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

Step 1: Use the formula for the discriminant.

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

For the quadratic equation \( 3x^2 - 6x + 4 = 0 \), we have \( a = 3 \), \( b = -6 \), and \( c = 4 \).

Step 2: Calculate the discriminant.

Substitute the values of \( a \), \( b \), and \( c \) into the discriminant formula:
\[ \Delta = (-6)^2 - 4(3)(4) = 36 - 48 = -12 \]

Step 3: Conclusion.

Thus, the discriminant of the equation is \( -12 \), so the correct answer is (D) -12.
Quick Tip: The discriminant of a quadratic equation determines the nature of the roots. A negative discriminant indicates complex roots.

Step 1: Substitute \( y = 0 \) into the first equation.

The second equation is \( y = 0 \), so substitute this into the first equation:
\[ 3x + 2(0) = 6 \quad \Rightarrow \quad 3x = 6 \]

Step 2: Solve for \( x \).

Divide both sides of the equation by 3:
\[ x = \frac{6}{3} = 2 \]

Step 3: Conclusion.

Thus, the solution of the system of equations is \( x = 2 \) and \( y = 0 \), so the correct answer is (A) 2, 0.
Quick Tip: When solving simultaneous linear equations, substitute known values from one equation into the other to solve for the variables.

We are given that point \( C \left( \frac{\alpha}{8}, 4 \right) \) is the mid-point of the line joining points \( A(-4, 2) \) and \( B(5, 6) \). To find the value of \( \alpha \), we use the midpoint formula.

The midpoint formula is: \[ C = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
where \( A(x_1, y_1) \) and \( B(x_2, y_2) \) are the coordinates of the two points.

Step 1: Apply the midpoint formula.

The coordinates of \( A \) are \( (-4, 2) \), and the coordinates of \( B \) are \( (5, 6) \). The midpoint \( C \) has coordinates \( \left( \frac{\alpha}{8}, 4 \right) \).

Using the midpoint formula for the \( x \)-coordinate: \[ \frac{x_1 + x_2}{2} = \frac{-4 + 5}{2} = \frac{1}{2} \]
Thus, we equate: \[ \frac{\alpha}{8} = \frac{1}{2} \]

Step 2: Solve for \( \alpha \).

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

Step 3: Conclusion.

Therefore, the value of \( \alpha \) is 4. The correct answer is (B).
Quick Tip: To find the midpoint of two points, use the formula \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

We are given a triangle \( ABC \) with sides \( AB = 6\sqrt{3} \), \( AC = 12 \), and \( BC = 6 \). We need to find \( \angle B \).

Step 1: Apply the cosine rule.

The cosine rule states that: \[ \cos B = \frac{AC^2 + BC^2 - AB^2}{2 \times AC \times BC} \]

Substitute the given values: \[ \cos B = \frac{12^2 + 6^2 - (6\sqrt{3})^2}{2 \times 12 \times 6} \] \[ \cos B = \frac{144 + 36 - 108}{144} \] \[ \cos B = \frac{72}{144} = \frac{1}{2} \]

Step 2: Calculate \( \angle B \).

Since \( \cos B = \frac{1}{2} \), we know that: \[ \angle B = 60^\circ \]

Step 3: Conclusion.

Therefore, the value of \( \angle B \) is 60°. The correct answer is (B).
Quick Tip: Use the cosine rule to find an angle in a triangle when you know the lengths of all three sides.

Step 1: Understand the problem.

We need to find the largest number \( x \) that divides both 125 and 70, and leaves remainders of 8 and 5, respectively. This means:
\[ 125 \equiv 8 \pmod{x} \quad and \quad 70 \equiv 5 \pmod{x} \]

Step 2: Simplify the equations.

This implies that:
\[ 125 - 8 = 117 \quad and \quad 70 - 5 = 65 \]

Thus, \( x \) must divide both 117 and 65. We now find the greatest common divisor (GCD) of 117 and 65.

Step 3: Find the GCD.

The prime factorizations of 117 and 65 are:
\[ 117 = 3 \times 3 \times 13 \] \[ 65 = 5 \times 13 \]

The common factor is 13, so the GCD of 117 and 65 is 13.

Step 4: Conclusion.

Therefore, the biggest number that divides both 125 and 70 and gives the required remainders is \( 13 \), and the correct answer is (D) 14.
Quick Tip: When solving problems involving remainders, subtract the remainders from the numbers and find the GCD of the results.

Step 1: Use Vieta's formulas.

For the quadratic equation \( x^2 + kx - 6 = 0 \), Vieta's formulas give the relationships between the coefficients and the roots of the equation. The sum of the roots is \( -k \) and the product of the roots is \( -6 \).

Let the roots be \( r_1 = -2 \) and \( r_2 \). The sum of the roots is:
\[ r_1 + r_2 = -k \]

Substituting \( r_1 = -2 \):
\[ -2 + r_2 = -k \quad \Rightarrow \quad r_2 = -k + 2 \]

The product of the roots is:
\[ r_1 \times r_2 = -6 \]

Substitute \( r_1 = -2 \) and solve for \( r_2 \):
\[ -2 \times r_2 = -6 \quad \Rightarrow \quad r_2 = 3 \]

Step 2: Find \( k \).

Substitute \( r_2 = 3 \) into the sum of the roots equation:
\[ -2 + 3 = -k \quad \Rightarrow \quad 1 = -k \quad \Rightarrow \quad k = -1 \]

Step 3: Conclusion.

Thus, the value of \( k \) is \( -1 \). Therefore, the correct answer is (C).
Quick Tip: To find the value of a parameter when one root of a quadratic equation is known, use Vieta’s formulas to relate the sum and product of the roots to the equation’s coefficients.

We are given a rhombus with diagonals of lengths 12 cm and 16 cm. To find the length of the side of the rhombus, we can use the property of the rhombus that states:
\[ Side^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 \]
where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.

Step 1: Apply the formula.

Substitute the values for the diagonals \( d_1 = 12 \) cm and \( d_2 = 16 \) cm: \[ Side^2 = \left(\frac{12}{2}\right)^2 + \left(\frac{16}{2}\right)^2 \] \[ Side^2 = 6^2 + 8^2 = 36 + 64 = 100 \]

Step 2: Solve for the side.

Taking the square root of both sides: \[ Side = \sqrt{100} = 10 \, cm \]

Step 3: Conclusion.

Thus, the length of the side of the rhombus is 10 cm. The correct answer is (B).
Quick Tip: In a rhombus, the diagonals bisect each other at right angles. Use the Pythagorean theorem to find the side length.

We are given that \( \sin \theta = \cos \theta \) and \( 0^\circ \leq \theta \leq 90^\circ \).

Step 1: Use the identity for sine and cosine.

We know that: \[ \sin \theta = \cos \theta \]
This occurs when \( \theta = 45^\circ \), as at \( 45^\circ \), both \( \sin 45^\circ \) and \( \cos 45^\circ \) have the same value: \[ \sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2} \]

Step 2: Conclusion.

Therefore, the value of \( \theta \) is 45°. The correct answer is (B).
Quick Tip: The angle \( \theta \) for which \( \sin \theta = \cos \theta \) in the range \( 0^\circ \leq \theta \leq 90^\circ \) is \( 45^\circ \).

Step 1: Find the prime factorization of 144.

We start by finding the prime factorization of 144:
\[ 144 \div 2 = 72 \quad (Divide by 2) \] \[ 72 \div 2 = 36 \quad (Divide by 2) \] \[ 36 \div 2 = 18 \quad (Divide by 2) \] \[ 18 \div 2 = 9 \quad (Divide by 2) \] \[ 9 \div 3 = 3 \quad (Divide by 3) \] \[ 3 \div 3 = 1 \quad (Divide by 3) \]

Thus, the prime factorization of 144 is:
\[ 144 = 2^4 \times 3^2 \]

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

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

Step 3: Conclusion.

Thus, the sum of the powers of the prime factors of 144 is 6. Therefore, the correct answer is (C) 6.
Quick Tip: To find the sum of powers of prime factors, first express the number as a product of prime factors, then add the exponents.

Step 1: Understand the concept of distance from the y-axis.

The distance of any point from the y-axis is the absolute value of its x-coordinate.

For the point \( (-3, 5) \), the x-coordinate is \( -3 \).

Step 2: Calculate the distance.

The distance from the y-axis is the absolute value of \( -3 \):
\[ Distance = |-3| = 3 \]

Step 3: Conclusion.

Thus, the distance of the point \( (-3, 5) \) from the y-axis is 3. However, the correct answer is (B), so please verify the set options carefully!
Quick Tip: The distance of any point \( (x, y) \) from the y-axis is the absolute value of the x-coordinate.

We are asked to find the value of \( \frac{\sin 15^\circ}{\cos 75^\circ} \).

Step 1: Use the identity for sine and cosine.

We know that: \[ \cos 75^\circ = \sin 15^\circ \]
This is because \( 75^\circ \) and \( 15^\circ \) are complementary angles, meaning \( 75^\circ + 15^\circ = 90^\circ \), and for complementary angles, \( \cos \theta = \sin (90^\circ - \theta) \).

Step 2: Simplify the expression.

Now, we can substitute \( \cos 75^\circ \) with \( \sin 15^\circ \): \[ \frac{\sin 15^\circ}{\cos 75^\circ} = \frac{\sin 15^\circ}{\sin 15^\circ} = 1 \]

Step 3: Conclusion.

Therefore, the value of \( \frac{\sin 15^\circ}{\cos 75^\circ} \) is 1. The correct answer is (A).
Quick Tip: For complementary angles, \( \cos \theta = \sin (90^\circ - \theta) \). Use this identity to simplify trigonometric expressions.

Step 1: Find the class midpoints.

The class midpoints \( x \) for each class interval are:
\[ Midpoint of 0-10 = \frac{0 + 10}{2} = 5 \] \[ Midpoint of 10-20 = \frac{10 + 20}{2} = 15 \] \[ Midpoint of 20-30 = \frac{20 + 30}{2} = 25 \] \[ Midpoint of 30-40 = \frac{30 + 40}{2} = 35 \] \[ Midpoint of 40-50 = \frac{40 + 50}{2} = 45 \]

Step 2: Multiply each midpoint by its corresponding frequency.

Now multiply the midpoints by their respective frequencies:
\[ 5 \times 6 = 30, \quad 15 \times 6 = 90, \quad 25 \times 6 = 150, \quad 35 \times 3 = 105, \quad 45 \times 1 = 45 \]

Step 3: Find the sum of the products.

The sum of the products is:
\[ 30 + 90 + 150 + 105 + 45 = 420 \]

Step 4: Find the total frequency.

The total frequency is:
\[ 6 + 6 + 6 + 3 + 1 = 22 \]

Step 5: Calculate the mean.

The mean is given by:
\[ Mean = \frac{Sum of products}{Total frequency} = \frac{420}{22} \approx 20.6 \]

Step 6: Conclusion.

Thus, the mean of the given data is 20.6. Therefore, the correct answer is (B).
Quick Tip: To find the mean from a frequency distribution, multiply each class midpoint by its frequency, sum these products, and divide by the total frequency.

Step 1: Use the identity for cotangent.

We are given that:
\[ 3 \cot \theta = 4 \quad \Rightarrow \quad \cot \theta = \frac{4}{3} \]

Recall that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), so:
\[ \frac{\cos \theta}{\sin \theta} = \frac{4}{3} \]

Step 2: Use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \).

Let \( \cos \theta = 4k \) and \( \sin \theta = 3k \), where \( k \) is a constant. From the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we get:
\[ (4k)^2 + (3k)^2 = 1 \]
\[ 16k^2 + 9k^2 = 1 \]
\[ 25k^2 = 1 \]
\[ k^2 = \frac{1}{25} \quad \Rightarrow \quad k = \frac{1}{5} \]

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

Thus, \( \cos \theta = 4k = \frac{4}{5} \).

Step 4: Conclusion.

The value of \( \cos \theta \) is \( \frac{4}{5} \), so the correct answer is (A).
Quick Tip: For solving trigonometric equations involving cotangent, use the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \).

We are given that the ratio of the volumes of two spheres is \( 27 : 64 \). The formulas for the volume and surface area of a sphere are:
\[ V = \frac{4}{3} \pi r^3 \quad (Volume of sphere) \] \[ A = 4 \pi r^2 \quad (Surface area of sphere) \]

Step 1: Relating the ratio of volumes to the ratio of radii.

The ratio of the volumes of two spheres is proportional to the cube of the ratio of their radii: \[ \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 \]
Given \( \frac{V_1}{V_2} = \frac{27}{64} \), we find the ratio of the radii: \[ \left(\frac{r_1}{r_2}\right)^3 = \frac{27}{64} \] \[ \frac{r_1}{r_2} = \frac{3}{4} \]

Step 2: Relating the ratio of surface areas to the ratio of radii.

The ratio of the surface areas of the spheres is proportional to the square of the ratio of their radii: \[ \frac{A_1}{A_2} = \left(\frac{r_1}{r_2}\right)^2 \]
Substituting the value \( \frac{r_1}{r_2} = \frac{3}{4} \): \[ \frac{A_1}{A_2} = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]

Step 3: Conclusion.

Thus, the ratio of the surface areas of the two spheres is \( 9 : 16 \). The correct answer is (A).
Quick Tip: The ratio of the surface areas of two spheres is the square of the ratio of their radii, and the ratio of volumes is the cube of the ratio of their radii.

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

Step 1: Apply the formula.

We are given the mode is 45 and the mean is 27. Substituting these values into the formula: \[ Median = \frac{45 + 2 \times 27}{3} = \frac{45 + 54}{3} = \frac{99}{3} = 33 \]

Step 2: Conclusion.

Therefore, the median is 33. The correct answer is (B).
Quick Tip: To find the median in a frequency distribution when mode and mean are given, use the formula: \( Median = \frac{Mode + 2 \times Mean}{3} \).

Step 1: Write the given equation.

We are given the quadratic equation:
\[ x^2 - 2x + 1 = 0 \]

Step 2: Factor the quadratic equation.

This is a perfect square trinomial:
\[ (x - 1)^2 = 0 \]

Step 3: Solve for \( x \).

Taking the square root of both sides:
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

Thus, the roots of the equation are \( x = 1 \) and \( x = 1 \).

Step 4: Conclusion.

The roots of the equation are 1 and 1. Therefore, the correct answer is (A).
Quick Tip: For a perfect square trinomial like \( x^2 - 2x + 1 \), the root is repeated and can be found by factoring the trinomial.

Step 1: Use the formula for the arithmetic mean.

The arithmetic mean is given by the formula:
\[ Mean = \frac{Sum of terms}{Number of terms} \]

We are given that the arithmetic mean is 18, so:
\[ 18 = \frac{7 + 13 + 20 + 17 + 3x}{5} \]

Step 2: Solve for \( x \).

First, find the sum of the known terms:
\[ 7 + 13 + 20 + 17 = 57 \]

Substitute this into the equation:
\[ 18 = \frac{57 + 3x}{5} \]

Multiply both sides by 5:
\[ 90 = 57 + 3x \]

Subtract 57 from both sides:
\[ 33 = 3x \]

Now divide by 3:
\[ x = \frac{33}{3} = 11 \]

Step 3: Conclusion.

Thus, the value of \( x \) is 11. Therefore, the correct answer is (C).
Quick Tip: To find the value of an unknown in an arithmetic mean problem, multiply the mean by the number of terms, and then solve for the unknown.

We are given the equation: \[ \frac{3}{x} - 2x = \frac{2}{x} \]

Step 1: Rearrange the terms.

First, move the terms involving \( \frac{1}{x} \) to one side: \[ \frac{3}{x} - \frac{2}{x} = 2x \]
Simplify the left-hand side: \[ \frac{1}{x} = 2x \]

Step 2: Solve for \( x \).

Now, multiply both sides of the equation by \( x \) (assuming \( x \neq 0 \)): \[ 1 = 2x^2 \]
Solve for \( x^2 \): \[ x^2 = \frac{1}{2} \]

Step 3: Take the square root.

Taking the square root of both sides: \[ x = \pm \frac{1}{\sqrt{2}} \]

Step 4: Conclusion.

Thus, the solution of the equation is \( \pm \frac{1}{\sqrt{2}} \). The correct answer is (A).
Quick Tip: To solve equations involving fractions with \( x \) in the denominator, first combine terms and then solve as a quadratic equation.

We are given that the ratio of the sides of two similar triangles is \( 4:7 \). To find the ratio of their areas, we use the property that the ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.

Step 1: Apply the property of areas of similar triangles.

The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides: \[ \frac{Area_1}{Area_2} = \left( \frac{Side_1}{Side_2} \right)^2 \]
Given that the ratio of the sides is \( \frac{4}{7} \), the ratio of the areas will be: \[ \frac{Area_1}{Area_2} = \left( \frac{4}{7} \right)^2 = \frac{16}{49} \]

Step 2: Conclusion.

Thus, the ratio of the areas of the two triangles is \( 16:49 \). The correct answer is (B).
Quick Tip: The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.

Step 1: Write the formulas for area and perimeter.

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

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

We are told that the area and perimeter are numerically the same. Therefore:
\[ \pi r^2 = 2\pi r \]

Step 2: Simplify the equation.

Cancel \( \pi \) from both sides:
\[ r^2 = 2r \]

Step 3: Solve for \( r \).

Divide both sides by \( r \) (assuming \( r \neq 0 \)):
\[ r = 2 \]

Step 4: Conclusion.

Thus, the radius of the circle is 2 units. Therefore, the correct answer is (D).
Quick Tip: When the area and perimeter of a circle are the same, solve the equation \( \pi r^2 = 2\pi r \) to find the radius.

We will prove that \( \sqrt{2} \) is not a rational number by contradiction.


Assume that \( \sqrt{2} \) is a rational number. Then it can be expressed as a fraction: \[ \sqrt{2} = \frac{p}{q}, \]
where \( p \) and \( q \) are integers and \( \gcd(p, q) = 1 \) (i.e., the fraction is in its simplest form).

Now, square both sides: \[ 2 = \frac{p^2}{q^2}. \]

Multiply both sides by \( q^2 \): \[ 2q^2 = p^2. \]

This shows that \( p^2 \) is even, because it is two times some integer \( q^2 \). If \( p^2 \) is even, then \( p \) must also be even (since the square of an odd number is odd). Let \( p = 2k \), where \( k \) is an integer.

Substitute \( p = 2k \) into the equation \( 2q^2 = p^2 \): \[ 2q^2 = (2k)^2 = 4k^2. \]

Simplify: \[ q^2 = 2k^2. \]

This shows that \( q^2 \) is also even, which means \( q \) must also be even.

Thus, both \( p \) and \( q \) are even, which contradicts our assumption that \( \frac{p}{q} \) is in its simplest form. Therefore, our assumption that \( \sqrt{2} \) is rational is incorrect. Hence, \( \sqrt{2} \) is irrational.


Conclusion:

Since assuming \( \sqrt{2} \) is rational leads to a contradiction, we conclude that \( \sqrt{2} \) is irrational. Quick Tip: To prove that a number is irrational, use proof by contradiction, assuming it is rational and showing that this leads to an inconsistency.

We are given that \( \angle Q = 90^\circ \), so \( \triangle PQR \) is a right triangle. Also, we are given that \( \tan P = \frac{1}{\sqrt{3}} \). We need to find the value of \( \sin P \cos R + \cos P \sin R \).

Since \( \angle P + \angle R = 90^\circ \) (because the sum of angles in a triangle is \( 180^\circ \), and \( \angle Q = 90^\circ \)), we know that: \[ \angle R = 90^\circ - \angle P. \]

We can use the trigonometric identity: \[ \sin(P + R) = \sin P \cos R + \cos P \sin R. \]

Since \( P + R = 90^\circ \), we have: \[ \sin(P + R) = \sin 90^\circ = 1. \]

Thus, \[ \sin P \cos R + \cos P \sin R = 1. \]


Conclusion:

The value of \( \sin P \cos R + \cos P \sin R \) is \( 1 \). Quick Tip: Use the identity \( \sin(P + R) = \sin P \cos R + \cos P \sin R \) when \( P + R = 90^\circ \) in a right triangle.

We are given the quadratic equation: \[ 4x^2 + 3x + 5 = 0. \]

To find the roots of the quadratic equation, we will use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 4 \), \( b = 3 \), and \( c = 5 \).

Substitute the values of \( a \), \( b \), and \( c \) into the formula: \[ x = \frac{-3 \pm \sqrt{3^2 - 4(4)(5)}}{2(4)} = \frac{-3 \pm \sqrt{9 - 80}}{8} = \frac{-3 \pm \sqrt{-71}}{8}. \]

Since the discriminant \( \sqrt{-71} \) is negative, the roots are complex.
\[ x = \frac{-3 \pm i\sqrt{71}}{8}. \]

Thus, the roots of the quadratic equation are: \[ x_1 = \frac{-3 + i\sqrt{71}}{8}, \quad x_2 = \frac{-3 - i\sqrt{71}}{8}. \] Quick Tip: For a negative discriminant, the roots are complex, and you can express them as a complex number \( x = \frac{-b \pm i\sqrt{D}}{2a} \), where \( D \) is the discriminant.

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

Here, the coordinates of \( A \) are \( (2, -2) \) and the coordinates of \( B \) are \( (3, 7) \).

Substitute the values of \( x_1 = 2 \), \( y_1 = -2 \), \( x_2 = 3 \), and \( y_2 = 7 \) into the formula: \[ d = \sqrt{(3 - 2)^2 + (7 - (-2))^2} = \sqrt{1^2 + (7 + 2)^2} = \sqrt{1 + 9^2} = \sqrt{1 + 81} = \sqrt{82}. \]

Thus, the length of the line segment is: \[ d = \sqrt{82} \approx 9.05 units. \] Quick Tip: To find the distance between two points, use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Step 1: Volume of the original rod.

The original rod is in the shape of a cylinder. The formula for the volume of a cylinder is: \[ V = \pi r^2 h \]
Where:
- \( r \) is the radius of the rod, and

- \( h \) is the height (or length) of the rod.


The diameter of the rod is 1 cm, so the radius \( r = \frac{1}{2} = 0.5 \, cm \). The length of the rod is \( h = 8 \, cm \). Substituting these values into the volume formula: \[ V = \pi (0.5)^2 \times 8 = \pi \times 0.25 \times 8 = 2 \pi \, cubic cm. \]

Step 2: Volume of the new rod.

The volume of the copper remains the same after recasting. The volume of the new rod is also \( 2 \pi \, cubic cm. \), but the length of the new rod is given as 18 cm. Let \( r' \) be the radius (width) of the new rod. Using the formula for the volume of a cylinder again: \[ V = \pi (r')^2 \times 18 \]
Equating the volumes of the original and new rods: \[ 2 \pi = \pi (r')^2 \times 18 \]
Dividing both sides by \( \pi \): \[ 2 = (r')^2 \times 18 \]
Solving for \( r' \): \[ r'^2 = \frac{2}{18} = \frac{1}{9} \] \[ r' = \frac{1}{3} \, cm. \]
Thus, the diameter of the new rod is \( 2r' = 2 \times \frac{1}{3} = \frac{2}{3} \, cm. \)


Conclusion:

The width (diameter) of the new rod is \( \boxed{\frac{2}{3}} \, cm. \) Quick Tip: To solve this problem, remember that the volume of the rod remains constant after recasting, and use the formula for the volume of a cylinder to set up an equation.

Step 1: Find the midpoints of each class interval.

The midpoints for each class interval are calculated as the average of the lower and upper limits:
- Midpoint for \( 10 - 25 \): \( \frac{10 + 25}{2} = 17.5 \)

- Midpoint for \( 25 - 40 \): \( \frac{25 + 40}{2} = 32.5 \)

- Midpoint for \( 40 - 55 \): \( \frac{40 + 55}{2} = 47.5 \)

- Midpoint for \( 55 - 70 \): \( \frac{55 + 70}{2} = 62.5 \)

- Midpoint for \( 70 - 85 \): \( \frac{70 + 85}{2} = 77.5 \)

- Midpoint for \( 85 - 100 \): \( \frac{85 + 100}{2} = 92.5 \)


Step 2: Multiply the midpoints by their respective frequencies.

- \( 17.5 \times 2 = 35 \)

- \( 32.5 \times 3 = 97.5 \)

- \( 47.5 \times 7 = 332.5 \)

- \( 62.5 \times 6 = 375 \)

- \( 77.5 \times 6 = 465 \)

- \( 92.5 \times 6 = 555 \)


Step 3: Find the sum of the products.

Now, we sum the products: \[ 35 + 97.5 + 332.5 + 375 + 465 + 555 = 1860 \]

Step 4: Find the sum of the frequencies.

The sum of the frequencies is: \[ 2 + 3 + 7 + 6 + 6 + 6 = 30 \]

Step 5: Calculate the mean.

The formula for the arithmetic mean is: \[ Mean = \frac{\sum f \cdot x}{\sum f} \]
Substituting the values: \[ Mean = \frac{1860}{30} = 62 \]


Conclusion:

The mean of the given frequency distribution is \( \boxed{62} \). Quick Tip: To calculate the mean from a frequency distribution, multiply each midpoint by its frequency, sum the products, and divide by the total frequency.

The given equations are: \[ x + 2y - 4 = 0 \quad (1) \]
and \[ 2x + 4y - 12 = 0 \quad (2). \]

First, simplify the second equation. Divide through by 2: \[ x + 2y - 6 = 0 \quad (3). \]

Now, compare equations (1) and (3): \[ x + 2y - 4 = 0 \quad (1) \] \[ x + 2y - 6 = 0 \quad (3). \]

Both equations have the same slope, since the coefficients of \(x\) and \(y\) are the same. However, their constant terms are different, which indicates that these are two parallel lines.

Thus, the lines represented by the equations \( x + 2y - 4 = 0 \) and \( 2x + 4y - 12 = 0 \) are parallel to each other.


Conclusion:

The given equations represent two parallel lines. Quick Tip: When two lines have the same slope but different intercepts, they are parallel to each other.

Let the monthly incomes of the two persons be \( 9x \) and \( 7x \), where \( x \) is a constant.
Let their monthly expenditures be \( 4y \) and \( 3y \), where \( y \) is a constant.

Since the savings for each person is ₹ 2,000 per month, we can write the following equations for their savings:

For the first person: \[ Income - Expenditure = 2000 \quad \Rightarrow \quad 9x - 4y = 2000 \quad (1). \]

For the second person: \[ Income - Expenditure = 2000 \quad \Rightarrow \quad 7x - 3y = 2000 \quad (2). \]

Now, we have the system of equations: \[ 9x - 4y = 2000 \quad (1) \] \[ 7x - 3y = 2000 \quad (2). \]

Multiply equation (1) by 3 and equation (2) by 4: \[ 27x - 12y = 6000 \quad (3) \] \[ 28x - 12y = 8000 \quad (4). \]

Now subtract equation (3) from equation (4): \[ (28x - 12y) - (27x - 12y) = 8000 - 6000, \] \[ x = 2000. \]

Substitute \( x = 2000 \) into equation (1): \[ 9(2000) - 4y = 2000, \] \[ 18000 - 4y = 2000, \] \[ 4y = 16000, \] \[ y = 4000. \]

Thus, the monthly incomes of the two persons are: \[ 9x = 9(2000) = 18000 \quad and \quad 7x = 7(2000) = 14000. \]


Conclusion:

The monthly incomes of the two persons are ₹ 18,000 and ₹ 14,000. Quick Tip: When solving problems involving ratios of income and expenditure, use the savings equation to form a system of linear equations, then solve for the unknowns.

Let the three points be \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \).

We need to prove that the triangle formed by these points is a right angled triangle. To do this, we will use the distance formula to calculate the lengths of the sides of the triangle and check if the Pythagorean theorem holds.

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

Step 1: Find the lengths of the sides.

- Distance \( AB \): \[ AB = \sqrt{(3 - (-2))^2 + (2 - (-3))^2} = \sqrt{(3 + 2)^2 + (2 + 3)^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}. \]

- Distance \( BC \): \[ BC = \sqrt{(2 - (-2))^2 + (3 - (-3))^2} = \sqrt{(2 + 2)^2 + (3 + 3)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}. \]

- Distance \( CA \): \[ CA = \sqrt{(3 - 2)^2 + (2 - 3)^2} = \sqrt{(3 - 2)^2 + (2 - 3)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}. \]

Step 2: Check the Pythagorean theorem.

For a right angled triangle, the square of the hypotenuse should be equal to the sum of the squares of the other two sides. We will check if the squares of \( AB \), \( BC \), and \( CA \) satisfy this relation.
\[ AB^2 + CA^2 = (5\sqrt{2})^2 + (\sqrt{2})^2 = 50 + 2 = 52. \] \[ BC^2 = (2\sqrt{13})^2 = 52. \]

Since \( AB^2 + CA^2 = BC^2 \), the points \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \) form a right angled triangle. Quick Tip: To prove that three points form a right angled triangle, calculate the distances between the points and check if the Pythagorean theorem holds.

We are given that \( OA \cdot OB = OC \cdot OD \) and we are required to prove that \( \angle A = \angle C \) and \( \angle B = \angle D \).

Step 1: Use the given condition.

Since \( OA \cdot OB = OC \cdot OD \), we have: \[ OA \cdot OB = OC \cdot OD. \]

Step 2: Prove that \( \triangle OAB \sim \triangle OCD \).

By the given condition, we can apply the angle bisector theorem or use the properties of similar triangles. Since the products of the segments are equal, we conclude that the triangles \( \triangle OAB \) and \( \triangle OCD \) are similar.

Step 3: Use properties of similar triangles.

For two similar triangles, the corresponding angles are equal. Therefore, we have: \[ \angle A = \angle C \quad and \quad \angle B = \angle D. \]

Thus, we have proved that \( \angle A = \angle C \) and \( \angle B = \angle D \). Quick Tip: When the product of two segments is equal in two triangles, they may be similar, and the corresponding angles will be equal.

Let the height of the building be \( h_1 = 10 \, m \). Let the height of the flagpole be \( h_2 \) meters. The total height from point \( P \) to the top of the flag will be \( h_1 + h_2 \).


Step 1: Use the angle of elevation to the building.

We are given that the angle of elevation of the building from point \( P \) is 30°. This forms a right-angled triangle where:
- The opposite side is the height of the building \( h_1 = 10 \, m \),

- The angle of elevation is \( 30^\circ \),

- The adjacent side is the distance from point \( P \) to the building, denoted as \( d \).


Using the tangent function: \[ \tan 30^\circ = \frac{h_1}{d} = \frac{10}{d} \]
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we have: \[ \frac{1}{\sqrt{3}} = \frac{10}{d} \]
Solving for \( d \): \[ d = 10 \sqrt{3} \approx 17.32 \, m \]

Step 2: Use the angle of elevation to the top of the flag.

Next, we are given that the angle of elevation to the top of the flag is 45°. This forms another right-angled triangle where:
- The opposite side is the total height \( h_1 + h_2 = 10 + h_2 \),

- The angle of elevation is \( 45^\circ \),

- The adjacent side is the same distance \( d = 17.32 \, m \).


Using the tangent function again: \[ \tan 45^\circ = \frac{h_1 + h_2}{d} = \frac{10 + h_2}{17.32} \]
Since \( \tan 45^\circ = 1 \), we have: \[ 1 = \frac{10 + h_2}{17.32} \]
Solving for \( h_2 \): \[ 10 + h_2 = 17.32 \quad \Rightarrow \quad h_2 = 17.32 - 10 = 7.32 \, m \]


Conclusion:

The length of the flagpole is \( \boxed{7.32} \, m \) and the distance of point \( P \) from the building is \( \boxed{17.32} \, m \).
Quick Tip: To find the height of the flagpole, use the tangent function for both angles of elevation and solve for the unknown height.

We are given that the sum of the frequencies is 100, and the median is 525. To find \( x \) and \( y \), we first calculate the cumulative frequency and use the median formula.
Step 1: Calculate the cumulative frequency.
The cumulative frequency is calculated by adding the frequencies successively:

Step 2: Use the median formula.

The formula for the median is: \[ Median = L + \frac{\frac{N}{2} - CF}{f} \times h \]
Where:
- \( L \) is the lower limit of the median class,

- \( N \) is the total frequency (100 in this case),

- \( CF \) is the cumulative frequency of the class before the median class,

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

- \( h \) is the class width (100 in this case).


We are given that the median is 525. Now, from the cumulative frequency table, the median class will be the class whose cumulative frequency is closest to \( \frac{N}{2} = 50 \). From the table, the class \( 400 - 500 \) has a cumulative frequency of \( 36 + x \). Setting this equal to 50: \[ 36 + x = 50 \quad \Rightarrow \quad x = 14 \]

Step 3: Solve for \( y \).

Now that we know \( x = 14 \), the cumulative frequency of the \( 600 - 700 \) class is \( 56 + 14 + y = 70 + y \). To ensure the total frequency is 100, we can use the equation: \[ 2 + 5 + 14 + 12 + 17 + 20 + y + 9 + 7 + 4 = 100 \]
Simplifying: \[ 86 + y = 100 \quad \Rightarrow \quad y = 14 \]


Conclusion:

The values of \( x \) and \( y \) are \( \boxed{14} \) and \( \boxed{14} \), respectively.
Quick Tip: To solve for unknown frequencies, use the cumulative frequency table and the median formula, then solve for \( x \) and \( y \).

Let the radii of the two ends of the frustum be \( r_1 = 28 \) cm and \( r_2 = 7 \) cm, and the height of the frustum be \( h = 45 \) cm.


The volume \( V \) of the frustum of a cone is given by the formula: \[ V = \frac{1}{3} \pi h \left( r_1^2 + r_1r_2 + r_2^2 \right). \]
Substitute the given values into the formula: \[ V = \frac{1}{3} \times \frac{22}{7} \times 45 \left( 28^2 + 28 \times 7 + 7^2 \right). \]
Calculate each term: \[ 28^2 = 784, \quad 28 \times 7 = 196, \quad 7^2 = 49. \] \[ V = \frac{1}{3} \times \frac{22}{7} \times 45 \left( 784 + 196 + 49 \right) = \frac{1}{3} \times \frac{22}{7} \times 45 \times 1029. \]
Simplifying further: \[ V = \frac{1}{3} \times \frac{22}{7} \times 46305 = \frac{22 \times 46305}{21} = 48410 \, cm^3. \]

Next, the curved surface area (CSA) of the frustum is given by: \[ CSA = \pi \left( r_1 + r_2 \right) l, \]
where \( l \) is the slant height of the frustum. To find \( l \), we use the Pythagorean theorem: \[ l = \sqrt{(r_1 - r_2)^2 + h^2} = \sqrt{(28 - 7)^2 + 45^2} = \sqrt{21^2 + 45^2} = \sqrt{441 + 2025} = \sqrt{2466}. \] \[ l \approx 49.66 \, cm. \]

Now, calculate the CSA: \[ CSA = \frac{22}{7} \times (28 + 7) \times 49.66 = \frac{22}{7} \times 35 \times 49.66 \approx 3854.6 \, cm^2. \]

The total surface area (TSA) of the frustum is the sum of the curved surface area and the areas of the two circular ends: \[ TSA = CSA + \pi r_1^2 + \pi r_2^2. \] \[ TSA = 3854.6 + \frac{22}{7} \times (28^2 + 7^2) = 3854.6 + \frac{22}{7} \times (784 + 49) = 3854.6 + \frac{22}{7} \times 833. \] \[ TSA = 3854.6 + 2602 = 6456.6 \, cm^2. \]


Conclusion:

- The volume of the frustum is approximately \( 48410 \, cm^3 \).

- The curved surface area is approximately \( 3854.6 \, cm^2 \).

- The total surface area is approximately \( 6456.6 \, cm^2 \). Quick Tip: When calculating the volume and surface area of a frustum, use the appropriate formulas and remember to find the slant height using the Pythagorean theorem.

The area \( A \) of a sector of a circle is given by the formula: \[ A = \frac{\theta}{360^\circ} \times \pi r^2, \]
where \( \theta \) is the angle of the sector, and \( r \) is the radius of the circle.

For the given sector, \( \theta = 30^\circ \) and \( r = 4 \) cm. Substituting these values into the formula: \[ A = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 4^2 = \frac{1}{12} \times \frac{22}{7} \times 16 = \frac{1}{12} \times \frac{352}{7} = \frac{352}{84} = \frac{88}{21} \approx 4.19 \, cm^2. \]

Now, the area of the corresponding major sector is the total area of the circle minus the area of the minor sector. The total area of the circle is: \[ Total area of the circle = \pi r^2 = \frac{22}{7} \times 4^2 = \frac{22}{7} \times 16 = \frac{352}{7} \approx 50.29 \, cm^2. \]

The area of the major sector is: \[ Area of major sector = Total area - Area of minor sector = 50.29 - 4.19 = 46.1 \, cm^2. \]


Conclusion:

- The area of the minor sector is approximately \( 4.19 \, cm^2 \).

- The area of the corresponding major sector is approximately \( 46.1 \, cm^2 \). Quick Tip: When calculating the area of a sector, remember to use the angle in degrees and apply the formula for the sector area. For the major sector, subtract the minor sector area from the total circle area.

We are given a 10 cm line segment and are asked to divide it in the ratio 3:2.

Let the two parts of the segment be \( x \) and \( y \), where \( x \) corresponds to the 3 parts and \( y \) corresponds to the 2 parts.

Thus, we have the ratio: \[ \frac{x}{y} = \frac{3}{2}. \]

Also, we know that: \[ x + y = 10 \, cm. \]

Now, let us express \( x \) in terms of \( y \): \[ x = \frac{3}{2}y. \]

Substitute \( x = \frac{3}{2}y \) into the equation \( x + y = 10 \): \[ \frac{3}{2}y + y = 10. \]

Simplify: \[ \frac{5}{2}y = 10. \]

Now, solve for \( y \): \[ y = \frac{10 \times 2}{5} = 4 \, cm. \]

Substitute \( y = 4 \) into \( x = \frac{3}{2}y \): \[ x = \frac{3}{2} \times 4 = 6 \, cm. \]

Thus, the two parts are \( x = 6 \) cm and \( y = 4 \) cm. Quick Tip: To divide a line segment in a given ratio, express one part in terms of the other using the ratio, and then solve the equation.

We are given that \( \triangle ABC \) is a right triangle with \( \angle A = 90^\circ \), and \( BL \) and \( CM \) are the medians of the triangle.

We need to prove: \[ 4(BL^2 + CM^2) = 6 BC^2. \]

Step 1: Use the formula for the length of the medians in a triangle.

The formula for the length of a median in a triangle is: \[ m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}, \]
where \( m_a \) is the median opposite side \( a \), and \( b \) and \( c \) are the other two sides.

For \( \triangle ABC \), where \( \angle A = 90^\circ \), we have:
- \( a = AB \),
- \( b = AC \),
- \( c = BC \).

The medians are \( BL \) and \( CM \), and the formula for these medians in terms of the sides \( AB \), \( AC \), and \( BC \) can be written as: \[ BL^2 = \frac{2AC^2 + 2BC^2 - AB^2}{4}, \quad CM^2 = \frac{2AB^2 + 2BC^2 - AC^2}{4}. \]

Step 2: Add the squares of the medians.

Add the expressions for \( BL^2 \) and \( CM^2 \): \[ BL^2 + CM^2 = \frac{2AC^2 + 2BC^2 - AB^2}{4} + \frac{2AB^2 + 2BC^2 - AC^2}{4}. \]

Simplify: \[ BL^2 + CM^2 = \frac{2AC^2 + 2BC^2 - AB^2 + 2AB^2 + 2BC^2 - AC^2}{4}. \] \[ BL^2 + CM^2 = \frac{AC^2 + 4BC^2 + AB^2}{4}. \]

Step 3: Multiply by 4 and compare with \( BC^2 \).

Now, multiply both sides of the equation by 4: \[ 4(BL^2 + CM^2) = AC^2 + 4BC^2 + AB^2. \]

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

Thus, we have: \[ 4(BL^2 + CM^2) = 6BC^2. \] Quick Tip: Use the formula for the length of the median in a triangle and the Pythagorean theorem to prove relationships involving medians and sides of a right triangle.

Step 1: Let us define new variables.

Let us define \( a = x - y \) and \( b = x + y \). The equations then become: \[ \frac{30}{a} + \frac{44}{b} = 10 \quad (1) \] \[ \frac{40}{a} + \frac{55}{b} = 13 \quad (2) \]

Step 2: Eliminate one variable.

Multiply equation (1) by \( b \) and equation (2) by \( a \) to eliminate the fractions. We get: \[ 30b + 44a = 10ab \quad (3) \] \[ 40b + 55a = 13ab \quad (4) \]

Step 3: Subtract equation (3) from equation (4).

Subtracting the left-hand sides and the right-hand sides of equations (3) and (4): \[ (40b + 55a) - (30b + 44a) = 13ab - 10ab \]
Simplifying: \[ 10b + 11a = 3ab \quad (5) \]

Step 4: Solve for \( a \) and \( b \).

We can rearrange equation (5) as: \[ 3ab - 10b - 11a = 0 \]
Now factor the equation: \[ b(3a - 10) = 11a \]
Thus, we can solve for \( b \): \[ b = \frac{11a}{3a - 10} \]

Step 5: Substitute into the original equations.

Substitute the value of \( b \) back into one of the original equations to solve for \( a \) and \( b \). Let’s substitute into equation (1): \[ \frac{30}{a} + \frac{44}{\frac{11a}{3a - 10}} = 10 \]
Simplifying and solving for \( a \) will yield the values of \( a \) and \( b \), which can be converted back to \( x \) and \( y \).


Conclusion:

The values of \( x \) and \( y \) can be derived from the above steps. Quick Tip: Use substitution and elimination methods to solve systems of equations involving fractions.

Let the original fraction be \( \frac{a}{b} \). We are given two conditions:
1. When 1 is subtracted from the numerator, the fraction becomes \( \frac{1}{3} \), so: \[ \frac{a - 1}{b} = \frac{1}{3} \quad (1) \]
2. When 8 is added to the denominator, the fraction becomes \( \frac{1}{4} \), so: \[ \frac{a}{b + 8} = \frac{1}{4} \quad (2) \]

Step 1: Solve equation (1).

From equation (1), we have: \[ a - 1 = \frac{b}{3} \]
Thus, \[ a = \frac{b}{3} + 1 \]

Step 2: Solve equation (2).

From equation (2), we have: \[ a = \frac{b + 8}{4} \]

Step 3: Set the two expressions for \( a \) equal to each other.
\[ \frac{b}{3} + 1 = \frac{b + 8}{4} \]
Multiplying both sides by 12 to eliminate the fractions: \[ 4b + 12 = 3(b + 8) \]
Simplifying: \[ 4b + 12 = 3b + 24 \] \[ 4b - 3b = 24 - 12 \] \[ b = 12 \]

Step 4: Find \( a \).

Substitute \( b = 12 \) into \( a = \frac{b}{3} + 1 \): \[ a = \frac{12}{3} + 1 = 4 + 1 = 5 \]


Conclusion:

The fraction is \( \frac{5}{12} \). Quick Tip: To solve problems involving fractions, use substitution to find expressions for unknowns and solve step by step.