Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 23 Graphical Representations

Exercise 23

Answer 1.

(i)

[Diagram: A data table showing Class Interval (0-10, 10-20, 20-30, 30-40, 40-50, 50-60) with corresponding Frequency values (8, 20, 34, 22, 10, 6), followed by a histogram with blue bars showing the frequency distribution.]

📝 Teacher's Note: Show students how to make equal spaces on x-axis for class intervals. Each bar touches the next bar in a histogram. This is different from a bar graph where bars have gaps.

🎯 Exam Tip: Always label both axes clearly. Write "Class Interval" on x-axis and "Frequency" on y-axis. Draw bars touching each other with no gaps.

(ii)

[Diagram: A data table showing Class Interval (130-140, 140-150, 150-160, 160-170, 170-180) with corresponding Frequency values (24, 16, 29, 20, 11), followed by a histogram with blue bars showing the frequency distribution.]

📝 Teacher's Note: Explain that class intervals must be equal in size. Here each interval covers 10 units (like 130-140 is 10 units wide). This makes the histogram fair and easy to read.

🎯 Exam Tip: Check that all class intervals have the same width before drawing. Mark the scale clearly on both axes using equal divisions.

(iii)

[Diagram: A histogram showing the continuous class intervals on x-axis and frequency on y-axis, with blue bars representing the data.]

📝 Teacher's Note: Explain that we subtract 0.5 from lower limit and add 0.5 to upper limit to make classes continuous. This removes gaps between classes like 10-11, 20-21, etc.

🎯 Exam Tip: Always show the step of making classes continuous. Write both the original and continuous intervals. This shows you understand the concept properly.

(iv)

[Diagram: A histogram showing the continuous class intervals on x-axis and frequency on y-axis, with blue bars representing the data.]

📝 Teacher's Note: Point out that the method is the same as part (iii). We always subtract and add 0.5 to make discrete classes continuous for histogram drawing.

🎯 Exam Tip: Show your working clearly. Write the continuous intervals in a table first, then draw the histogram. This gets you method marks even if the drawing has small errors.

(v)

📝 Teacher's Note: Teach students that class mark is the middle point of each class interval. We find class interval by going 5 units left and right of the class mark (since class width is 10).

🎯 Exam Tip: First find the class width by looking at the difference between class marks. Here 25-15=10, so width is 10. Then make intervals of width 10 around each class mark.

(vi)

📝 Teacher's Note: Here class width is 6 (since 12-6=6). So we go 3 units left and right of each class mark. Make sure students check that classes touch each other properly.

🎯 Exam Tip: Always check your class intervals are continuous with no gaps or overlaps. Each upper limit should equal the next lower limit.

Answer 2.

(i)

Steps for drawing frequency polygon:

1. On the x-axis, take 1 cm as 5 units and plot class interval.
2. On the y-axis, take 1 cm as 5 units and plot frequency.
3. Draw rectangles of histogram as per given data.
4. For each rectangle, mark the midpoint of its length at the top part. In this case, points are (5,8), (15,13), (25,27), (35,35), (45,14), (55,3).
5. Mark two more midpoints of zero frequency on x-axis at the start and at the end.
6. Now connect the points using straight lines.

📝 Teacher's Note: Explain that frequency polygon connects the midpoints of histogram bars. It shows the same data as histogram but as a line graph instead of bars.

🎯 Exam Tip: Always mark points at both ends with zero frequency to complete the polygon. Connect all points with straight lines, not curves. Label the axes clearly.

Question (i). Draw a frequency polygon for the given data:

Answer: Steps:

1. Organize the data into equal intervals by subtracting 0.5 from the lower limit of each class and add 0.5 to the upper limit of each class.
2. On the x-axis, take 1 cm as 5 units and plot class interval.
3. On the y-axis, take 1 cm as 5 units and plot frequency.
4. Draw rectangles of histogram as per given data.
5. For each rectangle, mark the midpoint of its length at the top part. In this case, points are (15.5,15),(25.5,28),(35.5,50),(45.5,35),(55.5,20),(65.5,12).
6. Mark two more midpoints of zero frequency on x-axis at the start and at the end.
7. Now connect the points using straight lines.

In simple words: A frequency polygon is made by joining the midpoints of each bar on a histogram with straight lines. This makes a shape like a mountain that shows how the data is spread.

[Diagram: The first image shows a histogram with bars for each class interval, with frequencies marked on top. A curved line connects the midpoints of each bar to form the frequency polygon.]

📝 Teacher's Note: Show students how to find midpoints by adding the two class boundaries and dividing by 2. For example, midpoint of 20-30 is (20+30)/2 = 25. Make them practice this calculation.

🎯 Exam Tip: Always mark the midpoints clearly on your graph. Connect them with straight lines, not curved ones. Add zero frequency points at both ends to complete the polygon.

Question (ii). Draw a frequency polygon for the given data:

Answer: Steps:

1. On the x-axis, take 1 cm as 5 units and plot class interval.
2. On the y-axis, take 1 cm as 5 units and plot frequency.
3. Draw rectangles of histogram as per given data.
4. For each rectangle, mark the midpoint of its length at the top part. In this case, points are (125,10),(175,15),(225,17),(275,12),(325,10).
5. Mark two more midpoints of zero frequency on x-axis at the start and at the end.
6. Now connect the points using straight lines.

In simple words: We follow the same steps as before. Plot the histogram first, then find midpoints of each bar and connect them with straight lines. This creates a polygon shape.

[Diagram: The second image shows a histogram with bars and a frequency polygon line connecting the midpoints of each bar.]

📝 Teacher's Note: Remind students that the class intervals here are already given in proper form. They don't need to adjust the boundaries like in the previous example.

🎯 Exam Tip: Calculate midpoints carefully. For 100-150, midpoint is (100+150)/2 = 125. For 150-200, midpoint is (150+200)/2 = 175. Always show your calculation.

Question (iii). Draw a frequency polygon for the given data:

Answer: Steps:

1. On the x-axis, take 1 cm as 5 units and plot class interval.
2. On the y-axis, take 1 cm as 5 units and plot frequency.
3. Mark the given data on the graph. (10,4),(15,20),(20,40),(25,45),(30,30),(35,25),(40,5)
4. Mark two more midpoints of zero frequency on x-axis at the start and at the end.
5. Now connect the points using straight lines.

In simple words: Here we already have the class marks (midpoints). We just plot these points on the graph and connect them with straight lines. This is the easiest way to draw a frequency polygon.

[Diagram: The third image shows a frequency polygon graph with points plotted at the given coordinates and connected with straight lines forming a bell-like curve.]

📝 Teacher's Note: This is the simplest case because class marks are already given. Students just need to plot the points and connect them. No calculation needed for midpoints.

🎯 Exam Tip: When class marks are given, directly plot them as coordinates (class mark, frequency). Start and end with zero frequency points at 5 and 45 on x-axis.

Question (iv). Draw a frequency polygon for the given data:

Answer: Steps:

1. Find class mark by calculating the average of the class interval.
2. On the x-axis, take 1 cm as 5 units and plot class interval.
3. On the y-axis, take 1 cm as 5 units and plot frequency.
4. Plot the points on the graph. (15,9),(30,17),(50,15),(70,20),(90,14).
5. Mark two more midpoints of zero frequency on x-axis at the start and at the end.
6. Now connect the points using straight lines.

In simple words: We calculate the middle point of each class interval by adding the two numbers and dividing by 2. Then we plot these points and connect them with lines.

[Diagram: The fourth image shows a frequency polygon with an irregular pattern, with the highest point at (70,20) and connecting lines between all plotted points.]

📝 Teacher's Note: Show students that class intervals can have different sizes (10-20 has size 10, but 20-40 has size 20). The midpoint calculation remains the same - add and divide by 2.

🎯 Exam Tip: Always calculate class marks first before plotting. Show your calculation clearly. Note that the class intervals here have unequal widths, but the method stays the same.

Question (v). Draw a frequency polygon for the given data:

Answer: Steps:

1. Make the class intervals continuous by subtracting 0.5 from the lower limit of each class and add 0.5 to the upper limit of each class.
2. Find class mark by calculating the average of the class interval.
3. On the x-axis, take 1 cm as 5 units and plot class interval.
4. On the y-axis, take 1 cm as 5 units and plot frequency.
5. Plot the points on the graph. (5.5,8),(15.5,12),(25.5,10),(35.5,16),(45.5,6).
6. Mark two more midpoints of zero frequency on x-axis at the start and at the end.

In simple words: Here the class intervals have gaps (1-10, then 11-20). We first make them continuous by adjusting boundaries, then find midpoints and plot the polygon.

[Diagram: The fifth image shows a frequency polygon with points connected by straight lines, with the highest peak at (35.5,16).]

📝 Teacher's Note: Explain to students why we need continuous intervals for proper graphs. Show them that 1-10 becomes 0.5-10.5, and 11-20 becomes 10.5-20.5, so there are no gaps.

🎯 Exam Tip: When class intervals are not continuous (have gaps), always convert them first. Subtract 0.5 from lower limit and add 0.5 to upper limit. Then calculate midpoints from these adjusted intervals.

Answer 4.
Answer:
Step 1: Calculate class marks for each interval.
Class mark = \( \frac{\text{Lower limit} + \text{Upper limit}}{2} \)

Step 2: Plot the frequency polygon by connecting the points using straight lines.

[Diagram: A frequency polygon graph showing class marks on x-axis (5.5, 15.5, 25.5, 35.5, 45.5) and frequencies on y-axis (8, 12, 10, 16, 6). Points are connected with straight blue lines forming a polygon shape.]

In simple words: A frequency polygon is made by plotting class marks and frequencies as points. Then we join these points with straight lines. It shows how the data is spread out.

📝 Teacher's Note: Show students that we use class marks (middle points) not class intervals. Connect the points with ruler to make straight lines. Start and end the polygon at the x-axis.

🎯 Exam Tip: Always calculate class marks first. Plot points carefully and use a ruler to connect them. Label both axes clearly with units.

Answer 4.
Answer:

Steps:

1. On the x-axis, take 1 cm as 5 units and plot class interval.
2. On the y-axis, take 1 cm as 5 units and plot frequency.
3. Plot the points with coordinates having abscissae as actual limits and ordinates as the cumulative frequencies. In this case (10,8),(20,20),(30,30),(40,44),(50,50).
4. Join the points plotted by a smooth curve.

[Diagram: A cumulative frequency curve (ogive) showing 'Less than' values on x-axis and cumulative frequency on y-axis. The curve starts at (10,8) and rises smoothly to (50,50).]

In simple words: A cumulative frequency curve shows how many students scored less than a certain mark. We add up frequencies step by step and plot the points.

📝 Teacher's Note: Explain that cumulative means "adding up". Each point shows total number of students up to that mark. The curve always goes upward.

🎯 Exam Tip: Make a "less than" table first. Add frequencies step by step. Plot points carefully and join with a smooth curve, not straight lines.

(ii)
Answer:

Steps:

1. On the x-axis, take 1 cm as 5 units and plot class interval.
2. On the y-axis, take 1 cm as 5 units and plot frequency.
3. Plot the points with coordinates having abscissae as actual limits and ordinates as the cumulative frequencies. In this case (150,10),(200,23),(250,40),(300,52),(350,62),(400,70).
4. Join the points plotted by a smooth curve.

[Diagram: A cumulative frequency curve showing 'Less than' values from 150 to 400 on x-axis and cumulative frequency from 10 to 70 on y-axis. The curve rises smoothly from (150,10) to (400,70).]

In simple words: This curve shows how many items cost less than a certain amount. As the price limit increases, more items are included.

📝 Teacher's Note: Use real examples like "How many items in a shop cost less than Rs 200?" This makes cumulative frequency easy to understand.

🎯 Exam Tip: Calculate cumulative frequency by adding each frequency to the previous total. Always check that your last cumulative frequency equals the total frequency.

(iii)
Answer:

Steps:

1. On the x-axis, take 1 cm as 5 units and plot class interval.
2. On the y-axis, take 1 cm as 5 units and plot frequency.
3. Plot the points with coordinates having abscissae as actual limits and ordinates as the cumulative frequencies. In this case (19,28),(29,51),(39,66),(49,86),(59,100).
4. Join the points plotted by a smooth curve.

[Diagram: A cumulative frequency curve showing 'Less than' values from 19 to 59 on x-axis and cumulative frequency from 28 to 100 on y-axis. The curve rises from (19,28) to (59,100).]

In simple words: This curve shows how many people are younger than a certain age. The total number of people is 100.

📝 Teacher's Note: Point out that the highest cumulative frequency (100) shows the total number of people in the survey. This is a good way to check if the calculations are correct.

🎯 Exam Tip: Make sure the last cumulative frequency equals the total of all frequencies. This proves your work is correct.

(iv)
Answer:

Steps:

1. On the x-axis, take 1 cm as 5 units and plot marks.
2. On the y-axis, take 1 cm as 5 units and plot frequency.
3. Plot the points with coordinates having abscissae as actual limits and ordinates as the cumulative frequencies. In this case (10,8),(20,22),(30,48),(40,60),(50,75).
4. Join the points plotted by a smooth curve.

[Diagram: A cumulative frequency curve showing marks less than certain values on x-axis and cumulative frequency on y-axis. The curve rises from (10,8) to (50,75).]

In simple words: This curve shows how many students scored less than different mark limits. For example, 22 students scored less than 20 marks.

📝 Teacher's Note: Help students understand that this type of data is already in cumulative form. They don't need to add frequencies - just plot the given points.

🎯 Exam Tip: When data is given as "less than" format, it's already cumulative. Just plot the points and draw a smooth curve.

(v)
Answer:

[Diagram: A cumulative frequency curve showing age 'less than' values on x-axis and cumulative frequency on y-axis. The curve starts at (10,0) and rises to (50,100).]

In simple words: This curve shows how many people are younger than different age limits. No one is less than 10 years old, and all 100 people are less than 50 years old.

📝 Teacher's Note: Point out that starting at 0 people under 10 years means the youngest person is 10 years old. Ending at 100 people under 50 years means no one is 50 or older.

🎯 Exam Tip: When plotting starts at 0, it means no data exists below that value. When it ends at the total, it means no data exists above that value.

Steps For Drawing Cumulative Frequency Curve (Less Than Type):

1. On the x-axis, take 1 cm as 5 units and plot age.

2. On the y-axis, take 1 cm as 5 units and plot frequency.

3. Plot the points with coordinates having abscissae as actual limits and ordinates as the cumulative frequencies. In this case (10,0),(20,17),(30,42),(40,67),(50,100).

4. Join the points plotted by a smooth curve.

[Diagram: A graph showing cumulative frequency curve (less than type) with age on x-axis (0 to 60) and cumulative frequency on y-axis (0 to 120). The curve starts at (10,0) and rises smoothly through points (20,17), (30,42), (40,67) to (50,100).]

Steps For Drawing Cumulative Frequency Curve (More Than Type):

1. Start with lower limits of class intervals and from cumulative frequency, subtract the frequency of each class to obtain c.f distribution.

2. Mark lower class limits along X-axis. 1cm = 5 units

3. Mark cumulative frequencies along Y-axis. 1 cm = 5 units

4. Plot points (x, f) where x is the lower limit of one class and f is the corresponding c.f. (10,188),(20,180),(30,155),(40,117),(50,67)

5. Join the points to get the ogive.

[Diagram: A graph showing cumulative frequency curve (more than type) with marks more than on x-axis (0 to 60) and cumulative frequency on y-axis (0 to 200). The curve starts at (10,188) and decreases smoothly through points (20,180), (30,155), (40,117) to (50,67).]

Steps:

1. Start with lower limits of class intervals and from cumulative frequency, subtract the frequency of each class to obtain c.f distribution.

2. Mark lower class limits along X-axis. 1cm = 5 units

3. Mark cumulative frequencies along Y-axis. 1 cm = 5 units

4. Plot points ( x, f) where x is the lower limit of one class and f is the corresponding c.f. (0,100),(10,87),(20,65),(30,55),(40,42),(50,36),(60,31),(70,21),(80,18), (90,7),(100,0).

5. Join the points to get the ogive.

[Diagram: A graph showing cumulative frequency curve (more than type) with marks more than on x-axis (0 to 120) and cumulative frequency on y-axis (0 to 120). The curve starts at (0,100) and decreases smoothly through all the plotted points to (100,0).]