Step 1: Concept Explanation:

This question involves understanding the nature of rational and irrational numbers.

A rational number can be expressed as \( \frac{p}{q} \), where \( p \) and \( q \) are integers, and \( q \neq 0 \) (for example, 2, \( \frac{3}{4} \), -5).

An irrational number, however, cannot be expressed in the form \( \frac{p}{q} \) (examples include \( \sqrt{2} \), \( \sqrt{3} \), \( \pi \)).

Step 2: Detailed Explanation:

When adding a rational number and an irrational number, the sum will always be irrational.

We can demonstrate this by proof by contradiction:

Let \( r \) be a rational number and \( i \) be an irrational number.

Assume that their sum, \( r + i = q \), is rational.

Then, we can express \( i = q - r \).

Since the difference of two rational numbers is rational, it would imply that \( i \) is rational.

However, this contradicts the assumption that \( i \) is irrational.

Therefore, the assumption that their sum is rational must be false.

Thus, the sum of a rational number and an irrational number is always irrational.

Example: 2 (rational) + \( \sqrt{3} \) (irrational) = 2 + \( \sqrt{3} \) (irrational).


Step 3: Final Conclusion:

The sum of a rational number and an irrational number results in an irrational number.

Therefore, the correct option is (D).
Quick Tip: Keep this in mind: the sum, difference, or product of a rational number and an irrational number is always irrational. Similarly, the quotient of a non-zero rational number and an irrational number is also irrational.

Step 1: Concept Explanation:

Prime factorization is the process of breaking down a number into its prime factors. This helps express the number as a product of primes.




Step 2: Approach for Factorization:

To perform the prime factorization, we divide the number by the smallest prime number repeatedly until we obtain 1.




Step 3: Detailed Calculation:

We begin by dividing 156 by the smallest prime number, 2:
\[ 156 \div 2 = 78 \]

Now, divide 78 by 2:
\[ 78 \div 2 = 39 \]

Next, divide 39 by 3:
\[ 39 \div 3 = 13 \]

Since 13 is a prime number, divide it by 13:
\[ 13 \div 13 = 1 \]

Thus, the prime factors of 156 are \(2, 2, 3, 13\).

This can be written as:
\[ 156 = 2 \times 2 \times 3 \times 13 \]

In exponential form, we have:
\[ 156 = 2^2 \times 3 \times 13 \]



Step 4: Final Conclusion:

The prime factorization of 156 is \(2^2 \times 3 \times 13\), which matches option (B).
Quick Tip: Always start factoring with the smallest prime number (2), and then proceed to higher primes (3, 5, 7, etc.). This ensures you don’t miss any prime factors and reduces the chances of errors.

Step 1: Conceptual Understanding:

We are given a rational equation involving \(x\) and \(y\), and our goal is to determine the ratio of \(x\) to \(y\). To solve this, we can either use cross-multiplication or apply a method called Componendo and Dividendo.




Step 2: Approach and Formula:

Method: Cross-multiplication

The first step is to cross-multiply the terms in the given equation to eliminate the fractions.

The equation is: \[ \frac{3x + 5y}{3x - 5y} = \frac{7}{3} \]
Now, we cross-multiply: \[ 3(3x + 5y) = 7(3x - 5y) \]



Step 3: Simplifying the Equation:

Simplifying both sides: \[ 9x + 15y = 21x - 35y \]
Now, move the terms involving \(x\) to one side and the terms involving \(y\) to the other side: \[ 15y + 35y = 21x - 9x \] \[ 50y = 12x \]
To find the ratio \(x : y\), divide both sides of the equation by \(y\): \[ \frac{x}{y} = \frac{50}{12} \]
Simplifying the fraction: \[ \frac{x}{y} = \frac{25}{6} \]
Hence, the ratio \(x : y = 25 : 6\).




Step 4: Conclusion:

The required ratio of \(x : y\) is \(25 : 6\), which corresponds to option (A).
Quick Tip: When solving such problems, the cross-multiplication method is effective for eliminating fractions. You can also use the Componendo and Dividendo rule, which simplifies the ratio calculation when given fractions.

Step 1: Concept of Discriminant:

The discriminant of a quadratic equation helps determine the nature of its roots.

For a general quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant \(D\) is given by the formula:




Step 2: Formula for Discriminant:

The formula for the discriminant is:
\[ D = b^2 - 4ac \]



Step 3: Applying the Formula:

For the given quadratic equation \(3x^2 - 2x + \frac{1}{3} = 0\), we can identify the coefficients as:
\[ a = 3, \quad b = -2, \quad c = \frac{1}{3} \]
Now, substitute these values into the discriminant formula:
\[ D = (-2)^2 - 4(3)\left(\frac{1}{3}\right) \]
\[ D = 4 - 4(1) \]
\[ D = 4 - 4 \]
\[ D = 0 \]



Step 4: Conclusion:

The value of the discriminant is 0, indicating that the equation has two real and equal roots.

Thus, option (A) is correct.
Quick Tip: When the discriminant \(D = 0\), the quadratic equation has exactly two real and equal roots. If \(D > 0\), the equation has two distinct real roots, and if \(D < 0\), there are no real roots.

Step 1: Concept Overview:

The mean of grouped data can be computed by using the class marks (mid-points) of each class interval. The formula for calculating the mean is based on the direct method.




Step 2: Formula for Mean:

The mean (\(\bar{x}\)) is given by the formula:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
Where \(f_i\) represents the frequency and \(x_i\) is the class mark for each class interval.


The class mark (\(x_i\)) is calculated as the average of the upper and lower class limits:
\[ x_i = \frac{Upper limit + Lower limit}{2} \]



Step 3: Detailed Solution:

To compute the mean, we first calculate the class marks and the products \(f_i \times x_i\) for each class:


\begin{tabular{|c|c|c|c|
\hline
Class-Interval & Frequency (\(f_i\)) & Class Mark (\(x_i\)) & \(f_i \times x_i\)

\hline
2 -- 4 & 2 & \(\frac{2 + 4}{2} = 3\) & \(2 \times 3 = 6\)

\hline
4 -- 6 & 4 & \(\frac{4 + 6}{2} = 5\) & \(4 \times 5 = 20\)

\hline
6 -- 8 & 7 & \(\frac{6 + 8}{2} = 7\) & \(7 \times 7 = 49\)

\hline
8 -- 10 & 3 & \(\frac{8 + 10}{2} = 9\) & \(3 \times 9 = 27\)

\hline
10 -- 12 & 4 & \(\frac{10 + 12}{2} = 11\) & \(4 \times 11 = 44\)

\hline
Total & \(\sum f_i = 20\) & & \(\sum f_i x_i = 146\)

\hline
\end{tabular


Now, applying the mean formula:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{146}{20} = 7.3 \]



Step 4: Conclusion:

The calculated mean of the data is 7.3. Hence, option (D) is correct.
Quick Tip: While calculating the mean, be sure to carefully compute the class marks and verify the products \(f_i \times x_i\). This will help prevent any mistakes in the final calculation. Double-check your sums to ensure accuracy.

Step 1: Conceptual Understanding:

This problem is based on the similarity of triangles and the Basic Proportionality Theorem (also known as Thales' Theorem).

The theorem states that if a line is drawn parallel to one side of a triangle and intersects the other two sides, then the two sides are divided in the same ratio.




Step 2: Formula for Basic Proportionality Theorem:

For \(\triangle ABC\), where \(DE \parallel BC\), the Basic Proportionality Theorem gives the relation:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]



Step 3: Detailed Calculation:

We are provided with the following information:
\(AD = 1.5\) cm, \(DB = 3\) cm, and \(AE = 2\) cm.


Using the formula from Step 2:
\[ \frac{1.5}{3} = \frac{2}{EC} \]
Now, solve for \(EC\):
\[ 1.5 \times EC = 3 \times 2 \] \[ 1.5 \times EC = 6 \] \[ EC = \frac{6}{1.5} \] \[ EC = 4 \, cm \]
To find the length of \(AC\), we add \(AE\) and \(EC\):
\[ AC = AE + EC = 2 \, cm + 4 \, cm = 6 \, cm \]



Step 4: Conclusion:

The length of \(AC\) is 6.0 cm. Therefore, option (A) is the correct answer.
Quick Tip: The Basic Proportionality Theorem is essential when working with triangles involving parallel lines. Ensure to use the correct ratios and substitute the given values accurately into the formula for solving such problems.

Step 1: Understanding the Concept:

This question is based on coordinate geometry and involves using the distance formula to find the distance between two points.




Step 2: Key Formula or Approach:

The formula for the distance (d) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]



Step 3: Detailed Explanation:

The given points are \((2, 3)\) and \((4, 1)\).

Let \( (x_1, y_1) = (2, 3) \) and \( (x_2, y_2) = (4, 1) \).

Substitute these values into the distance formula:
\[ d = \sqrt{(4 - 2)^2 + (1 - 3)^2} \] \[ d = \sqrt{(2)^2 + (-2)^2} \] \[ d = \sqrt{4 + 4} \] \[ d = \sqrt{8} \]
Simplifying \(\sqrt{8}\):
\[ d = \sqrt{4 \times 2} = 2\sqrt{2} \]
Hence, the distance is \(2\sqrt{2}\) units.




Step 4: Final Answer:

The distance between the points (2, 3) and (4, 1) is \(2\sqrt{2}\) units.

Therefore, option (B) is correct.
Quick Tip: When using the distance formula, be careful with the signs of the coordinates, especially during subtraction. Remember that the square of any number is always positive, so \( (-2)^2 = 4 \).

Step 1: Understanding the Concept:

The roots of a quadratic equation \(ax^2 + bx + c = 0\) are equal if and only if its discriminant is zero.




Step 2: Key Formula or Approach:

The condition for equal roots is:

Discriminant, \(D = b^2 - 4ac = 0\)




Step 3: Detailed Explanation:

Comparing the given equation \(3x^2 - 12x + m = 0\) with the standard quadratic equation \(ax^2 + bx + c = 0\):
\( a = 3 \)
\( b = -12 \)
\( c = m \)

Since the roots are equal, we use the condition \(D = 0\):
\[ b^2 - 4ac = 0 \] \[ (-12)^2 - 4(3)(m) = 0 \] \[ 144 - 12m = 0 \] \[ 144 = 12m \] \[ m = \frac{144}{12} \] \[ m = 12 \]



Step 4: Final Answer:

The value of m is 12.

Therefore, option (C) is correct.
Quick Tip: The nature of the roots of a quadratic equation depends on the discriminant (D):
- \(D > 0\): Two distinct real roots.
- \(D = 0\): Two equal real roots.
- \(D < 0\): No real roots.

Step 1: Understanding the Concept:

This question is based on the relationship between the HCF (Highest Common Factor) and LCM (Lowest Common Multiple) of two numbers.




Step 2: Key Formula or Approach:

For two positive integers 'a' and 'b', the formula for the relationship between their HCF and LCM is:
\[ HCF(a, b) \times LCM(a, b) = a \times b \]



Step 3: Detailed Explanation:

The given values are:

First number (a) = 15

Second number (b) = 25

HCF(15, 25) = 5


Using the formula:
\[ 5 \times LCM(15, 25) = 15 \times 25 \] \[ LCM(15, 25) = \frac{15 \times 25}{5} \] \[ LCM(15, 25) = 3 \times 25 \] \[ LCM(15, 25) = 75 \]



Step 4: Final Answer:

The LCM of 15 and 25 is 75.

Therefore, option (B) is correct.
Quick Tip: This formula (Product of two numbers = HCF \(\times\) LCM) is applicable only for two numbers. It is not valid for three or more numbers.

Step 1: Understanding the Concept:

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




Step 2: Key Formula or Approach:

The formula for the common difference (d) is:
\[ d = a_2 - a_1 \]
where \(a_2\) is the second term and \(a_1\) is the first term.




Step 3: Detailed Explanation:

The given A.P. is: 3, 1, -1, -3, ...

Here, the first term \(a_1 = 3\).

The second term \(a_2 = 1\).

Calculate the common difference using the formula:
\[ d = 1 - 3 \] \[ d = -2 \]
We can also use the next two terms to verify:
\[ d = a_3 - a_2 = -1 - 1 = -2 \]
Since the difference is constant, the common difference is -2.




Step 4: Final Answer:

The common difference for the given A.P. is -2.

Therefore, option (A) is correct.
Quick Tip: When calculating the common difference, always subtract the preceding term from the succeeding term (\(a_{n+1} - a_n\)). If you subtract the succeeding term from the preceding term, the sign will be incorrect.

Step 1: Understanding the Concept:

The median class is the class interval in which the median lies. To find it, we first need to calculate the cumulative frequency.




Step 2: Key Formula or Approach:

1. Find the sum of all frequencies (\(N = \sum f_i\)).

2. Calculate \(N/2\).

3. Find the class interval whose cumulative frequency is just greater than \(N/2\). This is the median class.




Step 3: Detailed Explanation:

We calculate the cumulative frequency (c.f.) for the given table:

\begin{tabular{|c|c|c|
\hline
Class-Interval & Frequency (\(f_i\)) & Cumulative Frequency (c.f.)

\hline
0 -- 5 & 2 & 2

\hline
5 -- 10 & 7 & 2 + 7 = 9

\hline
10 -- 15 & 3 & 9 + 3 = 12

\hline
15 -- 20 & 10 & 12 + 10 = 22

\hline
20 -- 25 & 4 & 22 + 4 = 26

\hline
\end{tabular


Total frequency, \(N = \sum f_i = 26\).

Now, we calculate \(N/2\):
\[ \frac{N}{2} = \frac{26}{2} = 13 \]
We need to find the class whose cumulative frequency is just greater than 13.

From the table, we see that the cumulative frequency 22 is just greater than 13.

The corresponding class interval for this cumulative frequency is 15 -- 20.




Step 4: Final Answer:

Hence, the median class is 15 -- 20.

Therefore, option (D) is correct.
Quick Tip: To find the median class, creating a cumulative frequency table is the first and most important step. Always compare the value of \(N/2\) with the cumulative frequency column, not the frequency column.

Step 1: Understanding the Concept:

This question is based on trigonometry. We need to find the value of angle \(A\) from the given value of \(\tan A\) and then substitute it into the given expression.




Step 2: Key Formula or Approach:

Method 1: By finding the value of the angle

We know that \( \tan 45^\circ = 1 \).

Therefore, if \(\tan A = 1\), then \(A = 45^\circ\).

Now we will substitute this value into the expression \(2 \sin A \cos A\).


Method 2: Using the identity

We can use the trigonometric identity \( \sin 2A = 2 \sin A \cos A \).




Step 3: Detailed Explanation:

Solution by Method 1:

Given, \(\tan A = 1\).

We know that \(\tan 45^\circ = 1\), so \(A = 45^\circ\).

Now, substituting \(A = 45^\circ\) into the expression \(2 \sin A \cos A\):
\[ 2 \sin 45^\circ \cos 45^\circ \]
We know that \(\sin 45^\circ = \frac{1}{\sqrt{2}}\) and \(\cos 45^\circ = \frac{1}{\sqrt{2}}\).
\[ = 2 \times \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{1}{\sqrt{2}}\right) \] \[ = 2 \times \frac{1}{2} \] \[ = 1 \]



Step 4: Final Answer:

The value of \(2 \sin A \cos A\) is 1.

Therefore, option (C) is correct.
Quick Tip: Remembering the identity \( \sin 2A = 2 \sin A \cos A \) helps solve such problems very quickly. From \(\tan A = 1\), we get \(A=45^\circ\), so \(\sin(2 \times 45^\circ) = \sin(90^\circ) = 1\).

Step 1: Understanding the Concept:

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

This property forms a right-angled triangle with the center, the external point, and the point of contact as vertices.




Step 2: Key Formula or Approach:

Use Pythagoras' theorem. If O is the center, Q is the external point, and P is the point of contact, then \(\triangle OPQ\) is a right-angled triangle with \(\angle OPQ = 90^\circ\).

According to Pythagoras' theorem:
\[ (Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2 \] \[ OQ^2 = OP^2 + PQ^2 \]
Where \(OQ\) is the distance of the external point from the center, \(OP\) is the radius, and \(PQ\) is the length of the tangent.




Step 3: Detailed Explanation:

The given values are:

Distance of point Q from the center, \(OQ = 25\) cm.

Length of the tangent, \(PQ = 24\) cm.

Radius of the circle, \(OP = r\) (to be found).


Substituting the values into the Pythagoras' theorem formula:
\[ 25^2 = r^2 + 24^2 \] \[ 625 = r^2 + 576 \]
Solve for \(r^2\):
\[ r^2 = 625 - 576 \] \[ r^2 = 49 \] \[ r = \sqrt{49} \] \[ r = 7 cm \]



Step 4: Final Answer:

The radius of the circle is 7 cm.

Therefore, option (A) is correct.
Quick Tip: Remembering Pythagorean triplets like (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17) saves time in competitive exams. Here, we saw the (7, 24, 25) triplet.

Step 1: Understanding the Concept:

This question is based on the applications of trigonometry (height and distance).

The angle of elevation is the angle between the horizontal line and the line of sight to an object above the horizontal.

Here, the height of the tower (perpendicular) and the length of the shadow (base) form a right-angled triangle.




Step 2: Key Formula or Approach:

We will use the trigonometric ratio \(\tan \theta\):
\[ \tan \theta = \frac{Perpendicular}{Base} = \frac{Height of the tower}{Length of the shadow} \]



Step 3: Detailed Explanation:

The given values are:

Height of the tower (Perpendicular) = 20 m

Length of the shadow (Base) = \(20\sqrt{3}\) m

Let the angle of elevation be \(\theta\).


Substituting the values into the formula:
\[ \tan \theta = \frac{20}{20\sqrt{3}} \] \[ \tan \theta = \frac{1}{\sqrt{3}} \]
We know that \(\tan 30^\circ = \frac{1}{\sqrt{3}}\).

Therefore, \(\theta = 30^\circ\).




Step 4: Final Answer:

The value of the angle of elevation is \(30^\circ\).

Therefore, option (A) is correct.
Quick Tip: For problems of height and distance, drawing a clear diagram is always helpful. It helps you identify which side is the perpendicular, base, and hypotenuse, and which trigonometric ratio to use.

Step 1: Understanding the Concept:

The area of a sector of a circle is the portion of the area of the circle enclosed by two radii and the corresponding arc.




Step 2: Key Formula or Approach:

The formula for the area of a sector with radius \(r\) and angle \(\theta\) (in degrees) is:
\[ Area of sector = \frac{\theta}{360^\circ} \times \pi r^2 \]



Step 3: Detailed Explanation:

The given values are:

Radius \(r = 14\) cm

Angle \(\theta = 30^\circ\)

Using the value of \(\pi\) as \(\frac{22}{7}\).


Substituting the values into the formula:
\[ Area = \frac{30}{360} \times \frac{22}{7} \times (14)^2 \] \[ Area = \frac{1}{12} \times \frac{22}{7} \times 14 \times 14 \] \[ Area = \frac{1}{12} \times 22 \times 2 \times 14 \] \[ Area = \frac{1}{6} \times 22 \times 14 \] \[ Area = \frac{1}{3} \times 11 \times 14 \] \[ Area = \frac{154}{3} cm^2 \]



Step 4: Final Answer:

The area of the corresponding sector is \(\frac{154}{3}\) cm\(^2\).

Therefore, option (C) is correct.
Quick Tip: When calculating the area of a sector, cancel out the numbers first to simplify the calculation, instead of multiplying them till the end. This avoids dealing with large numbers and reduces the chance of errors.

Step 1: Understanding the Concept:

This question is based on general knowledge and recognition of common 3D shapes.




Step 2: Detailed Explanation:

A joker's cap, also known as a birthday cap, is typically shaped like a cone.

A cone is a three-dimensional (3D) geometric shape that tapers smoothly from a flat base (often, though not necessarily, circular) to a point called the apex or vertex.

(A) Cylindrical: The shape of a cylinder (like a can).

(C) Triangular: This is a 2D shape.

(D) Rectangular: This is also a 2D shape.

Therefore, the correct 3D shape of a joker's cap is conical.




Step 3: Final Answer:

The shape of a joker's cap is conical.

Therefore, option (B) is correct.
Quick Tip: Practice identifying the geometric shapes of objects around you. This helps you differentiate between 2D (two-dimensional) and 3D (three-dimensional) shapes.

Step 1: Understanding the Concept:

This question is about understanding the relationship between the different methods of calculating the mean of grouped data.

There are three main methods for calculating the mean: the direct method, the assumed mean method, and the step-deviation method.




Step 2: Detailed Explanation:

Direct Method (\(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\)): This is the basic method of calculating the mean. However, when the values of \(x_i\) and \(f_i\) are large, the calculation becomes complex.

Assumed Mean Method (\(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\)): This is a simplified form of the direct method. In it, we choose an assumed mean ('a') and use deviations (\(d_i = x_i - a\)) to make the calculation easier.

Step-Deviation Method (\(\bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right)\)): This is an even more simplified form of the assumed mean method, used when the class size ('h') is uniform. It further simplifies the calculation by dividing the deviations by the class size (\(u_i = d_i/h\)).

Thus, both the assumed mean and step-deviation methods are ways to simplify the calculations of the basic Direct Method. They are simplified forms of the direct method.




Step 3: Final Answer:

The assumed mean method and the step-deviation method are simplified forms of the direct method.

Therefore, option (B) is correct.
Quick Tip: Think of the three methods as a progression: the direct method is the original, the assumed mean method simplifies it, and the step-deviation method simplifies it even further (when the class size is uniform).

Step 1: Understanding the Concept:

This question is based on the empirical relationship between mean, median, and mode. This relationship is particularly useful for moderately skewed distributions.




Step 2: Key Formula or Approach:

The formula for the empirical relationship between mean, median, and mode is:
\[ Mode = 3 \times Median - 2 \times Mean \]



Step 3: Detailed Explanation:

The given values are:

Mean = 15

Median = 24


Substituting these values into the formula:
\[ Mode = 3 \times (24) - 2 \times (15) \] \[ Mode = 72 - 30 \] \[ Mode = 42 \]



Step 4: Final Answer:

The value of the mode is 42.

Therefore, option (D) is correct.
Quick Tip: Remember this important empirical relationship (Mode = 3 Median - 2 Mean). It is often asked in competitive exams where two measures are given and the third has to be found.

Step 1: Understanding the Concept:

Probability is the measure of the likelihood of an event occurring.




Step 2: Key Formula or Approach:

The formula for the probability (P) of an event is:
\[ P(Event) = \frac{Number of favorable outcomes}{Total number of possible outcomes} \]



Step 3: Detailed Explanation:

When a die is thrown, the total possible outcomes are: \{1, 2, 3, 4, 5, 6\.

Therefore, the total number of possible outcomes = 6.

The favorable outcomes for the event of getting an even number are: \{2, 4, 6\.

Therefore, the number of favorable outcomes = 3.


Now, we calculate the probability:
\[ P(Even number) = \frac{3}{6} \] \[ P(Even number) = \frac{1}{2} \]



Step 4: Final Answer:

The probability of getting an even number is \(\frac{1}{2}\).

Therefore, option (B) is correct.
Quick Tip: For problems related to a die, always remember the total of 6 possible outcomes. Even numbers are (2, 4, 6), odd numbers are (1, 3, 5), and prime numbers are (2, 3, 5).

Step 1: Understanding the Concept:

This question is based on the probability of drawing a card from a standard deck of playing cards.




Step 2: Key Formula or Approach:

The formula for probability:
\[ P(Event) = \frac{Number of favorable outcomes}{Total number of possible outcomes} \]



Step 3: Detailed Explanation:

The total number of cards in a standard deck = 52.

Therefore, the total number of possible outcomes = 52.

The favorable outcomes for the event of drawing a queen.

A deck has 4 queens (one of spades, one of hearts, one of clubs, and one of diamonds).

Therefore, the number of favorable outcomes = 4.


Now, we calculate the probability:
\[ P(Queen) = \frac{4}{52} \]
Simplifying the fraction:
\[ P(Queen) = \frac{1}{13} \]



Step 4: Final Answer:

The probability of drawing a queen is \(\frac{1}{13}\).

Therefore, option (B) is correct.
Quick Tip: Remember the structure of a standard deck of cards: 52 cards, 4 suits (spades, hearts, clubs, diamonds), 13 cards in each suit (A, 2-10, J, Q, K). There are 26 red and 26 black cards.

Step 1: Understanding the Concept:

This question requires the calculation of the volume of a right circular cone using the given height and diameter.




Step 2: Key Formula or Approach:

The formula for the volume (V) of a cone is:
\[ V = \frac{1}{3} \pi r^2 h \]
where \(r\) is the radius of the base and \(h\) is the height of the cone.




Step 3: Detailed Explanation:

First, we need to find the radius from the given diameter.

Given:

Height (\(h\)) = 48 cm

Diameter = 28 cm


The radius (\(r\)) is half of the diameter:
\[ r = \frac{Diameter}{2} = \frac{28}{2} = 14 cm \]
Now, substitute the values of \(r\) and \(h\) into the volume formula (using \(\pi = \frac{22}{7}\)):
\[ V = \frac{1}{3} \times \frac{22}{7} \times (14)^2 \times 48 \] \[ V = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times 48 \]
To simplify the calculation, we can cancel the terms:
\[ V = \frac{1}{3} \times 22 \times ( \frac{14}{7} \times 14 ) \times 48 \] \[ V = \frac{1}{3} \times 22 \times (2 \times 14) \times 48 \] \[ V = 22 \times 28 \times (\frac{48}{3}) \] \[ V = 22 \times 28 \times 16 \] \[ V = 616 \times 16 \] \[ V = 9856 cm^3 \]



Step 4: Final Answer:

The volume of the cone is 9856 cm\(^3\).
Quick Tip: Always double-check if the question provides the radius or the diameter. A common mistake is to use the diameter value directly in the formula instead of the radius.

Step 1: Understanding the Concept:

The HCF (Highest Common Factor) of two numbers is the largest number that divides both of them. The prime factorization method involves breaking down each number into a product of its prime factors.




Step 2: Key Formula or Approach:

1. Find the prime factorization of each number.

2. Identify the common prime factors.

3. The HCF is the product of the lowest powers of these common prime factors.




Step 3: Detailed Explanation:

First, find the prime factorization of 96:
\[ 96 = 2 \times 48 = 2 \times 2 \times 24 = 2 \times 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \] \[ 96 = 2^5 \times 3^1 \]
Next, find the prime factorization of 404:
\[ 404 = 2 \times 202 = 2 \times 2 \times 101 \] \[ 404 = 2^2 \times 101^1 \]
(Note: 101 is a prime number).


Now, identify the common prime factors and their lowest powers:

The only common prime factor is 2.

The lowest power of 2 in both factorizations is \(2^2\).

Therefore, the HCF is the product of these lowest powers:
\[ HCF(96, 404) = 2^2 = 4 \]



Step 4: Final Answer:

The HCF of 96 and 404 is 4.
Quick Tip: To find the HCF, take the lowest power of common factors. To find the LCM (Lowest Common Multiple), take the highest power of all factors.

Step 1: Understanding the Concept:

This question requires the use of the section formula, which is used to find the coordinates of a point that divides a line segment in a given ratio.




Step 2: Key Formula or Approach:

The section formula for a point \(P(x, y)\) that divides the line segment joining \(A(x_1, y_1)\) and \(B(x_2, y_2)\) in the ratio \(m:n\) is:
\[ P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]



Step 3: Detailed Explanation:

Here, the given points and ratio are:
\( (x_1, y_1) = (-1, 7) \)
\( (x_2, y_2) = (4, -3) \)

Ratio \( m:n = 2:3 \), so \( m=2 \) and \( n=3 \).


Now, calculate the x-coordinate:
\[ x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1 \]
Next, calculate the y-coordinate:
\[ y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3 \]
So, the coordinates of the point are (1, 3).




Step 4: Final Answer:

The co-ordinates of the point are (1, 3).
Quick Tip: Be careful with the signs of the coordinates and the order of \(m, n, x_1, x_2, y_1, y_2\) when substituting into the section formula to avoid calculation errors.

Step 1: Understanding the Concept:

This question requires finding the values of angles \(\alpha\) and \(\beta\) from their given trigonometric ratios and then calculating their sum. We assume the angles are in the first quadrant (acute angles), which is standard for such problems.




Step 2: Key Formula or Approach:

We need to recall the standard values of trigonometric functions for common angles (like 0\(^\circ\), 30\(^\circ\), 45\(^\circ\), 60\(^\circ\), 90\(^\circ\)).




Step 3: Detailed Explanation:

First, find the value of \(\alpha\):

Given \(\sin \alpha = \frac{1}{2}\).

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

Therefore, \(\alpha = 30^\circ\).


Next, find the value of \(\beta\):

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

We know that \(\cos 60^\circ = \frac{1}{2}\).

Therefore, \(\beta = 60^\circ\).


Now, find the value of \((\alpha + \beta)\):
\[ \alpha + \beta = 30^\circ + 60^\circ = 90^\circ \]



Step 4: Final Answer:

The value of \((\alpha + \beta)\) is 90\(^\circ\).
Quick Tip: Memorizing the trigonometric values for standard angles (0\(^\circ\), 30\(^\circ\), 45\(^\circ\), 60\(^\circ\), 90\(^\circ\)) is essential for quickly solving such problems in exams.

Step 1: Understanding the Concept:

This question requires finding the distance between a given point and the origin (0, 0) in a Cartesian coordinate system using the distance formula.




Step 2: Key Formula or Approach:

The distance (d) of a point \((x, y)\) from the origin \((0, 0)\) is given by a simplified version of the distance formula:
\[ d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \]



Step 3: Detailed Explanation:

The given point is P(–6, 8).

Here, \(x = -6\) and \(y = 8\).

Substitute these values into the formula:
\[ d = \sqrt{(-6)^2 + (8)^2} \] \[ d = \sqrt{36 + 64} \] \[ d = \sqrt{100} \] \[ d = 10 \]
The distance is 10 units.




Step 4: Final Answer:

The distance of point P(–6, 8) from the origin is 10 units.
Quick Tip: The distance of any point \((x, y)\) from the origin is simply the hypotenuse of a right-angled triangle with base \(|x|\) and height \(|y|\). This is a direct application of the Pythagorean theorem.

Step 1: Understanding the Concept:

This question requires proving a trigonometric identity by manipulating one side of the equation (usually the more complex one) to show it is equivalent to the other side.




Step 2: Key Formula or Approach:

We will start with the Left Hand Side (LHS) and use the following fundamental identities:

1. \(\cot A = \frac{\cos A}{\sin A}\)

2. \(\csc A = \frac{1}{\sin A}\)




Step 3: Detailed Explanation:

Starting with the LHS:
\[ LHS = \frac{\cot A - \cos A}{\cot A + \cos A} \]
Substitute \(\cot A = \frac{\cos A}{\sin A}\) into the expression:
\[ LHS = \frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A} \]
Factor out \(\cos A\) from the numerator and the denominator:
\[ LHS = \frac{\cos A \left( \frac{1}{\sin A} - 1 \right)}{\cos A \left( \frac{1}{\sin A} + 1 \right)} \]
Cancel the common factor \(\cos A\):
\[ LHS = \frac{\frac{1}{\sin A} - 1}{\frac{1}{\sin A} + 1} \]
Now, substitute \(\csc A = \frac{1}{\sin A}\):
\[ LHS = \frac{\csc A - 1}{\csc A + 1} \]
This is equal to the Right Hand Side (RHS).
\[ LHS = RHS \]
Hence, proved.




Step 4: Final Answer:

The identity \(\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\csc A - 1}{\csc A + 1}\) is successfully proven by expressing \(\cot A\) in terms of \(\sin A\) and \(\cos A\) and simplifying.
Quick Tip: When proving trigonometric identities, a good strategy is often to express all terms in the form of \(\sin\) and \(\cos\). Then, use algebraic simplification and other identities to reach the desired expression.

Step 1: Understanding the Concept:

This question is based on the properties of an Arithmetic Progression (A.P.). We will use the formula for the n\(^{th}\) term of an A.P. to set up an equation and solve for the common difference.




Step 2: Key Formula or Approach:

The formula for the n\(^{th}\) term (\(a_n\)) of an A.P. is:
\[ a_n = a + (n-1)d \]
where \(a\) is the first term and \(d\) is the common difference.




Step 3: Detailed Explanation:

First, let's write the expressions for the 17\(^{th}\) term (\(a_{17}\)) and the 10\(^{th}\) term (\(a_{10}\)):
\[ a_{17} = a + (17-1)d = a + 16d \] \[ a_{10} = a + (10-1)d = a + 9d \]
According to the question, the 17\(^{th}\) term exceeds the 10\(^{th}\) term by 7. We can write this as an equation:
\[ a_{17} = a_{10} + 7 \]
Now, substitute the expressions for \(a_{17}\) and \(a_{10}\):
\[ a + 16d = (a + 9d) + 7 \]
Subtract 'a' from both sides:
\[ 16d = 9d + 7 \]
Solve for \(d\):
\[ 16d - 9d = 7 \] \[ 7d = 7 \] \[ d = \frac{7}{7} = 1 \]



Step 4: Final Answer:

The common difference of the A.P. is 1.
Quick Tip: For problems comparing two terms of an A.P., you can directly use the fact that the difference between the m\(^{th}\) and n\(^{th}\) term is \(a_m - a_n = (m-n)d\). Here, \(a_{17} - a_{10} = (17-10)d = 7d\). Since this difference is given as 7, we get \(7d = 7\), which gives \(d=1\).

Step 1: Understanding the Concept:

This is a quadratic equation of the form \(ax^2 + bx + c = 0\). We can find its roots by either using the quadratic formula or by factoring the equation (splitting the middle term).




Step 2: Key Formula or Approach:

We will use the factorization method. We need to find two numbers that multiply to \(ac\) and add up to \(b\).

Here, \(a = \sqrt{2}\), \(b = 7\), and \(c = 5\sqrt{2}\).

The product \(ac = (\sqrt{2})(5\sqrt{2}) = 5 \times 2 = 10\).

We need two numbers that multiply to 10 and add to 7. These numbers are 2 and 5.




Step 3: Detailed Explanation:

Split the middle term (7x) into 2x and 5x:
\[ \sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0 \]
Now, factor by grouping. We can write \(2\) as \((\sqrt{2} \times \sqrt{2})\):
\[ \sqrt{2}x^2 + \sqrt{2}\sqrt{2}x + 5x + 5\sqrt{2} = 0 \]
Factor out the common terms from the first two and the last two terms:
\[ \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0 \]
Now, factor out the common binomial term \((x + \sqrt{2})\):
\[ (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \]
To find the roots, set each factor to zero:

Either \(\sqrt{2}x + 5 = 0 \implies \sqrt{2}x = -5 \implies x = -\frac{5}{\sqrt{2}}\)

Or \(x + \sqrt{2} = 0 \implies x = -\sqrt{2}\)




Step 4: Final Answer:

The roots of the equation are \(-\sqrt{2}\) and \(-\frac{5}{\sqrt{2}}\).
Quick Tip: When factoring quadratic equations involving square roots, look for ways to express whole numbers as products of square roots (e.g., \(2 = \sqrt{2} \times \sqrt{2}\)). This often reveals the common factors.

Step 1: Understanding the Concept:

This problem requires the use of the Angle-Angle (AA) similarity criterion for triangles. If two angles of one triangle are equal to two corresponding angles of another triangle, then the two triangles are similar.




Step 2: Key Formula or Approach:

1. Identify two triangles that can be proven similar.

2. Use the AA similarity criterion.

3. Use the property that the ratios of corresponding sides of similar triangles are equal.




Step 3: Detailed Explanation:

Let's consider \(\triangle ADC\) and \(\triangle BAC\).

In these two triangles, we have:

1. \(\angle ADC = \angle BAC\) (This is given in the problem).

2. \(\angle ACD = \angle BCA\) (This is the common angle \(\angle C\)).


Since two angles of \(\triangle ADC\) are equal to two corresponding angles of \(\triangle BAC\), the two triangles are similar by the AA similarity criterion.
\[ \triangle ADC \sim \triangle BAC \]
Because the triangles are similar, the ratio of their corresponding sides must be equal. Let's match the corresponding vertices: A\(\leftrightarrow\)B, D\(\leftrightarrow\)A, C\(\leftrightarrow\)C.
\[ \frac{AD}{BA} = \frac{DC}{AC} = \frac{AC}{BC} \]
Taking the last two parts of the ratio:
\[ \frac{DC}{AC} = \frac{AC}{BC} \]
In a more intuitive order for this problem:
\[ \frac{CA}{CB} = \frac{CD}{CA} \]
Now, cross-multiply to get the desired result:
\[ CA \times CA = CB \times CD \] \[ CA^2 = CB \times CD \]
Hence, proved.




Step 4: Final Answer:

By proving that \(\triangle ADC \sim \triangle BAC\) using the AA similarity criterion, we establish that the ratio of corresponding sides is equal, which leads to the result \(CA^2 = CB \times CD\).
Quick Tip: When proving similarity, it is very important to write the similarity relation with the corresponding vertices in the correct order (e.g., \(\triangle ADC \sim \triangle BAC\)). This helps in correctly identifying the pairs of corresponding sides.

Step 1: Understanding the Concept:

This question is based on a property of tangents drawn to a circle from an external point. A quadrilateral that circumscribes a circle is called a tangential quadrilateral.




Step 2: Key Formula or Approach:

The key theorem is: The lengths of tangents drawn from an external point to a circle are equal.




Step 3: Detailed Explanation:

Let the circle touch the sides AB, BC, CD, and DA of the quadrilateral at points P, Q, R, and S, respectively.

ABCD is the quadrilateral and A, B, C, D are the external points from which tangents are drawn to the circle.


Using the theorem about tangents from an external point:

From point A: \(AP = AS\) ---(1)

From point B: \(BP = BQ\) ---(2)

From point C: \(CR = CQ\) ---(3)

From point D: \(DR = DS\) ---(4)


Now, add these four equations:
\[ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \]
By observing the figure, we can group the segments to form the sides of the quadrilateral:
\(AP + BP = AB\)
\(CR + DR = CD\)
\(AS + DS = AD\)
\(BQ + CQ = BC\)


Substitute these into the combined equation:
\[ AB + CD = AD + BC \]
Hence, proved. This is known as Pitot's theorem.




Step 4: Final Answer:

By applying the theorem that tangents from an external point to a circle are equal in length and summing up the lengths of the tangent segments, we have proven that for a quadrilateral circumscribing a circle, \(AB + CD = AD + BC\).
Quick Tip: This is a standard theorem for tangential quadrilaterals. The key to the proof is to label the points of tangency and add the four equations of equal tangent lengths. The rearrangement of terms is crucial to get the final result.

Step 1: Understanding the Concept:

We are given the median and the total frequency of a distribution with two missing frequencies. We will use the formula for the median of grouped data and the sum of frequencies to create two simultaneous equations to solve for x and y.




Step 2: Key Formula or Approach:

1. Sum of frequencies: \(\sum f_i = n\)

2. Median formula: \( Median = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
where \(l\) = lower limit of median class, \(n\) = total frequency, \(cf\) = cumulative frequency of the class preceding the median class, \(f\) = frequency of the median class, and \(h\) = class size.




Step 3: Detailed Explanation:

First, let's create the cumulative frequency (cf) table:

\begin{tabular{|c|c|c|
\hline
Class interval & Frequency (\(f_i\)) & Cumulative Frequency (cf)

\hline
0 -- 10 & 5 & 5

\hline
10 -- 20 & x & 5 + x

\hline
20 -- 30 & 20 & 25 + x

\hline
30 -- 40 & 15 & 40 + x

\hline
40 -- 50 & y & 40 + x + y

\hline
50 -- 60 & 5 & 45 + x + y

\hline
\end{tabular


Equation 1: From the total frequency

Given \(n = 60\). The sum of frequencies from our table is \(45 + x + y\).
\[ 45 + x + y = 60 \] \[ x + y = 15 \] ---(1)


Equation 2: From the median formula

Given Median = 28.5. This value lies in the class interval 20 -- 30.

Therefore, the median class is 20 -- 30.

From this class, we get:

Lower limit (\(l\)) = 20

Frequency of median class (\(f\)) = 20

Class size (\(h\)) = 10

Total frequency (\(n\)) = 60, so \(\frac{n}{2} = 30\).

Cumulative frequency of the class preceding the median class (\(cf\)) = \(5 + x\).


Substitute these values into the median formula:
\[ 28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10 \] \[ 28.5 - 20 = \left( \frac{30 - 5 - x}{20} \right) \times 10 \] \[ 8.5 = \left( \frac{25 - x}{2} \right) \] \[ 8.5 \times 2 = 25 - x \] \[ 17 = 25 - x \] \[ x = 25 - 17 = 8 \]

Now, substitute \(x=8\) into equation (1):
\[ 8 + y = 15 \] \[ y = 15 - 8 = 7 \]



Step 4: Final Answer:

The values of the missing frequencies are x = 8 and y = 7.
Quick Tip: When solving for two missing frequencies, one equation will almost always come from the sum of frequencies (\(\sum f_i = n\)). The second will come from the formula for the given measure of central tendency (mean, median, or mode).

Step 1: Understanding the Concept:

This is a basic probability problem. Probability is the ratio of the number of favorable outcomes to the total number of possible outcomes.




Step 2: Key Formula or Approach:

The formula for probability of an event (E) is:
\[ P(E) = \frac{Number of favorable outcomes}{Total number of possible outcomes} \]



Step 3: Detailed Explanation:

First, we determine the total number of pens.

Number of defective pens = 12

Number of good pens = 132

Total number of pens = Number of defective pens + Number of good pens
\[ Total number of pens = 12 + 132 = 144 \]
This is the total number of possible outcomes.


Next, we determine the number of favorable outcomes.

The event is "the pen taken out is a good one".

Number of good pens = 132

This is the number of favorable outcomes.


Now, calculate the probability:
\[ P(good pen) = \frac{Number of good pens}{Total number of pens} = \frac{132}{144} \]
Simplify the fraction. Both numbers are divisible by 12.
\[ P(good pen) = \frac{132 \div 12}{144 \div 12} = \frac{11}{12} \]



Step 4: Final Answer:

The probability that the pen taken out is a good one is \(\frac{11}{12}\).
Quick Tip: In probability questions, always start by identifying two key numbers: the total number of possible outcomes (the denominator) and the number of outcomes that satisfy the event's condition (the numerator).

Step 1: Understanding the Concept:

This is a word problem that can be solved by setting up a pair of linear equations in two variables representing the numerator and the denominator of the fraction.




Step 2: Key Formula or Approach:

Let the fraction be \(\frac{x}{y}\), where \(x\) is the numerator and \(y\) is the denominator.

We will translate the two given conditions into two separate linear equations.




Step 3: Detailed Explanation:

Condition 1: "A fraction becomes \(\frac{1}{3}\) when 1 is subtracted from the numerator."

This gives us the equation:
\[ \frac{x-1}{y} = \frac{1}{3} \]
Cross-multiplying, we get:
\[ 3(x-1) = y \] \[ 3x - 3 = y \] \[ 3x - y = 3 \] --- (Equation 1)


Condition 2: "it becomes \(\frac{1}{4}\) when 8 is added to its denominator."

This gives us the equation:
\[ \frac{x}{y+8} = \frac{1}{4} \]
Cross-multiplying, we get:
\[ 4x = y+8 \] \[ 4x - y = 8 \] --- (Equation 2)


Now we have a system of two linear equations:

1) \(3x - y = 3\)

2) \(4x - y = 8\)


To solve this, we can subtract Equation 1 from Equation 2:
\[ (4x - y) - (3x - y) = 8 - 3 \] \[ 4x - y - 3x + y = 5 \] \[ x = 5 \]
Now, substitute the value of \(x=5\) into Equation 1 to find \(y\):
\[ 3(5) - y = 3 \] \[ 15 - y = 3 \] \[ y = 15 - 3 = 12 \]
So, the numerator is 5 and the denominator is 12.




Step 4: Final Answer:

The required fraction is \(\frac{5}{12}\).
Quick Tip: Always check your answer by substituting the values back into the original conditions.
Condition 1: \(\frac{5-1}{12} = \frac{4}{12} = \frac{1}{3}\) (Correct).
Condition 2: \(\frac{5}{12+8} = \frac{5}{20} = \frac{1}{4}\) (Correct).

Step 1: Understanding the Concept:

This problem relates distance, speed, and time. We will form a quadratic equation based on the relationship \( Time = \frac{Distance}{Speed} \). The key is that the difference in the time taken at the usual speed and the increased speed is 30 minutes.




Step 2: Key Formula or Approach:

Let the usual speed of the plane be \(x\) km/hr.

Let the distance be \(D = 1500\) km.

Usual time taken (\(T_1\)) = \(\frac{1500}{x}\) hours.

Increased speed = \((x + 250)\) km/hr.

New time taken (\(T_2\)) = \(\frac{1500}{x+250}\) hours.

The plane is late by 30 minutes, which is \(\frac{30}{60} = \frac{1}{2}\) hour.

This means the usual time is longer than the new time by \(\frac{1}{2}\) hour.

So, the equation is: \(T_1 - T_2 = \frac{1}{2}\).




Step 3: Detailed Explanation:

Set up the equation based on the time difference:
\[ \frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2} \]
Factor out 1500:
\[ 1500 \left( \frac{1}{x} - \frac{1}{x+250} \right) = \frac{1}{2} \] \[ 1500 \left( \frac{(x+250) - x}{x(x+250)} \right) = \frac{1}{2} \] \[ 1500 \left( \frac{250}{x^2 + 250x} \right) = \frac{1}{2} \]
Cross-multiply:
\[ 1500 \times 250 \times 2 = x^2 + 250x \] \[ 750000 = x^2 + 250x \]
Rearrange into a standard quadratic equation:
\[ x^2 + 250x - 750000 = 0 \]
We can solve this by factoring. We need two numbers that multiply to -750000 and have a difference of 250. These numbers are 1000 and -750.
\[ x^2 + 1000x - 750x - 750000 = 0 \] \[ x(x+1000) - 750(x+1000) = 0 \] \[ (x-750)(x+1000) = 0 \]
This gives two possible values for \(x\): \(x = 750\) or \(x = -1000\).

Since speed cannot be negative, we discard \(x = -1000\).




Step 4: Final Answer:

The usual speed of the plane is 750 km/hr.
Quick Tip: In time-speed-distance problems, ensure all units are consistent. If speed is in km/hr, time must be in hours. Convert minutes to hours (e.g., 30 minutes = 0.5 hours) before setting up the equation.

Step 1: Understanding the Concept:

This problem is an application of trigonometry involving heights and distances. We will use the tangent trigonometric ratio to set up two equations with two variables (the height of the poles and the distance of the point from one pole).




Step 2: Key Formula or Approach:

Let \(h\) be the height of the two poles. Let AB and CD be the poles.

Let P be the point on the road between the poles. The width of the road is BD = 80 m.

Let the distance of the point P from pole AB be \(x\) meters (i.e., BP = \(x\)).

Then the distance of the point P from pole CD will be \((80 - x)\) meters (i.e., DP = \(80 - x\)).

The angles of elevation are \(\angle APB = 60^\circ\) and \(\angle CPD = 30^\circ\).

We will use the formula: \(\tan \theta = \frac{Perpendicular}{Base}\).




Step 3: Detailed Explanation:

From the right-angled \(\triangle APB\):
\[ \tan 60^\circ = \frac{AB}{BP} = \frac{h}{x} \] \[ \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \] --- (Equation 1)


From the right-angled \(\triangle CPD\):
\[ \tan 30^\circ = \frac{CD}{DP} = \frac{h}{80-x} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{80-x} \implies h = \frac{80-x}{\sqrt{3}} \] --- (Equation 2)


Now, we equate the two expressions for \(h\) from Equation 1 and Equation 2:
\[ x\sqrt{3} = \frac{80-x}{\sqrt{3}} \]
Cross-multiply:
\[ x\sqrt{3} \times \sqrt{3} = 80-x \] \[ 3x = 80-x \] \[ 4x = 80 \] \[ x = 20 m \]
This is the distance of the point from the first pole (AB).

The distance from the second pole (CD) is \(80 - x = 80 - 20 = 60 m\).


Now, find the height \(h\) using Equation 1:
\[ h = x\sqrt{3} = 20\sqrt{3} m \]



Step 4: Final Answer:

The height of the poles is 20\(\sqrt{3}\) m, and the distances of the point from the poles are 20 m and 60 m.
Quick Tip: Drawing a clear, labelled diagram is the most important first step for solving heights and distances problems. It helps you visualize the triangles and apply the correct trigonometric ratios.

Step 1: Understanding the Concept:

This is another problem on heights and distances. We have a building with a tower on top. We will use the tangent ratio for two different right-angled triangles to find the required height.




Step 2: Key Formula or Approach:

Let BC be the building and CD be the tower on top. Let A be the point on the ground.

Height of the building, BC = 20 m.

Let the height of the tower, CD = \(h\) m.

Let the distance of the point A from the base of the building, AB = \(x\) m.

The angle of elevation of the bottom of the tower is \(\angle CAB = 45^\circ\).

The angle of elevation of the top of the tower is \(\angle DAB = 60^\circ\).

We will use \(\tan \theta = \frac{Perpendicular}{Base}\).




Step 3: Detailed Explanation:

From the right-angled \(\triangle ABC\):
\[ \tan 45^\circ = \frac{BC}{AB} = \frac{20}{x} \]
Since \(\tan 45^\circ = 1\):
\[ 1 = \frac{20}{x} \implies x = 20 m \] --- (Equation 1)


From the right-angled \(\triangle ABD\):

The total height of the perpendicular is BD = BC + CD = \(20 + h\).
\[ \tan 60^\circ = \frac{BD}{AB} = \frac{20+h}{x} \]
Since \(\tan 60^\circ = \sqrt{3}\):
\[ \sqrt{3} = \frac{20+h}{x} \] \[ x\sqrt{3} = 20+h \] --- (Equation 2)


Now, substitute the value of \(x=20\) from Equation 1 into Equation 2:
\[ 20\sqrt{3} = 20+h \]
Solve for \(h\):
\[ h = 20\sqrt{3} - 20 \] \[ h = 20(\sqrt{3} - 1) m \]
We can use the value \(\sqrt{3} \approx 1.732\):
\[ h = 20(1.732 - 1) = 20(0.732) = 14.64 m \]



Step 4: Final Answer:

The height of the tower is 20(\(\sqrt{3}\) - 1) m, which is approximately 14.64 m.
Quick Tip: In problems with two angles of elevation from the same point, you will always get two right-angled triangles sharing a common base or height. Use this common side to equate the two equations you form.

Step 1: Understanding the Concept:

The total surface area of the toy is the sum of the curved surface area of the cylinder and the curved surface areas of the two hemispheres that have been scooped out. The flat circular bases of the cylinder have been removed, so we do not include their area.




Step 2: Key Formula or Approach:

Total Surface Area (TSA) of the toy = (Curved Surface Area of Cylinder) + 2 \(\times\) (Curved Surface Area of Hemisphere)

- CSA of Cylinder = \(2 \pi r h\)

- CSA of Hemisphere = \(2 \pi r^2\)

So, TSA = \(2 \pi r h + 2 \times (2 \pi r^2) = 2 \pi r h + 4 \pi r^2 = 2 \pi r (h + 2r)\)




Step 3: Detailed Explanation:

Given values:

Height of the cylinder (\(h\)) = 10 cm

Radius of the base (\(r\)) = 3.5 cm = \(\frac{7}{2}\) cm


Now, calculate the required areas:

Curved Surface Area of Cylinder:
\[ CSA_{cyl} = 2 \pi r h = 2 \times \frac{22}{7} \times 3.5 \times 10 = 2 \times \frac{22}{7} \times \frac{7}{2} \times 10 = 22 \times 10 = 220 cm^2 \]

Curved Surface Area of two Hemispheres:
\[ 2 \times CSA_{hemi} = 2 \times (2 \pi r^2) = 4 \pi r^2 = 4 \times \frac{22}{7} \times (3.5)^2 = 4 \times \frac{22}{7} \times 3.5 \times 3.5 \] \[ = 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 22 \times 7 = 154 cm^2 \]

Total Surface Area of the toy:
\[ TSA_{toy} = CSA_{cyl} + 2 \times CSA_{hemi} = 220 + 154 = 374 cm^2 \]
Alternatively, using the combined formula:
\[ TSA = 2 \pi r (h + 2r) = 2 \times \frac{22{7} \times 3.5 (10 + 2 \times 3.5) \] \[ = 2 \times \frac{22}{7} \times \frac{7}{2} (10 + 7) = 22 \times 17 = 374 cm^2 \]



Step 4: Final Answer:

The total surface area of the toy is 374 cm\(^2\).
Quick Tip: For combined solids, visualize the surfaces that are exposed. In this case, scooping out the hemispheres exposes their inner curved surfaces while removing the cylinder's flat top and bottom. So, we add the CSA of hemispheres and CSA of the cylinder.

Step 1: Understanding the Concept:

The solid is a combination of a cone and a hemisphere. The total volume of the solid will be the sum of the volume of the cone and the volume of the hemisphere.




Step 2: Key Formula or Approach:

Volume of the solid = Volume of Cone + Volume of Hemisphere

- Volume of Cone = \(\frac{1}{3} \pi r^2 h_{cone}\)

- Volume of Hemisphere = \(\frac{2}{3} \pi r^3\)




Step 3: Detailed Explanation:

Given values:

Radius of cone = Radius of hemisphere (\(r\)) = 3 cm

Total height of the solid = 6 cm


First, we need to find the height of the cone (\(h_{cone}\)).

The height of the hemispherical part is equal to its radius, so Height\(_{hemi}\) = 3 cm.

Height of the cone = Total height - Height of hemisphere
\[ h_{cone} = 6 - 3 = 3 cm \]

Now, calculate the volumes:

Volume of Cone:
\[ V_{cone} = \frac{1}{3} \pi r^2 h_{cone} = \frac{1}{3} \pi (3)^2 (3) = \frac{1}{3} \pi \times 9 \times 3 = 9\pi cm^3 \]

Volume of Hemisphere:
\[ V_{hemi} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3)^3 = \frac{2}{3} \pi \times 27 = 2 \times 9\pi = 18\pi cm^3 \]

Total Volume of the solid:
\[ V_{solid} = V_{cone} + V_{hemi} = 9\pi + 18\pi = 27\pi cm^3 \]
If asked to use \(\pi = \frac{22}{7}\):
\[ V_{solid} = 27 \times \frac{22}{7} = \frac{594}{7} \approx 84.86 cm^3 \]



Step 4: Final Answer:

The volume of the solid is 27\(\pi\) cm\(^3\).
Quick Tip: For combined solids, remember that heights add up but radii must be common. In a cone-on-hemisphere problem, the total height is the sum of the cone's height and the hemisphere's radius.