CBSE EXAMINATION PAPER-2023

Answer: Let the two numbers be 2x and 3x, where x is their HCF. The LCM of the two numbers is given by (First Number × Second Number) ÷ HCF. According to the question, LCM is 180. So, (2x) × (3x) ÷ x = 180, which simplifies to 6x = 180. Therefore, x = 180 ÷ 6 = 30. Hence, the HCF of the two numbers is 30.

Answer: Given the polynomial p(x) = 6x² + 37x – (k – 2), let the zeros be α and β. If one zero is the reciprocal of the other, then β = 1/α. From the relationships of zeros and coefficients of a quadratic equation ax² + bx + c = 0, we know that α + β = –b/a and αβ = c/a. Here, α + β = –37/6 and αβ = –(k – 2)/6. Since β = 1/α, αβ = α × (1/α) = 1. Therefore, the product of zeros is 1. Equating this to the expression, we get –(k – 2)/6 = 1. Solving for k, we have –(k – 2) = 6, which gives –k + 2 = 6, and thus k = –4. Hence, the value of k is –4.

Answer: For the quadratic equation 2x² – 9x + 4 = 0, the sum of the roots is given by -b/a = 9/2 = 4.5, and the product of the roots is c/a = 4/2 = 2.

Answer: The given quadratic equation is 4x² – 5 = 0. Here, a = 4, b = 0, and c = –5. The discriminant, D, is given by b² – 4ac = 0² – 4 × 4 × (–5) = 80. Since the discriminant is positive (D > 0), the quadratic equation has two distinct real roots.

Answer: When a fair coin is tossed twice, the possible outcomes are HH, HT, TH, and TT. The event ‘at most one head’ means getting zero or one head. The outcomes satisfying this are HT, TH, and TT. So, there are 3 favorable outcomes out of 4 possible outcomes. Therefore, the probability of getting at most one head is 3 by 4.

Answer: First, calculate the values of each trigonometric function: cos 60° = 1/2, so cos² 60° = (1/2)² = 1/4. sec 30° = 1/cos 30° = 1/(√3/2) = 2/√3, so sec² 30° = (2/√3)² = 4/3. tan 45° = 1, so tan² 45° = 1² = 1. sin 30° = 1/2, so sin² 30° = (1/2)² = 1/4. cos 30° = √3/2, so cos² 30° = (√3/2)² = 3/4. Now substitute these values into the expression: Numerator: 5 × (1/4) + 4 × (4/3) - 1 = 5/4 + 16/3 - 1 = 1.25 + 5.333 - 1 = 5.583. Denominator: 1/4 + 3/4 = 1. Therefore, the value of the expression is 5.583 / 1 = 5.583, which simplifies to 67/12.

Answer: Given sin(A – B) = 0, it implies that A – B = 0°, so A = B. Also, 2 cos(A + B) – 1 = 0 means cos(A + B) = 1/2. For acute angles, cos 60° = 1/2, so A + B = 60°. Since A = B, we get 2A = 60°, hence A = 30° and B = 30°. Therefore, both angles A and B are 30 degrees.

Answer: Given the first term a = -14 and the fifth term a₅ = 2, we can find the common difference d. The formula for the nth term of an AP is aₙ = a + (n - 1)d. For the fifth term, a₅ = a + 4d = 2, so 2 = -14 + 4d. Solving for d gives d = 4. Now, the last term l = 62, and l = a + (n - 1)d. Substituting the values, 62 = -14 + (n - 1) × 4. Simplifying: 62 + 14 = 4(n - 1), so 76 = 4(n - 1), and n - 1 = 19, giving n = 20. Therefore, the AP has 20 terms.

Answer: Given the AP: 65, 61, 57, 53, ... we identify the first term a = 65 and the common difference d = 61 - 65 = -4. The nth term of an AP is given by aₙ = a + (n-1)d. Substituting, aₙ = 65 + (n-1)(-4) = 65 - 4(n-1) which simplifies to aₙ = 69 - 4n. To find the first negative term, set aₙ < 0, so 69 - 4n < 0, which leads to 4n > 69, hence n > 17.25. Therefore, the first term with an integer value of n greater than 17.25 is for n = 18. Thus, the 18th term is the first negative term in the sequence.

Answer:

To prove that √5 is irrational, we use proof by contradiction. Assume that √5 is rational, which means it can be expressed as a fraction a/b in lowest terms, where a and b are integers with no common factors other than 1, and b ≠ 0. Then, √5 = a/b implies 5 = a²/b², so 5b² = a². This means a² is divisible by 5, so a must also be divisible by 5. Let a = 5k, where k is an integer. Substituting back, 5b² = (5k)² = 25k², thus b² = 5k², meaning b² is divisible by 5, and hence b is divisible by 5. But this contradicts the assumption that a and b have no common factors. Hence, our assumption that √5 is rational is false, so √5 is irrational.

Answer: Consider a circle with centre O and an external point T. From T, two tangents TP and TQ are drawn, touching the circle at points P and Q respectively. We need to prove that the angle between these tangents at T (i.e., ∠PTQ) and the angle subtended by the chord PQ at the centre O (i.e., ∠POQ) are supplementary. Since TP and TQ are tangents, OP and OQ are radii perpendicular to these tangents, thus ∠OPT = 90° and ∠OQT = 90°. In quadrilateral OPQT, the sum of interior angles is 360°. Therefore, ∠POQ + ∠PTQ + 90° + 90° = 360°, which simplifies to ∠POQ + ∠PTQ = 180°. This proves that the angle between the two tangents is supplementary to the angle subtended by the chord joining their points of contact at the centre.

Answer: To prove the identity sin A - 2 sin³ A divided by 2 cos³ A - cos A equals tan A, we start with the left-hand side (LHS). First, rewrite the numerator as sin A - 2 sin³ A, and the denominator as 2 cos³ A - cos A. Factor sin A from the numerator: sin A (1 - 2 sin² A). Similarly, factor cos A from the denominator: cos A (2 cos² A - 1). Using the Pythagorean identity sin² A + cos² A = 1, replace sin² A with (1 - cos² A) in the expression. This transforms 1 - 2 sin² A into 1 - 2(1 - cos² A) = 1 - 2 + 2 cos² A = 2 cos² A - 1. Now, both numerator and denominator contain the factor (2 cos² A - 1). Hence, the expression becomes (sin A)(2 cos² A - 1) divided by (cos A)(2 cos² A - 1), which simplifies by canceling (2 cos² A - 1). The result is sin A divided by cos A, which equals tan A. Thus, the given expression is proven to be equal to tan A.

Answer: We start with the expression sec A (1 – sin A)(sec A + tan A). Using the definitions, sec A = 1 / cos A and tan A = sin A / cos A. Substitute these into the expression: (1 / cos A) (1 - sin A) (1 / cos A + sin A / cos A). Simplifying, it becomes (1 / cos A) (1 - sin A) ((1 + sin A) / cos A). Multiply the terms: (1 - sin A)(1 + sin A) / (cos A * cos A) which is (1 - sin² A) / cos² A. Since 1 - sin² A = cos² A by the Pythagorean identity, the expression reduces to cos² A / cos² A = 1. Therefore, sec A (1 – sin A)(sec A + tan A) = 1, as required.

Answer: Given two concentric circles with radii 5 cm (larger circle) and 3 cm (smaller circle), we need to find the length of a chord of the larger circle that touches the smaller circle. Since the circles are concentric, the chord that touches the smaller circle is tangent to the smaller circle. The perpendicular distance from the centre to this chord is equal to the radius of the smaller circle, which is 3 cm. In the larger circle, the radius is 5 cm. Let the length of the chord be 2x. By applying the Pythagoras theorem to the right triangle formed by the radius, the perpendicular distance, and half the chord length, we write: (5)^2 = (3)^2 + x^2, giving 25 = 9 + x^2. Solving for x, we get x = 4 cm. Therefore, the length of the chord is 2x = 8 cm. So, the chord of the larger circle touching the smaller circle is 8 cm long.

Answer: Given the quadratic equation px(x - 2) + 6 = 0, first rewrite it in standard quadratic form as p(x² - 2x) + 6 = 0 or p x² - 2 p x + 6 = 0. Here, the coefficients are: a = p, b = -2p, and c = 6. For the equation to have two equal real roots, its discriminant must be zero. The discriminant, D, is given by D = b² - 4ac. Substituting in, D = (-2p)² - 4 × p × 6 = 4p² - 24p. Set the discriminant to zero for equal roots: 4p² - 24p = 0. Factor out 4p: 4p(p - 6) = 0. This gives two possible values for p: p = 0 or p = 6. However, p = 0 would make the equation non-quadratic (no x² term). Therefore, the quadratic equation has two equal real roots when p = 6.

Answer:

Given: Height of the tower = 75 m
Angle of depression to first car = 30°
Angle of depression to second car = 60°
Using the fact that the man is at the top of the tower, the angles of depression form two right angled triangles with the tower's height as one side and the horizontal distances of the cars from the base of the tower as the other sides.

For the car observed at 30°:
tan(30°) = height / distance => distance = height / tan(30°)
tan(30°) = 1 / √3 = 1 / 1.73 = 0.577
distance to first car = 75 / 0.577 = 130 m (approx.)

For the car observed at 60°:
tan(60°) = height / distance => distance = height / tan(60°)
tan(60°) = √3 = 1.73
distance to second car = 75 / 1.73 = 43.35 m (approx.)

Since one car is behind the other on the same side of the tower, the distance between the two cars is:
130 - 43.35 = 86.65 m (approx.)

Hence, the distance between the two cars is about 86.65 meters.

Answer: Let the height of the building be 7 m and the height of the tower be h meters. Let the horizontal distance between the building and the tower be x meters. From the top of the building, the angle of elevation of the top of the tower is 60°. Using the right triangle formed above the building, we have: Height difference from building top to tower top = h - 7 Using the tangent of the angle of elevation: Tan 60° = (h - 7) / x We know Tan 60° = √3, so: √3 = (h - 7) / x => (h - 7) = √3 * x Similarly, the angle of depression to the foot of the tower is 30°. Using the right triangle below the building: Tan 30° = 7 / x Since Tan 30° = 1/√3, 1/√3 = 7 / x => x = 7√3 Substitute the value of x in the first equation: (h - 7) = √3 * 7√3 = √3 * 7√3 = 7 * 3 = 21 Therefore, h - 7 = 21 => h = 21 + 7 = 28 m Hence, the height of the tower is 28 meters.

Answer:

Given a triangle ABC with point D on side BC such that angle ADC is equal to angle BAC, we need to prove that CA squared is equal to the product of CB and CD.

Construction: Join points A and D.

Proof:

Since ∠ADC = ∠BAC, triangles ADC and BAC share these equal angles.

Observe triangles ADC and BAC:

• They have ∠ADC = ∠BAC (Given)
• They share side AC.

Using the Law of Sines in triangle ADC and triangle BAC, we get:

In triangle ADC, applying the law of sines: AD / sin ∠ACD = CD / sin ∠ADC

In triangle BAC, applying the law of sines: AB / sin ∠ABC = BC / sin ∠BAC

But since ∠ADC = ∠BAC, it follows that the sides are proportional.

Alternatively, considering triangles ABD and ACD, we can write the sides to establish similarity, or use the properties of similar triangles.

Since angles ADC and BAC are equal, triangles ADC and BAC are similar by AA similarity criterion.

From the similarity, the sides are proportional as:

CA / CB = CD / CA

Cross-multiplied, this gives:

CA × CA = CB × CD

Thus, CA² = CB × CD, which is what we needed to prove.

Hence, the result is proved.

Answer:

Given that triangles ABC and PQR are similar, we have the corresponding sides proportional. That means, AB / PQ = BC / QR = AC / PR.

AD and PM are medians of triangles ABC and PQR respectively. By definition of a median, AD joins vertex A to the midpoint M of BC, and PM joins vertex P to the midpoint N of QR.

Since △ABC ~ △PQR, the midpoint M of BC corresponds to midpoint N of QR, so BM = MC and QN = NR. Therefore, BM / QN = BC / QR.

Because medians divide the triangles into two smaller triangles each, and similarity is maintained, the ratio of the lengths of medians AD and PM is equal to the ratio of the corresponding sides AB and PQ.

Hence, we conclude that AB / PQ = AD / PM. This shows that the ratio of the corresponding sides is equal to the ratio of their medians in two similar triangles.

Answer:

The model consists of a cylinder with two cones attached at its ends. The diameter of the base of the entire model is given as 3 cm, which means the radius of the cylinder and cones is 1.5 cm (since radius is half of the diameter). The total length of the model is 12 cm, which includes the length of the cylinder plus the heights of the two cones. Each cone has a height of 2 cm, so total height taken by the cones is 2 cm + 2 cm = 4 cm. Therefore, the height of the cylinder is 12 cm - 4 cm = 8 cm.

First, we find the volume of the cylinder using the formula:
Volume of cylinder = π × radius² × height
Volume of cylinder = 3.14 × (1.5)² × 8 = 3.14 × 2.25 × 8 = 56.52 cm³

Next, find the volume of one cone using the formula:
Volume of cone = (1/3) × π × radius² × height
Volume of cone = (1/3) × 3.14 × (1.5)² × 2 = (1/3) × 3.14 × 2.25 × 2 = 4.71 cm³

Since there are two cones, total volume of cones = 2 × 4.71 = 9.42 cm³

Finally, total volume of air contained in the model is the sum of volume of cylinder and cones:
Total volume = 56.52 + 9.42 = 65.94 cm³

Thus, the volume of air contained in the model is approximately 65.94 cubic centimeters.

Answer:

To find the value of x, use the fact that the total number of families is 200. So, sum the frequency of all intervals including 'x' and set equal to 200, then solve for x.

After finding x, calculate the median expenditure as follows: find the cumulative frequencies and identify the median class, then use the median formula: Median = L + [(N/2 - F)/f] × h, where L is lower boundary of median class, N total frequency, F cumulative frequency before median class, f frequency of median class, and h class width.

For mean expenditure, calculate the class marks for each class, multiply them by frequencies, sum the products and divide by total number of families (200). This gives the mean monthly expenditure.

Thus, by following these steps, x can be found, and the median and mean expenditure on milk can be determined.