Class 9 Maths Sample Paper Set II (Class 9 - Maths NCERT Exemplar Solutions)

Solution:

A line parallel to the x-axis is represented by an equation where the y-coordinate is constant for all values of $x$. Such an equation is of the general form $y = k$, where $k$ is a constant.

Let us examine the options:

(A) $x + y = 3$ can be written as $y = 3 - x$. This is a linear equation representing a slanting line.

(B) $2x + 3 = 7 \Rightarrow 2x = 4 \Rightarrow x = 2$. This represents a line parallel to the y-axis.

(C) $2y - 3 = y + 1$

Simplifying the equation:

Since the equation results in $y$ being a constant, it represents a line parallel to the x-axis.

(D) $x + 3 = 0 \Rightarrow x = -3$. This represents a line parallel to the y-axis.

Hence, the correct option is (C).

Given:

Polynomial $p(x) = 3x + 5$.

To Find:

The zero of the polynomial.

Solution:

To find the zero of a polynomial, we equate the polynomial to zero and solve for $x$.

Transposing 5 to the right-hand side:

Dividing by 3:

Hence, the zero of the polynomial is $-\frac{5}{3}$.

Correct option is (D).

Solution:

In a Cartesian coordinate system, any point $P$ is represented by coordinates $(x, y)$.

1. The abscissa (x-coordinate) represents the perpendicular distance of the point from the y-axis.

2. The ordinate (y-coordinate) represents the perpendicular distance of the point from the x-axis.

Since the question asks for the abscissa, it is the distance from the y-axis.

Hence, the correct option is (A).

Solution:

Angles are categorized based on their measure:

1. Acute Angle: $0^\circ < \theta < 90^\circ$

2. Right Angle: $\theta = 90^\circ$

3. Obtuse Angle: $90^\circ < \theta < 180^\circ$

4. Straight Angle: $\theta = 180^\circ$

5. Reflex Angle: An angle which is greater than $180^\circ$ but less than $360^\circ$.

Hence, a reflex angle is an angle greater than $180^\circ$.

Correct option is (D).

Given:

Three lines $l, m,$ and $n$.

$l \parallel m$ and $m \parallel n$.

Solution:

According to the theorem in geometry, lines which are parallel to the same line are parallel to each other.

From equations (i) and (ii), since both $l$ and $n$ are parallel to line $m$, we can conclude:

Hence, the correct option is (A).

Given:

In the given figure, $\angle B < \angle A$ and $\angle D > \angle C$ (or $\angle C < \angle D$).

To Find:

The relationship between sides $AD$ and $BC$.

Solution:

In $\triangle OAB$:

We know that in a triangle, the side opposite to the smaller angle is shorter.

Similarly, in $\triangle OCD$:

Since the side opposite to the smaller angle is shorter:

Adding inequations (i) and (ii):

$OA + OD < OB + OC$

From the figure, we observe that $OA + OD = AD$ and $OB + OC = BC$.

Therefore:

$AD < BC$

Hence, the correct option is (C) AD < BC.

Given:

A circle with points $A, B, C, D$ on the circumference. $\angle BAC = 30^\circ$ and $\angle CBD = 70^\circ$.

To Find:

The measure of $\angle BCD$.

Solution:

We know that angles in the same segment of a circle are equal.

Both $\angle BAC$ and $\angle BDC$ are subtended by the same arc $BC$.

Now, consider $\triangle BCD$. Using the angle sum property of a triangle:

$\angle CBD + \angle BDC + \angle BCD = 180^\circ$

Substituting the known values:

$70^\circ + 30^\circ + \angle BCD = 180^\circ$

$100^\circ + \angle BCD = 180^\circ$

Hence, the measure of $\angle BCD$ is 80°.

The correct option is (C).

Given:

Diameter of the cone ($d$) = 10 cm.

Slant height ($l$) = 13 cm.

To Find:

The height ($h$) of the cone.

Solution:

First, we find the radius ($r$):

We know the relationship between slant height, radius, and height of a cone:

Substituting the given values:

$13^2 = 5^2 + h^2$

$169 = 25 + h^2$

$h^2 = 169 - 25$

Taking square root on both sides:

Hence, the height of the cone is 12 cm.

The correct option is (B).

Solution:

A hemisphere is half of a sphere. For a solid hemisphere, the surface area consists of two parts:

1. The Curved Surface Area (CSA): $2\pi r^2$

2. The flat circular base area: $\pi r^2$

Therefore, the total surface area of a solid hemisphere is:

$\text{Total Surface Area} = \text{Curved Surface Area} + \text{Area of Base}$

$\text{Total Surface Area} = 2\pi r^2 + \pi r^2$

Hence, the correct option is (C) 3πr².

Given:

Data: 10, 11, 12, 10, 15, 14, 15, 13, 12, $x$, 9, 7.

Mode of the data = 15.

To Find:

The value of $x$.

Solution:

Mode is the observation that occurs with the highest frequency in a data set.

Let us count the current frequencies of the observations (excluding $x$):

10: appears 2 times

11: appears 1 time

12: appears 2 times

15: appears 2 times

14: appears 1 time

13: appears 1 time

9: appears 1 time

7: appears 1 time

Currently, 10, 12, and 15 all have the same maximum frequency of 2.

Since it is given that the mode is 15, the value 15 must occur more times than any other observation. For this to happen, the frequency of 15 must be at least 3.

This implies that $x$ must be 15.

When $x = 15$, the frequency of 15 becomes 3, making it the unique mode.

Hence, the correct option is (B) 15.

Given:

Rational number $\frac{1}{7} = 0.142857142857...$ (or $0.\overline{142857}$)

To Find:

An irrational number between $\frac{1}{7}$ and $\frac{2}{7}$.

Solution:

First, we find the decimal representation of $\frac{2}{7}$:

We need to find an irrational number $s$ such that:

An irrational number is a non-terminating and non-recurring decimal. We can pick any such decimal between $0.14$ and $0.28$.

Let the irrational number be $0.15015001500015...$

Justification:

1. The number $0.15015001500015...$ is clearly greater than $0.142857...$ and less than $0.285714...$.

2. It is irrational because its decimal expansion is non-terminating and it does not have a repeating block of digits (the number of zeros between the fifteen increases each time).

Given:

Dividend polynomial $p(x) = x^{4} + x^{3} - 2x^{2} + x + 1$

Divisor $g(x) = x - 1$

To Find:

The remainder using the Remainder Theorem.

Solution:

According to the Remainder Theorem, when a polynomial $p(x)$ is divided by $(x - a)$, the remainder is equal to $p(a)$.

Here, the divisor is $(x - 1)$, so $a = 1$.

To find the remainder, we calculate $p(1)$:

$p(1) = 1 + 1 - 2(1) + 1 + 1$

$p(1) = 1 + 1 - 2 + 1 + 1$

Hence, the remainder is 2.

Justification:

The Remainder Theorem provides a direct method to find the remainder of a division involving linear divisors without performing the long division process. Since $p(1) = 2$, it is mathematically certain that the remainder of the division is 2.

Given:

A point $(x, y) = (2, 10)$.

Solution:

To find equations of lines passing through $(2, 10)$, we need to form linear equations in $x$ and $y$ that are satisfied by $x = 2$ and $y = 10$.

Equation 1:

Let us consider the sum of $x$ and $y$:

$x + y = 2 + 10 = 12$

Equation 2:

Let us consider a relationship where $y$ is a multiple of $x$:

$y = 5x \implies 5x - y = 0$

Conclusion:

There are infinitely many such lines passing through the point $(2, 10)$.

Reason: Through a single point in a plane, an infinite number of straight lines can be drawn in different directions. Every such line will have a unique linear equation that is satisfied by the coordinates of that point.

Given:

Points $A(2, 3)$ and $B(2, -1)$.

Solution:

We observe that both points have the same x-coordinate, which is 2.

The equation of the line passing through these points is $x = 2$.

A line with the equation $x = k$ (where $k$ is a constant) is always a vertical line.

Hence, the line is parallel to the y-axis.

Justification:

1. In the Cartesian plane, all points with a constant x-coordinate form a vertical line.

2. A vertical line is perpendicular to the x-axis and therefore parallel to the y-axis.

3. Since the x-coordinates of both $(2, 3)$ and $(2, -1)$ are identical, the distance of every point on this line from the y-axis is a constant 2 units.

Given:

Total number of trials ($n$) = 100

Experimental probability of getting a 6, $P(E) = \frac{2}{5}$

To Find:

The frequency ($f$) of 6 coming up.

Solution:

The formula for experimental probability is:

Let $f$ be the number of times 6 came up. Substituting the values:

By cross-multiplication:

Hence, 6 came up 40 times.

Justification:

Probability represents the ratio of successes to total attempts. If $\frac{2}{5}$ of the total 100 rolls were 6s, then we simply multiply the total trials by the probability fraction: $100 \times \frac{2}{5} = 40$. This satisfies the experimental data provided.

Given:

Two rational numbers $\frac{2}{5}$ and $\frac{3}{5}$.

To Find:

Three rational numbers between them.

Solution:

To find three rational numbers, we multiply the numerator and denominator of both fractions by $(3 + 1) = 4$.

For the first number:

For the second number:

Now, we can choose any three integers between the numerators 8 and 12, which are 9, 10, and 11.

The rational numbers are $\frac{9}{20}$, $\frac{10}{20}$, and $\frac{11}{20}$.

Simplifying where possible:

$\frac{10}{20} = \frac{\cancel{10}^{1}}{\cancel{20}_{2}} = \frac{1}{2}$

The three rational numbers are $\frac{9}{20}$, $\frac{1}{2}$, and $\frac{11}{20}$.

To Factorise:

$54a^{3} - 250b^{3}$

Solution:

First, we take out the common factor 2 from both terms:

The expression inside the bracket can be written as a difference of two cubes:

Using the algebraic identity $x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2})$:

Here, $x = 3a$ and $y = 5b$.

$= 2(3a - 5b)[(3a)^{2} + (3a)(5b) + (5b)^{2}]$

$= 2(3a - 5b)(9a^{2} + 15ab + 25b^{2})$

Hence, the factorised form is $2(3a - 5b)(9a^{2} + 15ab + 25b^{2})$.

Given:

Polynomial $p(y) = 2y^{3} + y^{2} + 4y - 15$

Divisor = $(2y - 3)$

To Find:

Whether $p(y)$ is a multiple of $(2y - 3)$.

Solution:

A polynomial $p(y)$ is a multiple of $(2y - 3)$ if $(2y - 3)$ is a factor of $p(y)$. By Factor Theorem, this is true if the remainder is zero when $p(y)$ is divided by $(2y - 3)$.

First, find the zero of the divisor:

$2y - 3 = 0 \implies 2y = 3 \implies y = \frac{3}{2}$

Now, calculate $p\left(\frac{3}{2}\right)$:

$p\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^{3} + \left(\frac{3}{2}\right)^{2} + 4\left(\frac{3}{2}\right) - 15$

$p\left(\frac{3}{2}\right) = 2\left(\frac{27}{8}\right) + \left(\frac{9}{4}\right) + 6 - 15$

$p\left(\frac{3}{2}\right) = \frac{27}{4} + \frac{9}{4} - 9$

$p\left(\frac{3}{2}\right) = \frac{36}{4} - 9$

Since the remainder is zero, the given polynomial $p(y)$ is indeed a multiple of $(2y - 3)$.

Given:

Equation: $2y = ax + 6$.

Point $(3, 4)$ lies on the graph.

To Find:

Does the point $(6, 5)$ lie on the same graph?

Solution:

Step 1: Find the value of $a$.

Substitute $x = 3$ and $y = 4$ into the equation:

$8 = 3a + 6$

$3a = 8 - 6 = 2$

Now, the equation becomes $2y = \frac{2}{3}x + 6$.

Step 2: Check if $(6, 5)$ satisfies the equation.

Substitute $x = 6$ and $y = 5$ in the new equation:

$\text{RHS} = 2(2) + 6 = 4 + 6 = 10$

Since $\text{LHS} = \text{RHS}$, the point $(6, 5)$ also lies on the graph.

Given:

Points $A(-3, 0)$, $B(5, 0)$, and $C(0, 4)$.

Solution:

Step 1: Plotting and identifying the figure.

When we plot these points and join them, we obtain a Triangle ($\triangle ABC$).

Step 2: Calculating the Area.

The base of the triangle lies on the x-axis between $x = -3$ and $x = 5$.

The height of the triangle is the y-coordinate of vertex $C$ (which lies on the y-axis).

The area of a triangle is given by:

Hence, the figure formed is a triangle and its area is 16 sq units.

Part 1: Area of Triangles in a Trapezium

Given:

A trapezium $ABCD$ where $AB \parallel DC$. The diagonals $AC$ and $BD$ intersect at point $O$.

To Prove:

Proof:

We observe that $\triangle ADC$ and $\triangle BDC$ are on the same base $DC$ and lie between the same parallels $AB$ and $DC$.

We know that triangles on the same base and between the same parallels are equal in area.

From the figure, we can expand these areas as follows:

Substituting these relations into equation (i):

$\text{ar}(\triangle AOD) + \text{ar}(\triangle ODC) = \text{ar}(\triangle BOC) + \text{ar}(\triangle ODC)$

Subtracting the common area $\text{ar}(\triangle ODC)$ from both sides:

Hence proved.

OR (Alternate Question)

Given:

$ABCD$ is a rectangle. Diagonal $AC$ bisects $\angle A$ and $\angle C$.

To Prove:

$ABCD$ is a square.

Proof:

In rectangle $ABCD$, all angles are $90^\circ$.

Since $AC$ bisects $\angle A$:

Since $AC$ bisects $\angle C$:

Now, consider $\triangle ADC$. From (i) and (ii):

In a triangle, sides opposite to equal angles are equal. Therefore:

In a rectangle, opposite sides are already equal ($AB = DC$ and $AD = BC$). From equation (iii), we can conclude that all four sides are equal:

A rectangle with all sides equal is a square.

Hence, ABCD is a square.

Given:

Perimeter of $\triangle PQR$ = 11 cm.

$\angle Q = 60^\circ$ and $\angle R = 45^\circ$.

Construction Required:

1. Draw a line segment $XY$ equal to the perimeter, i.e., $11 \text{ cm}$.

2. At $X$, construct $\angle YXA = 60^\circ$ and at $Y$, construct $\angle XYB = 45^\circ$.

3. Bisect $\angle YXA$ and $\angle XYB$. Let these bisectors intersect at point $P$.

4. Draw the perpendicular bisector of $PX$ and $PY$ to intersect $XY$ at points $Q$ and $R$ respectively.

5. Join $PQ$ and $PR$. $\triangle PQR$ is the required triangle.

Solution:

The resulting triangle $PQR$ has $\angle PQR = 60^\circ$, $\angle PRQ = 45^\circ$ and the sum of its sides $PQ + QR + PR = 11 \text{ cm}$.

Given:

Two sides of the triangle: $a = 18 \text{ cm}$ and $b = 10 \text{ cm}$.

Perimeter ($P$) = $42 \text{ cm}$.

To Find:

Area of the triangle.

Solution:

First, find the third side $c$ of the triangle:

Now, calculate the semi-perimeter ($s$):

Using Heron's Formula for the area:

$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$

$\text{Area} = \sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

$\text{Area} = \sqrt{21 \times 3 \times 11 \times 7}$

$\text{Area} = \sqrt{(21 \times 1) \times (3 \times 7) \times 11}$

$\text{Area} = \sqrt{21 \times 21 \times 11}$

Hence, the area of the triangle is $21\sqrt{11} \text{ cm}^{2}$.

Given:

Diameter of the pillar ($d$) = $50 \text{ cm}$ = $0.5 \text{ m}$.

Height of the pillar ($h$) = $3.5 \text{ m}$.

Rate of painting = $\textsf{₹} 12.50$ per $m^{2}$.

To Find:

Total cost of painting the curved surface.

Solution:

First, we find the radius ($r$):

Curved Surface Area (CSA) of a cylinder = $2\pi rh$.

Now, calculating the total cost:

$\text{Cost} = \text{Area} \times \text{Rate}$

$\text{Cost} = 5.5 \times 12.50$

Hence, the cost of painting the pillar is $\textsf{₹} 68.75$.

OR (Alternate Question)

Given:

Height of the cone ($h$) = $16 \text{ cm}$.

Radius of the cone ($r$) = $12 \text{ cm}$.

To Find:

Total Surface Area (TSA) of the cone.

Solution:

First, we calculate the slant height ($l$):

$l = \sqrt{r^2 + h^2}$

$l = \sqrt{12^2 + 16^2} = \sqrt{144 + 256}$

Total Surface Area (TSA) of a cone = $\pi r(l + r)$.

Hence, the total surface area of the cone is approximately $1206.86 \text{ cm}^{2}$.

Given:

Total number of trials ($n$) = 400.

Frequencies of outcomes: $1 \to 70$, $2 \to 65$, $3 \to 60$, $4 \to 75$, $5 \to 63$, $6 \to 67$.

To Find:

Probability of getting an odd number.

Solution:

An odd number on a die includes the outcomes 1, 3, and 5.

Total frequency of favorable outcomes ($m$):

$m = \text{Freq(1)} + \text{Freq(3)} + \text{Freq(5)}$

$m = 70 + 60 + 63$

Experimental Probability $P(E)$ is given by:

Substituting values from (i) and the given trials:

Hence, the probability of the occurrence of an odd number is 0.4825.

Given:

Parallel sides of the trapezium ($a$ and $b$) = 25 m and 10 m.

Non-parallel sides = 14 m and 13 m.

To Find:

The area of the trapezium field.

Construction Required:

Let the trapezium be $ABCD$ where $AB \parallel CD$, $AB = 25 \text{ m}$ and $CD = 10 \text{ m}$. Draw $CE \parallel DA$ such that $E$ lies on $AB$. Drop a perpendicular $CF \perp AB$.

Solution:

In the construction, $AECD$ is a parallelogram because $AE \parallel DC$ and $AD \parallel EC$.

Opposite sides of a parallelogram are equal:

Now, calculate $EB$:

Consider $\triangle CEB$ with sides $13 \text{ m}$, $14 \text{ m}$, and $15 \text{ m}$.

Semi-perimeter ($s$) of $\triangle CEB$:

Using Heron's Formula for the area of $\triangle CEB$:

$\text{Area} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$

$\text{Area} = \sqrt{21 \times 8 \times 7 \times 6}$

$\text{Area} = \sqrt{7 \times 3 \times 4 \times 2 \times 7 \times 3 \times 2}$

We also know that $\text{Area}(\triangle CEB) = \frac{1}{2} \times \text{Base} \times \text{Height}$:

$84 = \frac{1}{2} \times 15 \times CF$

This $CF$ is the height ($h$) of the trapezium.

Area of Trapezium $= \frac{1}{2} \times (a + b) \times h$

$\text{Area} = \frac{1}{2} \times (25 + 10) \times 11.2$

$\text{Area} = \frac{1}{2} \times 35 \times 11.2 = 35 \times 5.6$

The area of the field is 196 m².

To Find: Represent the given data as a histogram and a frequency polygon.

Solution:

To draw the frequency polygon, we first determine the Class Marks (mid-points) for each interval.

Steps for Construction:

1. Represent Marks Obtained on the horizontal axis (x-axis) and No. of Students on the vertical axis (y-axis).

2. Choose a scale for y-axis: 1 unit = 2 students.

3. Draw the histogram rectangles based on class intervals and frequencies.

4. Join the mid-points of the top of the histogram bars sequentially to form the frequency polygon. Extend it to the imaginary class marks -5 and 85 with zero frequency.

Part 1: Theorem Proof (ASA Congruence Rule)

Given:

Two triangles $\triangle ABC$ and $\triangle DEF$ such that $\angle B = \angle E$, $\angle C = \angle F$ and the included side $BC = EF$.

To Prove:

$\triangle ABC \cong \triangle DEF$

Proof:

To prove the congruence, we consider three cases based on the lengths of $AB$ and $DE$.

Case 1: Let $AB = DE$.

In $\triangle ABC$ and $\triangle DEF$:

By SAS Congruence Rule, $\triangle ABC \cong \triangle DEF$.

Case 2: Let $AB > DE$.

Take a point $P$ on $AB$ such that $PB = DE$. Now, in $\triangle PBC$ and $\triangle DEF$, by SAS rule, $\triangle PBC \cong \triangle DEF$. This implies $\angle PCB = \angle DFE$ (by CPCT). But it is given that $\angle ACB = \angle DFE$. This is only possible if $P$ coincides with $A$. Thus, $AB$ must be equal to $DE$.

Case 3: Let $AB < DE$.

By choosing a point on $DE$ and using the same logic as Case 2, we find that this also leads to a contradiction unless $AB = DE$.

Thus, in all cases, the triangles are congruent. Hence proved.

Part 2: Application (Solution of Fig. 3)

Given:

$AD \perp AB$ and $BC \perp AB$. Also, $AD = BC$.

To Prove:

$CD$ bisects $AB$ (i.e., $OA = OB$).

Solution:

In $\triangle OAD$ and $\triangle OBC$:

By AAS (Angle-Angle-Side) Congruence Rule:

$\triangle OAD \cong \triangle OBC$

By CPCT (Corresponding Parts of Congruent Triangles):

Since $OA = OB$, point $O$ is the midpoint of $AB$. Therefore, $CD$ bisects $AB$.

Part 1: Theorem Proof

Given:

A circle with centre $O$ and two equal chords $AB$ and $CD$.

To Prove:

$\angle AOB = \angle COD$

Proof:

In $\triangle AOB$ and $\triangle COD$:

By SSS (Side-Side-Side) Congruence Rule:

$\triangle AOB \cong \triangle COD$

By CPCT (Corresponding Parts of Congruent Triangles):

Hence proved.

Part 2: Application (Solution of Fig. 4)

Given:

In Figure 4, $AB = CD$ and $\angle COD = 70^\circ$.

To Find:

The measure of $\angle ABO$.

Solution:

Since equal chords subtend equal angles at the centre:

Now, in $\triangle AOB$, $OA = OB$ (Radii of the same circle).

In an isosceles triangle, the angles opposite to equal sides are equal.

Using the Angle Sum Property in $\triangle AOB$:

$\angle AOB + \angle OAB + \angle OBA = 180^\circ$

$70^\circ + \angle ABO + \angle ABO = 180^\circ$

$2\angle ABO = 180^\circ - 70^\circ$

Hence, the measure of $\angle ABO$ is 55°.

Solution (Part 1):

To Factorise: $8x^{3} + 27 y^{3} + 36x^{2} y + 54xy^{2}$

We use the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}$.

We can rewrite the given expression as:

$= (2x)^{3} + (3y)^{3} + 3(2x)^{2}(3y) + 3(2x)(3y)^{2}$

Comparing this with the identity where $a = 2x$ and $b = 3y$:

So, the factorised form is $(2x + 3y)(2x + 3y)(2x + 3y)$.

OR (Alternate Solution):

To Find: Graph of $F = \frac{9}{5}C + 32$ and value of $F$ at $C = 30^\circ$.

Let $C$ be the x-axis and $F$ be the y-axis. We calculate coordinates:

From the graph:

When the temperature is 30°C ($x = 30$):

$F = \frac{9}{5} \times 30 + 32$

$F = 9 \times 6 + 32$

Hence, the temperature in Fahrenheit is 86°F.