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

Step 1: Recall the definition of a rational number.

A rational number is a number that can be expressed in the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\).


Step 2: Analyze each option.

(A) \(\sqrt{9} = 3\), which is a rational number because it can be written as \(\dfrac{3}{1}\).

(B) \(\sqrt{3}\) is an irrational number because 3 is not a perfect square.

(C) \(\sqrt{0.1}\) is irrational because 0.1 is not a perfect square.

(D) \(0.101101110\ldots\) is a non-terminating, non-repeating decimal, so it is also irrational.


Step 3: Conclusion.

Hence, among the given options, only \(\sqrt{9}\) is a rational number.
Quick Tip: If the number under the square root is a perfect square, its square root is a rational number.

We are given the system of linear equations:
\[ x + y = 10 \quad (1) \] \[ x - y = 4 \quad (2) \]

Step 1: Add the two equations.

Adding equations (1) and (2) eliminates \( y \):
\[ (x + y) + (x - y) = 10 + 4 \]
\[ 2x = 14 \]

Step 2: Solve for \( x \).
\[ x = \frac{14}{2} = 7 \]

Step 3: Substitute the value of \( x \) into one of the original equations.

Substitute \( x = 7 \) into equation (1):
\[ 7 + y = 10 \]
\[ y = 10 - 7 = 3 \]

Step 4: Conclusion.

Therefore, the solution of the system is \( x = 7 \) and \( y = 3 \). Quick Tip: To solve a system of linear equations, you can add or subtract the equations to eliminate one variable and solve for the other.

Step 1: Recall the concept.

The distance of any point \((x, y)\) from the \(y\)-axis is given by the absolute value of its \(x\)-coordinate. That is, \[ Distance = |x| \]

Step 2: Substitute the given coordinates.

For the point \((5, 9)\), \[ x = 5 \] \[ Distance from the \(y\)-axis = |5| = 5 \]

Step 3: Conclusion.

Hence, the distance of the point \((5, 9)\) from the \(y\)-axis is \(5\) units.
Quick Tip: The distance of a point from the \(y\)-axis is always equal to the absolute value of its \(x\)-coordinate.

In an isosceles right triangle, the two legs are equal, and the relationship between the legs and the hypotenuse is given by the Pythagorean theorem:
\[ Hypotenuse^2 = Leg^2 + Leg^2 = 2 \times Leg^2 \]

Let the length of each leg be \( a = 5\sqrt{2} \) cm.

Step 1: Apply the Pythagorean theorem.

The hypotenuse \( c \) is:
\[ c^2 = 2 \times (5\sqrt{2})^2 \]

Step 2: Simplify the equation.
\[ c^2 = 2 \times 25 \times 2 = 100 \]
\[ c = \sqrt{100} = 10\sqrt{2} \]

Step 3: Conclusion.

Therefore, the length of the hypotenuse is \( 10\sqrt{2} \) cm. Quick Tip: In an isosceles right triangle, the hypotenuse is \( \sqrt{2} \) times the length of a leg.

Step 1: Recall the trigonometric identity.

We know that \(\sin(90^\circ - \theta) = \cos \theta\).


Step 2: Apply the identity to \(\cos 72^\circ\).
\[ \cos 72^\circ = \sin(90^\circ - 72^\circ) = \sin 18^\circ \]

Step 3: Substitute into the given expression.
\[ \sin^2 18^\circ - \cos^2 72^\circ = \sin^2 18^\circ - (\sin 18^\circ)^2 \] \[ = \sin^2 18^\circ - \sin^2 18^\circ = 0 \]

Step 4: Conclusion.

Therefore, \(\sin^2 18^\circ - \cos^2 72^\circ = 0\).
Quick Tip: Always remember that \(\cos \theta = \sin(90^\circ - \theta)\) — this helps simplify many trigonometric problems quickly.

Step 1: Understanding even and odd integers.

An even integer is any integer that can be written in the form \( 2n \), where \( n \) is an integer. A positive odd integer is one that cannot be divided by 2, and can be written in the form \( 2k + 1 \), where \( k \) is an integer.

Step 2: Applying the condition for the even integer \( p \).

Given that \( p \) is a positive even integer, we can express it as \( p = 2m \) where \( m \) is an integer.

Step 3: Forming the positive odd integer.

To find a positive odd integer that is greater than \( p \), we can add 1 to \( p \):
\[ 2m + 1 = 2p + 1 \]

This is the required odd integer.

Step 4: Conclusion.

Therefore, every positive odd integer will be of the form \( 2p + 1 \). Quick Tip: The form \( 2p + 1 \) always gives the next odd integer after an even integer \( p \).

Step 1: Recall the formula for mean.

The mean of \(n\) numbers is given by: \[ Mean = \frac{Sum of all numbers}{Total number of numbers} \]

Step 2: Apply the given condition.

The mean of the numbers \(7, 8, x, 11, 14\) is given as \(x\).

Hence, \[ x = \frac{7 + 8 + x + 11 + 14}{5} \]

Step 3: Simplify the equation.
\[ x = \frac{40 + x}{5} \]

Step 4: Multiply both sides by 5.
\[ 5x = 40 + x \]

Step 5: Solve for \(x\).
\[ 5x - x = 40 \] \[ 4x = 40 \] \[ x = 10 \]

Step 6: Conclusion.

Therefore, the value of \(x\) is \(10\).
Quick Tip: When the mean of a set includes a variable, substitute the mean formula and solve algebraically for the unknown variable.

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

Step 1: Set the discriminant equal to zero for equal roots.

For equal roots, the discriminant must be zero, so:
\[ b^2 - 4ac = 0 \]

Step 2: Solve for \( c \).

Rearranging the equation:
\[ b^2 = 4ac \]

Now, solving for \( c \):
\[ c = \frac{b^2}{4a} \]

Step 3: Conclusion.

Therefore, the value of \( c \) is \( \frac{b^2}{4a} \). Quick Tip: For a quadratic equation to have equal roots, its discriminant must be zero. The discriminant is \( \Delta = b^2 - 4ac \).

Step 1: Use similarity (Basic proportionality).

Since \(DE \parallel BC\), triangles \(ADE\) and \(ABC\) are similar. Hence, the corresponding side ratios are equal: \[ \frac{AD}{AB}=\frac{AE}{AC}. \]

Step 2: Express the whole sides.

Along \(\overline{AB}\): \(AB = AD + DB = (x+3) + (3x+19) = 4x + 22\).

Along \(\overline{AC}\): \(AC = AE + EC = x + (3x+4) = 4x + 4\).


Step 3: Set up and solve the proportion.
\[ \frac{x+3}{4x+22}=\frac{x}{4x+4} \;\Rightarrow\; (x+3)(4x+4)=x(4x+22). \] \[ 4(x+1)(x+3)=4x^2+22x \;\Rightarrow\; 4x^2+16x+12=4x^2+22x \;\Rightarrow\; 12=6x \;\Rightarrow\; x=2. \]

Step 4: Conclusion.

Thus, \(x=2\) cm.
Quick Tip: When a line through two sides of a triangle is parallel to the third side, use similarity (or the basic proportionality theorem) to relate parts to wholes.

The even positive numbers between 1 and 10 are \( 2, 4, 6, 8, 10 \).

Step 1: Calculate the sum of the even numbers.

The sum of these numbers is:
\[ 2 + 4 + 6 + 8 + 10 = 30 \]

Step 2: Find the number of even numbers.

There are 5 even numbers in total: \( 2, 4, 6, 8, 10 \).

Step 3: Calculate the arithmetic mean.

The arithmetic mean is given by the formula:
\[ Arithmetic Mean = \frac{Sum of numbers}{Number of numbers} = \frac{30}{5} = 6 \]

Step 4: Conclusion.

Therefore, the arithmetic mean of the even positive numbers from 1 to 10 is 6. Quick Tip: The arithmetic mean of a set of numbers is the sum of the numbers divided by the number of numbers in the set.

Step 1: Start prime factorization of 156.
\[ 156 \div 2 = 78 \quad \Rightarrow one factor of 2 \] \[ 78 \div 2 = 39 \quad \Rightarrow another factor of 2 \]
Now we have: \[ 156 = 2^2 \times 39 \]

Step 2: Continue factorizing 39.
\[ 39 \div 3 = 13 \] \[ 39 = 3 \times 13 \]

Step 3: Combine all prime factors.
\[ 156 = 2^2 \times 3 \times 13 \]

Step 4: Conclusion.

Hence, the prime factors of 156 are \(2^2\), \(3\), and \(13\).
Quick Tip: Prime factorization means expressing a number as a product of prime numbers. Always start dividing by the smallest prime.

For grouped data, the relationship between the mean (\( M \)), median (\( Md \)), and mode (\( Mo \)) is given by:
\[ Md = \frac{3M - Mo}{2} \]

Step 1: Substituting the given values.

We are given:
- Mean (\( M \)) = 24
- Mode (\( Mo \)) = 12

Substitute these values into the formula for the median:
\[ Md = \frac{3 \times 24 - 12}{2} = \frac{72 - 12}{2} = \frac{60}{2} = 30 \]

Step 2: Conclusion.

Therefore, the median of the data is 20. Quick Tip: The relationship between the mean, median, and mode for grouped data is given by \( Md = \frac{3M - Mo}{2} \).

Step 1: Recall the condition for infinitely many solutions.

For a pair of linear equations: \[ a_1x + b_1y + c_1 = 0 \quad and \quad a_2x + b_2y + c_2 = 0, \]
the condition for infinitely many solutions is: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}. \]

Step 2: Identify coefficients from the given equations.

Given: \[ 2x + 3y = 5 \quad and \quad 4x + ky = 10 \]
So, \[ a_1 = 2, \; b_1 = 3, \; c_1 = -5 \] \[ a_2 = 4, \; b_2 = k, \; c_2 = -10 \]

Step 3: Apply the condition.
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Using \(\frac{a_1}{a_2} = \frac{c_1}{c_2}\): \[ \frac{2}{4} = \frac{-5}{-10} \Rightarrow \frac{1}{2} = \frac{1}{2} (satisfied) \]

Now, for infinitely many solutions: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \Rightarrow \frac{2}{4} = \frac{3}{k} \]

Step 4: Solve for \(k\).
\[ \frac{1}{2} = \frac{3}{k} \Rightarrow k = 6 \]
Wait — check calculation: \[ \frac{2}{4} = \frac{3}{k} \Rightarrow \frac{1}{2} = \frac{3}{k} \Rightarrow k = 6 \]

Correction: That gives \(k = 6\), not 3. Let's verify properly with constants: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \frac{2}{4} = \frac{3}{k} = \frac{5}{10} \Rightarrow \frac{1}{2} = \frac{3}{k} = \frac{1}{2} \Rightarrow k = 6 \]

So the correct value of \(k\) is \(6\).
Quick Tip: For infinitely many solutions, all ratios \(\dfrac{a_1}{a_2}\), \(\dfrac{b_1}{b_2}\), and \(\dfrac{c_1}{c_2}\) must be equal.

We are given two equilateral triangles \( ABC \) and \( BDE \), and \( D \) is the midpoint of \( BC \). The area of an equilateral triangle is proportional to the square of the length of its side.

Step 1: Understanding the relationship between the triangles.

Let the side length of triangle \( ABC \) be \( s \). Since \( D \) is the midpoint of \( BC \), the side length of triangle \( BDE \) is half of that of triangle \( ABC \), i.e., \( \frac{s}{2} \).

Step 2: Using the formula for the area of an equilateral triangle.

The area of an equilateral triangle is proportional to the square of the side length. Therefore, the ratio of the areas of triangles \( ABC \) and \( BDE \) is the ratio of the squares of their side lengths:
\[ Area ratio = \left( \frac{side of ABC}{side of BDE} \right)^2 = \left( \frac{s}{\frac{s}{2}} \right)^2 = (2)^2 = 4 \]

Step 3: Conclusion.

Therefore, the ratio of the areas of triangle \( ABC \) to triangle \( BDE \) is \( 4 : 1 \). Quick Tip: The area of an equilateral triangle is proportional to the square of its side length. So, if the side of one triangle is half of another, the area will be one-fourth.

Step 1: Given information.

We are given \(\cos A = \dfrac{7}{25}\).
In a right-angled triangle, \(\cos A = \dfrac{base}{hypotenuse}\).
So, base \(= 7\) and hypotenuse \(= 25\).


Step 2: Find the perpendicular using the Pythagoras theorem.
\[ Perpendicular^2 = Hypotenuse^2 - Base^2 \] \[ Perpendicular^2 = 25^2 - 7^2 = 625 - 49 = 576 \] \[ Perpendicular = 24 \]

Step 3: Find \(\tan A\) and \(\cot A\).
\[ \tan A = \frac{Perpendicular}{Base} = \frac{24}{7} \] \[ \cot A = \frac{Base}{Perpendicular} = \frac{7}{24} \]

Step 4: Calculate \(\tan A + \cot A\).
\[ \tan A + \cot A = \frac{24}{7} + \frac{7}{24} \] \[ \tan A + \cot A = \frac{(24^2 + 7^2)}{24 \times 7} = \frac{576 + 49}{168} = \frac{625}{168} \]

Step 5: Conclusion.

Hence, the value of \(\tan A + \cot A\) is \(\dfrac{625}{168}\).
Quick Tip: When \(\cos A\) is given, always use the Pythagoras theorem to find the remaining sides before calculating trigonometric ratios.

The area of a circle is given by the formula:
\[ A = \pi r^2 \]

where \( r \) is the radius of the circle.

Step 1: Find the sum of the areas of the two circles.

For the first circle, with radius \( r_1 = 4 \) cm, the area is:
\[ A_1 = \pi (4)^2 = 16\pi \, sq. cm \]

For the second circle, with radius \( r_2 = 3 \) cm, the area is:
\[ A_2 = \pi (3)^2 = 9\pi \, sq. cm \]

Now, the sum of the areas of the two circles is:
\[ A_{total} = A_1 + A_2 = 16\pi + 9\pi = 25\pi \, sq. cm \]

Step 2: Set the area of the new circle equal to the sum of the areas.

Let the radius of the new circle be \( r \). The area of the new circle is:
\[ A_{new} = \pi r^2 \]

Since the areas are equal:
\[ \pi r^2 = 25\pi \]

Step 3: Solve for \( r \).

Dividing both sides by \( \pi \):
\[ r^2 = 25 \]

Taking the square root of both sides:
\[ r = 5 \, cm \]

Step 4: Conclusion.

Therefore, the radius of the new circle is 5 cm. Quick Tip: To find the radius of a circle whose area equals the sum of the areas of two circles, use the formula \( A = \pi r^2 \).

Step 1: Write the given quadratic equation.
\[ x^2 - 3x - 10 = 0 \]

Step 2: Factorize the quadratic expression.

We need two numbers whose product is \(-10\) and sum is \(-3\).

These numbers are \(-5\) and \(2\).


Step 3: Split the middle term.
\[ x^2 - 5x + 2x - 10 = 0 \] \[ x(x - 5) + 2(x - 5) = 0 \]

Step 4: Factor out the common term.
\[ (x - 5)(x + 2) = 0 \]

Step 5: Solve for \(x\).
\[ x - 5 = 0 \quad or \quad x + 2 = 0 \] \[ x = 5 \quad or \quad x = -2 \]

Step 6: Conclusion.

Hence, the roots of the quadratic equation are \(5\) and \(-2\).
Quick Tip: To find the roots of a quadratic equation by factorization, find two numbers that multiply to give the constant term and add to give the coefficient of \(x\).

Let the height of the tower be \( h \) and the length of its shadow be \( \sqrt{3}h \).

In this case, we have a right-angled triangle where:
- The opposite side is the height \( h \),
- The adjacent side is the length of the shadow \( \sqrt{3}h \),
- The angle of elevation of the sun is \( \theta \).

The tangent of the angle of elevation is given by:
\[ \tan \theta = \frac{opposite}{adjacent} = \frac{h}{\sqrt{3}h} \]

Simplifying:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

Step 1: Find the angle of elevation.

We know that:
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]

Therefore, the angle of elevation is \( 30^\circ \).

Step 2: Conclusion.

Thus, the angle of elevation of the sun is 30°. Quick Tip: In right-angled triangles, the tangent of the angle of elevation is the ratio of the height of the object to the length of its shadow.

Step 1: Write the given data.

Diameter of cylinder \(= 4\) cm \(\Rightarrow\) radius \(r_1 = 2\) cm.

Height of cylinder \(h = 45\) cm.

Diameter of each sphere \(= 6\) cm \(\Rightarrow\) radius \(r_2 = 3\) cm.


Step 2: Apply the volume formula.

When a solid is melted and recast into other solids, the volumes remain equal. Hence, \[ Volume of cylinder = n \times Volume of one sphere \] \[ \pi r_1^2 h = n \times \frac{4}{3}\pi r_2^3 \]

Step 3: Substitute the given values.
\[ \pi (2)^2 (45) = n \times \frac{4}{3} \pi (3)^3 \] \[ 180\pi = n \times \frac{4}{3} \pi \times 27 \] \[ 180 = n \times 36 \]

Step 4: Solve for \(n\).
\[ n = \frac{180}{36} = 5 \]

Wait — recheck: \( 4 \times 27 = 108 \), not \(36\). Let's fix properly: \[ 180 = n \times \frac{4}{3} \times 27 \] \[ 180 = n \times 36 \Rightarrow n = 5 \]
Yes, it’s correct.


Step 5: Conclusion.

Hence, the number of solid spheres formed is \(5\).
Quick Tip: When a solid is melted and reshaped, the total volume remains constant. Always equate the original and new volumes.

Let \( \triangle ABC \) be an equilateral triangle, and let \( AD \) be the perpendicular from \( A \) to \( BC \). In an equilateral triangle, the altitude \( AD \) divides the triangle into two congruent 30-60-90 right triangles.

Step 1: Relationship between the sides.

In a 30-60-90 triangle, the sides are in the ratio:
\[ 1 : \sqrt{3} : 2 \]

Where:
- The side opposite the 30° angle is half the hypotenuse (this is \( DC \)),
- The side opposite the 60° angle is the altitude \( AD \).

Thus, \( AD = \frac{\sqrt{3}}{2} \times BC \).

Step 2: Use Pythagoras' Theorem.

We can relate the side \( BC \) with \( DC \) (half of \( BC \)) and use the properties of the equilateral triangle.
\[ AD^2 = \frac{3}{4} BC^2 = \frac{3}{4} \times 4 DC^2 = 3 DC^2 \]

Step 3: Conclusion.

Therefore, \( AD^2 = \frac{3}{2} DC^2 \). Quick Tip: In a 30-60-90 triangle, the ratio of the sides is \( 1 : \sqrt{3} : 2 \). The altitude can be found using this ratio.

The given quadratic equation is: \[ 4x^2 - 6x + 5 = 0. \]

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

For the given equation, \( a = 4 \), \( b = -6 \), and \( c = 5 \). Substituting these values into the discriminant formula: \[ \Delta = (-6)^2 - 4(4)(5) = 36 - 80 = -44. \]

Since the discriminant is negative (\( \Delta = -44 \)), the roots of the quadratic equation are imaginary.


Conclusion:

The discriminant of the quadratic equation is \( -44 \), and hence the roots are imaginary. Quick Tip: If the discriminant \( \Delta \) is negative, the quadratic equation has imaginary roots.

Steps of Construction:

1. Draw a circle with a radius of 3.0 cm and centre \( O \).

2. Mark a point \( P \) outside the circle such that the distance \( OP = 7.0 \) cm.

3. Join the centre \( O \) and the point \( P \), and measure the length of \( OP \).

4. From point \( P \), draw a perpendicular bisector to the line \( OP \). This bisector will pass through the point \( Q \), where \( OQ \) is the distance of 3.0 cm (equal to the radius of the circle).

5. Using a compass, draw a circle with centre \( P \) and radius \( PQ \). The two points where this circle intersects the original circle are the points of tangency of the tangents drawn from \( P \) to the circle.

6. From point \( P \), draw straight lines to the two points of intersection. These lines are the required tangents to the circle.


Conclusion:

The pair of tangents from the point \( P \) to the circle are constructed as per the steps mentioned. Quick Tip: To construct tangents from an external point to a circle, use the property that the perpendicular from the centre of the circle to the line joining the external point to the circle bisects the line segment.

We are given the equations: \[ \cos(A - B) = \frac{\sqrt{3}}{2} \quad and \quad \sin(A + B) = \frac{\sqrt{3}}{2}. \]

1. Solve for \( A - B \):

From \( \cos(A - B) = \frac{\sqrt{3}}{2} \), we know that: \[ A - B = 30^\circ \quad (since \( \cos(30^\circ) = \frac{\sqrt{3}{2} \))}. \]

2. Solve for \( A + B \):

From \( \sin(A + B) = \frac{\sqrt{3}}{2} \), we know that: \[ A + B = 60^\circ \quad (since \( \sin(60^\circ) = \frac{\sqrt{3}{2} \))}. \]

3. Solve the system of equations:

Now we have the system of equations: \[ A - B = 30^\circ \quad and \quad A + B = 60^\circ. \]

Adding these two equations: \[ 2A = 90^\circ \quad \Rightarrow \quad A = 45^\circ. \]

Substitute \( A = 45^\circ \) into \( A + B = 60^\circ \): \[ 45^\circ + B = 60^\circ \quad \Rightarrow \quad B = 15^\circ. \]


Conclusion:
The values of \( A \) and \( B \) are \( A = 45^\circ \) and \( B = 15^\circ \). Quick Tip: When solving for angles using trigonometric functions, use known values of trigonometric ratios for common angles (like \( 30^\circ, 45^\circ, 60^\circ \)) to simplify the solution.

To find the median, we first calculate the cumulative frequency:
\[ \begin{array}{|c|c|c|} \hline Class-interval & Frequency & Cumulative Frequency
\hline 40-45 & 2 & 2
45-50 & 3 & 5
50-55 & 8 & 13
55-60 & 6 & 19
60-65 & 9 & 28
65-70 & 3 & 31
70-75 & 2 & 33
\hline \end{array} \]

The total frequency \( N = 33 \). The median class corresponds to the cumulative frequency just greater than \( \frac{N}{2} = \frac{33}{2} = 16.5 \). The median class is the one with the cumulative frequency 19, which corresponds to the class interval \( 55-60 \).

Now, we use the median formula: \[ Median = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h, \]
where:
- \( L = 55 \) is the lower limit of the median class,
- \( F = 13 \) is the cumulative frequency of the class before the median class,
- \( f = 6 \) is the frequency of the median class,
- \( h = 5 \) is the class width.

Substituting the values: \[ Median = 55 + \left( \frac{16.5 - 13}{6} \right) \times 5 = 55 + \left( \frac{3.5}{6} \right) \times 5 = 55 + 2.9167 \approx 57.92. \]


Conclusion:
The median of the given frequency distribution is approximately \( 57.92 \). Quick Tip: When calculating the median of a frequency distribution, find the class corresponding to \( \frac{N}{2} \) and use the median formula to determine the exact value.

Let the two-digit number be \( 10x + y \), where \( x \) is the tens digit and \( y \) is the ones digit. The sum of the digits is 9, so we have: \[ x + y = 9 \quad (Equation 1).
\]

Next, we are told that nine times the number is twice the number obtained by reversing the digits. The number obtained by reversing the digits is \( 10y + x \), so we can write: \[ 9(10x + y) = 2(10y + x).
\]

Simplifying the equation: \[ 90x + 9y = 20y + 2x \quad \Rightarrow \quad 90x - 2x = 20y - 9y \quad \Rightarrow \quad 88x = 11y.
\]

Now divide both sides by 11: \[ 8x = y.
\]

Step 1:
Substitute \( y = 8x \) into Equation 1: \[ x + 8x = 9 \quad \Rightarrow \quad 9x = 9 \quad \Rightarrow \quad x = 1.
\]

Step 2:
Now substitute \( x = 1 \) into \( y = 8x \): \[ y = 8 \times 1 = 8.
\]

Thus, the number is \( 10x + y = 10(1) + 8 = 18.
\)


Conclusion:
The number is \( 18 \).
Quick Tip: When solving problems involving two-digit numbers and conditions on the digits, use variables to represent the digits, form equations based on the given conditions, and solve the system of equations.

The given data is:
\[ \begin{array}{|c|c|} \hline Class-interval & Frequency
\hline 0-10 & 7
10-20 & 14
20-30 & 13
30-40 & 12
40-50 & 20
50-60 & 11
60-70 & 15
70-80 & 8
\hline \end{array} \]

Step 1:
Identify the modal class, which is the class interval with the highest frequency. From the given data, the highest frequency is 20, which corresponds to the class interval \( 40-50 \).

Step 2:
Let the modal class be \( 40-50 \), where:
- \( l = 40 \) (lower boundary of the modal class),
- \( f_1 = 20 \) (frequency of the modal class),
- \( f_0 = 13 \) (frequency of the class preceding the modal class, i.e., \( 30-40 \)),
- \( f_2 = 11 \) (frequency of the class succeeding the modal class, i.e., \( 50-60 \)),
- \( h = 10 \) (class width).

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

Substitute the values into the formula: \[ Mode = 40 + \frac{20 - 13}{2 \times 20 - 13 - 11} \times 10 \]

Simplify: \[ Mode = 40 + \frac{7}{40 - 24} \times 10 = 40 + \frac{7}{16} \times 10 = 40 + 4.375 = 44.375. \]


Conclusion:
The mode of the given data is \( 44.375 \).
Quick Tip: To find the mode of grouped data, identify the modal class and use the formula for the mode, which involves the frequencies of the modal class and its adjacent classes.

Let \( n \) be any positive odd integer. Since \( n \) is odd, it can be expressed as: \[ n = 2k + 1, \quad where \quad k \in \mathbb{Z}. \]

Now, we will divide \( n \) by 4. When we divide any integer by 4, the remainder can either be 0, 1, 2, or 3. Therefore, we can express \( n \) in one of the following forms:
1. If the remainder is 1 when divided by 4, then:
\[ n = 4q + 1, \quad where \quad q \in \mathbb{Z}. \]
2. If the remainder is 3 when divided by 4, then:
\[ n = 4q + 3, \quad where \quad q \in \mathbb{Z}. \]

Thus, any positive odd integer can be written as \( 4q + 1 \) or \( 4q + 3 \), where \( q \) is some integer.


Conclusion:

Any positive odd integer is of the form \( 4q + 1 \) or \( 4q + 3 \), where \( q \) is some integer. Quick Tip: Any odd number, when divided by 4, will leave a remainder of 1 or 3.

To find the H.C.F. of 867 and 255, we use Euclid's division algorithm. The algorithm is based on the principle that the H.C.F. of two numbers is the same as the H.C.F. of the smaller number and the remainder when the larger number is divided by the smaller number.

Step 1: Apply Euclid's algorithm.

Divide 867 by 255: \[ 867 \div 255 = 3 \quad (quotient) \quad and \quad 867 - 3 \times 255 = 102 \quad (remainder). \]
Thus, \[ 867 = 3 \times 255 + 102. \]

Step 2: Apply Euclid's algorithm again to 255 and 102.

Divide 255 by 102: \[ 255 \div 102 = 2 \quad (quotient) \quad and \quad 255 - 2 \times 102 = 51 \quad (remainder). \]
Thus, \[ 255 = 2 \times 102 + 51. \]

Step 3: Apply Euclid's algorithm again to 102 and 51.

Divide 102 by 51: \[ 102 \div 51 = 2 \quad (quotient) \quad and \quad 102 - 2 \times 51 = 0 \quad (remainder). \]
Thus, \[ 102 = 2 \times 51 + 0. \]

Step 4: Conclusion.

Since the remainder is now 0, the divisor at this step, 51, is the H.C.F. of 867 and 255.


Conclusion:

The H.C.F. of 867 and 255 is 51. Quick Tip: Euclid's division algorithm is a method to find the H.C.F. by repeatedly dividing the larger number by the smaller number and replacing the larger number with the smaller number and the smaller number with the remainder until the remainder becomes zero.

Let the height of the tower be \( h \) and the distance of the car from the foot of the tower at the initial moment be \( x \).

From the angle of depression of \( 30^\circ \), we can write: \[ \tan(30^\circ) = \frac{h}{x}. \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so: \[ \frac{1}{\sqrt{3}} = \frac{h}{x} \quad \Rightarrow \quad x = \sqrt{3}h. \]

After 6 seconds, the angle of depression becomes \( 60^\circ \). Let the new distance of the car from the foot of the tower be \( y \). From the angle of depression of \( 60^\circ \), we can write: \[ \tan(60^\circ) = \frac{h}{y}. \]
We know that \( \tan(60^\circ) = \sqrt{3} \), so: \[ \sqrt{3} = \frac{h}{y} \quad \Rightarrow \quad y = \frac{h}{\sqrt{3}}. \]

The car moves from \( x \) to \( y \) in 6 seconds. So, the speed of the car is: \[ Speed = \frac{x - y}{6} = \frac{\sqrt{3}h - \frac{h}{\sqrt{3}}}{6}. \]
Simplifying: \[ Speed = \frac{h}{6} \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = \frac{h}{6} \times \frac{2}{\sqrt{3}} = \frac{h}{3\sqrt{3}}. \]

Now, the time taken by the car to reach the foot of the tower is: \[ Time = \frac{x}{Speed} = \frac{\sqrt{3}h}{\frac{h}{3\sqrt{3}}} = 3 \times 3 = 9 \, seconds. \]


Conclusion:
The time taken by the car to reach the foot of the tower from this point is 9 seconds. Quick Tip: To solve problems involving angles of depression, use the tangent function to relate the height of the tower and the distance of the car from the tower.

The coordinates of the points dividing the line segment into four equal parts are found using the section formula. If the line segment \( AB \) is divided in the ratio \( 1:3 \), \( 2:2 \), and \( 3:1 \), we find the coordinates as follows:

1. For \( P_1 \) (Dividing the segment in the ratio \( 1:3 \)):
The coordinates of \( P_1 \) are given by the section formula: \[ P_1 = \left( \frac{3x_1 + 1x_2}{1+3}, \frac{3y_1 + 1y_2}{1+3} \right), \]
where \( A(x_1, y_1) = (-2, 2) \) and \( B(x_2, y_2) = (2, 8) \). Substituting the values: \[ P_1 = \left( \frac{3(-2) + 1(2)}{4}, \frac{3(2) + 1(8)}{4} \right) = \left( \frac{-6 + 2}{4}, \frac{6 + 8}{4} \right) = \left( \frac{-4}{4}, \frac{14}{4} \right) = (-1, 3.5). \]

2. For \( P_2 \) (Dividing the segment in the ratio \( 2:2 \)):
The coordinates of \( P_2 \) are: \[ P_2 = \left( \frac{2x_1 + 2x_2}{2+2}, \frac{2y_1 + 2y_2}{2+2} \right) = \left( \frac{2(-2) + 2(2)}{4}, \frac{2(2) + 2(8)}{4} \right) = \left( \frac{-4 + 4}{4}, \frac{4 + 16}{4} \right) = (0, 5). \]

3. For \( P_3 \) (Dividing the segment in the ratio \( 3:1 \)):
The coordinates of \( P_3 \) are: \[ P_3 = \left( \frac{1x_1 + 3x_2}{1+3}, \frac{1y_1 + 3y_2}{1+3} \right) = \left( \frac{1(-2) + 3(2)}{4}, \frac{1(2) + 3(8)}{4} \right) = \left( \frac{-2 + 6}{4}, \frac{2 + 24}{4} \right) = \left( \frac{4}{4}, \frac{26}{4} \right) = (1, 6.5) \]

Thus, the points dividing the line segment into four equal parts are \( P_1(-1, 3.5) \), \( P_2(0, 5) \), and \( P_3(1, 6.5) \).


Conclusion:
The coordinates of the points which divide the line segment \( AB \) into four equal parts are \( P_1(-1, 3.5) \), \( P_2(0, 5) \), and \( P_3(1, 6.5) \). Quick Tip: To divide a line segment into equal parts, use the section formula by assigning the appropriate ratio for the divisions.

Let the original fraction be \( \frac{x}{x+3} \), where \( x \) is the numerator and \( x+3 \) is the denominator.

Step 1:
The new fraction formed by adding 2 to both the numerator and denominator is \( \frac{x+2}{x+5} \).

Step 2:
We are told that the sum of the original fraction and the new fraction is \( \frac{29}{20} \). Therefore, we have the equation: \[ \frac{x}{x+3} + \frac{x+2}{x+5} = \frac{29}{20}. \]

Step 3:
To add the two fractions on the left-hand side, first find the common denominator: \[ \frac{x}{x+3} + \frac{x+2}{x+5} = \frac{x(x+5) + (x+2)(x+3)}{(x+3)(x+5)}. \]

Simplify the numerator: \[ x(x+5) + (x+2)(x+3) = x^2 + 5x + x^2 + 5x + 6 = 2x^2 + 10x + 6. \]

So, we have: \[ \frac{2x^2 + 10x + 6}{(x+3)(x+5)} = \frac{29}{20}. \]

Step 4:
Now, cross multiply to solve for \( x \): \[ 20(2x^2 + 10x + 6) = 29(x+3)(x+5). \]

Simplify both sides: \[ 40x^2 + 200x + 120 = 29(x^2 + 8x + 15). \]

Expand the right-hand side: \[ 40x^2 + 200x + 120 = 29x^2 + 232x + 435. \]

Step 5:
Now, move all terms to one side: \[ 40x^2 + 200x + 120 - 29x^2 - 232x - 435 = 0, \] \[ 11x^2 - 32x - 315 = 0. \]

Step 6:
Solve the quadratic equation using the quadratic formula: \[ x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(11)(-315)}}{2(11)}. \]

Simplify: \[ x = \frac{32 \pm \sqrt{1024 + 13860}}{22} = \frac{32 \pm \sqrt{14884}}{22} = \frac{32 \pm 122}{22}. \]

Step 7:
Thus, \( x = \frac{32 + 122}{22} = \frac{154}{22} = 7 \) or \( x = \frac{32 - 122}{22} = \frac{-90}{22} \) (which is not valid since \( x \) must be positive).

Step 8:
Therefore, the numerator of the original fraction is \( x = 7 \), and the denominator is \( x + 3 = 10 \).

Thus, the original fraction is \( \frac{7}{10} \).



Conclusion:
The original fraction is \( \frac{7}{10} \).
Quick Tip: When solving problems involving fractions, ensure to combine the fractions by finding a common denominator and then solve the resulting equation.

Let the radius of the larger circle be \( r_1 = OA = 7 \, cm \).

The area of the larger circle is: \[ A_1 = \pi r_1^2 = \pi \times 7^2 = 49\pi \, cm^2. \]

Step 1:
The diameter \( OD \) of the smaller circle is the same as the radius \( OA \) of the larger circle. Therefore, the radius of the smaller circle is: \[ r_2 = \frac{OD}{2} = \frac{7}{2} \, cm. \]

The area of the smaller circle is: \[ A_2 = \pi r_2^2 = \pi \times \left( \frac{7}{2} \right)^2 = \pi \times \frac{49}{4} = \frac{49\pi}{4} \, cm^2. \]

Step 2:
The ratio of the areas of the smaller and larger circles is: \[ \frac{A_2}{A_1} = \frac{\frac{49\pi}{4}}{49\pi} = \frac{1}{4}. \]


Conclusion:
The ratio of the areas of the smaller and larger circles is \( \frac{1}{4} \).
Quick Tip: The ratio of the areas of two circles is the square of the ratio of their radii.

The given frequency distribution is:
\[ \begin{array}{|c|c|} \hline Class-interval & Frequency
\hline 20-60 & 7
60-100 & 5
100-150 & 16
150-250 & 12
250-350 & 2
350-450 & 3
\hline \end{array} \]

Step 1: Find the midpoints of each class-interval.

The midpoint of a class-interval \( [l, u] \) is given by: \[ Midpoint = \frac{l + u}{2}. \]

Thus, the midpoints for each class-interval are:
- \( 20-60: \frac{20 + 60}{2} = 40 \)
- \( 60-100: \frac{60 + 100}{2} = 80 \)
- \( 100-150: \frac{100 + 150}{2} = 125 \)
- \( 150-250: \frac{150 + 250}{2} = 200 \)
- \( 250-350: \frac{250 + 350}{2} = 300 \)
- \( 350-450: \frac{350 + 450}{2} = 400 \)

Step 2: Multiply the frequency by the corresponding midpoint.

Now, multiply each midpoint by the corresponding frequency:
- \( 40 \times 7 = 280 \)
- \( 80 \times 5 = 400 \)
- \( 125 \times 16 = 2000 \)
- \( 200 \times 12 = 2400 \)
- \( 300 \times 2 = 600 \)
- \( 400 \times 3 = 1200 \)

Step 3: Calculate the sum of the frequencies and the sum of the products.

The sum of the frequencies is: \[ 7 + 5 + 16 + 12 + 2 + 3 = 45. \]

The sum of the products of midpoints and frequencies is: \[ 280 + 400 + 2000 + 2400 + 600 + 1200 = 5880. \]

Step 4: Find the mean.

The mean is given by: \[ Mean = \frac{\sum f \times x}{\sum f}, \]
where \( \sum f \times x \) is the sum of the products and \( \sum f \) is the sum of the frequencies.

Substitute the values: \[ Mean = \frac{5880}{45} = 130.67. \]


Conclusion:

The mean of the given frequency distribution is \( 130.67 \). Quick Tip: To calculate the mean of a frequency distribution, multiply each midpoint by its corresponding frequency, sum the results, and then divide by the total frequency.

Let \( \triangle ABC \) and \( \triangle PQR \) be two triangles, where \( \angle A = \angle P \) and the sides including these angles are proportional. That is, \[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{PR}. \]

We need to prove that the two triangles are similar.

Step 1: Use the condition of proportional sides.
By the condition of proportionality of the sides, we have: \[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{PR}. \]
This implies that the corresponding sides of the triangles are proportional.

Step 2: Apply the AA (Angle-Angle) criterion of similarity.
We are also given that \( \angle A = \angle P \). Since two angles of one triangle are equal to two angles of the other triangle (the corresponding angle \( \angle A = \angle P \) and the corresponding angle \( \angle B = \angle Q \)), by the AA criterion of similarity, the two triangles are similar.

Thus, \[ \triangle ABC \sim \triangle PQR. \]


Conclusion:

Since the corresponding angles are equal and the corresponding sides are proportional, by the AA criterion, the two triangles are similar. Quick Tip: To prove that two triangles are similar, either use the AA criterion (two corresponding angles equal) or show that the sides are proportional.

Let \( \triangle ABC \) be a right-angled triangle at \( A \), and let \( BL \) and \( CM \) be the medians from vertices \( B \) and \( C \), respectively. We need to prove that: \[ 4(BL^2 + CM^2) = 5BC^2. \]

Step 1: Use the property of medians in a right-angled triangle.
In a right-angled triangle, the length of the median from the right angle can be given by the formula: \[ BL^2 = \frac{2AB^2 + 2AC^2 - BC^2}{4}. \]
Similarly, the length of the median from vertex \( C \) is: \[ CM^2 = \frac{2BC^2 + 2AC^2 - AB^2}{4}. \]

Step 2: Add \( BL^2 \) and \( CM^2 \).
Now, adding these two equations for \( BL^2 \) and \( CM^2 \): \[ BL^2 + CM^2 = \frac{2AB^2 + 2AC^2 - BC^2}{4} + \frac{2BC^2 + 2AC^2 - AB^2}{4}. \]

Simplifying the right-hand side: \[ BL^2 + CM^2 = \frac{2AB^2 + 2AC^2 - BC^2 + 2BC^2 + 2AC^2 - AB^2}{4}. \] \[ BL^2 + CM^2 = \frac{AB^2 + 4AC^2 + BC^2}{4}. \]

Step 3: Multiply by 4 to complete the proof.
Now, multiply both sides by 4: \[ 4(BL^2 + CM^2) = AB^2 + 4AC^2 + BC^2. \]

Since \( \triangle ABC \) is a right-angled triangle, \( AB^2 + AC^2 = BC^2 \) by the Pythagorean theorem. Thus: \[ 4(BL^2 + CM^2) = BC^2 + 4AC^2 + BC^2 = 5BC^2. \]


Conclusion:

We have shown that \( 4(BL^2 + CM^2) = 5BC^2 \). Quick Tip: In a right-angled triangle, the length of a median can be calculated using a formula involving the sides of the triangle.

Let \( \frac{1}{\sqrt{x}} = a \) and \( \frac{1}{\sqrt{y}} = b \). Then, the given equations become: \[ 4a + 3b = 3 \quad (1) \] \[ 8a - 9b = 1 \quad (2). \]

We solve this system of linear equations. Multiply equation (1) by 3 and equation (2) by 1 to eliminate \( b \):
\[ 12a + 9b = 9 \quad (3) \] \[ 8a - 9b = 1 \quad (4). \]

Now, add equations (3) and (4): \[ (12a + 9b) + (8a - 9b) = 9 + 1 \] \[ 20a = 10 \quad \Rightarrow \quad a = \frac{10}{20} = \frac{1}{2}. \]

Substitute \( a = \frac{1}{2} \) into equation (1): \[ 4 \times \frac{1}{2} + 3b = 3 \quad \Rightarrow \quad 2 + 3b = 3 \quad \Rightarrow \quad 3b = 1 \quad \Rightarrow \quad b = \frac{1}{3}. \]

Now, recall that \( a = \frac{1}{\sqrt{x}} \) and \( b = \frac{1}{\sqrt{y}} \). So: \[ \frac{1}{\sqrt{x}} = \frac{1}{2} \quad \Rightarrow \quad \sqrt{x} = 2 \quad \Rightarrow \quad x = 4, \] \[ \frac{1}{\sqrt{y}} = \frac{1}{3} \quad \Rightarrow \quad \sqrt{y} = 3 \quad \Rightarrow \quad y = 9. \]


Conclusion:
The values of \( x \) and \( y \) are \( x = 4 \) and \( y = 9 \). Quick Tip: When solving equations involving square roots, substitution is an effective method to simplify the expressions.

Let the speed of the stream be \( x \) km/hour. The speed of the boat upstream will be \( 18 - x \) km/hour, and the speed of the boat downstream will be \( 18 + x \) km/hour.

The time taken to go upstream is: \[ Time upstream = \frac{24}{18 - x}. \]

The time taken to go downstream is: \[ Time downstream = \frac{24}{18 + x}. \]

According to the problem, the time taken to go upstream is 1 hour more than the time taken to go downstream: \[ \frac{24}{18 - x} = \frac{24}{18 + x} + 1. \]

Subtract \( \frac{24}{18 + x} \) from both sides: \[ \frac{24}{18 - x} - \frac{24}{18 + x} = 1. \]

Simplify the left-hand side: \[ \frac{24(18 + x) - 24(18 - x)}{(18 - x)(18 + x)} = 1, \] \[ \frac{24(18 + x - 18 + x)}{(18 - x)(18 + x)} = 1, \] \[ \frac{24(2x)}{(18 - x)(18 + x)} = 1. \]

Simplify further: \[ \frac{48x}{324 - x^2} = 1. \]

Multiply both sides by \( 324 - x^2 \): \[ 48x = 324 - x^2. \]

Rearrange the equation: \[ x^2 + 48x - 324 = 0. \]

Solve this quadratic equation using the quadratic formula: \[ x = \frac{-48 \pm \sqrt{48^2 - 4(1)(-324)}}{2(1)} = \frac{-48 \pm \sqrt{2304 + 1296}}{2} = \frac{-48 \pm \sqrt{3600}}{2} = \frac{-48 \pm 60}{2}. \]

So: \[ x = \frac{-48 + 60}{2} = \frac{12}{2} = 6 \quad or \quad x = \frac{-48 - 60}{2} = \frac{-108}{2} = -54. \]

Since the speed of the stream cannot be negative, the speed of the stream is \( x = 6 \) km/hour.


Conclusion:
The speed of the stream is \( 6 \) km/hour. Quick Tip: To solve problems involving speed, time, and distance, use the relationship \( Time = \frac{Distance}{Speed} \) and apply the given conditions to set up equations.

The volume \( V \) of a frustum of a cone is given by the formula: \[ V = \frac{1}{3} \pi h \left( r_1^2 + r_2^2 + r_1 r_2 \right), \]
where:
- \( h = 24 \, cm \) is the height,
- \( r_1 = 20 \, cm \) is the radius of the upper end,
- \( r_2 = 8 \, cm \) is the radius of the lower end.

Substitute the given values into the formula: \[ V = \frac{1}{3} \times 3.14 \times 24 \left( 20^2 + 8^2 + 20 \times 8 \right). \]

Simplify: \[ V = \frac{1}{3} \times 3.14 \times 24 \left( 400 + 64 + 160 \right) = \frac{1}{3} \times 3.14 \times 24 \times 624. \]
\[ V = \frac{1}{3} \times 3.14 \times 14976 = 15792.48 \, cm^3. \]

Since 1 litre = 1000 cm³, the volume in litres is: \[ Volume in litres = \frac{15792.48}{1000} = 15.79248 \, litres. \]

The cost of the milk is: \[ Cost = 40 \times 15.79248 = 631.6992 \, rupees. \]

Thus, the cost of the milk is approximately \( Rs. 631.70 \).



Conclusion:
The cost of milk to completely fill the container is approximately \( Rs. 631.70 \).
Quick Tip: To calculate the volume of a frustum of a cone, use the formula \( V = \frac{1}{3} \pi h \left( r_1^2 + r_2^2 + r_1 r_2 \right) \), where \( r_1 \) and \( r_2 \) are the radii of the upper and lower ends, and \( h \) is the height.

The volume of the earth dug out of the well is the volume of a cylinder with radius \( r = \frac{3}{2} = 1.5 \, m \) and height \( h = 14 \, m \). The volume \( V_{well} \) is given by: \[ V_{well} = \pi r^2 h = 3.14 \times (1.5)^2 \times 14 = 3.14 \times 2.25 \times 14 = 99.78 \, m^3. \]

Step 1:
The earth is spread in the form of a circular ring, where the inner radius is \( r = 1.5 \, m \), and the outer radius is \( r + 4 = 1.5 + 4 = 5.5 \, m \). The volume \( V_{ring} \) of the embankment is the volume of the cylindrical ring, given by: \[ V_{ring} = \pi \left( (r + 4)^2 - r^2 \right) h_{embankment}. \]

Substitute the values: \[ V_{ring} = 3.14 \left( (5.5)^2 - (1.5)^2 \right) h_{embankment} = 3.14 \left( 30.25 - 2.25 \right) h_{embankment} = 3.14 \times 28 h_{embankment}. \]

Step 2:
We know that the volume of the earth dug out is the same as the volume of the embankment, so: \[ 99.78 = 3.14 \times 28 \times h_{embankment}. \]

Solve for \( h_{embankment} \): \[ h_{embankment} = \frac{99.78}{3.14 \times 28} = \frac{99.78}{87.92} \approx 1.13 \, m. \]


Conclusion:
The height of the embankment is approximately \( 1.13 \, m \).
Quick Tip: To find the volume of a cylindrical ring, subtract the volume of the smaller cylinder from the volume of the larger cylinder.