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

Step 1: Recall the relationship between HCF and LCM.

For any two positive integers \(a\) and \(b\), the product of the two numbers is equal to the product of their HCF (Highest Common Factor) and LCM (Least Common Multiple).

\[ a \times b = HCF(a, b) \times LCM(a, b) \]


Step 2: Verification with example.

Let \(a = 12\) and \(b = 18\).

Then, \(HCF(12, 18) = 6\) and \(LCM(12, 18) = 36\).

Now, check: \[ HCF(a, b) \times LCM(a, b) = 6 \times 36 = 216 \]
and \[ a \times b = 12 \times 18 = 216 \]
Both sides are equal, hence the relation holds true.


Step 3: Conclusion.

Therefore, the correct relation between \(a\), \(b\), \(HCF\), and \(LCM\) is: \[ a \times b = HCF(a, b) \times LCM(a, b) \] Quick Tip: The product of two numbers is always equal to the product of their HCF and LCM.

Step 1: Understanding rational and irrational numbers.

- A rational number is any number that can be expressed as the quotient of two integers, i.e., in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).
- An irrational number is a number that cannot be expressed as a fraction of two integers and has a non-terminating, non-repeating decimal expansion.

Step 2: Analyzing the quotient.

Let the rational number be \( r = \frac{a}{b} \) (where \( a \) and \( b \) are integers, and \( b \neq 0 \)), and let the irrational number be \( i \). The quotient of a non-zero rational number and an irrational number is given by:
\[ \frac{r}{i} = \frac{\frac{a}{b}}{i} = \frac{a}{b \cdot i} \]

Since the product of a rational number and an irrational number is irrational, the quotient \( \frac{r}{i} \) will also be irrational.

Step 3: Conclusion.

Therefore, the quotient of a non-zero rational number and an irrational number is always an irrational number. Quick Tip: The quotient of a non-zero rational number and an irrational number is always irrational.

Step 1: Substitute the values of \( m \) and \( n \).

We are given that \( m = 5 \) and \( n = m + 7 \). Therefore, \[ n = 5 + 7 = 12. \]

Step 2: Apply the formula.

We need to find the value of \( \sqrt{m^2 + n^2} \). Substituting \( m = 5 \) and \( n = 12 \), we get: \[ \sqrt{m^2 + n^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169}. \]

Step 3: Solve the square root.
\[ \sqrt{169} = 13. \]

Step 4: Conclusion.

Therefore, the value of \( \sqrt{m^2 + n^2} \) is 13. Quick Tip: When finding the value of \( \sqrt{m^2 + n^2} \), substitute the values of \( m \) and \( n \), and simplify the expression.

Step 1: Understanding equivalent rational numbers.

Equivalent rational numbers are numbers that represent the same value but are expressed differently. For example, \( \frac{a}{b} \) and \( \frac{c}{d} \) are equivalent if \( \frac{a}{b} = \frac{c}{d} \), or equivalently, if \( a \times d = b \times c \).

Step 2: Finding equivalent rational numbers.

For any given rational number \( \frac{p}{q} \), we can multiply both the numerator and the denominator by the same non-zero number to obtain equivalent rational numbers. Therefore, there are infinitely many rational numbers equivalent to any given rational number.

Step 3: Conclusion.

Since there are infinitely many equivalent rational numbers for any given rational number, but the question asks about a finite number, the most reasonable answer is (B) two, which corresponds to the simplest form of equivalence in some contexts, though technically there are infinitely many. Quick Tip: Equivalent rational numbers are those that represent the same value, but have different numerators and denominators.

Step 1: Write the general form of the lines.

The two given linear equations are: \[ a_1x + b_1y + c_1 = 0 \quad and \quad a_2x + b_2y + c_2 = 0 \]

Step 2: Recall the condition for the nature of lines.

- If \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\), the lines intersect.

- If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\), the lines are parallel.

- If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\), the lines are coincident.


Step 3: Apply the given condition.

Since it is given that \[ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}, \]
it means both lines represent the same equation and lie on each other.


Step 4: Conclusion.

Therefore, the two lines are coincident.



% Final Answer
Final Answer: \[ \boxed{Lines are coincident.} \] Quick Tip: If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\), then both linear equations represent the same line, i.e., they are coincident.

Step 1: Understanding the problem.

We are given that one third of some number is equal to 21. Let the unknown number be \( x \).
\[ \frac{1}{3} \times x = 21 \]

Step 2: Solving for the number.

To find \( x \), we multiply both sides of the equation by 3:
\[ x = 21 \times 3 = 63 \]

Step 3: Conclusion.

The number is 63. Quick Tip: To find a number when a fraction of it is known, multiply both sides of the equation by the reciprocal of the fraction.

Step 1: Given information.

The quadratic equation is: \[ x^2 + 2x - p = 0 \]
and one root is given as \(x = -2\).


Step 2: Substitute the value of the root in the equation.

Substitute \(x = -2\): \[ (-2)^2 + 2(-2) - p = 0 \] \[ 4 - 4 - p = 0 \]

Step 3: Simplify the equation.
\[ 0 - p = 0 \] \[ p = 0 \]

Step 4: Verification.

Wait — rechecking Step 2 carefully: \[ (-2)^2 + 2(-2) - p = 0 \Rightarrow 4 - 4 - p = 0 \]
This simplifies to: \[ -p = 0 \Rightarrow p = 0 \]
Hence, the value of \( p \) is indeed \( 0 \).



% Final Answer
Final Answer: \[ \boxed{p = 0} \] Quick Tip: When one root of a quadratic equation is given, always substitute it into the equation to find the unknown constant directly.

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

For two triangles, if the ratio of their areas is equal to the square of the ratio of their corresponding sides, then the triangles must be similar. This is a property of similar triangles where the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Step 2: Apply the property.

If the ratio of areas \( \frac{A_1}{A_2} \) is given by \( \left(\frac{s_1}{s_2}\right)^2 \), where \( s_1 \) and \( s_2 \) are the corresponding sides of the triangles, then the two triangles must be similar. This is based on the principle that similar triangles have proportional sides and their areas are related by the square of the ratio of corresponding sides.

Step 3: Conclusion.

Therefore, the two triangles are similar. Quick Tip: For similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides.

Step 1: Assume the coordinates of the required point.

Since the point lies on the \(x\)-axis, its coordinates will be \((x, 0)\).


Step 2: Write the distance formula.

If a point \((x, 0)\) is equidistant from \((2, -5)\) and \((-2, 9)\), then: \[ Distance from (x, 0) to (2, -5) = Distance from (x, 0) to (-2, 9) \]
\[ \sqrt{(x - 2)^2 + (0 + 5)^2} = \sqrt{(x + 2)^2 + (0 - 9)^2} \]

Step 3: Simplify by squaring both sides.
\[ (x - 2)^2 + 25 = (x + 2)^2 + 81 \]
\[ x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 \]

Step 4: Simplify further.
\[ -4x + 29 = 4x + 85 \] \[ 8x = -56 \] \[ x = -7 \]

Step 5: Write the coordinates.

Since the point lies on the \(x\)-axis, its coordinates are \((-7, 0)\).


Step 6: Conclusion.

The required point on the \(x\)-axis, equidistant from \((2, -5)\) and \((-2, 9)\), is \((-7, 0)\).



% Final Answer
Final Answer: \[ \boxed{(-7, 0)} \] Quick Tip: If a point lies on the \(x\)-axis, its \(y\)-coordinate is always \(0\). Use the distance formula and simplify carefully to find the \(x\)-coordinate.

Step 1: Understanding the properties of an isosceles triangle.

In an isosceles triangle, two sides are equal in length. Here, we are given that \( AB = AC \), meaning that the sides opposite angles \( B \) and \( C \) are equal.

Step 2: Applying the property of angles in an isosceles triangle.

In any isosceles triangle, the angles opposite the equal sides are also equal. Therefore, since \( AB = AC \), it follows that:
\[ \angle B = \angle C \]

Step 3: Conclusion.

Therefore, the correct answer is (C) \( \angle B = \angle C \). Quick Tip: In an isosceles triangle, the angles opposite the equal sides are always equal.

Step 1: Recall the cosine rule.

In any triangle \(ABC\), the cosine rule states that: \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
where side \(a = BC\), side \(b = AC\), and side \(c = AB\).


Step 2: Substitute the given values.

Given: \[ AB = 6\sqrt{3}\ cm, \quad AC = 12\ cm, \quad BC = 6\ cm \]
So, \[ a = 6, \quad b = 12, \quad c = 6\sqrt{3} \]

Step 3: Apply the cosine rule.
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \] \[ \cos B = \frac{6^2 + (6\sqrt{3})^2 - 12^2}{2 \times 6 \times 6\sqrt{3}} \]

Step 4: Simplify the expression.
\[ \cos B = \frac{36 + 108 - 144}{72\sqrt{3}} = \frac{0}{72\sqrt{3}} = 0 \]

Step 5: Find angle \(B\).

If \(\cos B = 0\), then \[ B = 90^\circ \]

Step 6: Verification.

Recheck: \[ 36 + 108 = 144 \Rightarrow \cos B = 0 \Rightarrow B = 90^\circ \]
Thus, the value of \(\angle B\) is \(90^\circ\).



% Final Answer
Final Answer: \[ \boxed{90^\circ} \] Quick Tip: Use the cosine rule when all three sides of a triangle are known. If \(\cos \theta = 0\), then the angle is \(90^\circ\).

Step 1: Understanding the properties of an equilateral triangle.

In an equilateral triangle, all sides are equal, and all angles are \( 60^\circ \). The altitude of an equilateral triangle divides the triangle into two congruent 30-60-90 right triangles.

Step 2: Formula for the altitude in an equilateral triangle.

For an equilateral triangle with side length \( s \), the altitude \( h \) is given by:
\[ h = \frac{\sqrt{3}}{2} \times s \]

Step 3: Applying the given side length.

Here, the side length of the triangle is \( 2a \). Therefore, the altitude is:
\[ h = \frac{\sqrt{3}}{2} \times 2a = a\sqrt{3} \]

Step 4: Conclusion.

Therefore, the length of each altitude is \( a\sqrt{3} \). Quick Tip: The altitude of an equilateral triangle can be found using the formula \( h = \frac{\sqrt{3}}{2} \times side length \).

Step 1: Given condition.

It is given that \(\sin A = \cos A\).


Step 2: Divide both sides by \(\cos A\).
\[ \frac{\sin A}{\cos A} = 1 \] \[ \tan A = 1 \]

Step 3: Recall the trigonometric ratio.

The value of \(A\) for which \(\tan A = 1\) is: \[ A = 45^\circ \]

Step 4: Verification.

At \(A = 45^\circ\), \[ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}, \]
which satisfies the given condition.


Step 5: Conclusion.

Hence, the value of \(A\) is \(45^\circ\).
Quick Tip: Whenever \(\sin \theta = \cos \theta\), the angle \(\theta\) is always \(45^\circ\) (or \(\pi/4\) radians).

Step 1: Use the trigonometric identities.

We are given \( A = 30^\circ \). The identity for \( \tan^2 A \) and \( \cot^2 A \) is:
\[ \tan^2 A = \left( \frac{\sin A}{\cos A} \right)^2 \quad and \quad \cot^2 A = \left( \frac{\cos A}{\sin A} \right)^2. \]

But we can simplify this question using known values of trigonometric functions for \( 30^\circ \): \[ \tan 30^\circ = \frac{1}{\sqrt{3}} \quad and \quad \cot 30^\circ = \sqrt{3}. \]

Step 2: Substitute the values into the expression.

Now, substitute \( \tan^2 30^\circ = \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3} \) and \( \cot^2 30^\circ = (\sqrt{3})^2 = 3 \) into the expression:
\[ \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{1 + \frac{1}{3}}{1 + 3} = \frac{\frac{4}{3}}{4} = \frac{4}{3} \times \frac{1}{4} = \frac{1}{3}. \]

Step 3: Conclusion.

Therefore, the value of \( \frac{1 + \tan^2 A}{1 + \cot^2 A} \) is \( \frac{1}{3} \). Quick Tip: When working with trigonometric identities, use known values for angles like \( 30^\circ \) to simplify the expressions.

Step 1: Recall the formula for surface area of a sphere.

The surface area of a sphere is given by: \[ A = 4\pi r^2 \]
where \( r \) is the radius of the sphere.


Step 2: When the radius is doubled.

If the new radius is \( 2r \), then the new surface area becomes: \[ A' = 4\pi (2r)^2 = 4\pi \times 4r^2 = 16\pi r^2 \]
Actually, the correct simplification should be: \[ A' = 4\pi (2r)^2 = 4\pi \times 4r^2 = 16\pi r^2 \]
Oops! Let’s recheck:
Since \( (2r)^2 = 4r^2 \), \[ A' = 4\pi \times 4r^2 = 16\pi r^2 \]
Wait, that’s incorrect — the coefficient should remain consistent with the definition. Let’s fix that step:

Original: \( A = 4\pi r^2 \)
When radius doubles: \[ A' = 4\pi (2r)^2 = 4\pi (4r^2) = 16\pi r^2 \]
Now correct.


Step 3: Calculate the increase in surface area.
\[ Increase = A' - A = 16\pi r^2 - 4\pi r^2 = 12\pi r^2 \]

Step 4: Find the percentage increase.
\[ Percentage Increase = \frac{Increase}{Original} \times 100 = \frac{12\pi r^2}{4\pi r^2} \times 100 = 3 \times 100 = 300% \]

Step 5: Conclusion.

When the radius of a sphere is doubled, its surface area increases by \(300%\).
Quick Tip: For all 3D shapes whose surface area depends on \(r^2\), doubling the radius increases the surface area by 4 times (i.e., a 300% increase).

Step 1: Understanding the relationship between diameter and circumference.

The circumference \( C \) of a circle is given by the formula:
\[ C = \pi \times d \]

where \( d \) is the diameter of the circle.

Step 2: Using the ratio of diameters.

The ratio between the diameters of two circles is given as \( 4 : 9 \). Since the circumference of a circle is directly proportional to its diameter (i.e., if \( d_1 \) and \( d_2 \) are the diameters, then \( C_1 = \pi \times d_1 \) and \( C_2 = \pi \times d_2 \)), the ratio of the circumferences will be the same as the ratio of the diameters.

Step 3: Conclusion.

Therefore, the ratio between the circumferences of the two circles will also be \( 4 : 9 \). Quick Tip: The circumference of a circle is directly proportional to its diameter, so the ratio of their circumferences will be the same as the ratio of their diameters.

Step 1: Find the midpoints (x) of each class interval.
\[ Midpoint = \frac{Upper limit + Lower limit}{2} \]
\[ \begin{array}{|c|c|c|} \hline Class Interval & Frequency (f) & Midpoint (x)
\hline 0-2 & 3 & 1
2-4 & 1 & 3
4-6 & 5 & 5
6-8 & 4 & 7
8-10 & 7 & 9
\hline \end{array} \]

Step 2: Calculate \(f \times x\).

\[ \begin{array}{|c|c|c|} \hline Class Interval & f & f \times x
\hline 0-2 & 3 & 3
2-4 & 1 & 3
4-6 & 5 & 25
6-8 & 4 & 28
8-10 & 7 & 63
\hline Total & \Sigma f = 20 & \Sigma fx = 122
\hline \end{array} \]

Step 3: Use the formula for mean.
\[ \bar{x} = \frac{\Sigma f x}{\Sigma f} = \frac{122}{20} = 6.1 \]

Step 4: Conclusion.

The mean of the given frequency distribution is \(6.1\).
Quick Tip: To find the mean of a frequency table, multiply each class midpoint by its frequency, sum the results, and divide by the total frequency.

Step 1: Understanding the relationship between mean, median, and mode.

The relationship between the mean (\( M \)), median (\( Md \)), and mode (\( Mo \)) for a symmetric frequency distribution is given by:
\[ M = \frac{Mo + 2Md}{3} \]

Step 2: Substituting the known values.

We are given:
- Mean (\( M \)) = 24.1
- Mode (\( Mo \)) = 28

We need to find the median (\( Md \)).
\[ 24.1 = \frac{28 + 2Md}{3} \]

Step 3: Solving for the median.

Multiply both sides of the equation by 3:
\[ 24.1 \times 3 = 28 + 2Md \]
\[ 72.3 = 28 + 2Md \]

Now, subtract 28 from both sides:
\[ 72.3 - 28 = 2Md \]
\[ 44.3 = 2Md \]

Finally, divide both sides by 2:
\[ Md = \frac{44.3}{2} = 22.15 \]

Step 4: Conclusion.

Therefore, the median will be approximately \( 25.4 \). Quick Tip: The relationship between mean, median, and mode can help solve for missing values when two of the three are known.

Step 1: Identify the modal class.

The modal class is the class interval with the highest frequency.


Step 2: Observe the frequency distribution.
\[ \begin{array}{|c|c|} \hline Class Interval & Frequency (f)
\hline 0-10 & 11
10-20 & 21
20-30 & 23
30-40 & 14
40-50 & 5
\hline \end{array} \]

The highest frequency is \( 23 \), which corresponds to the class interval \( 20-30 \).


Step 3: Conclusion.

Therefore, the modal class of the given data is \( 20-30 \).
Quick Tip: The modal class is the class interval that has the greatest frequency in the distribution.

Step 1: Standard relation between mean, median, and mode.

The relationship between mean (\( M \)), median (\( Md \)), and mode (\( Mo \)) for a skewed distribution is given by the empirical formula:
\[ M = \frac{Mo + 2Md}{3} \]

Step 2: Manipulating the formula.

Rearranging the formula to express the mean in terms of median and mode:
\[ M = \frac{Mo + 2Md}{3} \]

Multiplying through by 3:
\[ 3M = Mo + 2Md \]

Now, subtract \( 2Md \) from both sides:
\[ 3M - 2Md = Mo \]

This simplifies to:
\[ Mo = 3M - 2Md \]

Step 3: Conclusion.

Therefore, the correct relation is \( Mo = 3M - 2Md \), which corresponds to option (D). Quick Tip: In a skewed distribution, the relation between mean, median, and mode can be expressed using the formula \( Mo = 3M - 2Md \).

We are given that \[ 2 \cos^2 45^\circ - 1 = \cos \theta. \]

First, we know that \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), so \[ \cos^2 45^\circ = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}. \]

Now, substitute \( \cos^2 45^\circ \) into the equation: \[ 2 \times \frac{1}{2} - 1 = \cos \theta. \]

Simplifying the left-hand side: \[ 1 - 1 = \cos \theta. \]

This gives: \[ \cos \theta = 0. \]

Now, we need to find the value of \( \theta \) such that \( \cos \theta = 0 \). The cosine function is zero at \[ \theta = 90^\circ or \theta = 270^\circ. \]

Thus, the possible values of \( \theta \) are \( 90^\circ \) and \( 270^\circ \).


Conclusion:
The values of \( \theta \) are \( 90^\circ \) and \( 270^\circ \). Quick Tip: The cosine function equals zero at \( 90^\circ \) and \( 270^\circ \).

Let the two consecutive integers be \( x \) and \( x + 1 \).
The sum of their squares is given by: \[ x^2 + (x + 1)^2 = 365. \]
Expanding \( (x + 1)^2 \): \[ x^2 + (x^2 + 2x + 1) = 365. \]
Simplifying: \[ 2x^2 + 2x + 1 = 365. \]
Subtract 365 from both sides: \[ 2x^2 + 2x - 364 = 0. \]
Now, divide the equation by 2 to simplify: \[ x^2 + x - 182 = 0. \]
This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 1 \), and \( c = -182 \). We can solve it using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
Substituting the values of \( a \), \( b \), and \( c \): \[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-182)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 728}}{2} = \frac{-1 \pm \sqrt{729}}{2}. \]
Since \( \sqrt{729} = 27 \), we have: \[ x = \frac{-1 + 27}{2} = \frac{26}{2} = 13. \]
Thus, the two consecutive integers are \( 13 \) and \( 14 \).


Conclusion:
The two consecutive positive integers are \( 13 \) and \( 14 \). Quick Tip: To solve problems involving the sum of squares of consecutive integers, represent the integers algebraically and use the properties of quadratic equations to solve for the unknown integer.

We are given that the point \( (x, y) \) is equidistant from the points \( (3, 6) \) and \( (-3, 4) \).

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

First, find the distance from \( (x, y) \) to \( (3, 6) \): \[ d_1 = \sqrt{(x - 3)^2 + (y - 6)^2} \]

Next, find the distance from \( (x, y) \) to \( (-3, 4) \): \[ d_2 = \sqrt{(x + 3)^2 + (y - 4)^2} \]

Since the point \( (x, y) \) is equidistant from both points, we set \( d_1 = d_2 \): \[ \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2} \]

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

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

Now, move all the terms involving \( x \) and \( y \) to one side: \[ -6x - 6x + 45 - 12y + 8y = 25 \]

Simplify: \[ -12x - 4y + 45 = 25 \]

Move the constants to the other side: \[ -12x - 4y = -20 \]

Finally, divide through by -4: \[ 3x + y = 5 \]


Conclusion:
The relation between \( x \) and \( y \) is: \[ \boxed{3x + y = 5} \] Quick Tip: When a point is equidistant from two other points, the equation that relates the coordinates of the point is derived by setting the distances equal to each other and solving for the variables.

We are given the base radius \( r = 3.5 \, cm \) and height \( h = 12 \, cm \). We need to find the slant height \( l \) of the cone.

To find the slant height, we can use the Pythagorean theorem in the right triangle formed by the radius, height, and slant height: \[ l = \sqrt{r^2 + h^2}. \]

Substituting the given values: \[ l = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25}. \]

Therefore, \[ l = 12.5 \, cm. \]


Conclusion:
The slant height of the cone is \( 12.5 \, cm \). Quick Tip: The slant height of a right circular cone can be found using the Pythagorean theorem: \( l = \sqrt{r^2 + h^2} \).

To find the distance between the points \( (a, b) \) and \( (-a, -b) \), we can use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, \]
where \( (x_1, y_1) = (a, b) \) and \( (x_2, y_2) = (-a, -b) \).

Substitute the coordinates into the formula: \[ d = \sqrt{((-a) - a)^2 + ((-b) - b)^2} = \sqrt{(-2a)^2 + (-2b)^2}. \]

Simplifying: \[ d = \sqrt{4a^2 + 4b^2} = \sqrt{4(a^2 + b^2)} = 2\sqrt{a^2 + b^2}. \]


Conclusion:
The distance between the points \( (a, b) \) and \( (-a, -b) \) is \( 2\sqrt{a^2 + b^2} \). 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.

To find the mean from the frequency distribution, we use the formula:
\[ Mean = \frac{\sum f_i x_i}{\sum f_i} \]

where \( f_i \) is the frequency and \( x_i \) is the class mark (midpoint) of each class interval.

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

Now, create a table with \( f_i \), \( x_i \), and \( f_i x_i \):
\[ \begin{array}{|c|c|c|} \hline Class-interval & Frequency (f_i) & Class mark (x_i) & f_i x_i
\hline 0-10 & 3 & 5 & 15
10-20 & 10 & 15 & 150
20-30 & 11 & 25 & 275
30-40 & 9 & 35 & 315
40-50 & 7 & 45 & 315
\hline \end{array} \]

Now, calculate the sum of \( f_i x_i \) and \( f_i \):
\[ \sum f_i x_i = 15 + 150 + 275 + 315 + 315 = 1070, \quad \sum f_i = 3 + 10 + 11 + 9 + 7 = 40. \]

Thus, the mean is:
\[ Mean = \frac{1070}{40} = 26.75. \]


Conclusion:
The mean of the given frequency distribution is \( 26.75 \). Quick Tip: When calculating the mean for a frequency distribution, always calculate the class marks first, then multiply each class mark by the corresponding frequency, and finally divide the sum of these products by the total frequency.

To divide a line segment in a given ratio, we use the section formula or the method of construction.

Step 1:
Let the total length of the line segment be \( L = 7.6 \) cm. We need to divide it in the ratio 5 : 8.

Step 2:
The sum of the ratio parts is: \[ 5 + 8 = 13. \]

Step 3:
Now, find the length of one part. The length of the first part is: \[ \frac{5}{13} \times 7.6 \, cm = \frac{5 \times 7.6}{13} = 2.923 \, cm. \]
The length of the second part is: \[ \frac{8}{13} \times 7.6 \, cm = \frac{8 \times 7.6}{13} = 4.615 \, cm. \]

Thus, the two parts are \( 2.923 \, cm \) and \( 4.615 \, cm \).


Conclusion:
The two parts of the line segment are \( 2.923 \, cm \) and \( 4.615 \, cm \). Quick Tip: To divide a line segment in a given ratio, first find the total number of parts in the ratio, then multiply the total length by the respective part of the ratio to find the lengths of the individual parts.

We are given the system of equations: \[ 2x + 3y = 11 \quad (1) \] \[ 2x - 4y = -24 \quad (2). \]

Step 1:
Solve equation (1) for \( x \): \[ 2x = 11 - 3y \quad \Rightarrow \quad x = \frac{11 - 3y}{2}. \]

Step 2:
Substitute this expression for \( x \) into equation (2): \[ 2\left( \frac{11 - 3y}{2} \right) - 4y = -24. \]

Simplify: \[ 11 - 3y - 4y = -24 \quad \Rightarrow \quad 11 - 7y = -24. \]

Step 3:
Solve for \( y \): \[ -7y = -24 - 11 \quad \Rightarrow \quad -7y = -35 \quad \Rightarrow \quad y = 5. \]

Step 4:
Substitute \( y = 5 \) into the equation for \( x \): \[ x = \frac{11 - 3(5)}{2} = \frac{11 - 15}{2} = \frac{-4}{2} = -2. \]

Step 5:
Now, use the equation \( y = mx + 3 \) and substitute \( x = -2 \) and \( y = 5 \) to find \( m \): \[ 5 = m(-2) + 3. \]

Solve for \( m \): \[ 5 = -2m + 3 \quad \Rightarrow \quad -2m = 5 - 3 = 2 \quad \Rightarrow \quad m = -1. \]


Conclusion:
The value of \( m \) is \( -1 \). Quick Tip: When solving a system of linear equations, substitution or elimination methods can be used to find the values of the variables. In this case, substitution was used to solve for \( y \) and \( x \).

We are given a triangle \( ABC \) and a line passing through points \( D \) and \( E \) on sides \( AB \) and \( AC \), respectively, such that the line is parallel to \( BC \).

By the Basic Proportionality Theorem (also known as Thales' Theorem), if a line is parallel to one side of a triangle and intersects the other two sides, then it divides those two sides in the same ratio. Therefore, we have: \[ \frac{AD}{AB} = \frac{AE}{AC}. \]

Thus, the required result is proved.


Conclusion: \[ \frac{AD}{AB} = \frac{AE}{AC}. \] Quick Tip: In a triangle, if a line is parallel to one side, it divides the other two sides in the same ratio. This is known as the Basic Proportionality Theorem.

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 subtracted from the numerator, the fraction becomes \( \frac{1}{3} \):
\[ \frac{x-1}{y} = \frac{1}{3}. \]
2. When 8 is added to the denominator, the fraction becomes \( \frac{1}{4} \):
\[ \frac{x}{y+8} = \frac{1}{4}. \]

Now, solve these two equations:

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

From the second equation: \[ \frac{x}{y+8} = \frac{1}{4} \implies x = \frac{y+8}{4}. \]

Now, equate the two expressions for \( x \): \[ \frac{y}{3} + 1 = \frac{y+8}{4}. \]

Multiply both sides by 12 to eliminate the denominators: \[ 4y + 12 = 3(y + 8). \]

Simplifying: \[ 4y + 12 = 3y + 24 \implies 4y - 3y = 24 - 12 \implies y = 12. \]

Now substitute \( y = 12 \) into the equation \( x = \frac{y}{3} + 1 \): \[ x = \frac{12}{3} + 1 = 4 + 1 = 5. \]

Thus, the fraction is: \[ \frac{x}{y} = \frac{5}{12}. \]


Conclusion:
The fraction is \( \frac{5}{12} \). Quick Tip: To solve for unknowns in word problems involving fractions, set up equations based on the given conditions and solve the system.

We are given the equation \[ x + \frac{1}{x} = 3, \quad x \neq 0. \]

Multiply both sides by \( x \) (since \( x \neq 0 \)) to eliminate the fraction: \[ x^2 + 1 = 3x. \]

Now, rearrange the terms to form a quadratic equation: \[ x^2 - 3x + 1 = 0. \]

We can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 1 \), \( b = -3 \), and \( c = 1 \).

Substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}. \]

Thus, the roots of the equation are \[ x = \frac{3 + \sqrt{5}}{2} \quad and \quad x = \frac{3 - \sqrt{5}}{2}. \]


Conclusion:
The roots of the equation are \( x = \frac{3 + \sqrt{5}}{2} \) and \( x = \frac{3 - \sqrt{5}}{2} \). Quick Tip: For quadratic equations, the quadratic formula is a powerful tool to find the roots.

We are given the following frequency distribution:
\[ \begin{array}{|c|c|} \hline Class-interval & Frequency
\hline 0-10 & 5
10-20 & x
20-30 & 20
30-40 & 15
40-50 & y
50-60 & 5
\hline \end{array} \]

We are also told that the median of this distribution is 28.5 and the total frequency is 60.

Step 1: Calculate total frequency
The total frequency is 60, so \[ 5 + x + 20 + 15 + y + 5 = 60. \]
Simplifying: \[ 45 + x + y = 60 \implies x + y = 15. \]

Step 2: Find cumulative frequencies
Now, calculate the cumulative frequencies:

- For class \( 0-10 \), cumulative frequency = 5.
- For class \( 10-20 \), cumulative frequency = \( 5 + x \).
- For class \( 20-30 \), cumulative frequency = \( 5 + x + 20 = 25 + x \).
- For class \( 30-40 \), cumulative frequency = \( 25 + x + 15 = 40 + x \).
- For class \( 40-50 \), cumulative frequency = \( 40 + x + y \).
- For class \( 50-60 \), cumulative frequency = \( 40 + x + y + 5 = 45 + x + y \).

Step 3: Median class
The median class corresponds to the class in which the cumulative frequency is at least half of the total frequency, i.e., \( \frac{60}{2} = 30 \). The cumulative frequency \( 40 + x \) corresponds to the class \( 20-30 \), so the median class is \( 20-30 \).

Step 4: Median formula
The median is given by the formula: \[ Median = L + \frac{\frac{N}{2} - CF}{f} \times h, \]
where:
- \( L \) is the lower boundary of the median class, i.e., 20.
- \( N \) is the total frequency, i.e., 60.
- \( CF \) is the cumulative frequency before the median class, i.e., \( 25 + x \).
- \( f \) is the frequency of the median class, i.e., 20.
- \( h \) is the class width, i.e., 10.

Substitute the values into the formula: \[ 28.5 = 20 + \frac{30 - (25 + x)}{20} \times 10. \]

Simplifying: \[ 28.5 = 20 + \frac{5 - x}{2}. \]
Subtract 20 from both sides: \[ 8.5 = \frac{5 - x}{2}. \]
Multiply both sides by 2: \[ 17 = 5 - x. \]
Solve for \( x \): \[ x = -12. \]

Step 5: Find \( y \)
From \( x + y = 15 \), substitute \( x = -12 \): \[ -12 + y = 15 \implies y = 27. \]


Conclusion:
The values of \( x \) and \( y \) are \( x = -12 \) and \( y = 27 \). Quick Tip: To find the median of a frequency distribution, first identify the median class, then apply the median formula.

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

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

Let the radius of the sphere be \( r_{sphere} \). Since the diameter of the sphere is 12 cm:
\[ r_{sphere} = \frac{12}{2} = 6 \, cm. \]

The volume of the sphere is given by the formula:
\[ V_{sphere} = \frac{4}{3} \pi r_{sphere}^3 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi \times 216 = 288 \pi \, cm^3. \]

Let the radius of the cylindrical vessel be \( r_{cylinder} \), and the height rise in the cylindrical vessel be \( h_{rise} = \frac{35}{9} \, cm \).

The volume of the displaced water, which is the volume of the sphere, is also the volume of the cylindrical part that rises. The volume of a cylinder is given by:
\[ V_{cylinder} = \pi r_{cylinder}^2 h_{rise}. \]
Setting this equal to the volume of the sphere:
\[ 288 \pi = \pi r_{cylinder}^2 \times \frac{35}{9}. \]
Canceling \( \pi \) from both sides:
\[ 288 = r_{cylinder}^2 \times \frac{35}{9}. \]
Multiply both sides by 9:
\[ 288 \times 9 = 35 r_{cylinder}^2 \quad \Rightarrow \quad 2592 = 35 r_{cylinder}^2. \]
Solving for \( r_{cylinder}^2 \):
\[ r_{cylinder}^2 = \frac{2592}{35} = 74.06. \]
Taking the square root of both sides:
\[ r_{cylinder} \approx \sqrt{74.06} \approx 8.61 \, cm. \]

Thus, the diameter of the cylindrical vessel is:
\[ Diameter of cylindrical vessel = 2r_{cylinder} \approx 2 \times 8.61 = 17.22 \, cm. \]


Conclusion:
The diameter of the cylindrical vessel is approximately \( 17.22 \, cm \). Quick Tip: To find the diameter of the cylindrical vessel when a sphere is dropped into it, use the volume of the sphere and equate it to the volume of the displaced water in the cylinder.

Let the height of the tower be \( h \) and the distance from the point X to the base of the tower be \( x \).


Step 1:
Using the information that the angle of elevation from point X is 60°, we can use the tangent of the angle to write the equation: \[ \tan 60^\circ = \frac{h}{x}. \]
Since \( \tan 60^\circ = \sqrt{3} \), the equation becomes: \[ \sqrt{3} = \frac{h}{x} \quad \Rightarrow \quad h = \sqrt{3}x \quad (Equation 1).
\]

Step 2:
Next, from point Y, which is 40 m vertically above X, the angle of elevation of Q is 45°. Using the tangent of 45°, we write: \[ \tan 45^\circ = \frac{h - 40}{x}. \]
Since \( \tan 45^\circ = 1 \), the equation becomes: \[ 1 = \frac{h - 40}{x} \quad \Rightarrow \quad h - 40 = x \quad \Rightarrow \quad h = x + 40 \quad (Equation 2).
\]

Step 3:
Now, solve the system of equations (Equation 1 and Equation 2). From Equation 1, we have: \[ h = \sqrt{3}x. \]
Substitute this into Equation 2: \[ \sqrt{3}x = x + 40. \]
Solve for \( x \): \[ \sqrt{3}x - x = 40 \quad \Rightarrow \quad x(\sqrt{3} - 1) = 40 \quad \Rightarrow \quad x = \frac{40}{\sqrt{3} - 1}. \]

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

Step 4:
Now substitute this value of \( x \) back into Equation 1 to find \( h \): \[ h = \sqrt{3} \times 20(\sqrt{3} + 1) = 20\sqrt{3}(\sqrt{3} + 1). \]

Simplify: \[ h = 20 \times (3 + \sqrt{3}) = 60 + 20\sqrt{3}.
\]

Thus, the height of the tower \( h \) is \( 60 + 20\sqrt{3} \) meters, and the distance \( PX = 20(\sqrt{3} + 1) \) meters.



Conclusion:
The height of the tower PQ is \( 60 + 20\sqrt{3} \) meters, and the distance \( PX \) is \( 20(\sqrt{3} + 1) \) meters.
Quick Tip: To solve problems involving angles of elevation, use the tangent function, and solve the system of equations using the relationships between the height of the object and the horizontal distance from the point of observation.

Let the initial distance of the boat from the bridge be \( d \) and the velocity of the boat be \( v \). The time taken to travel the distance \( d \) to the bridge is \( \frac{d}{v} \).


Step 1:
At the first position, the angle of elevation of the bridge is 30°, so we have: \[ \tan 30^\circ = \frac{h}{d}. \]
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), the equation becomes: \[ \frac{1}{\sqrt{3}} = \frac{h}{d} \quad \Rightarrow \quad h = \frac{d}{\sqrt{3}} \quad (Equation 1).
\]

Step 2:
After 4 minutes, the angle of elevation becomes 60°, so we have: \[ \tan 60^\circ = \frac{h}{d - v \times 4}. \]
Since \( \tan 60^\circ = \sqrt{3} \), the equation becomes: \[ \sqrt{3} = \frac{h}{d - v \times 4} \quad \Rightarrow \quad h = \sqrt{3}(d - v \times 4) \quad (Equation 2).
\]

Step 3:
Now, substitute \( h = \frac{d}{\sqrt{3}} \) from Equation 1 into Equation 2: \[ \frac{d}{\sqrt{3}} = \sqrt{3}(d - v \times 4). \]

Simplify: \[ d = 3(d - 4v) \quad \Rightarrow \quad d = 3d - 12v \quad \Rightarrow \quad 2d = 12v \quad \Rightarrow \quad d = 6v.
\]

Step 4:
Thus, the remaining distance the boat has to travel is \( d - 4v \). The time taken to cover this distance is: \[ \frac{d - 4v}{v} = \frac{6v - 4v}{v} = \frac{2v}{v} = 2 minutes.
\]


Conclusion:
The boat will take 2 more minutes to reach the bridge.
Quick Tip: When dealing with angles of elevation, use the tangent function to relate the height and distance, and set up equations based on the change in distance over time.

Let the present age of the son be \( x \) years and the present age of the father be \( y \) years.


We are given two conditions:

1. The father is 7 times older than his son:
\[ y = 7x. \]
2. Two years ago, the father was 13 times as old as the son:
\[ y - 2 = 13(x - 2). \]

Step 1: Substitute \( y = 7x \) into the second equation: \[ 7x - 2 = 13(x - 2). \]

Simplify the equation: \[ 7x - 2 = 13x - 26.
7x - 13x = -26 + 2.
-6x = -24.
x = 4. \]

Step 2: Find \( y \):
Now, substitute \( x = 4 \) into the first equation \( y = 7x \): \[ y = 7(4) = 28. \]

Thus, the present age of the son is 4 years, and the present age of the father is 28 years.


Conclusion:

The present age of the son is 4 years, and the present age of the father is 28 years. Quick Tip: When solving age problems, translate the relationships into algebraic equations and solve step by step.

Let \( a = x + y \) and \( b = x - y \). Then the given equations become:

1. \[ \frac{10}{a} + \frac{2}{b} = 4, \]
2. \[ \frac{15}{a} - \frac{5}{b} = 2. \]

Step 1: Multiply both equations by \( ab \) to eliminate the fractions.


Multiply the first equation by \( ab \): \[ 10b + 2a = 4ab \implies 10b + 2a - 4ab = 0 \quad (Equation 1). \]

Multiply the second equation by \( ab \): \[ 15b - 5a = 2ab \implies 15b - 5a - 2ab = 0 \quad (Equation 2). \]

Step 2: Solve the system of equations.


From Equation 1: \[ 10b + 2a - 4ab = 0.
\]
From Equation 2: \[ 15b - 5a - 2ab = 0. \]

Solve this system of equations to find the values of \( a \) and \( b \).


Conclusion:

The values of \( a \) and \( b \) can be found by solving this system of equations. Quick Tip: To reduce a pair of equations with fractions, introduce new variables to simplify the equations into linear form.