Welcome dear students! Today we are going to learn about Introduction to Graphs from Class 8 Maths.
Have you seen graphs in the newspapers, television, magazines, books etc.? The purpose of the graph is to show numerical facts in visual form so that they can be understood quickly, easily and clearly. Thus graphs are visual representations of data collected. Data can also be presented in the form of a table; however a graphical presentation is easier to understand. This is true in particular when there is a trend or comparison to be shown. We have already seen some types of graphs. Let us quickly recall them here.
A line graph displays data that changes continuously over periods of time. When Renu fell sick, her doctor maintained a record of her body temperature, taken every four hours. It was in the form of a graph, which we may call a time-temperature graph. It is a pictorial representation of data given in tabular form. The table shows time at 6 a.m., 10 a.m., 2 p.m., and 6 p.m., with corresponding temperatures of 37, 40, 38, and 35 degrees Celsius. The horizontal line, usually called the x-axis, shows the timings at which the temperatures were recorded. The vertical line, usually called the y-axis, shows the temperature in degrees Celsius.
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In Figure 13.1 and Figure 13.2, each piece of data is shown by a point on a square grid. The points are then connected by line segments. The result is the line graph. What all does this graph tell you? For example, you can see the pattern of temperature; it is more at 10 a.m. and then decreasing till 6 p.m. Notice that the temperature increased by 3°C, calculated as 40°C – 37°C, during the period 6 a.m. to 10 a.m. There was no recording of temperature at 8 a.m., however the graph suggests that it was more than 37°C. Why? Because the line connecting 6 a.m. and 10 a.m. is continuously rising, so at any point between them, the temperature must be higher than at 6 a.m.
Now let us look at Example 1, which is a graph on performance. The given graph represents the total runs scored by two batsmen A and B, during each of the ten different matches in the year 2007. Study the graph and answer the following questions. First, what information is given on the two axes? The horizontal axis, or the x-axis, indicates the matches played during the year 2007. The vertical axis, or the y-axis, shows the total runs scored in each match. Second, which line shows the runs scored by batsman A? The dotted line shows the runs scored by batsman A, as already indicated at the top of the graph. Third, were the runs scored by them same in any match in 2007? If so, in which match? During the 4th match, both have scored the same number of 60 runs. This is indicated by the point at which both graphs meet. Fourth, among the two batsmen, who is steadier? How do you judge it? Batsman A has one great peak but many deep valleys. He does not appear to be consistent. B, on the other hand, has never scored below a total of 40 runs, even though his highest score is only 100 in comparison to 115 of A. Also, A has scored a zero in two matches and in a total of 5 matches he has scored less than 40 runs. Since A has a lot of ups and downs, B is a more consistent and reliable batsman.
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Let us move to Example 2. The given graph describes the distances of a car from a city P at different times when it is travelling from City P to City Q, which are 350 km apart. Study the graph and answer the following. First, what information is given on the two axes? The horizontal x-axis shows the time. The vertical y-axis shows the distance of the car from City P. Second, from where and when did the car begin its journey? The car started from City P at 8 a.m. Third, how far did the car go in the first hour? The car travelled 50 km during the first hour. At 8 a.m. it just started from City P. At 9 a.m. it was at the 50th km mark. Hence during the one-hour time between 8 a.m. and 9 a.m. the car travelled 50 km. Fourth, how far did the car go during the 2nd hour and the 3rd hour? The distance covered by the car during the 2nd hour, that is from 9 a.m. to 10 a.m., is 100 km, calculated as 150 – 50. During the 3rd hour, from 10 a.m. to 11 a.m., it is 50 km, calculated as 200 – 150. Fifth, was the speed same during the first three hours? How do you know it? From the answers to questions 3 and 4, we find that the speed of the car was not the same all the time. In fact the graph illustrates how the speed varied. Sixth, did the car stop for some duration at any place? Justify your answer. We find that the car was 200 km away from city P when the time was 11 a.m. and also at 12 noon. This shows that the car did not travel during the interval 11 a.m. to 12 noon. The horizontal line segment representing travel during this period is illustrative of this fact. Seventh, when did the car reach City Q? The car reached City Q at 2 p.m.
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Now it is time to practice with Exercise 13.1. I will solve each question step by step. Question 1: The following graph shows the temperature of a patient in a hospital, recorded every hour. The y-axis shows temperature in °C from 34 to 39, and the x-axis shows time from 9 a.m. to 3 p.m. Part a asks, what was the patient's temperature at 1 p.m.? Looking at the graph, at 1 p.m., the temperature is 38.5°C. Part b asks, when was the patient's temperature 38.5°C? From the graph, we see the temperature was 38.5°C at 12 noon and at 1 p.m. Part c asks, the patient's temperature was the same two times during the period given. What were these two times? The temperature was the same at 12 noon and 1 p.m. Part d asks, what was the temperature at 1.30 p.m.? How did you arrive at your answer? The temperature at 1.30 p.m. is approximately 38.25°C. We arrive at this by observing that the temperature at 1 p.m. is 38.5°C and at 2 p.m. is 38°C. Since 1.30 p.m. is exactly halfway between these two times, we take the midpoint of the temperatures, which is 38.25°C. Part e asks, during which periods did the patient's temperature show an upward trend? The temperature showed an upward trend from 9 a.m. to 11 a.m., and again from 2 p.m. to 3 p.m.
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Question 2: The line graph shows yearly sales figures for a manufacturing company from 2002 to 2006. The y-axis shows sales in ₹ crores from 0 to 12, and the x-axis shows the years. Part a asks for sales in 2002 and 2006. Sales in 2002 were ₹4 crore, and in 2006 were ₹8 crore. Part b asks for sales in 2003 and 2005. Sales in 2003 were ₹7 crore, and in 2005 were ₹10 crore. Part c asks to compute the difference between sales in 2002 and 2006. The difference is 8 – 4, which equals ₹4 crore. Part d asks, in which year was there the greatest difference between the sales as compared to its previous year? Let us calculate the yearly differences. 2003 minus 2002 is 7 – 4 = 3. 2004 minus 2003 is 6 – 7 = -1. 2005 minus 2004 is 10 – 6 = 4. 2006 minus 2005 is 8 – 10 = -2. The greatest difference occurred in 2005, with an increase of ₹4 crore compared to 2004.
Question 3: 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 graph shows height in cm on the y-axis and weeks on the x-axis. Plant A heights are 0 at start, 2 at week 1, 7 at week 2, and 9 at week 3. Plant B heights are 0 at start, 1 at week 1, 7 at week 2, and 10 at week 3. Part a asks, how high was Plant A after 2 weeks and 3 weeks? After 2 weeks, it was 7 cm. After 3 weeks, it was 9 cm. Part b asks, how high was Plant B after 2 weeks and 3 weeks? After 2 weeks, it was 7 cm. After 3 weeks, it was 10 cm. Part c asks, how much did Plant A grow during the 3rd week? It grew from 7 cm to 9 cm, so the growth was 2 cm. Part d asks, how much did Plant B grow from the end of the 2nd week to the end of the 3rd week? It grew from 7 cm to 10 cm, so the growth was 3 cm. Part e asks, during which week did Plant A grow most? From start to week 1: 2 cm. Week 1 to week 2: 5 cm. Week 2 to week 3: 2 cm. It grew most during the 2nd week.
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Part f asks, during which week did Plant B grow least? Start to week 1: 1 cm. Week 1 to week 2: 6 cm. Week 2 to week 3: 3 cm. It grew least during the 1st week. Part g asks, were the two plants of the same height during any week shown here? Specify. Yes, at the end of the 2nd week, both plants were exactly 7 cm tall.
Question 4: The graph shows temperature forecast and actual temperature for each day of a week. Part a asks, on which days was the forecast temperature the same as the actual temperature? Looking at the graph, the forecast and actual temperatures match on Tuesday, Friday, and Sunday. Part b asks, what was the maximum forecast temperature during the week? The maximum forecast temperature was 35°C on Sunday. Part c asks, what was the minimum actual temperature during the week? The minimum actual temperature was 15°C on Monday. Part d asks, on which day did the actual temperature differ the most from the forecast temperature? The greatest difference occurred on Thursday, where the forecast was 25°C but the actual temperature was 15°C, a difference of 10°C.
Question 5: Use the tables below to draw linear graphs. Part a gives the number of days a hill side city received snow in different years. Year 2003: 8 days. 2004: 10 days. 2005: 5 days. 2006: 12 days. To draw this, place years on the x-axis and days on the y-axis. Plot the points (2003, 8), (2004, 10), (2005, 5), and (2006, 12). Join them to form a line graph. Part b gives population in thousands of men and women in a village. For men: 2003 is 12, 2004 is 12.5, 2005 is 13, 2006 is 13.2, 2007 is 13.5. For women: 2003 is 11.3, 2004 is 11.9, 2005 is 13, 2006 is 13.6, 2007 is 12.8. Plot these two sets of points on the same axes and join them with two different lines to show the trends.
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Question 6: A courier person cycles from a town to a neighbouring suburban area. The graph shows distance in km on the y-axis from 0 to 22, and time on the x-axis from 8 a.m. to 12 noon. Points are (8 a.m., 0), (9 a.m., 10), (10 a.m., 16), (10:30 a.m., 16), (11:30 a.m., 22), and (12 noon, 22). Part a asks, what is the scale taken for the time axis? The scale is 1 hour per major division, with half hour marks in between. Part b asks, how much time did the person take for the travel? The travel started at 8 a.m. and the person reached the merchant at 11:30 a.m., so the travel time was 3 and a half hours. Part c asks, how far is the place of the merchant from the town? The maximum distance reached is 22 km. Part d asks, did the person stop on his way? Explain. Yes, the person stopped once on the way. The distance remained constant at 16 km from 10 a.m. to 10:30 a.m., indicating a stop. The horizontal line from 11:30 a.m. to 12 noon at 22 km does not represent a stop on the way, but rather shows that the person had already reached the destination and remained there. Part e asks, during which period did he ride fastest? Speed is distance divided by time. From 8 to 9 a.m.: 10 km/h. From 9 to 10 a.m.: 6 km/h. From 10 to 10:30 a.m.: 0. From 10:30 to 11:30 a.m.: 6 km in half an hour, which is 12 km/h. From 11:30 to 12 noon: 0. He rode fastest between 10:30 a.m. and 11:30 a.m.
Question 7: Can there be a time-temperature graph as follows? Justify your answer. Part 1 shows an upward sloping line. Yes, this is possible as temperature can increase over time. Part 2 shows a downward sloping line. Yes, this is possible as temperature can decrease over time. Part 3 shows a vertical line. No, this is not possible. A vertical line would mean that at a single instant of time, the temperature has multiple values, which is impossible. Part 4 shows a horizontal line. Yes, this is possible as it indicates the temperature remains constant over a period of time.
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Now let us move to section 13.2, Some Applications. In everyday life, you might have observed that the more you use a facility, the more you pay for it. If more electricity is consumed, the bill is bound to be high. If less electricity is used, then the bill will be easily manageable. This is an instance where one quantity affects another. Amount of electric bill depends on the quantity of electricity used. We say that the quantity of electricity is an independent variable, or sometimes control variable, and the amount of electric bill is the dependent variable. The relation between such variables can be shown through a graph.
Think, Discuss and Write: The number of litres of petrol you buy to fill a car petrol tank will decide the amount you have to pay. Which is the independent variable here? Think about it. The independent variable is the number of litres of petrol, because it determines the amount to be paid, which is the dependent variable.
Let us look at Example 3, Quantity and Cost. The table gives the quantity of petrol and its cost. 10 litres cost ₹500. 15 litres cost ₹750. 20 litres cost ₹1000. 25 litres cost ₹1250. We need to plot a graph to show the data. First, let us take a suitable scale on both axes. Mark number of litres along the horizontal axis. Mark cost of petrol along the vertical axis. Plot the points (10, 500), (15, 750), (20, 1000), and (25, 1250). Join the points. We find that the graph is a line. It is a linear graph. Why does this graph pass through the origin? Think about it. It passes through the origin because if 0 litres are purchased, the cost is ₹0. This graph can help us to estimate a few things. Suppose we want to find the amount needed to buy 12 litres of petrol. Locate 12 on the horizontal axis. Follow the vertical line through 12 till you meet the graph at a point P. From P, take a horizontal line to meet the vertical axis. This meeting point provides the answer. This is the graph of a situation in which two quantities are in direct variation. In such situations, the graphs will always be linear.
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Try These: In the above example, use the graph to find how much petrol can be purchased for ₹800. Locate 800 on the vertical axis. Move horizontally to meet the graph. From that intersection point, drop a vertical line to the horizontal axis. It meets at 16. So, 16 litres of petrol can be purchased for ₹800.
Example 4 is about Principal and Simple Interest. A bank gives 10% Simple Interest on deposits by senior citizens. Draw a graph to illustrate the relation between the sum deposited and simple interest earned. Find from your graph the annual interest obtainable for an investment of ₹250, and the investment one has to make to get an annual simple interest of ₹70. First, we calculate the simple interest for various deposits using the formula S.I. = (P × R × T) / 100. For ₹100, S.I. is (100 × 1 × 10) / 100 = ₹10. For ₹200, it is ₹20. For ₹300, it is ₹30. For ₹500, it is ₹50. For ₹1000, it is ₹100. We get a table of values. Deposit in ₹: 100, 200, 300, 500, 1000. Annual S.I. in ₹: 10, 20, 30, 50, 100. Scale: 1 unit = ₹100 on horizontal axis, and 1 unit = ₹10 on vertical axis. Mark deposits along horizontal axis. Mark simple interest along vertical axis. Plot the points (100, 10), (200, 20), (300, 30), (500, 50), and (1000, 100). Join the points. We get a graph that is a line. Part a: corresponding to ₹250 on horizontal axis, we get the interest to be ₹25 on vertical axis. Part b: corresponding to ₹70 on the vertical axis, we get the sum to be ₹700 on the horizontal axis.
Try These: Is Example 4 a case of direct variation? Yes, because as the deposit increases, the simple interest increases in the same ratio, and the graph is a straight line passing through the origin.
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Example 5 is about Time and Distance. 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 the time taken by Ajit to ride 75 km, and the distance covered by Ajit in three and a half hours. First, we create a table. 1 hour: 30 km. 2 hours: 2 × 30 = 60 km. 3 hours: 3 × 30 = 90 km. 4 hours: 4 × 30 = 120 km. We get a table of values. Time in hours: 1, 2, 3, 4. Distance covered in km: 30, 60, 90, 120. Scale: horizontal 2 units = 1 hour, vertical 1 unit = 10 km. Mark time on horizontal axis. Mark distance on vertical axis. Plot the points (1, 30), (2, 60), (3, 90), (4, 120). Join the points. We get a linear graph. Part a: corresponding to 75 km on the vertical axis, we get the time to be 2.5 hours on the horizontal axis. Thus 2.5 hours are needed to cover 75 km. Part b: corresponding to three and a half hours on the horizontal axis, the distance covered is 105 km on the vertical axis.
Now let us solve Exercise 13.2. Question 1: Draw the graphs for the following tables of values, with suitable scales on the axes. Part a: Cost of apples. Number of apples: 1, 2, 3, 4, 5. Cost in ₹: 5, 10, 15, 20, 25. Plot these on x and y axes respectively. The graph will be a straight line passing through the origin. Part b: Distance travelled by a car. Time: 6 a.m., 7 a.m., 8 a.m., 9 a.m. Distances in km: 40, 80, 120, 160. Plot these points and join them. Sub question i asks, how much distance did the car cover during the period 7.30 a.m. to 8 a.m.? At 7 a.m. the distance is 80 km, and at 8 a.m. it is 120 km. By linear interpolation, at 7.30 a.m., the distance is 100 km. Therefore, the distance covered from 7.30 a.m. to 8 a.m. is 120 minus 100, which equals 20 km. Sub question ii asks, what was the time when the car had covered a distance of 100 km since its start? Since 100 km is exactly halfway between 80 km and 120 km, the corresponding time is halfway between 7 a.m. and 8 a.m., which is 7.30 a.m. Part c: Interest on deposits for a year. Deposit in ₹: 1000, 2000, 3000, 4000, 5000. Simple interest in ₹: 80, 160, 240, 320, 400. Plot and join. Sub question i asks, does the graph pass through the origin? Yes, because 0 deposit gives 0 interest. Sub question ii asks, use the graph to find the interest on ₹2500 for a year. ₹2500 is halfway between 2000 and 3000, so the interest is halfway between 160 and 240, which is ₹200. Sub question iii asks, to get an interest of ₹280 per year, how much money should be deposited? ₹280 is halfway between 240 and 320, so the deposit should be halfway between 3000 and 4000, which is ₹3500.
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Question 2: Draw a graph for the following. Part i: Side of square in cm: 2, 3, 3.5, 5, 6. Perimeter in cm: 8, 12, 14, 20, 24. Plot these points and join them. Is it a linear graph? Yes, it is a linear graph because the perimeter is directly proportional to the side, following the formula Perimeter = 4 × side, and the points lie on a straight line. Part ii: Side of square in cm: 2, 3, 4, 5, 6. Area in cm²: 4, 9, 16, 25, 36. Plot these points and join them. Is it a linear graph? No, it is not a linear graph. The points form a curve because area is proportional to the square of the side, not directly to the side itself.
Let us quickly review what we have discussed in this chapter. First, graphical presentation of data is easier to understand. Second, a line graph displays data that changes continuously over periods of time. Third, a line graph which is a whole unbroken line is called a linear graph. Fourth, for fixing a point on the graph sheet we need x-coordinate and y-coordinate. Fifth, the relation between dependent variable and independent variable is shown through a graph.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]