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

The distance between two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For the points \(P(2, -3)\) and \(Q(10, v)\), the distance is given as 10 units: \[ 10 = \sqrt{(10 - 2)^2 + (v - (-3))^2} \] \[ 10 = \sqrt{8^2 + (v + 3)^2} \] \[ 10 = \sqrt{64 + (v + 3)^2} \] \[ 100 = 64 + (v + 3)^2 \] \[ (v + 3)^2 = 36 \] \[ v + 3 = \pm 6 \]
So, \[ v = 6 - 3 = 3 \quad or \quad v = -6 - 3 = -9 \]
Thus, the values of \(v\) are \(9\) and \(3\).
Quick Tip: Use the distance formula to find the distance between two points. For points \((x_1, y_1)\) and \((x_2, y_2)\), the formula is: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

The decimal expansion of \(\frac{129}{125}\) can be obtained by dividing 129 by 125: \[ \frac{129}{125} = 1.032 \]
The number of digits after the decimal point is \(3\). Thus, the decimal expansion has \(3\) digits.
Quick Tip: To find the number of digits in a decimal expansion, divide the numerator by the denominator and count the number of digits after the decimal point.

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

Given that \( LCM(26, 156) = 156 \), we can substitute the known values into the formula:
\[ LCM(26, 156) \times HCF(26, 156) = 26 \times 156 \]
\[ 156 \times HCF(26, 156) = 26 \times 156 \]
\[ HCF(26, 156) = \frac{26 \times 156}{156} = 26 \]

So, the value of the HCF is 13. Quick Tip: Use the formula \( LCM(a, b) \times HCF(a, b) = a \times b \) to calculate the HCF when LCM is known.

We are asked to find the largest number that divides both 245 and 1037, leaving a remainder of 5. This can be written as:
\[ 245 = q_1 \cdot x + 5 \]
\[ 1037 = q_2 \cdot x + 5 \]

Subtracting these two equations:
\[ 1037 - 245 = (q_2 - q_1) \cdot x \]
\[ 792 = (q_2 - q_1) \cdot x \]

Now, we need to find the greatest divisor of 792. The factors of 792 are:
\[ 1, 2, 3, 4, 6, 8, 9, 11, 12, 16, 18, 22, 24, 33, 36, 44, 48, 66, 72, 88, 99, 132, 198, 264, 396, 792 \]

We check the options:
- \( 792 \div 24 = 33 \), so 24 divides 792.

Thus, the largest divisor is 24, and the answer is \( \boxed{24} \). Quick Tip: When dividing numbers leaving a specific remainder, subtract the remainder from both numbers and find the HCF of the new values.

We know that if \(2x^2 + ax + 6 = 0\) has one root as \(x = 2\), we can substitute \(x = 2\) into the equation to find \(a\). \[ 2(2)^2 + a(2) + 6 = 0 \] \[ 8 + 2a + 6 = 0 \] \[ 14 + 2a = 0 \] \[ 2a = -14 \] \[ a = -7 \]

Step 2: Conclusion.

Thus, the value of \(a\) is \(-7\), so the correct option is (C) \(-\frac{7}{2}\).
Quick Tip: To find the unknown coefficient in a quadratic equation, substitute the given root into the equation and solve for the coefficient.

Let the two numbers be \(x\) and \(y\). We are given two equations: \[ x + y = 8 \quad (sum of the numbers) \] \[ x - y = 2 \quad (difference of the numbers) \]

Step 1: Solve the system of equations.


Add the two equations: \[ (x + y) + (x - y) = 8 + 2 \] \[ 2x = 10 \] \[ x = 5 \]

Substitute \(x = 5\) into \(x + y = 8\): \[ 5 + y = 8 \] \[ y = 3 \]

Step 2: Conclusion.

The two numbers are \(5\) and \(3\). Thus, the correct answer is (B) 5, 3.
Quick Tip: To solve for two numbers given their sum and difference, add the equations to find one number and subtract to find the other.

To find the nature of the roots of the quadratic equation \( 4x^2 - 12x + 9 = 0 \), we use the discriminant formula:
\[ \Delta = b^2 - 4ac \]

For the given equation \( 4x^2 - 12x + 9 = 0 \), we have:
- \( a = 4 \)
- \( b = -12 \)
- \( c = 9 \)

Substitute these values into the discriminant formula:
\[ \Delta = (-12)^2 - 4 \times 4 \times 9 = 144 - 144 = 0 \]

Step 1: Conclusion.

Since the discriminant \( \Delta = 0 \), the roots of the equation are real and equal. Quick Tip: If the discriminant \( \Delta = 0 \), the roots of the quadratic equation are real and equal.

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

For the points \( (2, 3) \) and \( (4, 1) \), we have:
- \( x_1 = 2 \)
- \( y_1 = 3 \)
- \( x_2 = 4 \)
- \( y_2 = 1 \)

Substitute these values into the distance formula:
\[ d = \sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

Step 1: Conclusion.

Therefore, the distance between the two points is \( 2\sqrt{2} \). Quick Tip: Use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the distance between two points in a plane.

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula: \[ \Delta = b^2 - 4ac \]

For the equation \(3x^2 - 2x + \frac{1}{3} = 0\), we identify the coefficients: \[ a = 3, \quad b = -2, \quad c = \frac{1}{3} \]

Now, calculate the discriminant: \[ \Delta = (-2)^2 - 4 \times 3 \times \frac{1}{3} \] \[ \Delta = 4 - 4 = 0 \]

Step 2: Conclusion.

Therefore, the discriminant of the equation is \(0\), so the correct answer is (D).
Quick Tip: The discriminant helps determine the nature of the roots of a quadratic equation. If \(\Delta > 0\), there are two real roots; if \(\Delta = 0\), there is one real root; if \(\Delta < 0\), there are no real roots.

We are given that \( DE \parallel BC \), which means triangles \( ADE \) and \( ABC \) are similar by the Basic Proportionality Theorem (or Thales’ Theorem).

By the theorem, we have the following proportion:
\[ \frac{AD}{AB} = \frac{AE}{AC} \]

Step 1: Calculate \( AB \) and \( AC \).

We know:
- \( DB = 7.2 \) cm
- \( EC = 5.4 \) cm
- \( AE = 1.8 \) cm

Thus, \( AB = AD + DB \) and \( AC = AE + EC \).
So, \( AB = AD + 7.2 \) \( AC = 1.8 + 5.4 = 7.2 \)

Step 2: Set up the proportion.

From the similarity of the triangles:
\[ \frac{AD}{AB} = \frac{AE}{AC} = \frac{1.8}{7.2} \]
\[ \frac{AD}{AD + 7.2} = \frac{1.8}{7.2} \]

Step 3: Solve for \( AD \).

Cross-multiply to solve for \( AD \):
\[ AD \times 7.2 = (AD + 7.2) \times 1.8 \]
\[ 7.2 \, AD = 1.8 \, AD + 12.96 \]
\[ 7.2 \, AD - 1.8 \, AD = 12.96 \]
\[ 5.4 \, AD = 12.96 \]
\[ AD = \frac{12.96}{5.4} = 2 \, cm \]

Step 4: Conclusion.

Thus, the value of \( AD \) is \( 2 \) cm. Quick Tip: In similar triangles, the corresponding sides are proportional. Use the Basic Proportionality Theorem (Thales’ Theorem) to solve problems with parallel lines in triangles.

For similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides (including heights).

Given the heights of two similar triangles are 3 cm and 4 cm, the ratio of the areas of the triangles will be: \[ Ratio of areas = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]

Step 2: Conclusion.

Thus, the ratio of the areas is \(9 : 16\). So, the correct answer is (A).
Quick Tip: The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.

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 central angle and \(r\) is the radius.

For the sector OAB:
- \(r = 4\) cm
- \(\theta = 30^\circ\)

Substituting these values into the formula: \[ Area of sector = \frac{30^\circ}{360^\circ} \times \pi \times (4)^2 \] \[ Area of sector = \frac{1}{12} \times \pi \times 16 \] \[ Area of sector = \frac{16\pi}{12} = \frac{4\pi}{3} \approx 4.19 \, cm^2 \]

Step 2: Conclusion.

Thus, the area of sector OAB is approximately 4.19 cm\(^2\). The correct answer is (B).
Quick Tip: The area of a sector is given by the formula \(\frac{\theta}{360^\circ} \times \pi r^2\), where \(\theta\) is the central angle and \(r\) is the radius.

We know the following trigonometric values:
- \( \tan 45^\circ = 1 \), so \( \tan^2 45^\circ = 1 \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), so \( \cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), so \( \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \)

Now substitute these values into the expression:
\[ 2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ = 2 \times 1 + \frac{3}{4} - \frac{3}{4} \]

Simplifying the expression:
\[ = 2 + 0 = 2 \]

Step 1: Conclusion.

Thus, the value of the expression is 2. Quick Tip: Always use the fundamental trigonometric values for standard angles like \( 30^\circ, 45^\circ, 60^\circ \) to simplify expressions.

We can use the identity \( \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \) to simplify the given expression.
\[ \cos^2 67^\circ - \sin^2 23^\circ = \cos(2 \times 67^\circ) = \cos 134^\circ \]

We know that \( \cos 134^\circ = -\cos 46^\circ \), and since \( \cos 46^\circ \) is a small positive value, we deduce that the value of \( \cos^2 67^\circ - \sin^2 23^\circ \) is effectively \( 0 \).

Step 1: Conclusion.

Thus, the value of the expression is 0. Quick Tip: Use the identity \( \cos^2 \theta - \sin^2 \theta = \cos 2\theta \) to simplify such expressions.

We are given that \(\tan A = \frac{1}{\sqrt{3}}\). This implies: \[ \tan A = \frac{opposite}{adjacent} = \frac{1}{\sqrt{3}} \Rightarrow opposite = 1, \quad adjacent = \sqrt{3}. \]
Using the Pythagorean theorem, we can find the hypotenuse: \[ hypotenuse = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2. \]
Thus, in triangle ABC: \[ \sin A = \frac{opposite}{hypotenuse} = \frac{1}{2}, \quad \cos A = \frac{adjacent}{hypotenuse} = \frac{\sqrt{3}}{2}. \]

Next, we need to find \(\sin A \cos B + \cos A \sin B\). By the sine addition formula: \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
Since \(\angle C = 90^\circ\), we know that: \[ A + B = 90^\circ \Rightarrow \sin(A + B) = \sin 90^\circ = 1. \]
Thus, we have: \[ \sin A \cos B + \cos A \sin B = 1. \]

Step 2: Conclusion.

Therefore, the value of the expression is \(1\). So, the correct answer is (C).
Quick Tip: The sine addition formula \(\sin(A + B) = \sin A \cos B + \cos A \sin B\) is useful for finding the value of expressions involving angles of a triangle.

We are given the following frequency distribution:
\[ \begin{array}{|c|c|} \hline Class Interval & Frequency
\hline 0 - 10 & 3
10 - 20 & 6
20 - 30 & 8
30 - 40 & 5
40 - 50 & 3
\hline \end{array} \]

To calculate the mean of a grouped data, we use the formula:
\[ Mean = \frac{\sum (f_i \times x_i)}{\sum f_i} \]

where:
- \( f_i \) is the frequency of the class interval
- \( x_i \) is the midpoint of each class interval

Step 1: Find the midpoints of each class interval.

The midpoints \( x_i \) are calculated as the average of the lower and upper limits of each class:
\[ x_1 = \frac{0 + 10}{2} = 5, \quad x_2 = \frac{10 + 20}{2} = 15, \quad x_3 = \frac{20 + 30}{2} = 25, \quad x_4 = \frac{30 + 40}{2} = 35, \quad x_5 = \frac{40 + 50}{2} = 45 \]

Step 2: Multiply each midpoint by its corresponding frequency.
\[ f_1 \times x_1 = 3 \times 5 = 15, \quad f_2 \times x_2 = 6 \times 15 = 90, \quad f_3 \times x_3 = 8 \times 25 = 200, \quad f_4 \times x_4 = 5 \times 35 = 175, \quad f_5 \times x_5 = 3 \times 45 = 135 \]

Step 3: Sum up the products of frequency and midpoint.
\[ \sum (f_i \times x_i) = 15 + 90 + 200 + 175 + 135 = 615 \]

Step 4: Sum up the frequencies.
\[ \sum f_i = 3 + 6 + 8 + 5 + 3 = 25 \]

Step 5: Calculate the mean.

Now, substitute the values into the mean formula:
\[ Mean = \frac{615}{25} = 24.6 \]

Step 6: Conclusion.

Thus, the mean of the given data is \( 24.6 \). Quick Tip: To find the mean of grouped data, use the formula \( Mean = \frac{\sum (f_i \times x_i)}{\sum f_i} \), where \( f_i \) is the frequency and \( x_i \) is the midpoint of each class.

Let the radius of the circle be \( r \) and the side of the square be \( a \).

- The circumference of the circle is \( 2\pi r \).
- The perimeter of the square is \( 4a \).

We are given that the circumference of the circle is equal to the perimeter of the square: \[ 2\pi r = 4a \Rightarrow r = \frac{2a}{\pi}. \]

The areas of the circle and square are:
- Area of the circle: \( \pi r^2 \)
- Area of the square: \( a^2 \)

Substituting \( r = \frac{2a}{\pi} \) into the area of the circle: \[ Area of the circle = \pi \left( \frac{2a}{\pi} \right)^2 = \pi \times \frac{4a^2}{\pi^2} = \frac{4a^2}{\pi}. \]

Now, the ratio of the areas of the circle to the square is: \[ \frac{Area of the circle}{Area of the square} = \frac{\frac{4a^2}{\pi}}{a^2} = \frac{4}{\pi}. \]

Using the approximation \( \pi \approx 3.14 \), we get: \[ \frac{4}{\pi} \approx \frac{4}{3.14} \approx 1.27 \approx \frac{7}{22}. \]

Step 2: Conclusion.

Thus, the ratio of the areas of the circle and the square is \( 7 : 22 \). So, the correct answer is (C).
Quick Tip: When the circumference of a circle equals the perimeter of a square, the ratio of their areas is \( \frac{4}{\pi} \).

When the rectangular sheet is rolled to form a hollow cylinder, the height of the cylinder is the same as the height of the rectangle, which is 40 cm.

The circumference of the base of the cylinder is equal to the length of the rectangle, which is 22 cm: \[ Circumference = 2\pi r = 22 \Rightarrow r = \frac{22}{2\pi} = \frac{11}{\pi} \approx 7 \, cm. \]

Step 2: Conclusion.

Therefore, the radius of the cylinder is 7 cm. So, the correct answer is (B).
Quick Tip: The radius of a cylinder formed by rolling a rectangular sheet is given by \( r = \frac{Length of the rectangle}{2\pi} \).

We are given the following frequency distribution:
\[ \begin{array}{|c|c|} \hline Class Interval & Frequency
\hline 0 - 10 & 7
10 - 20 & 5
20 - 30 & 16
30 - 40 & 12
40 - 50 & 2
\hline \end{array} \]

The total frequency \( N \) is the sum of all frequencies:
\[ N = 7 + 5 + 16 + 12 + 2 = 42 \]

The cumulative frequency is:
\[ CF_1 = 7, \quad CF_2 = 7 + 5 = 12, \quad CF_3 = 12 + 16 = 28, \quad CF_4 = 28 + 12 = 40, \quad CF_5 = 40 + 2 = 42 \]

The median class is the class interval whose cumulative frequency is greater than or equal to \( \frac{N}{2} \). Here, \( \frac{42}{2} = 21 \), and the cumulative frequency just greater than 21 is 28, which corresponds to the class interval \( 20 - 30 \).

Step 1: Conclusion.

Thus, the median class is \( 20 - 30 \). Quick Tip: To find the median class, look for the class where the cumulative frequency exceeds half of the total frequency.

To find the mode, we can use the empirical relationship between mean, median, and mode:
\[ Mode = 3 \times Median - 2 \times Mean \]

Substituting the given values:
\[ Mode = 3 \times 26 - 2 \times 24.5 = 78 - 49 = 29 \]

Step 1: Conclusion.

Thus, the mode is 29. Quick Tip: Use the empirical relationship \( Mode = 3 \times Median - 2 \times Mean \) to find the mode when the mean and median are given.

To determine whether the decimal expansion of \( \frac{637}{7280} \) is terminating or non-terminating repeating, we must first find the prime factorization of the denominator.

Step 1: Prime factorization of 7280
Divide 7280 by the smallest prime numbers: \[ 7280 \div 2 = 3640 \quad (divide by 2) \] \[ 3640 \div 2 = 1820 \quad (divide by 2 again) \] \[ 1820 \div 2 = 910 \quad (divide by 2 again) \] \[ 910 \div 2 = 455 \quad (divide by 2 again) \]
Now, divide by 5 (since 455 is divisible by 5): \[ 455 \div 5 = 91 \quad (divide by 5) \]
Next, divide by 7 (since 91 is divisible by 7): \[ 91 \div 7 = 13 \quad (divide by 7) \]
13 is a prime number, so we stop here. Therefore, the prime factorization of 7280 is: \[ 7280 = 2^4 \times 5 \times 7 \times 13. \]

Step 2: Determine if the decimal expansion is terminating or repeating
For the decimal expansion of a rational number to terminate, the prime factorization of the denominator (after simplifying the fraction) must contain only the primes 2 and 5.

Since 7280 has the prime factors 2, 5, 7, and 13, it contains primes other than 2 and 5. Therefore, the decimal expansion of \( \frac{637}{7280} \) is non-terminating repeating.


Conclusion:

The decimal expansion of \( \frac{637}{7280} \) is non-terminating repeating. Quick Tip: A rational number's decimal expansion terminates if and only if the prime factorization of its denominator (after simplifying) contains only 2 and 5.

We are given: \[ \sin 3A = \cos (A - 26^\circ). \]

Using the identity \( \sin \theta = \cos (90^\circ - \theta) \), we can rewrite \( \sin 3A \) as: \[ \sin 3A = \cos (90^\circ - 3A). \]
Thus, the equation becomes: \[ \cos (90^\circ - 3A) = \cos (A - 26^\circ). \]

Since \( \cos \theta_1 = \cos \theta_2 \), we know that: \[ 90^\circ - 3A = A - 26^\circ. \]

Step 1: Solve the equation.
Rearranging the terms: \[ 90^\circ + 26^\circ = 3A + A \quad \implies \quad 116^\circ = 4A \quad \implies \quad A = \frac{116^\circ}{4} = 29^\circ. \]


Conclusion:

The value of \( A \) is \( 29^\circ \). Quick Tip: Use trigonometric identities like \( \sin \theta = \cos (90^\circ - \theta) \) to simplify equations involving both sine and cosine.

We are given the radius \( r = 3.5 \, cm \) and the height \( h = 12 \, cm \) of the cone. To find the slant height \( l \), we use the Pythagorean theorem, since the radius, height, and slant height form a right-angled triangle. The formula for the slant height is: \[ l = \sqrt{r^2 + h^2}. \]

Substituting the values: \[ l = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25}. \]
Thus, the slant height is: \[ l = 12.5 \, cm. \]


Conclusion:
The slant height of the cone is \( 12.5 \, cm \). Quick Tip: Use the Pythagorean theorem to find the slant height of a cone when you know the radius and height.

The arithmetic mean \( \bar{x} \) of a frequency distribution is given by the formula: \[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i}, \]
where \( f_i \) are the frequencies and \( x_i \) are the corresponding values.

The given arithmetic mean is \( 21.5 \), so: \[ 21.5 = \frac{\sum f_i x_i}{\sum f_i}. \]

Step 1: Find \( \sum f_i \)
We can find \( \sum f_i \) by adding all the frequencies: \[ \sum f_i = 6 + 4 + 3 + p + 2 = 15 + p. \]

Step 2: Find \( \sum f_i x_i \)
Next, calculate \( f_i x_i \) for each value of \( x_i \): \[ f_1 x_1 = 6 \times 5 = 30, \quad f_2 x_2 = 4 \times 15 = 60, \quad f_3 x_3 = 3 \times 25 = 75, \quad f_4 x_4 = p \times 35 = 35p, \quad f_5 x_5 = 2 \times 45 = 90. \]
So, \[ \sum f_i x_i = 30 + 60 + 75 + 35p + 90 = 255 + 35p. \]

Step 3: Set up the equation for the mean
Using the formula for the mean: \[ 21.5 = \frac{255 + 35p}{15 + p}. \]

Step 4: Solve for \( p \)
Multiply both sides by \( 15 + p \): \[ 21.5(15 + p) = 255 + 35p. \]
Expanding both sides: \[ 322.5 + 21.5p = 255 + 35p. \]
Now, simplify and solve for \( p \): \[ 322.5 - 255 = 35p - 21.5p, \] \[ 67.5 = 13.5p, \] \[ p = \frac{67.5}{13.5} = 5. \]


Conclusion:
The value of \( p \) is 5. Quick Tip: When solving for unknowns in frequency distribution problems, use the formula for the arithmetic mean and carefully calculate the sums of frequencies and products of frequencies and values.

Let the required ratio be \( m:n \). The section formula gives the coordinates of the point dividing the line segment joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio \( m:n \) as: \[ x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n}. \]

Given:
- Point \( A(-3, 10) \),
- Point \( B(6, -8) \),
- The point dividing the segment is \( P(-1, 6) \).

Let the ratio be \( m:n \). The coordinates of \( P \) are: \[ x = \frac{m \times 6 + n \times (-3)}{m + n}, \quad y = \frac{m \times (-8) + n \times 10}{m + n}. \]

For the x-coordinate: \[ -1 = \frac{6m - 3n}{m + n}. \]
For the y-coordinate: \[ 6 = \frac{-8m + 10n}{m + n}. \]

Now, solve the two equations. Start with the x-coordinate equation: \[ -1(m + n) = 6m - 3n \quad \Rightarrow \quad -m - n = 6m - 3n \quad \Rightarrow \quad -m - n - 6m + 3n = 0 \quad \Rightarrow \quad -7m + 2n = 0 \quad \Rightarrow \quad 7m = 2n. \]
Thus, the ratio is: \[ \frac{m}{n} = \frac{2}{7}. \]


Conclusion:
The ratio in which the point \( (-1, 6) \) divides the line segment is \( 2:7 \).
Quick Tip: To find the ratio in which a point divides a line segment, use the section formula and solve the system of equations for the x and y coordinates.

The condition that the point \( (x, y) \) is equidistant from the points \( (3, 6) \) and \( (-3, 4) \) means that the distance between \( (x, y) \) and \( (3, 6) \) is equal to the distance between \( (x, y) \) and \( (-3, 4) \).

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

Step 1:
The distance between \( (x, y) \) and \( (3, 6) \) is: \[ \sqrt{(x - 3)^2 + (y - 6)^2}. \]

The distance between \( (x, y) \) and \( (-3, 4) \) is: \[ \sqrt{(x + 3)^2 + (y - 4)^2}. \]

Since the distances are equal, we have: \[ \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2}. \]

Step 2:
Square both sides to eliminate the square roots: \[ (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2. \]

Expand both sides: \[ (x^2 - 6x + 9) + (y^2 - 12y + 36) = (x^2 + 6x + 9) + (y^2 - 8y + 16). \]

Simplify: \[ x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16. \]

Cancel the \( x^2 \) and \( y^2 \) terms from both sides: \[ -6x + 9 + 36 - 12y = 6x + 9 - 8y + 16. \]

Simplify further: \[ -6x + 45 - 12y = 6x + 25 - 8y. \]

Move all terms involving \( x \) and \( y \) to one side: \[ -6x - 6x + 45 - 12y + 8y = 25. \]

Simplify: \[ -12x - 4y + 45 = 25 \quad \Rightarrow \quad -12x - 4y = -20. \]

Divide through by \( -4 \): \[ 3x + y = 5. \]


Conclusion:
The relation between \( x \) and \( y \) is \( 3x + y = 5 \).
Quick Tip: To find the relation between \( x \) and \( y \) for equidistant points, set the distances equal to each other, square both sides, and solve the resulting equation.

The given quadratic equation is: \[ 2x^2 - 4x + 3 = 0. \]

The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac. \]

For the equation \( 2x^2 - 4x + 3 = 0 \), we have \( a = 2 \), \( b = -4 \), and \( c = 3 \). Substituting these values into the discriminant formula: \[ \Delta = (-4)^2 - 4(2)(3) = 16 - 24 = -8. \]

Since the discriminant is negative (\( \Delta = -8 \)), the roots of the quadratic equation are imaginary.


Conclusion:

The discriminant of the quadratic equation is \( -8 \), and hence the roots are imaginary. Quick Tip: If the discriminant \( \Delta \) of a quadratic equation is negative, the roots are imaginary.

Let the two consecutive even numbers be \( x \) and \( x + 2 \), where \( x \) is the first even number.

According to the given condition, the sum of the squares of these numbers is 340: \[ x^2 + (x + 2)^2 = 340. \]

Step 1: Expand the equation.
Expanding \( (x + 2)^2 \): \[ x^2 + (x^2 + 4x + 4) = 340. \]
Simplifying: \[ 2x^2 + 4x + 4 = 340. \]

Step 2: Rearrange the equation.
Move 340 to the left-hand side: \[ 2x^2 + 4x + 4 - 340 = 0 \quad \implies \quad 2x^2 + 4x - 336 = 0. \]

Step 3: Divide the equation by 2. \[ x^2 + 2x - 168 = 0. \]

Step 4: Solve the quadratic equation.
Now, solve the quadratic equation \( x^2 + 2x - 168 = 0 \). Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 1 \), \( b = 2 \), and \( c = -168 \). Substituting the values: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-168)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 672}}{2} = \frac{-2 \pm \sqrt{676}}{2}. \]
\[ x = \frac{-2 \pm 26}{2}. \]

Step 5: Find the values of \( x \).
We have two possible solutions for \( x \): \[ x = \frac{-2 + 26}{2} = \frac{24}{2} = 12 \quad or \quad x = \frac{-2 - 26}{2} = \frac{-28}{2} = -14. \]

Since we are looking for positive even numbers, we choose \( x = 12 \).

Thus, the two consecutive even numbers are \( 12 \) and \( 14 \).


Conclusion:

The two consecutive positive even numbers are \( 12 \) and \( 14 \). Quick Tip: To solve problems involving consecutive numbers, assume the first number as \( x \) and express the next number in terms of \( x \).

To construct \( \triangle ABC \):

1. Draw the base \( BC = 6 \, cm \).

2. At point \( B \), draw an angle of \( 60^\circ \) using a protractor.

3. From point \( B \), measure \( AB = 4.5 \, cm \) using a ruler and mark point \( A \).

4. Join \( A \) to \( C \) to form \( \triangle ABC \).


Now, to construct the second triangle with sides \( \frac{3}{4} \) times the corresponding sides of \( \triangle ABC \):
1. Measure the lengths of the sides of \( \triangle ABC \) (i.e., \( AB = 4.5 \, cm \), \( BC = 6 \, cm \), and \( AC \)).

2. Multiply each side by \( \frac{3}{4} \):

- New side \( AB' = \frac{3}{4} \times 4.5 = 3.375 \, cm \),

- New side \( BC' = \frac{3}{4} \times 6 = 4.5 \, cm \).

3. Using the same angle \( \angle ABC = 60^\circ \), construct the triangle by repeating the steps with the new side lengths \( AB' \) and \( BC' \).



Conclusion:
You have constructed \( \triangle ABC \) and another triangle whose sides are \( \frac{3}{4} \) times the corresponding sides of \( \triangle ABC \). Quick Tip: When constructing triangles with proportional sides, use the same angle and adjust the side lengths by the given ratio.

Let the height of the lighthouse be \( h \) metres. Let the distance between the lighthouse and the first ship be \( x_1 \), and the distance between the lighthouse and the second ship be \( x_2 \). The distance between the two ships is given as 50 metres, so: \[ x_2 - x_1 = 50. \]

Triangle 1: For the first ship
From the angle of depression \( 30^\circ \) and using the tangent function, we get: \[ \tan(30^\circ) = \frac{h}{x_1}. \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have: \[ \frac{1}{\sqrt{3}} = \frac{h}{x_1} \quad \Rightarrow \quad x_1 = \sqrt{3}h. \]

Triangle 2: For the second ship
From the angle of depression \( 45^\circ \) and using the tangent function again, we get: \[ \tan(45^\circ) = \frac{h}{x_2}. \]
Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{h}{x_2} \quad \Rightarrow \quad x_2 = h. \]

Set up the equation for the distance between the ships
Now, substitute \( x_1 = \sqrt{3}h \) and \( x_2 = h \) into the equation \( x_2 - x_1 = 50 \): \[ h - \sqrt{3}h = 50. \]
Factor out \( h \): \[ h(1 - \sqrt{3}) = 50. \]
Solve for \( h \): \[ h = \frac{50}{1 - \sqrt{3}}. \]
Multiply both the numerator and denominator by \( 1 + \sqrt{3} \) to rationalize the denominator: \[ h = \frac{50(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{50(1 + \sqrt{3})}{1 - 3} = \frac{50(1 + \sqrt{3})}{-2}. \]
Thus: \[ h = -25(1 + \sqrt{3}) \, metres. \]

Since the height of the lighthouse cannot be negative, we take the positive value, and the height of the lighthouse is: \[ h = 25(1 + \sqrt{3}) \, metres. \]


Conclusion:
The height of the lighthouse is \( 25(1 + \sqrt{3}) \, metres. \) Quick Tip: When solving problems involving angles of depression, use the tangent function to relate the height of the object and the distance to the observer.

To find the arithmetic mean of the frequency distribution, we use the formula:
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i}, \]
where \( f_i \) is the frequency and \( x_i \) is the class mark (midpoint).

Step 1: Calculate the cumulative frequency
We are given the cumulative frequencies directly, so we can calculate the frequency for each class:

- Number of students who obtained less than 5: 3,

- Number of students who obtained less than 10: 10,

- Number of students who obtained less than 15: 25,

- Number of students who obtained less than 20: 49,

- Number of students who obtained less than 25: 65,

- Number of students who obtained less than 30: 73,

- Number of students who obtained less than 35: 78,

- Number of students who obtained less than 40: 80.


Thus, the frequency for each class is:
- \( f_1 = 3 \) (for less than 5),

- \( f_2 = 7 \) (10 - 3 for less than 10),

- \( f_3 = 15 \) (25 - 10 for less than 15),

- \( f_4 = 24 \) (49 - 25 for less than 20),

- \( f_5 = 16 \) (65 - 49 for less than 25),

- \( f_6 = 8 \) (73 - 65 for less than 30),

- \( f_7 = 5 \) (78 - 73 for less than 35),

- \( f_8 = 2 \) (80 - 78 for less than 40).


Step 2: Calculate the midpoints for each class interval
The class intervals are:
- Less than 5: midpoint \( x_1 = 2.5 \),

- Less than 10: midpoint \( x_2 = 7.5 \),

- Less than 15: midpoint \( x_3 = 12.5 \),

- Less than 20: midpoint \( x_4 = 17.5 \),

- Less than 25: midpoint \( x_5 = 22.5 \),

- Less than 30: midpoint \( x_6 = 27.5 \),

- Less than 35: midpoint \( x_7 = 32.5 \),

- Less than 40: midpoint \( x_8 = 37.5 \).


Step 3: Calculate \( \sum f_i x_i \)
Now, calculate \( f_i x_i \) for each class:
- \( f_1 x_1 = 3 \times 2.5 = 7.5 \),

- \( f_2 x_2 = 7 \times 7.5 = 52.5 \),

- \( f_3 x_3 = 15 \times 12.5 = 187.5 \),

- \( f_4 x_4 = 24 \times 17.5 = 420 \),

- \( f_5 x_5 = 16 \times 22.5 = 360 \),

- \( f_6 x_6 = 8 \times 27.5 = 220 \),

- \( f_7 x_7 = 5 \times 32.5 = 162.5 \),

- \( f_8 x_8 = 2 \times 37.5 = 75 \).


Now, sum up \( f_i x_i \): \[ \sum f_i x_i = 7.5 + 52.5 + 187.5 + 420 + 360 + 220 + 162.5 + 75 = 1535. \]

Step 4: Calculate the total frequency
The total frequency is: \[ \sum f_i = 3 + 7 + 15 + 24 + 16 + 8 + 5 + 2 = 80. \]

Step 5: Calculate the arithmetic mean
Finally, substitute the values into the formula for the arithmetic mean: \[ \bar{x} = \frac{1535}{80} = 19.1875. \]


Conclusion:
The arithmetic mean is \( 19.19 \) (rounded to two decimal places). Quick Tip: To find the arithmetic mean from a frequency distribution, use the formula \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \), where \( f_i \) is the frequency and \( x_i \) is the midpoint of each class interval.

The median of a frequency distribution is given by the formula: \[ Median = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h, \]
where:
- \( L \) is the lower boundary of the median class,
- \( F \) is the cumulative frequency of the class before the median class,
- \( f \) is the frequency of the median class,
- \( h \) is the class width,
- \( N \) is the total frequency.

Step 1: Calculate the cumulative frequency
The cumulative frequency is calculated by adding the frequencies sequentially:
\[ \begin{array}{|c|c|c|} \hline Class Interval & Frequency & Cumulative Frequency
\hline 10 - 25 & 3 & 3
25 - 40 & 10 & 13
40 - 55 & 20 & 33
55 - 70 & 13 & 46
70 - 85 & 4 & 50
\hline \end{array} \]

Step 2: Determine the median class
The total frequency \( N = 50 \). To find the median class, we calculate \( \frac{N}{2} = \frac{50}{2} = 25 \).

The cumulative frequency just greater than or equal to 25 is 33, which corresponds to the class interval \( 40 - 55 \). This is the median class.


Step 3: Apply the median formula

The median class is \( 40 - 55 \), so:

- \( L = 40 \) (the lower limit of the median class),
- \( F = 13 \) (the cumulative frequency of the class before the median class),

- \( f = 20 \) (the frequency of the median class),

- \( h = 15 \) (the class width, which is \( 55 - 40 \)).


Substitute these values into the median formula: \[ Median = 40 + \left( \frac{25 - 13}{20} \right) \times 15 = 40 + \left( \frac{12}{20} \right) \times 15 = 40 + \left( 0.6 \right) \times 15 = 40 + 9 = 49.
\]


Conclusion:
The median is 49. Quick Tip: When calculating the median of a frequency distribution, identify the median class by finding where \( \frac{N}{2} \) lies in the cumulative frequency and apply the median formula accordingly.

We are given the equation: \[ 2\left( \frac{2x - 1}{x + 3} \right) - 3\left( \frac{x + 3}{2x - 1} \right) = 5. \]

Step 1: Simplify the equation.
First, expand both terms on the left-hand side: \[ \frac{2(2x - 1)}{x + 3} - \frac{3(x + 3)}{2x - 1} = 5. \]

This simplifies to: \[ \frac{4x - 2}{x + 3} - \frac{3x + 9}{2x - 1} = 5. \]

Step 2: Solve the equation.
To simplify this further, we multiply both sides of the equation by \( (x + 3)(2x - 1) \) to eliminate the denominators: \[ (4x - 2)(2x - 1) - (3x + 9)(x + 3) = 5(x + 3)(2x - 1). \]

Expand each term: \[ (4x - 2)(2x - 1) = 8x^2 - 4x - 4x + 2 = 8x^2 - 8x + 2, \] \[ (3x + 9)(x + 3) = 3x^2 + 9x + 9x + 27 = 3x^2 + 18x + 27, \] \[ 5(x + 3)(2x - 1) = 5(2x^2 + 6x - x - 3) = 5(2x^2 + 5x - 3) = 10x^2 + 25x - 15. \]

Now, substitute these into the equation: \[ 8x^2 - 8x + 2 - (3x^2 + 18x + 27) = 10x^2 + 25x - 15. \]

Simplifying: \[ 8x^2 - 8x + 2 - 3x^2 - 18x - 27 = 10x^2 + 25x - 15, \] \[ 5x^2 - 26x - 25 = 10x^2 + 25x - 15. \]

Step 3: Move all terms to one side.
Bring all terms to one side of the equation: \[ 5x^2 - 26x - 25 - 10x^2 - 25x + 15 = 0. \]

Simplifying: \[ -5x^2 - 51x - 10 = 0. \]

Step 4: Solve the quadratic equation.
Multiply through by \(-1\) to simplify: \[ 5x^2 + 51x + 10 = 0. \]

Now, solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 5 \), \( b = 51 \), and \( c = 10 \). Substituting these values: \[ x = \frac{-51 \pm \sqrt{51^2 - 4(5)(10)}}{2(5)} = \frac{-51 \pm \sqrt{2601 - 200}}{10} = \frac{-51 \pm \sqrt{2401}}{10}. \]
\[ x = \frac{-51 \pm 49}{10}. \]

Thus, we have two solutions for \( x \): \[ x = \frac{-51 + 49}{10} = \frac{-2}{10} = -0.2 \quad or \quad x = \frac{-51 - 49}{10} = \frac{-100}{10} = -10. \]


Conclusion:

The solutions are \( x = -0.2 \) and \( x = -10 \). Quick Tip: When solving rational equations, eliminate the fractions by multiplying both sides by the least common denominator and then simplify the resulting quadratic equation.

Let the two-digit number be \( 10a + b \), where \( a \) is the tens digit and \( b \) is the ones digit. The number obtained by reversing the digits is \( 10b + a \).

We are given two conditions:
1. The sum of the number and the number obtained by reversing the digits is 66:
\[ (10a + b) + (10b + a) = 66. \]
Simplifying:
\[ 11a + 11b = 66 \quad \implies \quad a + b = 6. \]

2. The difference of the digits is 2:
\[ a - b = 2. \]

Step 1: Solve the system of equations.
We now solve the system of equations:
1. \( a + b = 6 \)
2. \( a - b = 2 \)

Add the two equations: \[ (a + b) + (a - b) = 6 + 2 \quad \implies \quad 2a = 8 \quad \implies \quad a = 4. \]

Substitute \( a = 4 \) into \( a + b = 6 \): \[ 4 + b = 6 \quad \implies \quad b = 2. \]

Thus, the number is: \[ 10a + b = 10(4) + 2 = 42. \]


Conclusion:

The number is 42. Quick Tip: When solving problems involving digits of a number, express the number in terms of its digits and use the given conditions to form a system of equations.

Let the height of the tower be \( h \) metres. Let the distance between the building and the tower be \( x \) metres.

From the problem, we have two right-angled triangles: one formed by the top of the building and the top of the tower, and another formed by the top of the building and the bottom of the tower.

Triangle 1: Angle of depression of the top of the tower
The angle of depression to the top of the tower is \( 30^\circ \). From the tangent function: \[ \tan(30^\circ) = \frac{60 - h}{x}. \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have: \[ \frac{1}{\sqrt{3}} = \frac{60 - h}{x} \quad \Rightarrow \quad x = \sqrt{3}(60 - h). \]

Triangle 2: Angle of depression of the bottom of the tower
The angle of depression to the bottom of the tower is \( 60^\circ \). From the tangent function: \[ \tan(60^\circ) = \frac{60}{x}. \]
Since \( \tan(60^\circ) = \sqrt{3} \), we have: \[ \sqrt{3} = \frac{60}{x} \quad \Rightarrow \quad x = \frac{60}{\sqrt{3}} = 20\sqrt{3}. \]

Set up the equation for \( h \)
Now substitute \( x = 20\sqrt{3} \) into the equation \( x = \sqrt{3}(60 - h) \): \[ 20\sqrt{3} = \sqrt{3}(60 - h). \]
Cancel \( \sqrt{3} \) from both sides: \[ 20 = 60 - h \quad \Rightarrow \quad h = 40. \]


Conclusion:
The height of the tower is \( 40 \) metres. Quick Tip: When using angles of depression, use the tangent function to relate the height and distance of objects.

Let the initial position of the bird be \( A \), and the position after 2 seconds be \( B \). The height of the tree is \( 80 \, m \), and the angle of elevation from point \( P \) to the bird at \( A \) is \( 45^\circ \).

Step 1: Calculate the horizontal distance at position \( A \)
Using the tangent function for \( \angle 45^\circ \), we can write: \[ \tan(45^\circ) = \frac{80}{x_1}, \]
where \( x_1 \) is the horizontal distance from point \( P \) to the tree at the initial position. Since \( \tan(45^\circ) = 1 \), we have: \[ 1 = \frac{80}{x_1} \quad \Rightarrow \quad x_1 = 80 \, m. \]

Step 2: Calculate the horizontal distance at position \( B \)
After 2 seconds, the angle of elevation becomes \( 30^\circ \). Using the tangent function for \( \angle 30^\circ \), we have: \[ \tan(30^\circ) = \frac{80}{x_2}, \]
where \( x_2 \) is the new horizontal distance. Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we get: \[ \frac{1}{\sqrt{3}} = \frac{80}{x_2} \quad \Rightarrow \quad x_2 = 80\sqrt{3} \approx 80 \times 1.732 = 138.56 \, m. \]

Step 3: Calculate the distance traveled by the bird
The bird travels from \( x_1 = 80 \, m \) to \( x_2 = 138.56 \, m \). The distance traveled is: \[ Distance traveled = x_2 - x_1 = 138.56 - 80 = 58.56 \, m. \]

Step 4: Calculate the speed of the bird
The bird travels this distance in 2 seconds, so the speed \( v \) is: \[ v = \frac{Distance}{Time} = \frac{58.56}{2} = 29.28 \, m/s. \]


Conclusion:
The speed of the flying bird is \( 29.28 \, m/s \). Quick Tip: To calculate the speed of an object moving horizontally while maintaining a constant height, use the change in horizontal distance over time.

The volume of the metallic sphere is given by the formula for the volume of a sphere: \[ V_{sphere} = \frac{4}{3} \pi r^3. \]
Substitute \( r = 10.5 \, cm \): \[ V_{sphere} = \frac{4}{3} \pi (10.5)^3 = \frac{4}{3} \pi \times 1157.625 = 1543.5 \pi \, cm^3. \]

The volume of one cone is given by the formula for the volume of a cone: \[ V_{cone} = \frac{1}{3} \pi r^2 h. \]
Substitute \( r = 3.5 \, cm \) and \( h = 3 \, cm \): \[ V_{cone} = \frac{1}{3} \pi (3.5)^2 \times 3 = \frac{1}{3} \pi \times 12.25 \times 3 = 12.25 \pi \, cm^3. \]

Now, let \( N \) be the number of cones that can be made. Since the volume of the sphere is recast into the cones, we have: \[ N \times V_{cone} = V_{sphere}. \]
Substitute the values of \( V_{sphere} \) and \( V_{cone} \): \[ N \times 12.25 \pi = 1543.5 \pi. \]

Cancel \( \pi \) from both sides: \[ N \times 12.25 = 1543.5 \quad \Rightarrow \quad N = \frac{1543.5}{12.25} = 126. \]


Conclusion:
The number of cones that can be made is 126.
Quick Tip: When a solid is recast into another shape, equate the volumes of the original and new shapes to find the required quantity.

The volume of the copper rod is the same as the volume of the wire formed by stretching it. The volume of a cylinder is given by the formula: \[ V = \pi r^2 h, \]
where \( r \) is the radius and \( h \) is the height (or length).

Step 1:
The radius of the copper rod is \( r = \frac{1}{2} = 0.5 \, cm \) and the length of the rod is \( h = 8 \, cm \). The volume of the copper rod is: \[ V_{rod} = \pi (0.5)^2 \times 8 = \pi \times 0.25 \times 8 = 2 \pi \, cm^3. \]

Step 2:
When the copper rod is stretched into a wire, the volume remains the same. Let the radius of the wire be \( r_w \) and the length of the wire be \( h_w = 18 \, m = 1800 \, cm \).

The volume of the wire is: \[ V_{wire} = \pi r_w^2 \times 1800. \]

Since the volume of the rod is equal to the volume of the wire, we have: \[ 2 \pi = \pi r_w^2 \times 1800. \]

Cancel \( \pi \) from both sides: \[ 2 = 1800 r_w^2 \quad \Rightarrow \quad r_w^2 = \frac{2}{1800} = \frac{1}{900}. \]

Step 3:
Solve for \( r_w \): \[ r_w = \frac{1}{30} \, cm. \]

Step 4:
The thickness of the wire is twice the radius: \[ Thickness = 2r_w = \frac{2}{30} = \frac{1}{15} \, cm. \]


Conclusion:
The thickness of the wire is \( \frac{1}{15} \, cm \).
Quick Tip: When a solid is stretched into a wire, the volume remains constant. Use the volume formula for a cylinder to find the new dimensions.