CBSE Class 8 Mathematics Introduction to Graphs Practice Paper Assignment Set 14 with Answers (Old NCERT Syllabus)

School Assignments for Class 8 Mathematics: Chapter 15 Introduction to Graphs

Access comprehensive school assignments for Chapter 15 Introduction to Graphs using the CBSE Class 8 Mathematics Introduction to Graphs Practice Paper Assignment Set 14 with Answers (Old NCERT Syllabus). Designed to align with the 2026-27 CBSE academic guidelines, these practice sets help Class 8 Mathematics students reinforce core concepts and improve their problem-solving accuracy.

Practice Class 8 Mathematics Assignments: Chapter 15 Introduction to Graphs

Access the complete assignment PDF for Class 8 Mathematics below. Regular practice with these targeted academic tasks builds familiarity with standard question patterns and helps secure higher marks in final school evaluations.

This Class 8 Maths Introduction to Graphs practice paper carries 40 marks and has 19 questions in five sections (A to E), and every question comes with a step-by-step answer and accurate graphs. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is useful for state-board students and for revising this chapter. The paper tests reading line graphs of distance, sales, temperature and speed, plotting points and linear graphs, finding where a line meets the axes, and using graphs for simple interest and time–distance problems.

Section A: Multiple Choice Questions (1 Mark Each)

Question 1: Comparison of parts of a whole may be done by a
(a) bar graph
(b) pie chart
(c) linear graph
(d) line graph
Show Answer & Explanation

Answer: (b) pie chart

Explanation:
1. A pie chart is a circle that stands for the whole.
2. Each sector shows what part of the whole one item makes up.
3. So parts of a whole are compared using a pie chart.

Teacher's Note:
1. Pie chart = parts of a whole; bar graph = comparing separate categories.
2. Line graphs show change over time.

Question 2: A graph that displays data that changes continuously over periods of time is
(a) bar graph
(b) pie chart
(c) histogram
(d) line graph
Show Answer & Explanation

Answer: (d) line graph

Explanation:
1. When a quantity changes continuously with time (like temperature or distance), we join the plotted points by line segments.
2. Such a graph is a line graph.
3. So the answer is a line graph.

Teacher's Note:
1. Time is usually taken on the horizontal axis in a line graph.
2. A histogram shows grouped data in class intervals, not change over time.

Question 3: Data was collected on a student's typing rate, and the graph was drawn as shown below. Approximately how many words had this student typed in 30 seconds?
Typing rate 10 20 30 40 50 60 0 10 20 30 40 50 Time (in seconds) → Number of words → (a) 20
(b) 24
(c) 28
(d) 34
Show Answer & Explanation

Answer: (c) 28

Explanation:
1. The line passes through (0, 0) and (55, 50), so the student types 50 words in 55 seconds.
2. Reading up from 30 seconds, the line is a little below 30 words: 30 × \( \frac{50}{55} \) ≈ 27.3 words.
3. The nearest option is 28.

Teacher's Note:
1. Go up from the time axis to the line, then across to the words axis.
2. The question says 'approximately', so choose the closest option.

Question 4: The following graph shows the progress of a cyclist over a period of 6 hours. The journey commenced at 9:00 hours. During which of the periods did the cyclist maintain his greatest speed?
Progress of a cyclist over a period of 6 hours 9:00 10:00 11:00 12:00 13:00 14:00 15:00 0 10 20 30 40 Time (in h) → Distance (in km) → A B C D E (a) AB
(b) BC
(c) CD
(d) DE
Show Answer & Explanation

Answer: (a) AB

Explanation:
1. AB: 20 km in 1 hour (9:00 to 10:00), speed = 20 km/h.
2. BC: no change in distance (10:00 to 10:30), speed = 0.
3. CD: 20 km in 1.5 hours (10:30 to 12:00), speed ≈ 13.3 km/h. DE: 40 km in 3 hours, speed ≈ 13.3 km/h.
4. So the greatest speed is during AB.

Teacher's Note:
1. The steepest part of a distance–time graph shows the greatest speed.
2. A horizontal part means the cyclist was resting.

Question 5: Which axis runs horizontally?
(a) x-axis
(b) y-axis
(c) z-axis
(d) None of these
Show Answer & Explanation

Answer: (a) x-axis

Explanation:
1. In a graph, the horizontal line is the x-axis and the vertical line is the y-axis.
2. So the x-axis runs horizontally.

Teacher's Note:
1. x-axis: left to right; y-axis: bottom to top.
2. The x-coordinate is always written first in (x, y).

Question 6: A point which lies on both the axes is:
(a) (0, 0)
(b) (0, 1)
(c) (1, 0)
(d) (1, 1)
Show Answer & Explanation

Answer: (a) (0, 0)

Explanation:
1. A point on the x-axis has y = 0; a point on the y-axis has x = 0.
2. A point on both axes must have x = 0 and y = 0.
3. That point is the origin, (0, 0).

Teacher's Note:
1. (0, 1) is on the y-axis only and (1, 0) is on the x-axis only.
2. The axes meet at the origin.

Question 7: The following line graph shows the yearly sales figures of a manufacturing company. Compute the difference between the sales in 2013 and 2015.
Sales of a manufacturing company 2012 2013 2014 2015 2016 0 2 4 6 8 10 Years → Sales (in crores rupees) → (a) ₹ 3 crores
(b) ₹ 4 crores
(c) ₹ 5 crores
(d) ₹ 6 crores
Show Answer & Explanation

Answer: (a) ₹ 3 crores

Explanation:
1. From the graph, sales in 2013 = ₹ 7 crores and in 2015 = ₹ 10 crores.
2. Difference = 10 − 7 = 3.
3. So the difference is ₹ 3 crores.

Teacher's Note:
1. Read each point by going straight across to the vertical axis.
2. Each big square on the vertical axis is ₹ 2 crores.

Question 8: In question 7 above, find the average sales from 2012 to 2016.
(a) ₹ 6 crores
(b) ₹ 7 crores
(c) ₹ 8 crores
(d) ₹ 9 crores
Show Answer & Explanation

Answer: (b) ₹ 7 crores

Explanation:
1. Sales from the graph: 2012: 4, 2013: 7, 2014: 6, 2015: 10, 2016: 8 (in ₹ crores).
2. Total = 4 + 7 + 6 + 10 + 8 = 35.
3. Average = 35 ÷ 5 = 7.
4. So the average sales are ₹ 7 crores.

Teacher's Note:
1. Average = sum of values ÷ number of values.
2. There are 5 years from 2012 to 2016, not 4.

Question 9:
Assertion (A): Variables whose values depend upon the values of other variables are called dependent variables.
Reason (R): The relation between a dependent and an independent variable is shown through a graph.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: this is the meaning of a dependent variable (for example, cost depends on quantity).
2. R is true: a graph can show how the dependent variable changes with the independent variable.
3. But R does not explain why such variables are called dependent; it only says how the relation is shown.
4. So the answer is (b).

Teacher's Note:
1. For (a), R must give the reason for A, not just another true fact.
2. The independent variable usually goes on the x-axis and the dependent one on the y-axis.

Section B: Short Answer Questions (2 Marks Each)

Question 10. The following graph shows the relationship between the time and distance travelled by a car.
Time-distance graph 6 am 7 am 8 am 9 am 0 20 40 60 80 100 120 140 160 Time (in hours) → Distance (in km) → Using this graph, find:
(a) the distance covered during the period 7:30 am to 8 am.
(b) the time when the car had covered a distance of 60 km since its start.
Answer:
(a) At 7:30 am the distance is 100 km and at 8 am it is 120 km.
Distance covered from 7:30 am to 8 am = 120 − 100 = 20 km.
(b) Going across from 60 km to the line and then down, we reach halfway between 6 am and 7 am.
So the car had covered 60 km at 6:30 am.
In simple words: The car does 40 km every hour from 5 am, so it does 20 km in half an hour and reaches 60 km at 6:30 am.

Teacher's Note:
1. The vertical axis shows distance in km.
2. The dashed red lines show how to read the graph in both directions.

 

Question 11. Observe the following graph and answer the questions that follow.
Sales of instruments Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec 0 10 20 30 40 50 60 70 80 90 100 110 Months → Number of instruments → (The x-axis gives the months under consideration.)
(a) What is the difference between the maximum and minimum amounts of sales?
(b) How many instruments have been sold in the last 5 months?
Answer:
(a) Maximum sales = 100 (July); minimum sales = 30 (March).
Difference = 100 − 30 = 70 instruments.
(b) Last 5 months: Aug 75, Sep 60, Oct 70, Nov 50, Dec 70.
Total = 75 + 60 + 70 + 50 + 70 = 325 instruments.
In simple words: The highest point is 100 and the lowest is 30; adding August to December gives 325.

Teacher's Note:
1. Aug's point is halfway between 70 and 80, so it is 75.
2. Count the last 5 months from December backwards.

 

Question 12. The given graph shows the distance travelled and the time taken by a train to reach city B from city A. Study the graph and answer the following questions.
5 6 7 8 9 10 11 12 1 2 3 4 0 50 100 150 200 250 300 350 400 Time → (5 am to 12 noon, then 1 pm to 4 pm) Distance (in km) → A B (a) At what time did the train leave city A?
(b) When did the train cover a distance of 200 km?
Answer:
(a) The graph starts at A, where the distance is 0, at 5 am.
(b) Going across from 200 km, the line first reaches it above 10 am (the train then waits there till 11 am).
In simple words: The train starts at 5 am and has covered 200 km by 10 am.

Teacher's Note:
1. Flat parts of the graph show where the train stopped.
2. Read the am/pm labels carefully: after 12 noon the times are 1, 2, 3, 4 pm.

 

Question 13. The graph below shows the maximum temperatures recorded for two consecutive weeks of a town. Study the graph and answer the questions that follow.
First week Second week Sun Mon Tue Wed Thu Fri Sat 0 30 32 34 36 38 40 42 Days of the week → Maximum temperature (in °C) → (a) In which week was the temperature higher on most of the days?
(b) On which day was the difference in temperatures the maximum for both the weeks?
Answer:
(a) The dashed line (first week) is above the solid line on Sunday, Tuesday, Thursday, Friday and Saturday (5 of 7 days).
So the temperature was higher on most days in the first week.
(b) On Friday the temperatures are 40 °C and 32 °C, a gap of 8 °C, the widest on the graph.
So the difference was maximum on Friday.
In simple words: The first-week line is on top most of the time, and the two lines are farthest apart on Friday.

Teacher's Note:
1. The zigzag on the vertical axis shows that the values from 0 to 30 have been skipped.
2. The day where the two lines are farthest apart has the biggest difference.

 

Section C: Short Answer Questions (3 Marks Each)

Question 14. Draw a graph for the following data.
Number of apples123456
Cost (in ₹)51015202530

Answer:
Draw the axes and choose a scale: horizontal axis 1 unit = 1 apple; vertical axis 1 unit = ₹ 5.
Mark the number of apples along the x-axis and the cost along the y-axis.
Plot the points (1, 5), (2, 10), (3, 15), (4, 20), (5, 25) and (6, 30), and join them.
1 2 3 4 5 6 0 5 10 15 20 25 30 Number of apples → Cost of apples (in ₹) → (1, 5) (2, 10) (3, 15) (4, 20) (5, 25) (6, 30)
The points lie on a straight line through the origin, since cost = 5 × number of apples.
In simple words: Each apple costs ₹ 5, so the points rise evenly and form a straight line from (0, 0).

Teacher's Note:
1. Choose a scale that lets all the points fit on the graph.
2. A straight line through the origin shows direct proportion.

 

Question 15. Draw a graph for the following data.
Side of the square (in cm)23456
Perimeter of the square (in cm)812162024

Answer:
Choose a scale: horizontal axis 1 unit = 1 cm (side); vertical axis 1 unit = 4 cm (perimeter).
Mark the side along the x-axis and the perimeter along the y-axis.
Plot the points (2, 8), (3, 12), (4, 16), (5, 20) and (6, 24), and join them.
Square (Side–Perimeter graph) 1 2 3 4 5 6 0 4 8 12 16 20 24 Side (in cm) → Perimeter (in cm) → (2, 8) (3, 12) (4, 16) (5, 20) (6, 24)
The graph is a straight line through the origin, because perimeter = 4 × side.
In simple words: The perimeter is always four times the side, so the points line up in a straight line.

Teacher's Note:
1. Extending the line backwards passes through (0, 0).
2. Label both axes with the quantity and its unit.

 

Question 16. Plot the points (2, 3) and (3, 2). Draw a line passing through these points. Find the coordinates of the points at which this line meets the x-axis and y-axis.
Answer:
Plot (2, 3) and (3, 2) and draw the line through them.
x y -1 1 2 3 4 5 6 -1 1 2 3 4 5 6 O x + y = 5 (2, 3) (3, 2) (5, 0) (0, 5)
For every point on this line, x + y = 5 (2 + 3 = 5 and 3 + 2 = 5).
On the x-axis y = 0, so x = 5: the line meets the x-axis at (5, 0).
On the y-axis x = 0, so y = 5: the line meets the y-axis at (0, 5).
In simple words: Extend the line both ways; it cuts the x-axis at 5 and the y-axis at 5.

Teacher's Note:
1. Moving 1 unit right takes the line 1 unit down, so you can step along it to the axes.
2. Points on the x-axis have y-coordinate 0.

 

Section D: Long Answer Questions (5 Marks Each)

Question 17. A bank gives 10% Simple Interest (S.I.) on deposits by senior citizens. Draw a graph to illustrate the relation between the sum deposited and simple interest earned. Find from your graph
(a) the annual interest obtainable for an investment of ₹ 250.
(b) the investment one has to make to get an annual simple interest of ₹ 70.
Answer:
Annual S.I. = 10% of the deposit. Table of values:
Deposit (in ₹)1002003005001000
Annual S.I. (in ₹)10203050100

Scale: 1 unit = ₹ 100 on the horizontal axis and 1 unit = ₹ 10 on the vertical axis.
Mark deposits on the x-axis and S.I. on the y-axis, plot the points and join them. The graph is a straight line through the origin.
100 200 300 400 500 600 700 800 900 1000 0 10 20 30 40 50 60 70 80 90 100 Deposit (in ₹) → Simple interest (in ₹) → (250, 25) (700, 70)
(a) From ₹ 250 on the horizontal axis, go up to the line and across: the interest is ₹ 25.
(b) From ₹ 70 on the vertical axis, go across to the line and down: the deposit is ₹ 700.
In simple words: Interest is one-tenth of the deposit, so ₹ 250 earns ₹ 25 and ₹ 70 interest needs ₹ 700.

Teacher's Note:
1. Check by calculation: 10% of 250 = 25 and 70 ÷ 0.1 = 700.
2. Use different scales on the two axes when the numbers are of very different sizes.

 

Question 18. Ajit can ride a scooter constantly at a speed of 30 km/hour. Draw a time-distance graph for this situation. Use it to find
(i) the time taken by Ajit to ride 75 km.
(ii) the distance covered by Ajit in 3½ hours.
Answer:
Distance = 30 × time. Table of values:
Time (in hours)1234
Distance (in km)306090120

Scale: horizontal 2 units = 1 hour; vertical 1 unit = 10 km.
Mark time on the x-axis and distance on the y-axis, plot (1, 30), (2, 60), (3, 90), (4, 120) and join them to the origin.
0.5 1 1.5 2 2.5 3 3.5 4 0 10 20 30 40 50 60 70 80 90 100 110 120 Time (in hours) → Distance (in km) → (2.5, 75) (3.5, 105)
(i) From 75 km on the vertical axis, go across to the line and down: time = 2.5 hours (2 hours 30 minutes).
(ii) From 3½ hours on the horizontal axis, go up to the line and across: distance = 105 km.
In simple words: Ajit covers 30 km each hour, so 75 km takes 2½ hours and 3½ hours gives 105 km.

Teacher's Note:
1. Check: 75 ÷ 30 = 2.5 and 30 × 3.5 = 105.
2. A constant speed always gives a straight-line distance–time graph.

 

Section E: Case Study Based Question (4 Marks)

Question 19. The given graph shows the speed–time graph for car A and car B. Read the graph and answer the following questions.
Car A Car B 7:00 am 7:15 am 7:30 am 7:45 am 8:00 am 8:15 am 8:30 am 8:45 am 9:00 am 10 20 30 40 50 60 70 Time → Speed (in km/h) →
(a) What is the difference in the initial speeds of car A and car B? (1)
(b) At what time is the speed of car A the maximum? (1)
(c) What is the difference in the speed of car B at 7 am and 9 am? (1)
(d) For how long does car B move at constant speed? (1)
Answer:
(a) At 7:00 am, car A's speed is 45 km/h and car B's is 30 km/h. Difference = 45 − 30 = 15 km/h.
(b) The highest point of car A's graph is 70 km/h at 8:15 am.
(c) Car B: 30 km/h at 7 am and 70 km/h at 9 am. Difference = 70 − 30 = 40 km/h.
(d) Car B's graph is flat at 60 km/h from 8:00 am to 8:15 am, so it moves at constant speed for 15 minutes.
In simple words: Read the starting points, the highest point of A, B's first and last points, and the flat part of B's graph.

Teacher's Note:
1. A horizontal piece on a speed–time graph means the speed is not changing.
2. Use the legend to tell which line belongs to which car.

 

Free study material for Mathematics

Chapter 15 Introduction to Graphs Printable Assignments & Solutions for Class 8 Mathematics

Revision Assignment: Chapter 15 Introduction to Graphs (CBSE)

Explore reliable practice questions for Chapter 15 Introduction to Graphs tailored for Class 8 learners. Use these structured worksheets to evaluate preparedness and strengthen core problem-solving skills.

Why Practice Class 8 Mathematics Assignments?

  • Syllabus Compliance: Sets reflect current CBSE evaluation criteria and official marking frameworks.
  • Multi-Format Practice: Includes varied problem types designed to deepen comprehension across all sub-topics.
  • Time Management: Routine practice optimizes pacing to finish school examinations comfortably within schedule.

Effective Strategy for Class 8 Mathematics Assignments

  1. Textbook Review: Always study the core NCERT book for Class 8 Mathematics prior to beginning the assignment.
  2. Independent Attempt: Solve Chapter 15 Introduction to Graphs questions on your own initially before cross-checking with expert solutions.
  3. Error Tracking: Record challenging concepts in a dedicated notebook and practice online MCQ tests for revision.

FAQs

Where can I download the latest CBSE Class 8 Mathematics Chapter 15 Introduction to Graphs assignments?

You can download free PDF assignments for Class 8 Mathematics Chapter 15 Introduction to Graphs from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.

Do these Mathematics Chapter 15 Introduction to Graphs assignments include solved questions?

Yes, our teachers have given solutions for all questions in the Class 8 Mathematics Chapter 15 Introduction to Graphs assignments. This will help you to understand step-by-step methodology to get full marks in school tests and exams.

Are the assignments for Class 8 Mathematics Chapter 15 Introduction to Graphs based on the 2026 exam pattern?

Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 15 Introduction to Graphs.

How can practicing Chapter 15 Introduction to Graphs assignments help in Mathematics preparation?

Practicing topicw wise assignments will help Class 8 students understand every sub-topic of Chapter 15 Introduction to Graphs. Daily practice will improve speed, accuracy and answering competency-based questions.

Can I download Mathematics Chapter 15 Introduction to Graphs assignments for free on mobile?

Yes, all printable assignments for Class 8 Mathematics Chapter 15 Introduction to Graphs are available for free download in mobile-friendly PDF format.