Key Concept: Line graph, Linear graph

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that a line graph displaying data that changes continuously over time will always be a linear graph. However, this is not necessarily true because a linear graph specifically refers to a graph where the relationship between the variables is represented by a straight line, not just any unbroken line. The reason defines a linear graph as a line graph that is a whole unbroken line, which is correct as per the syllabus. However, the assertion is incorrect because not all line graphs that display continuous data are linear graphs; they can have curves or other patterns. Therefore, the assertion is false, but the reason is true.

Key Concept: Comparison of data presentation

c) Graphs make trends and comparisons clearer

[Solution Description]

Graphs provide a visual representation of data that can quickly convey trends and comparisons, making them easier to understand than tables, which require more detailed reading and interpretation.

Key Concept: Purpose of Graphs

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that graphs are used to show numerical facts in visual form for quick and easy understanding. This is true as graphs are visual representations of data, making it easier to grasp trends or comparisons. The reason states that graphical presentation is easier to understand than tabular presentation, particularly when there is a trend or comparison. This is also true because graphs provide a visual summary that is often more intuitive than tables. Further, the reason correctly explains why graphs are used as stated in the assertion.

Key Concept: Purpose of Graphs, Data Visualization

c) Line Graph

[Solution Description]

The purpose of a graph is to show numerical facts visually for quick and clear understanding. When comparing sales performance over time, a line graph is most effective as it clearly shows trends and variations across different years. Other options like pie charts or bar graphs may not emphasize trends as effectively.

Key Concept: Visual Representation of Data

d) Graphs

[Solution Description]

Graphs are visual representations of data collected.

Key Concept: Graph Comparison

c) School B has more students than School A

[Solution Description]

School B consistently has more students enrolled than School A over the three years. In each year, the number of students in School B is higher than that in School A by 100 students.

Key Concept: Graphical Representation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that graphs are more effective than tables in presenting numerical data because they provide a visual representation that is easier to understand. This is true because graphs visually represent data, making it easier to grasp trends and comparisons at a glance. The reason given is that visual representations help in identifying trends and comparisons quickly, which is not as straightforward in tabular form. This reason correctly explains why graphs are more effective than tables. Both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Graph Interpretation

d) Seasonal variation in consumer spending

[Solution Description] The graph shows that sales are highest in December, which is typically associated with holiday shopping, and then drop in January, which is common as people tend to reduce spending after the holidays. This pattern is consistent with seasonal trends in consumer behavior.

Key Concept: Comparison between Tables and Graphs

b) Graphs may obscure exact numerical values.

[Solution Description]

Graphs may not always provide the exact numerical values of data points, as they focus on visual representation. This can be a limitation when precise data is required, which tables can provide more effectively.

Key Concept: Graphs vs Tables

c) When showing trends or comparisons.

[Solution Description]

A graphical presentation is particularly useful when there is a trend or comparison to be shown. It helps in visualizing the data and understanding the patterns or relationships more easily than a table.

Key Concept: Trend Analysis

b) Values are increasing

[Solution Description]

A rising trend in a line graph indicates that the values of the variable are increasing over time or another continuous variable.

Key Concept: Graphs, Trends

c) Sales showed fluctuation with an overall increase

[Solution Description]

The line graph shows a consistent upward trend from January to June, indicating that sales were increasing during this period. The sharp decline in July suggests a significant drop in sales for that month. Following July, the graph shows a steady increase from August to Dezember, indicating that sales recovered and continued to grow. This suggests that the company had a strong performance overall, with the exception of the month of July.

Key Concept: Line graph, Time-temperature graph

c) 65°C

[Solution Description]

The temperature increases from 20°C at 9 a.m. to 50°C at 1 p.m., which is a total increase of 30°C over 4 hours. Thus, the rate of increase is $30°C / 4 \text{ hours} = 7.5°C/\text{hour}$. From 1 p.m. to 3 p.m. is 2 hours, so the temperature will increase by $7.5°C/\text{hour} \times 2 \text{ hours} = 15°C$. Therefore, the temperature at 3 p.m. will be $50°C + 15°C = 65°C$.

Key Concept: Line Graph

b) Data that changes continuously over periods of time

[Solution Description]

A line graph displays data that changes continuously over periods of time. It is used to show trends or comparisons over time.

Key Concept: Line Graph Characteristics

b) It displays data that changes continuously over periods of time.

[Solution Description]

A line graph displays data that changes continuously over periods of time. It is a visual representation of data where points are connected by lines to show trends or patterns. The correct statement is: "A line graph displays data that changes continuously over periods of time."

Key Concept: Graph Components

b) Time intervals

[Solution Description] In a time-temperature graph, the horizontal line (x-axis) represents the timings at which the temperatures were recorded. The vertical line (y-axis) represents the temperature values.

Key Concept: Bar graphs, Data Interpretation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that in a bar graph, the height of each bar is proportional to the frequency of the corresponding category. This is true because bar graphs visually represent data by using bars whose lengths or heights correspond to the values they represent. The reason states that bar graphs are used to represent categorical data where the length or height of the bars represents the magnitude of the data. This is also true and correctly explains why the height of each bar is proportional to the frequency of the corresponding category. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Bar Graphs

d) City W

[Solution Description]

To determine which city received the least amount of rainfall, compare the rainfall values for all cities.

City X = 40 mm, City Y = 30 mm, City Z = 50 mm, City W = 20 mm.

The smallest value is 20 mm, which corresponds to City W.

Therefore, City W received the least amount of rainfall.

Key Concept: Pie Chart Data Interpretation

b) \$1000

[Solution Description] The angle for groceries is $120^\circ$, and the total angle of the pie chart is $360^\circ$. The fraction of the total expenses spent on groceries is $\frac{120}{360} = \frac{1}{3}$. Therefore, the amount spent on groceries is $\frac{1}{3} \times 3000 = 1000$.

Key Concept: Pie Charts, Percentages

b) 25\%

[Solution Description]

The total angle in a pie chart is $360^\circ$. The angles for categories \textit{A}, \textit{B}, and \textit{C} are given as $90^\circ$, $120^\circ$, and $60^\circ$ respectively. The angle for category \textit{D} can be calculated as:

$\text{Angle for D} = 360^\circ - (90^\circ + 120^\circ + 60^\circ) = 90^\circ$

The percentage of expenses allocated to category \textit{D} is calculated using the formula:

$\text{Percentage} = \left(\frac{\text{Angle}}{360^\circ}\right) \times 100$

Substituting the angle for \textit{D}:

$\text{Percentage} = \left(\frac{90^\circ}{360^\circ}\right) \times 100 = 25\%$

Key Concept: Line Graph, Temperature Trend

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that the temperature of a patient changed over time, which is a factual observation based on the given data. The reason explains that a line graph can display such continuous changes, which is also true. A line graph is indeed used to represent data that changes continuously over time, and it can clearly show trends like increases or decreases. However, the reason does not explain why the temperature specifically changed; it only explains the utility of a line graph in displaying such data. Therefore, both the assertion and reason are true, but the reason does not correctly explain the assertion.

Key Concept: Line Graph

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that a line graph is used to display data that changes continuously over time, which is true as per the syllabus. The reason states that a line graph shows numerical facts in visual form for quick and easy understanding, which is also true based on the syllabus. However, the reason does not directly explain why a line graph is used for continuous data over time; it only explains the general purpose of graphs. Therefore, both statements are true, but the reason is not the correct explanation of the assertion.

Key Concept: Line Graph Interpretation

b) 1°C

[Solution Description]

To find the increase in temperature between 6 a.m. and 2 p.m., subtract the temperature at 6 a.m. from the temperature at 2 p.m. The temperature at 6 a.m. is 37°C, and at 2 p.m., it is 38°C. So, the increase in temperature is $38°C - 37°C = 1°C$.

Key Concept: Time-Temperature Graph

b) Temperatures

[Solution Description] The time-temperature graph is a graphical representation of the relationship between time and temperature. Here, the horizontal line (x-axis) represents the time at which the temperatures were recorded. Since the vertical line (y-axis) typically represents the dependent variable in such graphs, it would be labelled with the temperature values. Therefore, the y-axis is labelled with the temperature values in degrees Celsius.

Key Concept: Time-Temperature Graph

c) 3°C

[Solution Description]

The total increase in temperature is calculated by subtracting the initial temperature from the final temperature:

$\text{Total Increase} = 40\degree C - 37\degree C = 3\degree C$

Therefore, the total increase in temperature between 10 a.m. and 12 p.m. is $3\degree C$.

Key Concept: Time-Temperature Graph, Temperature Estimation

b) $37^\circ C$

[Solution Description]

First, find the rate of decrease in temperature from 10 a.m. to 6 p.m. The temperature decreases from $40^\circ C$ to $34^\circ C$ in 8 hours, so the rate is $\frac{6^\circ C}{8 \text{ hours}} = 0.75^\circ C/\text{hour}$. At 2 p.m., which is 4 hours after 10 a.m., the temperature would have decreased by $0.75^\circ C/\text{hour} \times 4 \text{ hours} = 3^\circ C$. Therefore, the temperature at 2 p.m. would be $40^\circ C - 3^\circ C = 37^\circ C$.

Key Concept: Line Graph Comparison

b) 4th match

[Solution Description]

From the graph, we can observe that the lines representing the runs scored by batsmen A and B intersect at a point. This point indicates that both batsmen scored the same number of runs in that particular match. According to the graph, this intersection occurs during the 4th match, where both batsmen scored 60 runs each.

Key Concept: Identifying Trends in Graphs

d) No change in values

[Solution Description]

A horizontal line on a graph typically represents that there is no change in the values over time or across the measured variable.

Key Concept: Slope of a Line Graph

d) 60 km/h

[Solution Description]

The slope of a line graph is calculated using the formula: $slope = \frac{\Delta y}{\Delta x}$. Here, $\Delta y$ is the change in distance (120 km) and $\Delta x$ is the change in time (2 hours). Therefore, the slope is $\frac{120}{2} = 60$ km/h.

Key Concept: Line Graph, Temperature Trend Prediction

b) 36.5°C

[Solution Description]

From 2 p.m. to 6 p.m., the temperature decreased from 38°C to 35°C over a period of 4 hours. This means the temperature decreased at a rate of $\frac{38°C - 35°C}{4\ \text{hours}} = 0.75°C/\text{hour}$. At 4 p.m., which is 2 hours after 2 p.m., the temperature would have decreased by $0.75°C/\text{hour} \times 2\ \text{hours} = 1.5°C$. Therefore, the expected temperature at 4 p.m. is $38°C - 1.5°C = 36.5°C$.

Key Concept: Performance Graph of Two Batsmen

c) Assertion is true, but Reason is false.

[Solution Description]

From the given graph, it is clear that Batsman A has a higher peak score of 115 runs compared to Batsman B's highest score of 100 runs. Therefore, the Assertion is true. However, Batsman A has many deep "valleys" and scored zero in two matches, making him less consistent than Batsman B, who never scored below 40 runs. Hence, the Reason is false.

Key Concept: Performance Graph, Consistency

b) Batsman B

[Solution Description]

Batsman A has scored 0 runs in two matches and less than 40 runs in three other matches, totaling five matches with low scores. Batsman B has never scored below 40 runs. Since consistency is determined by the number of matches where a batsman scores below 40 runs, Batsman B is more consistent.

Key Concept: Analyzing cricket performance using a line graph

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion claims that Batsman A is more consistent than Batsman B because his highest score is higher. However, consistency in cricket performance is not determined solely by the highest score but by the ability to maintain a steady level of runs across matches. In the given example, Batsman A has significant variations in scores, including several low scores and zeroes, whereas Batsman B consistently scores above 40 runs in all matches. Therefore, Batsman B is more consistent. The assertion is false because a higher highest score does not imply consistency. The reason is true because consistency is indeed measured by steady performance across matches.

Key Concept: Analyzing cricket performance using a line graph

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that Batsman A scored a zero in two matches. This is true as per the given information in the example. The reason states that Batsman A has a lot of ups and downs in his performance. This is also true as per the given information. Moreover, the reason correctly explains why Batsman A scored a zero in two matches, as his inconsistent performance led to such low scores.

Key Concept: Comparing runs, differences

c) 50

[Solution Description]

The absolute difference between the runs scored in Match 1 and Match 2 is $|150 - 200| = 50$.

Key Concept: Comparing Runs Scored

b) 200

[Solution Description]

To find the average runs scored per match, add the runs from all three matches and divide by the number of matches. The calculation is as follows:

$\text{Average runs} = \frac{180 + 220 + 200}{3} = \frac{600}{3} = 200$

Therefore, the average runs scored per match by Team A is 200.

Key Concept: Identifying trends, Consistency in scores

b) Median

[Solution Description]

To find the most consistent measure of central tendency, we need to calculate the mean, median, and mode of the data set. The mean is the average of all scores. The median is the middle value when the scores are arranged in order. Since all scores are unique, there is no mode. The mean is calculated as (85 + 88 + 86 + 90 + 87) / 5 = 87.2. The median is the third score, which is 87. Since the mean and median are close, but the median is a whole number and more representative of the data, it is the most consistent measure of central tendency for this data.

Key Concept: Identifying trends and consistency in scores

d) Assertion is false, but Reason is true.

[Solution Description] To evaluate consistency, we need to analyze the variability in scores rather than just the peak score. The solution states that Batsman A has a lot of ups and downs, with zero runs in two matches and less than 40 runs in five matches, while Batsman B has never scored below 40 runs. This indicates that Batsman B is more consistent. Therefore, the Assertion that Batsman A is more consistent due to a higher peak score is false, but the Reason about judging consistency through variability is true.

Key Concept: Distance-Time Graph

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

If the distance-time graph is a straight line, it means that the car is covering equal distances in equal intervals of time. This implies that the speed of the car remains constant throughout the journey. The reason correctly explains why a straight line in a distance-time graph indicates constant speed. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Distance-Time Graph Interpretation

c) The object is speeding up

[Solution Description]

The slope of a distance-time graph represents the speed of the object. An increase in slope indicates that the object is speeding up.

Key Concept: Distance-Time Graph

c) 200 km

[Solution Description]

To find the total distance travelled by the car in the first three hours, add the distances covered in each hour: First hour = 50 km, second hour = 100 km, third hour = 50 km. Total distance = 50 + 100 + 50 = 200 km.

Key Concept: Distance-Time Graph

c) 2 p.m.

[Solution Description]

The car reached City Q at 2 p.m., as stated in the syllabus. This is directly derived from the graph.

Key Concept: Identifying changes in speed

b) 20 km/h

[Solution Description]

The initial speed of the bus is 40 km/h, and the final speed after increasing is 60 km/h. The change in speed is the difference between the final speed and the initial speed: $60\,km/h - 40\,km/h = 20\,km/h$.

Key Concept: Identifying Changes in Speed

c) 30 km/h

[Solution Description]

The initial speed is 90 km/h, and the final speed is 60 km/h. The change in speed is calculated by subtracting the final speed from the initial speed: $90 - 60 = 30$ km/h.

Key Concept: Graph Analysis, Distance-Time Graph

d) The train was not moving

[Solution Description]

In a distance-time graph, a horizontal line segment signifies that the distance from the location does not change over the given time interval. This suggests that the train was not in motion during this period. Thus, the correct conclusion is that the train was stopped between 12 p.m. and 1 p.m.

Key Concept: Analyzing stopping points using a horizontal segment

d) The car was stationary

[Solution Description]

A horizontal line segment in a distance-time graph indicates that the distance of the car from the starting point remains constant over the given time interval. This means the car did not move during the time period 10 a.m. to 11 a.m. Therefore, the car was stationary during this interval.

Key Concept: Graph interpretation, Time-Distance

c) 30 km/h

[Solution Description]

To calculate the average speed, we need to find the total distance traveled and divide it by the total time taken.

Total distance = 30 km (first hour) + 60 km (second hour) = 90 km

Total time = 1 hour (first hour) + 1 hour (second hour) + 1 hour (stop) = 3 hours

Average speed = $\frac{Total \: distance}{Total \: time} = \frac{90 \: km}{3 \: hours} = 30 \: km/h$

Key Concept: Graph Interpretation

a) Distance of the car from City P

[Solution Description]

The y-axis in the given graph represents the distance of the car from City P at different times. Since the car is traveling from City P to City Q, the y-axis shows how far the car is from its starting point (City P) at various times.

Key Concept: Independent and Dependent Variables, Graphs

b) $C = 5L$

[Solution Description]

The cost of petrol (\$C\$) increases by \$25 for every 5 litres purchased, which means the rate of increase is \$5 per litre. Since there is no fixed charge, the relationship can be expressed as:

$C = 5L$

Here, \$C\$ is the dependent variable and \$L\$ is the independent variable.

Key Concept: Independent and Dependent Variables

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that the dependent variable is always plotted on the x-axis, which is incorrect. In graphs, the independent variable is typically plotted on the x-axis, while the dependent variable is plotted on the y-axis. The reason correctly explains that the independent variable is manipulated to observe its effect on the dependent variable. Therefore, the assertion is false, but the reason is true.

Key Concept: Independent and Dependent Variables

b) Number of hours the borewell is used

[Solution Description]

In this scenario, the independent variable is the factor that is controlled or manipulated by the farmer, which is the number of hours the borewell is used. The dependent variable, which is influenced by the independent variable, is the cost of using the borewell.

Key Concept: Independent variable

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The independent variable is the number of litres of petrol bought, and the dependent variable is the amount paid. The assertion correctly identifies the independent variable, and the reason correctly explains why it is independent because the amount paid depends on how much petrol is bought. Both are true, and the reason explains the assertion.

Key Concept: Time and Distance, Linear Graph

c) 5 hours

[Solution Description]

First, confirm the speed: $\frac{60}{3} = 20$ km/h. Time taken to travel 100 km at this speed is $\frac{100}{20} = 5$ hours.

Key Concept: Independent and Dependent Variables

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

In the given scenario, the cost of petrol depends on the number of litres purchased. This means the number of litres is the independent variable, and the cost is the dependent variable. The relationship between these two variables is directly proportional, as shown in the graph where increasing the number of litres leads to a proportional increase in cost. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Quantity and Cost

c) \$10

[Solution Description]

First, find the cost of one apple. Since 10 apples cost \$20, one apple costs \$20 divided by 10, which equals \$2. Now, to find the cost of 5 apples, multiply the cost of one apple by 5. So, 5 apples cost \$2 multiplied by 5, which equals \$10.

Key Concept: Quantity and Cost

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that in a graph representing the cost of petrol against the quantity of petrol purchased, the graph will always be a straight line passing through the origin. The reason states that the cost of petrol is directly proportional to the quantity of petrol purchased.

According to the syllabus, when two quantities are in direct variation, their graph is linear and passes through the origin. Since the cost of petrol is directly proportional to the quantity of petrol purchased, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Graph interpretation, Direct variation

c) 50

[Solution Description]

The graph shows a direct variation between cost ($C$) and quantity ($Q$), given by $C = k \cdot Q$.

From the estimation, when $Q = 12$, $C = 600$. Plugging these values into the equation:

$600 = k \cdot 12$

Solving for $k$:

$k = \frac{600}{12} = 50$

Therefore, the constant of variation is 50.

Key Concept: Plotting a graph, Direct variation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The graph of cost versus quantity of petrol is plotted by taking the quantity of petrol on the horizontal axis and the cost on the vertical axis. The points (10,500), (15,750), (20,1000), and (25,1250) are plotted and joined to form a straight line that passes through the origin. This indicates a linear relationship between the two variables, where the cost (\$C) is directly proportional to the quantity (\$Q), i.e., \$C = kQ\$, where \$k\$ is the constant of proportionality. Since the graph passes through the origin, it confirms that when the quantity is zero, the cost is also zero, which is characteristic of direct variation.

Key Concept: Using the graph to estimate costs for intermediate values

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

In a situation where two quantities are in direct variation, the relationship between them is such that as one quantity increases, the other also increases proportionally. This means that the ratio of the two quantities remains constant throughout. Such relationships are represented graphically as straight lines (linear graphs) because the rate of change is constant. Therefore, both the Assertion and Reason are true, and the Reason correctly explains why the graph is linear in cases of direct variation.

Key Concept: Graphical estimation, Area under curve

b) 180 km

[Solution Description]

Since the speed is constant, the distance traveled can be calculated by multiplying the speed by the time. Distance = Speed × Time = 60 km/h × 3 h = 180 km. Therefore, the total distance traveled by the car is 180 km.

Key Concept: Principal and Simple Interest, Graph Interpretation

c) \$650

[Solution Description]

First, calculate the simple interest for one year:

$\text{Simple Interest} = \frac{10}{100} \times 500 = \$50$

Then, for three years:

$\text{Total Interest} = 50 \times 3 = \$150$

The total amount after three years will be:

$\text{Total Amount} = 500 + 150 = \$650$

Key Concept: Principal and Simple Interest

b) \$510

[Solution Description]

The formula for simple interest is:

$A = P + \frac{P \times R \times T}{100}$

Here, $A = \$560$, $R = 5\%$, and $T = 2$ years.

Rearranging the formula to solve for $P$:

$560 = P + \frac{P \times 5 \times 2}{100}$

$560 = P + \frac{10P}{100}$

$560 = P + 0.1P$

$560 = 1.1P$

$P = \frac{560}{1.1}$

$P = \$509.09$

Therefore, the principal amount is approximately \$509.09.

Key Concept: Simple Interest

b) 100

[Solution Description]

The formula for simple interest is $I = P \times r \times t$, where $P$ is the principal amount, $r$ is the rate of interest, and $t$ is the time period. Here, $P = 1000$, $r = 5\% = 0.05$, and $t = 2$ years. Plugging in the values, we get:

$I = 1000 \times 0.05 \times 2 = 100$

So, the interest earned after 2 years is \$100.

Key Concept: Simple Interest

c) \$9600

[Solution Description]

First, calculate the simple interest using the formula $I = P \times r \times t$, where $P = 8000$, $r = 0.04$, and $t = 5$. So, $I = 8000 \times 0.04 \times 5 = 1600$. Now add this interest to the principal to get the total amount: $A = P + I = 8000 + 1600 = 9600$. The total amount is \$9600.

Key Concept: Graph Interpretation, Time-Temperature Analysis

b) $38.5^\circ C$

[Solution Description]

The temperature increases linearly from 10 a.m. to 2 p.m. At 10 a.m., the temperature is $38^\circ C$, and at 2 p.m., it is $39^\circ C$. The time difference between 10 a.m. and 2 p.m. is 4 hours. The rate of increase in temperature per hour can be calculated as $\frac{39 - 38}{4} = \frac{1}{4}^\circ C/\text{hour}$. To find the temperature at 12 p.m., which is 2 hours after 10 a.m., we multiply the rate by 2 and add it to the initial temperature: $38 + \left(\frac{1}{4} \times 2\right) = 38 + \frac{1}{2} = 38.5^\circ C$.

Key Concept: Graph Interpretation, Distance-Time Analysis

b) 60 km

[Solution Description]

The distance covered in each hour is given as follows: first hour - 20 km, second hour - 15 km, third hour - 25 km. The total distance covered after 3 hours is the sum of these distances: $20 + 15 + 25 = 60$ km.

Key Concept: Time and Distance

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

First, let us verify the assertion. Given the speed of the car is 60 km/h and it travels for 2 hours, the distance covered can be calculated using the formula distance = speed × time. Substituting the values, we get distance = 60 km/h × 2 h = 120 km. Thus, the assertion is true. Now, let us analyze the reason. When an object moves at a constant speed, the relationship between distance and time is given by distance = speed × time. This shows that distance is directly proportional to time when speed is constant. Therefore, the reason is also true, and it correctly explains the assertion.

Key Concept: Time, Distance, Speed

b) 96 km/h

[Solution Description]

To find the average speed for the entire journey, we need to calculate the total distance divided by the total time taken. The total distance is 480 km. The first half of the journey is $\frac{480}{2} = 240$ km, which is covered at 80 km/h. The time taken for this part is $\frac{240}{80} = 3$ hours. The second half of the journey is also 240 km, but it is covered at 120 km/h. The time taken for this part is $\frac{240}{120} = 2$ hours. Therefore, the total time taken for the journey is $3 + 2 = 5$ hours. The average speed is then $\frac{480}{5} = 96$ km/h.

Key Concept: Distance-Time Graph

d) The vehicle is stationary

[Solution Description]

A horizontal line in a distance-time graph indicates that the vehicle is not moving. Since the distance does not change over time, the vehicle is stationary.

Key Concept: Distance-Time Graph

d) The vehicle is accelerating

[Solution Description]

A curve that becomes steeper over time in a distance-time graph indicates that the vehicle is accelerating. As time progresses, the vehicle covers more distance in each unit of time, which means its speed is increasing.

Key Concept: Interpreting speed variation

d) The speed varied over time.

[Solution Description] The speed of the car varied over time. It was 50 km/h in the first hour, 100 km/h in the second hour, and 50 km/h in the third hour. Thus, the speed was not consistent.

Key Concept: Calculating Speed

c) 100 km/h

[Solution Description]

Speed is calculated using the formula $Speed = \frac{Distance}{Time}$. Here, the distance is 100 km and the time is 1 hour. So, the speed is $\frac{100}{1} = 100$ km/h.

Key Concept: Linear Graphs, Direct Variation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that the graph of a situation where two quantities are in direct variation will always be a straight line passing through the origin. This is true because in direct variation, as one quantity increases, the other increases proportionally, resulting in a linear relationship. The reason states that when two quantities are in direct variation, their ratio remains constant. This is also true because direct variation implies that the ratio of the two quantities is always the same. However, the reason does not directly explain why the graph passes through the origin. The origin point (0,0) is included in the graph because if one quantity is zero, the other must also be zero in a direct variation relationship. Thus, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Linear Graph Application

b) 300 km

[Solution Description]

Given the speed is constant, the relationship between distance and time is linear. The formula for distance is $Distance = Speed \times Time$. Substituting the given values, $Distance = 60 \times 5 = 300$ km.

Key Concept: Drawing Graphs, Steps to Plot a Graph, Linear Graphs

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine the correctness of the assertion and reason, let's analyze the concepts step by step:

1. **Assertion Analysis**: The assertion states that the graph of distance covered versus time for a car traveling at a constant speed will always be a straight line passing through the origin. This is true because when a car travels at a constant speed, the distance covered is directly proportional to the time taken. Mathematically, this relationship can be represented as $d = vt$, where $d$ is the distance, $v$ is the constant speed, and $t$ is the time. This equation represents a straight line passing through the origin in a graph of distance versus time.

2. **Reason Analysis**: The reason states that in direct variation relationships, the graph is linear and always passes through the origin. This is also true because direct variation implies a proportional relationship where one quantity is a constant multiple of another. In such cases, the graph will indeed be a straight line passing through the origin.

3. **Conclusion**: Both the assertion and the reason are true, and the reason correctly explains why the assertion is true. Therefore, the correct answer is that both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Scales on Axes

b) Ensuring the data fits well within the graph

[Solution Description]

The primary consideration when choosing scales for the axes is to ensure that the data fits well within the graph without being too cramped or too spread out.

Key Concept: Choosing an appropriate scale, Plotting points

b) 1 unit = 0.1 hours

[Solution Description]

The train covers 240 km at 80 km/h. The time taken is $240 \div 80 = 3$ hours. The horizontal axis has 30 units. To determine the scale, divide the total time by the number of units: $3 \div 30 = 0.1$ hours per unit. Therefore, the appropriate scale for the horizontal axis is 1 unit = 0.1 hours.

Key Concept: Choosing an appropriate scale, Plotting points

b) 1 unit = 12.5 km

[Solution Description]

The car travels at 50 km/h. In 5 hours, it will cover $50 \times 5 = 250$ km. The vertical axis has 20 units available. To determine the scale, divide the total distance by the number of units: $250 \div 20 = 12.5$ km per unit. Therefore, the appropriate scale for the vertical axis is 1 unit = 12.5 km.

Key Concept: Labeling the axes

b) Horizontal axis

[Solution Description]

The horizontal axis (x-axis) is typically labeled with the independent variable, as it represents the variable that is controlled or manipulated.

Key Concept: Labeling the axes

d) Assertion is false, but Reason is true.

[Solution Description]

In a time-temperature graph, the x-axis typically represents the independent variable, which is time. The y-axis represents the dependent variable, which is temperature. Therefore, the assertion that the x-axis represents temperature readings is incorrect. However, the reason that the y-axis represents time is also incorrect because it actually represents temperature. Thus, both statements are false.

Key Concept: Interpreting plotted data

b) Linear

[Solution Description]

The data points $(1, 2)$, $(2, 4)$, and $(3, 6)$ show that as the x-coordinate increases by 1, the y-coordinate increases by 2. This indicates a linear relationship with a slope of 2. Therefore, the relationship is best described as linear.

Key Concept: Plotting given data points

b) $(4, 9)$

[Solution Description]

The given data points $(1, 3)$, $(2, 5)$, and $(3, 7)$ show a consistent pattern where the y-coordinate increases by 2 as the x-coordinate increases by 1. Therefore, for $x = 4$, the y-coordinate would be $7 + 2 = 9$. Thus, the next point in the sequence is $(4, 9)$.

Key Concept: Line Graph

b) 2

[Solution Description]

To find the slope of the line passing through the points (1, 3) and (2, 5), we use the formula for slope:

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Substituting the values, we get:

$m = \frac{5 - 3}{2 - 1} = \frac{2}{1} = 2$.

Key Concept: Drawing Graphs

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that in a line graph, points are connected by line segments to show the trend of the data. This is true because, in line graphs, the connection of points by line segments helps in understanding the pattern or trend of the data over time or across categories. The reason given is that line segments help in visualizing how the data changes over time or across different categories. This is also true and correctly explains why line segments are used in line graphs. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Quadratic Graph

c) Parabola

[Solution Description]

The graph of the equation $y = x^2$ is a parabola that opens upwards. This is a standard shape for quadratic equations of the form $y = ax^2 + bx + c$ where $a > 0$.

Key Concept: Linear Graph, Simple Interest

d) \$80

[Solution Description]

Simple interest is calculated using the formula $I = P \times r \times t$, where $I$ is the interest, $P$ is the principal amount, $r$ is the rate of interest, and $t$ is the time. Given that \$500 earns \$40 in interest, we can find the time period using $40 = 500 \times 0.08 \times t$, which gives $t = 1$ year. Now, for a deposit of \$1000, the interest earned would be $I = 1000 \times 0.08 \times 1 = \$80$.

Key Concept: Cost of Apples vs. Number of Apples

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the cost of apples increases linearly with the number of apples purchased. This is true because if the cost per apple is constant, then the total cost will increase proportionally with the number of apples. The reason states that this is because the cost per apple remains constant, which is also true and correctly explains why the cost increases linearly. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Cost of Apples vs. Number of Apples

c) \$15

[Solution Description]

The cost of 1 apple is given as \$5. To find the cost of 3 apples, multiply the cost of one apple by 3.

$\text{Cost of 3 apples} = 3 \times 5 = \$15$

Therefore, the cost of 3 apples is \$15.

Key Concept: Distance Travelled by a Car vs. Time

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] When a car is moving at a constant speed, the distance it travels increases uniformly with time. This results in a straight-line graph for distance versus time. The slope of this graph is calculated as the change in distance divided by the change in time, which is equal to the speed of the car. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Distance-Time Graph

b) 75 km

[Solution Description]

To find the distance covered by the car, we use the formula:

$\text{Distance} = \text{Speed} \times \text{Time}$

Given that the speed is 30 km/h and the time is 2.5 hours, we substitute these values into the formula:

$\text{Distance} = 30 \, \text{km/h} \times 2.5 \, \text{hours} = 75 \, \text{km}$

Thus, the distance covered by the car in 2.5 hours is 75 km.

Key Concept: Graph Plotting

c) (3, 90)

[Solution Description] The graph plots time on the horizontal axis and distance on the vertical axis. For the given data, the point to be plotted is (3, 90), where 3 represents the time in hours and 90 represents the distance in km.

Key Concept: Interest on Deposits

a) \$5788.13

[Solution Description]

The formula for compound interest is given by:

$A = P \left(1 + \frac{r}{n}\right)^{nt}$

Here, $P = 5000$, $r = 5\% = 0.05$, $n = 1$ (since it's compounded annually), and $t = 3$ years.

Plugging in the values:

$A = 5000 \left(1 + \frac{0.05}{1}\right)^{1 \times 3} = 5000 \left(1.05\right)^3$

Calculating $(1.05)^3$:

$1.05^3 = 1.157625$

Then:

$A = 5000 \times 1.157625 = 5788.125$

So, the amount after 3 years is \$5788.13.

Key Concept: Interest on Deposits

a) \$2164.86

[Solution Description]

Using the compound interest formula:

$A = P \left(1 + \frac{r}{n}\right)^{nt}$

Here, $P = 2000$, $r = 4\% = 0.04$, $n = 2$ (since it's compounded semi-annually), and $t = 2$ years.

Plugging in the values:

$A = 2000 \left(1 + \frac{0.04}{2}\right)^{2 \times 2} = 2000 \left(1.02\right)^4$

Calculating $(1.02)^4$:

$1.02^4 = 1.08243216$

Then:

$A = 2000 \times 1.08243216 = 2164.86432$

So, the amount after 2 years is \$2164.86.

Key Concept: Interest on Deposits for a Year, Graph Plotting

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine if the assertion and reason are correct, we need to understand the relationship between simple interest (\$S.I.\$) and the deposit amount (\$P\$). The formula for simple interest is given by:

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

where \$R\$ is the rate of interest and \$T\$ is the time period in years. For a year, \$T = 1\$, so the formula simplifies to:

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

This shows that \$S.I.\$ is directly proportional to \$P\$, as the reason states. When plotting \$S.I.\$ against \$P\$, since the relationship is linear and passes through the origin (when \$P = 0\$, \$S.I. = 0\$), the graph is a straight line passing through the origin, as the assertion states. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Linear Graph

c) 28 cm

[Solution Description]

The perimeter of a square is given by $P = 4 \times \text{side}$. Given the side is 7 cm, we multiply it by 4 to find the perimeter:

$P = 4 \times 7 = 28 \text{ cm}$

Therefore, the perimeter is 28 cm.

Key Concept: Side of a Square vs. Perimeter, Linear Graph

c) 24 cm

[Solution Description]

The original side length of the square is 4 cm. The new side length after increasing it by 2 cm is $4 + 2 = 6$ cm. The perimeter of the square is calculated using the formula $P = 4 \times \text{side}$. Therefore, the new perimeter is $4 \times 6 = 24$ cm.

Key Concept: Side of a Square vs. Perimeter, Linear Graph

b) 7 cm

[Solution Description]

The perimeter of a square is calculated using the formula $P = 4 \times \text{side}$. To find the side length, we rearrange the formula as $\text{side} = \frac{P}{4}$. Given the perimeter is 28 cm, the side length is $\frac{28}{4} = 7$ cm.

Key Concept: Side of Square vs. Area, Linear Graph

b) 125\%

[Solution Description]

Let the original side of the square be $x$. The original area is $x^2$.

When the side is increased by 50\%, the new side becomes $1.5x$.

The new area is $(1.5x)^2 = 2.25x^2$.

The increase in area is $2.25x^2 - x^2 = 1.25x^2$.

The percentage increase is $\frac{1.25x^2}{x^2} \times 100 = 125\%$.

Key Concept: Side of a Square vs Area

b) 69\%

[Solution Description]

Let the original side of the square be $s$. The area of the square is $A = s^2$. After increasing the side by 30\%, the new side becomes $1.3s$. The new area is $A_{\text{new}} = (1.3s)^2 = 1.69s^2$. The increase in area is $1.69s^2 - s^2 = 0.69s^2$. The percentage increase is calculated as $\frac{0.69s^2}{s^2} \times 100 = 69\%$.

Key Concept: Side of Square vs. Area

d) 25 cm$^2$

[Solution Description]

Using the formula for the area of a square $A = s^2$, we can find the area when the side length is 5 cm. Substituting the value into the formula gives $A = 5^2$. Calculating this results in $A = 25$. Thus, the area of the square is 25 cm$^2$.