.

Exam Tip: Always label both axes clearly and use a consistent scale - this helps anyone reading your graph understand it quickly.

 

Question 2. The number of books sold by a shopkeeper in a certain week was as follows:
Draw a bar graph for the above data.
Answer: To draw the bar graph, follow these steps:

1. Position days along the x-axis.

2. Use 1 cm on the y-axis to represent 50 books.

3. Draw rectangles matching the frequency table provided.

The bar graph is shown in the adjoining figure.
In simple words: Each day gets its own bar. If a day had more book sales, its bar is taller. If fewer books sold, the bar is shorter.

Exam Tip: Make sure your bars have equal width and are evenly spaced - uneven spacing will lose marks.

 

Question 3. Given below is the data of percentage of passes of a certain school in the ICSE for consecutive years:
Draw a bar graph to represent the above data.
Answer: To construct the bar graph, follow these steps:

1. Position years along the x-axis.

2. Use 1 cm on the y-axis to represent 10%.

3. Draw rectangles that match the data in the frequency table.

The bar graph is shown in the adjoining figure.
In simple words: The bar graph shows how pass rates changed year after year. You can see at a glance which years had higher or lower pass rates.

Exam Tip: When drawing multiple graphs, always check that your scale choices fit the data range and leave enough space on your page.

 

Question 4. Birth rate per thousand of different countries over a period is:
Draw a horizontal bar graph to represent the above data.
Answer: To draw the horizontal bar graph, follow these steps:

1. Position countries along the y-axis.

2. Position birth rate along the x-axis.

3. Draw rectangles based on the data shown in the frequency table.

The bar graph is shown in the adjoining figure.
In simple words: In a horizontal bar graph, the bars go left-to-right instead of bottom-to-top. This works well when category names (like country names) are long and would be hard to fit below a vertical bar.

Exam Tip: Horizontal bar graphs are easier to read when labels are lengthy - take advantage of this format.

 

Question 5. Given below is the data of number of students (boys and girls) in class IX of a certain school:
Draw a bar graph to represent the above data.
Answer: To create the bar graph, follow these steps:

1. Position classes along the x-axis.

2. Use 1 cm on the y-axis to represent 5 students.

3. Draw rectangles matching the data in the frequency table.

The bar graph is shown in the adjoining figure.
In simple words: The graph shows both boys and girls for each class side by side, so you can compare them easily. One bar type might be for boys and another for girls.

Exam Tip: When showing two related categories, use different colors or shading patterns - this makes it clear which bar represents which group.

 

Question 6. Draw a histogram to represent the following data:
Answer: To construct the histogram, follow these steps:

1. Use 2 cm on the x-axis to represent 10 marks.

2. Use 1 cm on the y-axis to represent 1 student.

3. Draw rectangles based on the continuous frequency distribution table.

The histogram is shown in the adjoining figure.
In simple words: A histogram is like a bar graph, but the bars touch each other with no gaps. This shows that the data is continuous (one range leads smoothly into the next).

Exam Tip: Remember - in a histogram, bars must touch each other. If they are separated, you have made a bar graph, not a histogram.

 

Question 7. Draw a histogram to represent the following frequency distribution of monthly wages of 255 workers of a factory.
Answer: To construct the histogram, follow these steps:

1. Because the x-axis scale starts at 850 (not zero), a break mark (kink) is shown near the origin on the x-axis. This tells readers that the graph starts at 850, not zero.

2. Use 2 cm on the x-axis to represent 100 rupees.

3. Use 1 cm on the y-axis to represent 10 workers.

4. Draw rectangles matching the continuous frequency distribution table.

The histogram is shown in the adjoining figure.
In simple words: When your data does not start at zero, draw a zigzag break near the start to show that you've jumped ahead. This saves paper and keeps your graph readable.

Exam Tip: Always include a kink or break symbol when your axis does not begin at zero - examiners specifically check for this.

 

Question 8. Draw a histogram for the following data:
Answer: Since class mark is the mid-point of a class, the frequency distribution table becomes:

To construct the histogram, follow these steps:

1. Because the x-axis scale starts at 10 (not zero), a break mark (kink) is shown near the origin. This signals that the graph begins at 10.

2. Use 2 cm on the x-axis to represent 5 units.

3. Use 1 cm on the y-axis to represent 5 units.

4. Draw rectangles matching the continuous frequency distribution table.

The histogram is shown in the adjoining figure.
In simple words: When you have class marks, first find the class range they represent. Then use those classes (not the marks alone) to draw your histogram.

Exam Tip: Always convert class marks back to class intervals before drawing - this is a common mistake students make.

 

Question 9. Draw a histogram for the following frequency distribution:
Answer: The given frequency distribution uses cumulative frequency (running total). Convert it to a regular frequency distribution first:

To construct the histogram, follow these steps:

1. Use 2 cm on the x-axis to represent 2 years.

2. Use 1 cm on the y-axis to represent 3 children.

3. Draw rectangles matching the continuous frequency distribution table.

The histogram is shown in the adjoining figure.
In simple words: Cumulative frequency means a running total (like adding up as you go). To get regular frequencies, subtract each row from the one before it.

Exam Tip: Watch out - if the question gives cumulative data, you must convert it first. Many students forget this step and draw the histogram incorrectly.

 

Question 10. Draw a histogram for the following data:
Answer: The given frequency distribution is discontinuous (there are gaps between classes). First, convert it to a continuous distribution.

To find the adjustment factor:

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

\( = \frac{66 - 65}{2} = \frac{1}{2} = 0.5 \)

Subtract 0.5 from all lower limits and add 0.5 to all upper limits:

To construct the histogram, follow these steps:

1. Because the x-axis scale starts at 58.5 (not zero), a break mark (kink) is shown near the origin. This tells readers that the scale begins at 58.5.

2. Use 2 cm on the x-axis to represent 7 units.

3. Use 1 cm on the y-axis to represent 5 units.

4. Draw rectangles matching the adjusted continuous frequency distribution table.

The histogram is shown in the adjoining figure.
In simple words: When classes have gaps (like 59-65, then 66-72), they are discontinuous. Adjust them by adding and subtracting 0.5 to close the gaps.

Exam Tip: Always check if classes are continuous or discontinuous before drawing. Failing to adjust will result in gaps in your histogram and lost marks.

 

Question 11. Draw a frequency polygon for the following data:
Answer: Create a frequency distribution table with class marks:

Steps to draw the frequency polygon:

1. Because the x-axis scale starts at 30, a break mark (kink) is shown near the origin. This shows that the graph begins at 30.

2. Use 1 cm on the x-axis to represent 10 units.

3. Use 1 cm on the y-axis to represent 5 units.

4. Find the mid-points (class marks) of each class interval.

5. Plot points where the class mark meets its frequency on the graph.

6. Join all consecutive points with straight line segments.

7. Extend the polygon by joining the first point to the mid-point of the class 30 - 40 (with zero frequency) and the last point to the mid-point of class 100 - 110 (with zero frequency).

The frequency polygon is shown in the adjoining figure.
In simple words: A frequency polygon is made by plotting points at class marks and frequencies, then connecting them with straight lines. Extend the ends to touch the x-axis at nearby classes with zero frequency.

Exam Tip: Always extend your polygon to the x-axis at both ends using imaginary classes with zero frequency - omitting this costs marks.

 

Question 12. In a class of 60 students, the marks obtained in a monthly test were as under:
Answer: Create a frequency distribution table with class marks:

Steps to draw the frequency polygon:

1. Use 2 cm on the x-axis to represent 10 marks.

2. Use 1 cm on the y-axis to represent 5 students.
In simple words: Plot points at each class mark with its student count, then connect the points with straight lines. This shows how student numbers change across mark ranges.

Exam Tip: In frequency polygon questions, always prepare the class marks table first - this is the foundation for accurate plotting.

 

Question 13. In a class of 90 students, the marks obtained in a weekly test were as under: Draw a frequency polygon for the above data.
Answer: The given frequency distribution is not continuous. To convert it into a continuous frequency distribution, calculate the adjustment factor.

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2 = (21 - 20) / 2 = 1/2 = 0.5

Reduce the adjustment factor (0.5) from all lower limits and increase it to all upper limits.

Continuous frequency distribution for the given data:

Steps to draw the frequency polygon:

1. Since the scale on the x-axis begins at 10.5, a break (or kink) is marked near the origin on the x-axis to show that the graph starts at 10.5 rather than at zero.

2. Use 1 cm along x-axis = 5 marks.

3. Use 1 cm along y-axis = 5 students.

4. Locate the middle points of each class interval.

5. Plot points using the frequencies and midpoints of the class intervals.

6. Link the plotted points with straight line segments.

7. Extend the first point to the midpoint of class 10.5 - 15.5 (frequency = 0) and extend the last point to the midpoint of class 50.5 - 55.5 (frequency = 0).
In simple words: First, adjust the class boundaries so there are no gaps. Find the middle of each class and plot it with its frequency. Connect all the points with straight lines, and extend the ends to imaginary classes with zero frequency.

Exam Tip: Always show the kink mark on the x-axis when the scale doesn't start at zero. Verify that you have extended the polygon to the zero-frequency classes on both ends.

 

Question 14. In a city, the weekly observations made in a study on the cost of living index are given in the following table: Draw a frequency polygon for the data given above.
Answer: Frequency distribution table:

Steps to draw the frequency polygon:

1. Since the scale on the x-axis starts at 130, a kink is shown near the origin on the x-axis to indicate that the graph begins at 130.

2. Use 2 cm along x-axis = 10 units (cost of living index).

3. Use 2 cm along y-axis = 5 weeks.

4. Find the midpoints of each class interval.

5. Locate points corresponding to the given frequencies and class midpoints, then plot them on the graph.

6. Join the consecutive points using straight line segments.

7. Connect the first point to the midpoint of class 130 - 140 (frequency = 0) and the last point to the midpoint of class 200 - 210 (frequency = 0).
In simple words: Mark the middle value of each class interval on the x-axis. Plot each midpoint against its frequency. Connect all the points in order with straight lines, and extend the ends to touch imaginary zero-frequency classes.

Exam Tip: Remember to include the zero-frequency class extensions on both ends - this is required for a complete frequency polygon. Always use the correct scale as given in the steps.

 

Question 15. Construct a combined histogram and frequency polygon for the following data:
Answer: Frequency distribution table:

Steps of construction of histogram:

1. Since the scale on the x-axis starts at 135, a break (kink) is shown near the origin on the x-axis to indicate that the graph is drawn starting at 135.

2. Use 2 cm along x-axis = 15 rupees.

3. Use 2 cm along y-axis = 5 workers.

4. Build rectangles matching the continuous frequency distribution shown above. The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon:

1. Mark the midpoints of the upper edges of the histogram rectangles.

2. Link the consecutive midpoints using straight line segments.

3. Connect the first point to the midpoint of class 135 - 150 (frequency = 0), and link the last point to the midpoint of class 240 - 255 (frequency = 0).

The required frequency polygon is shown by thick line segments in the diagram.
In simple words: Draw tall rectangles for each class showing how many workers earned that amount. Then mark the centre of the top of each rectangle and connect these centres with straight lines. Extend the lines to zero on both ends.

Exam Tip: The frequency polygon should pass through the midpoints of the upper sides of all the rectangles. Make sure the end extensions touch the zero-frequency classes; forgetting this is a common error.

 

Question 16. In a study of diabetic patients, the following data was obtained: Represent the above data by a histogram and a frequency polygon.
Answer: Frequency distribution table:

Steps of construction of histogram:

1. Use 1 cm along x-axis = 10 years.

2. Use 1 cm along y-axis = 6 patients.

3. Build rectangles matching the continuous frequency distribution shown above. The required histogram appears in the adjoining figure.

Steps of construction of frequency polygon:

1. Identify the midpoints of the upper sides of the histogram rectangles.

2. Join these midpoints using straight line segments.

3. Extend the first point to the midpoint of class 0 - 10 (frequency = 0), and extend the last point to the midpoint of class 80 - 90 (frequency = 0).

The required frequency polygon is shown by thick line segments in the diagram.
In simple words: Build rectangles for each age group showing how many patients are in it. Mark the middle of the top of each bar. Connect these marks with straight lines. Continue the lines to touch zero on both sides.

Exam Tip: Check that your polygon touches the top middle of every rectangle bar. The polygon must extend beyond the first and last classes to the zero-frequency classes - this is essential for full marks.

 

Question 17. The water bills (in rupees) of 32 houses in a locality are given below: Taking class intervals 30 - 40, 40 - 50, 50 - 60, ......, form frequency distribution table. Construct a combined histogram and frequency polygon.
Answer: First, show the given data in the form of a frequency distribution table.

In this case, class intervals show the water bill amounts and frequency shows the number of houses.

Steps of construction of histogram:

1. Use 1 cm along x-axis = 10 units.

2. Use 1 cm along y-axis = 1 unit.

3. Build rectangles matching the continuous frequency distribution shown above. The required histogram appears in the adjoining figure.

Steps of construction of frequency polygon:

1. Identify the midpoints of the upper edges of the histogram rectangles.

2. Join the consecutive midpoints using straight line segments.

3. Extend the first point to the midpoint of class 20 - 30 (frequency = 0), and extend the last point to the midpoint of class 110 - 120 (frequency = 0).

The required frequency polygon is shown by thick line segments in the diagram.
In simple words: Count how many houses fall into each bill range. Draw bars to show these counts. Mark the centre of each bar's top. Connect these centres in order with straight lines, and extend the lines to zero-frequency classes at both ends.

Exam Tip: When creating a frequency table from raw data, use tally marks for accuracy and count carefully. Always extend the polygon to the adjacent zero-frequency classes - this is part of the standard definition of a frequency polygon.

 

Question 18. The number of matchsticks in 40 boxes on counting was found as given below: Taking classes 40 - 42, 42 - 44 ......, construct the frequency distribution table for the above data. Also draw a combined histogram and frequency polygon to represent the distribution.
Answer: First, display the given data in a frequency distribution table as shown below:

Steps of construction of histogram:

1. Since the scale on the x-axis starts at 38, a kink is shown near the origin on the x-axis to show that the graph begins at 38.

2. Use 1 cm along x-axis = 2 units.

3. Use 1 cm along y-axis = 2 units.

4. Build rectangles matching the continuous frequency distribution shown above. The required histogram appears in the adjoining figure.

Steps of construction of frequency polygon:

1. Mark the midpoints of the upper sides of the histogram rectangles.

2. Join the consecutive midpoints using straight line segments.

3. Extend the first point to the midpoint of class 38 - 40 (frequency = 0), and extend the last point to the midpoint of class 54 - 56 (frequency = 0).

The required frequency polygon is shown by thick line segments in the diagram.
In simple words: Count how many boxes have each number of matchsticks. Draw rectangles showing these counts. Mark the top-centre of each rectangle. Connect these marks with straight lines. Extend the lines at the beginning and end to touch zero-frequency classes.

Exam Tip: With small class widths like 2 units, use a smaller scale unit to make the histogram and polygon readable. Always include the zero-frequency class extensions at both ends - this is required to complete a proper frequency polygon.

 

Question 19. The histogram showing the weekly wages (in rupees) of workers in a factory is given alongside. Answer the following about the frequency distribution: (i) What is the frequency of the class 400 - 425? (ii) What is the class having minimum frequency? (iii) What is the cumulative frequency of the class 425 - 450? (iv) Construct a frequency and cumulative frequency table for the given distribution.
Answer:
(i) By studying the graph, the frequency of class 400 - 425 is 18.

(ii) From the graph, the smallest frequency of any class is 4. Thus, the class having minimum frequency is 475 - 500.

(iii) Cumulative frequency of class 425 - 450 = Sum of frequency of class 425 - 450 and all earlier classes = 10 + 18 + 6 = 34. Thus, cumulative frequency of class 425 - 450 is 34.

(iv) The cumulative frequency distribution table for the given distribution is shown below:

In simple words: Read the bar heights from the histogram to find how many workers earn in each wage range. The cumulative frequency for any class means adding up the count from that class and all classes below it.

Exam Tip: When reading a histogram, be precise about the bar heights. For cumulative frequency, always add the current class frequency to the total of all previous classes - never reset the count. Verify your final cumulative frequency equals the sum of all individual frequencies.

 

Question 20. Marks scored by students of class 10A and 10B in a particular class test are as follows: Draw their frequency polygons on the same graph.
Answer: The given frequency distribution is not continuous. To convert it into a continuous frequency distribution:

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2 = (6 - 5) / 2 = 1/2 = 0.5

Reduce the adjustment factor (0.5) from all lower limits and increase it to all upper limits.

Continuous frequency distribution for the given data:

Steps to draw frequency polygon:

1. Use 1 cm along x-axis = 4 units.

2. Use 1 cm along y-axis = 1 unit.

3. Find the midpoints of each class interval.

4. Locate points matching the given frequencies and class midpoints, then plot them.

5. Join the consecutive points using straight line segments.

6. Extend the first point to the midpoint of class - 5.5 - 0.5 (frequency = 0) and join the other end to the midpoint of class 40.5 - 45.5 (frequency = 0).

Draw both polygons on the same graph - one for class 10A and one for class 10B - using different colours or line styles to distinguish between them.
In simple words: Adjust the class limits so they have no gaps. Find the middle of each class. Plot the number of students from each class at its midpoint. Draw two separate lines on the same graph, one for 10A and one for 10B. Extend each line to the zero-frequency classes at both ends.

Exam Tip: When drawing two polygons on the same graph, use clearly different line styles (solid versus dashed) or different colours to distinguish the two classes. Label each polygon clearly, and make sure both extend to the adjacent zero-frequency classes.

 

Question 1. The marks obtained by 17 students in a mathematics test (out of 100) are given below: 91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 66, 68, 48, 49. The range of data is
(a) 46
(b) 54
(c) 90
(d) 100
Answer: (b) 54
In simple words: To find the range, take the biggest score and subtract the smallest score. Here, 100 - 46 = 54.

Exam Tip: Range is always the difference between the highest and lowest values in a dataset. Make sure to identify both correctly before subtracting.

 

Question 2. The class mark of the class 90 - 120 is
(a) 90
(b) 105
(c) 115
(d) 120
Answer: (b) 105
In simple words: The class mark sits in the middle of a class interval. Add the lower and upper limits and divide by 2: (90 + 120) ÷ 2 = 105.

Exam Tip: Class mark is the midpoint of any class interval. Always use the formula: (Lower limit + Upper limit) ÷ 2.

 

Question 3. In a frequency distribution, the mid-value of a class is 10 and the width of the class is 6. The lower limit of the class is
(a) 6
(b) 7
(c) 8
(d) 12
Answer: (b) 7
In simple words: The mid-value is at the center of the class. To find the lower limit, subtract half the width from the mid-value: 10 - (6 ÷ 2) = 10 - 3 = 7.

Exam Tip: Remember that class width is split evenly on both sides of the mid-value. Half the width goes below the mid-value and half goes above it.

 

Question 4. The width of each of 5 continuous classes in a frequency distribution is 5 and the lower limit of the lowest class is 10. The upper limit of the highest class is
(a) 15
(b) 25
(c) 35
(d) 40
Answer: (c) 35
In simple words: Each class spans 5 units. With 5 classes starting at 10, the highest class ends at: 10 + (5 classes × 5 width) = 10 + 25 = 35.

Exam Tip: When dealing with continuous classes, multiply the number of classes by the class width, then add to the lower limit of the first class.

 

Question 5. The class marks of a frequency distribution are given as follows: 15, 20, 25, ........... The class corresponding to the class mark 20 is
(a) 12.5 - 17.5
(b) 17.5 - 22.5
(c) 18.5 - 21.5
(d) 19.5 - 20.5
Answer: (b) 17.5 - 22.5
In simple words: The class marks go up by 5 each time, so the class width is 5. Half of 5 is 2.5. The class with mark 20 has lower limit 20 - 2.5 = 17.5 and upper limit 20 + 2.5 = 22.5.

Exam Tip: When class marks are equally spaced, the gap between marks equals the class width. Use this to find the exact class interval containing any given mark.

 

Question 6. In the class intervals 10 - 20, 20 - 30, the number 20 is included in
(a) 10 - 20
(b) 20 - 30
(c) both the intervals
(d) none of these intervals
Answer: (b) 20 - 30
In simple words: In statistics, when we write a class like 10 - 20, the number 20 is the upper limit, and in standard notation, it belongs to the next class 20 - 30.

Exam Tip: Know the convention used in your course - typically the lower limit is included but the upper limit is not (10-20 means 10 is in, but 20 is not). Always check your textbook's notation.

 

Question 7. A grouped frequency distribution table with class intervals of equal size using 250 - 270 (270 not included in this interval) as one of the class intervals is constructed for the following data: 268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236. The frequency of class 310 - 330 is
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (c) 6
In simple words: Count how many values fall in the range 310 to 330 (not including 330). The values 310, 310, 320, 319, 318, and 316 belong to this class, so the frequency is 6.

Exam Tip: When counting frequency, be careful about the class boundary convention. The upper limit is typically not included, so 330 itself would not count.

 

Question 8. The mean of x - 1, x + 1, x + 3 and x + 5 is
(a) x + 1
(b) x + 2
(c) x + 3
(d) x + 4
Answer: (b) x + 2
In simple words: Add all four numbers: (x-1) + (x+1) + (x+3) + (x+5) = 4x + 8. Divide by 4 to get the mean: (4x + 8) ÷ 4 = x + 2.

Exam Tip: When finding the mean of expressions with variables, combine all like terms first, then divide by the count of items.

 

Question 9. The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is
(a) 28
(b) 30
(c) 35
(d) 38
Answer: (d) 38
In simple words: If five numbers have a mean of 30, their total is 5 × 30 = 150. After removing one number, four numbers have a mean of 28, so their total is 4 × 28 = 112. The removed number is 150 - 112 = 38.

Exam Tip: Use the relationship: Mean = Sum ÷ Count. Always find the total sum first when working with means.

 

Question 10. If the mean of x₁, x₂ is 7.5, and the mean of x₁, x₂, x₃ is 8, then the value of x₃ is
(a) 9
(b) 8
(c) 7.5
(d) 6
Answer: (a) 9
In simple words: From the first mean, x₁ + x₂ = 7.5 × 2 = 15. From the second mean, x₁ + x₂ + x₃ = 8 × 3 = 24. So x₃ = 24 - 15 = 9.

Exam Tip: When means change as new items are added or removed, use subtraction to isolate the new item's value.

 

Question 11. If each observation of the data is increased by 5, then their mean
(a) remains the same
(b) becomes 5 times the original mean
(c) is decreased by 5
(d) is increased by 5
Answer: (d) is increased by 5
In simple words: If every value goes up by 5, the total sum goes up by 5n (where n is the count). When we divide this new sum by n, the mean also goes up by 5.

Exam Tip: Adding or subtracting the same number to all observations shifts the mean by that same number. Multiplying all observations by a factor multiplies the mean by that factor.

 

Question 12. The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be
(a) 50.5
(b) 51
(c) 51.5
(d) 52
Answer: (b) 51
In simple words: The original sum is 100 × 50 = 5000. When we replace 50 with 150, the sum becomes 5000 - 50 + 150 = 5100. The new mean is 5100 ÷ 100 = 51.

Exam Tip: When replacing one value, find the difference (150 - 50 = 100) and add it to the old sum, then recalculate the mean.

 

Question 13. For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are respectively
(a) upper limits of the classes
(b) lower limits of the classes
(c) class marks of the classes
(d) upper limits of preceding classes
Answer: (c) class marks of the classes
In simple words: A frequency polygon is drawn by plotting frequency values (up the y-axis) against class marks (along the x-axis). The class mark is the center point of each class interval.

Exam Tip: Frequency polygons always use class marks on the horizontal axis, not limits. This represents the typical value for each class.

 

Question 14. Median of the numbers 4, 4, 5, 7, 6, 7, 7, 3, 12 is
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (c) 6
In simple words: First, arrange in order: 3, 4, 4, 5, 6, 7, 7, 7, 12. There are 9 numbers. The median is the middle number, which is the 5th number = 6.

Exam Tip: For an odd count of items, the median is the middle item after arranging in order. Its position is (n + 1) ÷ 2.

 

Question 15. The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is
(a) 45
(b) 49.5
(c) 54
(d) 56
Answer: (c) 54
In simple words: Arrange the 10 numbers in order: 22, 34, 39, 45, 54, 54, 56, 68, 78, 84. Since there are 10 numbers (even), the median is the average of the 5th and 6th values: (54 + 54) ÷ 2 = 54.

Exam Tip: For an even count, find the two middle values and take their average. The positions are n ÷ 2 and (n ÷ 2) + 1.

 

Question 16. In a data, 10 numbers are arranged in ascending order. If the 8th entry is increased by 6, then the median increases by
(a) 0
(b) 2
(c) 3
(d) 6
Answer: (a) 0
In simple words: The median of 10 numbers depends on the 5th and 6th values. Since 8 is much further along, changing the 8th number does not change which two middle values we use, so the median stays the same.

Exam Tip: The median only depends on the middle values. Changes to values far from the middle (near the start or end of a sorted list) do not affect the median.

 

Question 17. Consider the following two statements. Statement 1: A histogram consists of a set of adjacent rectangles whose bases are equal to class size, and heights are equal to class frequencies. Statement 2: In a bar graph, the breadth of a rectangle has no significance, whereas in a histogram, the breadth of a rectangle is meaningful and it represents the class size. Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (a) Both the statements are true.
In simple words: A histogram uses touching bars where the width shows class size and height shows frequency. A bar graph uses separate bars where width is just for looks, not meaningful data.

Exam Tip: The key difference: histograms are for continuous data with bars touching and width = class width; bar graphs are for categories with separate bars and arbitrary width.

 

Question. Assertion (A): Runs scored by batsman A are 55, 60, 30, 80 while runs scored by batsman B are 81, 87, 76, 92. Then B has higher range than A. Reason (R): Range is the difference between maximum and minimum values of a variable.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (b) Assertion (A) is false, Reason (R) is true.
In simple words: The reason correctly defines range. For batsman A: range = 80 - 30 = 50. For batsman B: range = 92 - 76 = 16. So A's range is larger, not B's.

Exam Tip: For assertion-reason questions, check each part separately first. Even if the reason is correct, it might not explain the assertion if the assertion is wrong.

 

Question. Assertion (A): Mean of first 5 odd natural numbers is 5. Reason (R): Mean is the middle value of a set of observations.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (a) Assertion (A) is true, Reason (R) is false.
In simple words: The first 5 odd numbers are 1, 3, 5, 7, 9. Their sum is 25, and 25 ÷ 5 = 5, so the assertion is true. However, the middle value (median) is 5, but the mean is found by adding and dividing, not by finding the middle. The reason incorrectly defines mean.

Exam Tip: Don't confuse mean (average) with median (middle value). Mean is calculated by summing all values and dividing by count. Median is the center position when arranged in order.

 

Question 3. Assertion (A): If 6 students score 78, 62, 91, 37, 80 and 66 marks in a subject, then median score is 72. Reason (R): If number of observations is even, then Median = \( \frac{\frac{n}{2} \text{th observation} + \left(\frac{n}{2} + 1\right) \text{th observation}}{2} \), after putting all observations in ascending or descending order.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: First, check the claim: if 6 students score 78, 62, 91, 37, 80 and 66 marks in a subject, then median score is 72. When the scores are arranged in ascending order: 37, 62, 66, 78, 80, 91. Since there are 6 scores, n = 6, which is an even number. Using the formula, Median = \( \frac{3\text{rd observation} + 4\text{th observation}}{2} = \frac{66 + 78}{2} = \frac{144}{2} = 72 \). So Assertion (A) is true. For Reason (R): The statement about the median formula when the count of observations is even is correct. Reason (R) is true, and it directly explains why Assertion (A) is correct.
In simple words: When you arrange the scores from smallest to largest, the middle two scores are 66 and 78. Their average is 72. The formula shown explains exactly how to find the median when you have an even number of values.

Exam Tip: Always arrange data in order before finding the median. For an even count of observations, identify the two middle values and take their average - this is the core concept examiners test.

 

Chapter Test

 

Question 1. Find the mean and the median of the following set of numbers: 8, 0, 5, 3, 2, 9, 1, 5, 4, 7, 2, 5.
Answer: Using the formula, Mean = \( \frac{\text{Sum of observations}}{\text{No. of observations}} \). Sum of observations = 8 + 0 + 5 + 3 + 2 + 9 + 1 + 5 + 4 + 7 + 2 + 5 = 51. Mean = \( \frac{51}{12} = 4.25 \). To find the median, arrange the data in ascending order: 0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9. Here, n = 12, which is even. Using the formula, Median = \( \frac{\frac{12}{2}\text{th observation} + \left(\frac{12}{2} + 1\right)\text{th observation}}{2} = \frac{6\text{th observation} + 7\text{th observation}}{2} = \frac{4 + 5}{2} = \frac{9}{2} = 4.5 \).
In simple words: Add all the numbers and divide by how many there are to get the mean. For the median, arrange the numbers from smallest to largest and find the average of the two middle numbers.

Exam Tip: Always verify that your sum is correct before dividing. When finding the median with an even count, make sure you identify the correct two middle positions and average them accurately.

 

Question 2. Find the mean and the median of all the (positive) factors of 48.
Answer: The positive factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Using the formula, Mean = \( \frac{\text{Sum of positive factors of 48}}{\text{No. of factors}} \). Sum of positive factors of 48 = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 16 + 24 + 48 = 124. Mean = \( \frac{124}{10} = 12.4 \). Here, n = 10, which is even. Using the formula, Median = \( \frac{\frac{10}{2}\text{th observation} + \left(\frac{10}{2} + 1\right)\text{th observation}}{2} = \frac{5\text{th observation} + 6\text{th observation}}{2} = \frac{6 + 8}{2} = \frac{14}{2} = 7 \).
In simple words: Find all numbers that divide evenly into 48. Add them together and divide by how many there are to get the mean. The median is the average of the two middle factors when listed in order.

Exam Tip: When finding factors, check each number systematically - start from 1 and work upward. List all factors in order to ensure you identify the correct middle values for the median.

 

Question 3. The mean weight of 60 students of a class is 52.75 kg. If the mean weight of 35 of them is 54 kg, find the mean weight of the remaining students.
Answer: Using the formula, Mean = \( \frac{\text{Sum of weight of students}}{\text{No. of students}} \). Given that the mean weight of 60 students is 52.75 kg: \( 52.75 = \frac{\text{Total weight}}{60} \). Therefore, Total weight = 60 × 52.75 = 3165 kg. The mean weight of 35 students is 54 kg, so: Total weight of 35 students = 54 × 35 = 1890 kg. The number of remaining students = 60 - 35 = 25. Total weight of 25 students = 3165 - 1890 = 1275 kg. Mean weight of 25 students = \( \frac{1275}{25} = 51 \) kg.
In simple words: Find the total weight of all students first. Then subtract the total weight of the first 35 students from this. Finally, divide what is left by the number of remaining students.

Exam Tip: Always calculate the total sum first before breaking it into parts. Double-check that your remaining count and total match the original numbers.

 

Question 4. The mean age of 18 students of a class is 14.5 years. Two more students of age 15 years and 16 years join the class. What is the new mean age?
Answer: Using the formula, Mean = \( \frac{\text{Total age}}{\text{No. of students}} \). Given that the mean age of 18 students is 14.5 years: \( 14.5 = \frac{\text{Total age}}{18} \). Therefore, Total age = 18 × 14.5 = 261 years. When two more students join: Total age of 20 students = 261 + 15 + 16 = 292 years. So, mean age of 20 students = \( \frac{292}{20} = 14.6 \) years.
In simple words: Find the total age of the original 18 students. Add the ages of the two new students. Then divide the new total by the new count of students.

Exam Tip: When students join or leave, recalculate the total and the count before finding the new mean. A small change in the total can lead to a small change in the mean, depending on whether the new values are above or below the original mean.

 

Question 5. If the mean of the five observations x + 1, x + 3, x + 5, 2x + 2, 3x + 3 is 14, find the mean of first three observations.
Answer: Using the formula, Mean = \( \frac{\text{Sum of observations}}{\text{No. of observations}} \). Sum of observations = (x + 1) + (x + 3) + (x + 5) + (2x + 2) + (3x + 3) = 8x + 14. Given that the mean of the five observations is 14: \( \frac{8x + 14}{5} = 14 \). Therefore, 8x + 14 = 70, which gives 8x = 56, so x = 7. The first three observations are: x + 1 = 7 + 1 = 8, x + 3 = 7 + 3 = 10, x + 5 = 7 + 5 = 12. Mean of first three observations = \( \frac{8 + 10 + 12}{3} = \frac{30}{3} = 10 \).
In simple words: Combine all the terms with x and all the numbers separately to get the total sum. Use the given mean to find the value of x. Then substitute that value back into the first three expressions and calculate their mean.

Exam Tip: When expressions contain variables, combine like terms first. Use the given mean condition to solve for the variable, then substitute back carefully to find the required answer.

 

Question 6. The mean height of 36 students of a class is 150.5 cm. Later on, it was detected that the height of one student was wrongly copied as 165 cm instead of 156 cm. Find the correct mean height.
Answer: Using the formula, Mean = \( \frac{\text{Total height}}{\text{No. of students}} \). Given that the mean height of 36 students is 150.5 cm: \( 150.5 = \frac{\text{Total height}}{36} \). Therefore, Total height = 150.5 × 36 = 5418 cm. The height of one student was wrongly copied as 165 cm instead of 156 cm. Actual total height = 5418 - 165 + 156 = 5409 cm. Actual Mean height = \( \frac{5409}{36} = 150.25 \) cm.
In simple words: Find the total height using the incorrect mean. Subtract the wrong height and add the correct height. Divide the corrected total by the number of students to get the actual mean.

Exam Tip: When correcting data, remove the incorrect value and add the correct one in sequence. Recalculate the mean using the corrected total - this type of correction question appears frequently in exams.

 

Question 7. The mean of 40 items is 35. Later on, it was discovered that two items were misread as 36 and 29 instead of 63 and 22. Find the correct mean.
Answer: Using the formula, Mean = \( \frac{\text{Sum of observations}}{\text{No. of observations}} \). Given that the mean of 40 items is 35: \( 35 = \frac{\text{Sum of observations}}{40} \). Therefore, Sum of observations = 40 × 35 = 1400. Two items were misread as 36 and 29 instead of 63 and 22. Actual sum of observations = 1400 - 36 - 29 + 63 + 22 = 1420. Correct mean = \( \frac{1420}{40} = 35.5 \).
In simple words: Find the incorrect total using the given mean. Remove the two wrong values and add the two correct values. Divide the corrected total by the number of items to find the actual mean.

Exam Tip: Keep track of which values are being removed and which are being added. A systematic approach - subtract the wrong values first, then add the correct ones - helps avoid arithmetic errors.

 

Question 8. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x. 29, 32, 48, 50, x, x + 2, 72, 75, 87, 91.
Answer: Here n = 10, which is even. Using the formula, Median = \( \frac{\frac{n}{2}\text{th observation} + \left(\frac{n}{2} + 1\right)\text{th observation}}{2} \). Therefore, \( 63 = \frac{\frac{10}{2}\text{th observation} + \left(\frac{10}{2} + 1\right)\text{th observation}}{2} = \frac{5\text{th observation} + 6\text{th observation}}{2} \). This gives \( 126 = 5\text{th observation} + 6\text{th observation} = x + (x + 2) = 2x + 2 \). Solving: 2x = 124, so x = 62.
In simple words: Count to find which two observations are in the middle. The median is their average. Set up an equation using this information and solve for x.

Exam Tip: In questions with unknowns, identify the middle positions first. Remember that with 10 items, the 5th and 6th values form the middle pair - use this to set up your equation.

 

Question 9. Draw a histogram showing marks obtained by the students of a school in a Mathematics paper carrying 60 marks.
Answer: Steps of construction of histogram:
1. Take 2 cm along the x-axis = 10 marks.
2. Take 1 cm along the y-axis = 5 students.
3. Construct rectangles corresponding to the continuous frequency distribution table.

The histogram displays the mark ranges on the horizontal axis and the number of students on the vertical axis. The bars show that the highest concentration of students scored between 50 and 60 marks (40 students), with smaller frequencies in the lower and middle mark ranges.
In simple words: A histogram uses rectangles of different heights to show how many students got each range of marks. Taller bars show mark ranges with more students.

Exam Tip: Label both axes clearly with the scale and units. Make sure the height of each bar exactly matches the frequency for that class interval - precision in drawing is important for full marks.

 

Question 10. In a class of 60 students, the marks obtained in a surprise test were as under: Represent the above data by a histogram and a frequency polygon.
Answer: Steps of construction of histogram:
1. Since the scale on the x-axis starts at 8, a break (kink) is shown near the origin on the x-axis to indicate that the graph is drawn to scale beginning at 8.
2. Take 1 cm along the x-axis = 10 marks.
3. Take 1 cm along the y-axis = 5 students.
4. Construct rectangles corresponding to the continuous frequency distribution table.

Steps of construction of frequency polygon:
1. Mark the mid-points of upper bases of rectangles of the histogram.
2. Join the consecutive mid-points by line segments.
3. Join the first end point with the mid-point of class 8 - 14 with zero frequency, and join the other end point with the mid-point of class 62 - 68 with zero frequency.

The frequency polygon is shown by thick line segments connecting these mid-points, creating a closed figure that gives a visual representation of the data distribution.
In simple words: Plot dots at the middle top of each bar in the histogram. Connect the dots with straight lines. Extend the lines to the zero-frequency classes on both ends to close the polygon.

Exam Tip: The frequency polygon and histogram display the same information in different forms. The polygon makes it easier to see trends. Always include the two zero-frequency classes at both ends to properly close the figure.

 

Question 11. Construct a combined histogram and frequency polygon for the following distribution:
Answer: The given frequency distribution is discontinuous. To convert it into continuous frequency distribution: Adjustment factor = \( \frac{\text{Lower limit of one class - Upper limit of previous class}}{2} = \frac{101 - 100}{2} = \frac{1}{2} = 0.5 \). Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits. The continuous frequency distribution becomes:

Steps of construction of histogram:
1. Since the scale on the x-axis starts at 80.5, a break (kink) is shown near the origin on the x-axis to indicate that the graph is drawn to scale beginning at 80.5.
2. Take 1 cm along the x-axis = 10 units.
3. Take 1 cm along the y-axis = 5 units.
4. Construct rectangles corresponding to the continuous frequency distribution table.

Steps of construction of frequency polygon:
1. Mark the mid-points of upper bases of rectangles of the histogram.
2. Join the consecutive mid-points by line segments.
3. Join the first end point with the mid-point of class 80.5 - 90.5 with zero frequency, and join the other end point with the mid-point of class 160.5 - 170.5 with zero frequency.

The frequency polygon is shown by thick line segments in the diagram.
In simple words: First adjust the class boundaries so there are no gaps between classes. Then draw rectangles for the histogram and connect the tops of these rectangles to form the frequency polygon.

Exam Tip: Always calculate the adjustment factor and apply it consistently to all boundaries. The frequency polygon should touch the zero-frequency classes at both ends - this closing step is often checked by examiners.

 

Question 12. The Water bills (in rupees) of 40 houses in a locality are given below: Form a frequency distribution table with a class size of 10. Also represent the above data with a histogram and frequency polygon.
Answer: From the data: Least term = 52 and Greatest term = 129. Range = Greatest term - Least term = 129 - 52 = 77.

Steps of construction of histogram:
1. Since the scale on the x-axis starts at 40, a break (kink) is shown near the origin on the x-axis to indicate that the graph is drawn to scale beginning at 40.
2. Take 1 cm along the x-axis = 10 units.
3. Take 1 cm along the y-axis = 2 units.
4. Construct rectangles corresponding to the continuous frequency distribution table.

Steps of construction of frequency polygon:
1. Mark the mid-points of upper bases of rectangles of the histogram.
2. Join the consecutive mid-points by line segments.
3. Join the first end point with the mid-point of class 40 - 50 with zero frequency, and join the other end point with the mid-point of class 130 - 140 with zero frequency.
In simple words: Count how many bills fall into each price range using tally marks. Draw bars for each range in a histogram. Then connect the tops of the bars to make a frequency polygon, extending to the zero-frequency classes on both ends.

Exam Tip: When forming a frequency distribution table, use tally marks to count systematically - this reduces errors. Verify that all frequencies add up to the total number of items before drawing the histogram.

 

Question 13. The data given below represent the marks obtained by 35 students:
21 26 21 20 23 24 22 19 24
26 25 23 26 29 21 24 19 25
26 25 22 23 23 27 26 24 25
30 25 23 28 28 24 28 28

Taking class intervals 19 - 20, 21 - 22 etc., make a frequency distribution for the above data. Construct a combined histogram and frequency polygon for the distribution.
Answer: Looking at the data provided, the smallest mark is 19 and the largest mark is 30, giving a range of 30 - 19 = 11. We establish class intervals as 19 - 20, 21 - 22, and so on.

Since the given frequency distribution is not continuous (there are gaps between classes), we need to make an adjustment. The adjustment factor is calculated as:

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2 = (21 - 20) / 2 = 0.5

We subtract 0.5 from each lower limit and add 0.5 to each upper limit to create a continuous distribution.

Steps to construct the histogram:
(1) Mark a break (kink) near the origin on the x-axis, since the scale begins at 16.5 rather than 0, showing that the graph starts at a scaled position.
(2) Use 2 cm along the x-axis to represent 2 units.
(3) Use 1 cm along the y-axis to represent 2 units.
(4) Draw rectangles matching the adjusted continuous frequency distribution table.

Steps to construct the frequency polygon:
(1) Mark the midpoints of the upper edges of the histogram rectangles.
(2) Connect these midpoints in order using straight line segments.
(3) Extend the polygon by joining the first point to the midpoint of the imaginary class 16.5 - 18.5 (with zero frequency) and the last point to the midpoint of the imaginary class 30.5 - 32.5 (with zero frequency).

In simple words: Count how many marks fall into each group (19-20, 21-22, etc.). Then adjust the class boundaries so there are no gaps. Draw rectangles with the correct heights based on how many marks are in each group. Finally, mark the centres of the rectangles and connect them with lines to show the shape of the data.

Exam Tip: Make sure to show the break symbol (kink) on the x-axis and use the adjusted class intervals for drawing the histogram. The frequency polygon should extend to zero-frequency classes on both ends to complete the shape properly.

 

Question 14. The given histogram and frequency polygon shows the ages of teachers in a school. Answer the following:
(i) What is the class size of each class?
(ii) What is the class whose class mark is 48?
(iii) What is the class whose frequency is maximum?
(iv) Construct a frequency table for the given distribution.
Answer:

(i) From the graph, we observe that the class marks of two consecutive classes are 36 and 30. Therefore, the class size can be found by subtracting:
Class size = 36 - 30 = 6
The class size for each class is 6.

(ii) To find the class with a class mark of 48, we use the relationship between class mark and class limits:
Lower limit = Class mark - (Class size / 2) = 48 - (6 / 2) = 48 - 3 = 45
Upper limit = Class mark + (Class size / 2) = 48 + (6 / 2) = 48 + 3 = 51
The class whose class mark is 48 is 45 - 51.

(iii) Looking at the distribution, the class mark with the highest frequency is 54. Using the same method:
Lower limit = Class mark - (Class size / 2) = 54 - (6 / 2) = 54 - 3 = 51
Upper limit = Class mark + (Class size / 2) = 54 + (6 / 2) = 54 + 3 = 57
The class with the highest frequency is 51 - 57.

(iv)

In simple words: The class size stays the same throughout (it is 6). The class mark tells us where the middle of each group is, so we can work out which class it belongs to. The highest bar on the histogram shows which class has the most teachers. A frequency table lists all the groups, their marks, and how many teachers are in each group.

Exam Tip: Always use the formula Class mark = (Lower limit + Upper limit) / 2 to check your class intervals. When reading from a histogram, identify the tallest rectangle carefully - that class has the maximum frequency. Present your frequency table clearly with all three columns aligned.