NCERT Solutions for Class 8 Maths Chapter 15 – Introduction to Graphs. Furthermore, here we’ve provided you with the latest solution for Class 8 Maths Chapter 15 – Introduction to Graphs. As a result here you’ll find solutions to all the exercises. This NCERT Class 8 solution will help you to score good marks in your exam.
Students can refer to our solution for NCERT Class 8 Maths Chapter 15 – Introduction to Graphs. The Chapter 15 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise-wise latest solution.
NCERT Solutions for Class 8 Maths Chapter 15 – Introduction to Graphs Exercise Wise Solution
Exercise 15.1 – Page 236 of NCERT
Exercise 15.2 – Page 243 of NCERT
Exercise 15.3 – Page 247 of NCERT
NCERT Solutions for Class 8 Maths Chapter 15 – Introduction to Graphs Exercise 15.1 Solution
Here you’ll find NCERT Chapter 14 – Factorisation Exercise 15.1 Solution
Exercise 15.1: Solutions of Questions on Page Number: 236
Q1: The following graph shows the temperature of a patient in a hospital, recorded every hour.
- What was the patient’s temperature at 1 p.m.?
- When was the patient’s temperature 38.5°C?
- The patient’s temperature was the same two times during the period given. What were these two times?
- What was the temperature at 1.30 p.m? How did you arrive at your answer?
- During which periods did the patient’s temperature show an upward trend?
Answer:
- At 1 p.m., the patient’s temperature was 36.5°C.
- The patient’s temperature was 38.5°C at 12 noon.
- The patient’s temperature was same at 1 p.m. and 2 p.m.
- The graph between the times 1 p.m. and 2 p.m. is parallel to the x-axis. The temperature at 1
p.m. and 2 p.m. is 36.5°C. So, the temperature at 1:30 p.m. is 36.5°C.
- During the following periods, the patient’s temperature showed an upward trend.
9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 2 p.m. to 3 p.m.
Q2: The following line graph shows the yearly sales figure for a manufacturing company.
- What were the sales in (i) 2002 (ii) 2006?
- What were the sales in (i) 2003 (ii) 2005?
- Compute the difference between the sales in 2002 and 2006.
- In which year was there the greatest difference between the sales as compared to its previous year?
Answer:
(a)
- In 2002, the sales were Rs 4 crores.
- In 2006, the sales were Rs 8 crores. (b)
- In 2003, the sales were Rs 7 crores.
- In 2005, the sales were Rs 10 crores.
(c)
(i) In 2002, the sales were Rs 4 crores and in 2006, the sales were Rs 8 crores.
Difference between the sales in 2002 and 2006
= Rs (8 – 4) crores = Rs 4 crores
(d) Difference between the sales of the year 2006 and 2005
= Rs (10 – 8) crores = Rs 2 crores
Difference between the sales of the year 2005 and 2004
= Rs (10 – 6) crores = Rs 4 crores
Difference between the sales of the year 2004 and 2003
= Rs (7 – 6) crore = Rs 1 crore
Difference between the sales of the year 2003 and 2002
= Rs (7 – 4) crores = Rs 3 crores
Hence, the difference was the maximum in the year 2005 as compared to its previous year 2004.
Q3: For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.
- How high was Plant A after (i) 2 weeks (ii) 3weeks?
- How high was Plant B after (i) 2 weeks (ii) 3weeks?
- How much did Plant A grow during the 3rd week?
- How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?
- During which week did Plant A grow most?
- During which week did Plant B grow least?
- Were the two plants of the same height during any week shown here? Specify.
Answer:
(a)
- After 2 weeks, the height of plant A was 7 cm.
- After 3 weeks, the height of plant A was 9 cm. (b)
- After 2 weeks, the height of plant B was 7 cm.
- After 3 weeks, the height of plant B was 10 cm.
- Growth of plant A during 3rd week = 9 cm – 7 cm = 2 cm
- Growth of plant B from the end of the 2nd week to the end of the 3rd week
= 10 cm – 7 cm = 3 cm
- Growth of plant A during 1st week = 2 cm – 0 cm = 2 cm Growth of plant A during 2nd week = 7 cm – 2 cm = 5 cm Growth of plant A during 3rd week = 9 cm – 7 cm = 2 cm
Therefore, plant A grew the most, i.e. 5 cm, during the 2nd week.
- Growth of plant B during 1st week = 1 cm – 0 cm = 1 cm
Growth of plant B during 2nd week = 7 cm – 1 cm = 6 cm Growth of plant B during 3rd week = 10 cm – 7 cm = 3 cm Therefore, plant B grew the least, i.e. 1 cm, during the 1st week.
- At the end of the 2nd week, the heights of both plants were same.
Q4: The following graph shows the temperature forecast and the actual temperature for each day of a week.
- On which days was the forecast temperature the same as the actual temperature?
- What was the maximum forecast temperature during the week?
- What was the minimum actual temperature during the week?
- On which day did the actual temperature differ the most from the forecast temperature?
Answer:
- The forecast temperature was same as the actual temperature on Tuesday, Friday, and Sunday.
- The maximum forecast temperature during the week was 35°C.
- The minimum actual temperature during the week was 15°C.
- The actual temperature differs the most from the forecast temperature on Thursday.
Q5: Use the tables below to draw linear graphs.
- The number of days a hill side city received snow in different years.
Year | 2003 | 2004 | 2005 | 2006 |
Days | 8 | 10 | 5 | 12 |
- Population (in thousands) of men and women in a village in different years.
Year | 2003 | 2004 | 2005 | 2006 | 2007 |
Number of men | 12 | 12.5 | 13 | 13.2 | 13.5 |
Number of women | 11.3 | 11.9 | 13 | 13.6 | 12.8 |
Answer:
- By taking the years on x-axis and the number of days on y-axis and taking scale as 1 unit = 2 days on y-axis and 2 unit = 1 year on x-axis, the linear graph of the given information can be drawn as follows.
- By taking the years on x-axis and population on y-axis and scale as 1 unit = 0.5 thousand on y– axis and 2 unit = 1 year on x-axis, the linear graph of the given information can be drawn as follows.
Q6: A courier-person cycles from a town to a neighboring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph.
- What is the scale taken for the time axis?
- How much time did the person take for the travel?
- How far is the place of the merchant from the town?
- Did the person stop on his way? Explain.
- During which period did he ride fastest?
Answer:
- Scale taken for the time axis is 4 units = 1 hour
- The person travelled during the time 8 a.m. – 11:30 a.m.
Therefore, the person took hours to travel.
- The merchant is 22 km far from the town.
- Yes, the person stopped on his way from 10 a.m. to 10: 30 a.m. This is indicated by the horizontal part of the graph.
- From the graph, it can be observed that during 8 a.m. to 9 a.m., the person travelled the maximum distance. Thus, the person’s ride was the fastest between 8 a.m. and 9 a.m.
Q7: Can there be a time temperature graph as follows? Justify you’re answer:
- This can be a time-temperature graph, as the temperature can increase with the increase in time.
- This can be a time-temperature graph, as the temperature can decrease with the decrease in time.
- This cannot be a time-temperature graph since different temperatures at the same time are not possible.
- This can be a time-temperature graph, as same temperature at different times is possible.
NCERT Solutions for Class 8 Maths Chapter 15 – Introduction to Graphs Exercise 15.2 Solution
Here you’ll find NCERT Chapter 14 – Factorisation Exercise 15.2 Solution
Exercise 15.2: Solutions of Questions on Page Number: 243
Q1: Plot the following points on a graph sheet. Verify if they lie on a line.
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5)
Answer:
- We can plot the given points and join the consecutive points on a graph paper as follows.
From the graph, it can be observed that the points A, B, C, and D lie on the same line.
- We can plot the given points and join the consecutive points on a graph paper as follows.
Hence, points P, Q, R, and S lie on the same line.
- We can plot the given points and join the consecutive points on a graph paper as follows.
Hence, points K, L, M, and N are not lying on the same line.
Q2: Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.
Answer:
From the graph, it can be observed that the line joining the points (2, 3) and (3, 2) meets the x– axis at the point (5, 0) and the y-axis at the point (0, 5).
Q3: Write the coordinates of the vertices of each of these adjoining figures.
Answer:
The coordinates of the vertices in the given figure are as follows. O (0, 0), A (2, 0), B (2, 3), C (0, 3)
P (4, 3), Q (6, 1), R (6, 5), S (4, 7)
K (10, 5), L (7, 7), M (10, 8)
Q4: State whether True or False. Correct those are false.
- A point whose x coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
- A point whose y coordinate is zero and x-coordinate is 5 will lie on y-axis.
- The coordinates of the origin are (0, 0).
Answer:
- True
- False
The point whose y-coordinate is zero and x-coordinate is 5 will lie on x-axis.
- True
NCERT Solutions for Class 8 Maths Chapter 15 – Introduction to Graphs Exercise 15.3 Solution
Here you’ll find NCERT Chapter 14 – Factorisation Exercise 15.3 Solution
Exercise 15.3: Solutions of Questions on Page Number: 247
Q1: Draw the graphs for the following tables of values, with suitable scales on the axes.
- Cost of apples
Number of apples | 1 | 2 | 3 | 4 | 5 |
Cost (in Rs) | 5 | 10 | 15 | 20 | 25 |
- Distance travelled by a car
Time (in hours) | 6 a.m. | 7 a.m. | 8 a.m. | 9 a.m. |
Distance (in km) | 40 | 80 | 120 | 160 |
- How much distance did the car cover during the period 7.30 a.m. to 8 a.m.?
- What was the time when the car had covered a distance of 100 km since its start?
- Interest on deposits for a year:
Answer:
- Taking a suitable scale (for x-axis, 1 unit = 1 apple and for y-axis, 1 unit = Rs 5), we can mark the number of apples on x-axis and the cost of apples on y-axis. A graph of the given data is as follows.
- Taking a suitable scale (for x-axis, 2 units = 1 hour and for y-axis, 2 units = 40 km), we can represent the time on x-axis and the distance covered by the car on y-axis. A graph of the given data is as follows.
- During the period 7:30 a.m. to 8 a.m., the car covered a distance of 20 km.
- The car covered a distance of 100 km at 7:30 a.m. since its start.
- Taking a suitable scale,
For x-axis, 1 unit = Rs 1000 and for y-axis, 1 unit = Rs 80
We can represent the deposit on x-axis and the interest earned on that deposit on y-axis. A graph of the given data is obtained as follows.
From the graph, the following points can be observed.
- Yes. The graph passes through the origin.
- The interest earned in a year on a deposit of Rs 2500 is Rs 200.
- To get an interest of Rs 280 per year, Rs 3500 should be deposited.
Q2: Draw a graph for the following.
(i)
Side of square (in cm) | 2 | 3 | 3.5 | 5 | 6 |
Perimeter (in cm) | 8 | 12 | 14 | 20 | 24 |
Is it a linear graph?
(ii)
Side of square (in cm) | 2 | 3 | 4 | 5 | 6 |
Area (in cm2) | 4 | 9 | 16 | 25 | 36 |
Is it a linear graph?
Answer:
(i) Choosing a suitable scale,
For x-axis, 1 unit = 1 cm and for y-axis, 1 unit = 4 cm
We can represent the side of a square on x-axis and the perimeter of that square on y-axis. A graph of the given data is drawn as follows.
It is a linear graph. (ii)Choosing a suitable scale,
For x-axis, 1 unit = 1 cm and for y-axis, 1 unit = 4 cm2
We can represent the side of a square on the x-axis and the area of that square on y-axis. A graph of the given data is as follows.
It is not a linear graph.
NCERT Class 8 Maths All Chapters Solution
Chapter 1: Rational Numbers
Chapter 2: Linear Equations in One Variable
Chapter 3: Understanding Quadrilaterals
Chapter 4: Practical Geometry
Chapter 5: Data Handling
Chapter 6: Squares and Square root
Chapter 7: Cubes and Cube Roots
Chapter 8: Comparing Quantities
Chapter 9: Arithmetic Expressions
Chapter 10: Visualising Solid Shapes
Chapter 11: Mensuration
Chapter 12: Exponents and Powers
Chapter 13: Direct and Inverse Proportions
Chapter 14: Factorisation
Chapter 15: Introduction to Graphs
Chapter 16: Playing With Numbers