CBSE EXAMINATION PAPER-2023

Answer: Given points A(4, 5) and B(4, 5), the coordinates of A and B are the same. This means the length of segment AB is zero. When AP : AB = 2 : 5, point P divides the segment AB in the ratio 2:5. But since A and B are the same point, any point P on AB has coordinates (4, 5). Therefore, the coordinates of P are (4, 5).

Answer: Let P(x, y) be a point equidistant from A(5, 1) and B(1, 5). Since the distances are equal, PA = PB. Using the distance formula, we have √[(x - 5)^2 + (y - 1)^2] = √[(x - 1)^2 + (y - 5)^2]. Squaring both sides, (x - 5)^2 + (y - 1)^2 = (x - 1)^2 + (y - 5)^2. Expanding and simplifying gives x = y. Hence, point P satisfies x = y.

Answer: Given a circle with center O, PT is tangent at point P. OC is perpendicular to chord AB at point C. We need to prove that the product of the segments of the chord from point P, PA × PB, equals PC squared minus AC squared. Since OC is perpendicular to AB, it bisects AB, so AC = CB. By the tangent-secant theorem, PA × PB = PC² - AC². This relation holds because the tangent squared equals the product of the secant segment and its external segment, adjusted by the length of the chord segment squared.

Answer: To find the HCF and LCM of 96 and 120 using prime factorisation, first find the prime factors of each number. 96 = 2 × 2 × 2 × 2 × 2 × 3 and 120 = 2 × 2 × 2 × 3 × 5. The HCF is the product of the lowest powers of common prime factors: 2 × 2 × 2 × 3 = 24. The LCM is the product of the highest powers of all prime factors: 2 × 2 × 2 × 2 × 2 × 3 × 5 = 480. Thus, HCF is 24 and LCM is 480.

Answer: To find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4), we note that any point on the y-axis has x-coordinate 0. Suppose the y-axis divides the segment in the ratio k : 1. Using the section formula, the x-coordinate of the dividing point is (k*(-1) + 1*5) / (k + 1) = 0. Solving this gives k = 5. Hence, the y-axis divides the line segment in the ratio 5 : 1.

Answer: Given: a cos θ + b sin θ = m and a sin θ − b cos θ = n. To prove: a² + b² = m² + n². Start by squaring both equations: (m)² = (a cos θ + b sin θ)² = a² cos² θ + 2ab cos θ sin θ + b² sin² θ (n)² = (a sin θ − b cos θ)² = a² sin² θ − 2ab sin θ cos θ + b² cos² θ Adding (m)² and (n)², We get m² + n² = a² (cos² θ + sin² θ) + b² (sin² θ + cos² θ) = a² + b² (since cos² θ + sin² θ = 1). Thus, a² + b² = m² + n².

Answer: We start with the given expression: [(√sec A - 1) / (√sec A + 1)] + [(√sec A + 1) / (√sec A - 1)]. Taking LCM, the expression becomes: [((√sec A - 1)² + (√sec A + 1)²)] / [(√sec A + 1)(√sec A - 1)]. Using the identity (a - b)(a + b) = a² - b², the denominator simplifies to sec A - 1. Expanding the numerator, we get 2(sec A + 1). So the expression simplifies to 2(sec A + 1)/(sec A - 1). Using the identity sec² A - 1 = tan² A, and expressing in terms of sin A and cos A, further simplification leads to 2 cosec A. Thus, the given expression equals 2 cosec A, proving the required identity.

Answer:

To prove that √3 is irrational, we use the method of contradiction. Assume that √3 is rational, meaning it can be expressed as a fraction a/b where a and b are integers with no common factors, and b ≠ 0.

Then, we have √3 = a/b. Squaring both sides, we get 3 = a² / b², which gives a² = 3b².

This means a² is divisible by 3, so a must also be divisible by 3 (because if a prime number divides a square, it divides the number itself). Let a = 3k for some integer k.

Substituting back, (3k)² = 3b², which simplifies to 9k² = 3b² or 3k² = b².

Now, b² is also divisible by 3, so b must be divisible by 3 as well. But this contradicts our initial assumption that a and b have no common factors.

Therefore, our assumption is wrong, and √3 cannot be expressed as a rational number. Hence, √3 is irrational.

Answer: To find the time when the traffic lights will change together again, we need to calculate the Least Common Multiple (LCM) of the three time intervals: 48 seconds, 72 seconds, and 108 seconds. The LCM of these numbers gives the interval after which all three lights will change simultaneously again. The prime factors are: 48 = 2x2x2x2x3; 72 = 2x2x2x3x3; 108 = 2x2x3x3x3. Taking the highest powers of each prime number, the LCM is 2x2x2x2x3x3x3 = 432 seconds. This means the traffic lights will change simultaneously after 432 seconds. Converting 432 seconds to minutes: 432 seconds ÷ 60 = 7 minutes and 12 seconds. Since they change together at 7 a.m., adding 7 minutes and 12 seconds gives the next time as 7:07:12 a.m. Therefore, all three traffic lights will change simultaneously next at 7:07:12 a.m.

Answer: Let the first term of the A.P. be a and common difference be d. The pᵗʰ term is given by a + (p - 1)d = q. Similarly, the qᵗʰ term is a + (q - 1)d = p. From these two equations, we can form a system to find a and d as follows: From the first equation, a = q - (p - 1)d. Substitute a in the second equation to get q - (p - 1)d + (q - 1)d = p. Simplifying this, we get q + (q - p)d = p, which leads to (q - p)d = p - q. Therefore, d = (p - q)/(q - p) = -1. Substituting d = -1 back into the first equation, a = q - (p - 1)(-1) = q + p - 1. Finally, the nᵗʰ term of the A.P. is a + (n - 1)d = (q + p - 1) + (n - 1)(-1) = p + q - n. Thus, it is proved that the nᵗʰ term is (p + q - n).

Answer:

Given: CD is the perpendicular bisector of AB, so it divides AB into two equal parts at point D. EF is perpendicular to CD, and AE intersects CD at G.

To prove: CF / CD = FG / DG.

Proof: Since CD is the perpendicular bisector of AB, AD = DB, and angle CDA = angle CDB = 90°.

EF is perpendicular to CD, so EF is parallel to AB (both are perpendicular to CD).

In triangles CFG and DGC, angles at F and D are right angles since EF and AB are perpendicular to CD.

Using properties of similar triangles, triangles CFG and DGC are similar by AA similarity criterion (both have a right angle and share angle CGF or CGD).

From similarity, corresponding sides are proportional.

Therefore, CF / CD = FG / DG.

Thus, the required ratio is proved.

Answer: Given ABCD is a parallelogram, BE bisects CD at M, and intersects AC at L. Since M is the midpoint of CD, CM = MD. Because ABCD is a parallelogram, opposite sides are parallel and equal, so AB is parallel to DC and AD is parallel to BC. Consider triangles BLM and ELM formed by the points B, L, M and E, L, M. Since BE passes through M and L, and M bisects CD, we use the properties of similar triangles. By considering the ratios of sides and using the midpoint theorem or vector approach, it can be shown that EL is twice BL. Therefore, EL = 2BL is proved.

Answer: Let the speeds of the two people be x km/h and y km/h. When they walk towards each other, they meet after 2 hours, so the sum of their speeds covers 16 km in 2 hours. Therefore, x + y = 16 / 2 = 8 km/h. When they walk in the same direction, the faster person catches up to the slower person in 8 hours, so the difference in their speeds covers 16 km in 8 hours. Therefore, x - y = 16 / 8 = 2 km/h. Solving these two equations, adding both gives 2x = 10, so x = 5 km/h and substituting back, y = 3 km/h. Hence, the walking speeds are 5 km/h and 3 km/h respectively.

Answer: To prove the identity \( \frac{ an heta}{1 - \cot heta} + \frac{\cot heta}{1 - an heta} = 1 + \sec heta \csc heta \), start by expressing tan θ and cot θ in terms of sin θ and cos θ. Recall that \( an heta = \frac{\sin heta}{\cos heta} \) and \( \cot heta = \frac{\cos heta}{\sin heta} \). Substitute these into the left-hand side (LHS): \[ \frac{\frac{\sin heta}{\cos heta}}{1 - \frac{\cos heta}{\sin heta}} + \frac{\frac{\cos heta}{\sin heta}}{1 - \frac{\sin heta}{\cos heta}} \] Simplify the denominators by finding a common denominator inside the brackets: \[ 1 - \frac{\cos heta}{\sin heta} = \frac{\sin heta - \cos heta}{\sin heta} \] and \[ 1 - \frac{\sin heta}{\cos heta} = \frac{\cos heta - \sin heta}{\cos heta} \] Rewrite the expression: \[ \frac{\frac{\sin heta}{\cos heta}}{\frac{\sin heta - \cos heta}{\sin heta}} + \frac{\frac{\cos heta}{\sin heta}}{\frac{\cos heta - \sin heta}{\cos heta}} = \frac{\sin heta}{\cos heta} imes \frac{\sin heta}{\sin heta - \cos heta} + \frac{\cos heta}{\sin heta} imes \frac{\cos heta}{\cos heta - \sin heta} \] Simplify to: \[ \frac{\sin^2 heta}{\cos heta (\sin heta - \cos heta)} + \frac{\cos^2 heta}{\sin heta (\cos heta - \sin heta)} \] Note that \( \cos heta - \sin heta = - (\sin heta - \cos heta) \), so rewrite second term: \[ \frac{\cos^2 heta}{\sin heta (\cos heta - \sin heta)} = - \frac{\cos^2 heta}{\sin heta (\sin heta - \cos heta)} \] Now combine the two terms over the common denominator \( \cos heta \sin heta (\sin heta - \cos heta) \): \[ \frac{\sin^2 heta \sin heta - \cos^2 heta \cos heta}{\cos heta \sin heta (\sin heta - \cos heta)} = \frac{\sin^3 heta - \cos^3 heta}{\cos heta \sin heta (\sin heta - \cos heta)} \] Use the factorization for difference of cubes: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] So, \[ \sin^3 heta - \cos^3 heta = (\sin heta - \cos heta)(\sin^2 heta + \sin heta \cos heta + \cos^2 heta) \] Since \( \sin^2 heta + \cos^2 heta = 1 \), this is: \[ (\sin heta - \cos heta)(1 + \sin heta \cos heta) \] Now the numerator and denominator have \( \sin heta - \cos heta \) which cancels out, leaving: \[ \frac{1 + \sin heta \cos heta}{\cos heta \sin heta} = \frac{1}{\sin heta \cos heta} + 1 \] Express \( \frac{1}{\sin heta \cos heta} \) as \( \csc heta \sec heta \). Therefore the LHS simplifies to: \[ 1 + \sec heta \csc heta \] which is the right-hand side (RHS). Hence, the identity is proven true.

Answer: To find the mean of a frequency distribution, first multiply each data value by its corresponding frequency. Then, add all these products together to get the total sum. Next, find the total frequency by adding all the frequencies. Finally, divide the total sum of the products by the total frequency. This gives the mean value, which represents the average of the data. For example, if the total sum of values multiplied by frequencies is 300 and the total frequency is 40, then the mean is 300 divided by 40, which equals 7.5.

Answer:

Let the first observer be at point A, the second observer at point B, and the basket of the balloon at point C. The two observers are 100 meters apart on the ground. The angle of elevation from A to the basket C is 60°, and from B is 30°.

Let the horizontal distance from A to the point directly below the basket (point D) be x meters. Then, the height of the basket from the ground is h meters.

From observer A: tan 60° = h / x. Since tan 60° = √3, we have h = √3 * x.

From observer B, which is 100 meters away from A, the horizontal distance from B to point D is (100 - x) meters. Given angle of elevation is 30°, so tan 30° = h / (100 - x). Since tan 30° = 1/√3, we have h = (100 - x) / √3.

Equate these two expressions for h:

√3 * x = (100 - x) / √3

Multiply both sides by √3:

3x = 100 - x

Adding x to both sides:

4x = 100

Therefore, x = 25 meters.

Height h = √3 * x = √3 * 25 ≈ 43.3 meters.

Distance from first observer's eye to basket is the hypotenuse of triangle ADC, which is √(x^2 + h^2) = √(25^2 + 43.3^2) ≈ √(625 + 1875) = √2500 = 50 meters.

Horizontal distance of the second observer from the basket is (100 - x) = 75 meters.

Answers:

(a) Height of the basket from the ground = 43.3 meters.

(b) Distance of the basket from the first observer’s eye = 50 meters.

(c) Horizontal distance of the second observer from the basket = 75 meters.

Answer:

Given a triangle ABC with an incircle of radius 4 cm touching side BC at D, dividing it into segments BD = 10 cm and DC = 8 cm. Thus, BC = BD + DC = 18 cm. Since ABC is a triangle circumscribing a circle, the tangents from each vertex are equal in length. Let the tangents from A be x, from B be y, and from C be z.

We know BD = 10 and DC = 8, so the tangents from B are BD and BA (say y), and tangents from C are DC and CA (say z). Hence, AB = y and AC = z.

Because the tangents from B are equal, BD = y = 10 cm, and from C, DC = z = 8 cm. Let the tangents from A be x. Now, the sides of triangle ABC are:

• AB = BD + DA = y + x = 10 + x
• AC = DC + DA = z + x = 8 + x
• BC = BD + DC = 18 cm

We use the semiperimeter s = (AB + BC + AC) / 2 = ((10 + x) + 18 + (8 + x)) / 2 = (36 + 2x)/2 = 18 + x.

The inradius r = 4 cm and area = r × s = 4 × (18 + x) = 90 cm² given.

So, 4(18 + x) = 90 → 18 + x = 22.5 → x = 4.5 cm.

Thus, AB = 10 + 4.5 = 14.5 cm and AC = 8 + 4.5 = 12.5 cm.

Answer: Given two circles with centers O and O' having radii 6 cm and 8 cm respectively, they intersect at points P and Q. It is given that OP and O'P act as tangents to the two circles. To find the length of the common chord PQ, we first note that the distance between the centers O and O' can be calculated by considering the right triangle formed by the radii and the chord. Since OP and O'P are tangents, the point P lies on both circles' tangents, and the radii OP and O'P are perpendicular to the tangent at point P. Using the radii, we calculate the distance between centers O and O'. By applying the Pythagoras theorem and properties of the intersecting chords, the length of the common chord PQ can be derived. The formula for the length of the common chord when two circles intersect is given by 2 times the square root of (radius1 squared minus half the distance between centers squared) or equivalent relations. Substituting values, the length PQ is found to be 8 cm.

Answer: Let the first average speed of the train be x km/h. For the first part of the journey, the train covers 54 km at speed x, so the time taken is 54 divided by x, or 54/x hours. For the second part, the train travels 63 km at an average speed of (x + 6) km/h, so the time taken is 63 divided by (x + 6), or 63/(x + 6) hours. The total time taken for the journey is given as 3 hours. Therefore, the sum of the two times is 3 hours. We have the equation: 54/x + 63/(x + 6) = 3. To solve this, multiply both sides of the equation by x(x + 6) to clear denominators: 54(x + 6) + 63x = 3x(x + 6). Simplify: 54x + 324 + 63x = 3x² + 18x. Combine like terms on the left: 117x + 324 = 3x² + 18x. Bring all terms to one side: 0 = 3x² + 18x - 117x - 324, which simplifies to 3x² - 99x - 324 = 0. Divide through by 3: x² - 33x - 108 = 0. Now, factor or use the quadratic formula to find x. Factors of -108 that sum to -33 are -36 and 3. So, x² - 36x + 3x -108 = 0. Group terms: (x² - 36x) + (3x - 108) = 0, x(x - 36) + 3(x - 36) = 0, (x + 3)(x - 36) = 0. So, x = -3 or 36. Since speed cannot be negative, the first average speed of the train is 36 km/h.

Answer:

Let the time taken by the pipe with smaller diameter to fill the tank be x hours. Then, the pipe with larger diameter will take (x - 2) hours.

The rate of filling the tank by the smaller pipe is 1/x of the tank per hour, and by the larger pipe is 1/(x - 2) of the tank per hour.

When both pipes work together, they fill the tank in 15/8 hours, so their combined rate is 8/15 of the tank per hour.

According to the problem: 1/x + 1/(x - 2) = 8/15

Multiply both sides by x(x - 2): (x - 2) + x = (8/15) * x(x - 2)

This simplifies to 2x - 2 = (8/15)(x^2 - 2x)

Multiply both sides by 15 to eliminate the fraction: 15(2x - 2) = 8(x^2 - 2x)

30x - 30 = 8x^2 - 16x

Bring all terms to one side: 8x^2 - 16x - 30x + 30 = 0

Which is 8x^2 - 46x + 30 = 0

Divide entire equation by 2 for simplicity: 4x^2 - 23x + 15 = 0

Use the quadratic formula to solve for x:

x = [23 ± sqrt(23^2 - 4*4*15)] / (2*4)

x = [23 ± sqrt(529 - 240)] / 8

x = [23 ± sqrt(289)] / 8

x = [23 ± 17] / 8

Two possible values: (23 + 17)/8 = 40/8 = 5, and (23 - 17)/8 = 6/8 = 0.75

Since time must be greater than 2 hours (due to x-2), we take x = 5 hours.

Therefore, the smaller pipe takes 5 hours, and the larger pipe takes 5 - 2 = 3 hours to fill the tank separately.

Answer:

Given, the horse is tied to a peg at one corner of a square grass field with side 15 m. The rope length initially is 5 m.

(i) To find the grazing area with 5 m rope:

The horse can graze in a quarter circle (as the rope is tied at the corner of the square, the horse can graze only in the quadrant within the field) with radius equal to the length of the rope.

Area of full circle = π × radius × radius = 3.14 × 5 × 5 = 78.5 m²

Area of quarter circle = 1/4 × 78.5 = 19.625 m²

Thus, the grazing area with 5 m rope = 19.625 m²

(ii) Now, if the rope length is increased to 10 m:

Area of full circle with radius 10 m = 3.14 × 10 × 10 = 314 m²

Area of quarter circle = 1/4 × 314 = 78.5 m²

Increase in grazing area = 78.5 m² - 19.625 m² = 58.875 m²

Answer:

The horse can graze over an area of 19.625 square meters with a 5 m rope. If the rope is increased to 10 m, the grazing area increases by 58.875 square meters, making the total grazing area 78.5 square meters.