Key Concept: Line Graph

c) x-axis

[Solution Description]

In a line graph, the horizontal line (usually called the x-axis) shows the timings at which the temperatures were recorded. Therefore, the x-axis typically represents time.

Key Concept: Introduction to Graphs

c) 240 km

[Solution Description]

To find the total distance traveled, add the distances covered in each hour.

Calculation:

Total Distance = $60 + 80 + 100$ km

Total Distance = $240$ km

Key Concept: Line Graph

c) $5^\circ$C

[Solution Description]

The highest temperature recorded is $40^\circ$C and the lowest temperature is $35^\circ$C. The difference between them is calculated as:

$40 - 35 = 5$

Thus, the difference is $5^\circ$C.

Key Concept: Linear Graph

b) $9$ cm$^2$

[Solution Description]

The area of a square is calculated using the formula:

$\text{Area} = \text{side length}^2$

Substituting the given side length:

$\text{Area} = 3^2 = 9$

Thus, the area is $9$ cm$^2$.

Key Concept: Graph Interpretation

d) Sales are increasing

[Solution Description]

The graph shows the sales increasing steadily over the five months. Starting from \$20,000 in January, the sales increase by \$5,000 each month, reaching \$40,000 in May. This indicates a consistent upward trend.

Key Concept: Graphs, Data Representation

c) Line graph

[Solution Description]

To represent the trends in the data effectively, a line graph is the most appropriate choice. A line graph can show the steady increase from January to June, the sharp decline in July and August, and the moderate increase from September to December. It connects data points with lines, making it easy to visualize the overall trend over time.

Key Concept: Graphs vs Tables

c) It is easier to understand trends and comparisons

[Solution Description]

A graphical presentation is easier to understand, especially when there is a trend or comparison to be shown. It helps in visualizing trends or comparisons more effectively than a table.

Key Concept: Data Presentation, Graphical Representation

c) April

[Solution Description]

To determine the month where City B received significantly more rainfall than City A, we need to compare the heights of the bars for each month in the bar graph. The month with the largest difference in bar height (City B being much higher than City A) would be the answer. For example, if in April, City B has a bar height representing 150 mm and City A has a bar height representing 50 mm, then April would be the month where City B received significantly more rainfall than City A.

Key Concept: Comparison between Tables and Graphs

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

[Solution Description]

The assertion states that tables are more effective than graphs for presenting precise numerical data, which is true because tables allow for the exact representation of numbers. The reason states that graphs are better for visualizing trends and patterns in data, which is also true as graphs provide a visual representation that can highlight trends and patterns. However, the reason does not explain the assertion directly; it provides a complementary point rather than an explanation.

Key Concept: Graphs vs Tables

d) To show numerical facts visually for quick understanding.

[Solution Description]

The main purpose of using graphs is to show numerical facts in a visual form so that they can be understood quickly, easily, and clearly. This is particularly useful when there is a trend or comparison to be shown.

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: Graphs, Relationships

c) Increased advertising leads to increased sales

[Solution Description]

The scatter plot shows a roughly linear pattern with a positive slope, indicating that there is a positive relationship between advertising expenditure and sales revenue. As advertising expenditure increases, sales revenue also tends to increase. This suggests that increased investment in advertising leads to higher sales revenue.

Key Concept: Line Graphs

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 periods of time, which is true as per the syllabus. The reason states that line graphs are useful for showing trends or comparisons, which is also true. However, the reason does not directly explain why a line graph is used to display continuous data over time. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Graph Interpretation

c) The population was increasing at a constant rate.

[Solution Description]

A straight line with a positive slope on a population graph indicates that the population of the city has been increasing at a constant rate over the 10-year period.

Key Concept: Time-temperature graph

b) 0.5°C/hour

[Solution Description]

The rate of change of temperature is calculated as $\frac{\Delta \text{Temperature}}{\Delta \text{Time}}$. The change in temperature is $39 - 38 = 1°C$. The time taken is $10 - 8 = 2$ hours. Therefore, the rate of change of temperature is $\frac{1}{2} = 0.5°C/\text{hour}$.

Key Concept: Line graph, dependent and independent variables

b) 50 km/h

[Solution Description]

To find the speed of the car, we use the formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$. The distance traveled between 10 a.m. and 12 p.m. is $150 - 50 = 100$ km. The time taken is $12 - 10 = 2$ hours. Therefore, the speed of the car is $\frac{100}{2} = 50$ km/h.

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: Bar Graphs

b) Categorical data[Solution Description]A bar graph is best suited for representing categorical data, where different categories are compared based on their values.

Key Concept: Pie Chart Basics

d) 360 degrees

[Solution Description]

A pie chart represents data as a circle, which is divided into sectors. The entire circle has a total angle of 360 degrees. Therefore, the total angle in a pie chart is 360 degrees.

Key Concept: Pie Chart Data Comparison

b) \$500

[Solution Description] The fraction of the total expenditure spent on salaries is $\frac{90}{360} = \frac{1}{4}$. The fraction spent on marketing is $\frac{60}{360} = \frac{1}{6}$. The amount spent on salaries is $\frac{1}{4} \times 6000 = 1500$. The amount spent on marketing is $\frac{1}{6} \times 6000 = 1000$. The difference between the two amounts is $1500 - 1000 = 500$.

Key Concept: Line graph trend

b) Temperature increases

[Solution Description]

From the graph, the temperature at 6 a.m. is 37$^\circ\text{C}$ and it rises to 40$^\circ\text{C}$ at 10 a.m. Therefore, the temperature increases from 6 a.m. to 10 a.m.

Key Concept: Line Graph Interpretation

b) 3$^\circ$C

[Solution Description]

To find the increase in temperature from 6 a.m. to 10 a.m., subtract the temperature at 6 a.m. from the temperature at 10 a.m. The temperature at 10 a.m. is 40$^\circ$C and at 6 a.m. it is 37$^\circ$C. So, the increase in temperature is $40^\circ C - 37^\circ C = 3^\circ C$.

Key Concept: Line Graph

b) Temperature

[Solution Description]

The y-axis in a time-temperature graph represents the temperature recorded at different timings. This is explicitly stated in the syllabus.

Key Concept: Line Graph Interpretation

c) 40°C

[Solution Description]

To determine the highest temperature recorded during the day, compare all the given temperatures: 37°C at 6 a.m., 40°C at 10 a.m., 38°C at 2 p.m., and 35°C at 6 p.m. The highest temperature among these is 40°C at 10 a.m.

Key Concept: Understanding a time-temperature graph

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that the temperature at 12 PM is always the highest point on a time-temperature graph. This is not necessarily true, as the highest temperature can occur at different times depending on various factors like location and weather conditions. However, the reason states that temperature generally increases from morning to afternoon and decreases thereafter, which is true and explains the typical trend observed in a time-temperature graph. Therefore, the assertion is false, but the reason is true.

Key Concept: Understanding a time-temperature graph, trend analysis

d) Assertion is false, but Reason is true.

[Solution Description]

According to the syllabus, the temperature increased from $37^\circ C$ at 6 a.m. to $40^\circ C$ at 10 a.m. This indicates a rising trend in temperature during this period. Since the temperature at 6 a.m. was $37^\circ C$ and it was increasing, the temperature at 8 a.m. must have been more than $37^\circ C$. Therefore, the Assertion is false, but the Reason is true.

Key Concept: Line Graph, Trend Analysis

b) 24.2%

[Solution Description]

Let the initial population be 100. After the first year, the population increases by 20%, so it becomes 120. In the

second year, it remains constant at 120. In the third year, it decreases by 10%, so it becomes 108. In the fourth year,

it increases by 15%, so it becomes 124.2. In the fifth year, it remains constant at 124.2. The overall change in

population is $124.2 - 100 = 24.2$. Therefore, the overall percentage change is 24.2%.

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: Understanding Line Graphs

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 is a pictorial representation of the relationship between dependent and independent variables, which is also true. However, the reason does not directly explain why a line graph is used for continuous data over time. Therefore, the assertion and reason are both true, but the reason is not the correct explanation of the assertion.

Key Concept: Line Graph

a) Time or categories

[Solution Description]

In a line graph, the horizontal line (x-axis) usually represents the independent variable, such as time or categories. It is used to display data that changes continuously over periods of time.

Key Concept: Performance Graph, Interpretation

b) Batsman B

[Solution Description]

Batsman A has a higher peak score of 115 runs but has also scored 0 runs in two matches, indicating inconsistency. Batsman B has a lower peak score of 100 runs but has never scored below 40 runs, indicating steadier performance. Therefore, Batsman B is considered steadier.

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: Line Graph Interpretation

a) Matches played[Solution Description]The horizontal axis in a line graph representing runs scored by batsmen typically represents the matches played.

Key Concept: Line Graph Analysis

c) 87.5

[Solution Description]

To find the median, we first arrange the runs scored in ascending order: 25, 50, 75, 100, 125, 150. Since there are 6 values, the median will be the average of the 3rd and 4th values.

$\text{Median} = \frac{75 + 100}{2} = \frac{175}{2} = 87.5$

Therefore, the median of the runs scored by the player is 87.5.

Key Concept: Comparing runs, percentages

c) 40%

[Solution Description]

First, calculate the total runs: $80 + 120 + 100 = 300$. The percentage of runs scored in the second match is $\frac{120}{300} \times 100 = 40\%$.

Key Concept: Comparing Runs Scored

b) 40

[Solution Description]

To find the difference in runs between the two matches, subtract the runs scored in the second match from the runs scored in the first match. The calculation is as follows:

$\text{Difference} = 320 - 280 = 40$

Therefore, the difference in runs between the two matches is 40.

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, Consistency in scores

b) 2

[Solution Description]

To find the interquartile range (IQR), first arrange the scores in ascending order: 21, 22, 22, 23, 23, 24, 25. The first quartile (Q1) is the median of the first half of the data: (21 + 22) / 2 = 21.5. The third quartile (Q3) is the median of the second half of the data: (23 + 24) / 2 = 23.5. The IQR is Q3 - Q1 = 23.5 - 21.5 = 2.

Key Concept: Distance-Time Graph

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

[Solution Description]

The assertion is true because the slope of a distance-time graph indeed represents the speed of the object. The reason is also true because speed is defined as the ratio of distance travelled to the time taken. Moreover, the reason correctly explains the assertion, as the slope of the graph is calculated using the same ratio of distance to time.

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

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

[Solution Description]

From the graph, it is clear that the car covered 100 km between 9 a.m. and 10 a.m. This corresponds to the change in distance from 50 km to 150 km. Since the slope of the distance-time graph represents speed, and the slope is constant during this interval, the car was moving at a constant speed. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

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]

To find the change in speed, first calculate the speed for each hour. Speed is calculated using the formula $speed = \frac{distance}{time}$. For the first hour, the speed is $\frac{60\,km}{1\,h} = 60\,km/h$. For the second hour, the speed is $\frac{80\,km}{1\,h} = 80\,km/h$. The change in speed is the difference between these two speeds: $80\,km/h - 60\,km/h = 20\,km/h$.

Key Concept: Identifying changes in speed

b) 10 km/h

[Solution Description]

The initial speed of the cyclist is 25 km/h, and the final speed after increasing is 35 km/h. The change in speed is the difference between the final speed and the initial speed: $35\,km/h - 25\,km/h = 10\,km/h$.

Key Concept: Analyzing stopping points using a horizontal segment

d) The bus was stationary

[Solution Description]

In a distance-time graph, a horizontal line segment indicates that the bus's distance from town P remained unchanged between 3 p.m. and 4 p.m. This implies that the bus did not move during this period and was stationary.

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

c) 70 km

[Solution Description]

From the graph, we observe that at time $t = 0$ hours, the car is at 0 km from City P. At $t = 1$ hour, the car has traveled a certain distance, say $d$ km. This distance $d$ can be read directly from the graph and represents the distance traveled during the first hour.

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 in a line graph, the x-axis represents time and the y-axis represents temperature. This is generally true for many line graphs, especially those depicting temperature changes over time. The reason states that line graphs are used to display data that changes continuously over time. This is also correct as line graphs are indeed suitable for such continuous data. However, while both statements are true, the reason does not directly explain why the x-axis and y-axis represent time and temperature respectively. Therefore, the correct answer is b).

Key Concept: Independent and Dependent Variables, Linear Graphs

a) $C = 5U + 100$

[Solution Description]

The relationship between the cost of electricity (\$C) and the number of units consumed (\$U) is linear. The cost increases by \$50 for every 10 units consumed, which means the rate of increase is \$5 per unit. The fixed charge is \$100. Therefore, the relationship can be expressed as:

$C = 5U + 100$

This is a linear equation where \$C\$ is the dependent variable and \$U\$ is the independent variable.

Key Concept: Independent and Dependent Variables

c) Student's performance

[Solution Description]

Here, the number of hours studied affects the student's performance. Therefore, the student's performance is the dependent variable, and the number of hours studied is the independent variable.

Key Concept: Independent and Dependent Variables

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

[Solution Description]

The assertion states that the cost of petrol depends on the number of litres purchased, which is correct because more litres result in a higher cost. The reason correctly identifies the number of litres as the independent variable, as it is the factor being controlled or changed. Since both are true and the reason explains the assertion, the correct answer is option a).

Key Concept: Independent Variable, Area Under Curve

b) The total distance traveled by the car

[Solution Description] The area under a velocity-time graph represents the displacement of the car over the given time interval. Here, an area of 200 meters indicates that the car has traveled a distance of 200 meters in 10 seconds.

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: Principal and Simple Interest, Linear Graph

c) Rs.8000[Solution Description]First, calculate the principal required for an interest of Rs.1: $\frac{500}{5000} = 0.10$. To earn an interest of Rs.800, the required deposit is $\frac{800}{0.10} = Rs.8000$.

Key Concept: Quantity and Cost

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

[Solution Description] The cost of petrol directly depends on the quantity purchased, and the relationship is linear as shown in the graph. The graph passes through the origin because when the quantity is zero, the cost is also zero.

Key Concept: Quantity and Cost

c) \$1000

[Solution Description]

First, find the cost per litre. The cost of 15 litres is \$750, so the cost per litre is \$750 divided by 15 litres, which equals \$50 per litre. To find the cost of 20 litres, multiply the cost per litre (\$50) by 20 litres. This gives \$1000.

Key Concept: Graph plotting, Direct variation

b) 750

[Solution Description]

Since the cost and quantity are in direct variation, we can write the relationship as $C = k \cdot Q$, where $C$ is the cost, $Q$ is the quantity, and $k$ is the constant of variation.

From the given data, for 8 litres, the cost is \$400. Therefore,

$400 = k \cdot 8$

Solving for $k$:

$k = \frac{400}{8} = 50$

Now, we need to find the cost for 15 litres:

$C = 50 \cdot 15 = 750$

So, the cost of 15 litres of petrol is \$750.

Key Concept: Graph plotting, Estimation

d) 1500

[Solution Description]

The graph represents a linear relationship where $C = k \cdot Q$.

For 20 litres, the cost is \$1000. Therefore:

$1000 = k \cdot 20$

Solving for $k$:

$k = \frac{1000}{20} = 50$

Now, we estimate the cost for 30 litres:

$C = 50 \cdot 30 = 1500$

So, the cost for 30 litres of petrol is \$1500.

Key Concept: Graph interpretation, Direct variation

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

[Solution Description] To determine if both Assertion and Reason are true and if Reason explains Assertion, we analyze the given situation. The graph plots points like (10,500), (15,750), (20,1000), and (25,1250), which indicates that as the number of litres increases, the cost increases proportionally. This confirms that the graph is linear and passes through the origin due to direct proportionality. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

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

b) 900 rupees

[Solution Description]

The graph is linear and represents a direct variation between the number of litres and the cost of petrol. To estimate the cost for 18 litres, we can use the slope of the line. The slope (cost per litre) can be calculated using two points from the graph, say (10, 500) and (20, 1000). The slope $m$ is given by:

$m = \frac{1000 - 500}{20 - 10} = \frac{500}{10} = 50 \text{ rupees per litre}$

Using the slope, the cost for 18 litres is:

$\text{Cost} = 18 \times 50 = 900 \text{ rupees}$

Therefore, the estimated cost for 18 litres of petrol is 900 rupees.

Key Concept: Principal and Simple Interest

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

[Solution Description]

The relationship between simple interest ($I$) and principal ($P$) for a fixed rate of interest ($r$) and time period ($t$) is given by $I = P \times r \times t$. Since $r$ and $t$ are constants, this equation represents a linear relationship between $I$ and $P$. A linear relationship in the form $y = mx$ always results in a straight line passing through the origin. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Principal and Simple Interest

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

[Solution Description]

To verify the assertion, we will use the formula for simple interest: $SI = P \times r \times t$. Here, $P = 1000$, $r = 5\% = 0.05$, and $t = 2$ years. Plugging in these values, we get:

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

Therefore, the simple interest calculated is \$100, which matches the assertion. The reason provided is also true as it correctly states the formula for calculating simple interest. Moreover, the reason accurately explains how the assertion was derived. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Simple Interest

b) 5%

[Solution Description]

The formula for simple interest is $I = P \times r \times t$. We are given $I = 300$, $P = 2000$, and $t = 3$. We need to find $r$. Rearranging the formula, we get:

$r = \frac{I}{P \times t} = \frac{300}{2000 \times 3} = \frac{300}{6000} = 0.05$

Converting this to a percentage, we get $r = 5\%$.

Key Concept: Simple Interest and Deposits

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

[Solution Description]

The assertion states that the graph of the relationship between the sum deposited and the simple interest earned is always a straight line. This is true because, for simple interest, the interest earned is calculated using the formula $SI = P \times R \times T / 100$, where $P$ is the principal (sum deposited), $R$ is the rate of interest, and $T$ is the time period. Since $R$ and $T$ are constants, $SI$ is directly proportional to $P$. Therefore, the graph of $SI$ versus $P$ will be a straight line passing through the origin.

The reason states that this is because the simple interest is directly proportional to the sum deposited. This is also true and correctly explains why the graph is a straight line. Thus, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Using the graph to determine missing values

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

[Solution Description]

The graph is a visual representation of the patient's temperature over time. Since the graph plots temperature against time, the temperature at any specific time, such as 1:30 p.m., can be directly read from the graph. This is because the graph provides a continuous and accurate depiction of the data. Therefore, both the Assertion and Reason are true, and the Reason correctly explains why the Assertion is true.

Key Concept: Graph Scale

c) 30 km/h

[Solution Description]

To find the speed of the car, we use the formula: speed = distance / time. Here, the distance covered is 60 km and the time taken is 2 hours. Plugging in these values, we get: speed = 60 km / 2 hours = 30 km/h. Therefore, the speed of the car is 30 km/h.

Key Concept: Speed Calculation

c) 60 km/h

[Solution Description]

The car travels a distance of 150 km in 2.5 hours. To find the average speed, we use the formula:

Speed = Distance / Time

Here, Distance = 150 km, Time = 2.5 hours.

Plugging in the values:

Speed = 150 km / 2.5 hours = 60 km/h.

Key Concept: Distance Calculation

c) 120 km

[Solution Description]

The car is traveling at a constant speed of 40 km/h. To find the distance covered in 3 hours, we use the formula:

Distance = Speed $\times$ Time

Here, Speed = 40 km/h, Time = 3 hours.

Plugging in the values:

Distance = 40 km/h $\times$ 3 hours = 120 km.

Key Concept: Distance-Time Graph

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion claims that the speed of the car during the first three hours was constant. To verify this, we analyze the distance-time graph. During the first hour, the car traveled 50 km, in the second hour it traveled 100 km, and in the third hour it traveled 50 km. Since the distances covered per hour vary, the speed was not constant. Therefore, the assertion is false. The reason states that the distance-time graph for the first three hours is a straight line with a constant slope. However, since the distances covered per hour are different (50 km, 100 km, 50 km), the slope of the graph changes, indicating varying speeds. Thus, the reason is also false.

Key Concept: Distance-Time Graph

c) 180 km

[Solution Description]

Distance = Speed × Time = $60 \text{ km/h} × 3 \text{ hours} = 180 \text{ km}$.

Key Concept: Identifying periods of rest and calculating speed

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

[Solution Description]

From the syllabus, it is clear that the car did not travel during the interval 11 a.m. to 12 noon, as the distance from City P remained constant at 200 km during this period. This indicates that the car was at rest. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Calculating speed

c) 75 km/h

[Solution Description] The total distance covered in two hours is $50 + 100 = 150$ km. The total time taken is 2 hours. Therefore, the average speed is $\frac{150}{2} = 75$ 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: Drawing Graphs

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

[Solution Description]

The Assertion states that the slope of a straight-line graph represents the rate of change between two variables. This is true because the slope directly indicates how one variable changes with respect to the other. The Reason explains that the slope is calculated as the ratio of the vertical change to the horizontal change between two points on the graph, which is also correct and provides the mathematical basis for understanding the slope. Since both the Assertion and Reason are true, and the Reason correctly explains the Assertion, the correct answer is (a).

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: Data Points

b) To ensure the graph correctly represents the relationship between variables

[Solution Description]

It is important to plot data points accurately on a graph because accurate plotting ensures that the graph correctly represents the relationship between variables.

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

b) (2, 4)

[Solution Description]

Given the horizontal scale is 1 unit = 2 hours and the vertical scale is 1 unit = 10 km, we need to find the coordinates for 4 hours and 40 km.

First, calculate the x-coordinate: Since 1 unit = 2 hours, then 4 hours = $4 / 2$ = 2 units.

Next, calculate the y-coordinate: Since 1 unit = 10 km, then 40 km = $40 / 10$ = 4 units.

Therefore, the coordinates are (2, 4).

Key Concept: Labeling the axes

b) Log (Number of Bacteria)

[Solution Description]

Since the population doubles every 2 hours, it is growing exponentially. Therefore, the y-axis should be labeled with a logarithmic scale to represent the exponential growth accurately. The correct label should indicate the number of bacteria on a logarithmic scale.

Key Concept: Labeling the axes

b) Temperature

[Solution Description]

The vertical axis in a graph representing temperature over time should be labeled "Temperature" since it represents the dependent variable.

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: Linear Graph, Direct Variation

c) 300 km

[Solution Description]

To find the distance covered by the car, we use the formula: Distance = Speed × Time. Given that the speed is 60 km/h and the time is 5 hours, we can calculate the distance as follows: Distance = 60 km/h × 5 h = 300 km.

Key Concept: Interpolation

b) 4

[Solution Description]

First, find the slope using the two points:

$m = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$. Then, use the point-slope form to find $y$ when $x = 2$:

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

$y - 2 = 2$

$y = 4$.

Key Concept: Linear Graphs

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

[Solution Description]

If the speed of the car remains constant, then the distance travelled by the car increases linearly with time. This means that the graph of distance versus time will be a straight line, i.e., a linear graph. Therefore, both the assertion and reason are true, and the reason correctly explains why the graph is linear.

Key Concept: Graph Interpretation

b) $(2, -1)$

[Solution Description]

To find the vertex of the parabola given by $y = x^2 - 4x + 3$, we can use the formula for the vertex of a quadratic function $y = ax^2 + bx + c$, where the x-coordinate of the vertex is given by $x = -\frac{b}{2a}$. In this case, $a = 1$ and $b = -4$. Substituting these values, we get $x = -\frac{-4}{2 \times 1} = 2$. Now, substitute $x = 2$ back into the original equation to find the y-coordinate: $y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$. Therefore, the vertex of the parabola is $(2, -1)$.

Key Concept: Linear Graph

c) y = 2x

[Solution Description]

A straight line passing through the origin has the equation of the form $y = mx$, where $m$ is the slope. Among the given options, only $y = 2x$ fits this form.

Key Concept: Cost of Apples vs. Number of Apples

c) 5

[Solution Description]

The cost of 1 apple is given as \$5. To find the number of apples that can be bought for \$25, divide the total amount by the cost of one apple.

$\text{Number of apples} = \frac{25}{5} = 5$

Therefore, 5 apples can be bought for \$25.

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 vs. Time Graph

b) 90 km

[Solution Description] Since the car travels at a constant speed, the distance covered is directly proportional to the time taken. Given that the car covers 30 km in 1 hour, in 3 hours it will cover $3 \times 30 = 90$ km.

Key Concept: Distance-Time Graph

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

[Solution Description]

From the given data, the car was 200 km away from City P at both 11 a.m. and 12 noon. This indicates that the car did not move during this interval because the distance remained unchanged. Therefore, the Assertion is true, and the Reason is also true. Additionally, the Reason correctly explains the Assertion since the constant distance implies no movement.

Key Concept: Simple Interest, Linear Graph

c) 0.1

[Solution Description]

The slope of the line can be calculated using the formula for simple interest: $SI = \frac{P \times R \times T}{100}$. Here, $R = 10\%$ and $T = 1$ year. So, $SI = \frac{P \times 10}{100} = 0.1P$. This implies that the slope of the line is 0.1.

Key Concept: Principal and Simple Interest

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

[Solution Description]

The formula for Simple Interest is given by $S.I. = P \times r \times t$, where $P$ is the principal amount, $r$ is the rate of interest, and $t$ is the time period. Here, the rate of interest and time period are constant, so Simple Interest is directly proportional to the principal amount $P$. Therefore, the graph of Simple Interest vs. Deposit will be a straight line, as it represents a direct variation. Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Interest on Deposits

a) \$13,468.55

[Solution Description]

Using the compound interest formula:

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

Here, $P = 10000$, $r = 6\% = 0.06$, $n = 4$ (since it's compounded quarterly), and $t = 5$ years.

Plugging in the values:

$A = 10000 \left(1 + \frac{0.06}{4}\right)^{4 \times 5} = 10000 \left(1.015\right)^{20}$

Calculating $(1.015)^{20}$:

$1.015^{20} = 1.346855007$

Then:

$A = 10000 \times 1.346855007 = 13468.55$

So, the amount after 5 years is \$13,468.55.

Key Concept: Side of a Square vs. Perimeter

b) 5 cm

[Solution Description]

The perimeter of a square is given by $P = 4 \times \text{side}$. To find the side, rearrange the formula: $\text{side} = P / 4$. Substituting the given perimeter of 20 cm, we get $\text{side} = 20 / 4 = 5$ cm. Hence, the side length is 5 cm.

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

c) 32 cm

[Solution Description]

The original perimeter of the square is 16 cm. Using the formula $P = 4 \times \text{side}$, we can find the original side length as $\frac{16}{4} = 4$ cm. If each side is doubled, the new side length becomes $4 \times 2 = 8$ cm. The new perimeter is then calculated as $4 \times 8 = 32$ cm.

Key Concept: Linear Graph

c) 10 cm

[Solution Description]

The perimeter of a square is given by $P = 4 \times \text{side}$. We can rearrange this formula to solve for the side when the perimeter is known:

$\text{side} = \frac{P}{4} = \frac{40}{4} = 10 \text{ cm}$

Therefore, the length of the side is 10 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 Square vs. Area

c) 6 cm

[Solution Description]

To find the side length of a square given its area, we use the formula $s = \sqrt{A}$, where $A$ is the area. For an area of 36 cm$^2$, substituting into the formula gives $s = \sqrt{36}$. Calculating this gives $s = 6$. Hence, the side length is 6 cm.

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 $\frac{0.69s^2}{s^2} \times 100 = 69\%$.