ICSE Class 10 Maths Graph Questions: Step-by-Step Guide
ICSE Class 10 Maths graph-based long answer solutions
ICSE Class 10 Maths graph-based long answer questions test two skills together: plotting points accurately on coordinate axes and reading statistical graphs such as ogives. This rewritten page gives clear step-by-step solutions for the graph-based long answer set on reflections, invariant points, ogives, median, percentages from cumulative frequency, modal height and mean-mode comparison.
Graph answers are partly visual, so a small difference in a graph-read value can occur because of scale and drawing accuracy. Wherever a value is read from an ogive, the answer below states the approximate value and also shows the calculation behind the reading.
Graph rules and methods
Use these rules before checking the worked solutions. For more related practice, use the ICSE Class 10 Maths solutions section.
| Topic | Rule or method | How it is used |
|---|---|---|
| Reflection in the origin | (x,y)\rightarrow(-x,-y) | Change both signs. |
| Reflection in the x-axis | (x,y)\rightarrow(x,-y) | Keep x, change the sign of y. |
| Reflection in the y-axis | (x,y)\rightarrow(-x,y) | Change the sign of x, keep y. |
| Reflection in y=k | (x,y)\rightarrow(x,2k-y) | Use equal vertical distance from the mirror line. |
| Invariant point in x=0 | The point lies on the y-axis. | The x-coordinate must be 0. |
| Less-than ogive | Plot upper class limit against cumulative frequency. | Read median and cumulative counts from the curve. |
| Mode of grouped data | \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h | Use the modal class and the two adjacent class frequencies. |
Concept snapshot: Treat reflection like folding graph paper. If the fold is the x-axis, every point lands the same distance on the other side of the x-axis. If the fold is x=0, only points already on the y-axis stay fixed. This is why an invariant point under reflection in x=0 must have x-coordinate 0.
Worked examples before the solutions
Worked Example 1: Reflect A(3,-4) in the origin
Step 1: For reflection in the origin, use (x,y)\rightarrow(-x,-y).
Step 2: Apply the rule to A(3,-4).
A(3,-4)\rightarrow A'(-3,4)
Final answer: A'(-3,4).
Worked Example 2: Reflect B(5,1) in y=-2
Step 1: For reflection in y=k, use (x,y)\rightarrow(x,2k-y).
Step 2: Here, k=-2, x=5, and y=1.
y'=2(-2)-1=-4-1=-5
Final answer: B(5,1) reflects to B'(5,-5).
Worked Example 3: Estimate cumulative frequency at x=75
Step 1: Suppose cumulative frequency is 40 at x=70 and 60 at x=80.
Step 2: Since 75 is halfway between 70 and 80, take half the increase in cumulative frequency.
60-40=20
40+\frac{20}{2}=50
Final answer: The estimated cumulative frequency is 50.
Question 106 solutions
Question: Plot A(2,2) and B(6,-2). Reflect A in the origin to get D, reflect A in y=-2 to get C, find the invariant point P on CD under reflection in x=0, name ABCD, and find the intersection of the diagonals.
Question 106(a): Coordinates of D
Step 1: Reflection in the origin changes (x,y) to (-x,-y).
A(2,2)\rightarrow D(-2,-2)
Final answer: D(-2,-2).
Question 106(b): Coordinates of C
Step 1: For reflection in y=k, use (x,y)\rightarrow(x,2k-y).
Step 2: Here k=-2, so
y'=2(-2)-2=-6
Final answer: C(2,-6).
Question 106(c): Point P on CD invariant in x=0
Step 1: An invariant point under reflection in x=0 lies on the y-axis, so x=0.
Step 2: Find the equation of CD, where C(2,-6) and D(-2,-2).
m=\frac{-2-(-6)}{-2-2}=\frac{4}{-4}=-1
y+6=-1(x-2)
y=-x-4
Step 3: Put x=0.
y=-4
Final answer: P(0,-4).
Question 106(d): Geometrical name of ABCD
Step 1: Compare adjacent side vectors.
\overrightarrow{AB}=(6-2,-2-2)=(4,-4)
\overrightarrow{BC}=(2-6,-6-(-2))=(-4,-4)
Step 2: Their lengths are equal and their dot product is zero.
AB=BC=\sqrt{4^2+(-4)^2}=\sqrt{32}
(4)(-4)+(-4)(-4)=0
Final answer: ABCD is a square.
Question 106(e): Intersection of diagonals
Step 1: Diagonals of a square bisect each other. Use the midpoint of A(2,2) and C(2,-6).
\left(\frac{2+2}{2},\frac{2+(-6)}{2}\right)=(2,-2)
Final answer: The diagonals intersect at (2,-2).
Question 107 solutions
Question: Plot A(0,3), B(4,0), C(6,2), and D(5,0). Reflect the points as directed and name the figure BCDC'.
Question 107(a): Reflect A in the x-axis
Step 1: Reflection in the x-axis changes (x,y) to (x,-y).
A(0,3)\rightarrow A'(0,-3)
Final answer: A'(0,-3).
Question 107(b): Reflect B in the y-axis
Step 1: Reflection in the y-axis changes (x,y) to (-x,y).
B(4,0)\rightarrow B'(-4,0)
Final answer: B'(-4,0).
Question 107(c): Reflect C in the x-axis
Step 1: Keep x and change the sign of y.
C(6,2)\rightarrow C'(6,-2)
Final answer: C'(6,-2).
Question 107(d): Line in which D is invariant
Step 1: A point remains invariant under reflection in a line if it lies on that line.
Step 2: D(5,0) lies on the x-axis.
\text{x-axis has equation }y=0
Final answer: The required line is y=0.
Question 107(e): Name of BCDC'
Step 1: The points are B(4,0), C(6,2), D(5,0), and C'(6,-2).
Step 2: The point D lies inward between C and C', so one angle is reflex.
Final answer: BCDC' is a concave quadrilateral, also called a dart-shaped quadrilateral.
Question 108 solutions
Question: The data gives daily wages and number of employees. Draw an ogive and find the median wage, the percentage earning more than ₹84, and the number earning ₹56 and below.
| Daily wages in ₹ | Frequency |
|---|---|
| 30-40 | 8 |
| 40-50 | 14 |
| 50-60 | 12 |
| 60-70 | 17 |
| 70-80 | 20 |
| 80-90 | 26 |
| 90-100 | 13 |
| 100-110 | 10 |
Question 108(a): Cumulative frequency table and ogive
Step 1: Prepare the less-than cumulative frequency table.
| Class | Frequency | Cumulative frequency | Point for ogive |
|---|---|---|---|
| 30-40 | 8 | 8 | (40,8) |
| 40-50 | 14 | 22 | (50,22) |
| 50-60 | 12 | 34 | (60,34) |
| 60-70 | 17 | 51 | (70,51) |
| 70-80 | 20 | 71 | (80,71) |
| 80-90 | 26 | 97 | (90,97) |
| 90-100 | 13 | 110 | (100,110) |
| 100-110 | 10 | 120 | (110,120) |
Step 2: Take daily wages on the x-axis and cumulative frequency on the y-axis. Plot the points and join them with a smooth increasing curve.
Final answer: The less-than ogive is drawn using the points from (40,8) to (110,120).
Question 108(b)(i): Median wage
Step 1: Total frequency is N=120.
\frac{N}{2}=\frac{120}{2}=60
Step 2: Locate 60 on the cumulative frequency axis, move to the ogive, and drop down to the wage axis.
Step 3: Interpolation between (70,51) and (80,71) gives:
\text{Median}=70+\frac{60-51}{20}\times 10=74.5
Final answer: Median wage \approx ₹74.5. A graph reading of about ₹74 to ₹75 is reasonable.
Question 108(b)(ii): Percentage earning more than ₹84
Step 1: Estimate the number earning ₹84 or less. Since 84 lies in 80-90:
\text{C.F. at }84=71+\frac{84-80}{10}\times 26
=71+10.4=81.4
Step 2: Number earning more than ₹84 is approximately:
120-81.4=38.6\approx 39
Step 3: Find the percentage using the nearest whole number of employees.
\frac{39}{120}\times 100=32.5\%
Final answer: Approximately 39 employees, or about 32.5\%, earn more than ₹84 per day.
Question 108(b)(iii): Employees earning ₹56 and below
Step 1: Since 56 lies in 50-60, use cumulative frequency at 50 and 60.
\text{C.F. at }56=22+\frac{56-50}{10}\times 12
=22+7.2=29.2
Final answer: About 29 employees earn ₹56 and below. A hand-drawn ogive may read this as about 30 employees.
Question 109 solutions
Question: Study the graph and form the frequency table. Then find the number of students below 150\text{ cm}, total students, modal height, and the difference between modal height and mean height when the average height is 145.5\text{ cm}.
Question 109(a): Frequency table
Step 1: Read each class interval and frequency from the graph.
| Height class in \text{cm} | Frequency | Cumulative frequency |
|---|---|---|
| 120-130 | 6 | 6 |
| 130-140 | 29 | 35 |
| 140-150 | 34 | 69 |
| 150-160 | 22 | 91 |
| 160-170 | 12 | 103 |
Final answer: The frequency table is shown above.
Question 109(b): Students below 150\text{ cm}
Step 1: Add frequencies up to the 140-150 class.
6+29+34=69
Final answer: 69 students are below 150\text{ cm}.
Question 109(c): Total number of students
Step 1: Add all frequencies.
6+29+34+22+12=103
Final answer: Total number of students is 103.
Question 109(d): Modal height
Step 1: The highest frequency is 34, so the modal class is 140-150.
Step 2: Use l=140, f_1=34, f_0=29, f_2=22, and h=10.
\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
=140+\frac{34-29}{2(34)-29-22}\times 10
=140+\frac{50}{17}\approx 142.94
Final answer: Modal height \approx 143\text{ cm}.
Question 109(e): Difference between mean and modal height
Step 1: Mean height is 145.5\text{ cm}, and modal height is approximately 143\text{ cm}.
145.5-143=2.5
Final answer: Difference =2.5\text{ cm}.
Examiner’s mindset for graph-based answers
In graph questions, the final coordinate or number is not enough. A good answer shows the rule, the graph construction, the plotted points, the cumulative frequency table where needed, and a final statement with units. For coordinate geometry, the examiner checks whether the plotted image agrees with the reflection rule. For statistics, the examiner checks whether the cumulative frequency table and graph reading are consistent.
Common mistakes students make
- Changing only one sign in origin reflection: Use (x,y)\rightarrow(-x,-y), so both signs change.
- Using lower class limits for a less-than ogive: Plot upper class limits with cumulative frequency.
- Writing graph-read answers as exact: Use approximately when reading from an ogive.
- Dropping units: Wages need ₹; heights need \text{cm}.
- Naming a figure only by sight: Check side lengths, slopes or angles before writing square, rectangle or quadrilateral.
Quick answer index
| Question | Part | Answer |
|---|---|---|
| 106 | (a) | D(-2,-2) |
| 106 | (b) | C(2,-6) |
| 106 | (c) | P(0,-4) |
| 106 | (d) | Square |
| 106 | (e) | (2,-2) |
| 107 | (a) | A'(0,-3) |
| 107 | (b) | B'(-4,0) |
| 107 | (c) | C'(6,-2) |
| 107 | (d) | y=0 |
| 107 | (e) | Concave quadrilateral |
| 108 | (b)(i) | Median \approx ₹74.5 |
| 108 | (b)(ii) | About 39 employees; 32.5\% |
| 108 | (b)(iii) | About 29 to 30 employees |
| 109 | (b) | 69 |
| 109 | (c) | 103 |
| 109 | (d) | 143\text{ cm} |
| 109 | (e) | 2.5\text{ cm} |
How to practise
Solve each graph first on paper, then compare the construction and final values. If your answer differs slightly in an ogive question, check the scale, the plotted point and whether the graph reading is reasonable. For broader revision, use Class 10 ICSE solutions, ICSE Class 10 Maths question papers and ICSE Class 10 specimen papers.
Frequently Asked Questions
How should I start an ICSE Class 10 Maths graph-based long answer?
Start by writing the scale, drawing the axes, plotting the given points or cumulative frequency points, and stating the rule you are using. In reflection questions, write the reflection rule before the final coordinates.
What is the rule for reflecting a point in the origin?
For reflection in the origin, (x,y) becomes (-x,-y). For example, (2,2) becomes (-2,-2).
How do I find the median from an ogive?
Find \frac{N}{2}, mark it on the cumulative frequency axis, draw a horizontal line to the ogive, and drop a perpendicular to the x-axis. The x-coordinate is the median.
Why can ogive answers differ slightly?
An ogive is read from a hand-drawn graph, so small differences occur because of scale and curve reading. A calculated value of about 29 may be read as about 30 from the graph.
What should I write for an invariant point under reflection in x=0?
The point must lie on the y-axis, so its x-coordinate is 0. Find where the required segment cuts the y-axis and write that coordinate.
Sources and syllabus alignment
This page follows the standard ICSE Class 10 Mathematics treatment of coordinate geometry transformations and statistics graph interpretation. For official syllabus framing, refer to the CISCE official website. For school-level statistics methods such as cumulative frequency and ogives, the NCERT textbook portal is an authoritative reference.