Chapter 12 Introduction to Graphs (Class 8 - Maths NCERT Exemplar Solutions)

Solution:

On the $x$-axis, the distance of any point from the $x$-axis itself is zero. Therefore, the $y$-coordinate (ordinate) of every point lying on the $x$-axis is always $0$.

Hence, any point on the $x$-axis is represented in the form $(x, 0)$.

Correct Option: (b)

Solution:

In the given distance-time graph, the horizontal axis represents the Time and the vertical axis represents the Distance.

Between $5\text{ pm}$ and $7\text{ pm}$, the graph is a horizontal straight line parallel to the time axis.

A horizontal line indicates that the distance from the starting point remains constant over that period of time. This means the car was stationary.

Therefore, Nisha was not driving during this interval.

Correct Option: (c)

Solution:

A graph in which all the plotted points lie on a single uninterrupted straight line is called a linear graph.

Answer: linear

Solution:

The origin is the point where the $x$-axis and $y$-axis intersect. At this point, both the horizontal distance ($x$) and the vertical distance ($y$) are zero.

Answer: $(0, 0)$

Solution:

In a coordinate pair $(x, y)$, the order of the numbers is crucial. For the point $(3, 4)$, the $x$-coordinate is $3$ and the $y$-coordinate is $4$. For the point $(4, 3)$, the $x$-coordinate is $4$ and the $y$-coordinate is $3$.

Since the coordinates are different, they represent two distinct positions on the Cartesian plane.

Answer: False (F)

Solution:

Any point on the $x$-axis has zero vertical displacement from the origin. Thus, its $y$-coordinate is always $0$.

Answer: True (T)

Given:

Points to be plotted: $A(4, 4)$, $B(1, 3)$, $C(4, 2)$, and $D(7, 3)$.

To Find:

The name of the geometric shape formed by joining these points.

Construction Required:

Plot the given points on a Cartesian plane and join them in the order $A \to B \to C \to D \to A$.

Solution:

Let the points be $A(4, 4)$, $B(1, 3)$, $C(4, 2)$, and $D(7, 3)$. After joining the points, we observe the following properties:

1. Length of sides: Using the distance formula, we find that $AB = BC = CD = DA = \sqrt{10}\text{ units}$. Since all sides are equal, the figure is either a square or a rhombus.

2. Length of diagonals: The horizontal diagonal $BD$ connects $(1, 3)$ and $(7, 3)$, so its length is $7 - 1 = 6\text{ units}$. The vertical diagonal $AC$ connects $(4, 4)$ and $(4, 2)$, so its length is $4 - 2 = 2\text{ units}$.

3. Relationship of diagonals: The diagonals are perpendicular to each other ($AC$ is vertical and $BD$ is horizontal) and they bisect each other at the point $(4, 3)$.

Since all sides are equal but the diagonals are not equal, the shape formed is a rhombus.

Answer: Rhombus

To Find:

Coordinates of points $A, B, C, D, E, F, G, H, I, J, K, L$.

Solution:

By observing the position of each point relative to the $x$-axis and $y$-axis, we determine the coordinates $(x, y)$ as follows:

The coordinates have been successfully identified from the graph.

Given:

A linear graph showing the relationship between Temperature in Celsius ($^\circ\text{C}$) on the $x$-axis and Temperature in Fahrenheit ($^\circ\text{F}$) on the $y$-axis.

To Find:

(a) Celsius equivalent of $140^\circ\text{F}$.

(b) Fahrenheit equivalent of $20^\circ\text{C}$.

Solution:

(a) Conversion of $140^\circ\text{F}$ to $^\circ\text{C}$:

Locate $140$ on the vertical axis (Fahrenheit scale). Move horizontally to meet the graph line, then move vertically downwards to the $x$-axis.

The corresponding value on the Celsius scale is $60$.

Thus, $140^\circ\text{F} = 60^\circ\text{C}$.

(b) Conversion of $20^\circ\text{C}$ to $^\circ\text{F}$:

Locate $20$ on the horizontal axis (Celsius scale). Move vertically upwards to meet the graph line, then move horizontally towards the $y$-axis.

The corresponding value on the Fahrenheit scale is $68$.

Thus, $20^\circ\text{C} = 68^\circ\text{F}$.

Given:

A double line graph showing sales over 12 months for two consecutive years.

Thin line indicates Sales in 1st year and Bold line indicates Sales in 2nd year.

Solution:

(a) Maximum difference:

By observing the gap between the two lines, the maximum vertical distance is seen in the month of June.

In June, Year 1 sales $\approx 4000$ and Year 2 sales $\approx 10000$. Difference $= 6000$.

(b) Stability:

Stability refers to fewer fluctuations. The thin line (Year 1) shows much smaller peaks and valleys compared to the bold line (Year 2).

Therefore, there was more stability in the 1st year.

(c) Same Sale:

The two lines intersect where the sales are equal. This happens in the month of August.

(d) Year 1 sales less than Year 2 sales:

We look for months where the thin line is below the bold line. This occurs in:

1. June (Year 1 $\approx 4000$, Year 2 $\approx 10000$)

2. November (Year 1 $\approx 6000$, Year 2 $\approx 8000$)

Solution:

Cyclist I:

The graph shows speed decreasing as time increases. The straight line sloping downwards indicates that the cyclist is decelerating (slowing down) at a constant rate until the speed eventually becomes zero.

Cyclist II:

The graph shows speed increasing as time increases. Initially, the cyclist accelerates quickly (steeper slope). After a certain point in time, the cyclist continues to accelerate but at a slower, constant rate (flatter slope).

Solution:

1. A double bar graph is useful for the comparison of two sets of data.

2. Data represented in a circular form is called a pie chart.

3. The graph of a linear equation is always a straight line.

4. The cartesian system used two axes which are perpendicular to each other.

Given:

The equation provided is:

$y = 2x$

The values of $x$ provided are $0, 1, 2, 3,$ and $4$.

To Find:

The corresponding values of $y$ and the resulting linear graph.

Solution:

By substituting each value of $x$ into the equation, we determine the values of $y$:

For $x = 0$: $y = 2(0) = 0$. The point is $(0, 0)$.

For $x = 1$: $y = 2(1) = 2$. The point is $(1, 2)$.

For $x = 2$: $y = 2(2) = 4$. The point is $(2, 4)$.

For $x = 3$: $y = 2(3) = 6$. The point is $(3, 6)$.

For $x = 4$: $y = 2(4) = 8$. The point is $(4, 8)$.

Completed Table:

Graph:

By plotting the ordered pairs $(0, 0), (1, 2), (2, 4), (3, 6),$ and $(4, 8)$ on the coordinate plane and joining them, we obtain a straight line passing through the origin.

The resulting graph is a linear graph, which demonstrates the direct variation between $x$ and $y$.

Solution:

A pie chart (also known as a circle graph) is specifically used to represent the relationship between parts and a whole. The entire circle represents the whole, and the various sectors represent the parts of that whole.

Correct Option: (b)

Solution:

A line graph is used to display data that changes continuously over a period of time. It is constructed by connecting a series of data points with straight line segments.

Correct Option: (d)

Solution:

To find the coordinates of point $x$, we observe its position relative to the axes:

1. Move horizontally along the $x$-axis from the origin to reach the line below point $x$. This distance is $3$ units.

2. Move vertically upwards from that position to reach point $x$. This distance is $2$ units.

Therefore, the coordinates are $(3, 2)$.

Correct Option: (c)

Solution:

The point $(0, 3)$ has an $x$-coordinate of $0$, which means it must lie on the $y$-axis.

Looking at the graph, point R is located on the $y$-axis at a height of $3$ units from the origin.

Thus, point R represents $(0, 3)$.

Correct Option: (c)

Solution:

For any point $(x, y)$:

1. The distance from the $x$-axis is the absolute value of the $y$-coordinate, which is $|y|$.

2. The distance from the $y$-axis is the absolute value of the $x$-coordinate, which is $|x|$.

For the point $(3, 4)$:

Distance from $x$-axis = $4$ units.

Distance from $y$-axis = $3$ units.

Correct Option: (c)

Solution:

A point lies on the $x$-axis if its $y$-coordinate is $0$. A point lies on the $y$-axis if its $x$-coordinate is $0$.

The only point that satisfies both conditions ($x=0$ and $y=0$) is the origin.

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

Correct Option: (a)

Solution:

1. Distance from the $x$-axis corresponds to the $y$-coordinate. So, $y = 3$.

2. Distance from the $y$-axis corresponds to the $x$-coordinate. So, $x = 6$.

The coordinates are $(x, y) = (6, 3)$.

Correct Option: (d)

Solution:

The position of an object in a 2D plane is given by $(x, y)$, where $x$ is the horizontal distance and $y$ is the vertical distance from the reference axes.

Looking at the arrows in the figure:

1. The horizontal distance from the $Y$-axis (along the $X$-axis) is $3$.

2. The vertical distance from the $X$-axis (along the $Y$-axis) is $7$.

Thus, the position is $(3, 7)$.

Correct Option: (b)

Given:

A linear graph representing the Typing Rate where the $x$-axis represents Time (in sec) and the $y$-axis represents the Number of Words.

To Find:

The approximate number of words typed in $30$ seconds.

Solution:

To find the number of words typed in $30$ seconds, follow these steps on the graph:

1. Locate the value $30$ on the horizontal axis (Time axis).

2. Move vertically upwards from the $30$ seconds mark until you intersect the line graph.

3. From that point of intersection, move horizontally to the left to reach the vertical axis (Number of Words axis).

4. Observing the grid, the point lies between $20$ and $30$. More precisely, it is slightly below the $30$ mark, approximately at $28$.

Correct Option: (c)

Given:

A table showing the relationship between the Length of Side of a Square ($x$) and its Perimeter ($y$):

$(1, 4), (2, 8), (3, 12), (4, 16), (5, 20)$

Solution:

Let's analyze the given graphs:

- Graph (a): Shows a horizontal line at $y = 20$. This would mean the perimeter is always $20$ regardless of the side length. This is incorrect.

- Graph (b): Shows a vertical line at $x = 1$. This would mean the side length is always $1$ regardless of the perimeter. This is incorrect.

- Graph (c): Shows the points $(4, 1), (8, 2), (12, 3) \dots$ which means the axes have been swapped. This is incorrect.

- Graph (d): Plotting the points $(1, 4), (2, 8), (3, 12), (4, 16),$ and $(5, 20)$ correctly results in a straight line passing through the origin. This exactly matches the table data.

Correct Option: (d)

Solution:

A line graph is used to show data that changes continuously over time.

Answer: Line graph

Solution:

The relationship between a dependent variable (usually on the $y$-axis) and an independent variable (usually on the $x$-axis) is represented using a graph.

Answer: graph

Solution:

To pinpoint the exact location of a point on a 2D plane, we require two coordinates, namely the $x$-coordinate (abscissa) and the $y$-coordinate (ordinate).

Answer: two

Solution:

If the $x$-coordinate is $0$, the point has no horizontal displacement from the origin. Therefore, it stays on the vertical axis.

Answer: y-axis

Solution:

In a standard coordinate system, the horizontal reference line is called the x-axis and the vertical reference line is called the y-axis.

Answer: x-axis and y-axis

Solution:

The act of marking a point at its designated coordinates $(x, y)$ on a graph paper is called plotting.

Answer: plotting

Solution:

The perpendicular distance of a point from the $y$-axis is measured along the horizontal direction. In the Cartesian coordinate system, this horizontal distance corresponds to the first value in the ordered pair $(x, y)$, which is the $x$-coordinate (also known as the abscissa).

Answer: x (or x-coordinate)

Solution:

Any point that has a $y$-coordinate of $0$ means it has no vertical displacement from the horizontal reference line. Therefore, all such points are located exactly on the $x$-axis.

Answer: x-axis

Solution:

The distance of a point $(x, y)$ from the $x$-axis is given by the absolute value of its $y$-coordinate. For the point $(5, 2)$, the $y$-coordinate is $2$.

Therefore, the distance from the $x$-axis is $2$ units.

Answer: 2

Solution:

A point lying on the $y$-axis has zero horizontal distance from the vertical axis. In the coordinate system, this horizontal distance is represented by the $x$-coordinate.

Hence, the $x$-coordinate of any point on the $y$-axis is always $0$.

Answer: 0

Solution:

In the ordered pair $(x, y)$, the first number represents the $x$-coordinate and the second number represents the $y$-coordinate. In the point $(2, 4)$, the second number is $4$.

Answer: 4

Solution:

In the coordinate system, a point is represented as $(x, y)$. The first number $4$ in the pair $(4, 7)$ represents the horizontal position of the point, which is called the $x$-coordinate or abscissa.

Answer: x-coordinate (or abscissa)

Solution:

Coordinates are always written in the form $(\text{x-coordinate}, \text{y-coordinate})$. Given that the $x$-coordinate is $5$ and the $y$-coordinate is $4$, we arrange them within parentheses.

Answer: $(5, 4)$

Solution:

The first number in a coordinate pair $(x, y)$ represents the $x$-coordinate, while the second number represents the $y$-coordinate (also known as the ordinate).

Answer: y-coordinate (or ordinate)

Solution:

The horizontal $x$-axis and the vertical $y$-axis cross each other at a specific reference point where both coordinates are zero. This intersection point is known as the origin.

Answer: origin

Solution:

In a two-dimensional Cartesian plane, the location of any point is uniquely determined by a pair of numerical coordinates, which are the distances from two fixed perpendicular lines (the $x$-axis and the $y$-axis). These are called the $x$-coordinate and the $y$-coordinate.

Answer: True (T)

Solution:

A line graph is formed by joining data points. If all the points lie on a single path and are connected, it can form a continuous unbroken line. Such a graph is specifically referred to as a linear graph.

Answer: True (T)

Solution:

The perpendicular distance of a point from the $x$-axis is represented by its vertical position, which is the $y$-coordinate (ordinate). The distance from the $y$-axis is what we call the $x$-coordinate (abscissa).

Answer: False (F)

Solution:

The distance of a point $(x, y)$ from the $y$-axis is equal to the absolute value of its $x$-coordinate. For the point $(3, 5)$, the $x$-coordinate is $3$. Therefore, the distance from the $y$-axis is $3$ units, not $5$.

Answer: False (F)

Solution:

In coordinate geometry, the ordinate refers to the $y$-coordinate, which is the distance of the point from the $x$-axis. The distance from the $y$-axis is called the abscissa ($x$-coordinate).

Answer: False (F)

Solution:

A point is written as an ordered pair $(x, y)$. In the given point $(2, 3)$, the first number $2$ is the $x$-coordinate and the second number $3$ is the $y$-coordinate.

Answer: True (T)

Solution:

The origin is the starting point of the coordinate system where the $x$-axis and $y$-axis intersect. At this location, both the horizontal and vertical distances are zero.

Answer: True (T)

Solution:

The order of numbers in a coordinate pair is essential. The point $(3, 5)$ means $3$ units on the $x$-axis and $5$ units on the $y$-axis, whereas $(5, 3)$ means $5$ units on the $x$-axis and $3$ units on the $y$-axis. These two represent different locations on the graph.

Answer: False (F)

Solution:

Any point on the $x$-axis has no vertical lift or drop from the axis itself. Since the $y$-coordinate measures this vertical displacement, it must be zero for all points on the $x$-axis.

Answer: True (T)

Solution:

By analyzing the properties of each coordinate pair in Column A, we can match them with the correct descriptions in Column B:

The matched pairs are: (1)-(d), (2)-(f), (3)-(e), (4)-(a), (5)-(b), (6)-(c).

Solution:

In a point $(x, y)$, $x$ is the abscissa and $y$ is the ordinate.

The matched pairs are: (a)-(ii), (b)-(iii), (c)-(i), (d)-(v), (e)-(vi), (f)-(iv).

Solution:

By identifying the coordinates $(x, y)$ of each lettered point on the provided graph:

(a) (2, 0): The point lies on the $x$-axis at distance $2$. This is indicated by letter F.

(b) (0, 4): The point lies on the $y$-axis at distance $4$. This is indicated by letter A.

(c) (5, 1): Move $5$ units on the $x$-axis and $1$ unit up. This is indicated by letter H.

(d) (2, 6): Move $2$ units on the $x$-axis and $6$ units up. This is indicated by letter C.

(e) (3, 3): Move $3$ units on the $x$-axis and $3$ units up. This is indicated by letter E.

Final Answers: (a) F, (b) A, (c) H, (d) C, (e) E.

Solution:

By observing the grid, where each division represents 1 unit, we find the coordinates for each letter as follows:

Note: The fractional values are determined by points lying exactly between two grid lines.

Solution:

To plot these points, we draw two perpendicular lines (the axes). The horizontal line is the $x$-axis and the vertical line is the $y$-axis.

1. Point (5, 4): Move $5$ units right and $4$ units up.

2. Point (2, 0): Move $2$ units right on the $x$-axis. Since $y$ is $0$, it stays on the axis.

3. Point (3, 1): Move $3$ units right and $1$ unit up.

4. Point (0, 4): Since $x$ is $0$, move $4$ units up on the $y$-axis.

5. Point (4, 5): Move $4$ units right and $5$ units up.

The points have been successfully plotted on the Cartesian plane.

Given:

A grid map where Road X is the horizontal axis and Road Y is the vertical axis. Horizontal positions are labeled $A, B, C, D, E, F$ and vertical positions are labeled $a, b, c, d, e, f$.

Solution:

By observing the intersections of the vertical and horizontal lines on the map:

(a) The Lions are located at the intersection of column A and row f. Thus, the location is (A, f).

(b) The coordinates (D, f) correspond to the Monkeys. The coordinates (C, d) correspond to the Elephant.

(c) The Toilets are located on Road Y at the point e. Its horizontal position is $0$. Thus, the location is (0, e).

(d) The Canteen is located at the intersection of column C and row c. Thus, the location is (C, c).

Solution:

In a coordinate pair $(x, y)$, the first number represents the $x$-coordinate (also called the abscissa).

(a) In $(7, 3)$, the $x$-coordinate is $7$.

(b) In $(5, 7)$, the $x$-coordinate is $5$.

(c) In $(0, 5)$, the $x$-coordinate is $0$.

Solution:

In a coordinate pair $(x, y)$, the second number represents the $y$-coordinate (also called the ordinate).

(a) In $(3, 5)$, the $y$-coordinate is $5$.

(b) In $(4, 0)$, the $y$-coordinate is $0$.

(c) In $(2, 7)$, the $y$-coordinate is $7$.

Solution (a):

Plot the points $A(1, 2), B(2, 4), C(3, 6),$ and $D(4, 8)$ on the graph. Here, we can observe that for every point, $y = 2x$.

Observation: All the points lie on a single straight line.

Solution (b):

Plot the points $P(1, 1), Q(1, 2), R(2, 2),$ and $S(2, 1)$. Join them in the order $P \to Q \to R \to S \to P$.

Observation: The points do not lie on a straight line. The shape formed is a Square.

Solution (c):

Plot the points $W(4, 2), X(2, 4), Y(3, 3),$ and $Z(5, 4)$. Join them in the order $W \to X \to Y \to Z \to W$.

Observation: The points $(2, 4), (3, 3),$ and $(4, 2)$ are collinear (lie on $x+y=6$). When joined in the specified order, the shape formed is a Quadrilateral (specifically a kite-like shape or triangle depending on the precise order of vertices).

Given:

Relationship: $y = 3x$

Solution:

We create a table by assuming values for $x$ and finding corresponding values for $y$.

The points $(1, 3), (2, 6), (3, 9),$ and $(4, 12)$ are plotted and joined to form a straight line graph.

Given:

Points to plot: $(1, 1), (2, 4), (3, 9), (4, 16)$

Solution:

The relationship between the side ($s$) and the area ($A$) of a square is $A = s^2$.

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

No, it is not a linear graph.

Given:

Cost of $1$ notebook = $\textsf{₹}$ $10$

Solution:

First, we prepare a cost table:

From the graph (by extending the line or following the pattern $y = 10x$):

(a) For $x = 7$ notebooks, the $y$-value is $10 \times 7 = 70$. Thus, the cost is $\textsf{₹}$ $70$.

(b) For a cost of $y = 50$, the $x$-value is $50 / 10 = 5$. Thus, $5$ notebooks can be purchased.

Solution:

(a) The graph is a straight line passing through the origin. This represents Uniform Motion, where the distance increases at a constant rate with respect to time. The object is moving with a constant speed.

(b) The graph initially shows a constant increase in distance (constant speed). Then, the line becomes horizontal (parallel to the time axis). This indicates that the distance remains constant as time progresses, meaning the object has stopped and is at rest.

(c) The graph is a curve whose slope decreases over time. Initially, the distance increases rapidly, but as time goes on, the rate of increase in distance slows down. This represents Non-Uniform Motion where the object is decelerating (slowing down).

Solution (a):

We calculate the values of $y$ using the equation $y = 3x + 1$:

For $x = 2$, $y = 3(2) + 1 = 6 + 1 = 7$.

For $x = 3$, $y = 3(3) + 1 = 9 + 1 = 10$.

Solution (b):

We calculate the values of $y$ using the equation $y = x - 1$:

For $x = 1$, $y = 1 - 1 = 0$.

For $x = 2$, $y = 2 - 1 = 1$.

For $x = 4$, $y = 4 - 1 = 3$.

For $x = 6$, $y = 6 - 1 = 5$.

Solution (a):

Based on the graph, the points follow the relationship where $y$ is equal to $x$.

Solution (b):

Based on the graph, the line starts at $y = 2$ when $x = 0$ and increases by $2$ units for every $1$ unit of $x$. The relationship is $y = 2x + 2$.

Given:

Radius ($r$) and Circumference ($C$) of a circle. Relationship: $C = 2\pi r$.

To Find:

(a) $C$ when $r = 42$

(b) $r$ when $C = 220$

Solution:

We use $\pi = \frac{22}{7}$ to prepare the table:

(a) When radius is 42 units:

By extending the graph line or using the formula:

$C = 2 \times \frac{22}{7} \times 42$

$C = 44 \times 6 = 264\text{ units}$.

(b) When circumference is 220 units:

Locate 220 on the Circumference axis ($y$-axis) and find the corresponding radius on the $x$-axis.

$220 = 2 \times \frac{22}{7} \times r$

$220 = \frac{44}{7} \times r$

$r = \frac{220 \times 7}{44} = 5 \times 7 = 35\text{ units}$.

Answers: (a) 264 units, (b) 35 units.

Solution:

By observing the given double line graph:

(a) The x-axis represents the days of the week and the y-axis represents the maximum temperature in degree Celsius ($^\circ\text{C}$).

(b) On comparing the two weeks:

First week (dotted line) was higher on Sun, Tue, Thu, Fri, and Sat ($5$ days).

Second week (solid line) was higher only on Mon ($1$ day).

Therefore, the temperature was higher on most days in the First week.

(c) The temperature was the same on Wednesday as both lines intersect at $36^\circ\text{C}$.

(d) The difference is maximum on Friday (First week: $39^\circ\text{C}$, Second week: $32^\circ\text{C}$; Difference = $7^\circ\text{C}$).

(e) On Thursday:

Temperature in the First week = $37^\circ\text{C}$.

Temperature in the Second week = $34^\circ\text{C}$.

(f) In the First week (dotted line), the temperature was $35^\circ\text{C}$ on Sunday.

(g) In the Second week (solid line), the highest temperature recorded was $36^\circ\text{C}$ on Wednesday.

Data Extraction:

Solution:

(a) The actual and expected sales were the same in April ($125$ cars).

(b) The difference was maximum in March (Difference = $150 - 75 = 75$ cars).

(c) The difference was least in April (Difference = $0$ cars).

(d) Total Actual Sales (Jan, Feb, Mar) = $75 + 100 + 75 = 250$ cars.

(e) Total Actual Sales in last three months (Apr, May, Jun) = $125 + 100 + 150 = 375$ cars.

$\text{Average} = \frac{375}{3} = 125$ cars.

(f) $\text{Ratio} = \frac{\text{Sales in first three months}}{\text{Sales in last three months}}$

$\text{Ratio} = \frac{250}{375} = \frac{\cancel{250}^2}{\cancel{375}_3} = 2 : 3$

Solution:

(a) The horizontal axis (x-axis) represents Subjects and the vertical axis (y-axis) represents the Marks obtained out of $10$.

(b) In Test I (solid line), the highest mark is $10$ in Maths.

(c) In Test II (dotted line), the least marks are $6$, scored in both English and Hindi.

(d) Sonia did not score the same marks in any subject across both tests (there is no intersection on the data points).

(e) In English Test II, she scored $6$ marks.

(f) Total marks in Test I = $7 + 8 + 10 + 7 + 5 = 37$

Total marks in Test II = $6 + 6 + 8 + 9 + 8 = 37$

Since the total marks in both tests are equal, her performance was equally good in both tests.

(g) She scored full marks ($10/10$) in Maths in Test I.

Solution:

By observing the vertices of each figure on the grid:

Figure I (Square): The vertices are $(1, 1), (3, 0), (4, 2),$ and $(2, 3)$.

Figure II (Triangle): The vertices are $(0, 5), (1, 2),$ and $(2, 4)$.

Figure III (Rhombus): The vertices are $(5, 1), (6, 3), (5, 5),$ and $(4, 3)$.

Figure IV (Polygon): The vertices are $(1, 5), (2, 6), (3, 6), (4, 5), (4, 4),$ and $(2, 4)$.

Solution:

(a) The person started from home at 10 am (Point A where distance is 0).

(b) The first four hours are from $10\text{ am}$ to $2\text{ pm}$. At $2\text{ pm}$, the graph shows a distance of $16\text{ km}$.

(c) From $3\text{ pm}$ to $5\text{ pm}$, the line is horizontal. This means he was at rest (stationary) at a distance of $20\text{ km}$ from home.

(d) He traveled $20\text{ km}$ to reach the farthest point and then $20\text{ km}$ back to return home. Total distance = $20 + 20 = 40\text{ km}$.

(e) The first 8 hours are from $10\text{ am}$ to $6\text{ pm}$. By $3\text{ pm}$ ($5$ hours), he covered $20\text{ km}$. He stayed there until $5\text{ pm}$. Between $5\text{ pm}$ and $6\text{ pm}$, he covered $20 - 16 = 4\text{ km}$ on the return journey. Total distance = $20 + 4 = 24\text{ km}$.

(f) He covered $16\text{ km}$ at 2 pm.

(g) Speeds and Return time:

(i) Speed from A to B: Distance = $20\text{ km}$, Time = $5$ hours ($10\text{ am}$ to $3\text{ pm}$). $\text{Speed} = \frac{20}{5} = 4\text{ km/h}$.

(ii) Speed from B to C: Distance = $0\text{ km}$ (stayed at same place), Time = $2$ hours. $\text{Speed} = 0\text{ km/h}$.

(iii) He returned home (Point D, distance 0) at 10 pm.

Given:

The relationship between variables $p$ and $q$ is defined by the linear equation:

$p = 2q$

Construction Required:

To plot the graph, we first determine some coordinate pairs $(q, p)$ by assuming values for the independent variable $q$:

Solution:

From the graph or by substituting values into the equation:

(a) When $q = 3$:

$p = 2 \times 3 = 6$

Looking at the graph, if we move vertically from $q = 3$, we meet the line at $p = 6$.

(b) When $p = 8$:

$8 = 2q$

$q = \frac{8}{2} = 4$

Looking at the graph, if we move horizontally from $p = 8$, we meet the line at $q = 4$.

Final Answer: (a) $p = 6$, (b) $q = 4$.

Solution:

By observing the given line graph, we can extract the following data:

Sun: $25^\circ\text{C}$, Mon: $28^\circ\text{C}$, Tue: $26^\circ\text{C}$, Wed: $32^\circ\text{C}$, Thu: $29^\circ\text{C}$, Fri: $34^\circ\text{C}$, Sat: $31^\circ\text{C}$.

(a) The graph represents the maximum temperature recorded in a town over the seven days of a specific week.

(b) The lowest point on the graph is at $25^\circ\text{C}$, which corresponds to Sunday.

(c) Moving horizontally from $31^\circ\text{C}$ on the vertical axis, we reach the data point for Saturday.

(d) The highest point on the graph is at $34^\circ\text{C}$. Therefore, the hottest day was Friday.

Solution:

From the graph, we observe points $P(1, 40), Q(3, 120), R(5, 200),$ and $S(6, 240)$. The car is moving at a uniform speed of $40\text{ km/h}$.

(a) To find the distance in 2 hours, move vertically from '$2$' on the time axis to the line. The corresponding distance is $80\text{ km}$.

(b) Point $R$ is at a distance of $200\text{ km}$. Looking down at the time axis, the car takes $5\text{ hours}$ to reach $R$.

(c) For a distance of $80\text{ km}$, we move horizontally from '$80$' on the distance axis. It corresponds to $2\text{ hours}$ on the time axis.

(d) Point $Q$ is at the coordinate $(3, 120)$. Thus, $Q$ is $120\text{ km}$ away from the starting point.

(e) Point $S$ is the last point plotted at $(6, 240)$. The car reaches $S$ after $6\text{ hours}$ of starting.

Given:

Three vertices of a rectangle: $A(1, 2)$, $B(4, 2)$, and $C(1, 4)$.

To Find:

The coordinates of the fourth vertex $D$ to complete rectangle $ABCD$.

Construction Required:

Solution:

In a rectangle, opposite sides are equal and parallel. The vertices are usually listed in order.

1. Side $AB$ is horizontal because both points have the same y-coordinate ($y=2$). Length of $AB = 4 - 1 = 3\text{ units}$.

2. Side $AC$ is vertical because both points have the same x-coordinate ($x=1$). Length of $AC = 4 - 2 = 2\text{ units}$.

3. To complete the rectangle, the fourth vertex $D$ must have the same x-coordinate as $B$ ($x=4$) and the same y-coordinate as $C$ ($y=4$).

Therefore, the coordinates of the fourth point are $D(4, 4)$.

Given:

Three vertices of a rhombus: $A(1, 2)$, $B(3, 4)$, and $C(5, 2)$.

To Find:

Coordinates of vertex $D$, lengths of diagonals, and comparison of diagonal lengths.

Construction Required:

Solution:

In a rhombus, the diagonals bisect each other at right angles.

1. Diagonal $AC$ connects $(1, 2)$ and $(5, 2)$. It is a horizontal line. Length of $AC = 5 - 1 = 4\text{ units}$.

2. The midpoint of $AC$ is $(\frac{1+5}{2}, 2) = (3, 2)$.

3. The vertex $B(3, 4)$ lies on the vertical line passing through the midpoint ($x=3$). It is $2$ units ($4 - 2$) above the midpoint.

4. Vertex $D$ must be $2$ units below the midpoint on the same vertical line. $y$-coordinate of $D = 2 - 2 = 0$.

Thus, the coordinates of $D$ are $(3, 0)$.

Measurement of Diagonals:

Diagonal $AC = 4\text{ units}$.

Diagonal $BD$ connects $(3, 4)$ and $(3, 0)$. Length $BD = 4 - 0 = 4\text{ units}$.

Conclusion:

The lengths of the diagonals are equal ($4\text{ units}$ each). Since the diagonals are equal and perpendicular, this rhombus is specifically a square.

Given:

The points to be located are $P(3, 4)$, $Q(1, 0)$, $R(0, 4)$, and $S(4, 1)$.

Construction Required:

Plot the given points on a Cartesian plane using a suitable scale. Join $P$ to $Q$ and $R$ to $S$ to form two line segments.

Solution:

After plotting the points $P, Q, R,$ and $S$ on the graph paper and drawing the line segments $PQ$ and $RS$, we observe the point where these two lines cross each other.

By observing the graph carefully, the point of intersection lies at a horizontal distance of $2$ units from the $y$-axis and a vertical distance of $2$ units from the $x$-axis.

Therefore, the coordinates of the point of intersection of line segments $PQ$ and $RS$ are $(2, 2)$.

Solution:

By observing the double line graph (Solid line for Vendor A, Dotted line for Vendor B):

(a) On Friday, the solid line (Vendor A) is at $35$ and the dotted line (Vendor B) is at $25$. Thus, Vendor A sold more ice creams.

(b) The sales were the same when the lines intersect. This occurs on Sunday (both sold $50$ ice creams).

(c) Vendor A's sales: Mon ($15$), Tue ($20$), Wed ($28$), Thu ($20$), Fri ($35$), Sat ($30$), Sun ($50$). The maximum increase occurred from Saturday to Sunday (increase of $20$).

(d) The difference is maximum on Thursday (Vendor B: $40$, Vendor A: $20$; Difference = $20$).

(e) For Vendor B (dotted line), the sales were the same on Tuesday and Wednesday (both days $30$ ice creams).

Solution:

(a) Looking at the graph at $8\text{ am}$, the corresponding temperature on the vertical axis is $7^\circ\text{C}$.

(b) Moving horizontally from $3^\circ\text{C}$ on the vertical axis, we meet the graph at $6\text{ am}$.

(c) The temperature fell between $5\text{ am}$ and $6\text{ am}$ (it dropped from $4^\circ\text{C}$ to $3^\circ\text{C}$).

(d) Temperature at $7\text{ am} = 5^\circ\text{C}$ and at $10\text{ am} = 8^\circ\text{C}$. The change is $8 - 5 =$ $3^\circ\text{C}$.

(e) The temperature was constant (horizontal line) between $8\text{ am}$ and $9\text{ am}$ (remained at $7^\circ\text{C}$).

Given:

Data points (Age, Height): $(2, 75), (4, 90), (6, 110), (8, 120), (10, 130)$.

Construction Required:

Solution:

(a) From the graph, by looking at $x = 5$ years (the midpoint between $4$ and $6$), the corresponding height is $100\text{ cm}$.

(b) Height at age $10 = 130\text{ cm}$. Height at age $6 = 110\text{ cm}$. Difference $= 130 - 110 =$ $20\text{ cm}$.

(c) Growth in intervals:

$2-4$ years: $90 - 75 = 15\text{ cm}$

$4-6$ years: $110 - 90 = 20\text{ cm}$

$6-8$ years: $120 - 110 = 10\text{ cm}$

$8-10$ years: $130 - 120 = 10\text{ cm}$

The child grew fastest between $4$ and $6$ years.

Solution:

(a) Sneha makes the least progress between $25$ and $40$ minutes. Reasoning: In this $15$-minute interval, the graph has the lowest slope (is most flat), indicating that she covered only $0.25\text{ km}$ ($1.5 - 1.25$), which is a lower rate than any other period.

(b) Average Speed $= \frac{\text{Total Distance}}{\text{Total Time}}$

Total Distance $= 2\text{ km}$

Total Time $= 55\text{ minutes} = \frac{55}{60}\text{ hours}$

Average Speed $= \frac{2}{55/60} = \frac{2 \times 60}{55} = \frac{120}{55} \approx$ $2.18\text{ km/h}$.

Given:

Table I provides the original coordinates for parallelogram ABCD:

$A(1, 1)$, $B(4, 4)$, $C(8, 4)$, and $D(5, 1)$.

Solution:

1. Completion of Table II (Coordinates for EFGH):

We multiply both the $x$ and $y$ coordinates of ABCD by $0.5$.

2. Completion of Table III (Coordinates for JKLM):

We multiply the $x$-coordinate by $2$ and the $y$-coordinate by $1.5$.

Graphing:

Plot the coordinates from all three tables on the same graph sheet. Additionally, plot the points $(2, 4)$ and $(4, 2)$ and join them to form a line segment.

The resulting graph shows how the original parallelogram ABCD is scaled down (EFGH) and stretched/scaled up (JKLM).

Given:

A line segment passes through points $P(2, 4)$ and $Q(4, 2)$.

Solution:

By extending the line segment $PQ$ on both sides using a ruler:

1. As we move towards the $y$-axis, the line follows the pattern where the sum of $x$ and $y$ is always $6$ ($x + y = 6$).

2. To find the point on the $y$-axis, we set $x = 0$. Thus, $y = 6$. The coordinate is $(0, 6)$.

3. To find the point on the $x$-axis, we set $y = 0$. Thus, $x = 6$. The coordinate is $(6, 0)$.

Final Answer: The line meets the $x$-axis at $(6, 0)$ and the $y$-axis at $(0, 6)$.

Solution:

By observing the graph (Image 37):

(a) The temperature is constant (horizontal line) from $0$ to $20$ seconds and from $50$ seconds onwards. In the initial phase, there was no change for $20$ seconds.

(b) The temperature was changing (rising line) between $20$ seconds and $50$ seconds. Duration $= 50 - 20 =$ $30$ seconds.

(c) The temperature is constant at $0^\circ\text{C}$ from $0$ to $20$ seconds. Thus, it was constant for the first $20$ seconds.

(d) At $20\text{s}$, temp $= 0^\circ\text{C}$. At $50\text{s}$, temp $= 100^\circ\text{C}$. The rate of increase is $\frac{100}{30} = 3.33^\circ\text{C/s}$. After $25$ seconds (which is $5$ seconds into the heating phase): $\text{Temp} = 5 \times 3.33 \approx$ $16.67^\circ\text{C}$.

(e) $1.5\text{ minutes} = 90\text{ seconds}$. The graph shows that after $50$ seconds, the temperature remains constant at $100^\circ\text{C}$. Therefore, the temperature will be $100^\circ\text{C}$. Justification: At $100^\circ\text{C}$, water begins to boil and turn into steam, and its temperature stays constant during this phase change.

Solution:

(a) This is a line graph.

(b) The graph shows the number of people present in a shop at various times throughout the day.

(c) The peak of the graph is at $25$ people, which occurs at $1\text{ pm}$. This is the busiest time.

(d) Assuming the shop opens at $9\text{ am}$, the number of people at the start is $2.5$ on the grid. Practically, this implies $2$ or $3$ people were present/entered at opening.

(e) At $1\text{ pm}$ there are $25$ people and at $2\text{ pm}$ there are $15$ people. At $1:30\text{ pm}$ (the midpoint), there are about $20$ people.

Solution:

(a) He started at $5:30\text{ am}$ and returned to point A (distance 0) at $6:00\text{ pm}$.

(b) From $5:30\text{ am}$ to $6:00\text{ pm}$ is $12.5\text{ hours}$.

(c) Forward journey (to reach $100\text{ km}$): $5:30\text{ am}$ to $2:00\text{ pm} = 8.5\text{ hours}$. Return journey (from $100\text{ km}$ to $0$): $2:00\text{ pm}$ to $6:00\text{ pm} = 4\text{ hours}$. The forward journey was longer.

(d) He was stationary (horizontal lines) from $6\text{ am}$ to $9\text{ am}$ ($3\text{ hrs}$) and from $10\text{ am}$ to $1\text{ pm}$ ($3\text{ hrs}$). Total = $6\text{ hours}$.

(e) Speed is the slope of the graph. Between $1\text{ pm}$ and $2\text{ pm}$, he traveled from $40\text{ km}$ to $100\text{ km}$ (covering $60\text{ km}$ in $1\text{ hour}$). This was his fastest speed ($60\text{ km/h}$).

Solution:

By observing the graph (Solid line = Cyclist I, Dotted line = Cyclist II):

(a) Cyclist II (dotted line) rested when the distance remained constant (horizontal line). This occurred between $8:45\text{ am}$ and $9:00\text{ am}$. The duration of rest was $15\text{ minutes}$.

(b) Let's compare the slopes (speed):

Before rest ($8:00$ to $8:45$): Distance = $10\text{ km}$, Time = $45\text{ min}$. Speed $\approx 0.22\text{ km/min}$.

After rest ($9:00$ to $10:00$): Distance = $20\text{ km}$, Time = $60\text{ min}$. Speed $\approx 0.33\text{ km/min}$.

Since the slope is steeper after the rest, Cyclist II was cycling faster after the rest.

(c) The two cyclists meet at the point of intersection of the two lines. They met at $9:00\text{ am}$.

(d) At $9:00\text{ am}$ (the meeting time), the vertical axis shows that Cyclist II was at a distance of $10\text{ km}$ from town A.

(e) Cyclist II started from Town A and reached Town B. The question likely refers to Cyclist I reaching Town A. Cyclist I reached town A at $9:30\text{ am}$. At this time ($9:30\text{ am}$), Cyclist II was at a distance of $20\text{ km}$ from town A. Since the total distance is $30\text{ km}$, Cyclist II was $30 - 20 =$ $10\text{ km}$ away from town B.

Given:

Departure from home = $07:00\text{ hours}$

Speed to school = $30\text{ km/h}$, Time taken = $0.5\text{ hours}$

Stay at school = $07:30$ to $13:30\text{ hours}$

Return speed = $10\text{ km/h}$

To Find:

Distance to school and the journey graph.

Solution:

Step 1: Calculate Distance to School

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

$\text{Distance} = 30 \times 0.5 = 15\text{ km}$

Step 2: Calculate Return Journey Time

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

$\text{Time} = \frac{15}{10} = 1.5\text{ hours}$

Return time $= 13:30 + 1.5\text{ hours} = 15:00\text{ hours}$

Graph Details:

Distance of school from house is $15\text{ km}$.

Solution:

We represent Year on the horizontal $x$-axis and Gross Profit (in $\textsf{₹}$) on the vertical $y$-axis. A suitable scale for the $y$-axis would be $1\text{ unit} = \textsf{₹} 2,00,000$.

The points to be plotted are: $(1, 17,00,000), (2, 15,50,000), (3, 11,40,000), (4, 12,10,000),$ and $(5, 14,90,000)$. These points are connected with straight line segments.

Construction Required:

Plot Age on the $x$-axis and Percentage of height on the $y$-axis. Use two different types of lines (solid and dotted) to represent Boys and Girls respectively.

Solution:

(a) Maximum height ($100\%$) is achieved by girls at the age of 17 years and by boys at the age of 18 years. Therefore, both have reached their maximum height by the age of 18 years.

(b) Growth between $14$ and $16$ years:

For Boys: $98\% - 92\% = 6\%$ increase.

For Girls: $99.5\% - 98\% = 1.5\%$ increase.

Since the percentage increase is higher for boys, Boys grow faster during this period of puberty.

Solution:

(a) Graph Plotting:

We plot Time (in min) on the $x$-axis and Distance (in km) on the $y$-axis using the coordinates: $(0, 0), (5, 0.5), (10, 1), (15, 1.25), (20, 1.5),$ and $(25, 1.75)$.

(b) Most Progress:

Progress is calculated as distance covered per 5-minute interval:

$0-5\text{ min}$: $0.5 - 0 = 0.5\text{ km}$

$5-10\text{ min}$: $1.0 - 0.5 = 0.5\text{ km}$

$10-15\text{ min}$: $1.25 - 1.0 = 0.25\text{ km}$

$15-20\text{ min}$: $1.5 - 1.25 = 0.25\text{ km}$

$20-25\text{ min}$: $1.75 - 1.5 = 0.25\text{ km}$

Dhruv made the most progress in the time periods $0$ to $5$ minutes and $5$ to $10$ minutes.

Given:

A graph showing the relationship between $x$ and $y$ with points plotted at $x = 1, 2, 3, 4$.

To Find:

The values of $y$ corresponding to $x = 1, 2, 3, 4, 5$ and complete the table.

Solution:

By observing the graph, we can determine the $y$-coordinates for each given $x$-coordinate:

1. At $x = 1$, the point is halfway between $0$ and $5$ on the $y$-axis. So, $y = 1.25$.

2. At $x = 2$, the point corresponds to $5$ on the $y$-axis. So, $y = 5$.

3. At $x = 3$, the point corresponds to $10$ on the $y$-axis. So, $y = 10$.

4. At $x = 4$, the point corresponds to $15$ on the $y$-axis. So, $y = 15$.

5. From $x = 2$ to $x = 4$, the graph is a straight line where $y$ increases by $5$ units for every $1$ unit increase in $x$. Following this linear pattern for $x = 5$:

$y = 15 + 5 = 20$.

The completed table is as follows:

Solution (a):

By observing the peaks and valleys of the line graph:

The highest point on the graph is at the year 1990, where the dropout rate was approximately $6.1\%$.

The lowest point on the graph is at the year 2000, where the dropout rate was $4\%$.

Solution (b):

We look for the first year where the $y$-value is less than $5$.

In 1994, the rate was exactly $5\%$. The next recorded point is in 1996, where the rate is $4.7\%$, which is below $5\%$.

Solution (c):

1. Dropout rate in 2007: The year 2007 lies between 2006 (approx. $4.8\%$) and 2008 (approx. $4.6\%$). Thus, the dropout rate in 2007 was about $4.7\%$.

2. Percentage of students who stayed in 2008: In 2008, the dropout rate was approximately $4.5\%$.

$\text{Percentage of students who stayed} = 100\% - \text{Dropout rate}$

$\text{Percentage of students who stayed} = 100\% - 4.5\% = 95.5\%$

So, about $95.5\%$ of students stayed in high school in 2008.

Solution (a):

By observing the pattern:

Pattern 1: $4$ toothpicks

Pattern 2: $4 + 3 = 7$ toothpicks

Pattern 3: $7 + 3 = 10$ toothpicks

Pattern 4: $10 + 3 = 13$ toothpicks

The rule for the number of toothpicks ($T$) for pattern $n$ is $T = 3n + 1$.

Pattern 5: $3(5) + 1 = 16$

Pattern 6: $3(6) + 1 = 19$

Solution (b):

To make the graph, plot the points $(1, 4), (2, 7), (3, 10), (4, 13), (5, 16), (6, 19)$.

Solution (c):

Using the rule $T = 3n + 1$ or extending the graph:

For Pattern 7: $T = 3(7) + 1 = 22$

For Pattern 8: $T = 3(8) + 1 = 25$

Verification: Drawing pattern 7 would involve 7 squares sharing walls, which requires $1 + (3 \times 7) = 22$ toothpicks.

Solution (d):

Yes, it makes sense to join the points on this graph.

Joining the points with a straight line helps to visualize the linear relationship between the pattern number and the number of toothpicks. It shows that the number of toothpicks increases at a constant rate of 3 for every unit increase in the pattern number. This trend line also helps in predicting the values for future patterns more easily.

Solution (a):

We plot the ordered pairs $(1, 2), (2, 5), (4, 11), (5, 14), (7, 20)$ on the graph.

Solution (b):

By observing the pattern, the relationship is $\text{Output} = (3 \times \text{Input}) - 1$.

1. For Input 3: $\text{Output} = (3 \times 3) - 1 = 8$.

2. For Input 8: $\text{Output} = (3 \times 8) - 1 = 23$.

From the graph, at $x = 3$, $y = 8$ and at $x = 8$, $y = 23$.

Solution (i):

Point $S$ is $3\text{ km}$ East and $5\text{ km}$ North, so its coordinates are $(3, 5)$.

By looking at the map, Point $S(3, 5)$ is on the island. It is not in the same place as Point $A(5, 3)$.

Solution (ii):

Coordinates of Point $C = (7, 5)$.

Coordinates of Point $D = (5, 7)$.

Points $C$ and $D$ are not in the same place as their coordinates are interchanged.

Solution (iii):

Looking at the grid:

Point $(2, 7)$ is on the island (the leftmost protruding part).

Point $(7, 2)$ is in the water (below the main body of the island). This point is labelled as $E$.

Solution (iv):

Point $A$ is at $(5, 3)$. Points $2\text{ km}$ away (horizontal or vertical) are $(3, 3)$, $(7, 3)$, $(5, 1)$, and $(5, 5)$.

Two points on the island are: $(7, 3)$ (which is Point $P$) and $(3, 3)$.

Solution (v):

Point $L$ is at $(9.5, 3)$ and Point $P$ is at $(7.5, 3)$.

The point halfway between them is $\left( \frac{9.5+7.5}{2}, \frac{3+3}{2} \right) = (8.5, 3)$.

The coordinates are $(8.5, 3)$.

Solution (vi):

Three points on the island with $x > 8$ are: $(9, 3)$, $(9, 4)$, and $(10, 5)$.

Solution (vii):

Three points on the island with $y < 4$ are: $(5, 3)$, $(7, 3)$, and $(9, 3)$.

Solution (a):

It makes sense to connect the points because temperature is a continuous variable. The temperature does not jump from one value to another instantly; it changes gradually every second between the recorded hours. Joining the points allows us to estimate the temperature at any given moment between the observations.

Solution (b):

The overall trend shows that the temperature gradually increases from $6\text{ am}$ ($8^\circ\text{C}$) until it reaches its peak of $21^\circ\text{C}$ at $1\text{ pm}$. After $2\text{ pm}$, the temperature starts gradually decreasing throughout the evening, reaching $8^\circ\text{C}$ again by midnight.

Solution (c):

By looking at the line segments connecting the known points:

1. At Noon: The point on the line between $11\text{ am}$ ($17^\circ\text{C}$) and $1\text{ pm}$ ($21^\circ\text{C}$) suggests a temperature of approximately $19^\circ\text{C}$.

2. At 8 pm: The point on the line between $7\text{ pm}$ ($11^\circ\text{C}$) and $9\text{ pm}$ ($9^\circ\text{C}$) suggests a temperature of approximately $10^\circ\text{C}$.

Analysis of the Graph:

Horizontal Axis represents Weight ($x$-coordinate).

Vertical Axis represents Price ($y$-coordinate).

Solution:

(a) To find brands costing more than $D$, we look for points higher than $D$ on the vertical axis. These are brands $F$ and $E$.

(b) The heaviest bag is the one furthest to the right on the horizontal axis. This is brand $D$.

(c) Brands that weigh the same will lie on the same vertical line. These are $B$ and $F$ and $C$ and $E$.

(d) Brands heavier than $B$ are those to the right of point $B$. These are $C, D, \text{ and } E$.

(e) The lightest bags are those furthest to the left.

(f) Bags of the same price lie on the same horizontal line. These are brands $A \text{ and } C$.

Solution (a):

The two variables represented in the graph are Height (on the horizontal axis) and Weight (on the vertical axis).

Solution (b):

By comparing the physical characteristics of the animals with the points on the graph:

1. Point A: Represents the Crocodile. It has a low height (low on $x$-axis) but is very heavy (high on $y$-axis).

2. Point B: Represents the Donkey. It has medium height and medium weight compared to the others.

3. Point C: Represents the Dog. It is the smallest in height and also the lightest in weight (lowest on both axes).

4. Point D: Represents the Ostrich. It is the tallest animal (furthest right on $x$-axis) but has a relatively lower weight than the crocodile or donkey because of its slender build.

Solution:

(a) False. According to the first graph, Car B is older (further right) but also more valuable (higher up) than Car A.

(b) True. In the second graph, Car B is further to the right (larger) and higher up (faster) than Car A.

(c) True. From the second graph, Car B is larger. From the first graph, Car B is older. Thus, the larger car is indeed older.

(d) True. From the second graph, Car B is faster. From the first graph, Car B is older. Thus, the faster car is older.

(e) False. From the first graph, Car B is more valuable. From the second graph, Car B is faster. So, the more valuable car is actually faster.

Given:

A sequence of tile designs where a purple square tile of side length $s$ is surrounded by a border of white tiles. By observing the provided designs:

1. For side length $s = 1$, the total grid is a $5 \times 5$ square. The number of white tiles is $5^2 - 1^2 = 25 - 1 = 24$.

2. For side length $s = 2$, the total grid is a $6 \times 6$ square. The number of white tiles is $6^2 - 2^2 = 36 - 4 = 32$.

3. For side length $s = 3$, the total grid is a $7 \times 7$ square. The number of white tiles is $7^2 - 3^2 = 49 - 9 = 40$.

To Find:

(a) Complete the table for side lengths $1, 2, 3, 4, 5, 10, \text{ and } 100$.

(b) Draw a graph for the first five pairs.

(c) Determine if the points are collinear.

Solution:

By observing the pattern, the total side length of the design is always 4 units more than the side length of the purple tile ($s + 4$).

The number of white tiles ($W$) can be calculated using the formula:

Expanding the expression:

$W = (s^2 + 8s + 16) - s^2$

Using this formula to complete the table:

For $s = 4: W = 8(4) + 16 = 32 + 16 = 48$

For $s = 5: W = 8(5) + 16 = 40 + 16 = 56$

For $s = 10: W = 8(10) + 16 = 80 + 16 = 96$

For $s = 100: W = 8(100) + 16 = 800 + 16 = 816$

(b) Graph:

We plot the points $(1, 24), (2, 32), (3, 40), (4, 48), (5, 56)$ on the coordinate plane where the horizontal axis represents the side length and the vertical axis represents the number of white tiles.

(c) Linearity:

Yes, the points lie on a line. Since the relationship $W = 8s + 16$ is of the form $y = mx + c$, it represents a linear function. The constant increase of $8$ white tiles for every $1$ unit increase in side length confirms that the graph is a straight line.

Final Answer:

The number of white tiles follows the linear rule $W = 8s + 16$, and all plotted points lie on a single straight line.

Solution (a):

Based on the patterns provided:

For Rows ($r$), White Tiles ($w$) follows the rule: $w = 3r - 3$.

For Rows ($r$), Purple Tiles ($p$) follows the rule: $p = \frac{(r-2)(r-3)}{2}$.

1. For $r = 6: w = 3(6) - 3 = 15$; $p = \frac{(4)(3)}{2} = 6$.

2. For $r = 8: w = 3(8) - 3 = 21$; $p = \frac{(6)(5)}{2} = 15$.

Solution (b):

Solution (c):

The graph for white tiles is linear. This is because the number of white tiles increases by a constant amount (6) for every increase of 2 rows. The graph for purple tiles is a curve (quadratic).