CBSE Class X Mathematics (Basic) Question Paper Set 2 2025

Accepted Answer:

We are given that $AB$ is a chord of the larger circle that touches the smaller circle at $C$. Since $AB$ touches the smaller circle at $C$, the radius $OC$ is perpendicular to $AB$. This means $OC$ bisects $AB$ at $C$.

Let $AB = 2x$, so that $AC = x$. In right triangle $OCA$, using the Pythagorean theorem:

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

Substituting the given values:

$(3.5)^2 = (2.1)^2 + x^2$

$12.25 = 4.41 + x^2$

$x^2 = 12.25 - 4.41 = 7.84$

$x = \sqrt{7.84} = 2.8$

Since $AB = 2x$, we get:

$AB = 2(2.8) = 5.6$ cm

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:

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

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

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

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

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

Accepted Answer:

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:

The given arithmetic progression (A.P.) is:

$\frac{13}{3}, \frac{9}{3}, \frac{5}{3}, \dots$

First term ($a$):
$a = \frac{13}{3}$

Common difference ($d$):
$d = \frac{9}{3} - \frac{13}{3} = \frac{-4}{3}$

The general formula for the $n$th term of an A.P. is:

$a_n = a + (n-1) d$

Substituting $n = 15$:

$a_{15} = \frac{13}{3} + (15-1) \times \frac{-4}{3}$

$= \frac{13}{3} + 14 \times \frac{-4}{3}$

$= \frac{13}{3} - \frac{56}{3}$

$= \frac{13 - 56}{3}$

$= \frac{-43}{3}$

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:

The standard form of a quadratic polynomial when its zeroes are $\alpha$ and $\beta$ is:

$p(x) = k(x - \alpha)(x - \beta)$, where $k$ is a constant.

Given zeroes: $\alpha = 0$, $\beta = -2$

Substituting the values:

$p(x) = k(x - 0)(x + 2)$

$p(x) = kx(x + 2)$

$p(x) = k(x^2 + 2x)$

For simplicity, let $k = 1$:

$p(x) = x^2 + 2x$

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:

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:

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:

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:

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

$6 \times 6 = 36$

We need to find the probability of getting an outcome $(a, b)$ such that $b = 2a$.

The possible values for $a$ and $b$ (both between 1 and 6) are:

- If $a = 1$, then $b = 2(1) = 2$ → Valid pair: $(1,2)$
- If $a = 2$, then $b = 2(2) = 4$ → Valid pair: $(2,4)$
- If $a = 3$, then $b = 2(3) = 6$ → Valid pair: $(3,6)$
- If $a = 4$, then $b = 2(4) = 8$ (not possible, since maximum value of $b$ is 6)

Thus, the valid pairs are:
$(1,2), (2,4), (3,6)$

Total favorable outcomes = 3

$\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} $

Accepted Answer:

We are given:

$\sin \theta = \frac{1}{9}$

Using the identity:

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$

Let the opposite side be $1$ and the hypotenuse be $9$.

Using the Pythagorean theorem:

$\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$

$9^2 = 1^2 + \text{adjacent}^2$

$81 = 1 + \text{adjacent}^2$

$\text{adjacent}^2 = 80$

$\text{adjacent} = \sqrt{80} = 4\sqrt{5}$

Now,

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{4\sqrt{5}}$

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:

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:

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:

Sum of angles in $\triangle ABC$:

$\angle A + \angle B + \angle C = 180^\circ$

Substituting given values:

$65^\circ + \angle B + 55^\circ = 180^\circ$

$\angle B = 180^\circ - 120^\circ = 60^\circ$

Since $\triangle ABC \sim \triangle PQR$, their corresponding angles are equal:

$\angle A = \angle P = 65^\circ$,
$\angle B = \angle Q = 60^\circ$,
$\angle C = \angle R = 55^\circ$.

Thus, Assertion (A) is true.

Reason (R) states that the sum of angles in a triangle is $180^\circ$, which is a true mathematical fact. 

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:

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 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 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 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:

Since $AB \parallel DE$ and $BD \parallel EF$, triangles formed by these parallel lines with transversals are similar by the basic proportionality theorem (Thales' theorem).

Using similarity properties in $\triangle DCF$ and $\triangle ACD$:

Since $\triangle DCF \sim \triangle ACD$, we have the property of similar triangles:

$ \frac{DC}{AC} = \frac{CF}{DC} $

Cross multiplying:

$ DC^2 = CF \times AC $

$\boxed{DC^2 = CF \times AC}$

Accepted Answer:

The three friends take steps of lengths $48$ cm, $52$ cm, and $56$ cm respectively.

To find the minimum distance each should walk so that they cover the same distance in complete steps, we need to find the Least Common Multiple (LCM) of $48$, $52$, and $56$.

First, we find the prime factorization:

$48 = 2^4 \times 3$

$52 = 2^2 \times 13$

$56 = 2^3 \times 7$

The LCM is given by taking the highest powers of all prime factors:

$LCM(48, 52, 56) = 2^4 \times 3 \times 7 \times 13$

Calculating step-by-step:

$2^4 = 16$

$16 \times 3 = 48$

$48 \times 7 = 336$

$336 \times 13 = 4368$

Thus, the minimum distance each should walk is $4368$ cm.

Since they need to cover this distance ten times, the total distance each friend should walk is:

$10 \times 4368 = 43680$ cm

Accepted Answer:

\textbf{To Prove:}
$ \left[1 + \frac{1}{\tan^2\theta} \right] + \left[1 + \frac{1}{\cot^2\theta} \right] = \frac{1}{\sin^2\theta - \sin^4\theta} $

\textbf{Proof:}

We start by simplifying each term inside the brackets:

Since $ \tan\theta = \frac{\sin\theta}{\cos\theta} $ and $ \cot\theta $

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:

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:

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 number be $x$.

According to the given condition:

$ x + \frac{1}{x} = \frac{13}{6} $

Multiplying both sides by $x$ to eliminate the fraction:

$ x^2 + 1 = \frac{13}{6} x $

Rearranging the equation:

$ 6x^2 - 13x + 6 = 0 $

Solving the quadratic equation $6x^2 - 13x + 6 = 0$ using the quadratic formula:

$ x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(6)(6)}}{2(6)} $

$ x = \frac{13 \pm \sqrt{169 - 144}}{12} $

$ x = \frac{13 \pm \sqrt{25}}{12} $

$ x = \frac{13 \pm 5}{12} $

Solving for $x$:

$ x = \frac{13 + 5}{12} = \frac{18}{12} = \frac{3}{2} $

$ x = \frac{13 - 5}{12} = \frac{8}{12} = \frac{2}{3} $

Accepted Answer:

Let the height of each pole be $h$ meters.
Let the point on the road be at distances $x$ meters and $(85 - x)$ meters from the two poles.

Using the tangent formula:

For the pole at distance $x$:

$ \tan 60^\circ = \frac{h}{x} $
Since $ \tan 60^\circ = \sqrt{3} $, we get:

$ \sqrt{3} = \frac{h}{x} $
$ h = \sqrt{3} x $ \quad (Equation 1)

For the pole at distance $(85 - x)$:

$ \tan 30^\circ = \frac{h}{85 - x} $
Since $ \tan 30^\circ = \frac{1}{\sqrt{3}} $, we get:

$ \frac{1}{\sqrt{3}} = \frac{h}{85 - x} $
$ h = \frac{(85 - x)}{\sqrt{3}} $ \quad (Equation 2)

Equating both expressions for $h$:

$ \sqrt{3} x = \frac{85 - x}{\sqrt{3}} $

Multiplying both sides by $\sqrt{3}$:

$ 3x = 85 - x $

$ 3x + x = 85 $

$ 4x = 85 $

$ x = \frac{85}{4} = 21.25 $ meters

Substituting $x$ in Equation 1:

$ h = \sqrt{3} \times 21.25 $

Using $ \sqrt{3} \approx 1.732 $:

$ h = 1.732 \times 21.25 = 36.8 $ meters

The height of each pole is $36.8$ meters.
The distances of the point from the poles are $21.25$ meters and $63.75$ meters.

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:

Class Intervals: $0-15, 15-30, 30-45, $

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:

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:

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:

(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} $