CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 2- 30/4/2) with Answer Key

We are given \(\sin \theta = 1\). This implies: \[ \theta = 90^\circ \quad (since \(\sin 90^\circ = 1\)). \]

Substitute \(\theta = 90^\circ\) into the expression \(\frac{1}{2} \sin\left(\frac{\theta}{2}\right)\): \[ \frac{1}{2} \sin\left(\frac{\theta}{2}\right) = \frac{1}{2} \sin\left(\frac{90^\circ}{2}\right) = \frac{1}{2} \sin(45^\circ). \]

We know \(\sin 45^\circ = \frac{1}{\sqrt{2}}\), so: \[ \frac{1}{2} \sin(45^\circ) = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}. \]

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Conclusion:

The value of \(\frac{1}{2} \sin\left(\frac{\theta}{2}\right)\) is \(\frac{1}{2\sqrt{2}}\). Quick Tip: Remember that \(\sin 45^\circ = \frac{1}{\sqrt{2}}\). Use trigonometric identities and angle relationships to simplify such expressions.

The total surface area of a solid hemisphere is the sum of its curved surface area and the area of its circular base: \[ Total Surface Area = 2\pi r^2 + \pi r^2 = 3\pi r^2. \]

The square of the radius is: \[ Square of radius = r^2. \]

The ratio of total surface area to the square of the radius is: \[ Ratio = \frac{3\pi r^2}{r^2} = 3\pi : 1. \]

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Conclusion:

The ratio of total surface area to the square of the radius is \(3\pi : 1\). Quick Tip: For solid hemispheres, remember that the total surface area includes both the curved surface and the base.

In the given figure, the tangent \(QR\) is divided into two equal segments at \(P\) because the circles are symmetric, and the tangent passes through the point of contact \(A\). Thus: \[ QR = 2 \times AP. \]

Substitute the given value \(AP = 4.2 \, cm\): \[ QR = 2 \times 4.2 = 8.4 \, cm. \]

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Conclusion:

The length of \(QR\) is \(8.4 \, cm\). Quick Tip: For tangents between two externally touching circles, the tangent length is twice the distance from the point of contact to the tangent's midpoint.

The shaded region represents a sector of a circle with a central angle of \(60^\circ\). The radius of the sector is \(3 \, cm\) (distance \(AP\)). The area of a sector is calculated as:
\[ Area of sector = \frac{\theta}{360^\circ} \pi r^2 \]

Substituting the values: \[ Area of sector = \frac{60^\circ}{360^\circ} \pi (3)^2 = \frac{1}{6} \pi (9) = 3\pi \, cm^2 \]

Thus, the area of the shaded region is \(3\pi \, cm^2\). Quick Tip: For sector problems, remember to use \(\frac{\theta}{360^\circ} \pi r^2\) for the area, where \(\theta\) is the central angle in degrees and \(r\) is the radius.

The formula for mean is: \[ Mean = \frac{Sum of all observations}{Number of observations} \]

Substituting the given values: \[ 9 = \frac{6 + 7 + p + 8 + q + 14}{6} \]

Simplify: \[ 9 \times 6 = 6 + 7 + p + 8 + q + 14 \] \[ 54 = 35 + p + q \] \[ p + q = 19 \]

Thus, the correct answer is \(p + q = 19\). Quick Tip: For mean calculations, always multiply the mean by the number of observations to find the total sum.

The shadow of the tower is equal to its height. Let the height of the tower be \(h\) and the length of its shadow also be \(h\). The altitude of the Sun forms a right triangle, where: \[ \tan(Sun's altitude) = \frac{Height of tower}{Length of shadow}. \]

Substitute: \[ \tan(Sun's altitude) = \frac{h}{h} = 1. \]

We know: \[ \tan(45^\circ) = 1. \]

Thus, the Sun's altitude is \(45^\circ\).

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Conclusion:

The Sun's altitude is \(45^\circ\). Quick Tip: When the height and shadow of an object are equal, the angle of elevation is always \(45^\circ\).

The length of an arc is proportional to the angle it subtends at the centre of the circle. The total circumference of a circle corresponds to an angle of \(360^\circ\). Hence, the ratio of the arc length to the circumference is: \[ Ratio = \frac{Angle subtended by arc}{Total angle of circle} = \frac{90^\circ}{360^\circ}. \]

Simplify: \[ Ratio = \frac{90}{360} = \frac{1}{4}. \]

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Conclusion:

The ratio of the arc length to the circumference is \(1:4\). Quick Tip: The arc length ratio can be determined by dividing the angle subtended by the arc by \(360^\circ\).

We are given two equations: \[ 1. \ ax + by = a^2 - b^2 \] \[ 2. \ bx + ay = 0 \]

From equation (2), solve for \( x \) or \( y \) in terms of the other variable. Assume \( ay = -bx \), so: \[ y = -\frac{b}{a}x \quad (substitute this into equation 1). \]

Substituting \( y = -\frac{b}{a}x \) into equation (1): \[ ax + b\left(-\frac{b}{a}x\right) = a^2 - b^2 \]

Simplify: \[ ax - \frac{b^2}{a}x = a^2 - b^2 \]

Combine terms: \[ \frac{a^2x - b^2x}{a} = a^2 - b^2 \]

Factorize \( x \): \[ x(a^2 - b^2) = a(a^2 - b^2) \]

Divide both sides by \( a^2 - b^2 \) (assuming \( a \neq b \)): \[ x = a \]

Substitute \( x = a \) into \( y = -\frac{b}{a}x \): \[ y = -\frac{b}{a}(a) = -b \]

Finally, calculate \( x + y \): \[ x + y = a - b \]

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Conclusion:

The value of \( x + y \) is \( a - b \). Quick Tip: When solving simultaneous equations, substitute one variable in terms of the other to simplify and solve step-by-step.

The tangents \(AB\) and \(AC\) meet at point \(A\), and the angle between the tangents at the external point is calculated as follows: \[ \angle BAC = 180^\circ - 2 \times \angle ABC \]

Substituting the values: \[ \angle BAC = 180^\circ - 2 \times 42^\circ = 180^\circ - 84^\circ = 96^\circ \]

Thus, the measure of \(\angle BAC\) is \(96^\circ\). Quick Tip: For angles formed by tangents and circles, use the formula \(\angle BAC = 180^\circ - 2 \times \angle ABC\) to find the desired angle.

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

For the equation \(3x^2 - 2x + c = 0\), \(a = 3\), \(b = -2\), and \(c\) is unknown. Substituting the given discriminant: \[ 16 = (-2)^2 - 4 \cdot 3 \cdot c \]

Simplify: \[ 16 = 4 - 12c \] \[ 12c = 4 - 16 = -12 \] \[ c = -1 \]

Thus, the value of \(c\) is \(-1\). Quick Tip: Remember, for quadratic equations, \(b^2 - 4ac\) determines the nature of roots: \(\Delta > 0\) for real and distinct roots, \(\Delta = 0\) for real and equal roots, and \(\Delta < 0\) for complex roots.

In a parallelogram, the diagonals bisect each other. Let the coordinates of the fourth vertex \(D\) be \((x, y)\). Using the midpoint formula, the midpoint of diagonal \(AC\) must be the same as the midpoint of diagonal \(BD\).

Midpoint of \(AC\): \[ \left(\frac{-2 + 8}{2}, \frac{3 + 3}{2}\right) = \left(3, 3\right). \]

Midpoint of \(BD\): \[ \left(\frac{6 + x}{2}, \frac{7 + y}{2}\right). \]

Equating the midpoints: \[ \frac{6 + x}{2} = 3 \quad \Rightarrow \quad 6 + x = 6 \quad \Rightarrow \quad x = 0, \] \[ \frac{7 + y}{2} = 3 \quad \Rightarrow \quad 7 + y = 6 \quad \Rightarrow \quad y = -1. \]

Thus, the coordinates of \(D\) are \((0, -1)\).

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Conclusion:

The fourth vertex \(D\) is at \((0, -1)\). Quick Tip: In a parallelogram, use the midpoint formula to find the unknown vertex by equating the midpoints of the diagonals.

Each die has \(6\) faces: \(1, 2, 3, 4, 5, 6\). Odd numbers are \(1, 3, 5\), so the probability of getting an odd number on one die is: \[ P(odd on one die) = \frac{3}{6} = \frac{1}{2}. \]

The two dice are independent, so the probability of getting odd numbers on both dice is: \[ P(odd on both dice) = P(odd on first die) \cdot P(odd on second die) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. \]

Convert to common fractions: \[ P(odd on both dice) = \frac{9}{36}. \]

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Conclusion:

The probability of getting odd numbers on both dice is \(9/36\). Quick Tip: For independent events, multiply individual probabilities to find the combined probability.

Two lines are parallel if their slopes are equal. The general equation of a line is given by: \[ Ax + By + C = 0. \]

The slope of the line is: \[ Slope = -\frac{A}{B}. \]

For the given line \(5x - 3y = 2\), the slope is: \[ Slope = -\frac{5}{-3} = \frac{5}{3}. \]

Now check the options for the second line to find which has the same slope:
- For \(-15x + 9y = 5\):
\[ Slope = -\frac{-15}{9} = \frac{5}{3}. \]

This matches the slope of the given line, so the two lines are parallel.

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Conclusion:

The equation of the second line is \(-15x + 9y = 5\). Quick Tip: Parallel lines have the same slope. To verify, compare \(-\frac{A}{B}\) for both equations.

Using the Basic Proportionality Theorem (Thales Theorem), which states that if a line is parallel to one side of a triangle and intersects the other two sides, it divides them in the same ratio:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Given: \[ AD = 4 \, cm, \, AB = 9 \, cm, \, AC = 13.5 \, cm \]

Calculate \(DB\): \[ DB = AB - AD = 9 - 4 = 5 \, cm \]

Substituting into the proportion: \[ \frac{4}{5} = \frac{AE}{EC} \]

Let \(AE = x\) and \(EC = y\). Since \(AE + EC = AC\): \[ x + y = 13.5 \, cm \]

Substitute \(\frac{4}{5} = \frac{x}{y}\): \[ x = \frac{4}{5} y \]

Substitute \(x = \frac{4}{5} y\) into \(x + y = 13.5\): \[ \frac{4}{5} y + y = 13.5 \]

Simplify: \[ \frac{9}{5} y = 13.5 \]
\[ y = \frac{13.5 \times 5}{9} = 7.5 \, cm \]

Thus, \(EC = 7.5 \, cm\). Quick Tip: For questions involving parallel lines in triangles, use the Basic Proportionality Theorem to establish the ratios of corresponding segments.

The word "MOBILE" consists of 6 letters: M, O, B, I, L, E. Among these, the vowels are O, I, and E (3 vowels).

The total number of letters is 6. The probability of selecting a vowel is:
\[ Probability = \frac{Number of vowels}{Total letters} = \frac{3}{6} = \frac{1}{2}. \] Quick Tip: Always count the vowels (A, E, I, O, U) carefully, and divide by the total number of letters for probability questions.

Factorize \(3825\):
\[ 3825 \div 3 = 1275, \quad 1275 \div 3 = 425 \quad \Rightarrow 3^2 \] \[ 425 \div 5 = 85, \quad 85 \div 5 = 17 \quad \Rightarrow 5^2 and 17^1 \]

Thus: \[ 3825 = 3^2 \times 5^2 \times 17^1 \]

Here: \[ x = 2, \, y = 2, \, z = 1 \]

Now calculate: \[ x + y - 2z = 2 + 2 - 2 \times 1 = 2. \] Quick Tip: For prime factorization questions, divide the number repeatedly by primes until it reduces to 1.

Let the zeroes of the polynomial be \(2 + \sqrt{5}\) and \(2 - \sqrt{5}\). Using the sum and product of roots: \[ Sum of roots = (2 + \sqrt{5}) + (2 - \sqrt{5}) = 4, \] \[ Product of roots = (2 + \sqrt{5})(2 - \sqrt{5}) = 4 - 5 = -1. \]

The quadratic polynomial is given by: \[ x^2 - (Sum of roots)x + (Product of roots) = x^2 - 4x - 1. \]

Thus, the required polynomial is \(x^2 - 4x - 1\). Quick Tip: For quadratic equations, use the sum and product of roots formulas: \[ x^2 - (Sum of roots)x + (Product of roots). \]

In an A.P., the difference between consecutive terms is constant. Hence: \[ (3p + 5) - (3p - 1) = (5p + 1) - (3p + 5). \]

Simplify both sides: \[ 3p + 5 - 3p + 1 = 5p + 1 - 3p - 5, \] \[ 6 = 2p - 4. \]

Solve for \(p\): \[ 2p = 6 + 4 = 10, \quad p = 5. \]

Thus, the value of \(p\) is \(5\). Quick Tip: In arithmetic progressions, verify the common difference by equating consecutive term differences.

1. The formula for the circumference of a circle is: \[ Circumference = 2\pi r. \]

2. Given the circumference is \(176 \, cm\), solve for the radius: \[ 176 = 2\pi r \quad \Rightarrow \quad r = \frac{176}{2\pi} = \frac{88}{\pi}. \]

Approximating \(\pi \approx 3.14\): \[ r = \frac{88}{3.14} = 28 \, cm. \]

3. Both the Assertion (A) and the Reason (R) are true, and Reason (R) correctly explains the Assertion (A).

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Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). Quick Tip: For problems involving circles, always use the formula \(Circumference = 2\pi r\) and substitute known values to find unknown parameters.

1. The mid-point of a line segment always divides the segment into two equal parts, which means the ratio is \(1 : 1\). Therefore, Assertion (A) is true.

2. To verify Reason (R), let the point \((-3, k)\) divide the segment joining \((-5, 4)\) and \((-2, 3)\). Using the section formula: \[ x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \quad y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}. \]
For the x-coordinate: \[ -3 = \frac{1 \cdot (-2) + 2 \cdot (-5)}{1 + 2} = \frac{-2 - 10}{3} = \frac{-12}{3} = -4 \, (which does not match -3). \]
Thus, the point does not divide the line segment in the ratio \(1 : 2\). Reason (R) is false.

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Conclusion:

Assertion (A) is true, but Reason (R) is false. Quick Tip: Use the section formula to determine the ratio in which a point divides a line segment.

N/A

We know: \[ \sin(A - B) = \sin 30^\circ \implies A - B = 30^\circ \tag{1} \] \[ \cos(A + B) = \cos 60^\circ \implies A + B = 60^\circ \tag{2} \]

Adding and subtracting (1) and (2): \[ (A + B) + (A - B) = 60^\circ + 30^\circ \implies 2A = 90^\circ \implies A = 45^\circ. \] \[ (A + B) - (A - B) = 60^\circ - 30^\circ \implies 2B = 30^\circ \implies B = 15^\circ. \]

Thus, \(\angle A = 45^\circ\) and \(\angle B = 15^\circ\). Quick Tip: To solve trigonometric equations involving angles, use known values of trigonometric ratios and logical constraints on angles.

Since \(\triangle AHK \sim \triangle ABC\), the corresponding sides are proportional: \[ \frac{AK}{AC} = \frac{HK}{BC}. \]

Substitute the known values: \[ \frac{8}{AC} = \frac{6.4}{3.2}. \]

Simplify: \[ \frac{8}{AC} = 2 \quad \Rightarrow \quad AC = \frac{8}{2} = 4 \, cm. \]

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Conclusion:

The length of \(AC\) is \(4 \, cm\). Quick Tip: For similar triangles, use the property that corresponding sides are proportional to find unknown lengths.

To find the minimum number of books, calculate the LCM of the number of students in both sections.
\[ 40 = 2^3 \times 5, \quad 48 = 2^4 \times 3 \]
\[ L.C.M.(40, 48) = 2^4 \times 3 \times 5 = 240 \]

Thus, the minimum number of books required in the library is: \[ \boxed{240}. \] Quick Tip: To find the minimum items required for equal distribution, always calculate the LCM of the given quantities.

The angle subtended by the minute hand in \(5\) minutes is: \[ Angle = \frac{360^\circ}{60} \times 5 = 30^\circ. \]

The area described by the minute hand is a sector of a circle, given by: \[ Area = \frac{\theta}{360^\circ} \cdot \pi r^2, \]
where \(\theta = 30^\circ\) and \(r = 14 \, cm\).

Substitute the values: \[ Area = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14. \]

Simplify: \[ Area = \frac{1}{12} \times \frac{22}{7} \times 196 = \frac{154}{3} \, cm^2 \approx 51.33 \, cm^2. \]

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Conclusion:

The area described by the minute hand is approximately \(51.33 \, cm^2\). Quick Tip: The area of a sector is proportional to the angle subtended by it at the center of the circle. Always express the angle in degrees when using the formula.

The formula for the length of an arc is: \[ Length of arc = 2\pi r \cdot \frac{\theta}{360}, \]
where \(r\) is the radius and \(\theta\) is the central angle in degrees.

Substitute the given values \(r = 42 \, cm\) and \(\theta = 60^\circ\): \[ Length of arc = 2 \times \frac{22}{7} \times 42 \times \frac{60}{360}. \]

Simplify step-by-step: \[ Length of arc = 2 \times \frac{22}{7} \times 42 \times \frac{1}{6}. \] \[ Length of arc = \frac{2 \times 22 \times 42}{7 \times 6}. \] \[ Length of arc = \frac{1848}{42} = 44 \, cm. \]

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Conclusion:

The length of the arc is \(44 \, cm\). Quick Tip: The length of an arc is proportional to the central angle it subtends. Use \(Length of arc = 2\pi r \cdot \frac{\theta}{360}\) for quick calculations.

In a circle, the angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the circumference.

1. Let \(P\) be a point on the circumference of the circle. Join \(AP\) and \(BP\).

2. The angle subtended by the arc \(AB\) at the circumference is: \[ \angle APB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 145^\circ = 72.5^\circ. \]

3. In \(\triangle APB\), the sum of angles \(\angle APB + \angle ACB = 180^\circ\) (angles on a straight line).

4. Solve for \(\angle ACB\): \[ \angle ACB = 180^\circ - \angle APB = 180^\circ - 72.5^\circ = 107.5^\circ. \]

Thus: \[ x = \angle ACB = 107.5^\circ. \]

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Conclusion:

The value of \(x\) is \(107.5^\circ\). Quick Tip: In a circle, the angle subtended by an arc at the center is always twice the angle subtended by the same arc at the circumference.

The total number of outcomes when three coins are tossed is: \[ Total outcomes = 2^3 = 8 \quad (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT). \]

(i) Probability of at least one head: \[ P(at least one head) = 1 - P(no heads) = 1 - \frac{1}{8} = \frac{7}{8}. \]

(ii) Probability of exactly two tails: \[ P(exactly two tails) = \frac{Number of favorable outcomes}{Total outcomes} = \frac{3}{8} \quad (favorable outcomes: HTT, THT, TTH). \]

(iii) Probability of at most one tail: \[ P(at most one tail) = \frac{Number of favorable outcomes}{Total outcomes} = \frac{4}{8} = \frac{1}{2} \] \quad (favorable outcomes: HHH, HHT, HTH, THH).

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Conclusion:


(i) \(P(\text{at least one head) = \frac{7}{8}\).
(ii) \(P(exactly two tails) = \frac{3}{8}\).
(iii) \(P(at most one tail) = \frac{1}{2}\). Quick Tip: For coin toss problems, count all possible outcomes carefully and identify favorable cases for each condition.

The total number of outcomes is: \[ Total outcomes = 90. \]

(i) Probability of a two-digit number less than \(40\):
Two-digit numbers less than \(40\) are \(10\) to \(39\) (inclusive). The total count is: \[ Count = 30 \quad (10, 11, 12, ..., 39). \] \[ P(two-digit number less than 40) = \frac{30}{90} = \frac{1}{3}. \]

(ii) Probability of a number divisible by \(5\) and greater than \(50\):
Numbers divisible by \(5\) and greater than \(50\) are \(55, 60, 65, 70, 75, 80, 85, 90\). The total count is: \[ Count = 8. \] \[ P(divisible by 5 and greater than 50) = \frac{8}{90} = \frac{4}{45}. \]

(iii) Probability of a perfect square number:
Perfect squares from \(1\) to \(90\) are \(1, 4, 9, 16, 25, 36, 49, 64, 81\). The total count is: \[ Count = 9. \] \[ P(perfect square number) = \frac{9}{90} = \frac{1}{10}. \]

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Conclusion:


(i) \(P(two-digit number less than 40) = \frac{1}{3}\).
(ii) \(P(divisible by 5 and greater than 50) = \frac{4}{45}\).
(iii) \(P(perfect square number) = \frac{1}{10}\). Quick Tip: For probability problems, list all favorable outcomes systematically and divide by the total possible outcomes.

From the given figure:

\(AP = AS \quad (i),\)
\(BP = BQ \quad (ii),\)
\(CR = CQ \quad (iii),\)
\(DR = DS \quad (iv).\)


Adding equations (i), (ii), (iii), and (iv): \[ AP + BP + CR + DR = AS + BQ + CQ + DS. \]

This simplifies to: \[ AB + CD = AD + BC. \]

Since \(ABCD\) is a parallelogram, the opposite sides are equal: \[ AB = CD \quad and \quad AD = BC. \]

Substitute these equalities: \[ 2AB = 2AD \quad \Rightarrow \quad AB = AD. \]

Thus, all sides of \(ABCD\) are equal, which proves it is a rhombus.

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Conclusion:

The parallelogram \(ABCD\) circumscribing a circle is a rhombus. Quick Tip: A parallelogram circumscribing a circle always has equal sides due to the tangential property of the circle, making it a rhombus.

Assume \(\frac{2 - \sqrt{3}}{5}\) to be a rational number.

Let: \[ \frac{2 - \sqrt{3}}{5} = \frac{p}{q}, \]
where \(p\) and \(q\) are integers, and \(q \neq 0\).

Rearrange the equation to isolate \(\sqrt{3}\): \[ \sqrt{3} = \frac{2q - 5p}{q}. \]

Here:

\(p\) and \(q\) are integers, so \(2q - 5p\) and \(q\) are also integers.
Therefore, \(\frac{2q - 5p}{q}\) is a rational number.


However, \(\sqrt{3}\) is known to be an irrational number. This creates a contradiction.

Thus, our assumption that \(\frac{2 - \sqrt{3}}{5}\) is rational is false.

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Conclusion:
\(\frac{2 - \sqrt{3}}{5}\) is an irrational number. Quick Tip: To prove a number is irrational, assume it to be rational and show that this assumption leads to a contradiction.

\[ L.H.S. = q(p^2 - 1) \] \[ = (\sec \theta + \csc \theta) \big[ (\sin \theta + \cos \theta)^2 - 1 \big] \] \[ = \left[\frac{1}{\cos \theta} + \frac{1}{\sin \theta}\right] \big[\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1 \big] \] \[ = \left[\frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}\right] \big[1 + 2 \sin \theta \cos \theta - 1\big] \] \[ = 2(\sin \theta + \cos \theta) \] \[ = 2p \] \[ = R.H.S. \] Quick Tip: Using trigonometric identities such as \(\sin^2 \theta + \cos^2 \theta = 1\) and factoring common terms simplifies proofs like this efficiently.

The given polynomial is: \[ P(x) = 4x^2 + 4x - 3 = (2x + 3)(2x - 1). \]

Zeroes of the polynomial: \[ x = -\frac{3}{2}, \quad x = \frac{1}{2}. \]

1. Verify the sum of the zeroes: \[ Sum of zeroes = -\frac{3}{2} + \frac{1}{2} = -\frac{4}{2} = -2. \]
This matches: \[ Sum of zeroes = \frac{-(coefficient of x)}{coefficient of x^2}. \]

2. Verify the product of the zeroes: \[ Product of zeroes = -\frac{3}{2} \times \frac{1}{2} = -\frac{3}{4}. \]
This matches: \[ Product of zeroes = \frac{constant term}{coefficient of x^2}. \]

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Conclusion:

The zeroes are \(-\frac{3}{2}\) and \(\frac{1}{2}\), and they satisfy the relationship between zeroes and coefficients of the polynomial. Quick Tip: To verify zeroes of a polynomial, use the relationships: \(Sum of zeroes = -\frac{coefficient of x}{coefficient of x^2}\) and \(Product of zeroes = \frac{constant term}{coefficient of x^2}\).

From the given polynomial: \[ \alpha + \beta = -1, \quad \alpha \beta = -2. \]

We need to find: \[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta}. \]

1. Using the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\): \[ \alpha^2 + \beta^2 = (-1)^2 - 2(-2) = 1 + 4 = 5. \]

2. Substitute into the expression for \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\): \[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{5}{-2} = -\frac{5}{2}. \]

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Conclusion:

The value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) is \(-\frac{5}{2}\). Quick Tip: For problems involving zeroes of polynomials, use the relationships: \(\alpha + \beta = -\frac{coefficient of x}{coefficient of x^2}\) and \(\alpha \beta = \frac{constant term}{coefficient of x^2}\).

Let his starting salary be ₹\(a\) and annual increment be ₹\(d\).

From the question: \[ A.T.Q. \] \[ a + 3d = 15000 \quad --------(i) \] \[ a + 9d = 18000 \quad --------(ii) \]

These equations represent the salary progression where:

\(a + 3d\) is the salary after 4 years of service.
\(a + 9d\) is the salary after 10 years of service.


To find \(a\) (starting salary) and \(d\) (annual increment), we solve the equations simultaneously.

Step 1: Subtract Equation (i) from Equation (ii): \[ (a + 9d) - (a + 3d) = 18000 - 15000 \] \[ 6d = 3000 \quad \implies \quad d = 500 \]

Step 2: Substitute \(d = 500\) into Equation (i): \[ a + 3(500) = 15000 \] \[ a + 1500 = 15000 \quad \implies \quad a = 13500 \]

Final Answer:

Starting salary = ₹13,500
Annual increment = ₹500


Explanation:
The problem uses the concept of arithmetic progression, where the salary increases each year by a fixed amount \(d\). By forming equations for the salary after a specific number of years and solving them, we find both the initial salary and the annual increment. Quick Tip: For problems involving salary increments, use arithmetic progression formulas and solve equations step-by-step for accurate results.

N/A

Let the required number be \(10x + y\), where \(x\) and \(y\) are the digits of the number.

Step 1: Write the given conditions as equations. \[ Condition 1: xy = 18 \quad --------(i) \] \[ Condition 2: (10x + y) - 63 = 10y + x \quad --------(ii) \]

Simplify Equation (ii): \[ 10x + y - 63 = 10y + x \] \[ 9x - 9y = 63 \] \[ x - y = 7 \quad --------(iii) \]

Step 2: Solve Equations (i) and (iii).
From Equation (iii): \[ x = y + 7 \]

Substitute \(x = y + 7\) into Equation (i): \[ (y+7)y = 18 \] \[ y^2 + 7y - 18 = 0 \]

Factorize: \[ (y + 9)(y - 2) = 0 \] \[ y = 2 \quad (since \(y > 0\)). \]

Substitute \(y = 2\) into Equation (iii): \[ x - 2 = 7 \quad \implies \quad x = 9 \]

Step 3: Find the number.
The required number is: \[ 10x + y = 10(9) + 2 = 92 \]

Final Answer: The required number is \(92\). Quick Tip: For digit-based problems, represent the number as \(10x + y\) and systematically use the given conditions to form equations.

The table with calculated midpoints (\(x_i\)) and \(x_i f_i\) is:
\[ \begin{array}{|c|c|c|c|} \hline \textbf{Age (in years)} & \textbf{No. of patients (\(f_i\))} & \textbf{Midpoint (\(x_i\))} & \textbf{\(x_i f_i\)}
\hline 5 - 15 & 6 & 10 & 60
\hline 15 - 25 & 11 & 20 & 220
\hline 25 - 35 & 21 & 30 & 630
\hline 35 - 45 & 23 & 40 & 920
\hline 45 - 55 & 14 & 50 & 700
\hline 55 - 65 & 5 & 60 & 300
\hline \textbf{Total} & \textbf{80} & & \textbf{2830}
\hline \end{array} \]

1. Mean Calculation: \[ Mean = \frac{\sum f_i x_i}{\sum f_i} = \frac{2830}{80} = 35.375 \, years. \]

2. Mode Calculation:

The modal class is \(35\)–\(45\), as it has the highest frequency (\(f = 23\)).

Using the formula for mode: \[ Mode = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) h, \]
where:

\(l = 35\),
\(f_m = 23\),
\(f_1 = 21\),
\(f_2 = 14\),
\(h = 10\).


Substitute the values: \[ Mode = 35 + \left( \frac{23 - 21}{2(23) - 21 - 14} \right) \times 10 = 35 + \left( \frac{2}{11} \right) \times 10 = 36.81 \, years. \]

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Conclusion:

The mode and mean of the data are \(36.81 \, years\) and \(35.375 \, years\), respectively. Quick Tip: For grouped data, use the formulas for mean and mode: \[ Mean = \frac{\sum f_i x_i}{\sum f_i}, \quad Mode = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) h. \]

Given: \[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM} \]

Step 1: Show proportionality of medians.

Since the medians AD and PM divide the opposite sides proportionally: \[ \frac{AB}{PQ} = \frac{BC}{QR} \quad and \quad \frac{AD}{PM} confirms proportionality. \]

Step 2: Use the SSS similarity criterion.

From the given proportionality: \[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \]

By the SSS (Side-Side-Side) similarity criterion: \[ \triangle ABC \sim \triangle PQR \]

Conclusion: \(\triangle ABC \sim \triangle PQR\). Quick Tip: For proving similarity using SSS, show that the corresponding sides of the two triangles are proportional.

Let \(AB\) be the tower of height \(H\) meters and \(CD\) be the building.

Step 1: Apply trigonometry in \(\triangle ABD\). \[ In \triangle ABD, \quad \tan 60^\circ = \sqrt{3} = \frac{H}{x} \] \[ \implies H = \sqrt{3} \cdot x \quad --------(i) \]

Step 2: Apply trigonometry in \(\triangle AEC\). \[ In \triangle AEC, \quad \tan 45^\circ = 1 = \frac{H - 50}{x} \] \[ \implies H - 50 = x \quad --------(ii) \]

Step 3: Solve equations (i) and (ii) to find \(H\).
From Equation (ii): \[ x = H - 50 \]

Substitute \(x = H - 50\) into Equation (i): \[ H = \sqrt{3} \cdot (H - 50) \] \[ H = \sqrt{3} \cdot H - 50\sqrt{3} \] \[ H - \sqrt{3} \cdot H = -50\sqrt{3} \] \[ H(1 - \sqrt{3}) = -50\sqrt{3} \] \[ H = \frac{50\sqrt{3}}{\sqrt{3} - 1} \]

Step 4: Rationalize the denominator. \[ H = \frac{50\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \] \[ H = \frac{50\sqrt{3}(\sqrt{3} + 1)}{3 - 1} \] \[ H = 25\sqrt{3}(\sqrt{3} + 1) \] \[ H = 25(3 + \sqrt{3}) = 75 + 25\sqrt{3} \]
Substitute \(\sqrt{3} = 1.73\): \[ H = 75 + 25(1.73) \] \[ H = 75 + 43.25 = 118.25 \]

Final Answer: The height of the tower is \(118.25 \, m\). Quick Tip: Use trigonometric ratios such as \(\tan\) for angle of depression problems, and ensure all calculations are consistent with given values like \(\sqrt{3}\).

The dimensions of the carton are 30 cm × 32 cm × 15 cm.

Using the volume formula for a cuboid:

Number of packets:

Hence, 24 packets can be filled in the carton.

• The volume of the cuboidal carton is 14400 cm³.
• The total surface area of the milk packet is 470 cm².
• 24 milk packets can be filled in the carton.
• The cup can hold 550 mL of milk.

N/A Quick Tip: For 3D geometry problems, use the volume and surface area formulas specific to the given shape (cuboid or cylinder). Pay attention to unit conversions if required.