Step 1: Understanding the Concept:

The zeroes of a polynomial function \( y = f(x) \) are represented geometrically by the x-coordinates of the points where the graph of the function intersects or touches the x-axis.


Step 2: Key Formula or Approach:

Identify all the unique points on the horizontal x-axis where the curve makes contact (either crosses or just touches).


Step 3: Detailed Explanation:

By observing the provided graph:

1. The curve crosses the x-axis once on the negative side (labeled point A).

2. The curve passes through the origin (0, 0), which is an intersection with the x-axis.

3. The curve crosses the x-axis again on the positive side.

Counting these points, we find three distinct points of intersection with the x-axis.

Therefore, the number of distinct zeroes is 3.


Step 4: Final Answer:

The number of distinct zeroes is 3.
Quick Tip: Always look for intersections with the horizontal axis only. Intersections with the vertical y-axis do not represent the zeroes of \( f(x) \).

Step 1: Understanding the Concept:

To distribute books equally among either 28 or 30 students, the total number of books must be a multiple of both 28 and 30. The "minimum" number implies finding the Least Common Multiple (LCM).


Step 2: Key Formula or Approach:

Find the LCM of 28 and 30 using the prime factorization method.


Step 3: Detailed Explanation:

Prime factorization of the numbers:
\[ 28 = 2 \times 2 \times 7 = 2^2 \times 7 \]
\[ 30 = 2 \times 3 \times 5 \]

To calculate the LCM, we take the highest power of every prime factor present in the factorizations:
\[ LCM = 2^2 \times 3 \times 5 \times 7 \]
\[ LCM = 4 \times 3 \times 5 \times 7 \]
\[ LCM = 12 \times 35 \]
\[ LCM = 420 \]

Thus, 420 is the smallest number divisible by both student counts.


Step 4: Final Answer:

The minimum number of books required is 420.
Quick Tip: Whenever a question asks for a "minimum" quantity that satisfies multiple distribution conditions, it is a direct application of LCM. If it asks for "maximum" size of a group/container, use HCF.

Step 1: Understanding the Concept:

A system of linear equations \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \) is inconsistent if the lines are parallel and never intersect. This happens when:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]


Step 2: Key Formula or Approach:

Simplify the fractional coefficients of the first equation to compare the ratios effectively.


Step 3: Detailed Explanation:

First equation: \( \frac{3}{2}x + \frac{5}{3}y = 7 \)

Multiply the entire equation by the LCM of denominators (6):
\[ 6 \left( \frac{3}{2}x \right) + 6 \left( \frac{5}{3}y \right) = 6(7) \]
\[ 9x + 10y = 42 \]

Comparing this with the second equation: \( 9x + 10y = 14 \).

Identify coefficients:
\( a_1 = 9, b_1 = 10, c_1 = 42 \)
\( a_2 = 9, b_2 = 10, c_2 = 14 \)

Calculate ratios:
\[ \frac{a_1}{a_2} = \frac{9}{9} = 1 \]
\[ \frac{b_1}{b_2} = \frac{10}{10} = 1 \]
\[ \frac{c_1}{c_2} = \frac{42}{14} = 3 \]

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the equations represent parallel lines.


Step 4: Final Answer:

The pair of equations is inconsistent.
Quick Tip: Always convert fractional equations to integer form first. If the left sides are identical but the constants on the right are different, the lines are parallel and the system is inconsistent.

Step 1: Understanding the Concept:

By mathematical definition:

- A prime number must have exactly two distinct factors (1 and the number itself).

- A composite number must have more than two factors.


Step 2: Detailed Explanation:

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

Since it doesn't have exactly two distinct factors, it cannot be prime.

Since it doesn't have more than two factors, it cannot be composite.

Therefore, 1 is excluded from both categories.


Step 3: Final Answer:

The natural number 1 is neither prime nor composite.
Quick Tip: Remember that 2 is the smallest prime number and the only even prime number. 4 is the smallest composite number. 1 is just a 'unit'.

Step 1: Understanding the Concept:

In an Arithmetic Progression (A.P.), the difference between any two consecutive terms is constant.


Step 2: Key Formula or Approach:

If \( a, b, c \) are in A.P., then \( b - a = c - b \), or more simply: \( 2b = a + c \).


Step 3: Detailed Explanation:

Let \( a = 2x, b = x + 10, c = 3x + 2 \).

Using the property \( 2b = a + c \):
\[ 2(x + 10) = 2x + (3x + 2) \]

Expand the left side and combine terms on the right:
\[ 2x + 20 = 5x + 2 \]

Shift terms involving \( x \) to one side and constants to the other:
\[ 20 - 2 = 5x - 2x \]
\[ 18 = 3x \]
\[ x = \frac{18}{3} \]
\[ x = 6 \]


Step 4: Final Answer:

The value of \( x \) is 6.
Quick Tip: The middle term of three consecutive A.P. terms is the arithmetic mean of the first and third terms. Use \( Middle = \frac{First + Third}{2} \) for quick solving.

Step 1: Understanding the Concept:

We need to determine the unit's digit of powers of 5. This relates to the concept of cyclicity in number systems.


Step 2: Detailed Explanation:

Consider the sequence of powers of 5:
\( 5^1 = 5 \)
\( 5^2 = 25 \)
\( 5^3 = 125 \)
\( 5^4 = 625 \)

Observe that the digit at the unit's place is always 5. This is because multiplying any number ending in 5 by 5 will result in a product that still ends in 5.


Step 3: Final Answer:

The unit digit for \( 5^n \) is always 5 for any natural number \( n \).
Quick Tip: Numbers with unit digits 0, 1, 5, or 6 have a cyclicity of 1, meaning their powers always end with the same digit.

Step 1: Understanding the Concept:

Triangles with two pairs of equal angles are similar by the AA (Angle-Angle) similarity criterion.


Step 2: Key Formula or Approach:

When two triangles are similar, the ratios of their corresponding sides are equal.


Step 3: Detailed Explanation:

Given: In \( \triangle ABC \) and \( \triangle QRP \):
\( \angle A = \angle Q \)
\( \angle B = \angle R \)

Therefore, \( \triangle ABC \sim \triangle QRP \) by AA similarity.

Corresponding sides property:
\[ \frac{AB}{QR} = \frac{BC}{RP} = \frac{AC}{QP} \]

From the first and third ratios:
\[ \frac{AB}{QR} = \frac{AC}{QP} \]

Rearranging to find the ratio \( AB : AC \):
\[ \frac{AB}{AC} = \frac{QR}{QP} \]

Thus, \( AB : AC = QR : QP \).


Step 4: Final Answer:

The correct ratio is \( QR : QP \).
Quick Tip: Be careful with the vertex order. Write down the similarity statement (like \( \triangle ABC \sim \triangle QRP \)) first based on equal angles to avoid pairing the wrong sides.

Step 1: Understanding the Concept:

For a quadratic polynomial \( ax^2 + bx + c \):

Sum of zeroes \( (\alpha + \beta) = - \frac{b}{a} \)

Product of zeroes \( (\alpha\beta) = \frac{c}{a} \)


Step 2: Key Formula or Approach:

Extract \( a, b, c \) from the given expression and apply the condition \( \alpha + \beta = \alpha\beta \).


Step 3: Detailed Explanation:

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

Coefficients: \( a = p, b = -2, c = 3p \).

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

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

Given: \( \alpha + \beta = \alpha\beta \).
\[ \frac{2}{p} = 3 \]
\[ 3p = 2 \]
\[ p = \frac{2}{3} \]


Step 4: Final Answer:

The value of \( p \) is \( \frac{2}{3} \).
Quick Tip: Notice that the variable \( p \) in the constant term \( 3p \) cancels out with the denominator \( p \) in the product of zeroes, simplifying the calculation significantly.

Step 1: Understanding the Concept:

For a moderately skewed distribution, there exists an empirical relationship connecting the three measures of central tendency.


Step 2: Key Formula or Approach:

The empirical formula is: \( Mode = 3 \times Median - 2 \times Mean \).


Step 3: Detailed Explanation:

Given values:

Mean \( = 43 \)

Median \( = 43.4 \)

Substituting into the 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: A simple mnemonic to remember the formula is "3 Medians minus 2 Means equals the Mode". The coefficients 3 and 2 are in descending order as you go from Median to Mean.

Step 1: Understanding the Concept:

The distance between two points in a 2D plane is found using the distance formula 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:

Points: \( (4, p) \) and \( (1, 0) \). Distance \( d = 5 \).
\[ \sqrt{(1 - 4)^2 + (0 - p)^2} = 5 \]

Square both sides to eliminate the square root:
\[ (-3)^2 + (-p)^2 = 5^2 \]
\[ 9 + p^2 = 25 \]
\[ p^2 = 25 - 9 \]
\[ p^2 = 16 \]
\[ p = \pm \sqrt{16} \]
\[ p = \pm 4 \]


Step 4: Final Answer:

The value of \( p \) is \( \pm 4 \).
Quick Tip: Always remember to include both positive and negative results when solving for a squared variable unless a geometric constraint (like a distance or height) is mentioned for the variable itself.

Step 1: Understanding the Concept:

The volume of the material used to make a hollow object is the difference between the external volume and the internal volume.


Step 2: Key Formula or Approach:

Volume of hemisphere \( = \frac{2}{3} \pi r^3 \).

Volume of steel \( = \frac{2}{3} \pi (R^3 - r^3) \), where \( R \) is outer radius and \( r \) is inner radius.


Step 3: Detailed Explanation:

Outer radius \( R = 6 \) cm.

Thickness \( = 1 \) cm.

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

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


Step 4: Final Answer:

The volume of steel used is \( \frac{182}{3} \pi \, cm^3 \).
Quick Tip: Do not confuse hemisphere volume with sphere volume. Sphere uses \( 4/3 \), while hemisphere uses \( 2/3 \). Double-check if the question provides diameter or radius.

Step 1: Understanding the Concept:

Trigonometric ratios relate the sides of a right-angled triangle to an angle.


Step 2: Key Formula or Approach:
\( \cos A = \frac{Base}{Hypotenuse} \) and \( \tan A = \frac{Perpendicular}{Base} \).

Use the Pythagorean identity: \( P^2 + B^2 = H^2 \).


Step 3: Detailed Explanation:

Given \( \cos A = \frac{4}{5} \). Let Base \( B = 4k \) and Hypotenuse \( H = 5k \).
\[ P^2 + (4k)^2 = (5k)^2 \]
\[ P^2 = 25k^2 - 16k^2 = 9k^2 \]
\[ P = 3k \]

Now, calculate \( \tan A \):
\[ \tan A = \frac{P}{B} = \frac{3k}{4k} = \frac{3}{4} \]


Step 4: Final Answer:

The value of \( \tan A \) is \( \frac{3}{4} \).
Quick Tip: Common Pythagorean triplets like (3, 4, 5) appear frequently in trig problems. Memorizing them helps skip the \( P^2 + B^2 = H^2 \) step.

Step 1: Understanding the Concept:

The area of a minor segment is defined as the area of the corresponding sector minus the area of the triangle formed by the radius and chord.


Step 2: Key Formula or Approach:

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

Area of triangle with central angle \( 60^\circ \) (equilateral triangle) \( = \frac{\sqrt{3}}{4} r^2 \).


Step 3: Detailed Explanation:

Given \( \theta = 60^\circ \).

Area of sector \( = \frac{60}{360} \pi r^2 = \frac{1}{6} \pi r^2 = \frac{\pi r^2}{6} \).

Since the triangle has two sides equal (radii) and an included angle of \( 60^\circ \), it is an equilateral triangle.

Area of triangle \( = \frac{\sqrt{3}}{4} r^2 \).

Area of segment \( = Area of sector - Area of triangle \)
\[ = \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2 \]


Step 4: Final Answer:

The area of the segment is \( \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2 \).
Quick Tip: For any angle \( \theta \), the area of the triangle is \( \frac{1}{2} r^2 \sin \theta \). For \( \theta = 60^\circ \), \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), which yields \( \frac{\sqrt{3}}{4} r^2 \).

Step 1: Understanding the Concept:

Find the value of angle A from the given ratio and substitute it into the required expression.


Step 2: Detailed Explanation:

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

From trigonometric tables, \( \sin 30^\circ = \frac{1}{2} \). So, \( A = 30^\circ \).

Now, find \( \tan A + \cot A \):
\( \tan 30^\circ = \frac{1}{\sqrt{3}} \)
\( \cot 30^\circ = \sqrt{3} \)

Expression \( = \frac{1}{\sqrt{3}} + \sqrt{3} \)

Take common denominator \( \sqrt{3} \):
\[ = \frac{1 + (\sqrt{3} \cdot \sqrt{3})}{\sqrt{3}} \]
\[ = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]


Step 3: Final Answer:

The value is \( \frac{4}{\sqrt{3}} \).
Quick Tip: An alternative way: \( \tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A} \). For \( A=30^\circ \), \( \frac{1}{\frac{1}{2} \cdot \frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}} \).

Step 1: Understanding the Concept:

The radius is perpendicular to the tangent at the point of contact. Also, tangents from an external point are equal.


Step 2: Detailed Explanation:

1. \( OA \perp PA \), so \( \angle OAP = 90^\circ \).

2. Given \( \angle OAB = 15^\circ \).

3. Therefore, \( \angle PAB = 90^\circ - 15^\circ = 75^\circ \).

4. Since \( PA = PB \) (tangents from point P), \( \triangle PAB \) is isosceles.

5. Thus, \( \angle PBA = \angle PAB = 75^\circ \).

6. Sum of angles in \( \triangle PAB \) is \( 180^\circ \):
\[ \angle APB + 75^\circ + 75^\circ = 180^\circ \]
\[ \angle APB = 180^\circ - 150^\circ = 30^\circ \]


Step 3: Final Answer:
\( \angle APB = 30^\circ \).
Quick Tip: In this geometric setup, the angle between the tangents is always twice the angle between the radius and the chord. \( \angle APB = 2 \times \angle OAB = 30^\circ \).

Step 1: Understanding the Concept:

This is a standard problem involving a right-angled triangle where we know the base and the angle.


Step 2: Detailed Explanation:

Let the height of the tower be \( h \) and the distance from the foot be \( d = 60 \) m.

Angle of elevation \( \theta = 45^\circ \).

In the right triangle:
\[ \tan 45^\circ = \frac{Height}{Distance} \]
\[ 1 = \frac{h}{60} \]
\[ h = 60 \, m \]


Step 3: Final Answer:

The height of the tower is 60 m.
Quick Tip: For a \( 45^\circ \) angle of elevation, the triangle is isosceles. This means the height is always exactly equal to the horizontal distance.

Step 1: Understanding the Concept:

Identify the total number of outcomes and the number of composite numbers in the given range.


Step 2: Detailed Explanation:

Total numbers = 25.

Composite numbers are those with more than two factors.

First, identify primes between 1 and 25: \{2, 3, 5, 7, 11, 13, 17, 19, 23\. There are 9 primes.

The number 1 is neither prime nor composite.

Number of composite numbers = Total - (Primes + 1)
\[ = 25 - (9 + 1) = 15 \]

Probability \( = \frac{15}{25} \).


Step 3: Final Answer:

The probability is \( \frac{15}{25} \).
Quick Tip: It is faster to count the primes and then subtract from the total. Always remember to subtract '1' as well because it's not composite.

Step 1: Understanding the Concept:

In quadrilateral OAPB, the angles at contact points A and B are right angles.


Step 2: Detailed Explanation:
\( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).

Sum of angles in quadrilateral OAPB is \( 360^\circ \).
\[ \angle AOB + \angle OAP + \angle OBP + \angle APB = 360^\circ \]
\[ 130^\circ + 90^\circ + 90^\circ + \angle APB = 360^\circ \]
\[ 310^\circ + \angle APB = 360^\circ \]
\[ \angle APB = 50^\circ \]


Step 3: Final Answer:
\( \angle APB = 50^\circ \).
Quick Tip: The angle between two tangents from an external point is supplementary to the angle subtended by the radii at the centre. Just subtract the given angle from \( 180^\circ \).

Step 1: Evaluate Reason (R):

The sum of the first \( n \) natural numbers (1, 2, 3, ..., n) is given by the formula \( S_n = \frac{n(n + 1)}{2} \). This is a standard identity.

So, Reason (R) is true.


Step 2: Evaluate Assertion (A):

Mean \( = \frac{Sum of terms}{Number of terms} \).
\[ Mean = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

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

So, Assertion (A) is false.


Step 3: Final Answer:

Since (A) is false and (R) is true, the correct choice is (D).
Quick Tip: Mean of an Arithmetic Progression is simply \( \frac{First term + Last term}{2} \). For natural numbers, it's \( \frac{1 + n}{2} \).

Step 1: Evaluate Assertion (A):

Joining two cubes of side 4 cm creates a cuboid with:

Length \( l = 4 + 4 = 8 \) cm, Breadth \( b = 4 \) cm, Height \( h = 4 \) cm.

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

So, Assertion (A) is true.


Step 2: Evaluate Reason (R):

The formula for the total surface area of a cuboid is \( 2(lb + bh + hl) \). The reason statement omits the factor of '2'.

So, Reason (R) is false.


Step 3: Final Answer:

Assertion is true, Reason is false. Choice (C).
Quick Tip: Check surface area formulas carefully in Assertion-Reason questions. A missing multiplier (like the '2' in cuboid surface area) is a common way to make a reason statement false.

Step 1: Understanding the Concept:

When two triangles are similar, the ratios of their corresponding sides are equal. This is known as the Basic Proportionality Theorem application in similarity.


Step 2: Key Formula or Approach:

For \(\Delta AHK \sim \Delta ABC\), the ratio of corresponding sides is:
\[ \frac{AH}{AB} = \frac{HK}{BC} = \frac{AK}{AC} \]


Step 3: Detailed Explanation:

Given values:
\(AK = 10 cm\)
\(BC = 3.5 cm\)
\(HK = 7 cm\)

Using the equality of ratios:
\[ \frac{HK}{BC} = \frac{AK}{AC} \]

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

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

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


Step 4: Final Answer:

The length of \(AC\) is 5 cm.
Quick Tip: To identify corresponding sides correctly, look at the order of vertices in the similarity statement: \(A \to A\), \(H \to B\), and \(K \to C\). Thus, \(HK\) corresponds to \(BC\) and \(AK\) corresponds to \(AC\).

Step 1: Understanding the Concept:

By 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.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

Given: \(\frac{PQ}{XQ} = \frac{7}{3}\) and \(PR = 6.3 cm\).

Using the ratio property of parallels:
\[ \frac{PQ}{XQ} = \frac{PR}{YR} \]

Substitute the given values:
\[ \frac{7}{3} = \frac{6.3}{YR} \]

Cross-multiply to find \(YR\):
\[ 7 \cdot YR = 3 \cdot 6.3 \]
\[ 7 \cdot YR = 18.9 \]
\[ YR = \frac{18.9}{7} \]
\[ YR = 2.7 cm \]


Step 4: Final Answer:

The length of \(YR\) is 2.7 cm.
Quick Tip: You can use the full-side ratio directly: \(\frac{Total Side}{Segment} = \frac{Total Side}{Segment}\). This saves the step of subtracting parts from the whole.

Step 1: Understanding the Concept:

This problem requires substituting standard trigonometric values into an expression and simplifying. We also use the identity \(\sin^2 \theta + \cos^2 \theta = 1\).


Step 2: Key Formula or Approach:

Use: \(\cos 60^\circ = \frac{1}{2}\), \(\sec 30^\circ = \frac{2}{\sqrt{3}}\), \(\tan 45^\circ = 1\), and \(\sin^2 \theta + \cos^2 \theta = 1\).


Step 3: Detailed Explanation:

First, observe the denominator:
\[ \sin^2 30^\circ + \cos^2 30^\circ = 1 \]

Now, simplify the numerator:
\[ 5 \left(\frac{1}{2}\right)^2 + 4 \left(\frac{2}{\sqrt{3}}\right)^2 - (1)^2 \]
\[ = 5 \left(\frac{1}{4}\right) + 4 \left(\frac{4}{3}\right) - 1 \]
\[ = \frac{5}{4} + \frac{16}{3} - 1 \]

Find the LCM of denominators (4 and 3) which is 12:
\[ = \frac{5 \times 3 + 16 \times 4 - 1 \times 12}{12} \]
\[ = \frac{15 + 64 - 12}{12} \]
\[ = \frac{79 - 12}{12} = \frac{67}{12} \]


Step 4: Final Answer:

The evaluated value is \(\frac{67}{12}\).
Quick Tip: Always look for the identity \(\sin^2 \theta + \cos^2 \theta = 1\) first; it usually simplifies the denominator to 1, saving a lot of complex calculation time.

Step 1: Understanding the Concept:

We use trigonometric identities to simplify the left-hand side (LHS) and match it with the right-hand side (RHS).


Step 2: Key Formula or Approach:

Use the identity: \(\cot^2 \alpha = \csc^2 \alpha - 1\).

Use algebraic identity: \(a^2 - b^2 = (a - b)(a + b)\).


Step 3: Detailed Explanation:

LHS: \(1 + \frac{\cot^2 \alpha}{1 + \csc \alpha}\)

Substitute \(\cot^2 \alpha\) with \(\csc^2 \alpha - 1\):
\[ = 1 + \frac{\csc^2 \alpha - 1}{\csc \alpha + 1} \]

Factorize the numerator using the difference of squares:
\[ = 1 + \frac{(\csc \alpha - 1)(\csc \alpha + 1)}{\csc \alpha + 1} \]

Cancel the common factor \((\csc \alpha + 1)\):
\[ = 1 + (\csc \alpha - 1) \]
\[ = 1 + \csc \alpha - 1 \]
\[ = \csc \alpha \]

This matches the RHS.


Step 4: Final Answer:

LHS = RHS. Hence proved.
Quick Tip: Whenever you see terms like \(\cot^2\) and \(\csc\), or \(\tan^2\) and \(\sec\), immediately think of the Pythagorean identities (\(1 + \cot^2 = \csc^2\)) to create common factors.

Step 1: Understanding the Concept:

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


Step 2: Key Formula or Approach:

Show that \(\angle Q + \angle R = 180^\circ\) or \(\angle P + \angle O = 180^\circ\).


Step 3: Detailed Explanation:

In quadrilateral PQOR:

1. \(OQ\) is the radius and \(PQ\) is the tangent at \(Q\). Therefore, \(OQ \perp PQ \Rightarrow \angle OQP = 90^\circ\).

2. \(OR\) is the radius and \(PR\) is the tangent at \(R\). Therefore, \(OR \perp PR \Rightarrow \angle ORP = 90^\circ\).

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

4. Since the sum of one pair of opposite angles is \(180^\circ\), the quadrilateral PQOR must be cyclic.


Step 4: Final Answer:

Quadrilateral PQOR is cyclic because its opposite angles are supplementary.
Quick Tip: In any tangent-radius geometry, the angle between the two tangents and the angle subtended by the radii at the center are always supplementary (\(180^\circ\)).

Step 1: Understanding the Concept:

For a polynomial \(ax^2 + bx + c\), the relationship between zeroes \(\alpha, \beta\) is:

Sum of zeroes: \(\alpha + \beta = -b/a\)

Product of zeroes: \(\alpha\beta = c/a\)


Step 2: Key Formula or Approach:

Let the zeroes be \(\alpha\) and \(6\alpha\).


Step 3: Detailed Explanation:

Given polynomial: \(px^2 - 14x + 8\).

Here \(a = p\), \(b = -14\), \(c = 8\).

1. Sum of zeroes:
\[ \alpha + 6\alpha = -\frac{(-14)}{p} \]
\[ 7\alpha = \frac{14}{p} \Rightarrow \alpha = \frac{2}{p} \]

2. Product of zeroes:
\[ \alpha \cdot (6\alpha) = \frac{8}{p} \]
\[ 6\alpha^2 = \frac{8}{p} \]

3. Substitute \(\alpha = 2/p\) into the product equation:
\[ 6 \left( \frac{2}{p} \right)^2 = \frac{8}{p} \]
\[ 6 \left( \frac{4}{p^2} \right) = \frac{8}{p} \]
\[ \frac{24}{p^2} = \frac{8}{p} \]

Since \(p \neq 0\), we can divide by \(8/p\):
\[ \frac{3}{p} = 1 \Rightarrow p = 3 \]


Step 4: Final Answer:

The value of \(p\) is 3.
Quick Tip: Always solve the "sum" equation for the variable zero (\(\alpha\)) first, then plug that result into the "product" equation to find the unknown coefficient.

Step 1: Understanding the Concept:

In a parallelogram, the diagonals bisect each other. This means the midpoint of diagonal \(AC\) is the same as the midpoint of diagonal \(BD\).


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:

1. Midpoint of \(AC\):
\[ M_{AC} = \left( \frac{4 + 4}{2}, \frac{5 + 3}{2} \right) = \left( \frac{8}{2}, \frac{8}{2} \right) = (4, 4) \]

2. Midpoint of \(BD\):
\[ M_{BD} = \left( \frac{m + 1}{2}, \frac{6 + n}{2} \right) \]

3. Equate the midpoints:

For x-coordinates:
\[ \frac{m + 1}{2} = 4 \Rightarrow m + 1 = 8 \Rightarrow m = 7 \]

For y-coordinates:
\[ \frac{6 + n}{2} = 4 \Rightarrow 6 + n = 8 \Rightarrow n = 2 \]


Step 4: Final Answer:

The values are \(m = 7\) and \(n = 2\).
Quick Tip: Using the midpoint of diagonals is the most efficient way to solve for unknown coordinates in a parallelogram. Avoid using the distance formula unless necessary.

Step 1: Understanding the Concept:

We transform terms into \(\sin\) and \(\cos\) to simplify complex trigonometric fractions.


Step 2: Key Formula or Approach:
\(\sec^2 \theta - 1 = \tan^2 \theta\) and \(\csc^2 \theta - 1 = \cot^2 \theta\).


Step 3: Detailed Explanation:

LHS: \(\frac{\sec^3 \theta}{\tan^2 \theta} + \frac{\csc^3 \theta}{\cot^2 \theta}\)

Convert to \(\sin, \cos\):
\[ = \frac{1/\cos^3 \theta}{\sin^2 \theta/\cos^2 \theta} + \frac{1/\sin^3 \theta}{\cos^2 \theta/\sin^2 \theta} \]
\[ = \frac{1}{\cos \theta \sin^2 \theta} + \frac{1}{\sin \theta \cos^2 \theta} \]

Take common denominator \(\sin^2 \theta \cos^2 \theta\):
\[ = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta} \]

RHS: \(\sec \theta \csc \theta (\sec \theta + \csc \theta)\)
\[ = \frac{1}{\cos \theta \sin \theta} \left( \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \right) \]
\[ = \frac{1}{\cos \theta \sin \theta} \left( \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \right) \]
\[ = \frac{\sin \theta + \cos \theta}{\sin^2 \theta \cos^2 \theta} \]

LHS = RHS.


Step 4: Final Answer:

Hence proved.
Quick Tip: When both sides look complex, simplify them independently to a basic form involving \(\sin\) and \(\cos\).

Step 1: Understanding the Concept:

Substitute the given values into the expression to be proved and use the identity \(\sec^2 \theta - \tan^2 \theta = 1\).


Step 2: Detailed Explanation:

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

Calculate \(p^2 - q^2\):
\[ p^2 - q^2 = \frac{\sec^2 \alpha}{\csc^2 \beta} - \frac{\tan^2 \alpha}{\csc^2 \beta} = \frac{\sec^2 \alpha - \tan^2 \alpha}{\csc^2 \beta} \]

Using the identity \(\sec^2 \alpha - \tan^2 \alpha = 1\):
\[ p^2 - q^2 = \frac{1}{\csc^2 \beta} \]

Now, calculate the LHS: \((p^2 - q^2) \sec^2 \alpha\)
\[ LHS = \frac{1}{\csc^2 \beta} \cdot \sec^2 \alpha = \frac{\sec^2 \alpha}{\csc^2 \beta} \]

From the given information, we know that \(\frac{\sec^2 \alpha}{\csc^2 \beta} = p^2\).

Thus, \(LHS = p^2 = RHS\).


Step 4: Final Answer:

LHS = RHS. Hence proved.
Quick Tip: Isolating common denominators (like \(\csc^2 \beta\)) makes it easier to apply standard trigonometric identities.

Step 1: Understanding the Concept:

Area of segment = Area of sector - Area of triangle.


Step 2: Key Formula or Approach:

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

Area of triangle \( = \frac{1}{2} r^2 \sin \theta \)


Step 3: Detailed Explanation:

Given: \(r = 21 cm\), \(\theta = 120^\circ\).

1. Area of sector AOBY:
\[ = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \]
\[ = \frac{1}{3} \times 22 \times 3 \times 21 = 22 \times 21 = 462 cm^2 \]

2. Area of \(\Delta AOB\):
\[ = \frac{1}{2} \times r^2 \times \sin 120^\circ \]

We know \(\sin 120^\circ = \sin(180 - 60) = \sin 60^\circ = \frac{\sqrt{3}}{2}\).
\[ = \frac{1}{2} \times 21 \times 21 \times \frac{\sqrt{3}}{2} = \frac{441\sqrt{3}}{4} cm^2 \]

3. Area of segment AYB:
\[ = 462 - \frac{441\sqrt{3}}{4} cm^2 \]


Step 4: Final Answer:

The area is \((462 - 110.25\sqrt{3}) cm^2\).
Quick Tip: For \(\theta > 90^\circ\), you can find the area of the triangle by using \(r^2 \sin(\theta/2) \cos(\theta/2)\) or simply \(\frac{1}{2}r^2 \sin \theta\).

Step 1: Understanding the Concept:

Total outcomes when two dice are thrown is \(6 \times 6 = 36\). Probability is the ratio of favorable outcomes to total outcomes.


Step 2: Detailed Explanation:

(i) Sum of numbers is 5:

Favorable outcomes are: (1, 4), (4, 1), (2, 3), (3, 2). Total = 4.
\[ P(sum 5) = \frac{4}{36} = \frac{1}{9} \]

(ii) Difference of numbers is 3:

Favorable outcomes (absolute difference) are: (1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3). Total = 6.
\[ P(diff 3) = \frac{6}{36} = \frac{1}{6} \]


Step 4: Final Answer:

The probabilities are 1/9 and 1/6 respectively.
Quick Tip: When dice are thrown, create a quick \(6 \times 6\) mental grid to avoid missing symmetric pairs like (1, 4) and (4, 1).

Step 1: Understanding the Concept:

We use proof by contradiction. Assume \(\sqrt{5}\) is rational and show that this leads to an impossibility.


Step 2: Detailed Explanation:

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

2. Squaring both sides: \(5 = p^2/q^2 \Rightarrow p^2 = 5q^2\).

3. This implies \(p^2\) is divisible by 5, so \(p\) must also be divisible by 5.

4. Let \(p = 5k\) for some integer \(k\).

5. Substitute \(p = 5k\): \((5k)^2 = 5q^2 \Rightarrow 25k^2 = 5q^2 \Rightarrow 5k^2 = q^2\).

6. This implies \(q^2\) is divisible by 5, so \(q\) must also be divisible by 5.

7. Since both \(p\) and \(q\) have a common factor 5, they are not co-prime.

8. This contradicts our initial assumption.


Step 4: Final Answer:

Therefore, \(\sqrt{5}\) is irrational.
Quick Tip: The "divisible by 5" step relies on Euclid's lemma: if a prime number divides a product, it must divide at least one factor.

Step 1: Understanding the Concept:

Points of trisection divide a segment into three equal parts. There are two such points, P (ratio 1:2) and Q (ratio 2:1).


Step 2: Key Formula or Approach:

Section formula: \(\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)\).


Step 3: Detailed Explanation:

Let \(A = (-1, 4)\) and \(B = (-3, -2)\).

1. Point P divides AB in ratio 1:2:
\[ P = \left( \frac{1(-3) + 2(-1)}{1+2}, \frac{1(-2) + 2(4)}{1+2} \right) = \left( \frac{-5}{3}, \frac{6}{3} \right) = (-5/3, 2) \]

2. Point Q divides AB in ratio 2:1:
\[ Q = \left( \frac{2(-3) + 1(-1)}{2+1}, \frac{2(-2) + 1(4)}{2+1} \right) = \left( \frac{-7}{3}, \frac{0}{3} \right) = (-7/3, 0) \]


Step 4: Final Answer:

The coordinates are \((-5/3, 2)\) and \((-7/3, 0)\).
Quick Tip: Alternatively, once you find P, Q is the midpoint of PB. This can sometimes simplify calculation.

Step 1: Understanding the Concept:

We need to prove that if \(XY\) is a tangent to a circle with center \(O\) at point \(P\), then \(OP \perp XY\).


Step 2: Key Formula or Approach:

We use the principle that the shortest distance from a point to a line is the perpendicular distance.


Step 3: Detailed Explanation:

1. Let \(XY\) be a tangent to a circle with center \(O\) at point \(P\).

2. Take any point \(Q\) on \(XY\) other than \(P\) and join \(OQ\).

3. The point \(Q\) must lie outside the circle. (If \(Q\) lies inside the circle, \(XY\) would be a secant and not a tangent).

4. Since \(Q\) lies outside the circle, \(OQ\) must be longer than the radius \(OP\).

5. Thus, \(OQ > OP\).

6. Since this is true for every point on the line \(XY\) except point \(P\), \(OP\) is the shortest distance from the center \(O\) to the line \(XY\).

7. Therefore, \(OP \perp XY\).


Step 4: Final Answer:

The tangent is perpendicular to the radius at the point of contact.
Quick Tip: Remember: Perpendicularity is equivalent to the shortest distance property. This theorem is the foundation for solving most circle-tangent problems.

Step 1: Understanding the Concept:

The lengths of tangents drawn from an external point to a circle are equal.


Step 2: Key Formula or Approach:

Let the circle touch the sides \(AB, BC, CD, DE, EF, FA\) at points \(P, Q, R, S, T, U\) respectively.


Step 3: Detailed Explanation:

By the property of tangents:
\(AP = AU\), \(BP = BQ\), \(CQ = CR\), \(DR = DS\), \(ES = ET\), \(FT = FU\).

Now, let's sum the segments for \(AB + CD + EF\):
\(AB + CD + EF = (AP + PB) + (CR + RD) + (ET + TF)\)

Substitute the equal tangent segments:
\(= (AU + BQ) + (CQ + DS) + (ES + FU)\)

Rearranging the terms:
\(= (BQ + CQ) + (DS + ES) + (FU + AU)\)
\(= BC + DE + FA\).

Thus, \(AB + CD + EF = BC + DE + FA\).


Step 4: Final Answer:

Hence proved.
Quick Tip: For any polygon circumscribing a circle, alternating side sums are often equal if the polygon has certain symmetries or an even number of sides.

Step 1: Understanding the Concept:

The median of a grouped distribution is calculated using the cumulative frequency and the median class formula.


Step 2: Key Formula or Approach:

Median \(= l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h\).

Also, the sum of all frequencies equals the total frequency (\(n\)).


Step 3: Detailed Explanation:

1. Cumulative Frequencies (\(cf\)):
\(x\), \(x+5\), \(x+14\), \(x+26\), \(x+26+y\), \(x+29+y\), \(x+31+y\).

2. Given Total Frequency \(= 40\):
\(x + y + 31 = 40 \implies x + y = 9\) (Equation 1).

3. Given Median \(= 32.5\). This value lies in the class 30 - 40.

Median Class: 30 - 40 \(\implies l = 30, f = 12, h = 10, cf = x+14, n/2 = 20\).

4. Apply Median Formula:
\(32.5 = 30 + \left( \frac{20 - (x + 14)}{12} \right) \times 10\)
\(2.5 = \frac{6 - x}{12} \times 10 \implies \frac{2.5}{10} = \frac{6 - x}{12}\)
\(0.25 = \frac{6 - x}{12} \implies 3 = 6 - x \implies x = 3\).

5. Substitute \(x=3\) in Equation 1:
\(3 + y = 9 \implies y = 6\).


Step 4: Final Answer:

The values are \(x = 3\) and \(y = 6\).
Quick Tip: Ensure you pick the cumulative frequency (\(cf\)) of the class PRECEDING the median class when applying the formula.

Step 1: Understanding the Concept:

We model the problem using two variables and solve by finding the intersection point of the two lines on a graph.


Step 2: Key Formula or Approach:

Let the price of 1 pencil be Rs \(x\) and 1 chocolate be Rs \(y\).

Equations: \(2x + 3y = 11\) and \(x + 2y = 7\).


Step 3: Detailed Explanation:

1. For \(2x + 3y = 11\):

If \(x=1, y=3\); If \(x=4, y=1\). Points: \((1, 3), (4, 1)\).

2. For \(x + 2y = 7\):

If \(x=1, y=3\); If \(x=3, y=2\). Points: \((1, 3), (3, 2)\).

3. Graphical Plotting:

Plot these points on a graph. The two lines intersect exactly at point \((1, 3)\).

4. Interpretation:
\(x = 1\) (price of 1 pencil) and \(y = 3\) (price of 1 chocolate).


Step 4: Final Answer:

The price of 1 pencil is Rs 1 and the price of 1 chocolate is Rs 3.
Quick Tip: When solving graphically, choose coordinates that yield integer values to make plotting easier and the intersection point clearer.

Step 1: Understanding the Concept:

Time taken is inversely proportional to speed for a fixed distance. We use the formula: Time = Distance / Speed.


Step 2: Key Formula or Approach:

Let usual speed \(= x\) km/h. Distance \(= 600\) km.

Usual time \(t_1 = 600/x\). Reduced speed time \(t_2 = 600/(x-200)\).

Condition: \(t_2 - t_1 = 30 min = 0.5 hours\).


Step 3: Detailed Explanation:
\[ \frac{600}{x - 200} - \frac{600}{x} = \frac{1}{2} \]

Multiply by \(2x(x-200)\):
\[ 1200x - 1200(x - 200) = x(x - 200) \]
\[ 1200x - 1200x + 240000 = x^2 - 200x \]
\[ x^2 - 200x - 240000 = 0 \]

Factoring the quadratic equation:
\[ x^2 - 600x + 400x - 240000 = 0 \]
\[ (x - 600)(x + 400) = 0 \implies x = 600 km/h (as speed > 0) \].

Scheduled duration \(= 600 / 600 = 1\) hour.


Step 4: Final Answer:

The scheduled duration of the flight is 1 hour.
Quick Tip: Always convert all time units (minutes to hours) to match the speed units (km/h) before setting up the equation.

Step 1: Understanding the Concept:

Rate of work is the reciprocal of the time taken to complete the job. Rate = 1 / Time.


Step 2: Key Formula or Approach:

Let large pipe take \(x\) hours. Small pipe takes \((x + 10)\) hours.

Work done in 1 hour: Large \(= 1/x\), Small \(= 1/(x+10)\).

Condition: \(4/x + 9/(x+10) = 1/2\).


Step 3: Detailed Explanation:
\[ \frac{4(x + 10) + 9x}{x(x + 10)} = \frac{1}{2} \]
\[ \frac{13x + 40}{x^2 + 10x} = \frac{1}{2} \]
\[ 26x + 80 = x^2 + 10x \implies x^2 - 16x - 80 = 0 \]

Factoring:
\[ x^2 - 20x + 4x - 80 = 0 \implies (x - 20)(x + 4) = 0 \]

Since \(x > 0\), \(x = 20\) hours.

Large pipe \(= 20\) hours. Small pipe \(= 20 + 10 = 30\) hours.


Step 4: Final Answer:

The larger pipe takes 20 hours and the smaller pipe takes 30 hours.
Quick Tip: In work-rate problems, remember that "half filled" means the equation should be set equal to \(1/2\), not \(1\).

Step 1: Understanding the Concept:

This is the Basic Proportionality Theorem (BPT) or Thales' Theorem. We use area ratios of triangles with common heights to prove it.


Step 2: Key Formula or Approach:

Area \((\Delta) = \frac{1}{2} \times base \times height\).


Step 3: Detailed Explanation:

1. Given \(\triangle ABC\) where \(DE \parallel BC\).

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

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

4. Area \((\triangle BDE) = \frac{1}{2} \times DB \times EN\).

5. \(\frac{Area(\triangle ADE)}{Area(\triangle BDE)} = \frac{AD}{DB}\). (Equation 1).

6. Similarly, \(\frac{Area(\triangle ADE)}{Area(\triangle CDE)} = \frac{AE}{EC}\). (Equation 2).

7. Note that \(\triangle BDE\) and \(\triangle CDE\) are on the same base \(DE\) and between the same parallels \(DE\) and \(BC\). Therefore, Area \((\triangle BDE) = Area (\triangle CDE)\).

8. From equations 1 and 2, the ratios are equal: \(\frac{AD}{DB} = \frac{AE}{EC}\).


Step 4: Final Answer:

Sides are divided in the same ratio.
Quick Tip: Visualizing the common altitudes (\(EN\) and \(DM\)) is key to understanding why the area ratios simplify to side ratios.

Step 1: Understanding the Concept:

The scenario forms two similar right-angled triangles because the girl and the post are both vertical to the ground.


Step 2: Key Formula or Approach:

Use AA similarity for \(\triangle ABE \sim \triangle CDE\).

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


Step 3: Detailed Explanation:

1. Height of post (\(AB\)) \(= 3.6\) m. Height of girl (\(CD\)) \(= 90\) cm \(= 0.9\) m.

2. Speed \(= 1.2\) m/s. Time \(= 4\) s.

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

4. Let shadow length (\(DE\)) \(= x\). Total distance (\(BE\)) \(= 4.8 + x\).

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


Step 4: Final Answer:

The length of her shadow after 4 seconds is 1.6 m.
Quick Tip: Ensure all units are consistent (meters vs centimeters) before plugging values into the similarity ratio.

Step 1: Understanding the Concept:

Angles of depression from the top are equal to angles of elevation from the ground point due to alternate interior angles.


Step 2: Key Formula or Approach:

In right \(\triangle ABC\) (at \(60^\circ\)): \(\tan 60^\circ = \frac{h}{25}\).


Step 3: Detailed Explanation:

Let height \(AB = h\). Distance of car from building at \(60^\circ\) is \(BC = 25\) m.
\[ \tan 60^\circ = \frac{h}{25} \implies \sqrt{3} = \frac{h}{25} \implies h = 25\sqrt{3} m \].


Step 4: Final Answer:

The height of the building is \(25\sqrt{3}\) meters.
Quick Tip: The closer the object is to the base, the larger the angle of elevation/depression.

Step 1: Understanding the Concept:

We use the larger triangle formed by the first observation point.


Step 2: Detailed Explanation:

In \(\triangle ABD\) (at \(30^\circ\)): \(\tan 30^\circ = \frac{h}{BD}\).
\[ \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{BD} \implies BD = 25 \times 3 = 75 m \].

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


Step 4: Final Answer:

The distance between the two positions is 50 meters.
Quick Tip: Always subtract the smaller horizontal distance from the larger one to find the distance moved by the object.

Step 1: Understanding the Concept:

Since speed is uniform, time is proportional to distance.


Step 2: Detailed Explanation:

Car covered distance \(CD = 50\) m in 6 seconds.

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

Time to cover remaining \(BC = 25\) m \(= \frac{25}{25/3} = 3\) seconds.

Total time \(= 6 + 3 = 9\) seconds.


Step 4: Final Answer:

The total time taken is 9 seconds.
Quick Tip: If the distance for the second part is half the first (\(25\) is half of \(50\)), the time taken will also be exactly half (\(3\) is half of \(6\)).

Step 1: Understanding the Concept:

The distance from the observer (at the top) to the car is the hypotenuse of the right triangle.


Step 2: Detailed Explanation:

In \(\triangle ABC\), the distance of observer from car is \(AC\).
\(\cos 60^\circ = \frac{BC}{AC} \implies \frac{1}{2} = \frac{25}{AC} \implies AC = 50\) m.


Step 4: Final Answer:

The distance of the observer from the car is 50 meters.
Quick Tip: Using cosine is the fastest way to relate the base distance and the hypotenuse distance.

Step 1: Understanding the Concept:

The space diagonal of a cube with edge '\(a\)' is given by \(a\sqrt{3}\).


Step 2: Detailed Explanation:

Given edge \(a = 6\) cm.

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


Step 4: Final Answer:

The length of the diagonal is \(6\sqrt{3}\) cm.
Quick Tip: The face diagonal is \(a\sqrt{2}\), but the main diagonal (space diagonal) of a cube is always \(a\sqrt{3}\).

Step 1: Understanding the Concept:

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


Step 2: Detailed Explanation:

Edge \(a = 7\) cm.

Volume \(= a^3 = 7^3 = 343\) cm³.


Step 4: Final Answer:

The volume of the cube is 343 cm³.
Quick Tip: Memorizing cubes up to 10 (\(7^3 = 343\), \(8^3 = 512\), etc.) helps solve calculation-heavy geometry problems faster.

Step 1: Understanding the Concept:

CSA of a hemisphere is \(2\pi r^2\).


Step 2: Detailed Explanation:

Radius \(r = 7\) cm.

CSA \(= 2 \times \frac{22}{7} \times 7 \times 7 = 44 \times 7 = 308\) cm².


Step 4: Final Answer:

The curved surface area is 308 cm².
Quick Tip: Total surface area of a hemisphere is \(3\pi r^2\), while curved surface area is \(2\pi r^2\). Don't mix them up !

Step 1: Understanding the Concept:

Joining two cubes changes only the length, while breadth and height remain the same.


Step 2: Detailed Explanation:

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

TSA \(= 2(lb + bh + hl) = 2(32 + 16 + 32) = 2(80) = 160\) cm².


Step 4: Final Answer:

The surface area of the cuboid is 160 cm².
Quick Tip: Alternatively, TSA \(= 2 \times\) (TSA of cube) \(- 2 \times\) (area of one face) \(= 2(6 \times 16) - 2(16) = 192 - 32 = 160\).

Step 1: Understanding the Concept:

The instalments form an Arithmetic Progression (AP) where \(a = 1000\) and \(d = 100\).


Step 2: Key Formula or Approach:
\(a_n = a + (n - 1)d\).


Step 3: Detailed Explanation:
\(a = 1000, d = 100, n = 30\).
\(a_{30} = 1000 + (30 - 1)100 = 1000 + 2900 = 3900\).


Step 4: Final Answer:

The amount paid in the 30th instalment is Rs 3,900.
Quick Tip: Verify the difference (\(d\)) is constant. Since he increases the amount by Rs 100 every month, it is a perfect AP.

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


Step 4: Final Answer:

The amount paid in the last instalment is Rs 4,900.
Quick Tip: The last term \(a_n\) is also known as \(l\) in AP summation formulas.

Step 1: Understanding the Concept:

Find the total amount paid in 30 months and subtract it from the total loan.


Step 2: Detailed Explanation:

Sum of first 30 instalments (\(S_{30}\)):
\(S_{30} = \frac{n}{2} [a + a_n] = \frac{30}{2} [1000 + 3900] = 15 \times 4900 = 73500\).

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


Step 4: Final Answer:

The amount left to pay is Rs 44,500.
Quick Tip: Using the formula \(\frac{n}{2}(a+l)\) is much faster than \(\frac{n}{2}(2a+(n-1)d)\) if you already solved for the \(n^{th}\) term in a previous step.

Step 1: Detailed Explanation:
\(10^{th}\) instalment (\(a_{10}\)) \(= 1000 + 9(100) = 1900\).

Last instalment (\(a_{40}\)) \(= 4900\) (calculated in part ii).

Ratio \(= \frac{1900}{4900} = \frac{19}{49}\).


Step 4: Final Answer:

The ratio is 19 : 49.
Quick Tip: Ratios should always be expressed in the simplest form by cancelling common factors (here, \(100\)).