Chapter 15 Introduction to Graphs

(i) What information is given on the two axes?

The graph has two axes, a horizontal axis (x-axis) and a vertical axis (y-axis).

• The horizontal axis (x-axis) represents the Matches played, from match 1 to match 10.
• The vertical axis (y-axis) represents the Runs scored by the batsmen in each match. The scale on this axis ranges from 0 to 120.

(ii) Which line shows the runs scored by batsman A?

According to the key provided at the top of the graph, the dashed line (---) represents the runs scored by Batsman A.

(iii) Were the run scored by them same in any match in 2007? If so, in which match?

To find out if they scored the same runs in any match, we need to look for a point where the two lines (the dashed line for Batsman A and the solid line for Batsman B) intersect.

Observing the graph, the two lines intersect at a point corresponding to the 4th match. At this point, the value on the "Runs scored" axis is 60.

Yes, they scored the same number of runs in the 4th match. Both scored 60 runs.

(iv) Among the two batsmen, who is steadier? How do you judge it?

To judge which batsman is steadier, we need to look at the consistency of their scores. A steadier batsman will have less variation in their scores, meaning their line on the graph will be flatter or have smaller peaks and troughs.

• Batsman A's scores (dashed line): The scores fluctuate wildly. For example, the scores are 20, 115, 0, 60, 20, 90, 35, 70, 0, 45. There are very high scores (115) and very low scores (0). The line has very high peaks and deep troughs.
• Batsman B's scores (solid line): The scores are more consistent. The scores are 60, 40, 50, 60, 45, 45, 100, 50, 50, 55. While there is one high score of 100, most of the scores are clustered between 40 and 60. The line on the graph is generally flatter with less extreme variations compared to Batsman A.

Therefore, Batsman B is steadier. We can judge this because his line on the graph has fewer and less extreme fluctuations, indicating a more consistent performance across the matches.

(i) What information is given on the two axes?

The graph shows the relationship between time and distance for a car's journey.

• The horizontal axis (x-axis) represents the Time of the day, from 8 a.m. to 3 p.m.
• The vertical axis (y-axis) represents the Distance of the car from City P (in km).

(ii) From where and when did the car begin its journey?

The journey begins at the first point plotted on the graph. Looking at the horizontal axis, the first point is at 8 a.m.. At this time, the vertical axis shows that the "Distance from P" is 0 km. This means the car started its journey from City P.

The car began its journey from City P at 8 a.m.

(iii) How far did the car go in the first hour?

The first hour of the journey is from 8 a.m. to 9 a.m.

• At 8 a.m., the distance from P was 0 km.
• At 9 a.m., we follow the vertical line up to the graph and then horizontally to the y-axis, which shows the distance is 50 km.

The distance covered in the first hour is $50 - 0 = 50$ km.

(iv) How far did the car go during (i) the 2nd hour? (ii) the 3rd hour?

(i) During the 2nd hour (from 9 a.m. to 10 a.m.):

• At 9 a.m., the distance from P was 50 km.
• At 10 a.m., the distance from P was 150 km.

Distance covered = $150 - 50 = 100$ km.

(ii) During the 3rd hour (from 10 a.m. to 11 a.m.):

• At 10 a.m., the distance from P was 150 km.
• At 11 a.m., the distance from P was 200 km.

Distance covered = $200 - 150 = 50$ km.

(v) Was the speed same during the first three hours? How do you know it?

No, the speed was not the same during the first three hours. We know this because the car covered different distances in each of the first three hours:

• 1st hour (8-9 a.m.): 50 km
• 2nd hour (9-10 a.m.): 100 km
• 3rd hour (10-11 a.m.): 50 km

Since the speed is the distance covered per unit of time, different distances covered in the same amount of time (1 hour) mean the speed was different. Also, the slope of the line graph represents the speed. The line is steepest between 9 a.m. and 10 a.m., indicating the highest speed in that interval.

(vi) Did the car stop for some duration at any place? Justify your answer.

Yes, the car stopped for some duration. We can tell this by looking for a horizontal segment on the line graph. A horizontal line means that time is passing (moving along the x-axis), but the distance from the starting point is not changing (the y-value is constant).

The graph shows a horizontal line between 11 a.m. and 12 noon. During this one-hour period, the distance from City P remained constant at 200 km, which means the car was stationary.

(vii) When did the car reach City Q?

City P and City Q are 350 km apart. To find when the car reached City Q, we need to find the time on the graph when the distance from City P is 350 km.

Looking at the vertical axis at 350 km and moving horizontally to the line graph, we find the corresponding point on the horizontal axis is at 2 p.m.

The car reached City Q at 2 p.m.

(a) What was the patient’s temperature at 1 p.m.?

To find the temperature at 1 p.m., we locate '1 p.m.' on the horizontal axis (Time). We then move vertically up to the point on the graph line. From that point, we move horizontally to the left to read the value on the vertical axis (Temperature). The value corresponding to 1 p.m. is 36.5° C.

(b) When was the patient’s temperature 38.5° C?

To find the time when the temperature was 38.5° C, we locate '38.5' on the vertical axis. We then move horizontally to the right to the point on the graph line. From that point, we move vertically down to read the value on the horizontal axis. The time corresponding to a temperature of 38.5° C is 12 noon.

(c) The patient’s temperature was the same two times during the period given. What were these two times?

To find when the temperature was the same, we look for a horizontal line segment or two points on the graph that have the same vertical value. We can see a horizontal line between 1 p.m. and 2 p.m. At both these times, the temperature reading is the same.

The two times were 1 p.m. and 2 p.m., and the temperature at both times was 36.5° C.

(d) What was the temperature at 1.30 p.m.? How did you arrive at your answer?

The time 1:30 p.m. is exactly halfway between 1 p.m. and 2 p.m. on the horizontal axis. We can observe that the line graph is a straight horizontal line between 1 p.m. and 2 p.m. This means the temperature did not change during this one-hour interval. Since the temperature was 36.5° C at 1 p.m. and also at 2 p.m., it must have been the same at any point in between.

The temperature at 1.30 p.m. was 36.5° C. We arrived at this answer by observing that the temperature was constant between 1 p.m. and 2 p.m.

(e) During which periods did the patients’ temperature showed an upward trend?

An upward trend is indicated by a line segment on the graph that rises from left to right. We can identify the following periods where the temperature was increasing:

• From 9 a.m. to 10 a.m. (Temperature increased from 35.5° C to 36° C)
• From 10 a.m. to 11 a.m. (Temperature increased from 36° C to 39° C)
• From 2 p.m. to 3 p.m. (Temperature increased from 36.5° C to 37° C)

(a) What were the sales in (i) 2002 (ii) 2006?

To find the sales for a particular year, we locate the year on the horizontal axis and find the corresponding value on the vertical axis.

• (i) 2002: Locate '2002' on the x-axis. The point on the graph above it corresponds to ₹ 4 crores on the y-axis.
• (ii) 2006: Locate '2006' on the x-axis. The point on the graph above it corresponds to ₹ 8 crores on the y-axis.

(b) What were the sales in (i) 2003 (ii) 2005?

• (i) 2003: Locate '2003' on the x-axis. The point on the graph above it corresponds to ₹ 7 crores on the y-axis.
• (ii) 2005: Locate '2005' on the x-axis. The point on the graph above it corresponds to ₹ 10 crores on the y-axis.

(c) Compute the difference between the sales in 2002 and 2006.

We need to find the difference between the sales figures for these two years.

Sales in 2006 = ₹ 8 crores

Sales in 2002 = ₹ 4 crores

Difference = Sales in 2006 - Sales in 2002

Difference = ₹ 8 crores - ₹ 4 crores = ₹ 4 crores.

(d) In which year was there the greatest difference between the sales as compared to its previous year?

To find this, we need to calculate the year-on-year difference in sales for each year from 2003 to 2006.

• 2003: Sales were ₹ 7 crores. Previous year (2002) sales were ₹ 4 crores.
  Difference = ₹ 7 cr - ₹ 4 cr = ₹ 3 crores (Increase)
• 2004: Sales were ₹ 6 crores. Previous year (2003) sales were ₹ 7 crores.
  Difference = ₹ 7 cr - ₹ 6 cr = ₹ 1 crore (Decrease)
• 2005: Sales were ₹ 10 crores. Previous year (2004) sales were ₹ 6 crores.
  Difference = ₹ 10 cr - ₹ 6 cr = ₹ 4 crores (Increase)
• 2006: Sales were ₹ 8 crores. Previous year (2005) sales were ₹ 10 crores.
  Difference = ₹ 10 cr - ₹ 8 cr = ₹ 2 crores (Decrease)

Comparing the differences (₹ 3 cr, ₹ 1 cr, ₹ 4 cr, ₹ 2 cr), the greatest difference is ₹ 4 crores, which occurred in the year 2005 compared to 2004. This is also visually represented by the steepest upward slope on the graph, which is between 2004 and 2005.

The graph shows the height of Plant A (dashed line) and Plant B (solid line) over a period of 3 weeks.

(a) How high was Plant A after (i) 2 weeks (ii) 3 weeks?

We look at the dashed line for Plant A.

• (i) After 2 weeks: Find '2' on the horizontal axis. Move up to the dashed line and then left to the vertical axis. The height is 7 cm.
• (ii) After 3 weeks: Find '3' on the horizontal axis. Move up to the dashed line and then left to the vertical axis. The height is 9 cm.

(b) How high was Plant B after (i) 2 weeks (ii) 3 weeks?

We look at the solid line for Plant B.

• (i) After 2 weeks: Find '2' on the horizontal axis. Move up to the solid line and then left to the vertical axis. The height is 7 cm.
• (ii) After 3 weeks: Find '3' on the horizontal axis. Move up to the solid line and then left to the vertical axis. The height is 10 cm.

(c) How much did Plant A grow during the 3rd week?

The 3rd week is the period between the end of week 2 and the end of week 3.

• Height of Plant A at the end of week 2 = 7 cm.
• Height of Plant A at the end of week 3 = 9 cm.

Growth = Height at end of week 3 - Height at end of week 2

Growth = $9 - 7 = 2$ cm.

Plant A grew 2 cm during the 3rd week.

(d) How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?

This is the same as asking for its growth during the 3rd week.

• Height of Plant B at the end of week 2 = 7 cm.
• Height of Plant B at the end of week 3 = 10 cm.

Growth = $10 - 7 = 3$ cm.

Plant B grew 3 cm during the 3rd week.

(e) During which week did Plant A grow most?

We calculate the growth of Plant A for each week.

• 1st week (Start to Week 1): Height at Start = 0 cm, Height at Week 1 = 2 cm. Growth = $2 - 0 = 2$ cm.
• 2nd week (Week 1 to Week 2): Height at Week 1 = 2 cm, Height at Week 2 = 7 cm. Growth = $7 - 2 = 5$ cm.
• 3rd week (Week 2 to Week 3): Height at Week 2 = 7 cm, Height at Week 3 = 9 cm. Growth = $9 - 7 = 2$ cm.

The greatest growth (5 cm) occurred during the 2nd week.

(f) During which week did Plant B grow least?

We calculate the growth of Plant B for each week.

• 1st week (Start to Week 1): Height at Start = 0 cm, Height at Week 1 = 1 cm. Growth = $1 - 0 = 1$ cm.
• 2nd week (Week 1 to Week 2): Height at Week 1 = 1 cm, Height at Week 2 = 7 cm. Growth = $7 - 1 = 6$ cm.
• 3rd week (Week 2 to Week 3): Height at Week 2 = 7 cm, Height at Week 3 = 10 cm. Growth = $10 - 7 = 3$ cm.

The least growth (1 cm) occurred during the 1st week.

(g) Were the two plants of the same height during any week shown here? Specify.

To find this, we look for a point where the two lines (dashed and solid) intersect.

Yes, the lines intersect at the end of the 2nd week. At this point, the height for both plants was 7 cm.

The graph compares the forecast temperature (dashed line) with the actual temperature (solid line) for each day of a week.

(a) On which days was the forecast temperature the same as the actual temperature?

To find this, we need to look for the days where the dashed line and the solid line intersect. By observing the graph, we can see that the two lines meet at the following points:

• On Tuesday, both temperatures were 20°C.
• On Friday, both temperatures were 15°C.
• On Sunday, both temperatures were 35°C.

(b) What was the maximum forecast temperature during the week?

To find the maximum forecast temperature, we need to find the highest point on the dashed line (Forecast). The highest point on the dashed line occurs on Sunday.

The maximum forecast temperature was 35°C on Sunday.

(c) What was the minimum actual temperature during the week?

To find the minimum actual temperature, we need to find the lowest point on the solid line (Actual). The lowest point on the solid line occurs on Friday.

The minimum actual temperature was 15°C on Friday.

(d) On which day did the actual temperature differ the most from the forecast temperature?

To find this, we need to find the day with the largest vertical gap between the dashed line and the solid line. Let's calculate the difference for each day:

• Monday: Actual = 17.5°C, Forecast = 15°C. Difference = 2.5°C.
• Tuesday: Difference = 0°C.
• Wednesday: Actual = 30°C, Forecast = 25°C. Difference = 5°C.
• Thursday: Actual = 15°C, Forecast = 22.5°C. Difference = 7.5°C.
• Friday: Difference = 0°C.
• Saturday: Actual = 25°C, Forecast = 30°C. Difference = 5°C.
• Sunday: Difference = 0°C.

The greatest difference between the actual and forecast temperatures was 7.5°C, which occurred on Thursday.

(a) Linear graph for the number of days a hill side city received snow.

To draw the linear graph, we will follow these steps:

1. Draw two axes: a horizontal axis (x-axis) for 'Year' and a vertical axis (y-axis) for 'Days'.
2. Choose a suitable scale. For the x-axis, we can mark the years 2003, 2004, 2005, and 2006 at equal intervals. For the y-axis, since the values range from 5 to 12, we can mark from 0 to 14, with each unit representing 1 day.
3. Plot the points corresponding to the data: (2003, 8), (2004, 10), (2005, 5), and (2006, 12).
4. Join the consecutive points with line segments to form the linear graph.

(b) Linear graph for the population of men and women in a village.

To draw this graph, we will have two lines on the same set of axes, one for men and one for women.

1. Draw two axes: a horizontal axis (x-axis) for 'Year' and a vertical axis (y-axis) for 'Population (in thousands)'.
2. Choose a suitable scale. For the x-axis, mark the years 2003 to 2007. For the y-axis, the values range from 11.3 to 13.6, so a scale from 10 to 14 would be appropriate, with each unit representing 0.5 thousand.
3. Create a key to distinguish between the lines for men and women (e.g., solid line for men, dashed line for women).
4. Plot the points for the male population: (2003, 12), (2004, 12.5), (2005, 13), (2006, 13.2), (2007, 13.5).
5. Join these points with line segments to create the graph for men.
6. Plot the points for the female population: (2003, 11.3), (2004, 11.9), (2005, 13), (2006, 13.6), (2007, 12.8).
7. Join these points with line segments to create the graph for women.

(a) What is the scale taken for the time axis?

Looking at the horizontal axis (time axis), we can see that the distance between consecutive hour marks (e.g., between 8 a.m. and 9 a.m.) is divided into 4 equal units (grid squares).

So, 4 units on the axis represent 1 hour (60 minutes).

Therefore, the scale for the time axis is 4 units = 1 hour.

(b) How much time did the person take for the travel?

The last data point is at 11:30 a.m. (2 units past 11 a.m.). So, the travel ended at 11:30 a.m.

Total time taken = 11:30 a.m. - 8:00 a.m. = 3 hours and 30 minutes.

(c) How far is the place of the merchant from the town?

The place of the merchant is the final destination. We need to find the maximum distance reached from the town. Looking at the vertical axis, the highest point on the graph corresponds to a distance of 22 km.

(d) Did the person stop on his way? Explain.

Yes, the person stopped on his way. We can tell this by looking for a horizontal segment on the line graph. A horizontal line indicates that time is passing, but the distance from the town is not changing, which means the person is stationary.

There is a horizontal line on the graph between 10:00 a.m. and 10:30 a.m. During this 30-minute period, the distance from the town remained constant at 16 km.

(e) During which period did he ride fastest?

The speed of the courier-person is represented by the steepness (or slope) of the line on the graph. A steeper line means a greater distance was covered in a given amount of time, indicating a higher speed. We can analyze the different segments of the journey:

• From 8 a.m. to 9 a.m.: The distance covered is 10 km in 1 hour. Speed = 10 km/hr.
• From 9 a.m. to 10 a.m.: The distance covered is from 10 km to 16 km (a total of 6 km) in 1 hour. Speed = 6 km/hr.
• From 10:30 a.m. to 12 noon: The distance covered is from 16 km to 22 km (a total of 6 km) in 1.5 hours. Speed = 6 km / 1.5 h = 4 km/hr.

By comparing the speeds, the highest speed (10 km/hr) was achieved in the first segment. Visually, the line segment between 8 a.m. and 9 a.m. is the steepest. Therefore, the person rode the fastest during this period.

Let's analyze each of the four graphs to determine if they can represent a valid time-temperature relationship.

Graph (i)

This graph shows that as time increases, the temperature also increases at a constant rate. This is a possible scenario, for example, when something is being heated steadily.

Yes, this graph is possible.

Graph (ii)

This graph shows that as time increases, the temperature decreases at a constant rate. This is also a possible scenario, for example, when a hot object is cooling down in a cooler environment.

Yes, this graph is possible.

Graph (iii)

This graph shows a vertical line. A vertical line means that at a single instant in time, the temperature has multiple different values. For example, it suggests that at one specific moment, the temperature is 10°C, 20°C, 30°C, and so on, all at the same time. This is physically impossible.

No, this graph is not possible because it is not possible for the temperature to have multiple values at the same instant in time.

Graph (iv)

This graph shows a horizontal line. A horizontal line means that as time passes, the temperature remains constant. This is a very common and possible scenario, for instance, the temperature of an object in a room that is at a stable temperature, or the boiling point of water as it turns to steam.

Yes, this graph is possible.

Solution:

To plot a point $(x, y)$ on a graph sheet, we use a Cartesian coordinate system. The first value, $x$, represents the horizontal distance from the origin (the x-coordinate or abscissa), and the second value, $y$, represents the vertical distance from the origin (the y-coordinate or ordinate).

Plotting the point $(4, 3)$:

1. Start at the origin, O$(0, 0)$.

2. Move 4 units along the positive x-axis (to the right).

3. From this position, move 3 units upwards, parallel to the positive y-axis.

4. Mark this point as P$(4, 3)$.

Plotting the point $(3, 4)$:

1. Start at the origin, O$(0, 0)$.

2. Move 3 units along the positive x-axis (to the right).

3. From this position, move 4 units upwards, parallel to the positive y-axis.

4. Mark this point as Q$(3, 4)$.

Now, let's compare the two points.

For the point $(4, 3)$, the x-coordinate is 4 and the y-coordinate is 3.

For the point $(3, 4)$, the x-coordinate is 3 and the y-coordinate is 4.

Since the x-coordinates and y-coordinates of the two points are interchanged, they represent two different positions on the graph sheet. As we can see from the graph, the points P$(4, 3)$ and Q$(3, 4)$ are distinct points.

Therefore, the point $(4, 3)$ is not the same as the point $(3, 4)$.

Solution:

By observing the given graph (Fig 15.14), we can determine the location of the points and their coordinates.

(i) To locate the point (2, 1), we start from the origin (0, 0), move 2 units to the right along the X-axis, and then move 1 unit upwards parallel to the Y-axis. The letter at this location is E.

(ii) To locate the point (0, 5), the x-coordinate is 0, which means the point lies on the Y-axis. We move 5 units upwards from the origin along the Y-axis. The letter at this location is B.

(iii) To locate the point (2, 0), the y-coordinate is 0, which means the point lies on the X-axis. We move 2 units to the right from the origin along the X-axis. The letter at this location is G.

(iv) To find the coordinates of A, we see that point A is 4 units to the right of the Y-axis (so its x-coordinate is 4) and 5 units above the X-axis (so its y-coordinate is 5). Thus, the coordinates of A are (4, 5).

(v) To find the coordinates of F, we see that point F lies on the X-axis, which means its y-coordinate is 0. On the X-axis, it is located exactly halfway between 5 and 6. Therefore, its x-coordinate is 5.5. Thus, the coordinates of F are (5.5, 0).

We will plot each set of points on a graph sheet and join them to see if they form a straight line.

(i) (0, 2), (0, 5), (0, 6), (0, 3.5)

We observe that the x-coordinate of all the given points is 0. Any point with an x-coordinate of 0 lies on the Y-axis. Therefore, all these points lie on the Y-axis.

Since the Y-axis is a straight line, the points lie on a line. This line is the Y-axis.

(ii) A (1, 1), B (1, 2), C (1, 3), D (1, 4)

We observe that the x-coordinate of all the given points is 1. This means all points are located at a constant horizontal distance of 1 unit from the Y-axis. When we plot these points and join them, they form a vertical line parallel to the Y-axis.

Yes, these points lie on a line. We can name this line AD.

(iii) K (1, 3), L (2, 3), M (3, 3), N (4, 3)

We observe that the y-coordinate of all the given points is 3. This means all points are located at a constant vertical distance of 3 units from the X-axis. When we plot these points and join them, they form a horizontal line parallel to the X-axis.

Yes, these points lie on a line. We can name this line KN.

(iv) W (2, 6), X (3, 5), Y (5, 3), Z (6, 2)

We plot the points W, X, Y, and Z on the graph sheet. After plotting, we join the points with a ruler.

We observe that all the points lie on the same line.

Yes, these points lie on a line. We can name this line WZ.

(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)

We plot the points A, B, C, and D on a graph sheet.

From the graph, we can observe that all the points A, B, C, and D have the same x-coordinate, which is 4. When we connect these points, they form a straight vertical line.

Therefore, yes, the points A, B, C, and D lie on a line.

(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)

We plot the points P, Q, R, and S on a graph sheet.

From the graph, we can observe that for each point, the x-coordinate is equal to the y-coordinate. When we connect these points, they form a straight line passing through the origin.

Therefore, yes, the points P, Q, R, and S lie on a line.

(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5)

We plot the points K, L, M, and N on a graph sheet.

From the graph, we can see that the points K, L, M, and N do not lie on a single straight line. Instead, they form the vertices of a closed figure, specifically a rectangle.

Therefore, no, the points K, L, M, and N do not lie on a line.

Solution:

First, we need to plot the given points A(2, 3) and B(3, 2) on a graph sheet.

1. To plot point A(2, 3), start from the origin, move 2 units along the positive x-axis, and then 3 units up parallel to the y-axis.

2. To plot point B(3, 2), start from the origin, move 3 units along the positive x-axis, and then 2 units up parallel to the y-axis.

3. After plotting the points, we draw a straight line passing through both points A and B. We then extend this line on both sides until it intersects the x-axis and the y-axis.

From the graph, we can observe the points where the line meets the axes:

1. The line meets the x-axis at the point where the y-coordinate is 0. By looking at the graph, this point is (5, 0).

2. The line meets the y-axis at the point where the x-coordinate is 0. By looking at the graph, this point is (0, 5).

Therefore, the coordinates of the points where the line meets the x-axis and y-axis are (5, 0) and (0, 5) respectively.

Alternate Solution (Using Equation of a Line)

We can also find the intersection points algebraically.

Let the given points be A(2, 3) and B(3, 2). The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$y - y_1 = m(x - x_1)$, where the slope $m = \frac{y_2 - y_1}{x_2 - x_1}$.

First, let's calculate the slope (m):

$m = \frac{2 - 3}{3 - 2} = \frac{-1}{1} = -1$

Now, using the point-slope form with point A(2, 3):

$y - 3 = -1(x - 2)$

$y - 3 = -x + 2$

$x + y = 2 + 3$

$x + y = 5$

This is the equation of the line passing through the given points.

To find the point where the line meets the x-axis, we set $y = 0$ in the equation:

$x + 0 = 5$

$x = 5$

So, the line meets the x-axis at (5, 0).

To find the point where the line meets the y-axis, we set $x = 0$ in the equation:

$0 + y = 5$

$y = 5$

So, the line meets the y-axis at (0, 5).

Solution:

By observing the given graph, we can determine the coordinates of the vertices for each figure.

For Figure OABC:

The vertices are O, A, B, and C.

O: This is the origin, so its coordinates are (0, 0).

A: This point is 2 units to the right on the x-axis and 0 units up. Its coordinates are (2, 0).

B: This point is 2 units to the right and 3 units up. Its coordinates are (2, 3).

C: This point is 0 units to the right and 3 units up on the y-axis. Its coordinates are (0, 3).

For Figure PQRS:

The vertices are P, Q, R, and S.

P: This point is 4 units to the right and 3 units up. Its coordinates are (4, 3).

Q: This point is 6 units to the right and 1 unit up. Its coordinates are (6, 1).

R: This point is 6 units to the right and 5 units up. Its coordinates are (6, 5).

S: This point is 4 units to the right and 7 units up. Its coordinates are (4, 7).

For Figure KLM:

The vertices are K, L, and M.

K: This point is 10 units to the right and 5 units up. Its coordinates are (10, 5).

L: This point is 7 units to the right and 7 units up. Its coordinates are (7, 7).

M: This point is 10 units to the right and 8 units up. Its coordinates are (10, 8).

(i) A point whose x coordinate is zero and y-coordinate is non-zero will lie on the y-axis.

True.

Any point on the y-axis has a horizontal distance of 0 from the y-axis itself, which means its x-coordinate must be 0. For example, the point (0, 4) lies on the y-axis.

(ii) A point whose y coordinate is zero and x-coordinate is 5 will lie on y-axis.

False.

A point with coordinates (5, 0) has a horizontal distance of 5 units from the y-axis and a vertical distance of 0 units from the x-axis. A point with a y-coordinate of 0 lies on the x-axis.

Correct statement: A point whose y-coordinate is zero and x-coordinate is 5 will lie on the x-axis.

(iii) The coordinates of the origin are (0, 0).

True.

The origin is the point where the x-axis and the y-axis intersect. At this point, both the x-coordinate and the y-coordinate are zero.

Solution:

To plot the given data, we will follow these steps:

Step 1: Choose the Axes

We need to represent the two quantities, "No. of Litres of petrol" and "Cost of petrol", on the two axes of the graph.

• Let's take the "No. of Litres of petrol" as the independent variable and represent it on the horizontal axis (x-axis).
• Let's take the "Cost of petrol in $\textsf{₹}$" as the dependent variable and represent it on the vertical axis (y-axis).

Step 2: Choose a suitable scale

• For the x-axis (Litres): The values range from 10 to 25. A suitable scale would be 1 unit = 5 litres.
• For the y-axis (Cost): The values range from 500 to 1250. A suitable scale would be 1 unit = $\textsf{₹}$ 250. This allows us to plot the points easily.

Step 3: Plot the points

We have the following pairs of (Litres, Cost) to plot as coordinates (x, y):

• (10, 500)
• (15, 750)
• (20, 1000)
• (25, 1250)

Step 4: Join the points

After plotting all the points on the graph paper, we join them with a straight line. We can observe that the graph is a straight line. It is a linear graph. The line can be extended to the origin (0,0) because 0 litres of petrol would cost $\textsf{₹}$ 0.

The resulting graph is shown below:

This graph visually represents the relationship between the quantity of petrol and its cost, showing that the cost increases linearly with the quantity.

Solution:

To draw the graph, we first need to find the relationship between the sum deposited (Principal) and the simple interest earned. We will prepare a table of values.

The formula for Simple Interest (S.I.) is:

$S.I. = \frac{P \times R \times T}{100}$

Here, P = Principal (Sum Deposited), R = Rate of Interest, T = Time.

Given:

Rate (R) = 10% per annum

Time (T) = 1 year (for annual interest)

So, the formula becomes:

$S.I. = \frac{P \times 10 \times 1}{100} = \frac{P}{10}$

Now, let's calculate the simple interest for different values of the principal:

Steps to draw the graph:

1. Choose Axes: We take the Sum Deposited on the x-axis and the Simple Interest on the y-axis.

2. Choose Scale:

On the x-axis, let 1 unit = $\textsf{₹}$ 100.

On the y-axis, let 1 unit = $\textsf{₹}$ 10.

3. Plot Points: We plot the points from the table: (100, 10), (200, 20), (500, 50), and (1000, 100).

4. Join Points: We join the plotted points with a straight line. The line will pass through the origin (0, 0), as zero deposit earns zero interest.

Now, we can find the required values from the graph.

(a) The annual interest obtainable for an investment of $\textsf{₹}$ 250.

To find this, we locate $\textsf{₹}$ 250 on the x-axis. From this point, we move vertically upwards to meet the graph line. From the point of intersection, we move horizontally to the left to meet the y-axis. The value we read on the y-axis is 25.

So, the annual interest for an investment of $\textsf{₹}$ 250 is $\textsf{₹}$ 25.

(b) The investment one has to make to get an annual simple interest of $\textsf{₹}$ 70.

To find this, we locate $\textsf{₹}$ 70 on the y-axis. From this point, we move horizontally to the right to meet the graph line. From the point of intersection, we move vertically downwards to meet the x-axis. The value we read on the x-axis is 700.

So, the investment required to get an annual simple interest of $\textsf{₹}$ 70 is $\textsf{₹}$ 700.

Solution:

To draw the time-distance graph, we first need a set of data points. We are given that Ajit's constant speed is 30 km/hour.

The relationship between distance, speed, and time is:

Distance = Speed × Time

So, for this situation, the relationship is:

Distance = 30 × Time

Let's create a table of values for different time intervals:

Steps to draw the graph:

1. Choose Axes: We will represent 'Time' on the horizontal axis (x-axis) and 'Distance' on the vertical axis (y-axis).

2. Choose Scale:

On the x-axis (Time): Let 2 units = 1 hour.

On the y-axis (Distance): Let 1 unit = 15 km.

3. Plot Points: We plot the points from the table: (0, 0), (1, 30), (2, 60), (3, 90), (4, 120).

4. Join Points: We join the points with a straight line. Since the speed is constant, the graph will be a straight line passing through the origin.

Now we use the graph to find the required information:

(i) The time taken by Ajit to ride 75 km.

To find the time taken to cover 75 km, we locate 75 on the y-axis (Distance). From this point, we draw a horizontal line to meet the graph. From the intersection point on the graph, we draw a vertical line down to the x-axis (Time). This vertical line meets the x-axis at 2.5.

Therefore, the time taken by Ajit to ride 75 km is 2.5 hours or $2\frac{1}{2}$ hours.

(ii) The distance covered by Ajit in $3\frac{1}{2}$ hours.

To find the distance covered in $3\frac{1}{2}$ hours (or 3.5 hours), we locate 3.5 on the x-axis (Time). From this point, we draw a vertical line up to meet the graph. From the intersection point on the graph, we draw a horizontal line to the y-axis (Distance). This horizontal line meets the y-axis at 105.

Therefore, the distance covered by Ajit in $3\frac{1}{2}$ hours is 105 km.

(a) Cost of apples

We are given the following data:

To draw the graph:

1. We take 'Number of apples' on the horizontal axis (x-axis) and 'Cost (in $\textsf{₹}$)' on the vertical axis (y-axis).

2. We choose a suitable scale:

On the x-axis: 2 units = 1 apple.

On the y-axis: 1 unit = $\textsf{₹}$ 5.

3. We plot the points: (1, 5), (2, 10), (3, 15), (4, 20), (5, 25).

4. We join the points to get the graph.

From the graph, we can see that the data forms a straight line. It is a linear graph.

(b) Distance travelled by a car

We are given the following data:

To draw the graph:

1. We take 'Time' on the x-axis and 'Distance (in km)' on the y-axis.

2. We choose a suitable scale:

On the x-axis: 2 units = 1 hour.

On the y-axis: 1 unit = 20 km.

3. We plot the points: (6 a.m., 40), (7 a.m., 80), (8 a.m., 120), (9 a.m., 160).

4. We join the points to get the graph.

Now, we answer the questions using the graph:

(i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m?

From the graph, at 7.30 a.m., the distance covered is 100 km. At 8 a.m., the distance covered is 120 km.

Distance covered during this period = Distance at 8 a.m. - Distance at 7.30 a.m.

= 120 km - 100 km = 20 km.

(ii) What was the time when the car had covered a distance of 100 km since its start?

We locate 100 km on the y-axis and draw a horizontal line to meet the graph. From the point of intersection, we draw a vertical line down to the x-axis. This line meets the x-axis at the midpoint of 7 a.m. and 8 a.m.

Therefore, the time was 7.30 a.m.

(c) Interest on deposits for a year

We are given the following data:

To draw the graph:

1. We take 'Deposit (in $\textsf{₹}$)' on the x-axis and 'Simple Interest (in $\textsf{₹}$)' on the y-axis.

2. We choose a suitable scale:

On the x-axis: 1 unit = $\textsf{₹}$ 1000.

On the y-axis: 1 unit = $\textsf{₹}$ 80.

3. We plot the points: (1000, 80), (2000, 160), (3000, 240), (4000, 320), (5000, 400).

4. We join the points with a line.

Now, we answer the questions:

(i) Does the graph pass through the origin?

Yes, if we extend the line, it will pass through the origin (0, 0). This is because a deposit of $\textsf{₹}$ 0 will earn an interest of $\textsf{₹}$ 0.

(ii) Use the graph to find the interest on $\textsf{₹}$ 2500 for a year.

We locate $\textsf{₹}$ 2500 on the x-axis (midway between 2000 and 3000). We move vertically up to the graph line and then horizontally to the y-axis. The value on the y-axis is midway between 160 and 240, which is 200.

The interest on $\textsf{₹}$ 2500 is $\textsf{₹}$ 200.

(iii) To get an interest of $\textsf{₹}$ 280 per year, how much money should be deposited?

We locate $\textsf{₹}$ 280 on the y-axis (midway between 240 and 320). We move horizontally to the graph line and then vertically down to the x-axis. The value on the x-axis is midway between 3000 and 4000, which is 3500.

The money to be deposited is $\textsf{₹}$ 3500.

(i) Graph of Side of a square vs. Perimeter

We are given the following data:

To draw the graph, we will plot the points (2, 8), (3, 12), (3.5, 14), (5, 20), and (6, 24). We will take the 'Side of square' on the x-axis and the 'Perimeter' on the y-axis.

Scale:

On x-axis: 1 unit = 1 cm

On y-axis: 1 unit = 4 cm

After plotting the points and joining them, we observe that all the points lie on a single straight line.

Therefore, yes, it is a linear graph.

(ii) Graph of Side of a square vs. Area

We are given the following data:

To draw the graph, we will plot the points (2, 4), (3, 9), (4, 16), (5, 25), and (6, 36). We will take the 'Side of square' on the x-axis and the 'Area' on the y-axis.

Scale:

On x-axis: 1 unit = 1 cm

On y-axis: 1 unit = 5 cm²

After plotting the points and joining them, we observe that the points do not lie on a single straight line. Instead, they form a curve.

Therefore, no, it is not a linear graph.