ICSE Class 10 Maths graph-based long answer solutions
ICSE Class 10 Maths graph-based long answer questions test two skills together: accurate construction on graph paper and correct interpretation of the graph. This page solves the Long Answer Questions 1 (Graph-based) practice set step by step, covering reflection of points, invariant points, ogives, cumulative frequency, modal height and graph reading.
Use these solutions to check the method, not just the final coordinates or numbers. In graph questions, a correct answer must usually be supported by a neat scale, labelled axes, plotted points, construction lines and the final reading written with the correct unit.
Formula and method reference for graph questions
Before solving the graph-based long answer questions, revise these rules. They are the rules used in the solutions below.
| Task | Rule or method | What to write in the answer |
|---|---|---|
| Reflection in the x-axis | \((x,y) \mapsto (x,-y)\) | The x-coordinate is unchanged; the sign of the y-coordinate changes. |
| Reflection in the y-axis | \((x,y) \mapsto (-x,y)\) | The y-coordinate is unchanged; the sign of the x-coordinate changes. |
| Reflection in the origin | \((x,y) \mapsto (-x,-y)\) | Both coordinates change signs. |
| Reflection in the line y=a | \((x,y) \mapsto (x,2a-y)\) | Keep the same perpendicular distance from the horizontal line. |
| Invariant under reflection in x=0 | The point lies on the line x=0, the y-axis. | Its x-coordinate must be 0. |
| Less-than ogive | Plot upper class limits against cumulative frequency. | Use a smooth increasing curve and read values by drawing horizontal or vertical guide lines. |
| Mode from grouped data | \text{Mode}=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h | Identify the modal class first, then substitute carefully. |
Concept snapshot: think of reflection as equal distance
A reflection is not a slide. It is a mirror image. If a point is reflected in the line y=-2, the original point and its image must be at equal perpendicular distances from y=-2, on opposite sides of that line. For example, \(A(2,2)\) is 4 units above y=-2, so its image is 4 units below y=-2, giving \(C(2,-6)\).
Question 106: reflection of points and square ABCD
Question: Plot the points \(A(2,2)\) and \(B(6,-2)\) on the graph and answer the following:
- Reflect point A in the origin to point D, and write the coordinates of D.
- Reflect point A in the line y=-2 to point C, and write the coordinates of C.
- Find a point P on CD which is invariant under reflection in x=0, and write its coordinates.
- Write the geometrical name of the closed figure ABCD.
- Write the coordinates of the point of intersection of the diagonals of ABCD.
Solution for Question 106
Step 1: Plot \(A(2,2)\) and \(B(6,-2)\) on the coordinate plane. Label the axes clearly and use the same scale on both axes, because the shape name depends on equal lengths and right angles.
Step 2: Reflect \(A(2,2)\) in the origin. Under reflection in the origin, \((x,y)\) becomes \((-x,-y)\).
A(2,2)\mapsto D(-2,-2)
Answer for (a): \(D=(-2,-2)\).
Step 3: Reflect \(A(2,2)\) in the line y=-2. For reflection in y=a, the image of \((x,y)\) is \((x,2a-y)\).
a=-2,\qquad y'=2(-2)-2=-4-2=-6
A(2,2)\mapsto C(2,-6)
Answer for (b): \(C=(2,-6)\).
Step 4: A point invariant under reflection in x=0 must lie on the y-axis, so its x-coordinate must be 0. Now find where CD meets the y-axis.
Step 5: Use \(C(2,-6)\) and \(D(-2,-2)\). The slope of CD is:
m=\frac{-2-(-6)}{-2-2}=\frac{4}{-4}=-1
Step 6: Equation of CD through \(C(2,-6)\):
y+6=-1(x-2)
y=-x-4
Step 7: Put x=0 because P lies on x=0.
y=-0-4=-4
Answer for (c): \(P=(0,-4)\).
Step 8: Check the type of quadrilateral ABCD. Compute two adjacent side lengths and their slopes.
AB=\sqrt{(6-2)^2+(-2-2)^2}=\sqrt{16+16}=4\sqrt{2}
BC=\sqrt{(2-6)^2+(-6-(-2))^2}=\sqrt{16+16}=4\sqrt{2}
m_{AB}=\frac{-2-2}{6-2}=-1,\qquad m_{BC}=\frac{-6-(-2)}{2-6}=1
Step 9: Since adjacent sides are equal and the slopes multiply to -1, adjacent sides are perpendicular. The closed figure is a square.
Answer for (d): ABCD is a square.
Step 10: The 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)=\left(2,-2\right)
Answer for (e): The diagonals intersect at \((2,-2)\).
Question 107: reflections in axes and closed figure
Question: Plot the points \(A(0,3)\), \(B(4,0)\), \(C(6,2)\) and \(D(5,0)\). Reflect the points as given below and write their coordinates:
- Reflect A in the x-axis to A'.
- Reflect B in the y-axis to B'.
- Reflect C in the x-axis to C'.
- D remains invariant when reflected in the line whose equation is to be found.
- Join A,B,C,D,C',B,A',B' and A to form a closed figure. Name the closed figure BCDC'.
Solution for Question 107
Step 1: Plot \(A(0,3)\), \(B(4,0)\), \(C(6,2)\) and \(D(5,0)\) on the graph. Points on the axes are useful here because they often remain unchanged under reflection in that axis.
Step 2: Reflect \(A(0,3)\) in the x-axis. The rule is \((x,y)\mapsto(x,-y)\).
A(0,3)\mapsto A'(0,-3)
Answer for (a): \(A’=(0,-3)\).
Step 3: Reflect \(B(4,0)\) in the y-axis. The rule is \((x,y)\mapsto(-x,y)\).
B(4,0)\mapsto B'(-4,0)
Answer for (b): \(B’=(-4,0)\).
Step 4: Reflect \(C(6,2)\) in the x-axis.
C(6,2)\mapsto C'(6,-2)
Answer for (c): \(C’=(6,-2)\).
Step 5: \(D(5,0)\) lies on the x-axis. Every point on the reflecting line remains invariant under reflection in that line.
\text{Equation of the }x\text{-axis is }y=0
Answer for (d): D remains invariant under reflection in y=0.
Step 6: Join the points in the given order. For the figure BCDC', take \(B(4,0)\), \(C(6,2)\), \(D(5,0)\) and \(C'(6,-2)\).
Step 7: The vertex \(D(5,0)\) bends inward between C and C'. Therefore, the quadrilateral has one reflex interior angle.
Answer for (e): BCDC' is a concave quadrilateral.
Question 108: ogive for daily wages
Question: The following data represents the daily wages in rupees of a certain number of employees of a company.
| Daily wages in rupees | Number of employees |
|---|---|
| 30-40 | 8 |
| 40-50 | 14 |
| 50-60 | 12 |
| 60-70 | 17 |
| 70-80 | 20 |
| 80-90 | 26 |
| 90-100 | 13 |
| 100-110 | 10 |
Use a graph to answer: draw an ogive, find the median wage, find the percentage of employees earning more than \text{₹}84 per day, and find the number of employees earning \text{₹}56 and below.
Solution for Question 108
Step 1: Prepare the less-than cumulative frequency table by adding the frequencies successively.
| Daily wages in rupees | Frequency f | Less-than cumulative frequency | Point to plot |
|---|---|---|---|
| 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: Draw the less-than ogive. Take daily wages on the x-axis and cumulative frequency on the y-axis. Plot the points \((40,8)\), \((50,22)\), \((60,34)\), \((70,51)\), \((80,71)\), \((90,97)\), \((100,110)\) and \((110,120)\). Join them by a smooth increasing curve.
Step 3: Find the total number of employees.
N=8+14+12+17+20+26+13+10=120
Step 4: The median corresponds to the \frac{N}{2}th observation.
\frac{N}{2}=\frac{120}{2}=60
Step 5: On the ogive, mark 60 on the y-axis, draw a horizontal line to the curve, and drop a perpendicular to the x-axis. From the table, the 60th value lies in the class 70-80, because the cumulative frequency just before it is 51 and the next cumulative frequency is 71.
Step 6: Interpolation gives the same reading expected from a well-scaled ogive:
\text{Median}=70+\frac{60-51}{20}\times 10
=70+\frac{9}{20}\times 10=70+4.5=74.5
Answer for (b)(i): Median wage \approx \text{₹}74.5. On a hand-drawn graph, a reading between \text{₹}74 and \text{₹}75 is acceptable if the construction is neat.
Step 7: To estimate the number earning \text{₹}84 or less, draw a vertical line from 84 on the x-axis to the ogive and then a horizontal line to the y-axis. By interpolation, 84 lies in the class 80-90.
\text{Cumulative frequency at }84=71+\frac{84-80}{10}\times 26
=71+\frac{4}{10}\times 26=71+10.4=81.4
Step 8: Hence, the number earning more than \text{₹}84 is approximately:
120-81.4=38.6\approx 39
\text{Percentage}=\frac{39}{120}\times 100=32.5\%
Answer for (b)(ii): Approximately 32.5\% of the employees earn more than \text{₹}84 per day, using the nearest whole-number graph reading.
Step 9: To find the number earning \text{₹}56 and below, draw a vertical line from 56 on the x-axis to the ogive and read the cumulative frequency. By interpolation, 56 lies in the class 50-60.
\text{Cumulative frequency at }56=22+\frac{56-50}{10}\times 12
=22+\frac{6}{10}\times 12=22+7.2=29.2
Answer for (b)(iii): About 29 to 30 employees earn \text{₹}56 and below. A graph reading rounded to the nearest employee gives 30.
Question 109: frequency table and modal height
Question: Study the given graph and answer the questions: make a frequency table, find the number of students whose height is less than 150 cm, find the total number of students, find the modal height, and find the difference between the modal height and the mean height if the average height is 145.5 cm.
Solution for Question 109
Step 1: Read the frequency for each class interval from the graph. The table formed from the given graph is:
| Height in cm | Frequency f | Cumulative frequency |
|---|---|---|
| 120-130 | 6 | 6 |
| 130-140 | 29 | 35 |
| 140-150 | 34 | 69 |
| 150-160 | 22 | 91 |
| 160-170 | 12 | 103 |
Answer for (a): The frequency table is shown above.
Step 2: Heights less than 150 cm include the classes 120-130, 130-140 and 140-150.
6+29+34=69
Answer for (b): 69 students have height less than 150 cm.
Step 3: The total number of students is the sum of all frequencies.
6+29+34+22+12=103
Answer for (c): Total number of students =103.
Step 4: The modal class is the class with the greatest frequency. Here the greatest frequency is 34, so the modal class is 140-150.
Step 5: Use the grouped-data mode formula. Here l=140, h=10, f_1=34, f_0=29, and f_2=22.
\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{5}{68-51}\times 10=140+\frac{50}{17}
=140+2.94=142.94
Answer for (d): Modal height \approx 143 cm.
Step 6: Given mean height =145.5 cm. Difference between mean height and modal height is:
145.5-143=2.5
Answer for (e): Difference =2.5 cm.
Extra worked examples for graph-based practice
These original examples are included to help you practise the same methods used in the ICSE Class 10 Maths graph-based questions above.
Worked example 1: reflection in a vertical line
Problem: Reflect \(R(1,4)\) in the line x=3. Write the coordinates of the image.
Step 1: The point \(R(1,4)\) is 2 units to the left of the line x=3, because 3-1=2.
Step 2: Its image must be 2 units to the right of x=3. Therefore the new x-coordinate is:
3+2=5
Step 3: Reflection in a vertical line does not change the y-coordinate.
Final answer: The image is \(R'(5,4)\).
Worked example 2: median from a less-than ogive table
Problem: A less-than ogive table has cumulative frequencies 5, 18, 32, 50, 60 at upper limits 20, 30, 40, 50, 60. Find the median by interpolation.
Step 1: The total frequency is N=60, so the median corresponds to:
\frac{N}{2}=\frac{60}{2}=30
Step 2: The 30th value lies between cumulative frequencies 18 and 32, so the median class is 30-40.
Step 3: Use l=30, f=32-18=14, h=10, and cumulative frequency before the class =18.
\text{Median}=30+\frac{30-18}{14}\times 10
=30+\frac{120}{14}=30+8.57=38.57
Final answer: Median \approx 38.6.
Worked example 3: mode from a histogram-type frequency table
Problem: For the classes 10-20, 20-30, 30-40, 40-50, the frequencies are 9, 15, 21, 13. Find the mode.
Step 1: The highest frequency is 21, so the modal class is 30-40.
Step 2: Use l=30, h=10, f_1=21, f_0=15, and f_2=13.
\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
=30+\frac{21-15}{2(21)-15-13}\times 10
=30+\frac{6}{42-28}\times 10=30+\frac{60}{14}
=30+4.29=34.29
Final answer: Mode \approx 34.3.
Quick answer index
| Question | Part | Final answer |
|---|---|---|
| 106 | \((a)\) | \(D=(-2,-2)\) |
| 106 | \((b)\) | \(C=(2,-6)\) |
| 106 | \((c)\) | \(P=(0,-4)\) |
| 106 | \((d)\) | ABCD is a square. |
| 106 | \((e)\) | Intersection of diagonals \(=(2,-2)\) |
| 107 | \((a)\) | \(A’=(0,-3)\) |
| 107 | \((b)\) | \(B’=(-4,0)\) |
| 107 | \((c)\) | \(C’=(6,-2)\) |
| 107 | \((d)\) | y=0 |
| 107 | \((e)\) | BCDC' is a concave quadrilateral. |
| 108 | \((a)\) | Less-than ogive using \((40,8)\), \((50,22)\), \((60,34)\), \((70,51)\), \((80,71)\), \((90,97)\), \((100,110)\), \((110,120)\) |
| 108 | \((b)(i)\) | Median wage \approx \text{₹}74.5 |
| 108 | \((b)(ii)\) | Percentage earning more than \text{₹}84 \approx 32.5\% using nearest whole-number graph reading |
| 108 | \((b)(iii)\) | Number earning \text{₹}56 and below \approx 29 to 30 |
| 109 | \((a)\) | Frequency table: 6,29,34,22,12 |
| 109 | \((b)\) | 69 students |
| 109 | \((c)\) | 103 students |
| 109 | \((d)\) | Modal height \approx 143 cm |
| 109 | \((e)\) | 2.5 cm |
Examiner’s mindset for graph-based questions
In graph-based ICSE Class 10 Maths answers, the examiner is not looking only at the final number. Credit is earned through the visible method: a suitable scale, labelled axes, correct plotting, a neat curve or figure, and a clear final reading. For reflection questions, the coordinate rule and the plotted image should agree. For ogives, show the cumulative frequency table before the graph; otherwise the graph reading has no support.
For graph readings such as median wage or number below a value, small differences can occur because of scale and hand drawing. Write an approximate symbol, such as \approx, when the value is read from a graph. If you use interpolation to support the graph reading, show the substitution clearly.
Common mistakes students make in graph-based Maths questions
- Changing the wrong coordinate in reflection: In reflection in the x-axis, only the sign of y changes. For example, \(C(6,2)\) becomes \(C'(6,-2)\), not \((-6,2)\).
- Confusing x=0 and y=0: The line x=0 is the y-axis, while y=0 is the x-axis. This matters for invariant points.
- Using frequency instead of cumulative frequency in an ogive: A less-than ogive uses upper class limits and cumulative frequencies, not the original frequencies.
- Forgetting units: Wage answers should be written in rupees and height answers in cm. A final answer such as 2.5 is incomplete for Question \(109(e)\); it must be 2.5 cm.
- Over-rounding too early: In Question 108, read or calculate first, then round. Rounding 81.4 before the final step changes the percentage slightly.
Related ICSE Class 10 Maths resources
Use these related pages to revise the same competency-focused practice set and supporting coordinate geometry topics:
- ICSE Maths competency-focused multiple-choice questions
- ICSE Class 10 Maths short answer competency questions
- More ICSE Class 10 graph-based long answer questions
- Class 10 equation of a straight line solutions
Syllabus and source note
This solution page is written for ICSE Class 10 Maths graph-based practice. For syllabus alignment, students should use the official CISCE website and the textbook or practice book prescribed by their school. The statistics methods also match the standard school treatment of cumulative frequency curves and grouped-data mode, which overlaps with NCERT Class 10 Mathematics statistics concepts.
Frequently Asked Questions
How should I start an ICSE Class 10 Maths graph-based long answer?
Start an ICSE Class 10 Maths graph-based long answer by choosing a clear scale, drawing labelled axes, and plotting all given points or data values before writing conclusions. For statistics graphs, prepare the frequency or cumulative frequency table first.
What is the rule for reflection of a point in the origin?
Reflection in the origin changes \((x,y)\) to \((-x,-y)\). For example, \(A(2,2)\) becomes \(D(-2,-2)\).
Why is \(P(0,-4)\) invariant under reflection in x=0?
\(P(0,-4)\) lies on the line x=0, which is the y-axis. A point on the mirror line does not move during reflection, so it remains invariant.
How do I find the median from an ogive in Maths?
Find N, mark \frac{N}{2} on the cumulative frequency axis, draw a horizontal line to the ogive, and drop a perpendicular to the x-axis. In Question 108, N=120, so the median is read at the 60th observation and is about \text{₹}74.5.
Why can graph answers be approximate in ICSE Maths?
Graph answers can be approximate because a hand-drawn scale and curve may give readings that differ slightly. Write \approx for values read from an ogive or histogram, and support the answer with construction lines or interpolation where possible.