CBSE Class X Mathematics (Basic) Question Paper Set 3 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:

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:

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 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 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 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:

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 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:

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:

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:

### 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:

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:

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:

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 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:

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:

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 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:

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} $

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 $