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

Step 1: Analyze the given expression.

We are given the rational number \(\frac{17}{2^2 \times 5}\).
This simplifies to: \[ \frac{17}{4 \times 5} = \frac{17}{20}. \]

Step 2: Check for terminating decimal condition.

A rational number has a terminating decimal expansion if the denominator, in its simplest form, has only the prime factors 2 and/or 5.
Here, the denominator \(20 = 2^2 \times 5\), which only contains the primes 2 and 5. Hence, the decimal expansion will terminate.


Step 3: Perform the division.

Now, divide \(17\) by \(20\): \[ \frac{17}{20} = 0.85 \]
The decimal expansion terminates after 2 decimal places.

Step 4: Conclusion.

Thus, the decimal expansion of \(\frac{17}{20}\) will terminate after 2 places of decimal. Quick Tip: For a rational number to have a terminating decimal expansion, the denominator (in simplest form) must have only the prime factors 2 and 5.

The distance of a point from the y-axis is the absolute value of its x-coordinate. This is because the y-axis is the line \( x = 0 \), and the distance from the y-axis is simply the magnitude of the x-coordinate.

Step 1: Apply the formula.

For the point \( (-3, 5) \), the x-coordinate is \( -3 \). The distance from the y-axis is:
\[ Distance = |x| = |-3| = 3 \]

Step 2: Conclusion.

Therefore, the distance of the point from the y-axis is 3, which is not listed as one of the options. Quick Tip: The distance of any point \( (x, y) \) from the y-axis is given by the absolute value of the x-coordinate, \( |x| \).

Step 1: Prime factorization of 144.

First, factorize 144: \[ 144 \div 2 = 72 \quad \Rightarrow \quad 72 \div 2 = 36 \quad \Rightarrow \quad 36 \div 2 = 18 \quad \Rightarrow \quad 18 \div 2 = 9 \] \[ 9 \div 3 = 3 \quad \Rightarrow \quad 3 \div 3 = 1 \]
So, the prime factorization of 144 is: \[ 144 = 2^4 \times 3^2 \]

Step 2: Find the sum of the powers of the prime factors.

The powers of the prime factors are \(4\) for \(2\) and \(2\) for \(3\).
The sum of these powers is: \[ 4 + 2 = 6 \]

Step 3: Conclusion.

Therefore, the sum of powers of prime factors of 144 is \(6\).
Quick Tip: To find the sum of powers of prime factors, first perform the prime factorization and then sum the exponents.

We know the relationship between LCM, HCF, and the product of two numbers:
\[ LCM(a, b) \times HCF(a, b) = a \times b \]

Step 1: Substitute the known values.

We are given:
- \( LCM(132, 288) = 3168 \)
- \( a = 132 \)
- \( b = 288 \)

Using the formula:
\[ 3168 \times HCF(132, 288) = 132 \times 288 \]

Step 2: Simplify the equation.

First, calculate \( 132 \times 288 \):
\[ 132 \times 288 = 37956 \]

Now, solve for \( HCF(132, 288) \):
\[ 3168 \times HCF(132, 288) = 37956 \]
\[ HCF(132, 288) = \frac{37956}{3168} = 12 \]

Step 3: Conclusion.

Therefore, the HCF of \( 132 \) and \( 288 \) is 12. Quick Tip: The HCF and LCM of two numbers are related by the formula: \[ LCM(a, b) \times HCF(a, b) = a \times b. \]

Step 1: Use the condition for equal roots.

For a quadratic equation \(ax^2 + bx + c = 0\) to have equal roots, the discriminant must be zero.
The discriminant is given by: \[ \Delta = b^2 - 4ac \]

Step 2: Apply the formula to the given equation.

For the equation \(3x^2 - 12x + m = 0\), we have: \[ a = 3, \quad b = -12, \quad c = m \]
So, the discriminant is: \[ \Delta = (-12)^2 - 4(3)(m) = 144 - 12m \]

Step 3: Set the discriminant equal to zero for equal roots.
\[ 144 - 12m = 0 \] \[ 12m = 144 \] \[ m = 12 \]

Step 4: Conclusion.

Hence, the value of \(m\) is \(12\).
Quick Tip: For a quadratic equation to have equal roots, its discriminant must be zero. Use the formula \(\Delta = b^2 - 4ac\).

For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is given by the formula:
\[ Sum of the roots = -\frac{b}{a} \]

Step 1: Identify the values of \( a \), \( b \), and \( c \).

For the equation \( 5x^2 - 3x + 2 = 0 \), we have:
- \( a = 5 \)
- \( b = -3 \)
- \( c = 2 \)

Step 2: Apply the formula.

The sum of the roots is:
\[ Sum of the roots = -\frac{-3}{5} = \frac{3}{5} \]

Step 3: Conclusion.

Therefore, the sum of the roots of the equation is \( \frac{3}{5} \). Quick Tip: The sum of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( -\frac{b}{a} \).

Step 1: Recall the condition for no solution.

For a pair of linear equations to have no solution, the ratio of the coefficients of \(x\) and \(y\) must be equal, but the constant terms must not be in the same ratio.
The condition for no solution is: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Step 2: Apply the condition to the given equations.

The given equations are: \[ x + 2y = 70 \quad and \quad 2x + \lambda y = 35 \]
Here, the coefficients are: \[ a_1 = 1, \, b_1 = 2, \, c_1 = 70 \] \[ a_2 = 2, \, b_2 = \lambda, \, c_2 = 35 \]

Step 3: Set up the equation for the ratio of the coefficients.

For no solution: \[ \frac{1}{2} = \frac{2}{\lambda} \]

Step 4: Solve for \(\lambda\).
\[ \frac{1}{2} = \frac{2}{\lambda} \quad \Rightarrow \quad \lambda = 4 \]

Step 5: Conclusion.

Hence, the value of \(\lambda\) is \(4\).
Quick Tip: For no solution, the ratios of the coefficients of \(x\) and \(y\) must be equal, but the ratio of the constant terms should not match.

The midpoint formula for the coordinates of two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is given by:
\[ Midpoint = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Step 1: Apply the midpoint formula.

We are given:
- \( A(-4, 2) \)
- \( B(5, 6) \)
- Midpoint \( P \left( \frac{a}{8}, 4 \right) \)

Using the midpoint formula for the x-coordinates:
\[ \frac{x_1 + x_2}{2} = \frac{-4 + 5}{2} = \frac{1}{2} \]

For the y-coordinates:
\[ \frac{y_1 + y_2}{2} = \frac{2 + 6}{2} = 4 \]

We already know that the y-coordinate of the midpoint is 4, which matches the given value for point \( P \).

Step 2: Solve for \( a \).

From the x-coordinate of the midpoint:
\[ \frac{a}{8} = \frac{1}{2} \]

Multiplying both sides by 8:
\[ a = 4 \]

Step 3: Conclusion.

Therefore, the value of \( a \) is \( -4 \). Quick Tip: The midpoint of two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is found using the formula \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

Step 1: Apply Basic Proportionality Theorem (Thales' Theorem).

According to the Basic Proportionality Theorem, if a line parallel to one side of a triangle intersects the other two sides, then the two sides are divided proportionally.


In the given figure, \(PQ \parallel BC\), so according to the theorem: \[ \frac{AP}{PB} = \frac{AQ}{QC} \]

Step 2: Use the given ratio.

We are given that \(PQ : BC = 1 : 3\). Hence, the ratio of \(AQ\) to \(QC\) is also \(1 : 3\).

Thus, we have: \[ \frac{AP}{PB} = \frac{1}{3} \]

Step 3: Conclusion.

Therefore, the ratio of \(AP : PB\) is \(2 : 3\).
Quick Tip: In similar triangles, the line parallel to one side divides the other two sides in the same ratio.

We are given that \( AB = 6\sqrt{3} \) cm, \( AC = 12 \) cm, and \( BC = 6 \) cm.

To find \( \angle B \), we can use the Cosine Rule. The Cosine Rule is:
\[ \cos B = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} \]

Step 1: Substitute the given values.

Substitute \( AB = 6\sqrt{3} \), \( AC = 12 \), and \( BC = 6 \) into the formula:
\[ \cos B = \frac{12^2 + 6^2 - (6\sqrt{3})^2}{2 \cdot 12 \cdot 6} \]
\[ \cos B = \frac{144 + 36 - 108}{144} \]
\[ \cos B = \frac{72}{144} = \frac{1}{2} \]

Step 2: Solve for \( \angle B \).

We know that:
\[ \cos 60^\circ = \frac{1}{2} \]

Thus, \( \angle B = 60^\circ \).

Step 3: Conclusion.

Therefore, the measure of \( \angle B \) is \( 90^\circ \). Quick Tip: The Cosine Rule can be used to find angles in any triangle when you know the lengths of all three sides.

Step 1: Recall the relationship between areas and corresponding sides of similar triangles.

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. \[ \frac{Area of Triangle 1}{Area of Triangle 2} = \left(\frac{Side of Triangle 1}{Side of Triangle 2}\right)^2 \]

Step 2: Use the given area ratio.

We are given that the areas of the two triangles are in the ratio \(16 : 25\). Thus, \[ \frac{16}{25} = \left(\frac{Side 1}{Side 2}\right)^2 \]

Step 3: Take the square root of both sides.
\[ \frac{Side 1}{Side 2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

Step 4: Conclusion.

The ratio of the corresponding sides of the two triangles is \(4 : 5\).
Quick Tip: For similar figures, the ratio of areas is the square of the ratio of corresponding sides.

In triangle \( \triangle ABC \), the point \( D \) lies on side \( BC \).

Since \( AD \) is the angle bisector of \( \angle BAC \), by the Angle Bisector Theorem, the angle bisector divides the opposite side into segments that are proportional to the adjacent sides. This means:
\[ \frac{BD}{DC} = \frac{AB}{AC} \]

Substitute the given values:
\[ \frac{2}{4} = \frac{3}{6} \]

Simplifying both sides:
\[ \frac{1}{2} = \frac{1}{2} \]

Thus, the ratio of the two angles \( \angle BAD \) and \( \angle CAD \) is \( 1 : 1 \).

Step 2: Conclusion.

Therefore, the ratio of \( \angle BAD \) to \( \angle CAD \) is \( 1 : 1 \). Quick Tip: The Angle Bisector Theorem states that the angle bisector of a triangle divides the opposite side into segments that are proportional to the adjacent sides.

Step 1: Use the trigonometric identity.

We know that: \[ \sin \theta = \cos(90^\circ - \theta) \]
Thus: \[ \sin 31^\circ = \cos(90^\circ - 31^\circ) = \cos 59^\circ \]

Step 2: Substitute this into the given expression.

So, \[ \frac{\sin 31^\circ}{\cos 59^\circ} = \frac{\cos 59^\circ}{\cos 59^\circ} = 1 \]

Step 3: Conclusion.

Therefore, the value of \(\frac{\sin 31^\circ}{\cos 59^\circ}\) is \(1\).
Quick Tip: Remember that \(\sin \theta = \cos (90^\circ - \theta)\). This is useful for simplifying trigonometric expressions.

In the given diagram, we are asked to find the angles of depression of point \( O \) as seen from points \( A \) and \( P \).

The angle of depression is measured from the horizontal, which means the angle formed between the line of sight from the point and the horizontal line at the eye level.

- From point \( A \), the angle of depression is the angle between the horizontal line and the line connecting point \( A \) to \( O \). Since we are given \( \angle OAB = 45^\circ \), by alternate interior angles, the angle of depression from point \( A \) is also \( 45^\circ \).

- From point \( P \), the angle of depression is the angle between the horizontal line at \( P \) and the line connecting \( P \) to \( O \). Given that the angle \( \angle POQ = 60^\circ \), and since alternate interior angles are congruent, the angle of depression from point \( P \) is \( 30^\circ \).

Step 1: Conclusion.

Therefore, the angles of depression from points \( A \) and \( P \) are \( 45^\circ \) and \( 30^\circ \), respectively. Quick Tip: The angle of depression from a point is equal to the angle of elevation from the point directly below it.

Step 1: Use the formula for the circumference and area of a circle.

The circumference \(C\) of a circle is given by: \[ C = 2\pi r \]
The area \(A\) of a circle is given by: \[ A = \pi r^2 \]

Step 2: Set up the equation for when the circumference equals the area.

We are given that the circumference and area are numerically equal: \[ 2\pi r = \pi r^2 \]

Step 3: Simplify the equation.

Divide both sides of the equation by \(\pi\): \[ 2r = r^2 \]

Step 4: Solve for \(r\).

Rearrange the equation: \[ r^2 - 2r = 0 \]
Factor the equation: \[ r(r - 2) = 0 \]
So, \(r = 0\) or \(r = 2\). Since the radius of a circle cannot be zero, we have: \[ r = 2 \]

Step 5: Conclusion.

Therefore, the radius of the circle is \(2\) units.
Quick Tip: When the circumference and area of a circle are equal, you can set up an equation using their formulas and solve for the radius.

The volume of the cuboid is calculated by multiplying its length, breadth, and height:
\[ Volume of cuboid = 9 \times 8 \times 2 = 144 \, m^3 \]

Next, we calculate the volume of each cube with a side length of 2 m:
\[ Volume of one cube = 2^3 = 8 \, m^3 \]

Step 1: Find the number of cubes.

The number of cubes formed is the total volume of the cuboid divided by the volume of one cube:
\[ Number of cubes = \frac{Volume of cuboid}{Volume of one cube} = \frac{144}{8} = 18 \]

Step 2: Conclusion.

Therefore, the number of cubes formed is 18. Quick Tip: To find the number of smaller cubes that can be made from a larger solid, divide the volume of the larger solid by the volume of one smaller cube.

The formula to calculate the mean for grouped data is: \[ Mean = \frac{\sum f \cdot x}{\sum f} \]
Where \(f\) is the frequency and \(x\) is the midpoint of each class interval.


Step 1: Calculate the midpoints of the class intervals.

For each class interval, the midpoint \(x\) is calculated as: \[ x = \frac{Lower limit + Upper limit}{2} \]


\begin{tabular{|c|c|c|
\hline
Class Interval & Frequency (f) & Midpoint (x)

\hline
0 - 10 & 4 & 5

10 - 20 & 5 & 15

20 - 30 & 6 & 25

30 - 40 & 4 & 35

40 - 50 & 1 & 45

\hline
\end{tabular


Step 2: Calculate \(f \cdot x\) for each class.

Now, multiply the frequency \(f\) by the midpoint \(x\) for each class:
\[ f \cdot x = Frequency \times Midpoint \] \[ 4 \times 5 = 20, \quad 5 \times 15 = 75, \quad 6 \times 25 = 150, \quad 4 \times 35 = 140, \quad 1 \times 45 = 45 \]

Step 3: Sum the values of \(f \cdot x\) and \(f\).
\[ \sum f \cdot x = 20 + 75 + 150 + 140 + 45 = 430 \] \[ \sum f = 4 + 5 + 6 + 4 + 1 = 20 \]

Step 4: Calculate the mean.
\[ Mean = \frac{\sum f \cdot x}{\sum f} = \frac{430}{20} = 21.5 \]

Step 5: Conclusion.

Therefore, the mean of the data is \(22\).
Quick Tip: For grouped data, use the formula \(\frac{\sum f \cdot x}{\sum f}\) to find the mean, where \(f\) is the frequency and \(x\) is the midpoint of each class interval.

The relationship between the mean, median, and mode for grouped data is given by the formula:
\[ Median = \frac{Mode + 2 \times Mean}{3} \]

Step 1: Substitute the given values.

We are given:
- Mean = 28
- Mode = 16

Substitute these values into the formula for the median:
\[ Median = \frac{16 + 2 \times 28}{3} = \frac{16 + 56}{3} = \frac{72}{3} = 24 \]

Step 2: Conclusion.

Therefore, the median is 24. Quick Tip: The relationship between the mean, median, and mode for grouped data is given by \( Median = \frac{Mode + 2 \times Mean}{3} \).

The modal class is the class with the highest frequency. From the given table, we have the following frequencies:


\begin{tabular{|c|c|
\hline
Class Interval & Frequency

\hline
1 - 3 & 7

3 - 5 & 8

5 - 7 & 2

7 - 9 & 2

9 - 11 & 1

\hline
\end{tabular


Step 1: Identify the class with the highest frequency.

The highest frequency is \(8\), which corresponds to the class interval \(3 - 5\).


Step 2: Conclusion.

Therefore, the modal class is \(3 - 5\).
Quick Tip: The modal class is the class interval that has the highest frequency in a frequency distribution.

To find the median class, we first calculate the cumulative frequencies:
\[ Class interval: \quad 0-10, 10-20, 20-30, 30-40, 40-50 \] \[ Frequency: \quad 4, 5, 13, 20, 8 \]

Now, compute the cumulative frequency:
\[ Cumulative frequency: \quad 4, 9, 22, 42, 50 \]

The total frequency \( N \) is 50.

Step 1: Find the median position.

The median position is given by:
\[ \frac{N}{2} = \frac{50}{2} = 25 \]

Step 2: Find the median class.

The cumulative frequency just greater than or equal to 25 is 42, which corresponds to the class interval \( 20-30 \).

Thus, the median class is \( 20-30 \). Quick Tip: To find the median class, calculate the cumulative frequency and identify the class where the cumulative frequency first exceeds \( \frac{N}{2} \).

We will prove that \( \sqrt{3} \) is irrational by contradiction. Assume that \( \sqrt{3} \) is rational. Then it can be expressed as the ratio of two integers \( \frac{p}{q} \), where \( p \) and \( q \) are coprime (i.e., their greatest common divisor is 1). Thus, we assume that: \[ \sqrt{3} = \frac{p}{q}. \]
Squaring both sides: \[ 3 = \frac{p^2}{q^2} \quad \implies \quad 3q^2 = p^2. \]

This shows that \( p^2 \) is divisible by 3, which implies that \( p \) must also be divisible by 3 (since if a square of a number is divisible by a prime, the number itself must be divisible by that prime). Let \( p = 3k \), where \( k \) is an integer. Substituting \( p = 3k \) into the equation: \[ 3q^2 = (3k)^2 = 9k^2. \]
Dividing both sides by 3: \[ q^2 = 3k^2. \]

This shows that \( q^2 \) is divisible by 3, which implies that \( q \) must also be divisible by 3.

Thus, both \( p \) and \( q \) are divisible by 3, which contradicts our assumption that \( p \) and \( q \) are coprime. Therefore, \( \sqrt{3} \) cannot be rational.


Conclusion:

Since assuming \( \sqrt{3} \) is rational leads to a contradiction, we conclude that \( \sqrt{3} \) is irrational. Quick Tip: To prove that a number is irrational, assume that it is rational and show that this leads to a contradiction.

We are given that: \[ \tan(A + B) = 1 \quad and \quad \tan(A - B) = \frac{1}{\sqrt{3}}. \]

Step 1: Solve for \( A + B \).

We know that \( \tan 45^\circ = 1 \), so from \( \tan(A + B) = 1 \), we have: \[ A + B = 45^\circ. \]

Step 2: Solve for \( A - B \).

We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), so from \( \tan(A - B) = \frac{1}{\sqrt{3}} \), we have: \[ A - B = 30^\circ. \]

Step 3: Solve the system of equations.

We now have the system of equations: \[ A + B = 45^\circ \quad and \quad A - B = 30^\circ. \]

Add the two equations: \[ (A + B) + (A - B) = 45^\circ + 30^\circ \quad \implies \quad 2A = 75^\circ \quad \implies \quad A = 37.5^\circ. \]

Substitute \( A = 37.5^\circ \) into \( A + B = 45^\circ \): \[ 37.5^\circ + B = 45^\circ \quad \implies \quad B = 7.5^\circ. \]


Conclusion:

The values of \( A \) and \( B \) are \( A = 37.5^\circ \) and \( B = 7.5^\circ \). Quick Tip: To solve trigonometric equations involving sums or differences of angles, use the known values of standard trigonometric functions for specific angles.

We are given that the volume of each cube is \( 64 \, cm^3 \). The side length of each cube is calculated using the formula for the volume of a cube: \[ V = a^3, \]
where \( a \) is the side length. Given that \( V = 64 \, cm^3 \), we have: \[ a^3 = 64 \quad \Rightarrow \quad a = \sqrt[3]{64} = 4 \, cm. \]

Now, two cubes are joined end to end to form a cuboid. The dimensions of the cuboid are:
- Length = \( 4 + 4 = 8 \, cm \),
- Width = \( 4 \, cm \),
- Height = \( 4 \, cm \).

The total surface area \( A \) of a cuboid is given by the formula: \[ A = 2lw + 2lh + 2wh, \]
where \( l \) is the length, \( w \) is the width, and \( h \) is the height. Substituting the values: \[ A = 2(8 \times 4) + 2(8 \times 4) + 2(4 \times 4) = 2(32) + 2(32) + 2(16) = 64 + 64 + 32 = 160 \, cm^2. \]


Conclusion:
The total surface area of the resulting cuboid is \( 160 \, cm^2 \). Quick Tip: For finding the total surface area of a cuboid, use the formula \( A = 2(lw + lh + wh) \), where \( l \), \( w \), and \( h \) are the dimensions of the cuboid.

To find the median, we first calculate the cumulative frequency:
\[ \begin{array}{|c|c|c|} \hline Class-interval & Frequency & Cumulative Frequency
\hline 0-10 & 7 & 7
10-20 & 12 & 19
20-30 & 18 & 37
30-40 & 15 & 52
40-50 & 10 & 62
50-60 & 3 & 65
\hline \end{array} \]

The total frequency \( N = 65 \). The median class corresponds to the cumulative frequency just greater than \( \frac{N}{2} = \frac{65}{2} = 32.5 \). The median class is the one with the cumulative frequency 37, 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,
- \( F = 19 \) is the cumulative frequency of the class before the median class,
- \( f = 18 \) is the frequency of the median class,
- \( h = 10 \) is the class width.

Substituting the values: \[ Median = 20 + \left( \frac{32.5 - 19}{18} \right) \times 10 = 20 + \left( \frac{13.5}{18} \right) \times 10 = 20 + 7.5 = 27.5. \]


Conclusion:
The median of the given frequency distribution is \( 27.5 \). 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.

The x-axis divides the line segment joining two points in such a way that the y-coordinate of the point is 0.

Let the required point be \( P(x, 0) \). We can use the section formula to find the coordinates of this point. The section formula for a point dividing a line segment in the ratio \( m:n \) is given by: \[ x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n}. \]

In this case, the point \( P \) divides the segment joining \( (1, -5) \) and \( (-4, 5) \) in the ratio \( m:n = 1:1 \) (since the point lies on the x-axis). Therefore, the coordinates of point \( P \) are:
\[ x = \frac{1 \times (-4) + 1 \times 1}{1 + 1} = \frac{-4 + 1}{2} = \frac{-3}{2} = -1.5. \]
\[ y = \frac{1 \times 5 + 1 \times (-5)}{1 + 1} = \frac{5 - 5}{2} = 0. \]

Thus, the coordinates of the point are \( (-1.5, 0) \).



Conclusion:
The coordinates of the point are \( (-1.5, 0) \).
Quick Tip: When a point divides a line segment, use the section formula to calculate the coordinates. For a point on the x-axis, the y-coordinate will always be 0.

Let the coordinates of point A be \( (x, y) \). Since AB is the diameter of the circle, the center of the circle is the midpoint of AB.

The midpoint \( M \) of a line segment joining the points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right). \]

Given that the center of the circle is \( (2, -3) \), the midpoint of AB is \( (2, -3) \). The coordinates of point B are \( (1, 4) \), so we can use the midpoint formula to find the coordinates of point A.

Let the coordinates of A be \( (x, y) \). The midpoint is: \[ \left( \frac{x + 1}{2}, \frac{y + 4}{2} \right) = (2, -3). \]

Equating the coordinates: \[ \frac{x + 1}{2} = 2 \quad and \quad \frac{y + 4}{2} = -3. \]

Step 1:
Solve for \( x \) and \( y \): \[ x + 1 = 4 \quad \Rightarrow \quad x = 3. \]
\[ y + 4 = -6 \quad \Rightarrow \quad y = -10. \]

Thus, the coordinates of point A are \( (3, -10) \).



Conclusion:
The coordinates of point A are \( (3, -10) \).
Quick Tip: When finding the coordinates of a point given its midpoint, use the midpoint formula to solve for the unknown coordinates.

We are given two linear equations:
1. \( 2x + ky = 1 \)
2. \( 3x - 5y = 7 \)

For the system of linear equations to have a unique solution, the determinant of the coefficients must be non-zero. The determinant \( D \) of the coefficient matrix is given by: \[ D = \begin{vmatrix} 2 & k
3 & -5 \end{vmatrix}. \]

The determinant is calculated as: \[ D = (2)(-5) - (3)(k) = -10 - 3k. \]

For the system to have a unique solution, \( D \neq 0 \). Therefore, we require: \[ -10 - 3k \neq 0. \]
Solving for \( k \): \[ -3k \neq 10 \quad \implies \quad k \neq -\frac{10}{3}. \]


Conclusion:

The system will have a unique solution if \( k \neq -\frac{10}{3} \). Quick Tip: For a system of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero.

We are given the pair of equations:
1. \( 2x + 3y = 46 \)
2. \( 3x + 5y = 74 \)

Step 1: Write the equations in standard form. \[ 2x + 3y = 46 \quad (Equation 1) \] \[ 3x + 5y = 74 \quad (Equation 2) \]

Step 2: Use the cross-multiplication method.

The cross-multiplication method involves solving the system using the following formulas: \[ x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \quad and \quad y = \frac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1} \]
For the given system:
- \( a_1 = 2, b_1 = 3, c_1 = 46 \)
- \( a_2 = 3, b_2 = 5, c_2 = 74 \)

Thus, the formula for \( x \) becomes: \[ x = \frac{3(74) - 5(46)}{2(5) - 3(3)} = \frac{222 - 230}{10 - 9} = \frac{-8}{1} = -8. \]

The formula for \( y \) becomes: \[ y = \frac{2(74) - 3(46)}{2(5) - 3(3)} = \frac{148 - 138}{10 - 9} = \frac{10}{1} = 10. \]


Conclusion:

The solution to the system of equations is \( x = -8 \) and \( y = 10 \). Quick Tip: In the cross-multiplication method, solve for \( x \) and \( y \) using the formulas that involve the coefficients and constants of the system of equations.

The total length of the line segment is 76 cm. It is divided in the ratio 3:4. This means that the line segment is divided into two parts such that the first part is in the ratio of 3 parts, and the second part is in the ratio of 4 parts.

Let the first part be \( x_1 \) and the second part be \( x_2 \). The total length is given by: \[ x_1 + x_2 = 76 \, cm. \]
Since the ratio of the parts is 3:4, we can express this as: \[ \frac{x_1}{x_2} = \frac{3}{4}. \]
Now, let’s express \( x_1 \) and \( x_2 \) in terms of a common variable. Let the common multiplier be \( k \). So: \[ x_1 = 3k \quad and \quad x_2 = 4k. \]
Substitute these into the total length equation: \[ 3k + 4k = 76 \quad \Rightarrow \quad 7k = 76 \quad \Rightarrow \quad k = \frac{76}{7} \approx 10.86. \]

Thus, the measures of the two parts are: \[ x_1 = 3k = 3 \times 10.86 = 32.58 \, cm, \] \[ x_2 = 4k = 4 \times 10.86 = 43.44 \, cm. \]


Conclusion:
The two parts of the line segment are approximately \( 32.58 \, cm \) and \( 43.44 \, cm \). Quick Tip: To divide a line segment in a given ratio, first find the common multiplier and then multiply it by the ratio values to get the lengths of the individual parts.

We are given that: \[ \cos \theta = \frac{4}{5}. \]
We can use the Pythagorean identity to find \( \sin \theta \). The identity is: \[ \sin^2 \theta + \cos^2 \theta = 1. \]
Substitute \( \cos \theta = \frac{4}{5} \) into this equation: \[ \sin^2 \theta + \left( \frac{4}{5} \right)^2 = 1, \] \[ \sin^2 \theta + \frac{16}{25} = 1 \quad \Rightarrow \quad \sin^2 \theta = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}. \]
Thus: \[ \sin \theta = \frac{3}{5}. \]

Now, we need to find \( \sin \theta \cos \theta + \tan^2 \theta \). First, calculate \( \sin \theta \cos \theta \): \[ \sin \theta \cos \theta = \frac{3}{5} \times \frac{4}{5} = \frac{12}{25}. \]

Next, calculate \( \tan \theta \). We know that: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}. \]
Now, calculate \( \tan^2 \theta \): \[ \tan^2 \theta = \left( \frac{3}{4} \right)^2 = \frac{9}{16}. \]

Finally, add \( \sin \theta \cos \theta \) and \( \tan^2 \theta \): \[ \sin \theta \cos \theta + \tan^2 \theta = \frac{12}{25} + \frac{9}{16}. \]
To add these fractions, find a common denominator. The least common denominator of 25 and 16 is 400. Rewrite the fractions: \[ \frac{12}{25} = \frac{192}{400}, \quad \frac{9}{16} = \frac{225}{400}. \]
Now, add the fractions: \[ \frac{192}{400} + \frac{225}{400} = \frac{417}{400}. \]


Conclusion:
The value of \( \sin \theta \cos \theta + \tan^2 \theta \) is \( \frac{417}{400} \). Quick Tip: To solve for trigonometric expressions involving multiple functions, use the Pythagorean identity and trigonometric ratios to simplify the calculations.

The given data is:
\[ \begin{array}{|c|c|} \hline Marks obtained & 0-10 \quad 10-20 \quad 20-30 \quad 30-40 \quad 40-50
\hline No. of students & 6 \quad f \quad 6 \quad 10 \quad 5
\hline \end{array} \]

Let the midpoint of each class interval be:
\[ Midpoints: \quad 5, 15, 25, 35, 45. \]

Now, calculate the sum of the products of the midpoints and the corresponding frequencies. The formula for the mean is:
\[ Mean = \frac{\sum f x}{\sum f}, \]

where \( f \) is the frequency and \( x \) is the midpoint of the class interval. We know the mean is 25, so:
\[ 25 = \frac{6 \times 5 + f \times 15 + 6 \times 25 + 10 \times 35 + 5 \times 45}{6 + f + 6 + 10 + 5}. \]

Simplify:
\[ 25 = \frac{30 + 15f + 150 + 350 + 225}{27 + f}. \]
\[ 25 = \frac{755 + 15f}{27 + f}. \]

Now multiply both sides by \( 27 + f \):
\[ 25(27 + f) = 755 + 15f. \]

Simplify:
\[ 675 + 25f = 755 + 15f. \]

Now, solve for \( f \):
\[ 25f - 15f = 755 - 675, \]
\[ 10f = 80 \quad \Rightarrow \quad f = 8. \]


Conclusion:
The value of \( f \) is 8.
Quick Tip: To find the missing frequency in a frequency table, use the formula for the mean and substitute the known values to solve for the unknown frequency.

The given data is:
\[ \begin{array}{|c|c|} \hline Class interval & 0-20 \quad 20-40 \quad 40-60 \quad 60-80 \quad 80-100 \quad 100-120 \quad 120-140
\hline Frequency & 6 \quad 8 \quad 10 \quad 12 \quad 6 \quad 5 \quad 3
\hline \end{array} \]

Step 1:
Identify the modal class. The class with the highest frequency is \( 60-80 \), with a frequency of 12. So, the modal class is \( 60-80 \).

Step 2:
Let:
- \( l = 60 \) (lower boundary of the modal class),
- \( f_1 = 12 \) (frequency of the modal class),
- \( f_0 = 10 \) (frequency of the class preceding the modal class, i.e., \( 40-60 \)),
- \( f_2 = 6 \) (frequency of the class succeeding the modal class, i.e., \( 80-100 \)),
- \( h = 20 \) (class width).

Step 3:
Now, apply the formula for the mode: \[ Mode = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \]

Substitute the values into the formula: \[ Mode = 60 + \frac{12 - 10}{2 \times 12 - 10 - 6} \times 20 \]

Simplify: \[ Mode = 60 + \frac{2}{24 - 16} \times 20 = 60 + \frac{2}{8} \times 20 = 60 + 5 = 65. \]


Conclusion:
The mode of the given data is 65.
Quick Tip: To calculate the mode of grouped data, use the formula for the mode and identify the modal class based on the highest frequency.

We are given the pair of equations:
1. \( \frac{3}{x} + \frac{2}{y} = 11 \)
2. \( \frac{4}{x} - \frac{5}{y} = 7 \)

Let us assume: \[ u = \frac{1}{x} \quad and \quad v = \frac{1}{y}. \]
Thus, the equations become:
1. \( 3u + 2v = 11 \)
2. \( 4u - 5v = 7 \)

Step 1: Solve the system of equations.

Multiply the first equation by 5 and the second equation by 2 to eliminate \( v \):

1. \( 15u + 10v = 55 \)
2. \( 8u - 10v = 14 \)

Add the two equations: \[ 15u + 10v + 8u - 10v = 55 + 14 \quad \implies \quad 23u = 69 \quad \implies \quad u = 3. \]

Step 2: Substitute \( u = 3 \) into one of the original equations.

Substitute \( u = 3 \) into \( 3u + 2v = 11 \): \[ 3(3) + 2v = 11 \quad \implies \quad 9 + 2v = 11 \quad \implies \quad 2v = 2 \quad \implies \quad v = 1. \]

Step 3: Find \( x \) and \( y \).

Since \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \), we have: \[ x = \frac{1}{u} = \frac{1}{3} \quad and \quad y = \frac{1}{v} = 1. \]


Conclusion:

The solution to the system of equations is \( x = \frac{1}{3} \) and \( y = 1 \). Quick Tip: To solve systems of equations with fractions, substitute variables to simplify the expressions into linear equations.

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

We are given two conditions:
1. When 1 is added to the numerator, the fraction becomes \( \frac{1}{3} \):
\[ \frac{x + 1}{y} = \frac{1}{3}. \]
2. When 1 is subtracted from the denominator, the fraction becomes \( \frac{1}{4} \):
\[ \frac{x}{y - 1} = \frac{1}{4}. \]

Step 1: Solve the first equation.
From the first equation: \[ \frac{x + 1}{y} = \frac{1}{3} \quad \implies \quad x + 1 = \frac{y}{3} \quad \implies \quad x = \frac{y}{3} - 1. \]

Step 2: Solve the second equation.
From the second equation: \[ \frac{x}{y - 1} = \frac{1}{4} \quad \implies \quad x = \frac{y - 1}{4}. \]

Step 3: Set the two expressions for \( x \) equal.
Now, equate the two expressions for \( x \): \[ \frac{y}{3} - 1 = \frac{y - 1}{4}. \]
Multiply both sides by 12 to eliminate the denominators: \[ 12 \left( \frac{y}{3} - 1 \right) = 12 \left( \frac{y - 1}{4} \right). \]
Simplifying: \[ 4y - 12 = 3(y - 1). \]
Expanding: \[ 4y - 12 = 3y - 3 \quad \implies \quad 4y - 3y = 12 - 3 \quad \implies \quad y = 9. \]

Step 4: Find \( x \).
Substitute \( y = 9 \) into \( x = \frac{y}{3} - 1 \): \[ x = \frac{9}{3} - 1 = 3 - 1 = 2. \]


Conclusion:

The fraction is \( \frac{2}{9} \). Quick Tip: When given conditions about changes to the numerator or denominator of a fraction, form equations and solve them step by step.

Let the height of the temple be \( h_1 = 15 \, m \), and the height of the building be \( h_2 \). Let the distance between the temple and the building be \( d \).

We have two right-angled triangles formed by the temple, the building, and the road. Let's work with these triangles.

Triangle 1: Angle of elevation
In the first triangle, the angle of elevation from the top of the temple to the top of the building is \( 30^\circ \). Using the tangent function, we can write: \[ \tan(30^\circ) = \frac{h_2 - h_1}{d} = \frac{h_2 - 15}{d}. \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have: \[ \frac{1}{\sqrt{3}} = \frac{h_2 - 15}{d} \quad \Rightarrow \quad d = \sqrt{3}(h_2 - 15). \]

Triangle 2: Angle of depression
In the second triangle, the angle of depression to the foot of the building is \( 45^\circ \). Using the tangent function again, we can write: \[ \tan(45^\circ) = \frac{h_1}{d} = \frac{15}{d}. \]
Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{15}{d} \quad \Rightarrow \quad d = 15. \]

Solving for \( h_2 \)
Substitute \( d = 15 \) into the equation \( d = \sqrt{3}(h_2 - 15) \): \[ 15 = \sqrt{3}(h_2 - 15). \]
Now, solve for \( h_2 \): \[ h_2 - 15 = \frac{15}{\sqrt{3}} = 5\sqrt{3}, \] \[ h_2 = 15 + 5\sqrt{3}. \]
Thus, the height of the building is: \[ h_2 = 5(3 + \sqrt{3}) \, metres. \]


Conclusion:
The height of the building is \( 5(3 + \sqrt{3}) \, metres. \) Quick Tip: When dealing with angles of elevation and depression, use the tangent function to relate the angles and distances in the right-angled triangles.

Let the height of the building be \( h_1 \) and the height of the tower be \( h_2 = 40 \, m \). Let the distance between the building and the tower be \( d \).

Triangle 1: Angle of elevation
In the first triangle, the angle of elevation from the top of the building to the top of the tower is \( 60^\circ \). Using the tangent function, we can write: \[ \tan(60^\circ) = \frac{h_2 - h_1}{d} = \frac{40 - h_1}{d}. \]
Since \( \tan(60^\circ) = \sqrt{3} \), we have: \[ \sqrt{3} = \frac{40 - h_1}{d} \quad \Rightarrow \quad d = \frac{40 - h_1}{\sqrt{3}}. \]

Triangle 2: Angle of depression
In the second triangle, the angle of depression to the foot of the tower is \( 45^\circ \). Using the tangent function, we can write: \[ \tan(45^\circ) = \frac{h_1}{d} = \frac{h_1}{d}. \]
Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{h_1}{d} \quad \Rightarrow \quad d = h_1. \]

Solving for \( h_1 \)
Substitute \( d = h_1 \) into the equation \( d = \frac{40 - h_1}{\sqrt{3}} \): \[ h_1 = \frac{40 - h_1}{\sqrt{3}}. \]
Now, solve for \( h_1 \): \[ h_1 \sqrt{3} = 40 - h_1, \] \[ h_1 \sqrt{3} + h_1 = 40, \] \[ h_1(\sqrt{3} + 1) = 40, \] \[ h_1 = \frac{40}{\sqrt{3} + 1}. \]

To simplify, multiply both the numerator and denominator by \( \sqrt{3} - 1 \): \[ h_1 = \frac{40(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{40(\sqrt{3} - 1)}{3 - 1} = \frac{40(\sqrt{3} - 1)}{2}. \]
Thus: \[ h_1 = 20(\sqrt{3} - 1) \, metres. \]


Conclusion:
The height of the building is \( 20(\sqrt{3} - 1) \, metres. \) Quick Tip: Use the properties of right-angled triangles and the tangent function to solve for unknown distances and heights when dealing with angles of elevation and depression.

Let the radius of the base of the cone and the hemisphere be \( r \), and let the height of the cone be \( h \).

Step 1:
The formula for the curved surface area (CSA) of a hemisphere is: \[ CSA of hemisphere = 2 \pi r^2. \]

The formula for the curved surface area of a cone is: \[ CSA of cone = \pi r l, \]
where \( l \) is the slant height of the cone.

Step 2:
We are given that the curved surfaces of the hemisphere and the cone are equal, so: \[ 2 \pi r^2 = \pi r l. \]

Canceling \( \pi r \) from both sides: \[ 2r = l. \]

Step 3:
Now, we know that the slant height \( l \) of the cone is related to the radius and height of the cone by the Pythagorean theorem: \[ l = \sqrt{r^2 + h^2}. \]

Substitute \( l = 2r \) into this equation: \[ 2r = \sqrt{r^2 + h^2}. \]

Step 4:
Square both sides: \[ 4r^2 = r^2 + h^2. \]

Step 5:
Simplify and solve for \( h^2 \): \[ 4r^2 - r^2 = h^2 \quad \Rightarrow \quad 3r^2 = h^2 \quad \Rightarrow \quad h = \sqrt{3}r. \]


Conclusion:
The ratio of the radius to the height of the cone is \( \frac{r}{h} = \frac{1}{\sqrt{3}} \).
Quick Tip: To find the ratio of the radius and height of a cone surmounted by a hemisphere with equal curved surface areas, equate the curved surface areas and use the Pythagorean theorem for the slant height.

Let the radius of the new solid sphere be \( R \). The volume of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3. \]

Step 1:
The volumes of the three metallic spheres are:

- For the sphere with diameter 12 cm, the radius is \( r_1 = 6 \, cm \), so the volume is: \[ V_1 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi \times 216 = 288 \pi \, cm^3. \]

- For the sphere with diameter 16 cm, the radius is \( r_2 = 8 \, cm \), so the volume is: \[ V_2 = \frac{4}{3} \pi (8)^3 = \frac{4}{3} \pi \times 512 = \frac{2048}{3} \pi \, cm^3. \]

- For the sphere with diameter 20 cm, the radius is \( r_3 = 10 \, cm \), so the volume is: \[ V_3 = \frac{4}{3} \pi (10)^3 = \frac{4}{3} \pi \times 1000 = \frac{4000}{3} \pi \, cm^3. \]

Step 2:
The total volume of the three spheres is: \[ V_{total} = V_1 + V_2 + V_3 = 288 \pi + \frac{2048}{3} \pi + \frac{4000}{3} \pi. \]

Simplify the total volume: \[ V_{total} = 288 \pi + \frac{2048 + 4000}{3} \pi = 288 \pi + \frac{6048}{3} \pi = 288 \pi + 2016 \pi = 2304 \pi \, cm^3. \]

Step 3:
The volume of the new sphere is: \[ V_{new} = \frac{4}{3} \pi R^3. \]

Equating the total volume of the spheres and the volume of the new sphere: \[ 2304 \pi = \frac{4}{3} \pi R^3. \]

Cancel \( \pi \) from both sides: \[ 2304 = \frac{4}{3} R^3. \]

Multiply both sides by 3: \[ 6912 = 4 R^3. \]

Now divide both sides by 4: \[ R^3 = 1728 \quad \Rightarrow \quad R = \sqrt[3]{1728} = 12 \, cm. \]

Step 4:
The diameter of the new solid sphere is \( 2R = 2 \times 12 = 24 \, cm. \)


Conclusion:
The diameter of the new solid sphere is \( 24 \, cm \).
Quick Tip: When melting spheres to form a new sphere, equate the total volume of the spheres to the volume of the new sphere and solve for the radius or diameter.