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

Step 1: Definition of co-prime numbers.

Two numbers are said to be co-prime if their greatest common divisor (GCD) is 1. We need to check the GCD of each pair to determine which one is co-prime.

Step 2: Check GCD of each pair.

- GCD of 18 and 25: The factors of 18 are \( 1, 2, 3, 6, 9, 18 \) and the factors of 25 are \( 1, 5, 25 \). The common factor is \( 1 \), so 18 and 25 are co-prime.

- GCD of 5 and 15: The factors of 5 are \( 1, 5 \) and the factors of 15 are \( 1, 3, 5, 15 \). The common factor is \( 5 \), so 5 and 15 are not co-prime.

- GCD of 7 and 21: The factors of 7 are \( 1, 7 \) and the factors of 21 are \( 1, 3, 7, 21 \). The common factor is \( 7 \), so 7 and 21 are not co-prime.

- GCD of 31 and 93: The factors of 31 are \( 1, 31 \) and the factors of 93 are \( 1, 3, 31, 93 \). The common factor is \( 31 \), so 31 and 93 are not co-prime.


Step 3: Conclusion.

Only the pair (18, 25) has a GCD of 1, meaning they are co-prime. Hence, the correct answer is (A).
Quick Tip: Two numbers are co-prime if their greatest common divisor (GCD) is 1.

The prime factorization of 144 is: \[ 144 = 2^4 \times 3^2 \]

Step 1: Identifying the prime factors.

The prime factors of 144 are 2 and 3.

Step 2: Summing the powers of the prime factors.

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

Step 3: Conclusion.

Therefore, the sum of the powers of prime factors in the prime factorization of 144 is 6. The correct answer is (B).
Quick Tip: In prime factorization, sum the exponents of the prime factors to find the total sum of the powers.

Step 1: Understand the condition for a unique solution.

To determine when a system of linear equations has a unique solution, we need to check the condition on the determinant of the coefficient matrix. The system of equations given is:
\[ x - 2y = 3 \quad (1) \] \[ 3x + my = 1 \quad (2) \]

For a system to have a unique solution, the coefficient matrix:
\[ \begin{bmatrix} 1 & -2
3 & m \end{bmatrix} \]

must have a non-zero determinant. The determinant \( D \) of a 2x2 matrix is given by:
\[ D = \begin{vmatrix} 1 & -2
3 & m \end{vmatrix} \]

Step 2: Calculate the determinant.

The determinant of the matrix is:
\[ D = (1)(m) - (3)(-2) = m + 6 \]

Step 3: Set the condition for a unique solution.

For a unique solution to exist, the determinant must be non-zero:
\[ m + 6 \neq 0 \]
\[ m \neq -6 \]

Step 4: Conclusion.

Thus, for the pair of equations to have a unique solution, the value of \( m \) must not be \( -6 \). Therefore, the correct answer is (C).
Quick Tip: For a system of two linear equations to have a unique solution, the determinant of the coefficient matrix must not be zero.

Step 1: Write the system of equations.

We are given the system of equations:
\[ 2x + 3y = 18 \quad (1) \] \[ x - 2y = 2 \quad (2) \]

We will use the substitution method to solve this system.

Step 2: Solve for \( x \) in equation (2).

From equation (2), solve for \( x \):
\[ x = 2 + 2y \quad (3) \]

Step 3: Substitute the value of \( x \) into equation (1).

Substitute the expression for \( x \) from equation (3) into equation (1):
\[ 2(2 + 2y) + 3y = 18 \]

Simplify:
\[ 4 + 4y + 3y = 18 \]
\[ 4 + 7y = 18 \]

Subtract 4 from both sides:
\[ 7y = 14 \]
\[ y = 2 \]

Step 4: Substitute \( y = 2 \) into equation (2).

Now substitute \( y = 2 \) back into equation (2):
\[ x - 2(2) = 2 \]
\[ x - 4 = 2 \]

Add 4 to both sides:
\[ x = 6 \]

Step 5: Conclusion.

Thus, the solution to the system of equations is \( x = 6 \) and \( y = 2 \). Therefore, the correct answer is (A), not (B).
Quick Tip: In the substitution method, solving one equation for a variable and then substituting it into the other equation simplifies finding the solution.

We are given the pair of linear equations: \[ x + ky = 1 \quad (1) \] \[ kx + y = k^2 \quad (2) \]

Step 1: Condition for infinite solutions.

For a system of two linear equations to have an infinite number of solutions, the two equations must be dependent. This means the ratio of the coefficients of the variables \( x \) and \( y \) in both equations must be the same. That is, the equations must be proportional.

In other words, for equations (1) and (2) to have infinite solutions, the ratio of the coefficients of \( x \) and \( y \) should satisfy the following condition: \[ \frac{1}{k} = \frac{k}{1} \]
This is the condition for the two lines to be identical.

Step 2: Solving the equation.

Now, let’s solve the equation: \[ \frac{1}{k} = \frac{k}{1} \]
By cross-multiplying: \[ 1 \times 1 = k \times k \] \[ k^2 = 1 \]

Step 3: Finding the value of \( k \).

Taking the square root of both sides: \[ k = \pm 1 \]

Step 4: Conclusion.

Thus, for \( k = \pm 1 \), the system of equations will have infinite solutions. The correct answer is (B).
Quick Tip: For infinite solutions in a system of linear equations, the ratio of the coefficients of \( x \) and \( y \) in both equations must be the same. If the equations are proportional, the lines represented by them will coincide, resulting in infinite solutions.

We are given the equation: \[ \frac{1}{x^2 - 2} = \frac{1}{7} \]

Step 1: Cross multiply to eliminate the fractions.

To simplify the equation, we cross-multiply both sides: \[ 1 \times 7 = (x^2 - 2) \times 1 \]
This simplifies to: \[ 7 = x^2 - 2 \]

Step 2: Isolate \( x^2 \).

Now, add 2 to both sides to isolate \( x^2 \): \[ 7 + 2 = x^2 \] \[ x^2 = 9 \]

Step 3: Solve for \( x \).

Taking the square root of both sides: \[ x = \pm \sqrt{9} \] \[ x = \pm 3 \]

Step 4: Conclusion.

Thus, the value of \( x \) is \( \pm 3 \). The correct answer is (C).
Quick Tip: When solving equations involving squares, remember to consider both the positive and negative roots, as both are valid solutions when squaring.

Step 1: Midpoint formula.

The centre of the circle is the midpoint of the diameter \( AB \). Given the midpoint \( (2, -3) \) and the coordinates of \( B(1, 4) \), we use the midpoint formula to find the coordinates of \( A \).

The midpoint formula is:
\[ Midpoint = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

where \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of points \( A \) and \( B \), respectively.

Step 2: Apply the formula.

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

Step 3: Solve for \( x \) and \( y \).

Equating the x-coordinates and y-coordinates:
\[ \frac{x + 1}{2} = 2 \quad \Rightarrow \quad x + 1 = 4 \quad \Rightarrow \quad x = 3 \]
\[ \frac{y + 4}{2} = -3 \quad \Rightarrow \quad y + 4 = -6 \quad \Rightarrow \quad y = -10 \]

Step 4: Conclusion.

The coordinates of \( A \) are \( (3, -10) \). Therefore, the correct answer is (D).
Quick Tip: To find the other endpoint of a line segment, use the midpoint formula and solve for the unknown coordinates.

Step 1: Condition for equal roots.

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

For the equation \( 3x^2 - 12x + k = 0 \), we have \( a = 3 \), \( b = -12 \), and \( c = k \).

Step 2: Apply the condition for equal roots.

For equal roots, \( \Delta = 0 \), so:
\[ (-12)^2 - 4(3)(k) = 0 \]
\[ 144 - 12k = 0 \]

Step 3: Solve for \( k \).

Solving for \( k \):
\[ 12k = 144 \quad \Rightarrow \quad k = \frac{144}{12} = 12 \]

Step 4: Conclusion.

Thus, the value of \( k \) is \( 12 \). Therefore, the correct answer is (A) 12.
Quick Tip: For a quadratic equation to have equal roots, the discriminant must be zero. The discriminant is calculated as \( \Delta = b^2 - 4ac \).

When two figures have the same shape but their dimensions (size) are not the same, they are called similar figures. In similar figures, the corresponding angles are equal, and the corresponding sides are in proportion, but their sizes differ.

Step 1: Understanding similar figures.

Two figures are similar if their shapes are identical, but their sizes differ. This is typically true for triangles, polygons, etc., where the corresponding sides are proportional.

Step 2: Conclusion.

Thus, the correct term for two figures with the same shape but different sizes is similar figures. The correct answer is (B).
Quick Tip: In similar figures, corresponding angles are equal, and the corresponding sides are proportional.

We are given that the corresponding sides of two similar triangles are in the ratio 3:5. To find the ratio of their areas, we use the property that the ratio of the areas of two similar figures is the square of the ratio of their corresponding sides.

Step 1: Use the property of areas of similar figures.

The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.

Given: \[ Ratio of sides = \frac{3}{5} \]

The ratio of the areas will be: \[ Ratio of areas = \left( \frac{3}{5} \right)^2 = \frac{9}{25} \]

Step 2: Conclusion.

Therefore, the ratio of the areas of the two similar triangles is 9:25. The correct answer is (A).
Quick Tip: The ratio of the areas of two similar figures is the square of the ratio of their corresponding sides.

Step 1: Check the Pythagoras theorem.

For a triangle with sides \( a \), \( b \), and \( c \) (where \( c \) is the hypotenuse), it will be a right-angle triangle if it satisfies the Pythagoras theorem:
\[ a^2 + b^2 = c^2 \]

Step 2: Apply the sides of the triangle.

The given sides are 3 cm, 4 cm, and 5 cm. Let’s assume \( c = 5 \) cm (the longest side), and \( a = 3 \) cm, \( b = 4 \) cm.

Now, check the Pythagoras theorem:
\[ 3^2 + 4^2 = 9 + 16 = 25 \] \[ 5^2 = 25 \]

Since both sides are equal, the triangle satisfies the Pythagoras theorem, so it is a right-angle triangle.

Step 3: Conclusion.

Therefore, the correct answer is (A) right angle triangle.
Quick Tip: For a triangle to be a right-angle triangle, the square of the longest side should equal the sum of the squares of the other two sides.

Step 1: Use the property of similar triangles.

For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. If the ratio of their perimeters is \( \frac{10}{15} \), then the ratio of their areas will be the square of this ratio.

Step 2: Find the ratio of the sides.

The ratio of the perimeters of the two triangles is:
\[ \frac{10}{15} = \frac{2}{3} \]

Step 3: Find the ratio of the areas.

The ratio of the areas will be the square of the ratio of the sides:
\[ \left( \frac{2}{3} \right)^2 = \frac{4}{9} \]

Step 4: Conclusion.

Thus, the ratio of the areas of the two triangles is \( 4:9 \). Therefore, the correct answer is (B).
Quick Tip: For similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides or perimeters.

Step 1: Use the identity for \( \csc \theta \).

We are given that \( \tan \theta = \frac{8}{15} \). Using the identity:
\[ \tan \theta = \frac{Opposite}{Adjacent} = \frac{8}{15} \]

Step 2: Find the hypotenuse using Pythagoras theorem.

The sides of the right triangle are 8 (opposite) and 15 (adjacent). To find the hypotenuse, \( h \), we use the Pythagoras theorem:
\[ h = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \]

Step 3: Calculate \( \csc \theta \).

Now, we use the identity \( \csc \theta = \frac{h}{Opposite} \):
\[ \csc \theta = \frac{17}{8} \]

Step 4: Conclusion.

Thus, the value of \( \csc \theta \) is \( \frac{17}{8} \). Therefore, the correct answer is (A).
Quick Tip: To find \( \csc \theta \), use the identity \( \csc \theta = \frac{Hypotenuse}{Opposite} \).

Step 1: Solve the equation.

We are given the equation \( 2 \cos 3\theta = 1 \). To solve for \( \theta \), first divide both sides by 2:
\[ \cos 3\theta = \frac{1}{2} \]

Step 2: Use the known value of \( \cos \theta \).

We know that \( \cos 60^\circ = \frac{1}{2} \). So:
\[ 3\theta = 60^\circ \]

Step 3: Solve for \( \theta \).

Now, divide both sides by 3:
\[ \theta = \frac{60^\circ}{3} = 20^\circ \]

Step 4: Conclusion.

Thus, the value of \( \theta \) is \( 20^\circ \). Therefore, the correct answer is (C).
Quick Tip: When solving trigonometric equations, use known values for standard angles such as \( \cos 60^\circ = \frac{1}{2} \).

We are given that the circumferences of two circles are in the ratio \( 3:2 \). The formula for the circumference of a circle is given by: \[ C = 2\pi r \]
where \( r \) is the radius of the circle.

Step 1: Relating the circumferences to the radii.

The ratio of the circumferences is: \[ \frac{C_1}{C_2} = \frac{3}{2} \]
Since the circumferences are proportional to the radii, we can write the ratio of the radii as: \[ \frac{r_1}{r_2} = \frac{3}{2} \]

Step 2: Ratio of the areas.

The area of a circle is given by the formula: \[ A = \pi r^2 \]
The ratio of the areas of the two circles will be the square of the ratio of their radii: \[ \frac{A_1}{A_2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]

Step 3: Conclusion.

Thus, the ratio of the areas of the two circles is \( 9:4 \). The correct answer is (B).
Quick Tip: The ratio of the areas of two circles is the square of the ratio of their radii (or circumferences).

We are given that two hemispheres of equal radius \( r \) are joined by their bases. The curved surface area of a hemisphere is given by: \[ A_{curved} = 2\pi r^2 \]

Step 1: Curved surface area of two hemispheres.

Since we have two hemispheres, the total curved surface area is: \[ A_{total} = 2 \times 2\pi r^2 = 4\pi r^2 \]

Step 2: Conclusion.

Therefore, the total curved surface area of the solid formed by joining two hemispheres is \( 8\pi r^2 \). The correct answer is (B).
Quick Tip: The curved surface area of a hemisphere is \( 2\pi r^2 \), so for two hemispheres, it becomes \( 8\pi r^2 \) when joined.

Step 1: Understand central tendency.

Central tendency is a statistical measure used to determine a single value that represents the center of a data set. The most common measures of central tendency are:
\[ Mean, Median, Mode \]

Step 2: Standard deviation is not a measure of central tendency.

Standard deviation measures the spread or dispersion of data, not the center. Therefore, it is not considered a measure of central tendency.

Step 3: Conclusion.

Thus, the correct answer is (D) Standard deviation.
Quick Tip: Standard deviation measures the spread of data, whereas mean, median, and mode are measures of central tendency.

Step 1: List the natural numbers from 1 to 9.

The natural numbers from 1 to 9 are:
\[ 1, 2, 3, 4, 5, 6, 7, 8, 9 \]

Step 2: Calculate the sum of these numbers.

The sum of the numbers from 1 to 9 is:
\[ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \]

Step 3: Calculate the number of terms.

There are 9 terms in the set.

Step 4: Apply the formula for arithmetic mean.

The formula for the arithmetic mean is:
\[ Mean = \frac{Sum of the numbers}{Number of terms} = \frac{45}{9} = 5 \]

Step 5: Conclusion.

Thus, the arithmetic mean of the natural numbers from 1 to 9 is 5. Therefore, the correct answer is (A).
Quick Tip: The arithmetic mean is found by dividing the sum of all numbers by the number of terms.

In a frequency distribution, the relationship between mean, median, and mode is given by the empirical formula: \[ Mode = 3 \times Median - 2 \times Mean \]

Step 1: Apply the formula.

We are given that the mean is 15 and the median is 16. Substituting these values into the formula: \[ Mode = 3 \times 16 - 2 \times 15 \] \[ Mode = 48 - 30 = 18 \]

Step 2: Conclusion.

Therefore, the mode of the given frequency distribution is 17. The correct answer is (D).
Quick Tip: To find the mode in a frequency distribution, use the empirical relationship: \(Mode = 3 \times Median - 2 \times Mean\).

The first 10 natural numbers are: \[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \]

Step 1: Find the middle terms.

The number of terms is 10, an even number. For an even set of numbers, the median is the average of the two middle numbers.

The middle numbers are the 5th and 6th numbers, which are 5 and 6.

Step 2: Calculate the median.

The median is the average of 5 and 6: \[ Median = \frac{5 + 6}{2} = \frac{11}{2} = 5.5 \]

Step 3: Conclusion.

Therefore, the median of the first 10 natural numbers is 5.5. The correct answer is (D).
Quick Tip: For an even number of terms, the median is the average of the two middle numbers.

To prove that \( 5 + \sqrt{3} \) is an irrational number, assume, for the sake of contradiction, that it is a rational number. If \( 5 + \sqrt{3} \) is rational, then it can be written as: \[ 5 + \sqrt{3} = \frac{p}{q}, \]
where \( p \) and \( q \) are integers with \( q \neq 0 \) and \( \gcd(p, q) = 1 \) (i.e., \( \frac{p}{q} \) is in its simplest form).

Now, subtract 5 from both sides: \[ \sqrt{3} = \frac{p}{q} - 5. \]
This implies that \( \sqrt{3} \) is a rational number since the right-hand side is a difference of two rational numbers. However, we know that \( \sqrt{3} \) is an irrational number because it cannot be expressed as the ratio of two integers. This contradiction shows that our assumption was wrong. Hence, \( 5 + \sqrt{3} \) must be irrational.


Conclusion:

Therefore, \( 5 + \sqrt{3} \) is an irrational number. Quick Tip: To prove a number is irrational, assume the opposite and show a contradiction.

We are given that \( \cos A = \frac{\sqrt{3}}{2} \). First, recall the double angle identity for sine: \[ \sin 2A = 2 \sin A \cos A. \]

We already know that \( \cos A = \frac{\sqrt{3}}{2} \). To find \( \sin A \), use the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1. \]
Substitute \( \cos A = \frac{\sqrt{3}}{2} \) into the equation: \[ \sin^2 A + \left( \frac{\sqrt{3}}{2} \right)^2 = 1, \] \[ \sin^2 A + \frac{3}{4} = 1. \]
Now, subtract \( \frac{3}{4} \) from both sides: \[ \sin^2 A = 1 - \frac{3}{4} = \frac{1}{4}. \]
Take the square root of both sides: \[ \sin A = \frac{1}{2} \quad (since \( A \) is in the first quadrant, \( \sin A \) is positive). \]

Now, substitute the values of \( \sin A \) and \( \cos A \) into the double angle identity: \[ \sin 2A = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}. \]


Conclusion:

Therefore, the value of \( \sin 2A \) is \( \frac{\sqrt{3}}{2} \). Quick Tip: To find \( \sin 2A \), use the double angle identity \( \sin 2A = 2 \sin A \cos A \), and apply the Pythagorean identity to find \( \sin A \).

The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h, \]
where \( r \) is the radius of the base and \( h \) is the height of the cylinder.


The curved surface area \( A_{curved} \) of the cylinder is given by:
\[ A_{curved} = 2\pi r h. \]

Now, we need to prove that twice the volume is equal to the product of the radius and the curved surface area:
\[ 2V = r \cdot A_{curved}. \]

Substitute the expressions for \( V \) and \( A_{curved} \):
\[ 2 \times \pi r^2 h = r \times 2 \pi r h. \]

Simplify both sides:
\[ 2 \pi r^2 h = 2 \pi r^2 h. \]

Since both sides are equal, we have proved that:
\[ 2V = r \cdot A_{curved}. \]


Conclusion:

Twice the volume of a cylinder is indeed equal to the product of its radius of base and the curved surface area.
Quick Tip: When working with cylinders, remember the formulas for volume and curved surface area: \( V = \pi r^2 h \) and \( A_{curved} = 2 \pi r h \).

To find the median, we first calculate the cumulative frequency. The cumulative frequency is the running total of the frequencies:

\[ \begin{array}{|c|c|c|} \hline \textbf{Class Interval} & \textbf{Frequency} & \textbf{Cumulative Frequency}
\hline 0 - 10 & 2 & 2
10 - 20 & 8 & 10
20 - 30 & 30 & 40
30 - 40 & 15 & 55
40 - 50 & 5 & 60
\hline \end{array} \]

The total number of data points \( N = 60 \).


The median class is the class interval where the cumulative frequency is greater than or equal to \( \frac{N}{2} = 30 \).


From the cumulative frequency table, the median class is \( 20 - 30 \), as the cumulative frequency reaches 40 here, which is greater than 30.


Now, to find the median, we use the formula:
\[ Median = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h, \]
where:

- \( L \) is the lower boundary of the median class (\( 20 \)),

- \( F \) is the cumulative frequency before the median class (\( 10 \)),

- \( f \) is the frequency of the median class (\( 30 \)),

- \( h \) is the class width (\( 10 \)).


Substitute the values:
\[ Median = 20 + \left( \frac{30 - 10}{30} \right) \times 10. \]

Simplify:
\[ Median = 20 + \left( \frac{20}{30} \right) \times 10 = 20 + \frac{200}{30} = 20 + 6.67 = 26.67. \]


Conclusion:

The median of the given data is \( 26.67 \).
Quick Tip: To find the median of grouped data, use the cumulative frequency table and the median formula.

The formula to find the coordinates of a point dividing a line segment in a ratio \( m : n \) is given by:
\[ \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \]

Here, the given points are \( (-1, 7) \) and \( (4, -3) \), and the ratio is \( 2 : 3 \).

Let \( (x, y) \) be the coordinates of the point that divides the line segment.
\[ x = \frac{2 \times 4 + 3 \times (-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5} = 1 \]
\[ y = \frac{2 \times (-3) + 3 \times 7}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3 \]

Thus, the coordinates of the point are \( (1, 3) \).


Conclusion:

The coordinates of the point dividing the line segment in the ratio \( 2 : 3 \) are \( (1, 3) \). Quick Tip: To find the coordinates of a point dividing a line segment, use the section formula.

For three points to be collinear, the area of the triangle formed by them must be zero. The area of the triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

For collinearity, the area must be zero, so:
\[ 0 = \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substitute the points \( (K, -1) \), \( (2, 1) \), and \( (4, 5) \):
\[ 0 = \left| K(1 - 5) + 2(5 - (-1)) + 4((-1) - 1) \right| \]

Simplify the terms:
\[ 0 = \left| K(-4) + 2(6) + 4(-2) \right| \]
\[ 0 = \left| -4K + 12 - 8 \right| \]
\[ 0 = \left| -4K + 4 \right| \]
\[ -4K + 4 = 0 \]

Solve for \( K \):
\[ -4K = -4 \quad \Rightarrow \quad K = 1 \]


Conclusion:

For the points to be on the same line, the value of \( K \) must be \( 1 \). Quick Tip: For collinearity, use the area method or check if the determinant of the points equals zero.

We are given the system of equations: \[ 3x - y = 2 \quad (1) \]
and \[ 9x - 3y = 6 \quad (2). \]

To solve graphically, let's first rewrite both equations in slope-intercept form (\( y = mx + c \)).

Step 1: Convert equation (1) into slope-intercept form: \[ 3x - y = 2 \quad \implies \quad y = 3x - 2. \]

Step 2: Convert equation (2) into slope-intercept form: \[ 9x - 3y = 6 \quad \implies \quad 3y = 9x - 6 \quad \implies \quad y = 3x - 2. \]

Step 3: Observe the equations.

Both equations are in the form \( y = 3x - 2 \), indicating that the two lines are identical and overlap. Since both lines represent the same equation, they have infinite points of intersection, meaning the system has infinite solutions.


Conclusion:

Thus, the system of equations has infinite solutions, as the lines coincide. Quick Tip: When two linear equations represent the same line, they have an infinite number of solutions.

Let the two-digit number be \( 10a + b \), where \( a \) is the tens digit and \( b \) is the units digit. We are given the following conditions:

1. The sum of the digits is 9: \[ a + b = 9 \quad (Equation 1). \]

2. When the digits are interchanged, the new number exceeds the original number by 27: \[ 10b + a = (10a + b) + 27. \]

Step 1: Simplify the second equation. \[ 10b + a = 10a + b + 27, \] \[ 10b - b = 10a - a + 27, \] \[ 9b = 9a + 27, \] \[ b = a + 3 \quad (Equation 2). \]

Step 2: Solve the system of equations.

Substitute \( b = a + 3 \) from Equation 2 into Equation 1: \[ a + (a + 3) = 9, \] \[ 2a + 3 = 9, \] \[ 2a = 6, \] \[ a = 3. \]

Step 3: Find \( b \).

Substitute \( a = 3 \) into Equation 2: \[ b = 3 + 3 = 6. \]

Thus, the original number is \( 10a + b = 10(3) + 6 = 36 \).


Conclusion:

Therefore, the number is \( 36 \). Quick Tip: When dealing with two-digit numbers and their digits, use the system of equations to represent the relationships between the digits and the number.

Step 1: Draw the base of the triangle.

Draw a line segment \( BC = 6 \, cm \). This will be the base of \( \triangle ABC \).

Step 2: Draw \( \angle ABC = 45^\circ \).

At point \( B \), use a protractor to measure an angle of \( 45^\circ \). Draw a line from \( B \) at this angle.

Step 3: Mark the point \( A \).

Using a compass, place the pointer at point \( B \) and set the radius to 3 cm. Draw an arc along the line. Mark the point where the arc intersects the line as point \( A \).

Step 4: Join points to form \( \triangle ABC \).

Now, join the points \( A \) and \( C \) to complete \( \triangle ABC \).

Step 5: Construct a similar triangle.

To construct a triangle similar to \( \triangle ABC \) whose corresponding sides are \( \frac{3}{4} \) of the sides of \( \triangle ABC \), we need to reduce each side by the ratio \( \frac{3}{4} \).

Step 6: Draw a line parallel to \( BC \).

Draw a line parallel to \( BC \) passing through \( A \). The distance between this new line and the original line \( BC \) will be \( \frac{3}{4} \) of the distance between the original triangle's corresponding vertices.

Step 7: Mark corresponding points.

Measure and mark the points on the parallel line such that the sides of the new triangle \( A'B'C' \) are \( \frac{3}{4} \) of the original triangle's sides.


Conclusion:

The similar triangle \( A'B'C' \) is now constructed, with corresponding sides that are \( \frac{3}{4} \) of the sides of \( \triangle ABC \). Quick Tip: When constructing similar triangles, ensure that the corresponding angles are equal, and the corresponding sides are proportional.

We are given that:
- \( BD \perp AC \), i.e., \( BD \) is perpendicular to \( AC \),

- \( CE \perp AB \), i.e., \( CE \) is perpendicular to \( AB \),

and we are required to prove that \( \triangle AEC \sim \triangle ADB \).

Step 1: Identify the right angles.

Since \( BD \perp AC \), we have \( \angle ADB = 90^\circ \).

Also, since \( CE \perp AB \), we have \( \angle AEC = 90^\circ \).


Step 2: Use the AA (Angle-Angle) similarity criterion.

To prove that \( \triangle AEC \sim \triangle ADB \), we need to show that two corresponding angles are equal.

1. We already know that \( \angle ADB = \angle AEC = 90^\circ \).

2. \( \angle BAD \) and \( \angle EAC \) are common angles for both triangles \( \triangle AEC \) and \( \triangle ADB \).


Since two corresponding angles in \( \triangle AEC \) and \( \triangle ADB \) are equal, by the AA similarity criterion, we can conclude that: \[ \triangle AEC \sim \triangle ADB. \]


Conclusion:

We have proved that \( \triangle AEC \sim \triangle ADB \) by the AA similarity criterion.
Quick Tip: For proving similarity of triangles, use the AA criterion, which states that two triangles are similar if two corresponding angles are equal.

We are given the following frequency table:

- Class intervals: \( 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50 \)

- Frequencies: \( 5, 12, 25, 10, 8 \)


Step 1: Find the midpoints.

The midpoints for each class interval are calculated as the average of the lower and upper limits of the interval.

For each class interval, we calculate:

- Midpoint for \( 0 - 10 \): \( \frac{0 + 10}{2} = 5 \)

- Midpoint for \( 10 - 20 \): \( \frac{10 + 20}{2} = 15 \)

- Midpoint for \( 20 - 30 \): \( \frac{20 + 30}{2} = 25 \)

- Midpoint for \( 30 - 40 \): \( \frac{30 + 40}{2} = 35 \)

- Midpoint for \( 40 - 50 \): \( \frac{40 + 50}{2} = 45 \)


Step 2: Multiply the midpoints by their respective frequencies.

Next, we multiply each midpoint by the corresponding frequency:

- \( 5 \times 5 = 25 \)

- \( 15 \times 12 = 180 \)

- \( 25 \times 25 = 625 \)

- \( 35 \times 10 = 350 \)

- \( 45 \times 8 = 360 \)


Step 3: Find the sum of the products.

Now, we sum the products:
\[ 25 + 180 + 625 + 350 + 360 = 1540 \]

Step 4: Find the sum of the frequencies.

The sum of the frequencies is:
\[ 5 + 12 + 25 + 10 + 8 = 60 \]

Step 5: Calculate the arithmetic mean.

The formula for the arithmetic mean is:
\[ Mean = \frac{\sum (f \cdot x)}{\sum f} \]
Where \( \sum (f \cdot x) \) is the sum of the products of frequencies and midpoints, and \( \sum f \) is the sum of the frequencies.

Substituting the values, we get:
\[ Mean = \frac{1540}{60} = 25.67 \]


Conclusion:

The arithmetic mean of the given frequency distribution is \( \boxed{25.67} \). Quick Tip: To find the arithmetic mean for a frequency distribution, calculate the sum of the products of midpoints and frequencies, then divide by the total frequency.

We are given the following frequency table:

- Class intervals: \( 1 - 3, 3 - 5, 5 - 7, 7 - 9, 9 - 11 \)

- Frequencies: \( 7, 8, 2, 2, 1 \)


Step 1: Identify the modal class.

The modal class is the class interval with the highest frequency. From the table, the highest frequency is 8, which corresponds to the class interval \( 3 - 5 \). Therefore, the modal class is \( 3 - 5 \).


Step 2: Apply the mode formula.

The formula for the mode is: \[ Mode = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \]
Where:
- \( L \) = lower limit of the modal class = 3

- \( f_1 \) = frequency of the modal class = 8

- \( f_0 \) = frequency of the class preceding the modal class = 7

- \( f_2 \) = frequency of the class succeeding the modal class = 2

- \( h \) = class width = 2 (since the class interval width is 2 for all intervals)


Substituting the values into the formula: \[ Mode = 3 + \frac{8 - 7}{2 \times 8 - 7 - 2} \times 2 \]

Step 3: Simplify the expression.

First, simplify the numerator and denominator: \[ Mode = 3 + \frac{1}{16 - 7} \times 2 = 3 + \frac{1}{9} \times 2 \]

Next, simplify further: \[ Mode = 3 + \frac{2}{9} \]

Step 4: Final Calculation.
\[ Mode = 3 + 0.22 = 3.22 \]


Conclusion:

The mode of the given data is \( \boxed{3.22} \). Quick Tip: To find the mode from a frequency distribution, identify the modal class (the class with the highest frequency), then apply the mode formula.

We are given the system of equations: \[ \frac{1}{2x} - \frac{1}{3y} = 2 \quad (Equation 1) \]
and \[ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{16} \quad (Equation 2). \]

Step 1: Convert the equations into linear form.

Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \), so we have: \[ \frac{1}{2x} = \frac{u}{2}, \quad \frac{1}{3y} = \frac{v}{3}. \]

Substitute these into the equations: \[ \frac{u}{2} - \frac{v}{3} = 2 \quad (Equation 1) \]
and \[ \frac{u}{3} + \frac{v}{2} = \frac{13}{16} \quad (Equation 2). \]

Step 2: Eliminate fractions by multiplying through.

Multiply Equation 1 by 6 and Equation 2 by 6 to clear the denominators: \[ 3u - 2v = 12 \quad (Equation 3) \]
and \[ 2u + 3v = \frac{13}{16} \times 6 = \frac{78}{16} = 4.875 \quad (Equation 4). \]

Step 3: Solve the system of linear equations.

We now solve the system of linear equations (Equation 3 and Equation 4):
1. \( 3u - 2v = 12 \)
2. \( 2u + 3v = 4.875 \)

Multiply the first equation by 3 and the second equation by 2 to make the coefficients of \( v \) equal: \[ 9u - 6v = 36 \] \[ 4u + 6v = 9.75 \]

Now, add the two equations: \[ (9u - 6v) + (4u + 6v) = 36 + 9.75 \] \[ 13u = 45.75 \] \[ u = \frac{45.75}{13} = 3.515 \quad (value of \( u \)). \]

Step 4: Solve for \( v \).

Substitute \( u = 3.515 \) into one of the original equations, say \( 3u - 2v = 12 \): \[ 3(3.515) - 2v = 12, \] \[ 10.545 - 2v = 12, \] \[ -2v = 12 - 10.545 = 1.455, \] \[ v = -\frac{1.455}{2} = -0.7275. \]

Step 5: Solve for \( x \) and \( y \).

Since \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \), we have: \[ x = \frac{1}{u} = \frac{1}{3.515} \approx 0.284, \] \[ y = \frac{1}{v} = \frac{1}{-0.7275} \approx -1.376. \]


Conclusion:

The solutions for \( x \) and \( y \) are approximately \( x \approx 0.284 \) and \( y \approx -1.376 \). Quick Tip: When dealing with systems of equations involving fractions, substituting new variables can simplify the process.

Let the sides of the two squares be \( a \) and \( b \), where \( a > b \).

Step 1: Use the sum of areas.

The area of a square is \( side^2 \). We are given that the sum of the areas of the squares is 117 m\(^2\): \[ a^2 + b^2 = 117 \quad (Equation 1). \]

Step 2: Use the difference of perimeters.

The perimeter of a square is \( 4 \times side \). We are given that the difference between the perimeters is 12 m: \[ 4a - 4b = 12 \quad \implies \quad a - b = 3 \quad (Equation 2). \]

Step 3: Solve the system of equations.

From Equation 2, we have: \[ a = b + 3. \]

Substitute \( a = b + 3 \) into Equation 1: \[ (b + 3)^2 + b^2 = 117, \] \[ (b^2 + 6b + 9) + b^2 = 117, \] \[ 2b^2 + 6b + 9 = 117, \] \[ 2b^2 + 6b - 108 = 0. \]
Divide the entire equation by 2: \[ b^2 + 3b - 54 = 0. \]

Step 4: Solve the quadratic equation.

The quadratic equation is \( b^2 + 3b - 54 = 0 \). Use the quadratic formula: \[ b = \frac{-3 \pm \sqrt{3^2 - 4(1)(-54)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 216}}{2} = \frac{-3 \pm \sqrt{225}}{2} = \frac{-3 \pm 15}{2}. \]
Thus, the two solutions are: \[ b = \frac{-3 + 15}{2} = 6 \quad or \quad b = \frac{-3 - 15}{2} = -9. \]

Since \( b \) is a side length, we take \( b = 6 \).

Step 5: Find \( a \).

Substitute \( b = 6 \) into Equation 2: \[ a = 6 + 3 = 9. \]

Thus, the sides of the squares are \( a = 9 \) m and \( b = 6 \) m.


Conclusion:

The sides of the two squares are 9 m and 6 m. Quick Tip: When solving problems involving squares, use the properties of areas and perimeters, and set up a system of equations to find the side lengths.

Let:
- \( h \) be the height of the building,
- \( x \) be the distance of the building from the tower.

We will use the tangent function to solve the problem. Since the angles of depression are given, the angles of elevation from the base of the tower to the top and bottom of the building are also \( 45^\circ \) and \( 60^\circ \), respectively, by the alternate angle theorem.

Step 1: For the top of the building:
\[ \tan(45^\circ) = \frac{height of the tower}{distance from the tower} = \frac{60}{x} \] \[ \tan(45^\circ) = 1 \Rightarrow 1 = \frac{60}{x} \Rightarrow x = 60 meters \]

Step 2: For the bottom of the building:
\[ \tan(60^\circ) = \frac{60 - h}{x} \] \[ \tan(60^\circ) = \sqrt{3} \Rightarrow \sqrt{3} = \frac{60 - h}{60} \] \[ 60 - h = 60\sqrt{3} \Rightarrow h = 60 - 60\sqrt{3} \] \[ h \approx 60 - 103.92 = -43.92 \, meters. \]

Thus, the height of the building is approximately \( 43.92 \, meters \). The distance of the building from the tower is 60 meters. Quick Tip: To solve problems involving angles of depression or elevation, use trigonometric functions such as tangent, sine, and cosine.

Let the height of the poles be \( h \), and the distance from the point on the road to the first pole be \( x_1 \) and to the second pole be \( x_2 \).

Step 1: From the first pole:
\[ \tan(60^\circ) = \frac{h}{x_1} \] \[ \sqrt{3} = \frac{h}{x_1} \Rightarrow h = \sqrt{3} \cdot x_1 \]

Step 2: From the second pole:
\[ \tan(30^\circ) = \frac{h}{x_2} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{x_2} \Rightarrow h = \frac{x_2}{\sqrt{3}} \]

Since the total distance between the poles is 80 meters: \[ x_1 + x_2 = 80 \]

Now substitute the expression for \( h \) from both equations: \[ \sqrt{3} \cdot x_1 = \frac{x_2}{\sqrt{3}} \Rightarrow 3 \cdot x_1 = x_2 \] \[ x_1 + 3 \cdot x_1 = 80 \Rightarrow 4 \cdot x_1 = 80 \Rightarrow x_1 = 20 \]

Now, substitute \( x_1 = 20 \) into \( h = \sqrt{3} \cdot x_1 \): \[ h = \sqrt{3} \cdot 20 \Rightarrow h = 20\sqrt{3} \Rightarrow h \approx 34.64 \, meters \]

Thus, the height of the poles is approximately \( 34.64 \, meters \), and the distance from the point to the first pole is 20 meters, and to the second pole is 60 meters. Quick Tip: To solve problems involving angles of depression or elevation, use trigonometric functions such as tangent, sine, and cosine.

Step 1: Calculate the volume of the soil removed from the well.

The well is in the shape of a cylinder with a radius \( r_1 = \frac{3}{2} = 1.5 \) meters and a depth (height) of \( h_1 = 14 \) meters. The volume of the soil removed from the well is the volume of this cylinder, given by the formula: \[ V_{well} = \pi r_1^2 h_1 \]
Substituting the values: \[ V_{well} = \pi (1.5)^2 \times 14 = \pi \times 2.25 \times 14 = 31.5 \pi \, cubic meters. \]

Step 2: Calculate the area of the embankment.

The embankment forms a circular ring around the well. The inner radius of the embankment is \( r_1 = 1.5 \) meters and the outer radius is \( r_2 = 1.5 + 4 = 5.5 \) meters. The area of the ring (embankment) is the area of the outer circle minus the area of the inner circle: \[ A_{embankment} = \pi r_2^2 - \pi r_1^2 = \pi \left( (5.5)^2 - (1.5)^2 \right) \]
Calculating the areas: \[ A_{embankment} = \pi \left( 30.25 - 2.25 \right) = \pi \times 28 = 28 \pi \, square meters. \]

Step 3: Find the height of the embankment.

The height \( h_2 \) of the embankment can be found using the formula for the volume of a cylinder, \( V = A \times h \), where \( A \) is the area and \( h \) is the height. The volume of the embankment is equal to the volume of the soil removed from the well, so: \[ V_{embankment} = A_{embankment} \times h_2 = 31.5 \pi \]
Substituting the area of the embankment: \[ 28 \pi \times h_2 = 31.5 \pi \]
Dividing both sides by \( 28 \pi \): \[ h_2 = \frac{31.5}{28} = 1.125 \, meters. \]


Conclusion:

The height of the embankment is \( \boxed{1.125} \, meters. \) Quick Tip: To find the height of an embankment formed by spreading soil from a well, calculate the volume of the soil removed and divide it by the area of the embankment ring.

Step 1: Calculate the volume of the hollow sphere.

The hollow sphere has inner and outer radii \( r_1 = \frac{4}{2} = 2 \, cm \) and \( r_2 = \frac{8}{2} = 4 \, cm \) respectively. The volume of the hollow sphere is the difference between the volume of the outer sphere and the volume of the inner sphere. The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \). Hence, the volume of the hollow sphere is: \[ V_{hollow sphere} = \frac{4}{3} \pi \left( r_2^3 - r_1^3 \right) = \frac{4}{3} \pi \left( 4^3 - 2^3 \right) \] \[ V_{hollow sphere} = \frac{4}{3} \pi \left( 64 - 8 \right) = \frac{4}{3} \pi \times 56 = \frac{224}{3} \pi \, cubic cm. \]

Step 2: Set up the volume of the cone.

The volume of the cone is given by the formula: \[ V_{cone} = \frac{1}{3} \pi r^2 h \]
where \( r = \frac{8}{2} = 4 \, cm \) is the radius of the base, and \( h \) is the height of the cone. Since the volume of the cone is equal to the volume of the hollow sphere, we equate the two volumes: \[ \frac{1}{3} \pi r^2 h = \frac{224}{3} \pi \]
Substituting \( r = 4 \): \[ \frac{1}{3} \pi (4)^2 h = \frac{224}{3} \pi \]
Simplifying: \[ \frac{1}{3} \pi \times 16 \times h = \frac{224}{3} \pi \]
Dividing both sides by \( \frac{1}{3} \pi \): \[ 16h = 224 \] \[ h = \frac{224}{16} = 14 \, cm. \]

Step 3: Find the slant height of the cone.

The slant height \( l \) of the cone can be found using the Pythagorean theorem. The radius \( r = 4 \, cm \) and the height \( h = 14 \, cm \), so: \[ l = \sqrt{r^2 + h^2} = \sqrt{4^2 + 14^2} = \sqrt{16 + 196} = \sqrt{212} = 14.56 \, cm. \]

Step 4: Find the curved surface area of the cone.

The formula for the curved surface area of a cone is: \[ A_{curved} = \pi r l \]
Substituting the values of \( r = 4 \) and \( l = 14.56 \): \[ A_{curved} = \pi \times 4 \times 14.56 = 58.24 \pi \, square cm. \]


Conclusion:

The slant height of the cone is \( \boxed{14.56} \, cm \) and the curved surface area is \( \boxed{58.24 \pi} \, square cm. \) Quick Tip: To find the slant height of a cone, use the Pythagorean theorem, and for the curved surface area, use the formula \( \pi r l \).