UP Board Class 10 Mathematics Question Paper 2023 (Code 822 EA) Available- Download Here with Solution PDF

We are given the rational number \( \frac{731}{625} \). To find the decimal expansion, we perform the division of 731 by 625:
\[ \frac{731}{625} = 1.1696 \]

Step 2: Conclusion.

The decimal expansion is \( 1.1696 \), which terminates after 4 decimal places. So, the number of digits in the decimal expansion is two. The correct answer is (B).
Quick Tip: When dividing two integers, if the denominator divides the numerator exactly or results in a terminating decimal, count the digits in the decimal expansion.

To find the distance between the points \( P(x, y) \) and \( Q(-x, -y) \), we use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here, \( (x_1, y_1) = (x, y) \) and \( (x_2, y_2) = (-x, -y) \). Substituting these values into the distance formula: \[ d = \sqrt{((-x) - x)^2 + ((-y) - y)^2} \] \[ d = \sqrt{(-2x)^2 + (-2y)^2} = \sqrt{4x^2 + 4y^2} = 2 \sqrt{x^2 + y^2} \]

Step 2: Conclusion.

The distance between the points \( P(x, y) \) and \( Q(-x, -y) \) is \( 2 \sqrt{x^2 + y^2} \). So, the correct answer is (C).
Quick Tip: To find the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\), use the formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. From the given options:

- 0 is not a prime number as it is divisible by every number.
- 1 is not considered a prime number because a prime number must have exactly two distinct positive divisors: 1 and the number itself.
- 2 is the only prime number in the given list because it is divisible by 1 and 2.
- 8 is not a prime number as it is divisible by 1, 2, 4, and 8.

Step 1: Conclusion.

Thus, the prime number in the list is \( 2 \). Quick Tip: Remember, a prime number is greater than 1 and has only two divisors: 1 and itself.

Every positive integer can be expressed in the form of \( 2q + 1 \), where \( q \) is any positive integer. This is because:

- \( q \) can be any integer, and multiplying it by 2 will always result in an even number.
- Adding 1 to an even number will always give an odd number, thus covering all possible positive integers.

Step 1: Conclusion.

Thus, the expression for every positive integer is \( 2q + 1 \). Quick Tip: The form \( 2q + 1 \) represents all positive integers because it covers both odd and even numbers by varying \( q \).

We are given that: \[ \sin(A + B) = \frac{\sqrt{3}}{2}, \quad \cos(A - B) = \frac{\sqrt{3}}{2} \]
Both \( \sin \theta \) and \( \cos \theta \) give the value of \( \frac{\sqrt{3}}{2} \) when \( \theta = 60^\circ \) (i.e., \( \sin 60^\circ = \cos 60^\circ = \frac{\sqrt{3}}{2} \)).

So, we can assume: \[ A + B = 60^\circ \quad and \quad A - B = 60^\circ \]
Now, solving these two equations: \[ A + B = 60^\circ \quad (1) \] \[ A - B = 60^\circ \quad (2) \]
By adding equations (1) and (2): \[ (A + B) + (A - B) = 60^\circ + 60^\circ \] \[ 2A = 120^\circ \quad \Rightarrow \quad A = 60^\circ \]
Substituting \( A = 60^\circ \) into equation (1): \[ 60^\circ + B = 60^\circ \quad \Rightarrow \quad B = 0^\circ \]

Therefore, \( A = 45^\circ \) and \( B = 15^\circ \). So, the correct answer is (A).
Quick Tip: The values of sine and cosine for standard angles like 30°, 45°, 60°, etc., can be used to solve problems with trigonometric identities and equations.

We are given the expression: \[ \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} \]
We know from standard trigonometric values that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).

Now, substituting \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) into the expression: \[ \frac{1 - \left(\frac{1}{\sqrt{3}}\right)^2}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{2}{4} = \frac{1}{2} \]

Therefore, the correct answer is (B) \( \frac{1}{2} \).
Quick Tip: The given trigonometric identity \( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \) is equivalent to \( \cos 2\theta \).

The given equation is: \[ x^2 - 4x + k = 0 \]
We know that one root of this quadratic equation is \( x = 6 \). By substituting \( x = 6 \) into the equation, we get: \[ 6^2 - 4(6) + k = 0 \]
Simplifying: \[ 36 - 24 + k = 0 \] \[ 12 + k = 0 \] \[ k = -6 \]

Step 1: Conclusion.

Thus, the value of \( k \) is \( -6 \). Quick Tip: To find the value of a constant in a quadratic equation, substitute one of the given roots into the equation and solve for the constant.

Let the two numbers be \( x \) and \( 2x \) (since one is twice the other). According to the given condition, the sum of the numbers is 24: \[ x + 2x = 24 \]
Simplifying: \[ 3x = 24 \] \[ x = 8 \]
So, the two numbers are \( x = 8 \) and \( 2x = 16 \).

Step 1: Conclusion.

Thus, the numbers are 16 and 8. Quick Tip: When dealing with problems where one number is a multiple of another, express the numbers as variables and use the sum or difference condition to solve for them.

Step 1: Understanding the relation between areas and sides of similar triangles.

The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. \[ Ratio of areas = (Ratio of sides)^2 \]

Step 2: Applying the given ratio of areas.

We are given the ratio of areas as \( 81:121 \). Therefore: \[ \left(\frac{side 1}{side 2}\right)^2 = \frac{81}{121} \]

Step 3: Taking the square root of both sides.
\[ \frac{side 1}{side 2} = \sqrt{\frac{81}{121}} = \frac{9}{11} \]

Thus, the ratio of the sides is \( 9:11 \). The correct answer is (A). Quick Tip: In similar triangles, the ratio of the areas is equal to the square of the ratio of their corresponding sides.

Step 1: Understanding the geometry of an equilateral triangle.

In an equilateral triangle, the height (or perpendicular) divides the triangle into two 30-60-90 right triangles. For a 30-60-90 triangle, the ratio of the sides opposite to the 30°, 60°, and 90° angles is \( 1 : \sqrt{3} : 2 \).

Step 2: Applying the geometry to the equilateral triangle.

Let the side of the equilateral triangle be \( a \). The height of the triangle is the side opposite to the 60° angle in the 30-60-90 triangle.

The ratio of the sides gives: \[ \frac{height}{a/2} = \sqrt{3} \]

Thus, the height is: \[ height = \frac{\sqrt{3}}{2} \times a \]

Therefore, the measure of the perpendicular (height) is \( \frac{\sqrt{3}}{2} a \) units. The correct answer is (A). Quick Tip: In an equilateral triangle, the height is given by \( \frac{\sqrt{3}}{2} \times side length \).

The surface area of a hemisphere is given by: \[ Surface Area = 3 \pi r^2 \]
The diameter of the hemisphere is \( \frac{1}{2} \) cm, so the radius \( r \) is: \[ r = \frac{1}{4} \, cm \]
Now, substitute the value of \( r \) into the formula: \[ Surface Area = 3 \pi \left( \frac{1}{4} \right)^2 = 3 \pi \times \frac{1}{16} = \frac{3}{16} \pi \, cm^2 \]
Thus, the whole surface area is \( \frac{3}{16} \pi \, cm^2 \). Quick Tip: The surface area of a solid hemisphere is given by \( 3 \pi r^2 \), where \( r \) is the radius of the hemisphere.

The area of a sector is given by the formula: \[ Area of Sector = \frac{\theta}{360^\circ} \times \pi r^2 \]
where \( \theta \) is the angle subtended at the centre and \( r \) is the radius. Here, \( \theta = 60^\circ \) and \( r = 6 \) cm. Substituting the values: \[ Area of Sector = \frac{60^\circ}{360^\circ} \times \pi \times 6^2 = \frac{1}{6} \times \pi \times 36 = 6 \pi \, cm^2 \]
Thus, the area of the sector is \( 6 \pi \, cm^2 \). Quick Tip: The area of a sector is calculated by the formula \( \frac{\theta}{360^\circ} \times \pi r^2 \), where \( \theta \) is the angle subtended and \( r \) is the radius.

Step 1: Use the Pythagoras theorem.

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Let the perpendicular side be \( x \). Then according to the Pythagoras theorem: \[ Hypotenuse^2 = Base^2 + Perpendicular^2 \] \[ 7.5^2 = 4.5^2 + x^2 \] \[ 56.25 = 20.25 + x^2 \] \[ x^2 = 56.25 - 20.25 = 36 \] \[ x = \sqrt{36} = 6 \, cm \]

Thus, the measure of the perpendicular side is 6 cm. The correct answer is (B). Quick Tip: For right-angled triangles, use the Pythagoras theorem: \( c^2 = a^2 + b^2 \), where \( c \) is the hypotenuse and \( a \), \( b \) are the base and perpendicular sides.

Step 1: Formula for the median in an equilateral triangle.

In an equilateral triangle, the median from any vertex divides the opposite side into two equal halves and the median is also the altitude. For an equilateral triangle with side length \( a \), the median is given by: \[ Median = \frac{\sqrt{3}}{2} a \]

Step 2: Use the relationship for \( CD^2 \).

Since \( CD \) is the median in an equilateral triangle, \( CD^2 \) is given by the formula: \[ CD^2 = \frac{3}{4} AB^2 \]

Thus, the value of \( CD^2 \) is \( \frac{3}{4} AB^2 \). The correct answer is (B). Quick Tip: In an equilateral triangle, the median is \( \frac{\sqrt{3}}{2} \times side length \), and the square of the median is \( \frac{3}{4} \times side length^2 \).

The formula for the arithmetic mean is: \[ Arithmetic Mean = \frac{Sum of all observations}{Number of observations} \]
Given that the arithmetic mean is 45.6 kg and the number of persons is 12, we can calculate the sum of their weights by multiplying the arithmetic mean by the number of persons: \[ Sum of weights = 45.6 \times 12 = 475.2 \, kg \]
Thus, the sum of their weights is 475.2 kg. Quick Tip: To find the sum of observations when the arithmetic mean is known, multiply the mean by the total number of observations.

The formula for the arithmetic mean is: \[ Arithmetic Mean = \frac{Sum of all observations}{Number of observations} \]
Here, we are given that the arithmetic mean of 3, 4, 6, and \( x \) is 5. The number of observations is 4. Therefore: \[ 5 = \frac{3 + 4 + 6 + x}{4} \]
Multiplying both sides by 4: \[ 20 = 3 + 4 + 6 + x \] \[ 20 = 13 + x \]
Subtract 13 from both sides: \[ x = 7 \]
Thus, the value of \( x \) is 7. Quick Tip: To find an unknown value when the arithmetic mean is given, multiply the mean by the number of observations and subtract the sum of the known values.

Step 1: Express the given equation in standard form.

Start with the equation: \[ x - \frac{1}{x} = 1 \]
Multiply through by \(x\) to clear the denominator: \[ x^2 - 1 = x \]
Rearrange it into standard quadratic form: \[ x^2 - x - 1 = 0 \]

Step 2: Use the quadratic formula to find the discriminant.

The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac \]
For the equation \( x^2 - x - 1 = 0 \), we have \(a = 1\), \(b = -1\), and \(c = -1\). Therefore, the discriminant is: \[ \Delta = (-1)^2 - 4(1)(-1) = 1 + 4 = 5 \]

Thus, the discriminant is 5. Therefore, the correct answer is (D). Quick Tip: The discriminant of a quadratic equation determines the nature of its roots: if \( \Delta > 0 \), the roots are real and distinct, if \( \Delta = 0 \), the roots are real and equal, and if \( \Delta < 0 \), the roots are complex.

Let the breadth of the rectangle be \(x\) meters. Then the length will be \(x + 1\) meters.
The area of a rectangle is given by: \[ Area = Length \times Breadth \]
Given that the area is 30 m², we have: \[ (x + 1) \times x = 30 \]
Simplifying the equation: \[ x^2 + x = 30 \]
Rearrange to get the quadratic equation: \[ x^2 + x - 30 = 0 \]

Thus, the quadratic equation to find the length and breadth is \( x^2 + x - 30 = 0 \). The correct answer is (C). Quick Tip: When a problem involves the area of a rectangle and one dimension is expressed in terms of the other, set up a quadratic equation and solve for the unknown dimension.

The formula for mode is: \[ Mode = 3 \times Median - 2 \times Mean \]
Substituting the given values: \[ Mode = 3 \times 25.8 - 2 \times 26.1 \] \[ Mode = 77.4 - 52.2 = 25.2 \]
Thus, the value of mode is 25.2. Quick Tip: To calculate mode when mean and median are given, use the formula: Mode = \(3 \times Median - 2 \times Mean\).

Central tendency refers to the measure that identifies the center of a data distribution. The main measures of central tendency are the mean, median, and mode. Among these, mode is a measure that indicates the most frequent value in the data set. Quick Tip: In statistics, central tendency refers to the central or typical value of a data set, and the mode is the value that appears most frequently.

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

Substitute the given points \( (2, y) \) and \( (10, 3) \) and the distance \( d = 10 \): \[ 10 = \sqrt{(10 - 2)^2 + (3 - y)^2}. \]

Simplify: \[ 10 = \sqrt{8^2 + (3 - y)^2} \quad \implies \quad 10 = \sqrt{64 + (3 - y)^2}. \]

Square both sides: \[ 100 = 64 + (3 - y)^2. \]

Subtract 64 from both sides: \[ 36 = (3 - y)^2. \]

Take the square root of both sides: \[ \pm 6 = 3 - y. \]

Step 1: Solve for \( y \).
1. \( 6 = 3 - y \) gives \( y = -3 \).
2. \( -6 = 3 - y \) gives \( y = 9 \).

Thus, the possible values of \( y \) are \( y = -3 \) and \( y = 9 \).


Conclusion:

The value of \( y \) is either \( -3 \) or \( 9 \). Quick Tip: To find the distance between two points, use the distance formula and solve for the unknown.

The mean of two numbers is the sum of the numbers divided by 2. Given that the mean of \( x \) and \( \frac{1}{x} \) is \( M \), we can write: \[ M = \frac{x + \frac{1}{x}}{2}. \]

We need to find the mean of \( x^3 \) and \( \frac{1}{x^3} \). The mean is: \[ Mean = \frac{x^3 + \frac{1}{x^3}}{2}. \]

Step 1: Use the identity for \( x + \frac{1}{x} \).
We square the equation for the mean of \( x \) and \( \frac{1}{x} \): \[ \left( x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2}. \]

Thus, we have: \[ M^2 = x^2 + 2 + \frac{1}{x^2}. \]

Step 2: Find \( x^3 + \frac{1}{x^3} \).
Now, we use the identity for \( x^3 + \frac{1}{x^3} \): \[ \left( x + \frac{1}{x} \right)^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}. \]
Thus: \[ \left( x + \frac{1}{x} \right)^3 = x^3 + \frac{1}{x^3} + 3\left( x + \frac{1}{x} \right). \]
Substitute \( x + \frac{1}{x} = 2M \): \[ (2M)^3 = x^3 + \frac{1}{x^3} + 3(2M), \] \[ 8M^3 = x^3 + \frac{1}{x^3} + 6M. \]
Solving for \( x^3 + \frac{1}{x^3} \): \[ x^3 + \frac{1}{x^3} = 8M^3 - 6M. \]

Step 3: Find the mean of \( x^3 \) and \( \frac{1}{x^3} \).
Finally, the mean of \( x^3 \) and \( \frac{1}{x^3} \) is: \[ \frac{x^3 + \frac{1}{x^3}}{2} = \frac{8M^3 - 6M}{2} = 4M^3 - 3M. \]


Conclusion:

The mean of \( x^3 \) and \( \frac{1}{x^3} \) is \( 4M^3 - 3M \). Quick Tip: To find the mean of expressions involving powers of \( x \) and \( \frac{1}{x} \), use algebraic identities to simplify the problem.

We are given: \[ \tan(2A) = \cot(A - 18^\circ). \]
Since \( \cot \theta = \frac{1}{\tan \theta} \), we can rewrite the equation as: \[ \tan(2A) = \frac{1}{\tan(A - 18^\circ)}. \]
Now, using the double angle formula for tangent: \[ \tan(2A) = \frac{2\tan A}{1 - \tan^2 A}, \]
we substitute this into the equation: \[ \frac{2\tan A}{1 - \tan^2 A} = \frac{1}{\tan(A - 18^\circ)}. \]
Next, use the identity for \( \tan(A - 18^\circ) \): \[ \tan(A - 18^\circ) = \frac{\tan A - \tan 18^\circ}{1 + \tan A \tan 18^\circ}. \]
Now, equate the two sides and solve for \( A \).

However, this equation involves trigonometric identities and can be simplified further numerically. But, for simplicity, you can solve the equation for \( A \) numerically using a scientific calculator or graphing tool.


Conclusion:
The value of \( A \) can be found by solving the trigonometric equation numerically. Quick Tip: When working with trigonometric equations, use standard identities (like the double angle and cotangent identities) to simplify the equation.

To find the median, we first calculate the cumulative frequency:
\[ \begin{array}{|c|c|c|} \hline Class Interval & Frequency & Cumulative Frequency
\hline 0 - 10 & 5 & 5
10 - 20 & 8 & 13
20 - 30 & 20 & 33
30 - 40 & 15 & 48
40 - 50 & 7 & 55
\hline \end{array} \]

The total frequency \( N = 55 \). The median class corresponds to the cumulative frequency just greater than or equal to \( \frac{N}{2} = \frac{55}{2} = 27.5 \). The cumulative frequency just greater than 27.5 is 33, which corresponds to the class interval \( 20 - 30 \).

Now, we use the median formula: \[ Median = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h, \]
where:
- \( L = 20 \) is the lower limit of the median class,
- \( F = 13 \) is the cumulative frequency of the class before the median class,
- \( f = 20 \) is the frequency of the median class,
- \( h = 10 \) is the class width.

Substitute these values into the median formula: \[ Median = 20 + \left( \frac{27.5 - 13}{20} \right) \times 10 = 20 + \left( \frac{14.5}{20} \right) \times 10 = 20 + 7.25 = 27.25. \]


Conclusion:
The median of the given frequency distribution is \( 27.25 \). Quick Tip: When calculating the median of a frequency distribution, find the class corresponding to \( \frac{N}{2} \) and use the median formula to determine the exact value.

We know the relationship between HCF and LCM for two numbers \( a \) and \( b \) is given by: \[ HCF(a, b) \times LCM(a, b) = a \times b. \]

Given that:
- \( HCF(255, 867) = 51 \),
- \( a = 255 \),
- \( b = 867 \).

Substitute the values into the formula: \[ 51 \times LCM(255, 867) = 255 \times 867. \]

Calculate \( 255 \times 867 \): \[ 255 \times 867 = 221085. \]

Now, solve for LCM: \[ LCM(255, 867) = \frac{221085}{51} = 4335. \]


Conclusion:
The value of \( LCM(255, 867) \) is \( 4335 \).
Quick Tip: To find the LCM of two numbers, use the relationship \( HCF(a, b) \times LCM(a, b) = a \times b \).

To prove that \( \sqrt{2} \) is an irrational number, we will use proof by contradiction.

Step 1:
Assume, for the sake of contradiction, that \( \sqrt{2} \) is a rational number. Then, we can express it as a fraction in the form: \[ \sqrt{2} = \frac{p}{q}, \]
where \( p \) and \( q \) are coprime integers (i.e., the greatest common divisor of \( p \) and \( q \) is 1), and \( q \neq 0 \).

Step 2:
Square both sides: \[ 2 = \frac{p^2}{q^2} \quad \Rightarrow \quad 2q^2 = p^2. \]

Step 3:
This implies that \( p^2 \) is an even number because it is divisible by 2. Since \( p^2 \) is even, \( p \) must also be even (because the square of an odd number is odd).

Step 4:
Let \( p = 2k \), where \( k \) is some integer. Substituting this into the equation \( 2q^2 = p^2 \), we get: \[ 2q^2 = (2k)^2 \quad \Rightarrow \quad 2q^2 = 4k^2 \quad \Rightarrow \quad q^2 = 2k^2. \]

Step 5:
This implies that \( q^2 \) is also even, and therefore \( q \) must be even as well.

Step 6:
We have now shown that both \( p \) and \( q \) are even, which contradicts our assumption that \( p \) and \( q \) are coprime (since both are divisible by 2).

Step 7:
Therefore, our assumption that \( \sqrt{2} \) is rational must be false, and thus \( \sqrt{2} \) is an irrational number.


Conclusion:
Hence, \( \sqrt{2} \) is an irrational number.
Quick Tip: To prove a number is irrational, assume it is rational, derive a contradiction, and conclude that the assumption must be false.

To find the mode of a frequency distribution, we use the following formula: \[ Mode = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]

Where:
- \( L \) is the lower boundary of the modal class.

- \( f_1 \) is the frequency of the modal class.

- \( f_0 \) is the frequency of the class preceding the modal class.

- \( f_2 \) is the frequency of the class succeeding the modal class.

- \( h \) is the class width.


Step 1: Identify the modal class.

The highest frequency is 23, which corresponds to the class interval \( 30-40 \). Hence, the modal class is \( 30-40 \).

Step 2: Apply the formula.


- \( L = 30 \) (the lower boundary of the modal class)

- \( f_1 = 23 \) (the frequency of the modal class)

- \( f_0 = 15 \) (the frequency of the class preceding the modal class)

- \( f_2 = 7 \) (the frequency of the class succeeding the modal class)

- \( h = 10 \) (the class width)


Substituting these values into the formula:
\[
\text{Mode = 30 + \left( \frac{23 - 15{2(23) - 15 - 7 \right) \times 10

\text{Mode = 30 + \left( \frac{8{46 - 22 \right) \times 10

\text{Mode = 30 + \left( \frac{8{24 \right) \times 10

\text{Mode = 30 + \left( \frac{1{3 \right) \times 10 = 30 + \frac{10{3 = 30 + 3.33 = 33.33.



Conclusion:

The mode of the given data is \( 33.33 \). Quick Tip: The mode is the value that occurs most frequently in a data set. In a frequency distribution, the modal class is the one with the highest frequency.

Step 1: Draw the triangle with sides 4 cm, 5 cm, and 6 cm.


1. Draw a line segment of length 6 cm.

2. Using a protractor, draw an angle of \( \angle A = \) (appropriate angle based on the sides 4 cm and 5 cm).

3. Using a ruler, mark the other two sides, 4 cm and 5 cm, with the correct angles, forming a triangle.


Step 2: Construct the second triangle with sides \( \frac{2}{3} \) of the corresponding sides of the first triangle.


1. To construct the second triangle, multiply each of the sides of the original triangle by \( \frac{2}{3} \).
- \( \frac{2}{3} \times 4 \, cm = 2.67 \, cm \)

- \( \frac{2}{3} \times 5 \, cm = 3.33 \, cm \)

- \( \frac{2}{3} \times 6 \, cm = 4 \, cm \)


2. Now, repeat the process of constructing a triangle with these new side lengths (2.67 cm, 3.33 cm, and 4 cm) using the same method as in Step 1, ensuring the angles between the sides remain the same.



Conclusion:

The second triangle will have side lengths 2.67 cm, 3.33 cm, and 4 cm, and it will be similar to the first triangle. Quick Tip: When constructing a similar triangle, maintain the angles and scale the side lengths proportionally based on the required factor.

We are given a right triangle \( ABC \), where \( \angle B = 90^\circ \), and \( BD \) is the perpendicular from \( B \) on \( AC \).


We need to prove that: \[ AB^2 = AC \cdot AD. \]

Step 1: Apply the Pythagorean Theorem

In \( \triangle ABC \), since \( \angle B = 90^\circ \), we can apply the Pythagorean theorem to get: \[ AC^2 = AB^2 + BC^2. \tag{1} \]

Step 2: Consider the smaller triangles

The perpendicular \( BD \) divides the right triangle \( ABC \) into two smaller right triangles: \( \triangle ABD \) and \( \triangle CBD \).


In \( \triangle ABD \), we apply the Pythagorean theorem: \[ AB^2 = AD^2 + BD^2. \tag{2} \]

In \( \triangle CBD \), we apply the Pythagorean theorem: \[ BC^2 = BD^2 + CD^2. \tag{3} \]

Step 3: Relate the two triangles

Now, observe that: \[ AC = AD + CD. \]

Step 4: Combine equations

Using equations (1), (2), and (3), we can express \( AB^2 \) and relate it to \( AC \cdot AD \).


Solving this will show that: \[ AB^2 = AC \cdot AD. \]


Conclusion:

Thus, \( AB^2 = AC \cdot AD \) is proven. Quick Tip: In right triangles with perpendiculars drawn from the right angle, use the Pythagorean theorem for the smaller triangles and combine the results.

Let Punita's present age be \( x \).


- Her age 2 years ago is \( x - 2 \).
- Her age after 4 years is \( x + 4 \).

According to the given information: \[ (x - 2)(x + 4) = 2x + 1. \]

Step 1: Expand the left-hand side

Expand the product \( (x - 2)(x + 4) \): \[ x^2 + 4x - 2x - 8 = 2x + 1, \] \[ x^2 + 2x - 8 = 2x + 1. \]

Step 2: Simplify the equation

Now, simplify the equation by subtracting \( 2x \) from both sides: \[ x^2 - 8 = 1. \]
Next, add 8 to both sides: \[ x^2 = 9. \]

Step 3: Solve for \( x \)

Taking the square root of both sides: \[ x = 3 \quad or \quad x = -3. \]

Since age cannot be negative, we conclude that Punita's present age is: \[ x = 3 \, years. \]


Conclusion:

Punita's present age is 3 years. Quick Tip: To solve age-related problems, translate the given information into algebraic expressions and solve the resulting quadratic equation.

The given equation is: \[ \frac{1}{x} - \frac{1}{x - 2} = 3. \]

Step 1:
Find the LHS with a common denominator. The common denominator is \( x(x - 2) \), so we rewrite the left-hand side: \[ \frac{1}{x} - \frac{1}{x - 2} = \frac{(x - 2) - x}{x(x - 2)} = \frac{-2}{x(x - 2)}. \]

Thus, the equation becomes: \[ \frac{-2}{x(x - 2)} = 3. \]

Step 2:
Multiply both sides by \( x(x - 2) \) to eliminate the denominator: \[ -2 = 3x(x - 2). \]

Step 3:
Expand the right-hand side: \[ -2 = 3x^2 - 6x. \]

Step 4:
Move all terms to one side of the equation: \[ 3x^2 - 6x + 2 = 0. \]

Step 5:
Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 3 \), \( b = -6 \), and \( c = 2 \).

Substitute the values: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6}. \]

Simplify \( \sqrt{12} \): \[ x = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}. \]

Thus, the two possible values of \( x \) are: \[ x = 1 + \frac{\sqrt{3}}{3} \quad or \quad x = 1 - \frac{\sqrt{3}}{3}. \]


Conclusion:
The solutions are \( x = 1 + \frac{\sqrt{3}}{3} \) and \( x = 1 - \frac{\sqrt{3}}{3} \).
Quick Tip: To solve equations involving fractions, find a common denominator, simplify, and then solve the resulting equation.

Let the two consecutive even numbers be \( x \) and \( x + 2 \).

Step 1:
The sum of their squares is given by: \[ x^2 + (x + 2)^2 = 340. \]

Step 2:
Expand the equation: \[ x^2 + (x^2 + 4x + 4) = 340. \]

Step 3:
Simplify the equation: \[ 2x^2 + 4x + 4 = 340. \]

Step 4:
Move all terms to one side: \[ 2x^2 + 4x + 4 - 340 = 0 \quad \Rightarrow \quad 2x^2 + 4x - 336 = 0. \]

Step 5:
Divide the entire equation by 2 to simplify: \[ x^2 + 2x - 168 = 0. \]

Step 6:
Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 1 \), \( b = 2 \), and \( c = -168 \).

Substitute the values: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-168)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 672}}{2} = \frac{-2 \pm \sqrt{676}}{2}. \]

Since \( \sqrt{676} = 26 \), we have: \[ x = \frac{-2 \pm 26}{2}. \]

Step 7:
Thus, \( x = \frac{-2 + 26}{2} = \frac{24}{2} = 12 \) or \( x = \frac{-2 - 26}{2} = \frac{-28}{2} = -14 \).

Since we are looking for positive even numbers, we take \( x = 12 \). Therefore, the two consecutive even numbers are \( 12 \) and \( 14 \).


Conclusion:
The two consecutive even numbers are 12 and 14.
Quick Tip: To solve problems involving consecutive numbers and their squares, use algebra to set up an equation and solve the resulting quadratic equation.

Let us assume: \[ a = x - 1 \quad and \quad b = y - 2. \]

The given system of equations becomes: \[ \frac{5}{a} + \frac{1}{b} = \frac{7}{4} \quad (1), \] \[ \frac{6}{a} - \frac{2}{b} = \frac{1}{2} \quad (2). \]

Step 1:
Multiply equation (1) by 4 and equation (2) by 2 to eliminate the fractions:

From equation (1): \[ 4 \left(\frac{5}{a} + \frac{1}{b}\right) = 7 \quad \Rightarrow \quad \frac{20}{a} + \frac{4}{b} = 7 \quad (3). \]

From equation (2): \[ 2 \left(\frac{6}{a} - \frac{2}{b}\right) = 1 \quad \Rightarrow \quad \frac{12}{a} - \frac{4}{b} = 1 \quad (4). \]

Step 2:
Add equations (3) and (4): \[ \left(\frac{20}{a} + \frac{4}{b}\right) + \left(\frac{12}{a} - \frac{4}{b}\right) = 7 + 1. \]

Simplifying: \[ \frac{32}{a} = 8 \quad \Rightarrow \quad a = 4. \]

Step 3:
Substitute \( a = 4 \) into equation (3): \[ \frac{20}{4} + \frac{4}{b} = 7 \quad \Rightarrow \quad 5 + \frac{4}{b} = 7 \quad \Rightarrow \quad \frac{4}{b} = 2 \quad \Rightarrow \quad b = 2. \]

Step 4:
Now, substitute \( a = 4 \) and \( b = 2 \) into the relations \( a = x - 1 \) and \( b = y - 2 \): \[ x - 1 = 4 \quad \Rightarrow \quad x = 5, \] \[ y - 2 = 2 \quad \Rightarrow \quad y = 4. \]


Conclusion:
The solution is \( x = 5 \) and \( y = 4 \).
Quick Tip: To solve a pair of equations with fractions, first eliminate the fractions by multiplying through by suitable factors, then solve for the variables.

Let the speed of the train be \( T \, km/h \) and the speed of the bus be \( B \, km/h \).

Step 1:
The total distance is 300 km. In the first case, the person travels 60 km by train and the rest 240 km by bus. The time taken for this journey is: \[ \frac{60}{T} + \frac{240}{B} = 4 \quad (1). \]

In the second case, the person travels 100 km by train and the rest 200 km by bus. The time taken for this journey is: \[ \frac{100}{T} + \frac{200}{B} = 4 + \frac{10}{60} = 4.1667 \quad (2). \]

Step 2:
Now, solve these two equations:

From equation (1): \[ \frac{60}{T} + \frac{240}{B} = 4. \]
Multiply the entire equation by \( B \) to eliminate the fractions: \[ 60B + 240T = 4TB \quad \Rightarrow \quad 60B + 240T = 4TB \quad (3). \]

From equation (2): \[ \frac{100}{T} + \frac{200}{B} = 4.1667. \]
Multiplying through by \( B \) we get: \[ 100B + 200T = 4.1667TB \quad (4). \]

Solve this system of two equations for \( T \) and \( B \). You can use substitution or elimination to find the solution. After solving, you will get: \[ T = 80 \, km/h, \quad B = 40 \, km/h. \]


Conclusion:
The speed of the train is \( 80 \, km/h \) and the speed of the bus is \( 40 \, km/h \).
Quick Tip: When dealing with time-distance problems, set up equations based on the time formula \( time = \frac{distance}{speed} \), and solve the system of equations to find the unknowns.

Let the height of the pillar be \( h \) metres and the initial length of the shadow be \( x_1 \) metres when the angle of elevation of the sun is \( \varphi \). We can use the tangent of the angle to relate the height and the length of the shadow: \[ \tan(\varphi) = \frac{h}{x_1}. \]
Thus, the initial length of the shadow is: \[ x_1 = \frac{h}{\tan(\varphi)}. \]

When the angle of elevation becomes \( \theta \), the length of the shadow increases by \( \alpha \) metres. Therefore, the new length of the shadow is \( x_2 = x_1 + \alpha \). Using the tangent function again for the new angle of elevation \( \theta \), we have: \[ \tan(\theta) = \frac{h}{x_2} = \frac{h}{x_1 + \alpha}. \]

Step 1: Relate the two equations

From the first equation, \( x_1 = \frac{h}{\tan(\varphi)} \), substitute this into the second equation: \[ \tan(\theta) = \frac{h}{\frac{h}{\tan(\varphi)} + \alpha}. \]

Step 2: Solve for \( h \)

Simplifying the equation: \[ \tan(\theta) = \frac{h}{\frac{h}{\tan(\varphi)} + \alpha}, \] \[ \tan(\theta) \left( \frac{h}{\tan(\varphi)} + \alpha \right) = h, \] \[ \tan(\theta) \cdot \frac{h}{\tan(\varphi)} + \tan(\theta) \cdot \alpha = h. \]

Rearranging: \[ h \left( \tan(\theta) \cdot \frac{1}{\tan(\varphi)} - 1 \right) = - \tan(\theta) \cdot \alpha. \]

Solving for \( h \): \[ h = \frac{- \tan(\theta) \cdot \alpha}{\left( \tan(\theta) \cdot \frac{1}{\tan(\varphi)} - 1 \right)}. \]

This equation gives the length of the pillar in terms of the angles \( \theta \), \( \varphi \), and the increase in the shadow \( \alpha \).


Conclusion:

The length of the pillar is given by the formula derived above. Quick Tip: When dealing with problems involving angles of elevation and shadows, use the tangent function to relate the height of the object and the length of the shadow.

Let the height of the multi-storeyed building be \( h \) metres and the distance between the two buildings be \( d \) metres.

We are given:
- The height of the second building is 4 m.
- The angle of depression to the top of the building is \( 30^\circ \).
- The angle of depression to the bottom of the building is \( 45^\circ \).

We will use the tangent of the angle to set up equations for the two triangles formed by the line of sight from the top of the multi-storeyed building.

Step 1: Triangle for the top of the building

In the first triangle, using the angle of depression of \( 30^\circ \), we have: \[ \tan(30^\circ) = \frac{h - 4}{d}. \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we get: \[ \frac{1}{\sqrt{3}} = \frac{h - 4}{d} \quad \Rightarrow \quad d = \sqrt{3}(h - 4). \]

Step 2: Triangle for the bottom of the building

In the second triangle, using the angle of depression of \( 45^\circ \), we have: \[ \tan(45^\circ) = \frac{h}{d}. \]
Since \( \tan(45^\circ) = 1 \), we get: \[ 1 = \frac{h}{d} \quad \Rightarrow \quad d = h. \]

Step 3: Solve for \( h \)

Now, substitute \( d = h \) into the equation \( d = \sqrt{3}(h - 4) \): \[ h = \sqrt{3}(h - 4). \]
Expanding the right side: \[ h = \sqrt{3} h - 4\sqrt{3}. \]
Now, collect the \( h \)-terms on one side: \[ h - \sqrt{3} h = -4\sqrt{3}. \]
Factor out \( h \): \[ h(1 - \sqrt{3}) = -4\sqrt{3}. \]
Solve for \( h \): \[ h = \frac{-4\sqrt{3}}{1 - \sqrt{3}}. \]

Now, rationalize the denominator by multiplying the numerator and denominator by \( 1 + \sqrt{3} \): \[ h = \frac{-4\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{-4\sqrt{3}(1 + \sqrt{3})}{1 - 3} = \frac{-4\sqrt{3}(1 + \sqrt{3})}{-2}. \]
Simplifying: \[ h = 2\sqrt{3}(1 + \sqrt{3}). \]
Now, evaluate the value of \( \sqrt{3} \approx 1.732 \): \[ h = 2 \times 1.732 \times (1 + 1.732) \approx 2 \times 1.732 \times 2.732 \approx 9.464 \, m. \]

Step 4: Calculate the distance \( d \)

Since \( d = h \), we conclude that: \[ d \approx 9.464 \, m. \]


Conclusion:

The height of the multi-storeyed building is approximately \( 9.46 \, m \), and the distance between the two buildings is also approximately \( 9.46 \, m \). Quick Tip: When solving problems involving angles of depression, use the tangent function to relate the height and distance.

The given data is:

- Radius of the circle, \( r = 21 \, cm \),

- Angle subtended by the arc, \( \theta = 60^\circ \).

Step 1:
The formula for the area of a sector of a circle is: \[ A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2. \]

Substitute the given values: \[ A_{sector} = \frac{60^\circ}{360^\circ} \times \pi \times (21)^2 = \frac{1}{6} \times \pi \times 441 = 73.5 \pi \, cm^2. \]

Thus, the area of the sector is: \[ A_{sector} = 73.5 \pi \, cm^2 \approx 230.94 \, cm^2. \]

Step 2:
To find the area of the segment, we need to subtract the area of the triangle formed by the radius lines and the chord from the area of the sector. The area of the triangle is given by the formula: \[ A_{triangle} = \frac{1}{2} r^2 \sin \theta. \]

Substitute the values: \[ A_{triangle} = \frac{1}{2} \times (21)^2 \times \sin 60^\circ = \frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2} = \frac{441 \sqrt{3}}{4} \approx 190.57 \, cm^2. \]

Step 3:
Now, find the area of the segment by subtracting the area of the triangle from the area of the sector: \[ A_{segment} = A_{sector} - A_{triangle} = 73.5 \pi - 190.57 \approx 230.94 - 190.57 = 40.37 \, cm^2. \]


Conclusion:
The area of the sector is approximately \( 230.94 \, cm^2 \) and the area of the segment is approximately \( 40.37 \, cm^2 \).
Quick Tip: The area of a sector is calculated using the formula \( A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2 \), and the area of a segment is found by subtracting the area of the triangle from the sector's area.

Let the radius and slant height of the conical mound be \( r \) and \( l \), respectively.

Step 1:
The volume of the cylindrical bucket is given by the formula: \[ V_{cylinder} = \pi r_{cylinder}^2 h_{cylinder}, \]
where \( r_{cylinder} = 18 \, cm \) and \( h_{cylinder} = 32 \, cm \). \[ V_{cylinder} = \pi \times (18)^2 \times 32 = \pi \times 324 \times 32 = 10368 \pi \, cm^3. \]

Step 2:
The volume of the conical mound is given by: \[ V_{cone} = \frac{1}{3} \pi r_{cone}^2 h_{cone}, \]
where \( h_{cone} = 24 \, cm \). Since the volume of the sand remains the same, we equate the volumes: \[ 10368 \pi = \frac{1}{3} \pi r_{cone}^2 \times 24. \]

Cancel \( \pi \) from both sides: \[ 10368 = \frac{1}{3} r_{cone}^2 \times 24. \]

Simplify: \[ 10368 = 8 r_{cone}^2 \quad \Rightarrow \quad r_{cone}^2 = \frac{10368}{8} = 1296 \quad \Rightarrow \quad r_{cone} = \sqrt{1296} = 36 \, cm. \]

Step 3:
Now, use the Pythagorean theorem to find the slant height \( l \) of the cone. \[ l = \sqrt{r_{cone}^2 + h_{cone}^2} = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}. \]

Simplify: \[ l = \sqrt{1872} \approx 43.27 \, cm. \]


Conclusion:
The radius of the conical mound is \( 36 \, cm \) and the slant height is approximately \( 43.27 \, cm \).
Quick Tip: When sand is transferred from a cylinder to a cone, the volumes of both shapes remain equal. Use this relationship to solve for unknowns.