ICSE Solutions Frank Brothers Class 9 Mathematics Chapter 23 Graphical Representation Of Statistical Data have been provided below and is also available in Pdf for free download. The Frank Brothers ICSE solutions for Class 9 Mathematics have been prepared as per the latest syllabus and ICSE books and examination pattern suggested in Class 9. Questions given in ICSE Frank Brothers book for Class 9 Mathematics are an important part of exams for Class 9 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for ICSE Class 9 Mathematics and also download more latest study material for all subjects. Chapter 23 Graphical Representation Of Statistical Data is an important topic in Class 9, please refer to answers provided below to help you score better in exams
Frank Brothers Chapter 23 Graphical Representation Of Statistical Data Class 9 Mathematics ICSE Solutions
Class 9 Mathematics students should refer to the following ICSE questions with answers for Chapter 23 Graphical Representation Of Statistical Data in Class 9. These ICSE Solutions with answers for Class 9 Mathematics will come in exams and help you to score good marks
Chapter 23 Graphical Representation Of Statistical Data Frank Brothers ICSE Solutions Class 9 Mathematics
Question 6. Draw a histogram for the following frequency distribution:
| Train fare (Rs.) | No. of travellers |
|---|---|
| 0 - 50 | 25 |
| 50 - 100 | 40 |
| 100 - 150 | 36 |
| 150 - 200 | 20 |
| 200 - 250 | 17 |
| 250 - 300 | 12 |
Answer: The frequency distribution provided is in exclusive form. To plot the histogram, we place the class boundaries along the horizontal axis and the corresponding frequencies along the vertical axis, using appropriate scales. With the class intervals forming the bases and the frequencies representing the heights, we draw rectangles to complete the graph. In simple words: This graph uses tall blocks to show how many travelers paid different ranges of train fares. The height of each block corresponds to the number of travelers in that range.
Exam Tip: Always make sure to write down the axis titles and define the scale used on both axes to score complete marks.
Question 7. The following frequency distribution table shows the cost of living in a city in a period of 2 years. Draw a histogram for this frequency distribution:
| Cost of living | No. of months |
|---|---|
| 2000 - 2500 | 3 |
| 2500 - 3000 | 4 |
| 3000 - 3500 | 2 |
| 3500 - 4000 | 5 |
| 4000 - 4500 | 3 |
| 4500 - 5000 | 2 |
| 5000 - 5500 | 4 |
| 5500 - 6000 | 1 |
Answer: We plot the class limits along the horizontal axis and the frequencies along the vertical axis, choosing a convenient scale for both. By using the class intervals as the base and their respective frequencies as the height, we construct adjacent rectangles to form the histogram. Since the class boundaries do not begin at zero, we insert a break (kink) on the x-axis between zero and the lower limit of our initial class. In simple words: A histogram displays the number of months falling into each range of cost of living. A small break is put at the beginning of the scale because the numbers start far from zero.
Exam Tip: Do not forget to construct a zigzag kink near the origin on the horizontal axis whenever the first class boundary is not zero.
Question 8. Draw a histogram for the following frequency table:
| Class interval | Frequency |
|---|---|
| 5 - 9 | 5 |
| 10 - 14 | 9 |
| 15 - 19 | 12 |
| 20 - 24 | 10 |
| 25 - 29 | 16 |
| 30 - 34 | 12 |
Answer: Since the provided class intervals are inclusive, we must convert them to an exclusive form. We then place these adjusted class boundaries on the x-axis and the frequencies on the y-axis, selecting a proper scale. The rectangles are drawn with class intervals as bases and their frequencies as heights to construct the histogram. Because the values do not commence at zero, we place a zigzag kink on the x-axis before the first class limit.
| Class interval | Frequency |
|---|---|
| 4.5 - 9.5 | 5 |
| 9.5 - 14.5 | 9 |
| 14.5 - 19.5 | 12 |
| 19.5 - 24.5 | 10 |
| 24.5 - 29.5 | 16 |
| 29.5 - 34.5 | 12 |
In simple words: We first make the intervals continuous by shifting the class boundaries by 0.5. Then, we construct adjacent rectangles where heights represent the frequency of each range.
Exam Tip: First check whether the intervals are inclusive or exclusive. If they are inclusive, they must be converted into exclusive limits before drawing the histogram.
Question 10. Draw a histogram for the following cumulative frequency table:
| Class interval | Cumulative Frequency |
|---|---|
| Less than 10 | 6 |
| Less than 20 | 10 |
| Less than 30 | 18 |
| Less than 40 | 32 |
| Less than 50 | 40 |
Answer: We begin by changing the given cumulative data into a standard exclusive frequency table. Next, we map the class boundaries on the horizontal axis and the frequencies on the vertical axis using matching scales. Rectangles are drawn with the class intervals as their widths and the corresponding frequencies as their vertical heights.
| Class interval | Frequency |
|---|---|
| 0 - 10 | 6 |
| 10 - 20 | 4 |
| 20 - 30 | 8 |
| 30 - 40 | 14 |
| 40 - 50 | 8 |
In simple words: We first convert the cumulative data to find the actual frequency for each class. Then, we draw a standard histogram showing those frequencies as bar heights.
Exam Tip: Convert cumulative frequencies to standard interval frequencies first by subtracting successive terms before drawing the bar rectangles.
Question 11. Draw a histogram and a frequency polygon for the following frequency distribution:
| Marks | No. of Students |
|---|---|
| 0 - 20 | 12 |
| 20 - 40 | 18 |
| 40 - 60 | 30 |
| 60 - 80 | 25 |
| 80 - 100 | 15 |
Answer: We plot the class intervals on the horizontal axis and the frequencies on the vertical axis, choosing a proper scale. By drawing rectangles where the bases are the class limits and the heights are the frequencies, we construct the histogram. To draw the frequency polygon, we locate the midpoints of the top edge of each rectangle. These points, along with the midpoints of two hypothetical classes added at both ends of the range, are connected sequentially using straight lines. In simple words: First, we draw bars for each class interval. Then, we find the middle of each bar's top and connect those middle points with straight lines to form the polygon.
Exam Tip: To draw a frequency polygon along with a histogram, mark midpoints on top of each rectangle, then connect them along with two additional zero-frequency endpoints on both sides.
Question 12. Draw a histogram and a frequency polygon for the following frequency distribution:
| Wages (Rs.) | No. of Workers |
|---|---|
| 150 - 200 | 25 |
| 200 - 250 | 40 |
| 250 - 300 | 35 |
| 300 - 350 | 28 |
| 350 - 400 | 30 |
| 400 - 450 | 22 |
Answer: We set up the class limits on the x-axis and the frequencies on the y-axis with appropriate scaling. Using the intervals as bases and their frequencies as heights, we build the rectangles for the histogram. Following this, we mark the midpoints of the top horizontal edge of each bar. We then connect these midpoints, including the midpoints of two imaginary outer intervals on each end, with consecutive straight line segments. Since our class boundaries do not start at zero, a kink is inserted on the horizontal axis between zero and the initial class limit. In simple words: We plot bars for each wage range and then join the top centers of these bars with straight lines. The line is also extended to the bottom line at both ends.
Exam Tip: Ensure that both the histogram rectangles and the polygon lines are neatly plotted and clearly distinct on the same coordinate axes.
Question 13. Draw a frequency polygon for the following data:
| Expenses (Rs.) | No. of families |
|---|---|
| 100 - 150 | 22 |
| 150 - 200 | 37 |
| 200 - 250 | 26 |
| 250 - 300 | 18 |
| 300 - 350 | 10 |
| 350 - 400 | 5 |
Answer: The class boundaries are plotted on the horizontal axis and the frequencies on the vertical axis, using convenient scales. We determine the class mark (midpoint) for every interval. Next, we plot coordinates \( (x_1, y_1) \) on our graph, where \( x_1 \) is the midpoint and \( y_1 \) is the associated frequency. These points are connected together with straight lines. Finally, we extend the line on both ends to meet the horizontal axis at the midpoints of the imaginary class intervals immediately preceding and succeeding our data set. Since the intervals do not start at zero, we insert a break (kink) between zero and our first boundary limit. In simple words: To make a frequency polygon alone, we calculate the middle value of each range, plot these values against their frequencies, and join them with lines.
Exam Tip: When drawing a frequency polygon without a histogram, calculate the class marks first and map them directly, extending the polygon line to meet the x-axis on both ends.
Question 14. Draw a frequency polygon for the following data:
| Class | Frequency |
|---|---|
| 15 - 20 | 5 |
| 20 - 25 | 12 |
| 25 - 30 | 15 |
| 30 - 35 | 26 |
| 35 - 40 | 18 |
| 40 - 45 | 7 |
Answer: We assign the class limits to the horizontal axis and the frequencies to the vertical axis, select suitable scales, and then find the midpoints (class marks) for each interval. We plot each coordinate pair \( (x_1, y_1) \), where \( x_1 \) represents the midpoint and \( y_1 \) represents the frequency, and connect them sequentially with straight lines. The endpoints of this line graph are joined to the horizontal axis at the midpoints of imaginary classes placed at both ends. Additionally, we draw a zigzag kink near the origin on the horizontal axis because the intervals do not commence at zero. In simple words: We find the midpoint of each class interval, plot those points against their frequencies, and then connect them with a continuous straight-line graph.
Exam Tip: Draw the kink on the x-axis clearly to indicate the discontinuity between zero and the initial class mark.
Question 15. Draw a frequency polygon for the following data:
| Marks | No. of students |
|---|---|
| 5 - 9 | 7 |
| 10 - 14 | 11 |
| 15 - 19 | 15 |
| 20 - 24 | 22 |
| 25 - 29 | 18 |
| 30 - 34 | 5 |
Answer: Since the class intervals are given in an inclusive form, we first convert them to a continuous exclusive range. We then represent the modified limits along the horizontal axis and the frequencies along the vertical axis using appropriate scales. After finding the class mark for each group, we plot the points \( (x_1, y_1) \) on the grid, with \( x_1 \) being the midpoint and \( y_1 \) being the frequency. These plotted points are connected in succession with straight lines. We extend these segments to the horizontal axis by connecting them to the midpoints of imaginary intervals on either side. Since the class boundaries do not begin at zero, a kink is placed on the horizontal axis near the origin.
| Marks | No. of students |
|---|---|
| 4.5 - 9.5 | 7 |
| 9.5 - 14.5 | 11 |
| 14.5 - 19.5 | 15 |
| 19.5 - 24.5 | 22 |
| 24.5 - 29.5 | 18 |
| 29.5 - 34.5 | 5 |
In simple words: First, we adjust the class limits to make them continuous, find the middle point of each adjusted range, plot them, and connect them with straight lines.
Exam Tip: For inclusive data, first convert the ranges into continuous limits and then find their true midpoints before plotting the frequency polygon.
ICSE Frank Brothers Solutions Class 9 Mathematics Chapter 23 Graphical Representation Of Statistical Data
Students can now access the detailed Frank Brothers Solutions for Chapter 23 Graphical Representation Of Statistical Data on our portal. These solutions have been carefully prepared as per latest ICSE Class 9 syllabus. Each solution given above has been updated based on the current year pattern to ensure Class 9 students have the most updated Mathematics content.
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