Step 1: Understanding the Concept:

The distance of a point \( P(x, y) \) from the x-axis is defined as the absolute value of its y-coordinate, represented as \( |y| \).

This represents the vertical displacement of the point from the horizontal axis.


Step 2: Key Formula or Approach:

Distance from x-axis \( = |y-coordinate| \).


Step 3: Detailed Explanation:

Given the point \( A(4a, 3a) \):

The x-coordinate is \( 4a \).

The y-coordinate is \( 3a \).

By applying the definition, the distance from the x-axis is the magnitude of the y-coordinate.
\[ Distance = |3a| \]

Since distances are scalar magnitudes and in such geometric contexts \( a \) is typically considered a positive parameter unless specified, the distance is \( 3a \).


Step 4: Final Answer:

The distance of the point A from the x-axis is \( 3a \).
Quick Tip: To avoid confusion: The distance from the \textbf{x-axis} is the \textbf{y-value}, and the distance from the \textbf{y-axis} is the \textbf{x-value}.

Step 1: Understanding the Concept:

Natural numbers are classified based on the number of factors they possess.

A prime number is a natural number \( > 1 \) that has exactly two distinct factors: 1 and itself.

A composite number is a natural number \( > 1 \) that has more than two factors.


Step 2: Detailed Explanation:

The number 1 has only one factor, which is 1 itself.

Because it does not have exactly two distinct factors, it fails the definition of a prime number.

Because it does not have more than two factors, it fails the definition of a composite number.

Historically and mathematically, 1 is treated as a unit, serving as the multiplicative identity, and is excluded from both categories.


Step 3: Final Answer:

The number 1 is neither prime nor composite.
Quick Tip: The number 2 is the smallest prime number and the only even prime number. 1 is often called a "unit" in number theory.

Step 1: Understanding the Concept:

In a right-angled triangle, the cotangent ratio is the ratio of the adjacent side to the opposite side. The cosine ratio is the ratio of the adjacent side to the hypotenuse.


Step 2: Key Formula or Approach:

1. \(\cot \theta = \frac{Base (B)}{Perpendicular (P)}\)

2. Pythagoras Theorem: \(H = \sqrt{P^2 + B^2}\)

3. \(\cos \theta = \frac{B}{H}\)


Step 3: Detailed Explanation:

Given \(\cot \theta = 3\), we can write this as \(\frac{3}{1}\).

Let Base (\(B\)) \( = 3k \) and Perpendicular (\(P\)) \( = 1k \).

Calculate Hypotenuse (\(H\)):
\[ H = \sqrt{(3k)^2 + (1k)^2} = \sqrt{9k^2 + k^2} = \sqrt{10k^2} = k\sqrt{10} \]

Now, find \(\cos \theta\):
\[ \cos \theta = \frac{Base}{Hypotenuse} = \frac{3k}{k\sqrt{10}} = \frac{3}{\sqrt{10}} \]


Step 4: Final Answer:

The value of \(\cos \theta\) is \(\frac{3}{\sqrt{10}}\).
Quick Tip: Alternatively, use the identity \(cosec^2 \theta = 1 + \cot^2 \theta\).
\(cosec^2 \theta = 1 + 3^2 = 10 \implies \sin \theta = \frac{1}{\sqrt{10}}\).
Then, \(\cos \theta = \cot \theta \times \sin \theta = 3 \times \frac{1}{\sqrt{10}} = \frac{3}{\sqrt{10}}\).

Step 1: Understanding the Concept:

The last digit (unit digit) of the power of a number depends on the cyclic property of that base number's units digit.


Step 2: Detailed Explanation:

Consider the powers of 5 for natural numbers \( n \):

For \( n = 1 \), \( 5^1 = 5 \)

For \( n = 2 \), \( 5^2 = 25 \)

For \( n = 3 \), \( 5^3 = 125 \)

For \( n = 4 \), \( 5^4 = 625 \)

Observation: Regardless of the power, the last digit is always 5.

This is because when we multiply a number ending in 5 by another 5, the result always ends in \( 5 \times 5 = 25 \), which has 5 as its unit digit.


Step 3: Final Answer:

For any natural number \( n \), \( 5^n \) ends with the digit 5.
Quick Tip: Numbers ending in 0, 1, 5, and 6 will always result in the same unit digit (0, 1, 5, and 6 respectively) when raised to any positive integer power.

Step 1: Understanding the Concept:

Find the specific value of angle \( A \) using the given equation and substitute it into the required expression.


Step 2: Detailed Explanation:

Given: \( 2 \sin A = 1 \implies \sin A = \frac{1}{2} \).

We know from the trigonometric table that \( \sin 30^\circ = \frac{1}{2} \).

So, \( A = 30^\circ \).

Now, we need to evaluate \( \tan A + \cot A \):
\[ Value = \tan 30^\circ + \cot 30^\circ \]

Substitute the standard values: \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) and \( \cot 30^\circ = \sqrt{3} \).
\[ Value = \frac{1}{\sqrt{3}} + \sqrt{3} \]

Taking LCM:
\[ Value = \frac{1 + (\sqrt{3} \times \sqrt{3})}{\sqrt{3}} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]


Step 3: Final Answer:

The value is \( \frac{4}{\sqrt{3}} \).
Quick Tip: Use the identity \( \tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A} \) to solve it without finding the angle directly if preferred.

Step 1: Understanding the Concept:

The Least Common Multiple (LCM) of two numbers is the smallest number that is a multiple of both. If one number is a multiple of the other, then the larger number is the LCM.


Step 2: Detailed Explanation:

Let's check the relationship between 960 and 240.

Divide 960 by 240:
\[ \frac{960}{240} = 4 \]

Since 240 exactly divides 960 (it goes in 4 times), 960 is a multiple of 240.

Alternatively, by prime factorization:
\( 240 = 2^4 \times 3 \times 5 \)
\( 960 = 2^6 \times 3 \times 5 \)

LCM is the product of prime factors with the highest powers:
\[ LCM = 2^6 \times 3 \times 5 = 64 \times 15 = 960 \]


Step 3: Final Answer:

The LCM is 960.
Quick Tip: Shortcut: If \( a \) is a multiple of \( b \), then \( LCM(a, b) = a \) and \( HCF(a, b) = b \).

Step 1: Understanding the Concept:

This problem involves basic applications of trigonometry in a right-angled triangle where the tower is the perpendicular side and the distance on the ground is the base.


Step 2: Key Formula or Approach:
\[ \tan \theta = \frac{Perpendicular (Height)}{Base (Distance)} \]


Step 3: Detailed Explanation:

Let \( h \) be the height of the tower.

The distance from the foot of the tower is given as 60 m.

The angle of elevation \( \theta = 45^\circ \).

In the right-angled triangle formed:
\[ \tan 45^\circ = \frac{h}{60} \]

We know that \( \tan 45^\circ = 1 \).
\[ 1 = \frac{h}{60} \implies h = 60 m \]


Step 4: Final Answer:

The height of the tower is 60 m.
Quick Tip: In any right-angled triangle, if the angle is \( 45^\circ \), the perpendicular and base are always equal because \( \tan 45^\circ = 1 \).

Step 1: Understanding the Concept:

The zeroes of a polynomial are the values of \( x \) for which the value of the polynomial becomes zero. The number of zeroes corresponds to the degree of the polynomial.


Step 2: Detailed Explanation:

The given polynomial is \( p(x) = (x - 2)(x + 3) \).

If we multiply the factors, we get:
\[ p(x) = x^2 + 3x - 2x - 6 = x^2 + x - 6 \]

The degree of this polynomial is 2, which indicates it is a quadratic polynomial.

To find the zeroes, set \( p(x) = 0 \):
\( (x - 2)(x + 3) = 0 \)

This implies either \( x - 2 = 0 \) or \( x + 3 = 0 \).

So, \( x = 2 \) and \( x = -3 \) are the two distinct zeroes.


Step 3: Final Answer:

The polynomial has two zeroes.
Quick Tip: For a polynomial in factored form \( (x-a)(x-b)(x-c)... \), the number of factors gives the number of zeroes.

Step 1: Understanding the Concept:

Tangents from an external point are equal. Radius is perpendicular to the tangent at the point of contact. Triangle formed by the radii and the chord is isosceles.


Step 2: Detailed Explanation:

In \(\triangle OAB\), \(OA = OB\) (radii of the same circle).

Therefore, \(\triangle OAB\) is an isosceles triangle, so \(\angle OBA = \angle OAB = 15^\circ\).

In \(\triangle OAB\), the sum of angles is \(180^\circ\):
\[ \angle AOB + \angle OAB + \angle OBA = 180^\circ \]
\[ \angle AOB + 15^\circ + 15^\circ = 180^\circ \implies \angle AOB = 150^\circ \]

Radius \(OA\) is perpendicular to tangent \(PA\), so \(\angle OAP = 90^\circ\). Similarly, \(\angle OBP = 90^\circ\).

In quadrilateral \(OAPB\), the sum of interior angles is \(360^\circ\):
\[ \angle APB + \angle OAP + \angle OBP + \angle AOB = 360^\circ \]
\[ \angle APB + 90^\circ + 90^\circ + 150^\circ = 360^\circ \]
\[ \angle APB + 330^\circ = 360^\circ \implies \angle APB = 30^\circ \]


Step 3: Final Answer:
\(\angle APB\) equals \(30^\circ\).
Quick Tip: Use the property that the angle between two tangents is supplementary to the angle between the radii through the points of contact: \(\angle APB + \angle AOB = 180^\circ\).

Step 1: Understanding the Concept:

For a quadratic polynomial \( ax^2 + bx + c \), the sum of zeroes \( \alpha + \beta = -\frac{b}{a} \) and product of zeroes \( \alpha\beta = \frac{c}{a} \).


Step 2: Detailed Explanation:

Given polynomial: \( f(x) = px^2 - 2x + 3p \).

Comparing with \( ax^2 + bx + c \): \( a = p, b = -2, c = 3p \).

Sum of zeroes, \( \alpha + \beta = -\frac{(-2)}{p} = \frac{2}{p} \).

Product of zeroes, \( \alpha\beta = \frac{3p}{p} = 3 \).

Given condition: \( \alpha + \beta = \alpha\beta \).

Substitute the expressions found:
\[ \frac{2}{p} = 3 \]
\[ 3p = 2 \implies p = \frac{2}{3} \]


Step 3: Final Answer:

The value of \( p \) is \( \frac{2}{3} \).
Quick Tip: Notice how \( p \) cancels out in the product calculation. This simplifies the equation significantly.

Step 1: Understanding the Concept:

The radii through the points of contact and the tangents at those points form a cyclic quadrilateral where opposite angles are supplementary.


Step 2: Detailed Explanation:

In quadrilateral \(OAPB\):
\(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\) (radius is perpendicular to tangent).

The sum of angles in a quadrilateral is \(360^\circ\):
\[ \angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ \]
\[ 130^\circ + 90^\circ + \angle APB + 90^\circ = 360^\circ \]
\[ \angle APB + 310^\circ = 360^\circ \]
\[ \angle APB = 360^\circ - 310^\circ = 50^\circ \]


Step 3: Final Answer:
\(\angle APB\) is equal to \(50^\circ\).
Quick Tip: Shortcut: \(\angle APB = 180^\circ - \angle AOB\). Just subtract the given central angle from \(180^\circ\).

Step 1: Understanding the Concept:

Consistent equations have solutions. Dependent consistent equations have infinitely many solutions because the lines are coincident.


Step 2: Detailed Explanation:

- If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines intersect at one point (Consistent and Independent).

- If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel (Inconsistent).

- If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines coincide, meaning every point on one line is a solution for both (Consistent and Dependent).


Step 3: Final Answer:

For the system to be consistent and dependent, the condition is \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
Quick Tip: Think of "Dependent" as the two equations being identical or one being a multiple of the other.

Step 1: Understanding the Concept:

The volume of the material used in a hollow object is the difference between the outer volume and the inner volume.


Step 2: Key Formula or Approach:
\[ Volume of hemisphere = \frac{2}{3}\pi r^3 \]
\[ Volume of material = \frac{2}{3}\pi (R^3 - r^3) \]


Step 3: Detailed Explanation:

Outer radius (\(R\)) \( = 6 \) cm.

Thickness \( = 1 \) cm.

Inner radius (\(r\)) \( = Outer radius - thickness = 6 - 1 = 5 \) cm.

Volume of steel:
\[ V = \frac{2}{3}\pi(6^3 - 5^3) \]
\[ V = \frac{2}{3}\pi(216 - 125) \]
\[ V = \frac{2}{3}\pi(91) = \frac{182}{3}\pi cm^3 \]


Step 4: Final Answer:

The volume of steel used is \( \frac{182}{3}\pi cm^3 \).
Quick Tip: Be careful to find the inner radius correctly by subtracting thickness from outer radius, not adding it.

Step 1: Understanding the Concept:

A sequence is an Arithmetic Progression (A.P.) if the difference between consecutive terms is constant.


Step 2: Detailed Explanation:

Check differences for each option:

(A) \( d = \frac{5}{2} - 2 = 0.5 \); \( 3 - \frac{5}{2} = 0.5 \). Constant \( d = 0.5 \). (It is an A.P.)

(B) \( d = -3.2 - (-1.2) = -2 \); \( -5.2 - (-3.2) = -2 \). Constant \( d = -2 \). (It is an A.P.)

(C) \(\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, \dots \) differences are \( \sqrt{2}, \sqrt{2}, \dots \) (It is an A.P.)

(D) \( 1, 9, 25, 49, \dots \)
\( 9 - 1 = 8 \)
\( 25 - 9 = 16 \)

The differences are not constant (\( 8 \neq 16 \)).


Step 3: Final Answer:

The sequence \( 1^2, 3^2, 5^2, 7^2, \dots \) is not an A.P.
Quick Tip: In option (C), simplify radical terms like \( \sqrt{8} = 2\sqrt{2} \) to see the common difference more easily.

Step 1: Understanding the Concept:

Area of a circle is \( \pi r^2 \). Area of a semicircle is half of the circle's area.


Step 2: Key Formula or Approach:
\[ Radius (r) = \frac{d}{2} \]
\[ Area of semicircle = \frac{1}{2} \pi r^2 \]


Step 3: Detailed Explanation:

Substitute \( r = \frac{d}{2} \) into the area formula:
\[ Area = \frac{1}{2} \pi \left(\frac{d}{2}\right)^2 \]
\[ Area = \frac{1}{2} \pi \left(\frac{d^2}{4}\right) \]
\[ Area = \frac{\pi d^2}{8} \]


Step 4: Final Answer:

The area of a semicircle with diameter 'd' is \( \frac{\pi d^2}{8} \).
Quick Tip: Area of a full circle in terms of diameter is \( \frac{\pi d^2}{4} \). For a semicircle, just multiply by \( \frac{1}{2} \).

Step 1: Understanding the Concept:

In a triangle, if a line is parallel to one side, the smaller triangle formed is similar to the larger triangle (by AA similarity).


Step 2: Key Formula or Approach:

Similarity of triangles \( \triangle ADE \sim \triangle ABC \):
\[ \frac{AD}{AB} = \frac{DE}{BC} \]


Step 3: Detailed Explanation:

Given: \( AD = 5 \) cm, \( DB = 2.5 \) cm.

So, \( AB = AD + DB = 5 + 2.5 = 7.5 \) cm.

Using the similarity property:
\[ \frac{5}{7.5} = \frac{8}{BC} \]

Convert \( \frac{5}{7.5} \) to a simpler fraction: \( \frac{5 \times 10}{7.5 \times 10} = \frac{50}{75} = \frac{2}{3} \).
\[ \frac{2}{3} = \frac{8}{BC} \]
\[ 2 \times BC = 24 \]
\[ BC = 12 cm \]


Step 4: Final Answer:

The length of BC is 12 cm.
Quick Tip: Avoid using \( \frac{AD}{DB} = \frac{DE}{BC} \). This is a common mistake. BPT relates segments of sides, but Similarity relates the full sides of the triangles.

Step 1: Understanding the Concept:

There is an empirical relationship between the three measures of central tendency: Mean, Median, and Mode.


Step 2: Key Formula or Approach:
\[ Mode = 3 \times Median - 2 \times Mean \]


Step 3: Detailed Explanation:

Given: Mean \( = 43 \), Median \( = 43.4 \).

Applying the empirical formula:
\[ Mode = 3(43.4) - 2(43) \]
\[ Mode = 130.2 - 86 \]
\[ Mode = 44.2 \]


Step 4: Final Answer:

The mode is 44.2.
Quick Tip: Remember the formula by noting the order: 3 (Median) comes first, then subtract 2 (Mean).

Step 1: Understanding the Concept:

A composite number is a natural number greater than 1 that is not prime.


Step 2: Detailed Explanation:

Total outcomes \( = 25 \) (numbers from 1 to 25).

Prime numbers in this range are: 2, 3, 5, 7, 11, 13, 17, 19, 23. (Total \( = 9 \) prime numbers).

The number 1 is neither prime nor composite.

Total numbers that are NOT composite \( = Primes + \{1\} = 9 + 1 = 10 \).

Number of composite numbers \( = 25 - 10 = 15 \).

The composite numbers are: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25.
\[ Probability = \frac{Number of composite numbers}{Total numbers} = \frac{15}{25} \]


Step 3: Final Answer:

The probability is \( \frac{15}{25} \).
Quick Tip: Don't forget to exclude "1" when identifying composite numbers. 1 is the "odd one out" in probability problems involving primes/composites.

Step 1: Understanding the Concept:

Join two cubes to form a cuboid and calculate its surface area using the correct formula. Check if the given formula in the Reason is correct.


Step 2: Detailed Explanation:

Evaluating Assertion (A):

When two cubes of side 4 cm are joined end-to-end:

Length (\(l\)) \( = 4 + 4 = 8 \) cm.

Breadth (\(b\)) \( = 4 \) cm.

Height (\(h\)) \( = 4 \) cm.

Surface area of cuboid \( = 2(lb + bh + hl) \)
\[ Area = 2(8 \times 4 + 4 \times 4 + 4 \times 8) \]
\[ Area = 2(32 + 16 + 32) = 2(80) = 160 cm^2 \]

Assertion (A) is True.

Evaluating Reason (R):

The standard formula for surface area of a cuboid is \( 2(lb + bh + hl) \).

The reason states the area is \( (lb + bh + hl) \), which is missing the factor of 2.

Reason (R) is False.


Step 3: Final Answer:

Assertion (A) is true but Reason (R) is false.
Quick Tip: Look very closely at formulas in Reason statements. Missing a small constant factor like "2" makes the entire statement false.

Step 1: Understanding the Concept:

The mean of a sequence of numbers is the sum of the numbers divided by the count.


Step 2: Detailed Explanation:

Evaluating Reason (R):

The sum of the first \( n \) natural numbers (1, 2, 3, ..., n) is given by the arithmetic progression sum formula:
\[ S_n = \frac{n(n + 1)}{2} \]

This is a standard true mathematical fact. Reason (R) is True.

Evaluating Assertion (A):

Mean \( = \frac{Sum of first n natural numbers}{n} \)
\[ Mean = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

The assertion claims the mean is \( \frac{n-1}{2} \), which is incorrect.

Assertion (A) is False.


Step 3: Final Answer:

Assertion (A) is false but Reason (R) is true.
Quick Tip: Mean of natural numbers can also be thought of as the middle value. For numbers 1, 2, ..., n, the middle value is \( \frac{1+n}{2} \).

Step 1: Understanding the Concept:

The distance between two points in a Cartesian plane is calculated using the distance formula, which is derived from the Pythagorean theorem.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

Given points are \( (x_1, y_1) = (4, p) \) and \( (x_2, y_2) = (1, 0) \).

The distance \( d \) is given as 5.

Substituting the values into the formula:
\[ 5 = \sqrt{(1 - 4)^2 + (0 - p)^2} \]
\[ 5 = \sqrt{(-3)^2 + (-p)^2} \]
\[ 5 = \sqrt{9 + p^2} \]

Squaring both sides of the equation:
\[ 25 = 9 + p^2 \]
\[ p^2 = 25 - 9 \]
\[ p^2 = 16 \]

Taking the square root:
\[ p = \pm \sqrt{16} \implies p = \pm 4 \]


Step 4: Final Answer:

The possible values of \( p \) are 4 or \(-4\).
Quick Tip: Always remember to consider both positive and negative roots when squaring both sides of an equation involving variables.

Step 1: Understanding the Concept:

A quadrilateral is cyclic if the sum of its opposite angles is \( 180^\circ \). In a circle, the tangent at any point is perpendicular to the radius through the point of contact.


Step 2: Key Formula or Approach:

Radius \( \perp \) Tangent at point of contact.

Sum of opposite angles in cyclic quadrilateral \( = 180^\circ \).


Step 3: Detailed Explanation:

In quadrilateral \( PQOR \):
\( OQ \) is the radius and \( PQ \) is the tangent at point \( Q \).

Therefore, \( \angle OQP = 90^\circ \) (Radius \( \perp \) tangent).

Similarly, \( OR \) is the radius and \( PR \) is the tangent at point \( R \).

Therefore, \( \angle ORP = 90^\circ \).

Now, let's find the sum of opposite angles \( \angle OQP \) and \( \angle ORP \):
\[ \angle OQP + \angle ORP = 90^\circ + 90^\circ = 180^\circ \]

We know that in any quadrilateral, the sum of all interior angles is \( 360^\circ \).

So, \( \angle QOR + \angle QPR = 360^\circ - (\angle OQP + \angle ORP) = 360^\circ - 180^\circ = 180^\circ \).

Since both pairs of opposite angles sum up to \( 180^\circ \), the quadrilateral \( PQOR \) is cyclic.


Step 4: Final Answer:

The quadrilateral \( PQOR \) is cyclic because the sum of its opposite angles is supplementary.
Quick Tip: For any quadrilateral where the vertices lie on a circle, or where the opposite angles are supplementary, the cyclic property holds.

Step 1: Understanding the Concept:

For a quadratic polynomial \( ax^2 + bx + c \), the sum of zeroes is \( -b/a \) and the product of zeroes is \( c/a \). We use algebraic identities to transform the given expression into forms involving the sum and product of zeroes.


Step 2: Key Formula or Approach:

1. \(\alpha + \beta = -\frac{q}{p}\)

2. \(\alpha\beta = \frac{r}{p}\)

3. \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)


Step 3: Detailed Explanation:

Given polynomial is \( px^2 + qx + r \).

Sum of zeroes \( \alpha + \beta = -\frac{q}{p} \).

Product of zeroes \( \alpha\beta = \frac{r}{p} \).

We need to find the value of \( \alpha^3\beta + \beta^3\alpha \).

Factoring out common terms:
\[ \alpha^3\beta + \beta^3\alpha = \alpha\beta(\alpha^2 + \beta^2) \]

We substitute \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\[ Value = \alpha\beta [(\alpha + \beta)^2 - 2\alpha\beta] \]

Substitute the sum and product values:
\[ Value = \left(\frac{r}{p}\right) \left[ \left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right) \right] \]
\[ Value = \left(\frac{r}{p}\right) \left[ \frac{q^2}{p^2} - \frac{2r}{p} \right] \]
\[ Value = \left(\frac{r}{p}\right) \left[ \frac{q^2 - 2pr}{p^2} \right] \]
\[ Value = \frac{r(q^2 - 2pr)}{p^3} \]


Step 4: Final Answer:

The value is \( \frac{r(q^2 - 2pr)}{p^3} \).
Quick Tip: Symmetric functions of zeroes like \( \alpha^2+\beta^2 \) or \( \alpha^3+\beta^3 \) are always expressible in terms of \( (\alpha+\beta) \) and \( \alpha\beta \).

Step 1: Understanding the Concept:

When two triangles are similar, their corresponding sides are in the same ratio.


Step 2: Key Formula or Approach:

For \( \triangle AHK \sim \triangle ABC \):
\[ \frac{AH}{AB} = \frac{HK}{BC} = \frac{AK}{AC} \]


Step 3: Detailed Explanation:

Given: \( AK = 10 \) cm, \( BC = 3.5 \) cm, and \( HK = 7 \) cm.

Using the similarity property:
\[ \frac{HK}{BC} = \frac{AK}{AC} \]

Substitute the given values:
\[ \frac{7}{3.5} = \frac{10}{AC} \]

Simplify the left side:
\[ 2 = \frac{10}{AC} \]

Solving for \( AC \):
\[ AC = \frac{10}{2} = 5 cm \]


Step 4: Final Answer:

The length of AC is 5 cm.
Quick Tip: Always list corresponding vertices in order to correctly identify the ratios of the sides.

Step 1: Understanding the Concept:

According to the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio. Also, using properties of ratios, we can find parts of segments.


Step 2: Key Formula or Approach:

If \( XY || QR \), then \( \frac{PX}{XQ} = \frac{PY}{YR} \) or \( \frac{PQ}{XQ} = \frac{PR}{YR} \).


Step 3: Detailed Explanation:

Given: \( \frac{PQ}{XQ} = \frac{7}{3} \).

Since \( XY || QR \), the segments are proportional:
\[ \frac{PQ}{XQ} = \frac{PR}{YR} \]

Substitute the given values \( \frac{PQ}{XQ} = \frac{7}{3} \) and \( PR = 6.3 \):
\[ \frac{7}{3} = \frac{6.3}{YR} \]
\[ 7 \times YR = 3 \times 6.3 \]
\[ 7 \times YR = 18.9 \]
\[ YR = \frac{18.9}{7} \]
\[ YR = 2.7 cm \]


Step 4: Final Answer:

The length of YR is 2.7 cm.
Quick Tip: Using the ratio \( \frac{Total Side}{Segment} \) directly can often save time compared to finding intermediate segment lengths.

Step 1: Understanding the Concept:

Trigonometric ratios are based on the sides of a right triangle. If one ratio is known, others can be found using the Pythagorean theorem.


Step 2: Key Formula or Approach:

1. \( \tan A = \frac{Opposite}{Adjacent} = \frac{P}{B} \)

2. \( H = \sqrt{P^2 + B^2} \)

3. \( \sin A = \frac{P}{H}, \cos A = \frac{B}{H} \)


Step 3: Detailed Explanation:

Given \( \tan A = \frac{4}{3} \).

Let Perpendicular (\( P \)) \( = 4k \) and Base (\( B \)) \( = 3k \).

Hypotenuse (\( H \)) \( = \sqrt{(4k)^2 + (3k)^2} = \sqrt{16k^2 + 9k^2} = \sqrt{25k^2} = 5k \).

Now, calculate \( \sin A \):
\[ \sin A = \frac{P}{H} = \frac{4k}{5k} = \frac{4}{5} \]

Calculate \( \cos A \):
\[ \cos A = \frac{B}{H} = \frac{3k}{5k} = \frac{3}{5} \]


Step 4: Final Answer:
\( \sin A = \frac{4}{5} \) and \( \cos A = \frac{3}{5} \).
Quick Tip: Memorize Pythagorean triplets like (3, 4, 5), (5, 12, 13) to solve these problems instantly.

Step 1: Understanding the Concept:

Trigonometric identities allow us to express one ratio in terms of another.


Step 2: Key Formula or Approach:

1. \( \sin^2 A + \cos^2 A = 1 \)

2. \( \tan A = \frac{\sin A}{\cos A} \)


Step 3: Detailed Explanation:

To express \( \cos A \) in terms of \( \sin A \):

From the identity \( \sin^2 A + \cos^2 A = 1 \):
\[ \cos^2 A = 1 - \sin^2 A \]

Taking the square root (assuming \( A \) is an acute angle):
\[ \cos A = \sqrt{1 - \sin^2 A} \]

Now, to express \( \tan A \) in terms of \( \sin A \):

We know \( \tan A = \frac{\sin A}{\cos A} \).

Substitute the expression for \( \cos A \) derived above:
\[ \tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}} \]


Step 4: Final Answer:
\( \cos A = \sqrt{1 - \sin^2 A} \) and \( \tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}} \).
Quick Tip: Using identities is often faster and more general than building triangles for variable-based proofs.

Step 1: Understanding the Concept:

We use the congruency of triangles created by radii and tangents to prove the equality of tangent segments.


Step 2: Key Formula or Approach:

RHS congruency criterion in right-angled triangles.


Step 3: Detailed Explanation:

Consider a circle with center \( O \). Let \( P \) be an external point.

Let \( PA \) and \( PB \) be two tangents drawn from \( P \) to the circle at points \( A \) and \( B \) respectively.

Join \( OA, OB, \) and \( OP \).

In \( \triangle OAP \) and \( \triangle OBP \):

1. \( OA = OB \) (Radii of the same circle).

2. \( \angle OAP = \angle OBP = 90^\circ \) (Radius is perpendicular to the tangent at the point of contact).

3. \( OP = OP \) (Common side).

Therefore, \( \triangle OAP \cong \triangle OBP \) by RHS congruence criterion.

By CPCT (Corresponding Parts of Congruent Triangles):
\[ PA = PB \]


Step 4: Final Answer:

The lengths of tangents drawn from an external point to a circle are equal.
Quick Tip: This is a core theorem in Circles. Many numerical problems are solved using this property directly.

Step 1: Understanding the Concept:

This proof involves the isosceles property of the triangle formed by the chord and tangents, and the radius-tangent perpendicularity.


Step 2: Detailed Explanation:

Let \( \angle PTQ = \theta \).

Since tangents from an external point are equal, \( TP = TQ \).

Thus, \( \triangle TPQ \) is an isosceles triangle.

In \( \triangle TPQ \):
\[ \angle TPQ = \angle TQP = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{\theta}{2} \]

We know that the radius \( OP \) is perpendicular to the tangent \( TP \).

So, \( \angle OPT = 90^\circ \).

From the figure, \( \angle OPT = \angle OPQ + \angle TPQ \).
\[ 90^\circ = \angle OPQ + (90^\circ - \frac{\theta}{2}) \]
\[ \angle OPQ = \frac{\theta}{2} \]
\[ 2 \angle OPQ = \theta \]

Substituting \( \theta = \angle PTQ \):
\[ 2 \angle OPQ = \angle PTQ \]


Step 4: Final Answer:

Hence proved that \( \angle PTQ = 2 \angle OPQ \).
Quick Tip: Alternatively, use the fact that \( \angle POQ = 180^\circ - \theta \) and in isosceles \( \triangle OPQ \), \( \angle OPQ = (180^\circ - \angle POQ)/2 \).

Step 1: Understanding the Concept:

We use the method of contradiction. We assume \( \sqrt{5} \) is rational and then show that this leads to a logical inconsistency.


Step 2: Detailed Explanation:

Assume \( \sqrt{5} \) is a rational number.

Then \( \sqrt{5} = \frac{p}{q} \), where \( p \) and \( q \) are co-prime integers and \( q \neq 0 \).

Squaring both sides:
\[ 5 = \frac{p^2}{q^2} \implies p^2 = 5q^2 \]

This means \( 5 \) divides \( p^2 \). By theorem, \( 5 \) also divides \( p \).

Let \( p = 5c \) for some integer \( c \).

Substituting this into \( p^2 = 5q^2 \):
\[ (5c)^2 = 5q^2 \implies 25c^2 = 5q^2 \implies q^2 = 5c^2 \]

This means \( 5 \) divides \( q^2 \). By theorem, \( 5 \) also divides \( q \).

Now, \( 5 \) is a common factor of both \( p \) and \( q \).

This contradicts our initial assumption that \( p \) and \( q \) are co-prime.

Therefore, our assumption is wrong.


Step 4: Final Answer:

Hence, \( \sqrt{5} \) is an irrational number.
Quick Tip: This standard proof applies to any \( \sqrt{n} \) where \( n \) is a prime number.

Step 1: Understanding the Concept:

The area of a sector depends on the radius and the angle subtended at the center. The major sector is the remaining portion of the circle.


Step 2: Key Formula or Approach:

1. Area of sector \( = \frac{\theta}{360} \times \pi r^2 \)

2. Area of major sector \( = Area of circle - Area of minor sector \)


Step 3: Detailed Explanation:

Given: \( r = 42 \) cm, \( \theta = 30^\circ \).

Area of minor sector:
\[ Area_{minor} = \frac{30}{360} \times \frac{22}{7} \times 42 \times 42 \]
\[ Area_{minor} = \frac{1}{12} \times 22 \times 6 \times 42 \]
\[ Area_{minor} = \frac{1}{2} \times 22 \times 42 = 11 \times 42 = 462 cm^2 \]

Now, find the area of the circle:
\[ Area_{circle} = \frac{22}{7} \times 42 \times 42 = 22 \times 6 \times 42 = 132 \times 42 = 5544 cm^2 \]

Area of major sector:
\[ Area_{major} = 5544 - 462 = 5082 cm^2 \]

Alternatively, angle of major sector \( = 360^\circ - 30^\circ = 330^\circ \).
\[ Area_{major} = \frac{330}{360} \times \frac{22}{7} \times 42 \times 42 = \frac{11}{12} \times 22 \times 6 \times 42 = 11 \times 11 \times 42 = 5082 cm^2 \]


Step 4: Final Answer:

Area of minor sector is 462 \(cm^2\) and major sector is 5082 \(cm^2\).
Quick Tip: Using \( Area_{major} = \frac{360-\theta}{360} \times \pi r^2 \) directly can be more straightforward.

Step 1: Understanding the Concept:

The diagonals of a rhombus bisect each other. This means they have the same midpoint.


Step 2: Key Formula or Approach:

Midpoint formula: \( M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \).


Step 3: Detailed Explanation:

Let the intersection point be \( M(x, y) \). It is the midpoint of diagonal \( PR \).

Vertices \( P(2, -3) \) and \( R(-2, 1) \).
\[ M = \left(\frac{2 + (-2)}{2}, \frac{-3 + 1}{2}\right) = \left(\frac{0}{2}, \frac{-2}{2}\right) = (0, -1) \]

Now, \( M(0, -1) \) is also the midpoint of diagonal \( QS \).

Let \( S = (x_s, y_s) \). Vertex \( Q(6, 5) \).

Using the midpoint formula for \( QS \):
\[ \frac{6 + x_s}{2} = 0 \implies 6 + x_s = 0 \implies x_s = -6 \]
\[ \frac{5 + y_s}{2} = -1 \implies 5 + y_s = -2 \implies y_s = -7 \]


Step 4: Final Answer:

Coordinates of intersection point are (0, \(-1\)) and the fourth vertex S is (\(-6\), \(-7\)).
Quick Tip: The property of diagonals bisecting each other is true for all parallelograms, including rectangles, rhombuses, and squares.

Step 1: Understanding the Concept:

When two dice are thrown, there are \( 6 \times 6 = 36 \) total outcomes. We need to identify specific outcomes that satisfy the given conditions.


Step 2: Detailed Explanation:

Total possible outcomes \( = 36 \).

(i) For an even sum:

The sum is even if both dice show even numbers or both show odd numbers.

Odd numbers on one die: {1, 3, 5 (3 choices).

Even numbers on one die: {2, 4, 6 (3 choices).

Case 1: Both Odd \( \to 3 \times 3 = 9 \) outcomes.

Case 2: Both Even \( \to 3 \times 3 = 9 \) outcomes.

Total favorable outcomes \( = 9 + 9 = 18 \).
\[ P(even sum) = \frac{18}{36} = \frac{1}{2} \]

(ii) For an even product:

The product is even if at least one die shows an even number.

The product is odd ONLY if both dice show odd numbers.

Total odd product outcomes (both odd) \( = 3 \times 3 = 9 \).

Favorable outcomes (even product) \( = Total - Odd product = 36 - 9 = 27 \).
\[ P(even product) = \frac{27}{36} = \frac{3}{4} \]


Step 4: Final Answer:

(i) Probability of even sum is \(\frac{1}{2}\). (ii) Probability of even product is \(\frac{3}{4}\).
Quick Tip: For product problems, it is usually easier to find the "odd" case first and subtract from total, since \( odd \times odd = odd \) is the only way to get an odd product.

Step 1: Understanding the Concept:

This identity proof requires the use of fundamental trigonometric identities: \( \sec^2 \theta - 1 = \tan^2 \theta \) and \( cosec^2 \theta - 1 = \cot^2 \theta \). Converting terms into sine and cosine usually simplifies the expressions.


Step 2: Key Formula or Approach:

1. Use \( 1 + \tan^2 \theta = \sec^2 \theta \) and \( 1 + \cot^2 \theta = cosec^2 \theta \).

2. Convert all terms to \( \sin \theta \) and \( \cos \theta \).


Step 3: Detailed Explanation:

Starting with LHS:
\[ LHS = \frac{\sec^3 \theta}{\tan^2 \theta} + \frac{cosec^3 \theta}{\cot^2 \theta} \]

Convert to sine and cosine:
\[ = \frac{\frac{1}{\cos^3 \theta}}{\frac{\sin^2 \theta}{\cos^2 \theta}} + \frac{\frac{1}{\sin^3 \theta}}{\frac{\cos^2 \theta}{\sin^2 \theta}} \]

Simplify the complex fractions:
\[ = \frac{1}{\cos \theta \sin^2 \theta} + \frac{1}{\sin \theta \cos^2 \theta} \]

Take the LCM of the denominators:
\[ = \frac{\cos \theta + \sin \theta}{\sin^2 \theta \cos^2 \theta} \]

Now, look at RHS:
\[ RHS = \sec \theta cosec \theta (\sec \theta + cosec \theta) = \frac{1}{\cos \theta \sin \theta} \left( \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \right) \]
\[ = \frac{1}{\cos \theta \sin \theta} \left( \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \right) = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta} \]

Since LHS = RHS, the identity is proved.


Step 4: Final Answer:

Hence, \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{cosec^3 \theta}{cosec^2 \theta - 1} = \sec \theta \cdot cosec \theta (\sec \theta + cosec \theta) \).
Quick Tip: When stuck on an identity proof, always try converting everything into sine and cosine. It often makes the structure of the equation clearer.

Step 1: Understanding the Concept:

We use the given expressions for \( p \) and \( q \) and substitute them into the target equation. The identity \( \sec^2 \alpha - \tan^2 \alpha = 1 \) is central to this proof.


Step 2: Detailed Explanation:

Given: \( p = \frac{\sec \alpha}{cosec \beta} \) and \( q = \frac{\tan \alpha}{cosec \beta} \).

First, find \( p^2 - q^2 \):
\[ p^2 - q^2 = \left( \frac{\sec \alpha}{cosec \beta} \right)^2 - \left( \frac{\tan \alpha}{cosec \beta} \right)^2 = \frac{\sec^2 \alpha - \tan^2 \alpha}{cosec^2 \beta} \]

Since \( \sec^2 \alpha - \tan^2 \alpha = 1 \):
\[ p^2 - q^2 = \frac{1}{cosec^2 \beta} = \sin^2 \beta \]

Now evaluate the LHS of the identity to prove:
\[ LHS = (p^2 - q^2) \sec^2 \alpha = (\sin^2 \beta) \sec^2 \alpha \]

From the given \( p \):
\[ p = \sec \alpha \sin \beta \implies p^2 = \sec^2 \alpha \sin^2 \beta \]

Therefore, LHS \( = p^2 \).


Step 3: Final Answer:

Hence proved that \( (p^2 - q^2) \sec^2 \alpha = p^2 \).
Quick Tip: Isolating the denominator \(cosec \beta\) as a common factor in \(p^2 - q^2\) makes the cancellation using the identity \( \sec^2 \alpha - \tan^2 \alpha = 1 \) obvious.

Step 1: Understanding the Concept:

This is the Basic Proportionality Theorem (Thales Theorem). We prove it using the ratio of areas of triangles having common heights.


Step 2: Detailed Explanation:

Given: In \( \triangle ABC \), \( DE \parallel BC \) where \( D \) and \( E \) are points on \( AB \) and \( AC \).

To prove: \( \frac{AD}{DB} = \frac{AE}{EC} \).

Construction: Join \( BE, CD \). Draw \( DM \perp AC \) and \( EN \perp AB \).

Proof:

Area(\( \triangle ADE \)) \( = \frac{1}{2} \times AD \times EN \)

Area(\( \triangle BDE \)) \( = \frac{1}{2} \times DB \times EN \)
\[ \implies \frac{Area(\triangle ADE)}{Area(\triangle BDE)} = \frac{AD}{DB} \quad ---(i) \]

Similarly,

Area(\( \triangle ADE \)) \( = \frac{1}{2} \times AE \times DM \)

Area(\( \triangle CDE \)) \( = \frac{1}{2} \times EC \times DM \)
\[ \implies \frac{Area(\triangle ADE)}{Area(\triangle CDE)} = \frac{AE}{EC} \quad ---(ii) \]

Triangles \( BDE \) and \( CDE \) are on the same base \( DE \) and between the same parallels \( DE \) and \( BC \).
\[ \implies Area(\triangle BDE) = Area(\triangle CDE) \]

Substituting this in (ii), we find that the LHS of (i) and (ii) are equal.

Therefore, \( \frac{AD}{DB} = \frac{AE}{EC} \).


Step 3: Final Answer:

Hence, the line parallel to one side divides the other two sides in the same ratio.
Quick Tip: Remember: Area of a triangle is \(\frac{1}{2} \times base \times height\). Obtuse-angled triangles (\(\triangle BDE, \triangle CDE\)) have their heights falling outside the base.

Step 1: Understanding the Concept:

This problem uses the similarity of triangles formed by the lamp post, the girl, and the shadow. The rays of light form the hypotenuse.


Step 2: Key Formula or Approach:

1. Distance \( = Speed \times Time \)

2. Similarity of triangles: Ratio of corresponding sides are equal.


Step 3: Detailed Explanation:

Let \( AB \) be the lamp post (3.6 m) and \( CD \) be the girl (90 cm = 0.9 m).

Let \( x \) be the length of the shadow \( DE \).

Distance walked in 4 seconds \( BD = 1.2 \times 4 = 4.8 m \).

Total distance from lamp base to shadow tip \( BE = 4.8 + x \).

In \( \triangle ABE \) and \( \triangle CDE \), \( \angle B = \angle D = 90^\circ \) and \( \angle E = \angle E \) (common).

Thus, \( \triangle ABE \sim \triangle CDE \) (AA similarity).
\[ \frac{AB}{CD} = \frac{BE}{DE} \implies \frac{3.6}{0.9} = \frac{4.8 + x}{x} \]
\[ 4 = \frac{4.8 + x}{x} \implies 4x = 4.8 + x \]
\[ 3x = 4.8 \implies x = 1.6 m \]


Step 4: Final Answer:

The length of her shadow after 4 seconds is 1.6 m.
Quick Tip: Always ensure units are consistent. Here, convert 90 cm to 0.9 m before starting the ratio calculation.

Step 1: Understanding the Concept:

Median involves finding the middle observation using cumulative frequencies. Mode is the value in the class with the highest frequency.


Step 2: Detailed Explanation:

Calculating Median:

Total frequency \( N = 100 \). \( N/2 = 50 \).

Cumulative frequencies: 2, 6, 24, 45, 78, 89, 92, 98, 100.

The 50th observation lies in the class 35-40 (Median Class).
\( l = 35, cf = 45, f = 33, h = 5 \).
\[ Median = l + \left( \frac{N/2 - cf}{f} \right) \times h = 35 + \left( \frac{50 - 45}{33} \right) \times 5 = 35 + \frac{25}{33} \approx 35.76 \]

Calculating Mode:

Highest frequency is 33, so the Modal Class is 35-40.
\( l = 35, f_1 = 33, f_0 = 21, f_2 = 11, h = 5 \).
\[ Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 35 + \left( \frac{33 - 21}{66 - 21 - 11} \right) \times 5 = 35 + \frac{12 \times 5}{34} \approx 36.76 \]


Step 3: Final Answer:

Median age is 35.76 years and Modal age is 36.76 years.
Quick Tip: Double check cumulative frequency additions as a single mistake here will make the median calculation wrong.

Step 1: Understanding the Concept:

Consistency is determined by the ratios of coefficients. If the ratios of all coefficients (\( a, b, c \)) are equal, the lines are coincident.


Step 2: Detailed Explanation:

For \( x - 5y = 6 \): Points (6, 0), (1, -1), (11, 1).

For \( 2x - 10y = 12 \): Points (6, 0), (1, -1), (11, 1).

When plotted on a graph, both equations represent the exact same straight line.

Ratio analysis:
\( \frac{a_1}{a_2} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2} \)
\( \frac{c_1}{c_2} = \frac{6}{12} = \frac{1}{2} \)

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident.


Step 3: Final Answer:

The pair is consistent and dependent, having infinitely many solutions.
Quick Tip: If the second equation is just a multiple of the first (here, multiplied by 2), they will always be coincident lines.

Step 1: Understanding the Concept:

Set up a variable for one subject and express the other in terms of it. Form a quadratic equation from the given product condition.


Step 2: Detailed Explanation:

Let marks in Maths be \( x \). Then marks in Science \( = 30 - x \).

New Maths marks \( = x + 2 \).

New Science marks \( = (30 - x) - 3 = 27 - x \).

Product \( = 210 \):
\[ (x + 2)(27 - x) = 210 \]
\[ 27x - x^2 + 54 - 2x = 210 \implies -x^2 + 25x + 54 = 210 \]
\[ x^2 - 25x + 156 = 0 \]

Solving the quadratic:
\[ (x - 12)(x - 13) = 0 \implies x = 12 or 13 \]

If Maths \( = 12 \), Science \( = 18 \).

If Maths \( = 13 \), Science \( = 17 \).


Step 3: Final Answer:

Her marks are (12, 18) or (13, 17).
Quick Tip: To factor \(x^2 - 25x + 156\), look for factors of 156 that sum to 25. \(12 \times 13 = 156\) and \(12 + 13 = 25\).

Step 1: Understanding the Concept:

Use Pythagoras Theorem: \( Base^2 + Perpendicular^2 = Hypotenuse^2 \). Define all sides in terms of the shortest side.


Step 2: Detailed Explanation:

Let shortest side be \( x \).

Hypotenuse \( = 2x + 6 \).

Third side \( = 3x - 6 \).

By Pythagoras Theorem:
\[ x^2 + (3x - 6)^2 = (2x + 6)^2 \]
\[ x^2 + (9x^2 - 36x + 36) = (4x^2 + 24x + 36) \]
\[ 10x^2 - 36x + 36 = 4x^2 + 24x + 36 \]
\[ 6x^2 - 60x = 0 \implies 6x(x - 10) = 0 \]

Since side length cannot be 0, \( x = 10 \).

Shortest side \( = 10 \) cm.

Third side \( = 3(10) - 6 = 24 \) cm.

Hypotenuse \( = 2(10) + 6 = 26 \) cm.


Step 3: Final Answer:

Dimensions are 10 cm, 24 cm, and 26 cm.
Quick Tip: Check with Pythagoras triplet: \( 10^2 + 24^2 = 100 + 576 = 676 \), which is indeed \( 26^2 \).

Step 1: Key Formula or Approach:

Length of diagonal of a cube with side '\( a \)' is \( a\sqrt{3} \).


Step 2: Detailed Explanation:

Given edge \( a = 6 \) cm.

Diagonal \( = 6 \times \sqrt{3} = 6\sqrt{3} cm \).


Step 3: Final Answer:

Diagonal length is \( 6\sqrt{3} \) cm.
Quick Tip: Don't confuse the diagonal of a cube face (\(a\sqrt{2}\)) with the body diagonal of the cube (\(a\sqrt{3}\)).

Step 1: Key Formula or Approach:

Volume of cube \( = a^3 \).


Step 2: Detailed Explanation:

Given edge \( a = 7 \) cm.

Volume \( = 7 \times 7 \times 7 = 343 cm^3 \).


Step 3: Final Answer:

Volume is 343 \( cm^3 \).
Quick Tip: Memorizing cubes of numbers from 1 to 10 helps speed up these calculations. \( 7^3 = 343 \).

Step 1: Key Formula or Approach:

Curved Surface Area (CSA) of hemisphere \( = 2\pi r^2 \).


Step 2: Detailed Explanation:

Radius \( r = 7 \) cm.

CSA \( = 2 \times \frac{22}{7} \times 7 \times 7 = 2 \times 22 \times 7 = 44 \times 7 = 308 cm^2 \).


Step 3: Final Answer:

CSA is 308 \( cm^2 \).
Quick Tip: CSA only covers the curved part. Total Surface Area (TSA) of a hemisphere is \(3\pi r^2\).

Step 1: Detailed Explanation:

New cuboid dimensions: \( l = 4+4=8 cm \), \( b = 4 cm \), \( h = 4 cm \).

Surface area \( = 2(lb + bh + hl) \)
\( = 2(8\times4 + 4\times4 + 4\times8) = 2(32 + 16 + 32) = 2(80) = 160 cm^2 \).


Step 2: Final Answer:

Surface area is 160 \( cm^2 \).
Quick Tip: When joining cubes, the breadth and height remain the same, only the length changes.

Step 1: Key Formula or Approach:

AP term: \( a_n = a + (n-1)d \).


Step 2: Detailed Explanation:
\( a = 1000, d = 100, n = 30 \).
\( a_{30} = 1000 + (30 - 1)100 = 1000 + 2900 = 3900 \).


Step 3: Final Answer:

Amount is ₹ 3,900.
Quick Tip: Monthly increment acts as the 'common difference' (d) in an AP.

Step 1: Detailed Explanation:
\( n = 40 \).
\( a_{40} = 1000 + (40 - 1)100 = 1000 + 3900 = 4900 \).


Step 2: Final Answer:

Amount is ₹ 4,900.
Quick Tip: Always use (n-1) for the multiplier of 'd'. For the 40th term, use 39.

Step 1: Key Formula or Approach:

Sum of AP: \( S_n = \frac{n}{2}[2a + (n-1)d] \).


Step 2: Detailed Explanation:

Sum paid in 30 months:
\( S_{30} = \frac{30}{2}[2(1000) + 29(100)] = 15[2000 + 2900] = 15 \times 4900 = 73,500 \).

Remaining amount \( = 1,18,000 - 73,500 = 44,500 \).


Step 3: Final Answer:

Remaining amount is ₹ 44,500.
Quick Tip: Calculate \(S_{30}\) carefully, then subtract it from the total loan amount.

Step 1: Detailed Explanation:

10th instalment: \( a_{10} = 1000 + 9(100) = 1900 \).

Last (40th) instalment: \( a_{40} = 4900 \).

Ratio \( = 1900 : 4900 = 19 : 49 \).


Step 2: Final Answer:

Ratio is 19 : 49.
Quick Tip: Cancel common zeros to simplify ratios immediately.

Step 1: Detailed Explanation:

Let building height be \( h \). Car at \( C \) is 25 m from base \( B \). Angle of elevation from \( C \) is \( 60^\circ \).

In right \(\triangle ABC\):
\( \tan 60^\circ = \frac{h}{25} \implies \sqrt{3} = \frac{h}{25} \implies h = 25\sqrt{3} m \).


Step 2: Final Answer:

Height is \( 25\sqrt{3} \) m.
Quick Tip: Angle of depression from top = Angle of elevation from bottom.

Step 1: Detailed Explanation:

Let the previous position be \( D \). Distance \( BD = x \). Angle at \( D \) is \( 30^\circ \).

In right \(\triangle ABD\):
\( \tan 30^\circ = \frac{25\sqrt{3}}{x} \implies \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{x} \implies x = 75 m \).

Distance between positions \( CD = BD - BC = 75 - 25 = 50 m \).


Step 2: Final Answer:

Distance is 50 m.
Quick Tip: Total base distance for \(30^\circ\) is always \( \sqrt{3} \) times the height.

Step 1: Detailed Explanation:

Distance 50 m (from D to C) covered in 6 seconds.

Speed \( = \frac{50}{6} = \frac{25}{3} m/s \).

Total distance from starting point D to building base B is 75 m.

Total time \( = \frac{Total Distance}{Speed} = \frac{75}{25/3} = 75 \times \frac{3}{25} = 9 seconds \).


Step 2: Final Answer:

Total time is 9 seconds.
Quick Tip: Speed is uniform, so you can use simple ratios: if 50m takes 6s, then 25m takes 3s. Total \(6+3=9\)s.

Step 1: Detailed Explanation:

Observer is at A, car at C. Distance is hypotenuse AC.

In \(\triangle ABC\):
\( \cos 60^\circ = \frac{BC}{AC} \implies \frac{1}{2} = \frac{25}{AC} \implies AC = 50 m \).


Step 2: Final Answer:

Distance is 50 m.
Quick Tip: In a \(30^\circ-60^\circ-90^\circ\) triangle, the hypotenuse is twice the side opposite to the \(30^\circ\) angle.