Step 1: Understanding the Concept:

The unit digit of a number raised to any power depends on the cycle of the unit digit of the base.

For any natural number \(n\), we need to find the last digit of \(6^{n}\).


Step 2: Key Formula or Approach:

We observe the pattern of powers of 6:
\[ 6^{1} = 6 \]
\[ 6^{2} = 36 \]
\[ 6^{3} = 216 \]
\[ 6^{4} = 1296 \]


Step 3: Detailed Explanation:

From the pattern above, it is clear that for any power of 6, the unit digit is always 6.

Mathematically, if the unit digit of a number is \(k\), then the unit digit of \(k \times k\) will determine the next power's unit digit.

Since \(6 \times 6 = 36\), which again ends in 6, the product will always end in 6 regardless of how many times it is multiplied by itself.


Step 4: Final Answer:

Therefore, for any natural number \(n\), \(6^{n}\) always ends with the digit 6.
Quick Tip: The digits 0, 1, 5, and 6 always result in the same unit digit (0, 1, 5, and 6 respectively) when raised to any positive integer power.
For example, \(5^{n}\) always ends in 5, and \(6^{n}\) always ends in 6.

Step 1: Understanding the Concept:

The zeroes of a polynomial function \(f(x)\) are the values of \(x\) for which \(f(x) = 0\).

Geometrically, these are the points where the graph of \(y = f(x)\) intersects or touches the \(x\)-axis.


Step 2: Key Formula or Approach:

Count the total number of distinct points where the curve crosses or meets the horizontal \(x\)-axis.


Step 3: Detailed Explanation:

Looking at the provided graph:

1. The curve crosses the \(x\)-axis once on the negative side (left of the origin).

2. The curve passes through the origin \((0,0)\), which is a point on the \(x\)-axis.

3. The curve crosses the \(x\)-axis once on the positive side (right of the origin).

Total points of intersection = \(1 + 1 + 1 = 3\).


Step 4: Final Answer:

Since the graph intersects the \(x\)-axis at 3 distinct points, the number of zeroes of \(f(x)\) is 3.
Quick Tip: Always look specifically at the \(x\)-axis intersections. Intersections with the \(y\)-axis represent the value \(f(0)\) and are not considered zeroes of the function.

Step 1: Understanding the Concept:

A solution to a system of linear equations corresponds to a point that lies on both lines.

If the lines are coincident, it means one line lies exactly on top of the other, effectively making them the same line.


Step 2: Detailed Explanation:

- If lines intersect at a single point, there is a unique solution.

- If lines are parallel, they never meet, so there is no solution.

- If lines are coincident, every point on one line is also on the other line.

Since a line consists of infinitely many points, there are infinitely many common points.


Step 3: Final Answer:

Therefore, a pair of coincident lines has an infinite number of solutions.
Quick Tip: For equations \(a_{1}x + b_{1}y + c_{1} = 0\) and \(a_{2}x + b_{2}y + c_{2} = 0\), the condition for coincident lines is:
\[ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \]

Step 1: Understanding the Concept:

In an Arithmetic Progression (AP), the common difference (\(d\)) is the constant value obtained by subtracting any term from its succeeding term.


Step 2: Key Formula or Approach:

Common difference \(d = a_{2} - a_{1} = a_{3} - a_{2}\).


Step 3: Detailed Explanation:

Given AP: \(\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots\)

Here, first term \(a_{1} = \sqrt{2}\) and second term \(a_{2} = 2\sqrt{2}\).
\[ d = a_{2} - a_{1} \]
\[ d = 2\sqrt{2} - \sqrt{2} \]
\[ d = \sqrt{2}(2 - 1) \]
\[ d = \sqrt{2} \]


Step 4: Final Answer:

The common difference of the given AP is \(\sqrt{2}\).
Quick Tip: Treat radicals like variables. Just as \(2x - x = x\), \(2\sqrt{2} - \sqrt{2} = \sqrt{2}\). Always verify with the third term: \(3\sqrt{2} - 2\sqrt{2} = \sqrt{2}\).

Step 1: Understanding the Concept:

When two triangles are similar (\(\Delta ABC \sim \Delta DEF\)), the ratios of their corresponding sides are equal.


Step 2: Key Formula or Approach:
\[ \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \]


Step 3: Detailed Explanation:

Given that \(2 AB = DE\), we can write the ratio of corresponding sides as:
\[ \frac{AB}{DE} = \frac{1}{2} \]

Since the triangles are similar, the ratio of \(BC\) to \(EF\) must be the same:
\[ \frac{BC}{EF} = \frac{1}{2} \]

Substitute the given value \(BC = 8\) cm:
\[ \frac{8}{EF} = \frac{1}{2} \]

Cross-multiplying gives:
\[ EF = 8 \times 2 \]
\[ EF = 16 cm \]


Step 4: Final Answer:

The length of \(EF\) is 16 cm.
Quick Tip: Similarity is about "scaling". If \(DE\) is twice \(AB\), then every side of \(\Delta DEF\) is twice the corresponding side of \(\Delta ABC\). So \(EF = 2 \times BC\).

Step 1: Understanding the Concept:

The mid-point of a line segment joining \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) is given by the average of the coordinates.

- A point lies on the \(x\)-axis if its \(y\)-coordinate is 0.

- A point lies on the \(y\)-axis if its \(x\)-coordinate is 0.


Step 2: Key Formula or Approach:

Midpoint \(M = \left( \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2} \right) \)


Step 3: Detailed Explanation:

Let the points be \(A(5, -4)\) and \(B(6, 4)\).
\[ x_{M} = \frac{5 + 6}{2} = \frac{11}{2} = 5.5 \]
\[ y_{M} = \frac{-4 + 4}{2} = \frac{0}{2} = 0 \]

The coordinates of the mid-point are \((5.5, 0)\).

Since the \(y\)-coordinate is 0, this point lies on the \(x\)-axis.


Step 4: Final Answer:

The mid-point lies on the \(x\)-axis.
Quick Tip: If you see two \(y\)-coordinates that are negatives of each other (like -4 and 4), their average will always be 0, meaning the midpoint will always lie on the \(x\)-axis (provided they are not both zero).

Step 1: Understanding the Concept:

Trigonometric ratios are related by the identity \(\sin^{2} \theta + \cos^{2} \theta = 1\).


Step 2: Key Formula or Approach:
\[ \cos \theta = \sqrt{1 - \sin^{2} \theta} \]


Step 3: Detailed Explanation:

Given \(\sin \theta = \frac{a}{b}\).

Substitute this into the identity:
\[ \cos^{2} \theta = 1 - \left( \frac{a}{b} \right)^{2} \]
\[ \cos^{2} \theta = 1 - \frac{a^{2}}{b^{2}} \]

Taking the LCM:
\[ \cos^{2} \theta = \frac{b^{2} - a^{2}}{b^{2}} \]

Taking the square root:
\[ \cos \theta = \sqrt{\frac{b^{2} - a^{2}}{b^{2}}} = \frac{\sqrt{b^{2} - a^{2}}}{b} \]


Step 4: Final Answer:
\(\cos \theta = \frac{\sqrt{b^{2} - a^{2}}}{b}\).
Quick Tip: Alternatively, think of a right triangle where opposite side = \(a\) and hypotenuse = \(b\).
By Pythagoras theorem, adjacent side = \(\sqrt{hypotenuse^{2} - opposite^{2}} = \sqrt{b^{2} - a^{2}}\).
Thus, \(\cos \theta = \frac{Adjacent}{Hypotenuse} = \frac{\sqrt{b^{2} - a^{2}}}{b}\).

Step 1: Understanding the Concept:

We can solve this by either finding the angle \(A\) or using trigonometric identities.


Step 2: Key Formula or Approach:

Identity: \(\sin^{2} A = 1 - \cos^{2} A\).


Step 3: Detailed Explanation:

Given \(\cos A = \frac{1}{2}\).

Then \(\cos^{2} A = \left(\frac{1}{2}\right)^{2} = \frac{1}{4}\).

Using the identity, \(\sin^{2} A = 1 - \frac{1}{4} = \frac{3}{4}\).

Now, calculate the required expression:
\[ Value = \sin^{2} A + 2 \cos^{2} A \]
\[ Value = \frac{3}{4} + 2 \left( \frac{1}{4} \right) \]
\[ Value = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \]


Step 4: Final Answer:

The value of the expression is \(\frac{5}{4}\).
Quick Tip: You can also identify that if \(\cos A = 1/2\), then \(A = 60^{\circ}\).
Then \(\sin^{2} 60^{\circ} + 2 \cos^{2} 60^{\circ} = (\sqrt{3}/2)^{2} + 2(1/2)^{2} = 3/4 + 2/4 = 5/4\).

Step 1: Understanding the Concept:

This is a right-angled triangle problem where the string is the hypotenuse, and the height of the kite is the perpendicular side relative to the angle with the ground.


Step 2: Key Formula or Approach:
\[ \sin \theta = \frac{Perpendicular}{Hypotenuse} \]


Step 3: Detailed Explanation:

Let \(h\) be the height of the kite and \(L = 80\) m be the length of the string.

The angle \(\theta = 30^{\circ}\).
\[ \sin 30^{\circ} = \frac{h}{80} \]

We know that \(\sin 30^{\circ} = \frac{1}{2}\).
\[ \frac{1}{2} = \frac{h}{80} \]
\[ h = \frac{80}{2} = 40 m \]


Step 4: Final Answer:

The height of the kite above the ground is 40 m.
Quick Tip: In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the side opposite the \(30^{\circ}\) angle is always half the hypotenuse. Since the string (hypotenuse) is 80 m, the height is simply 40 m.

Step 1: Understanding the Concept:

Tangents to a circle from an external point are perpendicular to the radii at the point of contact. Therefore, \(\angle OPT = \angle OQT = 90^{\circ}\).


Step 2: Key Formula or Approach:

In quadrilateral \(OPTQ\), the sum of all interior angles is \(360^{\circ}\).


Step 3: Detailed Explanation:
\[ \angle OPT + \angle PTQ + \angle TQO + \angle POQ = 360^{\circ} \]

Substituting the known values:
\[ 90^{\circ} + \angle PTQ + 90^{\circ} + 120^{\circ} = 360^{\circ} \]
\[ \angle PTQ + 300^{\circ} = 360^{\circ} \]
\[ \angle PTQ = 360^{\circ} - 300^{\circ} = 60^{\circ} \]


Step 4: Final Answer:

The measure of \(\angle PTQ\) is \(60^{\circ}\).
Quick Tip: For any external point \(T\) and center \(O\), the angle between the tangents (\(\angle PTQ\)) and the angle subtended by the radii at the center (\(\angle POQ\)) are supplementary.
Simply compute: \(\angle PTQ = 180^{\circ} - 120^{\circ} = 60^{\circ}\).

Step 1: Understanding the Concept:

A tangent is perpendicular to the radius at the point of contact. Thus, in \(\Delta OAP\), \(\angle OAP = 90^{\circ}\).


Step 2: Detailed Explanation:

From the figure, \(P-O-B\) is a straight line segment.

Thus, \(\angle POA\) and \(\angle AOB\) are related, or specifically, \(\angle POB\) is the exterior angle for \(\Delta OAP\) at vertex \(O\).

Using the property that an exterior angle of a triangle is equal to the sum of its two interior opposite angles:
\[ \angle POB = \angle OAP + \angle APO \]
\[ 125^{\circ} = 90^{\circ} + \angle APO \]
\[ \angle APO = 125^{\circ} - 90^{\circ} = 35^{\circ} \]


Step 4: Final Answer:

The measure of \(\angle APO\) is \(35^{\circ}\).
Quick Tip: Alternatively, find \(\angle POA\) using linear pair: \(\angle POA = 180^{\circ} - 125^{\circ} = 55^{\circ}\).
Then in \(\Delta OAP\), sum of angles is \(180^{\circ}\): \(90^{\circ} + 55^{\circ} + \angle APO = 180^{\circ} \Rightarrow \angle APO = 35^{\circ}\).

Step 1: Understanding the Concept:

The area of a sector is a fraction of the total area of the circle, proportional to the central angle \(\theta\).


Step 2: Key Formula or Approach:
\[ Area of Sector = \frac{\theta}{360^{\circ}} \times Area of Circle \]


Step 3: Detailed Explanation:

From the figure, the central angle \(\theta = 30^{\circ}\).

Given Area of Sector = 7 cm\^{2.
\[ 7 = \frac{30^{\circ}{360^{\circ}} \times Area of Circle \]
\[ 7 = \frac{1}{12} \times Area of Circle \]

Multiply both sides by 12:
\[ Area of Circle = 7 \times 12 = 84 cm^{2} \]


Step 4: Final Answer:

The total area of the circle is 84 cm\^{2.
Quick Tip: If the sector angle is \(30^{\circ\), there are \(360/30 = 12\) such sectors in a circle. Just multiply the sector area by 12 to get the full circle area.

Step 1: Understanding the Concept:

The length of an arc is determined by the radius and the angle it subtends at the center.


Step 2: Key Formula or Approach:
\[ Arc Length = \frac{\theta}{360^{\circ}} \times 2\pi r \]


Step 3: Detailed Explanation:

Given \(\theta = 45^{\circ}\) and \(r = 32\) cm.
\[ Length of arc XYZ = \frac{45^{\circ}}{360^{\circ}} \times 2\pi(32) \]

Simplify the fraction: \( \frac{45}{360} = \frac{1}{8} \).
\[ Length = \frac{1}{8} \times 64\pi \]
\[ Length = 8\pi cm \]


Step 4: Final Answer:

The length of the arc \(XYZ\) is \(8\pi\) cm.
Quick Tip: Remember that \(45^{\circ}\) is exactly \(\frac{1}{8}\) of a full circle (\(360^{\circ}\)). So the arc length is simply \(\frac{1}{8}\) of the circumference (\(2\pi r\)).

Step 1: Understanding the Concept:

Volume of a sphere is a function of its radius \(r\). Given the volume, we can solve for \(r\).


Step 2: Key Formula or Approach:
\[ Volume of Sphere V = \frac{4}{3} \pi r^{3} \]


Step 3: Detailed Explanation:

Given \(V = 36 \pi\).

Set the formula equal to the given value:
\[ \frac{4}{3} \pi r^{3} = 36 \pi \]

Divide both sides by \(\pi\):
\[ \frac{4}{3} r^{3} = 36 \]

Isolate \(r^{3}\):
\[ r^{3} = 36 \times \frac{3}{4} \]
\[ r^{3} = 9 \times 3 = 27 \]

Take the cube root of both sides:
\[ r = \sqrt[3]{27} = 3 \]


Step 4: Final Answer:

The radius of the sphere is 3 cm.
Quick Tip: Keep an eye out for perfect cubes like 1, 8, 27, 64 in these problems. Once you reach \(r^{3} = 27\), it is easy to see the answer is 3.

Step 1: Understanding the Concept:

In statistics, there is an empirical relationship that connects the three measures of central tendency: Mean, Median, and Mode.

This relationship is particularly useful for moderately skewed distributions where you are given two values and need to find the third.


Step 2: Key Formula or Approach:

The empirical formula is given by:
\[ Mode = 3 \times Median - 2 \times Mean \]


Step 3: Detailed Explanation:

Given:

Mean = 12

Mode = 21

Let the Median be \(M\).

Substituting the values into the formula:
\[ 21 = 3 \times M - 2 \times 12 \]
\[ 21 = 3M - 24 \]

Add 24 to both sides:
\[ 21 + 24 = 3M \]
\[ 45 = 3M \]

Divide by 3:
\[ M = \frac{45}{3} \]
\[ M = 15 \]


Step 4: Final Answer:

The median of the data is 15.
Quick Tip: A simple way to remember the formula is to arrange the terms alphabetically: Mean, Median, Mode.
The formula uses the coefficients 3 and 2. The larger coefficient (3) goes with the word having more letters (Median), and the smaller coefficient (2) goes with the word having fewer letters (Mean).
Formula: \(Mode = 3(Median) - 2(Mean)\).

Step 1: Understanding the Concept:

The probability of an event \(E\) is the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space.


Step 2: Key Formula or Approach:
\[ P(E) = \frac{Number of favorable outcomes}{Total number of outcomes} \]


Step 3: Detailed Explanation:

When a die is thrown once, the sample space \(S\) is:
\[ S = \{1, 2, 3, 4, 5, 6\} \]

Total number of outcomes, \(n(S) = 6\).

We need the probability of getting a number other than 3.

Let \(E\) be the event of getting a number other than 3.

The favorable outcomes are:
\[ E = \{1, 2, 4, 5, 6\} \]

Number of favorable outcomes, \(n(E) = 5\).
\[ P(E) = \frac{n(E)}{n(S)} = \frac{5}{6} \]


Step 4: Final Answer:

The probability of getting a number other than 3 is \(\frac{5}{6}\).
Quick Tip: You can also use the complement rule: \(P(Not A) = 1 - P(A)\).
The probability of getting exactly 3 is \(P(3) = \frac{1}{6}\).
So, \(P(Other than 3) = 1 - \frac{1}{6} = \frac{5}{6}\).

Step 1: Understanding the Concept:

The Highest Common Factor (HCF) is the largest positive integer that divides each of the integers without leaving a remainder. We can find it using Euclid's Division Algorithm or Prime Factorization.


Step 2: Key Formula or Approach:

We will use Euclid's Division Algorithm: \(a = bq + r\).


Step 3: Detailed Explanation:

Step 1: Since \(960 > 432\), apply Euclid's division lemma:
\[ 960 = 432 \times 2 + 96 \]

Step 2: Since the remainder \(96 \neq 0\), apply the lemma to 432 and 96:
\[ 432 = 96 \times 4 + 48 \]

Step 3: Since the remainder \(48 \neq 0\), apply the lemma to 96 and 48:
\[ 96 = 48 \times 2 + 0 \]

The remainder has now become zero. The divisor at this stage is 48.


Step 4: Final Answer:

Therefore, the HCF of 960 and 432 is 48.
Quick Tip: In multiple-choice questions, you can quickly check which option divides both numbers.
Try the largest option first if looking for HCF. 72 doesn't divide 960 exactly (\(960/72 \approx 13.3\)). 48 divides 960 (\(48 \times 20\)) and 432 (\(48 \times 9\)).

Step 1: Understanding the Concept:

- A prime number is a natural number greater than 1 that has exactly two factors: 1 and itself.

- A composite number is a natural number greater than 1 that has more than two factors.

- The number 1 is neither prime nor composite.


Step 2: Detailed Explanation:

The number 2 is a natural number.

Its factors are 1 and 2.

Since it has exactly two distinct factors (1 and itself), it fits the definition of a prime number.

Furthermore, 2 is the only even prime number.


Step 3: Final Answer:

The natural number 2 is a prime number.
Quick Tip: Always remember that 2 is the smallest prime number and the only even prime number. All other prime numbers are odd.

Step 1: Understanding the Concept:

The degree of a polynomial determines the maximum number of zeroes it can have. A quadratic polynomial is of degree 2.


Step 2: Detailed Explanation:

Evaluating Assertion (A):

The given polynomial is \(p(y) = y^{2} + 4y + 3\).

To find the zeroes, solve \(y^{2} + 4y + 3 = 0\):
\[ y^{2} + 3y + y + 3 = 0 \]
\[ y(y + 3) + 1(y + 3) = 0 \]
\[ (y + 1)(y + 3) = 0 \]

Zeroes are \(y = -1\) and \(y = -3\).

The polynomial has exactly two zeroes. Thus, Assertion (A) is true.


Evaluating Reason (R):

By the fundamental theorem of algebra, a polynomial of degree \(n\) has at most \(n\) real zeroes. For a quadratic polynomial (\(n=2\)), it can have 0, 1, or 2 zeroes. Thus, "at most two zeroes" is a true mathematical statement. Reason (R) is true.


Relating (A) and (R):

Since \(p(y)\) is a quadratic polynomial, it follows the rule stated in Reason (R). The fact that a quadratic polynomial can have at most two zeroes is the reason why this specific quadratic polynomial has two zeroes (and not more).


Step 3: Final Answer:

Both (A) and (R) are true and (R) is the correct explanation of (A).
Quick Tip: For a quadratic polynomial \(ax^{2} + bx + c\), calculate the discriminant \(D = b^{2} - 4ac\).
If \(D > 0\), there are 2 distinct zeroes.
If \(D = 0\), there is 1 repeated zero.
If \(D < 0\), there are no real zeroes.
In all cases, the count is \(\le 2\).

Step 1: Understanding the Concept:

A year consists of 52 full weeks plus some extra days. A leap year has 366 days, and a non-leap year has 365 days.


Step 2: Detailed Explanation:

Evaluating Assertion (A):

Leap year = 366 days = 52 weeks + 2 extra days.

52 weeks contain 52 Mondays. We get a 53rd Monday if one of the 2 extra days is a Monday.

Possible pairs for the 2 extra days:

1. (Sun, Mon)

2. (Mon, Tue)

3. (Tue, Wed)

4. (Wed, Thu)

5. (Thu, Fri)

6. (Fri, Sat)

7. (Sat, Sun)

Total outcomes = 7.

Favorable outcomes (containing Monday) = 2 (Pairs 1 and 2).
\(P(53 Mondays in leap year) = \frac{2}{7}\).

Thus, Assertion (A) is true.


Evaluating Reason (R):

Non-leap year = 365 days = 52 weeks + 1 extra day.

52 weeks contain 52 Mondays. We get a 53rd Monday if the extra day is a Monday.

Possible extra days: {Sun, Mon, Tue, Wed, Thu, Fri, Sat.

Total outcomes = 7. Favorable outcome = 1 (Monday).
\(P(53 Mondays in non-leap year) = \frac{1}{7}\).

The reason states the probability is \(\frac{5}{7}\), which is incorrect.

Thus, Reason (R) is false.


Step 3: Final Answer:

Assertion (A) is true but Reason (R) is false.
Quick Tip: In any year, there are at least 52 of every day of the week.
The probability of having a 53rd occurrence of any specific day is:
- \(\frac{1}{7}\) for a non-leap year (because of 1 extra day).
- \(\frac{2}{7}\) for a leap year (because of 2 extra days).

Step 1: Understanding the Concept:

To determine if three points form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

If a triangle is formed, we can identify its type (Scalene, Isosceles, or Equilateral) based on the lengths of its sides.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

Let the points be \(P(1, 0)\), \(Q(-5, 0)\), and \(R(-2, 5)\).

Calculate the length of side \(PQ\):
\[ PQ = \sqrt{(-5 - 1)^2 + (0 - 0)^2} = \sqrt{(-6)^2 + 0} = 6 \]

Calculate the length of side \(QR\):
\[ QR = \sqrt{(-2 - (-5))^2 + (5 - 0)^2} = \sqrt{(3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \]

Calculate the length of side \(RP\):
\[ RP = \sqrt{(1 - (-2))^2 + (0 - 5)^2} = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \]

Checking the triangle inequality:
\(PQ + QR = 6 + 5.83 = 11.83 > 5.83\) (\(RP\))
\(QR + RP = 5.83 + 5.83 = 11.66 > 6\) (\(PQ\))
\(RP + PQ = 5.83 + 6 = 11.83 > 5.83\) (\(QR\))

Since the sum of any two sides is greater than the third side, the points form a triangle.

Since \(QR = RP = \sqrt{34}\), two sides are equal in length.


Step 4: Final Answer:

The points form an Isosceles triangle.
Quick Tip: If the points were collinear, the area of the triangle calculated by the coordinates would be zero.
In coordinate geometry, check for equal side lengths first to quickly identify Isosceles or Equilateral types.

Step 1: Understanding the Concept:

In a right-angled triangle, \(\tan \theta\) is the ratio of the opposite side to the adjacent side. We can find the hypotenuse using the Pythagorean theorem to determine \(\sin \theta\) and \(\cos \theta\).


Step 2: Key Formula or Approach:
\[ Hypotenuse^2 = Opposite^2 + Adjacent^2 \]
\[ \sin \theta = \frac{Opposite}{Hypotenuse}, \quad \cos \theta = \frac{Adjacent}{Hypotenuse} \]


Step 3: Detailed Explanation:

Given \(\tan \theta = \frac{24}{7}\).

Let Opposite side = \(24k\) and Adjacent side = \(7k\).
\[ Hypotenuse = \sqrt{(24k)^2 + (7k)^2} = \sqrt{576k^2 + 49k^2} = \sqrt{625k^2} = 25k \]

Now, calculate \(\sin \theta\) and \(\cos \theta\):
\[ \sin \theta = \frac{24k}{25k} = \frac{24}{25} \]
\[ \cos \theta = \frac{7k}{25k} = \frac{7}{25} \]

Adding the values:
\[ \sin \theta + \cos \theta = \frac{24}{25} + \frac{7}{25} = \frac{31}{25} \]


Step 4: Final Answer:

The value of \(\sin \theta + \cos \theta\) is \(\frac{31}{25}\).
Quick Tip: Memorizing common Pythagorean triplets like (7, 24, 25) helps save time in competitive exams.
If \(\tan \theta = \frac{a}{b}\), then \(\sin \theta + \cos \theta = \frac{a+b}{\sqrt{a^2+b^2}}\).

Step 1: Understanding the Concept:

The given expression can be simplified using basic trigonometric identities before substituting the values.


Step 2: Key Formula or Approach:

Algebraic identity: \((a+b)(a-b) = a^2 - b^2\)

Trigonometric identities: \(1 - \sin^2 \theta = \cos^2 \theta\) and \(1 - \cos^2 \theta = \sin^2 \theta\)

Relationship: \(\cot \theta = \frac{\cos \theta}{\sin \theta}\)


Step 3: Detailed Explanation:

The expression is:
\[ E = \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} \]

Using \((a+b)(a-b) = a^2 - b^2\):
\[ E = \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} \]

Applying trigonometric identities:
\[ E = \frac{\cos^2 \theta}{\sin^2 \theta} = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \cot^2 \theta \]

Given \(\cot \theta = \frac{7}{8}\).
\[ E = \left(\frac{7}{8}\right)^2 = \frac{49}{64} \]


Step 4: Final Answer:

The value of the expression is \(\frac{49}{64}\).
Quick Tip: Always look to simplify expressions using identities like \(\sin^2 \theta + \cos^2 \theta = 1\) before calculating individual sine or cosine values. It usually saves a lot of arithmetic effort.

Step 1: Understanding the Concept:

A chord of the larger circle that touches the smaller circle is a tangent to the smaller circle. The radius of the smaller circle to the point of contact is perpendicular to this chord.


Step 2: Key Formula or Approach:

Pythagorean theorem in the right triangle formed by the radius of the small circle, the radius of the large circle, and half of the chord.


Step 3: Detailed Explanation:

Let \(O\) be the common center of the two circles.

Let \(AB\) be the chord of the larger circle which touches the smaller circle at point \(P\).

Since \(P\) is the point of contact, \(OP \perp AB\) and \(OP\) is the radius of the smaller circle (\(OP = 4\) cm).

Join \(OA\). \(OA\) is the radius of the larger circle (\(OA = 5\) cm).

In right-angled triangle \(OPA\), by Pythagoras theorem:
\[ OA^2 = OP^2 + AP^2 \]
\[ 5^2 = 4^2 + AP^2 \]
\[ 25 = 16 + AP^2 \]
\[ AP^2 = 25 - 16 = 9 \implies AP = 3 cm \]

Since the perpendicular from the center to a chord bisects the chord:
\[ AB = 2 \times AP = 2 \times 3 = 6 cm \]


Step 4: Final Answer:

The length of the chord is 6 cm.
Quick Tip: The configuration always creates a right triangle with the larger radius as the hypotenuse (\(R\)), the smaller radius as one side (\(r\)), and half the chord length as the other side (\(L/2\)).
Formula: \(L = 2\sqrt{R^2 - r^2}\).

Step 1: Understanding the Concept:

A quadratic polynomial with zeroes \(\alpha\) and \(\beta\) can be written as \(k[x^2 - (\alpha + \beta)x + \alpha\beta]\), where \(k\) is a constant.


Step 2: Key Formula or Approach:

Sum of zeroes (\(S\)) = \(\alpha + \beta\)

Product of zeroes (\(P\)) = \(\alpha\beta\)

Polynomial \(p(x) = x^2 - Sx + P\)


Step 3: Detailed Explanation:

Let \(\alpha = 5 - 2\sqrt{3}\) and \(\beta = 5 + 2\sqrt{3}\).

Find the Sum of zeroes (\(S\)):
\[ S = \alpha + \beta = (5 - 2\sqrt{3}) + (5 + 2\sqrt{3}) = 10 \]

Find the Product of zeroes (\(P\)):
\[ P = \alpha\beta = (5 - 2\sqrt{3})(5 + 2\sqrt{3}) \]

Using identity \((a-b)(a+b) = a^2 - b^2\):
\[ P = (5)^2 - (2\sqrt{3})^2 = 25 - (4 \times 3) = 25 - 12 = 13 \]

The quadratic polynomial is:
\[ p(x) = x^2 - (10)x + 13 \]


Step 4: Final Answer:

The required quadratic polynomial is \(x^2 - 10x + 13\).
Quick Tip: Irrational zeroes always occur in conjugate pairs for polynomials with rational coefficients. If one zero is \(a - \sqrt{b}\), the other must be \(a + \sqrt{b}\).
Sum is always \(2a\) and product is \(a^2 - b\).

Step 1: Understanding the Concept:

According to the Basic Proportionality Theorem (Thales Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.


Step 2: Key Formula or Approach:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]


Step 3: Detailed Explanation:

Given \(DE \parallel BC\), we have:
\[ \frac{x}{x - 2} = \frac{x + 2}{x - 1} \]

Cross-multiplying the terms:
\[ x(x - 1) = (x + 2)(x - 2) \]

Expanding both sides:
\[ x^2 - x = x^2 - 4 \]

Subtracting \(x^2\) from both sides:
\[ -x = -4 \implies x = 4 \]


Step 4: Final Answer:

The value of \(x\) is 4.
Quick Tip: When solving ratios involving \(x\), check the final answer. Lengths like \(x-2\) and \(x-1\) must be positive, which is true here as \(4-2=2\) and \(4-1=3\).

Step 1: Understanding the Concept:

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


Step 2: Key Formula or Approach:
\[ \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ} \]


Step 3: Detailed Explanation:

Substituting the given side lengths into the ratio:
\[ \frac{4}{x} = \frac{6}{7.2} = \frac{y}{6} \]

First, simplify the known ratio:
\[ \frac{6}{7.2} = \frac{60}{72} = \frac{5}{6} \]

To find \(x\):
\[ \frac{4}{x} = \frac{5}{6} \implies 5x = 24 \implies x = \frac{24}{5} = 4.8 \]

To find \(y\):
\[ \frac{y}{6} = \frac{5}{6} \implies y = 5 \]


Step 4: Final Answer:

The values are \(x = 4.8\) and \(y = 5\).
Quick Tip: Always ensure you are pairing corresponding sides correctly. Look at the order of vertices in the similarity statement (\(ABC \sim XYZ\)) to match sides.

Step 1: Understanding the Concept:

We need to isolate the trigonometric terms and use the fundamental identity \(\sin^2 \theta + \cos^2 \theta = 1\).


Step 2: Key Formula or Approach:

Transform the equations to find \(\cos \theta\) and \(\sin \theta\), then square and add them.


Step 3: Detailed Explanation:

Given:
\(x = h + a\cos\theta \implies x - h = a\cos\theta \implies \frac{x-h}{a} = \cos\theta\)
\(y = k + b\sin\theta \implies y - k = b\sin\theta \implies \frac{y-k}{b} = \sin\theta\)

Now, square both equations:
\[ \left(\frac{x-h}{a}\right)^2 = \cos^2\theta \]
\[ \left(\frac{y-k}{b}\right)^2 = \sin^2\theta \]

Add the two results:
\[ \left(\frac{x-h}{a}\right)^2 + \left(\frac{y-k}{b}\right)^2 = \cos^2\theta + \sin^2\theta \]

Using identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \left(\frac{x-h}{a}\right)^2 + \left(\frac{y-k}{b}\right)^2 = 1 \]


Step 4: Final Answer:

Hence proved.
Quick Tip: This derivation represents the equation of an ellipse centered at \((h, k)\). Whenever you see \(\cos\) and \(\sin\) in parametric form, squaring and adding is the most common technique to eliminate \(\theta\).

Step 1: Understanding the Concept:

Simplifying complex trigonometric fractions often involves finding a common denominator and using fundamental identities like \(1 - \sec^2 A = -\tan^2 A\).


Step 2: Key Formula or Approach:
\[ \sec^2 A - \tan^2 A = 1 \implies 1 - \sec^2 A = -\tan^2 A \]


Step 3: Detailed Explanation:

LHS = \(\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A}\)

Factor out \(\tan A\):
\[ = \tan A \left[ \frac{1}{1 + \sec A} - \frac{1}{1 - \sec A} \right] \]

Taking common denominator:
\[ = \tan A \left[ \frac{(1 - \sec A) - (1 + \sec A)}{(1 + \sec A)(1 - \sec A)} \right] \]
\[ = \tan A \left[ \frac{1 - \sec A - 1 - \sec A}{1 - \sec^2 A} \right] \]
\[ = \tan A \left[ \frac{-2 \sec A}{-\tan^2 A} \right] \]
\[ = \frac{2 \tan A \sec A}{\tan^2 A} = \frac{2 \sec A}{\tan A} \]

Convert to sine and cosine:
\[ = \frac{2 / \cos A}{\sin A / \cos A} = \frac{2}{\sin A} = 2 \csc A = RHS \]


Step 4: Final Answer:

Hence proved.
Quick Tip: Be very careful with signs when applying identities like \(1 - \sec^2 \theta\). Since \(\sec^2 \theta = 1 + \tan^2 \theta\), then \(1 - \sec^2 \theta = -\tan^2 \theta\). Missing a negative sign is a common error here.

Step 1: Understanding the Concept:

The inradius (\(r\)) of any triangle can be calculated using the formula \(r = \frac{Area}{Semi-perimeter}\). For a right-angled triangle, there is a specific shortcut formula.


Step 2: Key Formula or Approach:

Shortcut for right triangle: \(r = \frac{P + B - H}{2}\) where \(P\) is perpendicular, \(B\) is base, and \(H\) is hypotenuse.


Step 3: Detailed Explanation:

In \(\Delta ABC\), \(AB = 4\) cm and \(BC = 3\) cm.

Using Pythagoras theorem to find hypotenuse \(AC\):
\[ AC = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 cm \]

Using the inradius formula for right triangle:
\[ r = \frac{AB + BC - AC}{2} \]
\[ r = \frac{4 + 3 - 5}{2} = \frac{2}{2} = 1 cm \]


Step 4: Final Answer:

The radius of the inscribed circle is 1 cm.
Quick Tip: For a right-angled triangle with sides \(a, b\) and hypotenuse \(c\), the inradius is always \(\frac{a+b-c}{2}\). This is much faster than using \(\frac{Area}{s}\) during competitive exams.

Step 1: Understanding the Concept:

The lengths of tangents drawn from an external point to a circle are equal. We will identify these points and their corresponding tangents.


Step 2: Key Formula or Approach:

From point \(P\): \(PM = PN\)

From point \(Q\): \(QM = QS\)

From point \(R\): \(RN = RS\)


Step 3: Detailed Explanation:

Let's find the perimeter of \(\Delta PQR\):
\[ Perimeter = PQ + QR + PR \]

We can split side \(QR\) at point \(S\):
\[ = PQ + (QS + SR) + PR \]

Substituting equal tangents (\(QS = QM\) and \(SR = RN\)):
\[ = PQ + QM + RN + PR \]

Observing the segments: \(PQ + QM = PM\) and \(PR + RN = PN\).
\[ = PM + PN \]

Since tangents from external point \(P\) are equal (\(PM = PN\)):
\[ Perimeter = PM + PM = 2PM \]

Therefore:
\[ PM = \frac{1}{2}(Perimeter of \Delta PQR) = \frac{1}{2}(PQ + QR + PR) \]


Step 4: Final Answer:

Hence proved.
Quick Tip: The segment \(PM\) is called the semi-perimeter of triangle \(PQR\). In such configurations involving excircles, the distance from the vertex to the point of contact on the extended side always equals the semi-perimeter.

Step 1: Understanding the Concept:

Let \(r\) be the radius and \(h\) be the height for both solids. The curved surface area (CSA) of a cylinder is \(2\pi rh\) and the CSA of a cone is \(\pi rl\), where \(l = \sqrt{r^2 + h^2}\).


Step 2: Key Formula or Approach:
\[ \frac{CSA of Cylinder}{CSA of Cone} = \frac{8}{5} \]


Step 3: Detailed Explanation:
\[ \frac{2\pi rh}{\pi r \sqrt{r^2 + h^2}} = \frac{8}{5} \]

Simplifying by cancelling \(\pi r\):
\[ \frac{2h}{\sqrt{r^2 + h^2}} = \frac{8}{5} \implies \frac{h}{\sqrt{r^2 + h^2}} = \frac{4}{5} \]

Squaring both sides:
\[ \frac{h^2}{r^2 + h^2} = \frac{16}{25} \]

Cross-multiplying:
\[ 25h^2 = 16r^2 + 16h^2 \]
\[ 9h^2 = 16r^2 \]

Taking the ratio \(r/h\):
\[ \frac{r^2}{h^2} = \frac{9}{16} \]

Taking the square root:
\[ \frac{r}{h} = \frac{3}{4} \]


Step 4: Final Answer:

The ratio between the radius and height is \(3 : 4\).
Quick Tip: When you see a ratio resulting in \(h/\sqrt{r^2+h^2} = 4/5\), recognize the 3-4-5 triplet. Here \(h=4\) corresponds to the hypotenuse 5, meaning the other side \(r\) must be 3. Thus, \(r/h = 3/4\).

Step 1: Understanding the Concept:

When two coins are tossed, the sample space \(S\) consists of all possible outcomes.
\(S = \{HH, HT, TH, TT\}\)

Total number of outcomes \(n(S) = 4\).


Step 3: Detailed Explanation:

(i) At least one head:

Favorable outcomes: \(\{HH, HT, TH\}\).

Number of favorable outcomes = 3.

Probability = \(3/4\).


(ii) At most one tail:

"At most one" means 0 tails or 1 tail.

Favorable outcomes: \(\{HH, HT, TH\}\).

Number of favorable outcomes = 3.

Probability = \(3/4\).


(iii) A head and a tail:

Favorable outcomes: \(\{HT, TH\}\).

Number of favorable outcomes = 2.

Probability = \(2/4 = 1/2\).


Step 4: Final Answer:

(i) \(3/4\), (ii) \(3/4\), (iii) \(1/2\).
Quick Tip: "At least one head" and "At most one tail" in a two-coin toss describe the exact same set of outcomes. Writing down the sample space clearly is the safest way to avoid confusion with words like "at least" and "at most".

Step 1: Understanding the Concept:

We use the method of contradiction. We assume \(\sqrt{3}\) is rational and show that this leads to a logical impossibility regarding its co-prime factors.


Step 3: Detailed Explanation:

1. Assume \(\sqrt{3}\) is rational. Then \(\sqrt{3} = \frac{p}{q}\) where \(p\) and \(q\) are co-prime integers (\(q \neq 0\)).

2. Squaring both sides: \(3 = \frac{p^2}{q^2} \implies p^2 = 3q^2\).

3. This implies that \(p^2\) is divisible by 3. By theorem, if \(p^2\) is divisible by 3, then \(p\) is also divisible by 3.

4. Let \(p = 3k\) for some integer \(k\). Substitute this in \(p^2 = 3q^2\):
\[ (3k)^2 = 3q^2 \implies 9k^2 = 3q^2 \implies 3k^2 = q^2 \]

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

6. This means \(p\) and \(q\) have at least 3 as a common factor.

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


Step 4: Final Answer:

Since the assumption is wrong, \(\sqrt{3}\) is an irrational number.
Quick Tip: This proof relies on the property: "If a prime number \(p\) divides \(a^2\), then \(p\) divides \(a\)." It is a standard derivation frequently asked in board exams.

Step 1: Understanding the Concept:

Any point on the \(x\)-axis has a \(y\)-coordinate of 0. Let the ratio be \(k : 1\). We use the section formula to find \(k\).


Step 2: Key Formula or Approach:

Section formula: \(x = \frac{mx_2 + nx_1}{m+n}, y = \frac{my_2 + ny_1}{m+n}\)


Step 3: Detailed Explanation:

Let the ratio be \(k : 1\). The points are \(A(6, 5)\) and \(B(-4, -1)\).

The point of intersection on \(x\)-axis is \((x, 0)\).

Using the \(y\)-coordinate:
\[ 0 = \frac{k(-1) + 1(5)}{k + 1} \]
\[ -k + 5 = 0 \implies k = 5 \]

So the ratio is \(5 : 1\).

Now, find the \(x\)-coordinate:
\[ x = \frac{5(-4) + 1(6)}{5 + 1} = \frac{-20 + 6}{6} = \frac{-14}{6} = -\frac{7}{3} \]


Step 4: Final Answer:

The ratio is \(5 : 1\) and the point of intersection is \((-7/3, 0)\).
Quick Tip: A quick formula for the ratio in which \(x\)-axis divides \((x_1, y_1)\) and \((x_2, y_2)\) is \(-y_1 : y_2\).
Here, \(-5 : -1\) which simplifies to \(5 : 1\).

Step 1: Understanding the Concept:

The Basic Proportionality Theorem (BPT), also known as Thales Theorem, states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.


Step 2: Key Formula or Approach:

We use the concept of the area of triangles.

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

If two triangles have the same base and lie between the same parallel lines, their areas are equal.


Step 3: Detailed Explanation:

Given: In \(\Delta ABC\), a line \(DE\) is parallel to \(BC\) (\(DE \parallel BC\)), intersecting \(AB\) at \(D\) and \(AC\) at \(E\).

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

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

Proof:

Calculate Area(\(\Delta ADE\)) taking \(AD\) as base:
\[ Area(\Delta ADE) = \frac{1}{2} \times AD \times EN \]

Calculate Area(\(\Delta BDE\)) taking \(DB\) as base:
\[ Area(\Delta BDE) = \frac{1}{2} \times DB \times EN \]

Divide these two areas:
\[ \frac{Area(\Delta ADE)}{Area(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \dots(1) \]

Similarly, calculate Area(\(\Delta ADE\)) taking \(AE\) as base and Area(\(\Delta CDE\)) taking \(EC\) as base:
\[ \frac{Area(\Delta ADE)}{Area(\Delta CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \dots(2) \]

Notice that \(\Delta BDE\) and \(\Delta CDE\) are on the same base \(DE\) and between the same parallels \(DE\) and \(BC\).

Therefore, Area(\(\Delta BDE\)) = Area(\(\Delta CDE\)) \(\dots(3)\).

From (1), (2), and (3), we can conclude that the ratios of the areas are equal.


Step 4: Final Answer:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Hence proved.
Quick Tip: While proving BPT, remember that the altitude for an obtuse-angled triangle (like \(\Delta BDE\)) lies outside the triangle. This is why \(EN\) is the height for both \(\Delta ADE\) and \(\Delta BDE\).

Step 1: Understanding the Concept:

When two triangles are similar, their corresponding angles are equal and their corresponding sides are in the same ratio. Medians divide the side they are drawn to into two equal parts.


Step 2: Key Formula or Approach:

SAS (Side-Angle-Side) Similarity Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the triangles are similar.


Step 3: Detailed Explanation:

Part (i): Prove \(\Delta AMC \sim \Delta PNR\)

Given \(\Delta ABC \sim \Delta PQR\).

Therefore, \(\angle A = \angle P\) and \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \dots(1)\).

Since \(CM\) is the median of \(\Delta ABC\), \(M\) is the mid-point of \(AB\), so \(AB = 2AM\).

Since \(RN\) is the median of \(\Delta PQR\), \(N\) is the mid-point of \(PQ\), so \(PQ = 2PN\).

Substitute these into the ratio from (1):
\[ \frac{2AM}{2PN} = \frac{AC}{PR} \implies \frac{AM}{PN} = \frac{AC}{PR} \]

In \(\Delta AMC\) and \(\Delta PNR\):

1. \(\frac{AM}{PN} = \frac{AC}{PR}\) (Proved above)

2. \(\angle A = \angle P\) (Given)

By SAS similarity criterion, \(\Delta AMC \sim \Delta PNR\).


Part (ii): Prove \(\Delta CMB \sim \Delta RNQ\)

Similarly, \(AB = 2MB\) and \(PQ = 2NQ\).

From (1), \(\frac{2MB}{2NQ} = \frac{BC}{QR} \implies \frac{MB}{NQ} = \frac{BC}{QR}\).

In \(\Delta CMB\) and \(\Delta RNQ\):

1. \(\frac{MB}{NQ} = \frac{BC}{QR}\) (Proved above)

2. \(\angle B = \angle Q\) (Corresponding angles of similar \(\Delta ABC\) and \(\Delta PQR\))

By SAS similarity criterion, \(\Delta CMB \sim \Delta RNQ\).


Step 4: Final Answer:

Both (i) and (ii) are proved using the SAS similarity criterion and properties of medians.
Quick Tip: When working with medians in similar triangles, remember that the ratio of corresponding medians is the same as the ratio of corresponding sides. In this case, \(CM/RN = AB/PQ = AC/PR = BC/QR\).

Step 1: Understanding the Concept:

The given data is in "more than" cumulative frequency format. First, convert it into a standard frequency distribution table with class intervals.


Step 2: Key Formula or Approach:

Class Interval Frequency = (CF of current class) - (CF of next class).

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

Mode \(= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)


Step 3: Detailed Explanation:

Convert to standard frequency table:





Finding Median:
\(N/2 = 40\). The cumulative frequency just greater than 40 is 52, corresponding to class interval 50-60.
\(l = 50\), \(cf = 37\) (of preceding class), \(f = 15\), \(h = 10\).

Median \(= 50 + \left( \frac{40 - 37}{15} \right) \times 10 = 50 + \left( \frac{3}{15} \right) \times 10 = 50 + 2 = 52\).

(Correction: Checking calculation again, \(50 + 2 = 52\)).


Finding Mode:

Maximum frequency is 15 in class 50-60.
\(l = 50\), \(f_1 = 15\), \(f_0 = 12\), \(f_2 = 12\), \(h = 10\).

Mode \(= 50 + \left( \frac{15 - 12}{2(15) - 12 - 12} \right) \times 10 = 50 + \left( \frac{3}{30 - 24} \right) \times 10 = 50 + \left( \frac{3}{6} \right) \times 10 = 50 + 5 = 55\).


Step 4: Final Answer:

The Median is 52 and the Mode is 55.
Quick Tip: When given "more than" or "less than" data, always cross-check that the sum of your calculated frequencies equals the total number of students. Here, \(3+5+7+10+12+15+12+6+2+8 = 80\). Correct!

Step 1: Understanding the Concept:

To draw the graph, find at least two solutions for each equation. The area of the triangle formed by two lines and the \(x\)-axis is given by \(\frac{1}{2} \times base \times height\).


Step 2: Key Formula or Approach:

1. Find intercepts on \(x\)-axis for both lines (\(y=0\)).

2. Find intersection point of the two lines (\(x, y\)).

3. Area \(= \frac{1}{2} \times |x_1 - x_2| \times |y_{intersection}|\).


Step 3: Detailed Explanation:

Line 1: \(x - y + 2 = 0\)

If \(y = 0\), \(x = -2\). Point \(A(-2, 0)\).

If \(x = 0\), \(y = 2\). Point \(B(0, 2)\).


Line 2: \(4x - y - 4 = 0\)

If \(y = 0\), \(4x = 4 \implies x = 1\). Point \(C(1, 0)\).

If \(x = 0\), \(y = -4\). Point \(D(0, -4)\).


Intersection Point:

Subtract Line 1 from Line 2:
\((4x - y - 4) - (x - y + 2) = 0\)
\(3x - 6 = 0 \implies x = 2\).

Substitute \(x=2\) in Line 1: \(2 - y + 2 = 0 \implies y = 4\).

Intersection Point \(P(2, 4)\).


Area of Triangle formed by \(A, C,\) and \(P\):

Base is on the \(x\)-axis between \(A(-2, 0)\) and \(C(1, 0)\).

Length of base \(= |1 - (-2)| = 3\) units.

Height is the \(y\)-coordinate of the intersection point \(P(2, 4)\).

Height \(= 4\) units.

Area \(= \frac{1}{2} \times 3 \times 4 = 6\) square units.


Step 4: Final Answer:

The area of the triangle formed is 6 sq units.
Quick Tip: Always check your intersection point by plugging it into both equations. For \((2, 4)\):
Line 1: \(2 - 4 + 2 = 0\) (Correct).
Line 2: \(4(2) - 4 - 4 = 8 - 8 = 0\) (Correct).

Step 1: Understanding the Concept:

This problem involves the relationship Time \(= \frac{Distance}{Speed}\). We setup a quadratic equation based on the difference in travel times.


Step 2: Key Formula or Approach:

Let speed of faster train \(= x\) km/hr.

Speed of slower train \(= (x - 10)\) km/hr.

Time difference \(= 1\) hour.


Step 3: Detailed Explanation:

Distance \(= 200\) km.

Time taken by slower train \(= \frac{200}{x - 10}\).

Time taken by faster train \(= \frac{200}{x}\).

According to the problem:
\[ \frac{200}{x - 10} - \frac{200}{x} = 1 \]

Take the common denominator:
\[ 200 \left( \frac{x - (x - 10)}{x(x - 10)} \right) = 1 \]
\[ 200 \left( \frac{10}{x^2 - 10x} \right) = 1 \]
\[ 2000 = x^2 - 10x \implies x^2 - 10x - 2000 = 0 \]

Solve the quadratic equation by splitting the middle term (\(50 \times 40 = 2000\)):
\[ x^2 - 50x + 40x - 2000 = 0 \]
\[ x(x - 50) + 40(x - 50) = 0 \]
\[ (x - 50)(x + 40) = 0 \]

Since speed cannot be negative, \(x = 50\).

Faster train speed \(= 50\) km/hr.

Slower train speed \(= 50 - 10 = 40\) km/hr.


Step 4: Final Answer:

The speeds of the trains are 50 km/hr and 40 km/hr.
Quick Tip: In competitive exams, check the factors of (Distance \(\times\) Speed Difference). Here \(200 \times 10 = 2000\). We need two numbers with a product of 2000 and a difference of 10. These are 50 and 40.

Step 1: Understanding the Concept:

Area of a square \(= s^2\) and Perimeter \(= 4s\). We are given two conditions involving two unknown side lengths.


Step 2: Key Formula or Approach:

Let the sides of the two squares be \(x\) and \(y\) (with \(x > y\)).

1. \(x^2 + y^2 = 640\)

2. \(4x - 4y = 64 \implies x - y = 16\)


Step 3: Detailed Explanation:

From the second equation, \(x = y + 16\).

Substitute this value of \(x\) into the first equation:
\[ (y + 16)^2 + y^2 = 640 \]
\[ y^2 + 32y + 256 + y^2 = 640 \]
\[ 2y^2 + 32y - 384 = 0 \]

Divide by 2:
\[ y^2 + 16y - 192 = 0 \]

Factor the equation (numbers with product 192 and difference 16 are 24 and 8):
\[ y^2 + 24y - 8y - 192 = 0 \]
\[ y(y + 24) - 8(y + 24) = 0 \]
\[ (y + 24)(y - 8) = 0 \]

Since side length cannot be negative, \(y = 8\).

Then \(x = y + 16 = 8 + 16 = 24\).


Step 4: Final Answer:

The sides of the two squares are 24 m and 8 m.
Quick Tip: If the difference in perimeters is \(P_{diff}\), then the difference in sides is \(P_{diff}/4\). This simplifies the problem immediately to \(x - y = 16\).

Step 1: Understanding the Concept:

The radius of a circle is half of its diameter.


Step 2: Key Formula or Approach:
\[ r = \frac{d}{2} \]


Step 3: Detailed Explanation:

The given diameter (\(d\)) of the circular brooch is 35 mm.

Using the formula for radius:
\[ r = \frac{35}{2} mm \]
\[ r = 17.5 mm \]


Step 4: Final Answer:

The radius of the circle is 17.5 mm.
Quick Tip: Always double-check the units given in the question. In some versions of this problem, it might be mm or cm. Ensure you maintain consistency.

Step 1: Understanding the Concept:

The circumference of a circle is the total distance around the edge of the circle.


Step 2: Key Formula or Approach:
\[ C = \pi d \quad or \quad C = 2\pi r \]

Take \( \pi = \frac{22}{7} \).


Step 3: Detailed Explanation:

Diameter \(d = 35\) mm.
\[ C = \frac{22}{7} \times 35 \]
\[ C = 22 \times 5 \]
\[ C = 110 mm \]


Step 4: Final Answer:

The circumference of the brooch is 110 mm.
Quick Tip: If the diameter is a multiple of 7, using \( \pi = \frac{22}{7} \) simplifies calculations significantly.

Step 1: Understanding the Concept:

The total wire length consists of the length used for the circular boundary plus the length used for the internal diameters.


Step 2: Key Formula or Approach:
\[ Total Length = Circumference + (5 \times Diameter) \]


Step 3: Detailed Explanation:

Length used for circumference \( = 110 mm \) (as calculated in part ii).

Length used for 5 diameters \( = 5 \times 35 mm = 175 mm \).

Total wire required:
\[ L = 110 + 175 \]
\[ L = 285 mm \]


Step 4: Final Answer:

The total length of silver wire required is 285 mm.
Quick Tip: Don't forget to add the circumference! Many students only calculate the diameter lengths.

Step 1: Understanding the Concept:

The circle is divided into 10 equal sectors. The area of each sector is one-tenth of the total area of the circle.


Step 2: Key Formula or Approach:
\[ Area of sector = \frac{\pi r^2}{10} \]


Step 3: Detailed Explanation:

Total Area of Circle \( = \pi r^2 \).
\[ Area = \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \]
\[ Area = \frac{11 \times 5 \times 35}{2} \]
\[ Area = \frac{1925}{2} = 962.5 mm^2 \]

Since there are 10 equal sectors:
\[ Area of each sector = \frac{962.5}{10} \]
\[ Area of each sector = 96.25 mm^2 \]


Step 4: Final Answer:

The area of each sector of the brooch is 96.25 mm\^2.
Quick Tip: The central angle of each sector is \( \frac{360^{\circ}{10} = 36^{\circ} \). You can use the sector formula \( \frac{\theta}{360} \times \pi r^2 \), which will yield the same result.

Step 1: Understanding the Concept:

The competitor has to travel to the potato and return to the bucket. The distance covered is twice the displacement of the potato from the bucket.


Step 3: Detailed Explanation:

The first potato is 5 m from the bucket.

Distance covered to pick it up \( = 5 m \).

Distance covered to drop it back \( = 5 m \).

Total distance \( = 5 + 5 = 10 m \).


Step 4: Final Answer:

The distance covered for the first potato is 10 m.
Quick Tip: Remember to double the distance for every potato because of the "to-and-fro" movement.

Step 1: Understanding the Concept:

Each subsequent potato is 3 m further than the previous one.


Step 3: Detailed Explanation:

Distance of 1st potato from bucket \( = 5 m \).

Distance of 2nd potato from bucket \( = 5 + 3 = 8 m \).

Total distance for 2nd potato \( = 2 \times 8 m = 16 m \).


Step 4: Final Answer:

The distance covered for the second potato is 16 m.
Quick Tip: The sequence of distances for individual potatoes forms an Arithmetic Progression (AP) with common difference \( d = 2 \times 3 = 6 m \).

Step 1: Understanding the Concept:

The distances covered for each potato form an Arithmetic Progression. We need to find the sum of the first 10 terms.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

First term (\(a\)) \( = 10 m \).

Second term \( = 16 m \).

Common difference (\(d\)) \( = 16 - 10 = 6 m \).

Number of potatoes (\(n\)) \( = 10 \).
\[ S_{10} = \frac{10}{2} [2(10) + (10-1)6] \]
\[ S_{10} = 5 [20 + 9 \times 6] \]
\[ S_{10} = 5 [20 + 54] \]
\[ S_{10} = 5 \times 74 \]
\[ S_{10} = 370 m \]


Step 4: Final Answer:

The total distance the competitor has to run is 370 m.
Quick Tip: Alternatively, calculate the sum of distances to the potatoes \( \sum = (5 + 8 + 11 + \dots) \) and then double the final result.

Step 1: Understanding the Concept:

Time taken is the ratio of total distance covered to the average speed.


Step 2: Key Formula or Approach:
\[ Time = \frac{Distance}{Speed} \]


Step 3: Detailed Explanation:

Total distance covered \( = 370 m \) (from part iii-a).

Average speed \( = 5 m/s \).
\[ Time = \frac{370}{5} \]
\[ Time = 74 s \]


Step 4: Final Answer:

The average time taken is 74 seconds.
Quick Tip: Ensure units are consistent (meters and meters/second) before dividing.

Step 1: Understanding the Concept:

In a right-angled triangle, the wire represents the hypotenuse connecting the ground point to the top of the section.


Step 2: Key Formula or Approach:

Use trigonometric ratio: \( \cos \theta = \frac{Base}{Hypotenuse} \).


Step 3: Detailed Explanation:

Let \( P \) be the base of the tower. In right \(\Delta OPB\):

Base \( OP = 6 m \).

Angle of elevation \( \angle BOP = 30^{\circ} \).

Let the length of the wire be \( OB \).
\[ \cos 30^{\circ} = \frac{OP}{OB} \]
\[ \frac{\sqrt{3}}{2} = \frac{6}{OB} \]
\[ OB = \frac{12}{\sqrt{3}} \]

Rationalizing:
\[ OB = \frac{12\sqrt{3}}{3} = 4\sqrt{3} m \]


Step 4: Final Answer:

The length of the wire to the top of section 'B' is \(4\sqrt{3}\) m.
Quick Tip: In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the hypotenuse is \( \frac{2}{\sqrt{3}} \) times the base adjacent to the \(30^{\circ}\) angle.

Step 1: Understanding the Concept:

Similarly, the wire to the top of section 'A' is the hypotenuse of the larger right triangle.


Step 2: Key Formula or Approach:
\[ \cos \theta = \frac{Base}{Hypotenuse} \]


Step 3: Detailed Explanation:

In right \(\Delta OPA\):

Base \( OP = 6 m \).

Angle of elevation \( \angle AOP = 60^{\circ} \).

Let the length of the wire be \( OA \).
\[ \cos 60^{\circ} = \frac{OP}{OA} \]
\[ \frac{1}{2} = \frac{6}{OA} \]
\[ OA = 12 m \]


Step 4: Final Answer:

The length of the wire to the top of section 'A' is 12 m.
Quick Tip: If the angle of elevation is \(60^{\circ}\), the hypotenuse is exactly double the adjacent base side.

Step 1: Understanding the Concept:

The distance \( AB \) is the difference between the heights of the two sections.


Step 2: Key Formula or Approach:
\[ \tan \theta = \frac{Perpendicular}{Base} \]
\[ AB = AP - BP \]


Step 3: Detailed Explanation:

In \(\Delta OPB\):
\[ \tan 30^{\circ} = \frac{BP}{6} \implies BP = \frac{6}{\sqrt{3}} = 2\sqrt{3} m \]

In \(\Delta OPA\):
\[ \tan 60^{\circ} = \frac{AP}{6} \implies AP = 6\sqrt{3} m \]

Now, calculate \( AB \):
\[ AB = 6\sqrt{3} - 2\sqrt{3} \]
\[ AB = 4\sqrt{3} m \]


Step 4: Final Answer:

The distance \( AB \) is \(4\sqrt{3}\) m.
Quick Tip: Height \( = Base \times \tan(elevation angle) \). Subtract the heights to get the length of the upper section.

Step 1: Understanding the Concept:

The area of a right-angled triangle is given by half the product of its base and perpendicular.


Step 2: Key Formula or Approach:
\[ Area = \frac{1}{2} \times Base \times Height \]


Step 3: Detailed Explanation:

For \(\Delta OPB\):

Base \( OP = 6 m \).

Height \( BP = 2\sqrt{3} m \) (calculated in part iii-a).
\[ Area = \frac{1}{2} \times 6 \times 2\sqrt{3} \]
\[ Area = 6\sqrt{3} m^2 \]


Step 4: Final Answer:

The area of \(\Delta OPB\) is \(6\sqrt{3}\) m\^2.
Quick Tip: Always use the exact radical form (\(\sqrt{3\)) unless the question asks for a decimal approximation.