CBSE Class X Mathematics (Basic) Question Paper Set 1 2025

Accepted Answer:

Given A.P.: $ \displaystyle \frac{3}{2}, \frac{1}{2}, \frac{-1}{2}, \frac{-3}{2}, \dots $

First term: $ \displaystyle a = \frac{3}{2} $

Common difference:
$ \displaystyle d = \frac{1}{2} - \frac{3}{2} = -1 $

The general formula for the $ \displaystyle n $th term of an A.P. is:
$ \displaystyle a_n = a + (n-1) d $

Substituting $ \displaystyle n = 22 $:
$ \displaystyle a_{22} = \frac{3}{2} + (22-1)(-1) $

$ \displaystyle = \frac{3}{2} + 21(-1) $

$ \displaystyle = \frac{3}{2} - 21 $

$ \displaystyle = \frac{3}{2} - \frac{42}{2} $

$ \displaystyle = \frac{-39}{2} $

Thus, the 22nd term is $ \displaystyle \frac{-39}{2} $.

Accepted Answer:

To find the x-intercept of the line $ \displaystyle 2x - 3y = 6 $, we set $ \displaystyle y = 0 $ and solve for $ \displaystyle x $.

Substituting $ \displaystyle y = 0 $ into the equation:

$ \displaystyle 2x - 3(0) = 6 $

$ \displaystyle 2x = 6 $

$ \displaystyle x = 3 $

Thus, the line intersects the x-axis at $ \displaystyle (3,0) $.

Accepted Answer:

For a system of equations to be inconsistent, the two lines must be parallel, meaning they have the same slope but different constants.

Given equations:

$ \displaystyle 3x - 7y = 1 $
$ \displaystyle kx + 14y = 6 $

Rewrite in the form $ \displaystyle a_1x + b_1y + c_1 = 0 $ and $ \displaystyle a_2x + b_2y + c_2 = 0 $:

$ \displaystyle a_1 = 3, \quad b_1 = -7, \quad c_1 = 1 $
$ \displaystyle a_2 = k, \quad b_2 = 14, \quad c_2 = 6 $

For inconsistency, the condition is:

$ \displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} $

Substituting values:

$ \displaystyle \frac{3}{k} = \frac{-7}{14} $

$ \displaystyle \frac{3}{k} = -\frac{1}{2} $

Cross multiply:

$ \displaystyle 3 \times 2 = -1 \times k $

$ \displaystyle k = -6 $

Thus, the value of $ \displaystyle k $ for which the system is inconsistent is $ \displaystyle -6 $.

Accepted Answer:

When rolling two dice, the total number of possible outcomes is:

$ \displaystyle 6 \times 6 = 36 $

To find the probability of getting a sum more than 9, we count the favorable outcomes where the sum is 10, 11, or 12.

Possible outcomes:

- Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes
- Sum = 11: (5,6), (6,5) → 2 outcomes
- Sum = 12: (6,6) → 1 outcome

Total favorable outcomes = $ \displaystyle 3 + 2 + 1 = 6 $

Probability:

$ \displaystyle P(\text{sum} > 9) = \frac{6}{36} = \frac{1}{6} $

Thus, the probability of getting a sum more than 9 is $ \displaystyle \frac{1}{6} $.

Accepted Answer:

We are given a rectangle $ABCD$ with its vertices:

$ A(2, -2), B(8,4), C(4,8), D(-2,2) $

To find the length of its diagonal, we use the distance formula:

$ \displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $

Taking diagonal $AC$:

$ \displaystyle AC = \sqrt{(4 - 2)^2 + (8 - (-2))^2} $

$ \displaystyle = \sqrt{(2)^2 + (10)^2} $

$ \displaystyle = \sqrt{4 + 100} $

$ \displaystyle = \sqrt{104} $

$ \displaystyle = 2\sqrt{26} $

Thus, the length of the diagonal is $ 2\sqrt{26} $ units.

Accepted Answer:

Given:
- $ \displaystyle PA $ is a tangent to the circle at point $ \displaystyle A $.
- $ \displaystyle O $ is the center of the circle.
- $ \displaystyle \angle APO = 30^\circ $.
- $ \displaystyle OA = 2.5 $ cm.

Since the radius is perpendicular to the tangent at the point of contact, we have:

$ \displaystyle \angle OAP = 90^\circ $.

In right-angled triangle $ \displaystyle OAP $, using the cosine function:

$ \displaystyle \cos \angle APO = \frac{OA}{OP} $

Substituting the values:

$ \displaystyle \cos 30^\circ = \frac{2.5}{OP} $

Since $ \displaystyle \cos 30^\circ = \frac{\sqrt{3}}{2} $, we get:

$ \displaystyle \frac{\sqrt{3}}{2} = \frac{2.5}{OP} $

Cross multiply:

$ \displaystyle OP = \frac{2.5 \times 2}{\sqrt{3}} $

$ \displaystyle OP = \frac{5}{\sqrt{3}} $

Rationalizing the denominator:

$ \displaystyle OP = \frac{5\sqrt{3}}{3} $ cm.

Thus, the length of $ \displaystyle OP $ is **$ \displaystyle \frac{5}{\sqrt{3}} $ cm**.

Accepted Answer:

The probability of an event happening is given as $ \displaystyle 57\% $.

Since the total probability of all possible outcomes is always $ \displaystyle 1 $ (or $ \displaystyle 100\% $), the probability of the event not happening is:

$ \displaystyle P(\text{not happening}) = 1 - P(\text{happening}) $

$ \displaystyle = 1 - 0.57 $

$ \displaystyle = 0.43 $

Accepted Answer:

Given:
- Radius of the sector, $ \displaystyle r = 7 $ cm
- Arc length, $ \displaystyle l = \frac{22}{3} $ cm
- We need to find the central angle $ \displaystyle \angle AOB $.

The formula for the length of an arc is:

$ \displaystyle l = \frac{\theta}{360^\circ} \times 2\pi r $

Substituting the given values:

$ \displaystyle \frac{22}{3} = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 7 $

Simplifying:

$ \displaystyle \frac{22}{3} = \frac{\theta}{360} \times 44 $

Multiplying both sides by 3:

$ \displaystyle 22 = \frac{\theta}{360} \times 132 $

Dividing by 132:

$ \displaystyle \frac{\theta}{360} = \frac{22}{132} $

$ \displaystyle \frac{\theta}{360} = \frac{1}{6} $

Multiplying by 360:

$ \displaystyle \theta = 60^\circ $

Thus, the central angle $ \displaystyle \angle AOB $ is $ \displaystyle 60^\circ $.

Accepted Answer:

Given that $ \displaystyle DE \parallel BC $, by the **Basic Proportionality Theorem (Thales' Theorem)**, we have:

$ \displaystyle \frac{AD}{DB} = \frac{AE}{EC} $

Substituting the given values:

$ \displaystyle \frac{x + 1}{3} = \frac{2x + 1}{4} $

Cross multiplying:

$ \displaystyle (x + 1) \times 4 = (2x + 1) \times 3 $

$ \displaystyle 4x + 4 = 6x + 3 $

Rearranging:

$ \displaystyle 4 + 1 = 6x - 4x $

$ \displaystyle 5 = 2x $

$ \displaystyle x = \frac{5}{2} $

Thus, the value of $ \displaystyle x $ is **$ \displaystyle \frac{5}{2} $**.

Accepted Answer:

When three coins are tossed together, the total number of possible outcomes is:

$ \displaystyle 2^3 = 8 $

The sample space is:

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

We need to find the probability of getting exactly one head. The favorable outcomes are:

$ \{HTT, THT, TTH\} $

There are $3$ such outcomes.

Thus, the required probability is:

$ \displaystyle P(\text{exactly one head}) = \frac{3}{8} $

Accepted Answer:

We are given two concentric circles centered at $O$, where the chord $AB$ of the larger circle touches the smaller circle at $C$. Given:

$OA = 3.5$ cm,
$OC = 2.1$ cm.

We need to find the length of $AB$.

### Step 1: Identify Right Triangle
Since $AB$ is a chord of the larger circle and touches the smaller circle at $C$, the radius $OC$ is perpendicular to $AB$ at $C$. This means that $OC$ bisects $AB$, so:

$AC = CB = \frac{AB}{2}$.

In the right triangle $OAC$, using the Pythagorean theorem:

$ \displaystyle OA^2 = AC^2 + OC^2 $

Substituting the given values:

$ \displaystyle (3.5)^2 = AC^2 + (2.1)^2 $

$ \displaystyle 12.25 = AC^2 + 4.41 $

$ \displaystyle AC^2 = 12.25 - 4.41 = 7.84 $

$ \displaystyle AC = \sqrt{7.84} = 2.8 $

### Step 2: Find $AB$
Since $AB = 2AC$, we get:

$ \displaystyle AB = 2 \times 2.8 = 5.6 $

### Final Answer:
The length of $AB$ is $5.6$ cm.

Accepted Answer:

We are given the equation:

$ \displaystyle \sqrt{3} \sin \theta = \cos \theta $

### Step 1: Express in Terms of $\tan \theta$
Dividing both sides by $\cos \theta$:

$ \displaystyle \frac{\sqrt{3} \sin \theta}{\cos \theta} = \frac{\cos \theta}{\cos \theta} $

$ \displaystyle \sqrt{3} \tan \theta = 1 $

$ \displaystyle \tan \theta = \frac{1}{\sqrt{3}} $

### Step 2: Solve for $\theta$
We know that:

$ \displaystyle \tan 30^\circ = \frac{1}{\sqrt{3}} $

Thus,

$ \displaystyle \theta = 30^\circ $

Accepted Answer:

In the assumed mean method, the mean $ \displaystyle \bar{x} $ is given by:

$ \displaystyle \bar{x} = A + \frac{\sum fd}{\sum f} $

where:
- $ \displaystyle A $ is the assumed mean,
- $ \displaystyle d = x - A $ (deviation of class mark $ \displaystyle x $ from assumed mean),
- $ \displaystyle f $ is the frequency of the class,
- $ \displaystyle \sum fd $ is the summation of the product of frequency and deviation,
- $ \displaystyle \sum f $ is the total frequency.

Since $ \displaystyle \bar{d} = \frac{\sum fd}{\sum f} $, we can rewrite the formula as:

$ \displaystyle \bar{x} = A + \bar{d} $

If the class interval has a common width $ \displaystyle h $, the deviation is taken as $ \displaystyle d = \frac{x - A}{h} $. In such cases, the mean formula becomes:

$ \displaystyle \bar{x} = A + h \bar{d} $

Thus, the correct answer is:

**(b) $ \displaystyle A + h \bar{d} $**

Accepted Answer:

Given the sum of the first $ \displaystyle n $ terms of an A.P.:

$ \displaystyle S_n = \frac{n}{2} (3n + 1) $

The first term $ \displaystyle a $ is given by:

$ \displaystyle a = S_1 $

Substituting $ \displaystyle n = 1 $ in the given sum formula:

$ \displaystyle S_1 = \frac{1}{2} (3(1) + 1) $

$ \displaystyle = \frac{1}{2} (3 + 1) $

$ \displaystyle = \frac{1}{2} \times 4 $

$ \displaystyle = 2 $

Thus, the first term of the A.P. is $ \displaystyle 2 $.

Accepted Answer:

In $\triangle ABC$, given that $\angle B = 90^\circ$, and

$ \displaystyle \frac{AB}{AC} = \frac{1}{2} $,

we need to find $\cos C$.

### Step 1: Define Trigonometric Ratio
By definition,

$ \displaystyle \cos C = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{AB}{AC} $

### Step 2: Substitute Given Values
Since we are given

$ \displaystyle \frac{AB}{AC} = \frac{1}{2} $,

it follows that:

$ \displaystyle \cos C = \frac{1}{2} $.

### Final Answer:
$ \cos C = \frac{1}{2} $.

Accepted Answer:

$V_{\text{cylinder}} = 450 \, \text{cm}^3$

$V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}} = \frac{1}{3} \times 450 = 150 \, \text{cm}^3$

$V_{\text{empty space}} = V_{\text{cylinder}} - V_{\text{cone}}$

$= 450 - 150 = 300 \, \text{cm}^3$

$\text{Answer: } 300 \, \text{cm}^3$

Accepted Answer:

### Assertion (A):
$ (a + \sqrt{b}) \cdot (a - \sqrt{b}) $ is a rational number, where $a$ and $b$ are positive integers.

### Proof:
Using the identity for the difference of squares:

$ (a + \sqrt{b}) \cdot (a - \sqrt{b}) = a^2 - (\sqrt{b})^2 $

$ = a^2 - b $

Since $a$ and $b$ are positive integers, $a^2 - b$ is also an integer, which is a rational number.

Thus, Assertion (A) is **true**.

---

### Reason (R):
"Product of two irrationals is always rational."

This statement is **false**. The product of two irrational numbers is **not always** rational. For example:

$ \sqrt{2} \times \sqrt{3} = \sqrt{6} $

which is irrational.

Thus, Reason (R) is **false**.

---

### Conclusion:
- Assertion (A) is **true**.
- Reason (R) is **false**.
- Since Reason (R) is incorrect, it does not explain Assertion (A).

**Final Answer:** Assertion is true, but Reason is false.

Accepted Answer:

### Given:
Assertion (A): $\triangle ABC \sim \triangle PQR$, and given angles:

$ \angle A = 65^\circ, \quad \angle C = 60^\circ $

We need to determine whether $\angle Q = 55^\circ$.

### Step 1: Find $\angle B$
Since the sum of the angles in a triangle is $180^\circ$:

$ \angle B = 180^\circ - (\angle A + \angle C) $

$ = 180^\circ - (65^\circ + 60^\circ) $

$ = 180^\circ - 125^\circ $

$ = 55^\circ $

### Step 2: Find $\angle Q$
Since $\triangle ABC \sim \triangle PQR$, their corresponding angles are equal:

$ \angle A = \angle P, \quad \angle B = \angle Q, \quad \angle C = \angle R $

Thus, $\angle Q = \angle B = 55^\circ$, which verifies the assertion.

### Step 3: Verify the Reason
The reason states that the sum of all angles in a triangle is $180^\circ$, which is a fundamental property of triangles.

### Conclusion:

Accepted Answer:

We are given the quadratic equation:

$ \displaystyle 4x^2 - 9x + 3 = 0 $

### Step 1: Identify Coefficients
Comparing with the standard quadratic equation $ax^2 + bx + c = 0$:

$ a = 4, \quad b = -9, \quad c = 3 $

### Step 2: Apply the Quadratic Formula
The quadratic formula is:

$ \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

Substituting the values of $a$, $b$, and $c$:

$ \displaystyle x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(4)(3)}}{2(4)} $

$ \displaystyle = \frac{9 \pm \sqrt{81 - 48}}{8} $

$ \displaystyle = \frac{9 \pm \sqrt{33}}{8} $

### Final Answer: $ \displaystyle x = \frac{9 + \sqrt{33}}{8}, \quad x = \frac{9 - \sqrt{33}}{8} $

Accepted Answer:

We are given the quadratic equation:

$ \displaystyle 3x^2 - 4\sqrt{3}x + 4 = 0 $

### Step 1: Calculate the Discriminant
The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by:

$ \displaystyle \Delta = b^2 - 4ac $

Substituting $a = 3$, $b = -4\sqrt{3}$, and $c = 4$:

$ \displaystyle \Delta = (-4\sqrt{3})^2 - 4(3)(4) $

$ \displaystyle = 48 - 48 = 0 $

### Step 2: Determine the Nature of Roots
Since the discriminant $\Delta = 0$, the quadratic equation has real and equal roots.

Accepted Answer:

### Given:
In trapezium $ABCD$, $AB \parallel DC$, and diagonals $AC$ and $BD$ intersect at $O$.

### To Prove:
$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $

### Proof:
Since $AB \parallel DC$, the diagonals $AC$ and $BD$ intersect each other at $O$. We apply the **Basic Proportionality Theorem (BPT)** or **Thales' theorem**, which states:

"If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio."

Consider $\triangle ADC$:

Since $AB \parallel DC$ and $O$ lies on diagonals $AC$ and $BD$, by the Basic Proportionality Theorem:

$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $

### Conclusion:
Hence, we have proved that:

$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $

Accepted Answer:

We are given a box containing $120$ discs numbered from $1$ to $120$. A disc is drawn at random, and we need to find the probability of the following events:

### (a) Probability of Drawing a Two-Digit Number

A two-digit number ranges from $10$ to $99$. The total count of such numbers is:

$99 - 10 + 1 = 90$

Since there are $120$ discs, the probability is:

$ \displaystyle P(\text{two-digit number}) = \frac{90}{120} = \frac{3}{4} $

### (b) Probability of Drawing a Perfect Square

Perfect squares between $1$ and $120$ are:

$1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121$

Since $121$ is out of range, the valid perfect squares are:

$1, 4, 9, 16, 25, 36, 49, 64, 81, 100$

Total count = $10$

The probability is:

$ \displaystyle P(\text{perfect square}) = \frac{10}{120} = \frac{1}{12} $

### Final Answer:
(a) $ \frac{3}{4} $
(b) $ \frac{1}{12} $

Accepted Answer:

We need to evaluate the expression:

$ \displaystyle \frac{\cos 45^\circ }{\tan 30^\circ + \sin 60^\circ} $

### Step 1: Substitute Trigonometric Values
From trigonometric ratios:

$ \cos 45^\circ = \frac{1}{\sqrt{2}} $
$ \tan 30^\circ = \frac{1}{\sqrt{3}} $
$ \sin 60^\circ = \frac{\sqrt{3}}{2} $

Substituting these values:

$ \displaystyle \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2}} $

### Step 2: Simplify the Denominator
Taking LCM in the denominator:

$ \displaystyle \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \frac{2 + 3}{2\sqrt{3}} = \frac{5}{2\sqrt{3}} $

Thus, the expression simplifies to:

$ \displaystyle \frac{\frac{1}{\sqrt{2}}}{\frac{5}{2\sqrt{3}}} $

### Step 3: Multiply by the Reciprocal
Rewriting:

$ \displaystyle \frac{1}{\sqrt{2}} \times \frac{2\sqrt{3}}{5} = \frac{2\sqrt{3}}{5\sqrt{2}} $

Rationalizing the denominator:

$ \displaystyle \frac{2\sqrt{3} \times \sqrt{2}}{5\sqrt{2} \times \sqrt{2}} = \frac{2\sqrt{6}}{10} $

$ \displaystyle = \frac{\sqrt{6}}{5} $

### Final Answer:
$ \displaystyle \frac{\sqrt{6}}{5} $

Accepted Answer:

We need to verify that:

$ \displaystyle \sin 2A = \frac{2 \tan A}{1 + \tan^2 A} $

for $ \displaystyle A = 30^\circ $.

### Step 1: Compute Left-Hand Side (LHS)
The LHS is:

$ \sin 2A = \sin (2 \times 30^\circ) = \sin 60^\circ $

Using the known trigonometric value:

$ \sin 60^\circ = \frac{\sqrt{3}}{2} $

### Step 2: Compute Right-Hand Side (RHS)
The RHS is:

$ \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} $

Using the known values:

$ \tan 30^\circ = \frac{1}{\sqrt{3}} $, so:

$ \tan^2 30^\circ = \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3} $

Substituting these values into the RHS:

$ \displaystyle \frac{2 \times \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}} $

Simplify the denominator:

$ \displaystyle 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} $

Now the expression becomes:

$ \displaystyle \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} $

Multiplying by the reciprocal:

$ \displaystyle \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} $

Rationalizing the denominator:

$ \displaystyle \frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2} $

### Step 3: Compare LHS and RHS
Since both sides are equal:

$ \displaystyle \sin 2A = \frac{2 \tan A}{1 + \tan^2 A} $

Thus, the identity is verified for $ A = 30^\circ $.

Accepted Answer:

We need to find the HCF of $180, 140$ and $210$ using prime factorization.

### Step 1: Prime Factorization of Each Number

Factorizing $180$:
$180 = 2^2 \times 3^2 \times 5$

Factorizing $140$:
$140 = 2^2 \times 5 \times 7$

Factorizing $210$:
$210 = 2 \times 3 \times 5 \times 7$

### Step 2: Identify Common Factors
The common prime factors among all three numbers are:

- $2$ appears in $180$ as $2^2$, in $140$ as $2^2$, and in $210$ as $2$. The minimum power is $2^1 = 2$.
- $5$ appears in all three numbers as $5^1 = 5$.

### Step 3: Compute the HCF
The HCF is the product of the lowest powers of the common prime factors:

$HCF = 2^1 \times 5^1 = 2 \times 5 = 10$

### Final Answer:
The HCF of $180, 140,$ and $210$ is $10$.

Accepted Answer:

We are given that $ \alpha $ and $ \beta $ are the zeroes of the polynomial:

$ 8x^2 - 5x - 1 = 0 $

### Step 1: Find Sum and Product of Zeroes
Using Vieta’s formulas:

- Sum of the zeroes: $ \alpha + \beta = \frac{-(-5)}{8} = \frac{5}{8} $
- Product of the zeroes: $ \alpha \beta = \frac{-1}{8} $

We need to form a quadratic polynomial whose zeroes are $ \frac{2}{\alpha} $ and $ \frac{2}{\beta} $.

### Step 2: Find the Sum and Product of New Zeroes
Sum of the new zeroes:

$ \frac{2}{\alpha} + \frac{2}{\beta} = 2 \left( \frac{\alpha + \beta}{\alpha \beta} \right) $

Substituting the values:

$ 2 \times \frac{\frac{5}{8}}{\frac{-1}{8}} = 2 \times \left( -\frac{5}{1} \right) = -10 $

Product of the new zeroes:

$ \frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha \beta} $

Substituting the value:

$ \frac{4}{\frac{-1}{8}} = 4 \times (-8) = -32 $

### Step 3: Form the Quadratic Polynomial
A quadratic polynomial with zeroes $ p $ and $ q $ is given by:

$ x^2 - (p+q)x + pq = 0 $

Substituting $ p+q = -10 $ and $ pq = -32 $:

$ x^2 + 10x - 32 = 0 $

### Final Answer:
The required quadratic polynomial is:

$ x^2 + 10x - 32 $

Accepted Answer:

We are given the quadratic polynomial:

$ \displaystyle p(x) = 3x^2 + x - 10 $

### Step 1: Find the Zeroes Using the Quadratic Formula
The roots of a quadratic equation $ ax^2 + bx + c = 0 $ are given by:

$ \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

For $ p(x) = 3x^2 + x - 10 $, we have:

- $ a = 3 $, $ b = 1 $, $ c = -10 $

Substituting these values:

$ \displaystyle x = \frac{-1 \pm \sqrt{(1)^2 - 4(3)(-10)}}{2(3)} $

$ \displaystyle x = \frac{-1 \pm \sqrt{1 + 120}}{6} $

$ \displaystyle x = \frac{-1 \pm \sqrt{121}}{6} $

$ \displaystyle x = \frac{-1 \pm 11}{6} $

Solving for both values:

1. $ \displaystyle x = \frac{-1 + 11}{6} = \frac{10}{6} = \frac{5}{3} $
2. $ \displaystyle x = \frac{-1 - 11}{6} = \frac{-12}{6} = -2 $

Thus, the zeroes of the polynomial are:

$ \displaystyle \frac{5}{3} \quad \text{and} \quad -2 $

### Step 2: Verify the Relationship Between Zeroes and Coefficients
Using Vieta’s formulas:

- Sum of zeroes: $ \displaystyle \alpha + \beta = \frac{-b}{a} $
- Product of zeroes: $ \displaystyle \alpha \beta = \frac{c}{a} $

Substituting the values:

- $ \displaystyle \frac{5}{3} + (-2) = \frac{5}{3} - \frac{6}{3} = -\frac{1}{3} $
- $ \displaystyle \frac{-b}{a} = \frac{-1}{3} $

Since both are equal, the sum is verified.

For the product:

- $ \displaystyle \left( \frac{5}{3} \right) \times (-2) = \frac{-10}{3} $
- $ \displaystyle \frac{c}{a} = \frac{-10}{3} $

Since both are equal, the product is verified.

### Final Answer:
The zeroes of $ p(x) = 3x^2 + x - 10 $ are $ \displaystyle \frac{5}{3} $ and $ -2 $, and the relationships between zeroes and coefficients are verified.

Accepted Answer:

Let the length and breadth of the rectangular park be $ l $ and $ b $ respectively.

### Step 1: Use the Perimeter Equation: The perimeter of a rectangle is given by:

$ 2(l + b) = 100 $

Dividing by 2:

$ l + b = 50 $ ----(1)

### Step 2: Use the Area Equation: The area of a rectangle is given by:

$ l \times b = 600 $ ----(2)

### Step 3: Solve for $ l $ and $ b $: From equation (1), express $ l $ in terms of $ b $:

$ l = 50 - b $

Substituting in equation (2):

$ (50 - b) \times b = 600 $

Expanding:

$ 50b - b^2 = 600 $

Rearranging:

$ b^2 - 50b + 600 = 0 $

### Step 4: Solve the Quadratic Equation: Using the quadratic formula where:

$ a = 1, \quad b = -50, \quad c = 600 $

The roots are given by:

$ b = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(1)(600)}}{2(1)} $

$ b = \frac{50 \pm \sqrt{2500 - 2400}}{2} $

$ b = \frac{50 \pm \sqrt{100}}{2} $

$ b = \frac{50 \pm 10}{2} $

Solving for $ b $:

$ b = \frac{50 + 10}{2} = \frac{60}{2} = 30 $

or

$ b = \frac{50 - 10}{2} = \frac{40}{2} = 20 $

Thus, the possible dimensions are:
- If $ b = 30 $, then $ l = 20 $.
- If $ b = 20 $, then $ l = 30 $.

### Final Answer: The length and breadth of the rectangular park are $ 30 $ m and $ 20 $ m (or vice versa).

Accepted Answer:

To find the least length of the fence that can be measured exactly using any of the given rods, we need to determine the **Least Common Multiple (LCM)** of the rod lengths.

Given rod lengths:
$ \displaystyle 120 $ cm, $ \displaystyle 100 $ cm, $ \displaystyle 150 $ cm

### Step 1: Find the LCM
Prime factorizations:

$ \displaystyle 120 = 2^3 \times 3 \times 5 $
$ \displaystyle 100 = 2^2 \times 5^2 $
$ \displaystyle 150 = 2 \times 3 \times 5^2 $

The LCM is obtained by taking the highest power of each prime factor:

$ \displaystyle LCM = 2^3 \times 3 \times 5^2 $

$ \displaystyle = 8 \times 3 \times 25 $

$ \displaystyle = 600 $ cm

Thus, the least length of the fence that can be measured exactly using any of the rods is **600 cm**.

### Step 2: Number of times each rod is used
- Using the $ \displaystyle 120 $ cm rod: $ \displaystyle \frac{600}{120} = 5 $ times
- Using the $ \displaystyle 100 $ cm rod: $ \displaystyle \frac{600}{100} = 6 $ times
- Using the $ \displaystyle 150 $ cm rod: $ \displaystyle \frac{600}{150} = 4 $ times

### Final Answer:
The least length of the fence is **600 cm**.
- The $ \displaystyle 120 $ cm rod is used **5 times**.
- The $ \displaystyle 100 $ cm rod is used **6 times**.
- The $ \displaystyle 150 $ cm rod is used **4 times**.

Accepted Answer:

We are given a circle with center $O$ and radius $7$ cm. $AB$ and $CD$ are diameters, and $\angle BOD = 30^\circ$. We need to find the area and perimeter of the shaded region.

### Step 1: Area of Sector $BOD$
The area of a sector is given by:

$ \displaystyle A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 $

Substituting values:

$ \displaystyle A_{\text{sector}} = \frac{30^\circ}{360^\circ} \times 3.14 \times (7)^2 $

$ \displaystyle = \frac{1}{12} \times 3.14 \times 49 $

$ \displaystyle = \frac{153.86}{12} = 12.82 \text{ cm}^2 $

### Step 2: Area of $\triangle BOD$
Since $\angle BOD = 30^\circ$, $\triangle BOD$ is an isosceles triangle with sides $OB = OD = 7$ cm. The area of a triangle is:

$ \displaystyle A_{\triangle} = \frac{1}{2} \times OB \times OD \times \sin 30^\circ $

Since $\sin 30^\circ = \frac{1}{2}$:

$ \displaystyle A_{\triangle} = \frac{1}{2} \times 7 \times 7 \times \frac{1}{2} $

$ \displaystyle = \frac{49}{4} = 12.25 \text{ cm}^2 $

### Step 3: Shaded Region Area
Shaded area = Area of sector $BOD$ - Area of $\triangle BOD$:

$ \displaystyle A_{\text{shaded}} = 12.82 - 12.25 = 0.57 \text{ cm}^2 $

### Step 4: Perimeter of Shaded Region
Perimeter consists of arc $BD$ and the two line segments $OB$ and $OD$.

Arc length is given by:

$ \displaystyle L_{\text{arc}} = \frac{\theta}{360^\circ} \times 2\pi r $

$ \displaystyle L_{\text{arc}} = \frac{30^\circ}{360^\circ} \times 2 \times 3.14 \times 7 $

$ \displaystyle = \frac{1}{12} \times 43.96 = 3.66 \text{ cm} $

Total perimeter:

$ \displaystyle P_{\text{shaded}} = OB + OD + \text{Arc } BD $

$ \displaystyle = 7 + 7 + 3.66 = 17.66 \text{ cm} $

Area of shaded region = $0.57$ cm$^2$
Perimeter of shaded region = $17.66$ cm.

Accepted Answer:

To prove:
$ \displaystyle \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} $

Accepted Answer:

Given that the third term of an arithmetic progression (A.P.) is 16:
$ a_3 = a + 2d = 16 \quad \cdots (1) $

It is also given that the seventh term exceeds the fifth term by 12:
$ a_7 = a + 6d $
$ a_5 = a + 4d $

From the given condition:
$ (a + 6d) - (a + 4d) = 12 $
$ 2d = 12 $
$ d = 6 $

Substituting $ d = 6 $ in equation (1):
$ a + 2(6) = 16 $
$ a + 12 = 16 $
$ a = 4 $

Thus, the A.P. has first term $ a = 4 $ and common difference $ d = 6 $.

Now, finding the sum of the first 29 terms:
The sum of the first $ n $ terms of an A.P. is given by:
$ S_n = \frac{n}{2} (2a + (n-1)d) $

Substituting $ n = 29 $, $ a = 4 $, and $ d = 6 $:
$ S_{29} = \frac{29}{2} (2(4) + (29-1)(6)) $
$ = \frac{29}{2} (8 + 168) $
$ = \frac{29}{2} \times 176 $
$ = 29 \times 88 $
$ = 2552 $

Thus, the required A.P. is $ 4, 10, 16, 22, \dots $ and the sum of the first 29 terms is $ 2552 $.

Accepted Answer:

Given the nth term of an arithmetic progression:
$ a_n = 5 + 2n $

To find the sum of the first 20 terms, we use the sum formula:
$ S_n = \frac{n}{2} (2a + (n-1)d) $

Here,
First term $ a = a_1 = 5 + 2(1) = 7 $,
Common difference $ d = a_2 - a_1 $

Accepted Answer:

$ Given equations: $
$ 2x + y = 9 \quad \cdots (1) $
$ x - 2y = 2 \quad \cdots (2) $

To solve graphically, we find two points for each equation.

For the first equation $ 2x + y = 9 $, express $ y $ in terms of $ x $:
$ y = 9 - 2x $

Substituting values:
If $ x = 0 $, then $ y = 9 $ (Point: (0,9))
If $ x = 3 $, then $ y = 3 $ (Point: (3,3))

For the second equation $ x - 2y = 2 $, express $ x $ in terms of $ y $:
$ x = 2 + 2y $

Substituting values:
If $ y = 0 $, then $ x = 2 $ (Point: (2,0))
If $ y = 3 $, then $ x = 8 $ (Point: (8,3))

Now, plot these points on a graph and draw the lines. The point where the two lines intersect gives the solution to the equations.

Accepted Answer:

Let the amounts invested by Nidhi be x and y in Rs.
The formula for simple interest is:
$ SI = \frac{P \times R \times T}{100} $
where P is the principal, R is the rate, and T is the time in years.

$ \text{Given conditions:} $

$ \frac{x \times 6 \times 1}{100} + \frac{y \times 5 \times 1}{100} = 1200 $
$ \Rightarrow \frac{6x}{100} + \frac{5y}{100} = 1200 $
$ \Rightarrow 6x + 5y = 120000 $ \quad \cdots (1)

$ \text{Similarly, for the second case:} $

$ \frac{x \times 3 \times 1}{100} + \frac{y \times 8 \times 1}{100} = 1260 $
$ \Rightarrow \frac{3x}{100} + \frac{8y}{100} = 1260 $
$ \Rightarrow 3x + 8y = 126000 $ \quad \cdots (2)

$ \text{Solving equations (1) and (2):} $

$ \text{Multiply (1) by 3 and (2) by 6:} $

$ 18x + 15y = 360000 $
$ 18x + 48y = 756000 $

$ \text{Subtracting:} $

$ (18x + 48y) - (18x + 15y)$

Accepted Answer:

$ \text{Given:} $
A circle with center O  and radius 4  cm is inscribed in $ \triangle ABC. $
BC is tangent to the circle at D, $\text{ with } BD = 6 \text{ cm and } DC = 10 \text{ cm}. $

Let the tangents drawn from A, B, and C to the incircle be AE, AF, BD, BE, CF, and  CD.
Since tangents drawn from an external point to a circle are equal:

$ \text{Let } AE = AF = x, $ $\quad BD = BE = 6 \text{ cm}, $ $\quad DC = CF = 10 \text{ cm}. $

$ \text{The perimeter of } \triangle ABC \text{ is:} $
AB + BC + CA = (AE + EB) + (BD + DC) + (CF + FA)
$ = (x + 6) + (6 + 10) + (10 + x) $
$ = x + 6 + 16 + 10 + x $
$ = 2x + 32. $

Using the formula for the semi-perimeter:
$ s = \frac{\text{Perimeter}}{2} = \frac{2x + 32}{2} = x + 16. $

$ \text{Since } AE = s - (BD + DC), $
$ AE = (x + 16) - (6 + 10)$

Accepted Answer:

Given:
PA and PB are tangents to a circle with center O.
∠AOB = 120°, OA = OB = 10 cm
Since PA and PB are tangents, they are perpendicular to the radii at the points of tangency.

(a) Find ∠OPA:

Since the tangents from an external point are equal, PA = PB.
Triangle OAP is an isosceles triangle with OA = OP.
∠OAP = ∠OPB.

In triangle AOB, the exterior angle theorem states:
∠OPA = (180° - ∠AOB) / 2 = (180° - 120°) / 2 = 60° / 2 = 30°.

Thus, ∠OPA = 30°.

(b) Find the perimeter of triangle OAP:

We know that OA = OP = 10 cm.
Since PA is a tangent, we use the sine rule:
PA = OA × sin ∠OPA = 10 × sin 30° = 10 × 0.5 = 5 cm.

Perimeter of triangle OAP = OA + OP + PA = 10 + 10 + 5 = 25 cm.

(c) Find the length of chord AB:

In triangle AOB, drop a perpendicular from O to AB at M.
OM bisects AB because OA = OB.
∠AOM = ∠AOB / 2 = 120° / 2 = 60°.

Using trigonometry:
AM = OA × sin ∠AOM = 10 × sin 60° = 10 × √3 / 2 = 5√3 cm.

AB = 2 × AM = 2 × 5√3 = 10√3 cm.

Final Answers:

1. ∠OPA = 30°
2. Perimeter of triangle OAP = 25 cm
3. AB = 10√3 cm

Accepted Answer:

Let the height of the multi-storeyed building be H  meters.
Let the distance between the two buildings be } d meters.

$ \text{Given:} $
$ \text{Height of the smaller building } = 9 \text{ m} $
Angle of depression to the top of the smaller building = \theta_1
Angle of depression to the foot of the smaller building = \theta_2
$ \text{Using } \sqrt{3} \approx 1.73 $

From the top of the multi-storeyed building, two right-angled triangles are formed.

$ \text{In } \triangle ABC, \tan \theta_1 = \frac{(H - 9)}{d} $
$ \Rightarrow H - 9 = d \tan \theta_1 $
$ \Rightarrow H = 9 + d \tan \theta_1 $

$ \text{In } \triangle ABD, \tan \theta_2 = \frac{H}{d} $
$ \Rightarrow H = d \tan \theta_2 $

$ \text{Equating both equations:} $
$ d \tan \theta_2 = 9 + d \tan \theta_1 $
$ d (\tan \theta_2 - \tan \theta_1) = 9 $
$ d = \frac{9}{\tan \theta_2 - \tan \theta_1} $

$ \text{Now, substituting } d \text{ in } H = d \tan \theta_2, $
$ H = \frac{9 \tan \theta_2}{\tan \theta_2 - \tan \theta_1} $

Thus, the height of the multi-storeyed building is H meters, and the distance between the two buildings is } d meters.

Accepted Answer:

$ \text{Given data:} $

Class Intervals: 

Accepted Answer:

Given:
- $\triangle ABC$ is a right-angled triangle with $\angle A = 90^\circ$ and $AB = AC$.
- Points $P$ and $R$ trisect $AB$, and $PQ \parallel RS \parallel AC$.

(a) Show that $\triangle BPQ \sim \triangle BAC$:

Since $PQ \parallel AC$, corresponding angles are equal:
$\angle BPQ = \angle BAC$ and $\angle BQP = \angle BCA$.
Thus, by the AA similarity criterion,
$\triangle BPQ \sim \triangle BAC$.

(b) Prove that $PQ = \frac{1}{3} AC$:

Since $P$ trisects $AB$, we have $AP = \frac{1}{3} AB$.
As $PQ \parallel AC$ and $\triangle BPQ \sim \triangle BAC$,
$ \frac{PQ}{AC} = \frac{BP}{BA} $.
Since $BP = \frac{1}{3} AB = \frac{1}{3} AC$,
$ PQ = \frac{1}{3} AC $.

(c) If $AB = 3$ m, find length $BQ$ and $BS$. Verify that $BQ = \frac{1}{2}BS$.

Since $AB = 3$ m and $P$ trisects $AB$,
$ AP = PB = \frac{3}{3} = 1 $ m.

Since $\triangle BPQ \sim \triangle BAC$,
$ \frac{BQ}{BC} = \frac{BP}{BA} = \frac{1}{3} $.
Thus, $ BQ = \frac{1}{3} BC $.

Similarly, since $\triangle BRS \sim \triangle BAC$,
$ \frac{BS}{BC} = \frac{BR}{BA} = \frac{2}{3} $.
Thus, $ BS = \frac{2}{3} BC $.

Now, verifying $BQ = \frac{1}{2}BS$:
$ BQ = \frac{1}{3} BC $ and $ BS = \frac{2}{3} BC $,
so $ \frac{BQ}{BS} = \frac{\frac{1}{3} BC}{\frac{2}{3} BC} = \frac{1}{2} $.
Hence, $ BQ = \frac{1}{2} BS $, which is verified.

OR

(c) Prove that $BR^2 + RS^2 = \frac{4}{9} BC^2$:

Since $\triangle BRS \sim \triangle BAC$,
$ \frac{BR}{BA} = \frac{2}{3} $, so $ BR = \frac{2}{3} BA $.

Also, $ \frac{RS}{AC} = \frac{2}{3} $, so $ RS = \frac{2}{3} AC $.

Since $BA = AC$, we get
$ BR = RS = \frac{2}{3} BC $.

Now,
$ BR^2 + RS^2 = \left(\frac{2}{3} BC\right)^2 + \left(\frac{2}{3} BC\right)^2 $
$ = \frac{4}{9} BC^2 + \frac{4}{9} BC^2 $
$ = \frac{4}{9} BC^2 $.

Hence, proved.

Accepted Answer:

Given:
- Inner radius of the hemispherical bowl, $r = 10$ cm
- Outer radius of the hemispherical bowl, $R = 10.5$ cm
- The bowl just fits in the cubical box, so its diameter equals the side of the cube

(a) Dimensions of the cubical box:
The side length of the cube is equal to the diameter of the bowl.
$ \text{Side of the cube} = 2R = 2 \times 10.5 $

Accepted Answer:

(i) Determine the distance $OP$:

The coordinates of $O$ are $(0, 0)$ and the coordinates of $P$ are $(8, 6)$. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ on a plane is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For points $O(0, 0)$ and $P(8, 6)$, the distance $OP$ is:

$OP = \sqrt{(8 - 0)^2 + (6 - 0)^2} $