ICSE Class 10 Maths Histograms & Ogives Selina Solutions
ICSE Class 10 Maths Chapter 23: what this page covers
ICSE Class 10 Maths Chapter 23, Graphical Representation (Histograms and Ogives), is about converting grouped data into clear graphs and reading cumulative information correctly. These Concise Mathematics Selina Solutions Class 10 ICSE Chapter 23 notes show how to prepare frequency tables, cumulative frequency tables, histograms and less-than ogives with full working.
Use the worked solutions below to check each table and graph before drawing it on graph paper. For related revision, students may also use the ICSE solutions library, the Class 10 ICSE solutions page, and the ICSE Class 10 Maths study resources.
For graph questions, one suitable scale is suggested. A different scale is also acceptable if the graph is neat, the axes are labelled, and all points or bars fit on the graph paper.
Concept snapshot: Think of cumulative frequency as a running total on a scoreboard. After the first class, the scoreboard shows only that class. After the second class, it shows the first two classes together. An ogive plots these running totals against the upper limits of the class intervals.
Concise Mathematics Selina Solutions Class 10 ICSE Chapter 23 method sheet
The same small set of methods is used throughout this chapter. Before drawing a graph, prepare the table correctly. A neat graph with a wrong table will still give a wrong answer.
| Task | Method | Important point |
|---|---|---|
| Class mark | \text{Class mark}=\dfrac{\text{lower limit}+\text{upper limit}}{2} | Use the two limits of the same class interval. |
| Class interval from class mark | \text{Lower limit}=\text{class mark}-\dfrac{\text{class width}}{2}, \text{Upper limit}=\text{class mark}+\dfrac{\text{class width}}{2} | The class width is usually the difference between two consecutive class marks. |
| Less-than cumulative frequency | Add frequencies successively from the first class onward. | The ogive point uses the upper class limit and the cumulative frequency. |
| Frequency from less-than cumulative frequency | \text{frequency}=\text{current cumulative frequency}-\text{previous cumulative frequency} | For the first class, subtract 0. |
| Frequency from more-than cumulative frequency | \text{frequency}=\text{current more-than value}-\text{next more-than value} | The total should match the first more-than cumulative frequency. |
| Histogram with equal class widths | Height of each rectangle is the frequency. | Adjacent rectangles touch each other. |
| Histogram with unequal class widths | Use frequency density: \text{height}=\dfrac{\text{frequency}}{\text{class width}}. | Do this only when class widths are unequal. |
Worked examples before Exercise 23
Worked example 1: Convert class marks into class intervals
Class marks are 18, 26, 34, 42. Write the class intervals.
Step 1: Find the difference between consecutive class marks.
26-18=8
Step 2: Half of the class width is \dfrac{8}{2}=4.
Step 3: Subtract 4 from each class mark for the lower limit and add 4 for the upper limit.
18\Rightarrow 14\text{--}22,\quad26\Rightarrow 22\text{--}30,\quad34\Rightarrow 30\text{--}38,\quad42\Rightarrow 38\text{--}46
Final answer: The class intervals are 14\text{--}22, 22\text{--}30, 30\text{--}38 and 38\text{--}46.
Worked example 2: Prepare a less-than cumulative frequency table
For the classes 0\text{--}10, 10\text{--}20, 20\text{--}30, 30\text{--}40, the frequencies are 6, 14, 20, 10. Find the cumulative frequencies.
Step 1: The first cumulative frequency is the first frequency.
6
Step 2: Add the next frequency successively.
6+14=20,\quad 20+20=40,\quad 40+10=50
Final answer: The cumulative frequencies are 6, 20, 40, 50. The ogive points are (10,6), (20,20), (30,40) and (40,50), with the starting point (0,0).
Worked example 3: Convert more-than cumulative frequency into frequency
A more-than table gives 100, 76, 45, 18, 0 for marks more than 0, 20, 40, 60, 80. Find the frequency distribution.
Step 1: Subtract the next more-than value from the current more-than value.
100-76=24,\quad 76-45=31,\quad 45-18=27,\quad 18-0=18
Step 2: Attach each frequency to its class interval.
| Class interval | Frequency |
|---|---|
| 0\text{--}20 | 24 |
| 20\text{--}40 | 31 |
| 40\text{--}60 | 27 |
| 60\text{--}80 | 18 |
Final answer: The frequencies are 24, 31, 27, 18. Their sum is 100, which checks with the first more-than cumulative frequency.
Exercise 23 solutions: Histograms and Ogives
Question 1(a): Find a, b and c from the cumulative frequency table
| Class interval | Frequency | Cumulative frequency |
|---|---|---|
| 30\text{--}40 | 24 | 24 |
| 40\text{--}50 | a | 40 |
| 50\text{--}60 | 12 | b |
| 60\text{--}70 | c | 60 |
Step 1: The cumulative frequency after the second class is 40, so
24+a=40
a=40-24=16
Step 2: Add the third-class frequency to get b.
b=40+12=52
Step 3: The last cumulative frequency is 60, so
52+c=60
c=60-52=8
Final answer: a=16, b=52 and c=8. Hence, option (a) is correct.
Question 1(b): Cumulative frequency of the class 70\text{--}90
The histogram in this question gives frequencies 50, 30, 40, 60 for the classes 30\text{--}50, 50\text{--}70, 70\text{--}90 and 90\text{--}110, respectively.
Step 1: Write the cumulative frequency up to 70\text{--}90.
50+30+40=120
| Class interval | Frequency | Cumulative frequency |
|---|---|---|
| 30\text{--}50 | 50 | 50 |
| 50\text{--}70 | 30 | 80 |
| 70\text{--}90 | 40 | 120 |
| 90\text{--}110 | 60 | 180 |
Final answer: The cumulative frequency of the class 70\text{--}90 is 120. Hence, option (b) is correct.
Question 1(c): Choose the correct class mark
For the distribution 30\text{--}34, 34\text{--}38, 38\text{--}42, choose the expression for the class mark of the first class.
Step 1: Use the class mark formula.
\text{Class mark}=\frac{\text{lower limit}+\text{upper limit}}{2}
Step 2: For the first class 30\text{--}34, substitute the limits.
\text{Class mark}=\frac{30+34}{2}=32
Final answer: The correct expression is \dfrac{30+34}{2}. Hence, option (b) is correct.
Question 1(d): Starting point of a cumulative frequency curve
Step 1: In a less-than ogive, the curve begins where the cumulative frequency is 0.
Step 2: This point is at the lower limit of the first class interval.
Final answer: The cumulative frequency curve starts from the lower limit of the first class. Hence, option (c) is correct.
Question 1(e): End point of a cumulative frequency curve
Step 1: A less-than ogive uses the upper limits of the class intervals.
Step 2: The last plotted point is at the upper limit of the last class, where the cumulative frequency equals the total frequency.
Final answer: The curve terminates at the upper limit of the last class. Hence, option (d) is correct.
Question 2(i): Draw a histogram for the given frequency distribution
| Class interval | Frequency |
|---|---|
| 0\text{--}10 | 12 |
| 10\text{--}20 | 20 |
| 20\text{--}30 | 26 |
| 30\text{--}40 | 18 |
| 40\text{--}50 | 10 |
| 50\text{--}60 | 6 |
Step 1: The class widths are equal: 10 units each.
Step 2: Take a convenient scale, for example 2\text{ cm}=10\text{ units} on the x-axis and 1\text{ cm}=5\text{ frequency units} on the y-axis.
Step 3: Draw adjacent rectangles over the intervals 0\text{--}10, 10\text{--}20, 20\text{--}30, 30\text{--}40, 40\text{--}50 and 50\text{--}60.
Step 4: The heights of the rectangles are 12, 20, 26, 18, 10 and 6, respectively.
Final answer: The required histogram has six touching rectangles with heights 12, 20, 26, 18, 10, 6 on the given class intervals.
Question 2(ii): Draw a histogram for classes 10\text{--}16 to 34\text{--}40
| Class interval | Frequency |
|---|---|
| 10\text{--}16 | 15 |
| 16\text{--}22 | 23 |
| 22\text{--}28 | 30 |
| 28\text{--}34 | 20 |
| 34\text{--}40 | 16 |
Step 1: Each class has width 6, so the rectangle heights are the given frequencies.
Step 2: Since the first class starts at 10, show a kink near the origin on the x-axis if the graph paper begins at 0.
Step 3: One suitable scale is 2\text{ cm}=6\text{ units} on the x-axis and 1\text{ cm}=5\text{ frequency units} on the y-axis.
Step 4: Draw five adjacent rectangles with heights 15, 23, 30, 20 and 16.
Final answer: The histogram is formed by five touching rectangles over 10\text{--}16, 16\text{--}22, 22\text{--}28, 28\text{--}34 and 34\text{--}40, with the stated frequencies as heights.
Question 2(iii): Draw a histogram when class marks are given
| Class mark | Frequency |
|---|---|
| 16 | 8 |
| 24 | 12 |
| 32 | 15 |
| 40 | 18 |
| 48 | 25 |
| 56 | 19 |
| 64 | 10 |
Step 1: Find the class width from consecutive class marks.
24-16=8
Step 2: Half of the class width is 4. Subtract 4 and add 4 to each class mark.
| Class mark | Class interval | Frequency |
|---|---|---|
| 16 | 12\text{--}20 | 8 |
| 24 | 20\text{--}28 | 12 |
| 32 | 28\text{--}36 | 15 |
| 40 | 36\text{--}44 | 18 |
| 48 | 44\text{--}52 | 25 |
| 56 | 52\text{--}60 | 19 |
| 64 | 60\text{--}68 | 10 |
Step 3: Use a kink on the x-axis because the first class starts at 12. A suitable scale is 1\text{ cm}=8\text{ units} on the x-axis and 1\text{ cm}=5\text{ frequency units} on the y-axis.
Step 4: Draw adjacent rectangles for the intervals shown in the table, using their frequencies as heights.
Final answer: The histogram uses the class intervals 12\text{--}20, 20\text{--}28, 28\text{--}36, 36\text{--}44, 44\text{--}52, 52\text{--}60, 60\text{--}68 with heights 8, 12, 15, 18, 25, 19, 10.
Question 3(i): Draw an ogive for a continuous distribution
| Class interval | Frequency |
|---|---|
| 10\text{--}15 | 10 |
| 15\text{--}20 | 15 |
| 20\text{--}25 | 17 |
| 25\text{--}30 | 12 |
| 30\text{--}35 | 10 |
| 35\text{--}40 | 8 |
Step 1: Prepare the less-than cumulative frequency table.
| Class interval | Frequency | Cumulative frequency |
|---|---|---|
| 10\text{--}15 | 10 | 10 |
| 15\text{--}20 | 15 | 25 |
| 20\text{--}25 | 17 | 42 |
| 25\text{--}30 | 12 | 54 |
| 30\text{--}35 | 10 | 64 |
| 35\text{--}40 | 8 | 72 |
Step 2: Plot the starting point (10,0).
Step 3: Plot the upper limits against cumulative frequencies.
(15,10),\ (20,25),\ (25,42),\ (30,54),\ (35,64),\ (40,72)
Step 4: Join the points with a smooth freehand curve.
Final answer: The required less-than ogive starts at (10,0) and passes through (15,10), (20,25), (25,42), (30,54), (35,64) and (40,72).
Question 3(ii): Draw an ogive after converting discontinuous classes into continuous classes
| Class interval | Frequency |
|---|---|
| 10\text{--}19 | 23 |
| 20\text{--}29 | 16 |
| 30\text{--}39 | 15 |
| 40\text{--}49 | 20 |
| 50\text{--}59 | 12 |
Step 1: Find the adjustment factor.
\text{Adjustment factor}=\frac{20-19}{2}=0.5
Step 2: Subtract 0.5 from each lower limit and add 0.5 to each upper limit.
| Original class | Continuous class | Frequency | Cumulative frequency |
|---|---|---|---|
| 10\text{--}19 | 9.5\text{--}19.5 | 23 | 23 |
| 20\text{--}29 | 19.5\text{--}29.5 | 16 | 39 |
| 30\text{--}39 | 29.5\text{--}39.5 | 15 | 54 |
| 40\text{--}49 | 39.5\text{--}49.5 | 20 | 74 |
| 50\text{--}59 | 49.5\text{--}59.5 | 12 | 86 |
Step 3: Use the starting point (9.5,0).
Step 4: Plot the ogive points.
(19.5,23),\ (29.5,39),\ (39.5,54),\ (49.5,74),\ (59.5,86)
Final answer: The less-than ogive starts at (9.5,0) and passes through (19.5,23), (29.5,39), (39.5,54), (49.5,74) and (59.5,86).
Question 4(i): Draw an ogive from less-than cumulative data for marks
| Marks obtained | Number of students |
|---|---|
| Less than 10 | 8 |
| Less than 20 | 25 |
| Less than 30 | 38 |
| Less than 40 | 50 |
| Less than 50 | 67 |
Step 1: The given data is already in less-than cumulative form.
| Class interval | Cumulative frequency |
|---|---|
| 0\text{--}10 | 8 |
| 10\text{--}20 | 25 |
| 20\text{--}30 | 38 |
| 30\text{--}40 | 50 |
| 40\text{--}50 | 67 |
Step 2: Start from (0,0).
Step 3: Plot the upper limits with their cumulative frequencies.
(10,8),\ (20,25),\ (30,38),\ (40,50),\ (50,67)
Final answer: The ogive starts at (0,0) and passes through (10,8), (20,25), (30,38), (40,50) and (50,67).
Question 4(ii): Draw an ogive from less-than cumulative data for ages
| Age in years, less than | Cumulative frequency |
|---|---|
| 10 | 0 |
| 20 | 17 |
| 30 | 32 |
| 40 | 37 |
| 50 | 53 |
| 60 | 58 |
| 70 | 65 |
Step 1: This is a less-than cumulative frequency table.
Step 2: Since the cumulative frequency for less than 10 years is 0, the curve begins at (10,0).
Step 3: Plot the points.
(10,0),\ (20,17),\ (30,32),\ (40,37),\ (50,53),\ (60,58),\ (70,65)
Step 4: Join these points with a smooth curve.
Final answer: The ogive passes through (10,0), (20,17), (30,32), (40,37), (50,53), (60,58) and (70,65).
Question 5: Construct a frequency table from the histogram and then draw an ogive
Step 1: Read the frequencies from the given histogram and form the table.
| Class interval | Frequency | Cumulative frequency |
|---|---|---|
| 8\text{--}12 | 9 | 9 |
| 12\text{--}16 | 16 | 25 |
| 16\text{--}20 | 22 | 47 |
| 20\text{--}24 | 18 | 65 |
| 24\text{--}28 | 12 | 77 |
| 28\text{--}32 | 4 | 81 |
Step 2: Start the ogive at the lower limit of the first class.
(8,0)
Step 3: Plot the upper limits against cumulative frequencies.
(12,9),\ (16,25),\ (20,47),\ (24,65),\ (28,77),\ (32,81)
Final answer: The frequency table is shown above, and the ogive starts at (8,0) and passes through (12,9), (16,25), (20,47), (24,65), (28,77) and (32,81).
Question 6: Monthly wages of factory workers
| Wages in rupees | Number of workers |
|---|---|
| 6500\text{--}7000 | 10 |
| 7000\text{--}7500 | 18 |
| 7500\text{--}8000 | 22 |
| 8000\text{--}8500 | 25 |
| 8500\text{--}9000 | 17 |
| 9000\text{--}9500 | 10 |
| 9500\text{--}10000 | 8 |
Step 1: Calculate cumulative frequencies.
| Wages in rupees | Frequency | Cumulative frequency |
|---|---|---|
| 6500\text{--}7000 | 10 | 10 |
| 7000\text{--}7500 | 18 | 28 |
| 7500\text{--}8000 | 22 | 50 |
| 8000\text{--}8500 | 25 | 75 |
| 8500\text{--}9000 | 17 | 92 |
| 9000\text{--}9500 | 10 | 102 |
| 9500\text{--}10000 | 8 | 110 |
Step 2: Start at (6500,0). A kink on the x-axis is suitable because the scale begins at 6500.
Step 3: Plot the points.
(7000,10),\ (7500,28),\ (8000,50),\ (8500,75),\ (9000,92),\ (9500,102),\ (10000,110)
Final answer: The cumulative frequencies are 10, 28, 50, 75, 92, 102, 110. The ogive starts at (6500,0) and passes through the plotted points listed above.
Question 7: Heights of factory workers
| Height in cm | Number of workers |
|---|---|
| 150\text{--}155 | 6 |
| 155\text{--}160 | 12 |
| 160\text{--}165 | 18 |
| 165\text{--}170 | 20 |
| 170\text{--}175 | 13 |
| 175\text{--}180 | 8 |
| 180\text{--}185 | 6 |
Step 1: Calculate cumulative frequencies.
| Height in cm | Frequency | Cumulative frequency |
|---|---|---|
| 150\text{--}155 | 6 | 6 |
| 155\text{--}160 | 12 | 18 |
| 160\text{--}165 | 18 | 36 |
| 165\text{--}170 | 20 | 56 |
| 170\text{--}175 | 13 | 69 |
| 175\text{--}180 | 8 | 77 |
| 180\text{--}185 | 6 | 83 |
Step 2: Start the less-than ogive at (150,0).
Step 3: Plot the upper class limits with cumulative frequencies.
(155,6),\ (160,18),\ (165,36),\ (170,56),\ (175,69),\ (180,77),\ (185,83)
Final answer: The cumulative frequencies are 6, 18, 36, 56, 69, 77, 83. The required ogive starts at (150,0) and passes through the seven listed points.
Question 8(i): Construct a frequency table from less-than cumulative frequency
| Marks less than | Cumulative frequency |
|---|---|
| 0 | 0 |
| 10 | 7 |
| 20 | 28 |
| 30 | 54 |
| 40 | 71 |
| 50 | 84 |
| 60 | 105 |
| 70 | 147 |
| 80 | 180 |
| 90 | 196 |
| 100 | 200 |
Step 1: Subtract the previous cumulative frequency from the current cumulative frequency.
7-0=7,\quad 28-7=21,\quad 54-28=26,\quad 71-54=17,\quad 84-71=13
105-84=21,\quad 147-105=42,\quad 180-147=33,\quad 196-180=16,\quad 200-196=4
| Marks | Frequency |
|---|---|
| 0\text{--}10 | 7 |
| 10\text{--}20 | 21 |
| 20\text{--}30 | 26 |
| 30\text{--}40 | 17 |
| 40\text{--}50 | 13 |
| 50\text{--}60 | 21 |
| 60\text{--}70 | 42 |
| 70\text{--}80 | 33 |
| 80\text{--}90 | 16 |
| 90\text{--}100 | 4 |
Final answer: The frequency distribution is 7, 21, 26, 17, 13, 21, 42, 33, 16, 4 for the intervals 0\text{--}10 to 90\text{--}100.
Question 8(ii): Construct a frequency table from more-than cumulative frequency
| Marks more than | Cumulative frequency |
|---|---|
| 0 | 100 |
| 10 | 87 |
| 20 | 65 |
| 30 | 55 |
| 40 | 42 |
| 50 | 36 |
| 60 | 31 |
| 70 | 21 |
| 80 | 18 |
| 90 | 7 |
| 100 | 0 |
Step 1: In a more-than table, subtract the next cumulative frequency from the current one.
100-87=13,\quad 87-65=22,\quad 65-55=10,\quad 55-42=13,\quad 42-36=6
36-31=5,\quad 31-21=10,\quad 21-18=3,\quad 18-7=11,\quad 7-0=7
| Marks | Frequency |
|---|---|
| 0\text{--}10 | 13 |
| 10\text{--}20 | 22 |
| 20\text{--}30 | 10 |
| 30\text{--}40 | 13 |
| 40\text{--}50 | 6 |
| 50\text{--}60 | 5 |
| 60\text{--}70 | 10 |
| 70\text{--}80 | 3 |
| 80\text{--}90 | 11 |
| 90\text{--}100 | 7 |
Step 2: Check the total frequency.
13+22+10+13+6+5+10+3+11+7=100
Final answer: The frequency distribution is 13, 22, 10, 13, 6, 5, 10, 3, 11, 7, and the total 100 confirms the table.
Quick answer index
| Exercise | Question | Answer / graph data |
|---|---|---|
| 23 | 1(a) | a=16,\ b=52,\ c=8 |
| 23 | 1(b) | Cumulative frequency of 70\text{--}90 is 120 |
| 23 | 1(c) | \dfrac{30+34}{2}=32, option (b) |
| 23 | 1(d) | Lower limit of the first class, option (c) |
| 23 | 1(e) | Upper limit of the last class, option (d) |
| 23 | 2(i) | Histogram heights: 12,20,26,18,10,6 |
| 23 | 2(ii) | Histogram heights: 15,23,30,20,16 |
| 23 | 2(iii) | Intervals 12\text{--}20 to 60\text{--}68; heights 8,12,15,18,25,19,10 |
| 23 | 3(i) | Ogive points from (10,0) to (40,72) |
| 23 | 3(ii) | Ogive points from (9.5,0) to (59.5,86) |
| 23 | 4(i) | Ogive points (0,0),(10,8),(20,25),(30,38),(40,50),(50,67) |
| 23 | 4(ii) | Ogive points (10,0),(20,17),(30,32),(40,37),(50,53),(60,58),(70,65) |
| 23 | 5 | Frequency table: 9,16,22,18,12,4; ogive total 81 |
| 23 | 6 | Cumulative frequencies: 10,28,50,75,92,102,110 |
| 23 | 7 | Cumulative frequencies: 6,18,36,56,69,77,83 |
| 23 | 8(i) | Frequencies: 7,21,26,17,13,21,42,33,16,4 |
| 23 | 8(ii) | Frequencies: 13,22,10,13,6,5,10,3,11,7 |
Examiner’s mindset for graph questions
In ICSE Class 10 Maths graph work, the answer is not judged only by the final curve. The table, axes, scale, plotting and interpretation all matter. If a question gives grouped data, first show the frequency or cumulative frequency table. Then label both axes, mark the scale clearly, and plot the exact coordinates or rectangle heights.
For histogram questions, do not leave gaps between rectangles when the class intervals are continuous. For ogive questions, do not join the plotted points with straight line segments as if it were a polygon; use a smooth freehand cumulative curve. The exact mark split can vary by question paper, so do not assume a fixed split unless it is printed in the paper.
Common mistakes students make
- Using lower limits for a less-than ogive: The plotted points should use upper class limits with cumulative frequencies. The only extra starting point is at the lower limit with cumulative frequency 0.
- Forgetting to convert discontinuous classes: Classes such as 10\text{--}19, 20\text{--}29 must be converted to 9.5\text{--}19.5, 19.5\text{--}29.5 before drawing a continuous ogive.
- Using frequency instead of cumulative frequency: A histogram uses frequencies as heights, but an ogive uses running totals.
- Assuming every histogram height is frequency: This is true only when class widths are equal. If class widths are unequal, use frequency density \dfrac{\text{frequency}}{\text{class width}}.
- Writing no scale: A correct-looking graph can lose value if the scale is missing or inconsistent.
Sources used for accuracy
This page is aligned with the standard ICSE Class 10 Mathematics treatment of grouped data, histograms and ogives, and with the Selina Concise Mathematics Class 10 chapter topic named in the page title. For board-level alignment, students can refer to the CISCE official website. For overlapping statistics concepts, the NCERT official website is also a useful reference source.
Frequently Asked Questions
How do I draw a less-than ogive in ICSE Class 10 Maths?
Prepare the cumulative frequency table, plot the upper class limits on the x-axis and the cumulative frequencies on the y-axis, start with the lower limit of the first class at cumulative frequency 0, and join the points with a smooth curve.
When should I use frequency density in a histogram?
Use frequency density when class intervals are unequal. The height is \dfrac{\text{frequency}}{\text{class width}}. In the Exercise 23 histogram questions solved above, the listed class widths are equal, so the given frequencies can be used as rectangle heights.
How do I convert less-than cumulative frequency into a frequency table?
Subtract the previous cumulative frequency from the current cumulative frequency. For example, if the less-than cumulative frequencies are 7 and 28, the frequency for the class between those limits is 28-7=21.
Why is a kink shown on the x-axis in histograms and ogives?
A kink shows that the scale does not start from 0. It is used when the first class begins at a much larger value, such as 150 cm or 6500 rupees, so the graph can fit neatly on the page.
Is the graph scale fixed in Selina Chapter 23 questions?
No single scale is compulsory unless the question specifies one. Choose a scale that fits the data, label it clearly, and keep it uniform along each axis. This is why the solutions above give suitable scales rather than claiming that only one scale is correct.