Key Concept: Introduction to Graphs

c) 38.33$^\circ$C

[Solution Description]

To find the average temperature, add the temperatures at 6 a.m., 10 a.m., and 2 p.m. and divide by the number of data points (3).

Calculation:

Average Temperature = $(37 + 40 + 38) / 3$

Average Temperature = $115 / 3$

Average Temperature = $38.33^\circ$C

Key Concept: Line Graph

b) Values being measured

[Solution Description]

The vertical line (usually called the y-axis) in a line graph typically represents the values being measured or observed, such as temperature, height, or any other quantity.

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: Graphical Presentation

c) Graphical presentation makes trends and comparisons easier to understand

[Solution Description]

Graphical presentation is preferred because it makes trends and comparisons easier to understand than tables. Visual representation allows for quicker comprehension of the data.

Key Concept: Graphical Presentation of Data

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

[Solution Description]

The assertion states that graphical presentation of data is easier to understand than tabular presentation. This is true because graphs visually represent data, which helps in quickly identifying trends and comparisons. The reason provided is that graphs provide a visual representation of data, making trends and comparisons clearer. This is also true and directly explains why graphical presentation is easier to understand. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Introduction to Graphs

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

[Solution Description]

The assertion states that graphs are visual representations of data, which is correct as per the syllabus. The reason explains that graphs make it easier to understand numerical facts quickly and clearly, which is also true. Additionally, the reason correctly explains the purpose of graphs as described in the syllabus. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Data Comparison

c) City B experienced faster population growth

[Solution Description] City B's population grew at a faster rate compared to City A, as indicated by the higher percentage increase. This suggests that City B experienced more significant population growth over the same period.

Key Concept: Introduction to Graphs

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

[Solution Description]

The assertion states that graphs are visual representations of data collected, which is true as graphs are indeed used to represent data visually. The reason states that graphs help in understanding numerical facts quickly and clearly, which is also true and directly explains why graphs are used as visual representations of data. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

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: Tables vs Graphs, Data Comparison

b) Graph because it allows for easy comparison of trends

[Solution Description]

A graph is more effective for comparing quarterly performance because it allows for a visual comparison of trends and patterns across the quarters. Tables, while detailed, require more effort to compare the data visually.

Key Concept: Line Graphs, Data Trends

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that a line graph is always more effective in presenting data trends than a table. This is not entirely true because the effectiveness depends on the type of data and the purpose of presentation. For example, tables are better for precise numerical values. The reason states that line graphs provide a visual representation that makes it easier to identify patterns like increases, decreases, or constant trends over time. This is true as line graphs are particularly useful for showing trends over time. Therefore, the assertion is false, but the reason is true.

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 Graphs, Linear Graphs

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion claims that a line graph showing the relationship between time and temperature is always a linear graph. However, this is not necessarily true. A line graph can display various types of relationships, including non-linear ones, depending on how the data changes over time. For example, if the temperature fluctuates irregularly, the graph may not be a straight line. On the other hand, the reason correctly defines a linear graph as one where the relationship between the dependent and independent variables is represented by a straight line. Therefore, the assertion is false, but the reason is true.

Key Concept: Graphical Presentation

b) It is easier to understand

[Solution Description]

Graphical presentation of data is easier to understand as it shows numerical facts visually, making trends and comparisons clearer.

Key Concept: Line Graph

b) To display data continuously over time

[Solution Description] A line graph is used to display data that changes continuously over time. Its primary purpose is to show numerical facts in visual form so that they can be understood quickly, easily, and clearly.

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

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

[Solution Description]

The assertion states that bar graphs are used to represent categorical data. This is true because bar graphs are specifically designed to compare different categories. The reason states that bar graphs use vertical or horizontal bars to show the frequency of data in each category. This is also true as the length or height of the bars corresponds to the frequency of the data in each category. Additionally, the reason correctly explains the assertion because the use of bars is how bar graphs represent categorical data.

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: 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: Line graph interpretation

b) 10 a.m.

[Solution Description]

The line graph provides temperatures at four different times. The highest temperature recorded is 40\$^\circ\text{C}\$ at 10 a.\,m. Thus, the time when the temperature was highest is 10 a.\,m.

Key Concept: Line graph basics

b) Time

[Solution Description]

In a line graph, the x-axis typically represents the time intervals at which the data is recorded. This is because the graph shows how data changes continuously over time.

Key Concept: Time-Temperature Graph

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

[Solution Description]

The assertion states that a time-temperature graph is used to display data that changes continuously over periods of time. This is true because a time-temperature graph, which is a type of line graph, is specifically designed to show how temperature varies over time. The reason states that a line graph is a visual representation of data collected, and it is easier to understand trends or comparisons. This is also true because line graphs are commonly used to visualize continuous data and identify patterns or trends. However, the reason does not directly explain why a time-temperature graph is used for continuous data; it only explains the general purpose of a line graph. Therefore, both the assertion and reason 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) 5.41\%

[Solution Description]

The percentage increase in temperature is calculated using the formula:

$\text{Percentage Increase} = \frac{\text{Increase in Temperature}}{\text{Initial Temperature}} \times 100$

The increase in temperature is $39\degree C - 37\degree C = 2\degree C$. Therefore,

$\text{Percentage Increase} = \frac{2\degree C}{37\degree C} \times 100 \approx 5.41\%$

Therefore, the percentage increase in temperature between 6 a.m. and 8 a.m. is approximately $5.41\%$.

Key Concept: Time-Temperature Graph

b) -0.5°C/hour

[Solution Description]

The average rate of change in temperature can be calculated using the formula:

$\text{Average Rate of Change} = \frac{\text{Change in Temperature}}{\text{Change in Time}}$

Here, the change in temperature is $37.5\degree C - 38\degree C = -0.5\degree C$. The change in time is 1 hour. Thus,

$\text{Average Rate of Change} = \frac{-0.5\degree C}{1 \text{ hour}} = -0.5\degree C/\text{hour}$

Therefore, the average rate of change in temperature between 1 p.m. and 2 p.m. is $-0.5\degree C/\text{hour}$.

Key Concept: Line Graph Interpretation

c) $3^\circ$C

[Solution Description]

To find the increase in temperature from 6 a.m. to 10 a.m., we need to subtract the temperature at 6 a.m. from the temperature at 10 a.m. According to the graph, the temperature at 6 a.m. is $37^\circ$C and at 10 a.m. is $40^\circ$C. Therefore, the increase in temperature is $40^\circ$C - $37^\circ$C = $3^\circ$C.

Key Concept: Identifying Trends in Graphs

c) Values are increasing

[Solution Description]

An upward trend in a graph usually indicates that the values are increasing over time or across the measured variable.

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: 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: Consistency Analysis

c) 5

[Solution Description] According to the graph, Batsman A scored less than 40 runs in 5 matches. This is explicitly mentioned in the description.

Key Concept: Performance Graph of Two Batsmen

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

[Solution Description]

The assertion states that the performance graph of Batsman A shows a consistently increasing trend, which implies that his batting skills are steadily improving. The reason explains that a consistently increasing trend indicates that the batsman is adapting well to different bowling strategies and improving with each match. Both the assertion and reason are true, and the reason correctly explains the assertion.

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: Line Graph Interpretation

c) Dotted line

[Solution Description]

To identify which line represents the total runs scored by Batsman A, check the key or label provided on the graph. Typically, the dotted line indicates Batsman A's 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

b) 170

[Solution Description]

To find the difference between the runs scored in the first and second innings, subtract the runs scored in the second innings from the runs scored in the first innings.

$\text{Difference} = 320 - 150 = 170$

Key Concept: Identifying trends and consistency in scores

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion is false because the mean is not solely determined by the highest score. The mean is affected by all scores in the dataset. The reason is true because the mean is indeed calculated by dividing the sum of all scores by the number of scores. Therefore, the assertion is false, but the reason is true.

Key Concept: Identifying trends and consistency in scores

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

[Solution Description]

The assertion states that the standard deviation of a dataset is zero if all the data points are identical. This is true because standard deviation measures how much the data points deviate from the mean. If all data points are identical, there is no deviation, so the standard deviation is zero. The reason states that standard deviation measures the variability or dispersion of a dataset. This is also true as it is the definition of standard deviation. Moreover, the reason correctly explains why the assertion is true because zero variability implies zero standard deviation.

Key Concept: Distance-Time Graph Basics

d) The object is at rest

[Solution Description]

A horizontal line on a distance-time graph indicates that the object is not moving. The distance remains constant over time, which means the object is at rest.

Key Concept: Distance-Time Graph

d) Assertion is false, but Reason is true.

[Solution Description]

The given data shows that the car did not move at a constant speed throughout the journey. During the first hour, it covered 50 km, in the second hour 100 km, and in the third hour 50 km. The car also remained stationary between 11 a.m. and 12 noon. Therefore, the speed of the car varied, and the distance-time graph cannot be a straight line with a constant slope. Hence, the Assertion is false, but the Reason is true as a straight line with a constant slope would imply constant speed.

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: Distance-time graph, speed variation

b) 4 hours

[Solution Description]

The car travels 50 km in the first hour, stops for one hour (from hour 1 to hour 2), and then travels 150 km in the next two hours at a constant speed.

Time taken for the first hour = $1 \text{ hour}$

Time taken for stopping = $1 \text{ hour}$

Time taken to travel 150 km at constant speed = $2 \text{ hours}$

Total time taken = $1 \text{ hour} + 1 \text{ hour} + 2 \text{ hours} = 4 \text{ hours}$

Key Concept: Identifying changes in speed

c) $18 \text{ m/s}$

[Solution Description]

Initial velocity ($u$) is $0 \text{ m/s}$, acceleration ($a$) is $3 \text{ m/s}^2$, and distance ($s$) is $54 \text{ m}$. Using the equation of motion:

$v^2 = u^2 + 2as$

Substitute the known values:

$v^2 = 0^2 + 2(3)(54) = 324$

Take the square root of both sides to find $v$:

$v = \sqrt{324} = 18 \text{ m/s}$

Key Concept: Speed, Acceleration

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

[Solution Description]

The assertion states that if the speed of an object increases uniformly over time, its acceleration is constant. This is true because acceleration is defined as the rate of change of velocity with respect to time. If speed increases uniformly, the rate of change of velocity (which includes both speed and direction) remains constant, hence the acceleration is constant. The reason correctly explains the assertion by defining uniform increase in speed as a constant rate of change in velocity, which directly relates to acceleration.

Key Concept: Analyzing Stopping Points

c) The object is at rest

[Solution Description] A horizontal segment on a distance vs. time graph represents no change in distance over time, indicating that the object is stationary during that period.

Key Concept: Analyzing stopping points using a horizontal segment in the graph

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

[Solution Description]

The horizontal segment in the graph between 11 a.m. and 12 noon shows that the distance of the car from City P did not change during this period, indicating that the car was stationary. This is because the distance remained constant at 200 km from 11 a.m. to 12 noon. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Graph interpretation, Simple Interest

d) \$550

[Solution Description]

To calculate the total amount after one year, we need to compute the simple interest earned and add it to the principal.

Simple Interest = Principal × Rate × Time

Given: Principal = \$500, Rate = 10\%, Time = 1 year

Simple Interest = \$500 × $\frac{10}{100}$ × 1 = \$50

Total Amount = Principal + Simple Interest = \$500 + \$50 = \$550

Key Concept: Simple Interest Calculation

b) Rs 25

[Solution Description]

Simple Interest (S.I.) formula is:

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

Here, $P = 250$, $R = 10 \%$, $T = 1$ year.

Substituting the values:

$S.I. = \frac{250 \times 10 \times 1}{100} = 25$

Therefore, the annual simple interest earned is Rs 25.

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

b) Amount of water used

[Solution Description]

In this scenario, the quantity of water used determines the water bill. Therefore, the amount of water used is the independent variable, and the water bill is the dependent variable.

Key Concept: Graph Plotting

b) \$1000

[Solution Description] The cost of petrol is directly proportional to the quantity of petrol. If 10 litres cost \$500, then 20 litres will cost twice as much. Therefore, the cost of 20 litres = 2 * \$500 = \$1000.

Key Concept: Independent Variable, Graph Analysis

c) The instantaneous speed of the car at that point

[Solution Description] The slope of a graph representing distance versus time gives the speed of the car at that point. Since the slope is 60 km/h, it means the car is traveling at a constant speed of 60 km/h at that point.

Key Concept: Dependent Variable

c) \$250

[Solution Description]

To find the cost when 20 units are produced, substitute \$x = 20 into the equation \$C = 50 + 10x. So, \$C = 50 + 10(20) = 50 + 200 = \$250.

Key Concept: Dependent Variable

b) Cost of petrol

[Solution Description]

In this scenario, the cost of petrol depends on the quantity of petrol purchased. Therefore, the dependent variable is the cost of petrol.

Key Concept: Quantity and Cost, Graphs

c) 167 units

[Solution Description]

To find the break-even point, set the total cost equal to the total revenue. The total cost is given by $TC = 5000 + 20q$, where $q$ is the number of units. The total revenue is given by $TR = 50q$. Set $TC = TR$ and solve for $q$:

$5000 + 20q = 50q$

Subtract $20q$ from both sides:

$5000 = 30q$

Divide both sides by 30:

$q = \frac{5000}{30} \approx 166.67$

Since the number of units must be a whole number, the break-even point is 167 units.

Key Concept: Quantity and Cost

c) \$1500

[Solution Description]

The cost per litre is given as \$50. To find the total cost to fill the tank completely, multiply the cost per litre (\$50) by the capacity of the tank (30 litres). This gives \$1500.

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: Independent Variable Identification

b) Quantity of petrol

[Solution Description]

The independent variable is the one that you control or change, which in this case is the number of litres of petrol you decide to buy. The cost depends on the quantity of petrol purchased.

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, Slope interpretation

b) 170

[Solution Description]

The slope of the graph represents the hourly wage, which is \$15 per hour. The base pay is \$50. Therefore, the total earnings for working 8 hours can be calculated as Total Earnings = Base Pay + (Hourly Wage × Number of Hours Worked) = \$50 + (\$15 × 8) = \$50 + \$120 = \$170. Thus, the total earnings for working 8 hours are \$170.

Key Concept: Principal and Simple Interest

b) 2 years

[Solution Description]

The formula for simple interest is $I = P \times r \times t$. We can rearrange this to solve for $t$:\

$t = \frac{I}{P \times r}$

Here, $P = 1500$, $r = 6\% = 0.06$, and $I = 180$.\

$t = \frac{180}{1500 \times 0.06} = \frac{180}{90} = 2$

The money has been invested for 2 years.

Key Concept: Principal and Simple Interest

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.[Solution Description]The assertion can be verified using the formula for simple interest. Given $P = \$400$, $R = 10\%$ or $0.10$, and $T = 1$ year, the simple interest $I$ is calculated as follows:

$I = P \times R \times T = 400 \times 0.10 \times 1 = \$40$

Thus, the assertion is true. The reason provided is also true as it correctly states the formula for simple interest. Moreover, the reason explains why the assertion holds true.

Key Concept: Simple Interest, Graph Interpretation

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

[Solution Description]

The simple interest (\$I) is calculated using the formula $\text{I} = \text{P} \times \text{R} \times \text{T}$, where P is the principal amount, R is the rate of interest, and T is the time period. Here, R = 10\% and T = 1 year, so $\text{I} = \text{P} \times 0.10$. This shows that I is directly proportional to P. When we plot P on the x-axis and I on the y-axis, the relationship will be a straight line passing through the origin because when P = 0, I = 0.

Both the assertion and reason are true, and the reason correctly explains why the graph is a straight line passing through the origin.

Key Concept: Simple Interest

c) \$5750

[Solution Description]

To calculate the total amount, we use the formula for simple interest: $A = P(1 + rt)$, where $P$ is the principal, $r$ is the rate, and $t$ is the time in years. Here, \$P = 5000\$, \$r = 0.05\$, and \$t = 3\$. Plugging in the values, we get $A = 5000(1 + 0.05 \times 3)$. Calculating inside the parentheses first: $1 + 0.15 = 1.15$. Now multiply by the principal: $A = 5000 \times 1.15 = 5750$. The total amount after 3 years is \$5750.

Key Concept: Graph Interpretation

c) 120 km

[Solution Description]

From the given data, the distance covered increases by 30 km each hour. In the first hour, the car covers 30 km. In the second hour, it covers 60 km, which is an increase of 30 km from the first hour. Similarly, in the third hour, it covers 90 km, which is an increase of 30 km from the second hour. Therefore, in the fourth hour, the car will cover 120 km, which is an increase of 30 km from the third hour.

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: 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: Time, Distance, Speed

c) 210 minutes

[Solution Description]

To calculate the total time taken for the journey, we need to consider both the driving time and the rest time. The car travels a total distance of 180 km at a speed of 60 km/h. The time taken to cover this distance is $\frac{180}{60} = 3$ hours. Additionally, the car stops for 30 minutes (which is equivalent to 0.5 hours). Therefore, the total time taken for the journey is $3 + 0.5 = 3.5$ hours. Converted into minutes, this is $210$ minutes.

Key Concept: Distance-Time Graph, Total Journey Time

b) 6 hours

[Solution Description]

The car started at 8 a.m. and reached City Q at 2 p.m. To find the total time taken, we calculate the difference between these two times. From 8 a.m. to 2 p.m. is 6 hours. Therefore, the total time taken for the journey is 6 hours.

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: Identifying periods of rest and calculating speed

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

[Solution Description]

To determine if the courier-person cycled fastest between 10 a.m. and 11 a.m., we need to analyze the slope of the distance-time graph during this period. The steepest slope indicates the highest speed because speed is directly proportional to the slope of the distance-time graph.

Let’s assume the graph shows the following data points:

- At 10 a.m., the distance from the town was $d_1$ km.

- At 11 a.m., the distance from the town was $d_2$ km.

The speed between 10 a.m. and 11 a.m. can be calculated using the formula:

$\text{Speed} = \frac{d_2 - d_1}{1 \text{ hour}}$

If the slope is the steepest during this period, then the speed is the highest, which means the person cycled fastest between 10 a.m. and 11 a.m.

Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Identifying periods of rest, calculating speed

b) 20 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 traveled = Distance in first 30 minutes + Distance in next 45 minutes = 10 km + 20 km = 30 km.

Total time taken = Time for first 30 minutes + Rest period + Time for next 45 minutes = 0.5 hours + 0.25 hours + 0.75 hours = 1.5 hours.

Average speed = Total distance / Total time = 30 km / 1.5 hours = 20 km/h.

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: Graph Interpretation

d) \$900

[Solution Description]

Since the graph is linear, the cost per litre remains constant. From the given data, cost per litre = \$600 / 12 litres = \$50 per litre. Therefore, cost for 18 litres = \$50 $\times$ 18 = \$900.

Key Concept: Plotting Points

a) Decide on the scale for the axes

[Solution Description]

The first step in plotting a graph is to decide on the scale for the axes. This involves determining how many units on the graph paper will represent one unit of the variable being plotted.

Key Concept: Graph Plotting, Linear Graph

d) (25, 1250)

[Solution Description]

The cost of petrol per litre is \$50. For any number of litres, say $x$, the total cost will be $50x$. Let’s check each option:

For option a) (10, 500): Total cost = $50 \times 10 = \$500$. This matches.

For option b) (15, 700): Total cost = $50 \times 15 = \$750$. This does not match.

For option c) (20, 800): Total cost = $50 \times 20 = \$1000$. This does not match.

For option d) (25, 1250): Total cost = $50 \times 25 = \$1250$. This matches.

Therefore, options a) and d) lie on the graph. However, since only one option is to be selected, option d) (25, 1250) is more appropriate as it represents a higher quantity.

Key Concept: Choosing an appropriate scale

b) (4, 5)

[Solution Description]

Given the horizontal scale is 1 unit = 5 minutes and the vertical scale is 1 unit = 15 meters, we need to find the number of units for 20 minutes and 75 meters.

First, calculate the x-axis units: Since 1 unit = 5 minutes, then 20 minutes = $20 / 5$ = 4 units.

Next, calculate the y-axis units: Since 1 unit = 15 meters, then 75 meters = $75 / 15$ = 5 units.

Therefore, the number of units on the graph is (4, 5).

Key Concept: Choosing an appropriate scale

b) Horizontal: 1 unit = 1 hour; Vertical: 1 unit = 20 km

[Solution Description]

To plot this data effectively, we need to choose a scale that allows easy plotting of points. The cyclist covers 20 km every hour, so using increments of 1 hour on the horizontal axis and 20 km on the vertical axis is appropriate. For example:

- Horizontal: 1 hour is represented by 1 unit.

- Vertical: 20 km is represented by 1 unit.

This ensures that the graph is neither too compressed nor too spread out.

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: Labeling the axes

b) Time (Seconds)

[Solution Description]

Since the graph shows velocity over time, the x-axis should represent time. The label should clearly indicate that it represents time in seconds, minutes, or hours, depending on the unit used.

Key Concept: Drawing Graphs, Plotting given data points

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

[Solution Description]

The assertion is true because the graph of a linear equation in two variables is always a straight line. The reason is also true because a straight line represents a constant rate of change, which is a fundamental property of linear equations. The reason correctly explains the assertion.

Key Concept: Linear Graph, Data Plotting

c) \$216

[Solution Description]

Since the cost is directly proportional to the number of litres, we can set up a proportion: Cost₁ / Litres₁ = Cost₂ / Litres₂. Given that 12 litres cost \$144, we can find the cost for 18 litres as follows: \$144 / 12 = Cost₂ / 18. Solving for Cost₂, we get Cost₂ = (\$144 / 12) × 18 = \$216.

Key Concept: Linear Graph, Temperature-Time Relationship

c) 39°C

[Solution Description]

The temperature changes linearly over time. Between 10 a.m. and 2 p.m., the temperature decreases from 40°C to 38°C. This indicates a decrease of 1°C every 2 hours. At 12 p.m., which is 2 hours after 10 a.m., the temperature would be $40 - 1 = 39$°C.

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: 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: Graph Equation, Cost Prediction

c) 35 rupees

[Solution Description]

The equation of the line is given as $y = 5x$, where $y$ is the cost in rupees and $x$ is the number of apples. To find the cost of 7 apples, substitute $x = 7$ into the equation: $y = 5 \times 7 = 35$ rupees.

Key Concept: Graph Interpretation, Cost Calculation

c) 30 rupees

[Solution Description]

First, determine the slope (cost per apple) using the two given points (2,10) and (4,20). The slope $m$ is calculated as $m = \frac{20 - 10}{4 - 2} = \frac{10}{2} = 5$ rupees per apple. Now, to find the cost of 6 apples, multiply the number of apples by the cost per apple: $6 \times 5 = 30$ rupees.

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: Interpreting Graphs

b) 2.5 hours

[Solution Description] From the table provided in the syllabus, for every 1 hour, the car travels 30 km. To find the time corresponding to 75 km, we use the formula $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$. Here, speed is 30 km/h, so $\text{Time} = \frac{75}{30} = 2.5$ hours.

Key Concept: Graph Interpretation, Rate of Change

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

[Solution Description]

The assertion states that the slope of the distance-time graph for a car moving with uniform acceleration increases linearly with time. This is true because the slope of the distance-time graph represents the velocity of the car. For a car with uniform acceleration, the velocity $v$ at any time $t$ can be expressed as $$v = u + at$$ where $u$ is the initial velocity and $a$ is the acceleration. Since $a$ is constant, $v$ increases linearly with time, making the slope of the distance-time graph increase linearly with time.

The reason states that the velocity of the car increases uniformly with time when the acceleration is constant. This is also true because, from the equation $$v = u + at$$, it is clear that velocity increases linearly with time when acceleration is constant.

However, the reason does not explain why the slope of the distance-time graph increases linearly with time. It only explains the behavior of velocity with time. Therefore, the reason is not the correct explanation of the assertion.

Key Concept: Simple Interest, Linear Graph

b) \$400

[Solution Description]

First, find the rate of interest using the formula $SI = \frac{P \times R \times T}{100}$. Given $P = \$3000$, $SI = \$240$, and $T = 1$ year, we can calculate $R$ as follows: $240 = \frac{3000 \times R}{100}$. Solving for $R$, we get $R = 8\%$. Now, calculate the interest for \$5000: $SI = \frac{5000 \times 8}{100} = \$400$.

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: Interest on Deposits for a Year

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

[Solution Description]

The assertion states that the graph of interest earned on a fixed deposit over one year is a straight line. This is true because simple interest is calculated as $I = P \times r \times t$, where $P$ is the principal amount, $r$ is the rate of interest, and $t$ is the time period. Since the principal and rate are constant, the interest earned increases linearly with time, resulting in a straight-line graph. The reason given is that the interest is calculated using simple interest, which is linear. This is also true and correctly explains why the graph is a straight line. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Linear Graph

c) A straight line passing through the origin

[Solution Description]

The relationship between the side and perimeter of a square is linear because the perimeter increases proportionally with the side. This means the graph should be a straight line passing through the origin. The correct graph is a straight line that goes upwards as the side increases, indicating a direct proportional relationship.

Key Concept: Side of a Square vs. Perimeter

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

[Solution Description]

The relationship between the side length ($s$) of a square and its perimeter ($P$) is given by $P = 4s$. This is a linear equation, meaning that as the side length increases, the perimeter increases at a constant rate. Therefore, the graph of side length versus perimeter will be a straight line. Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

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: Side of square vs. Area

d) Assertion is false, but Reason is true.

[Solution Description]

The area of a square is calculated using the formula $A = s^2$, where $A$ is the area and $s$ is the side length. For the graph to be linear, the relationship between $s$ and $A$ must be proportional, i.e., $A = ks$, where $k$ is a constant. However, $A = s^2$ is a quadratic relationship, not proportional. Hence, the Assertion is false. On the other hand, the Reason states that the relationship is linear, which is also false. Therefore, both the Assertion and Reason are false.

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

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$^2$.