(A) 13 (B) 26 (C) 39 (D) 65

Answer and Solution

(B) 26

134 - 4 = 130 and 188 - 6 = 182

HCF (130, 182) = 26

Marking Scheme

1 mark for the correct option (no part marks).

(A) q/p (B) r/p (C) − r/p (D) − q/p

Answer and Solution

(C) − r/p

Substituting q = p + r in f(x)

f(x) = px² + (p + r)x + r = (x + 1)(px + r)

Zeroes are - 1 or − r/p.

Marking Scheme

1 mark for the correct option (no part marks).

(A) I only (B) II only (C) I and II (D) II and III

Answer and Solution

(B) II only

(5, - 1) is a solution of I: 3x - 3y = 18 also. But this will make the system of equations as consistent and dependent.

Only II: 2x - 3y = 13 fulfills the given condition of having only one point of intersection.

Marking Scheme

1 mark for the correct option (no part marks).

(A) 0, 1 (B) - 1, 1 (C) 0, - 1 (D) - 1, 2

Answer and Solution

(C) 0, - 1

As (1 - p) is a root of the quadratic equation x² + px + 1 - p = 0,

∴ (1 - p)² + p (1 - p) + 1 - p = 0 ⇒ p = 1

Now the quadratic equation will be:

x² + x = 0 ⇒ x (x + 1) = 0

So the roots are 0, - 1.

Marking Scheme

1 mark for the correct option (no part marks).

(A) - 26 (B) - 29 (C) - 32 (D) - 35

Answer and Solution

(A) - 26

- 62 = 10 + (n - 1) (- 3) ⇒ n = 25

Thus (25 + 1)/2 = 13th term is the middle term.

∴ a13 = 10 + 12 (- 3) = - 26

Marking Scheme

1 mark for the correct option (no part marks).

(A) 5 (B) 7.5 (C) 10 (D) 17.5

Answer and Solution

(D) 17.5

Ratio of perimeters of similar triangles = ratio of their corresponding sides

∴ 56/70 = 14/(corresponding side)

corresponding side = 17.5

Marking Scheme

1 mark for the correct option (no part marks).

(A) (0, 0) (B) (0, 3) (C) (3, 0) (D) (0, 4)

Answer and Solution

(A) (0, 0)

Mid-point of AB will lie on the perpendicular bisector of the line segment joining the points A (- 3, - 4) and B (3, 4).

∴ Coordinates of the mid-point = ((- 3 + 3)/2 , (- 4 + 4)/2) = (0, 0)

Marking Scheme

1 mark for the correct option (no part marks).

(A) 60°, 30° (B) 60°, 45° (C) 45°, 15° (D) 60°, 15°

Answer and Solution

(C) 45°, 15°

tan (A + B) = √3 ⇒ (A + B) = 60° …. (i)

tan (A - B) = 1/√3 ⇒ (A - B) = 30° …. (ii)

Solving equations (i) and (ii), we get A = 45° and B = 15°.

Marking Scheme

1 mark for the correct option (no part marks).

(A) 1/12 (B) 1/32 (C) 11/96 (D) 5/48

Answer and Solution

(D) 5/48

Total outcomes = 96

Perfect square numbers - 4, 9, 16, 25, 36, 49, 64, 81

Perfect cubes are 8, 27, 64; common number is 64

Number of favourable outcomes = 10

P(either a perfect square or a perfect cube) = 10/96 = 5/48

Marking Scheme

1 mark for the correct option (no part marks).

(A) 10⁴ (B) 10⁵ (C) 10⁶ (D) 10⁷

Answer and Solution

(B) 10⁵

Total revolutions = (66 × 10⁵ × 7)/(22 × 21) = 10⁵

Marking Scheme

1 mark for the correct option (no part marks).

(A) 192 cm² (B) 160 cm² (C) 96 cm² (D) 80 cm²

Answer and Solution

(B) 160 cm²

Side of cube = 4 cm

Dimensions of resulting cuboid: length = 8 cm, breadth = 4 cm, height = 4 cm.

Total surface area = 2 (8 × 4 + 4 × 4 + 8 × 4) = 160 cm²

Marking Scheme

1 mark for the correct option (no part marks).

(A) 15 (B) 25.5 (C) 35 (D) 37.5

Answer and Solution

(D) 37.5

Let the number of men be x and the number of women be y.

(38x + 30y)/(x + y) = 35 ⇒ 38x + 30y = 35x + 35y ⇒ 3x = 5y

∴ x/y = 5/3

Percentage of women = 3/8 × 100 = 37.5

Marking Scheme

1 mark for the correct option (no part marks).

(A) 1/4 (B) 1/9 (C) 1/36 (D) 1/6

Answer and Solution

(A) 1/4

Total outcomes = 36

Favourable outcomes = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)

P (E) = 9/36 = 1/4

Marking Scheme

1 mark for the correct option (no part marks).

(A) 2√17 units (B) 2√15 units (C) √101 units (D) 10 units

Answer and Solution

(A) 2√17 units

Coordinates of the midpoint M of QR = ((6 + 0)/2 , (4 + 0)/2) = (3, 2)

Median PM = √((5 − 3)² + (− 6 − 2)²) = √68 = 2√17 units

Marking Scheme

1 mark for the correct option (no part marks).

(A) 4 cm (B) 8 cm (C) 16 cm (D) 18 cm

Answer and Solution

(B) 8 cm

A = ½ l × r ⇒ 24π = ½ × 6π × r ⇒ r = 8 cm

Marking Scheme

1 mark for the correct option (no part marks).

(A) 23 (B) 26 (C) 25 (D) 30

Answer and Solution

(C) 25

Mode = 3 Median - 2 Mean = 3 (32) - 2 (35.5) = 25

Marking Scheme

1 mark for the correct option (no part marks).

(A) 40 m (B) 45.5 m (C) 48 m (D) 50 m

Answer and Solution

(C) 48 m

Not to scale.

tan θ = 12/5 ∴ sin θ = 12/13

⇒ h/52 = 12/13 ⇒ h = 48 m

Marking Scheme

1 mark for the correct option (no part marks).

(A) 13/52 (B) 13/46 (C) 11/52 (D) 11/46

Answer and Solution

(B) 13/46

Total remaining cards = 52 - (2 + 4) = 46

P(a heart card) = 13/46

Marking Scheme

1 mark for the correct option (no part marks).

Assertion-Reason Based Questions

Directions: Questions number 19 and 20 are Assertion and Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the options (A), (B), (C) and (D) as given below:

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(C) Assertion (A) is true but Reason (R) is false.

(D) Assertion (A) is false but Reason (R) is true.

Reason (R): Diagonals of a parallelogram bisect each other.

Answer and Solution

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

The diagonals of a parallelogram bisect each other, so the mid-point of one diagonal is the mid-point of the other.

Mid-point of the diagonal joining (-1, -6) and (7, 2) = ((-1 + 7)/2 , (-6 + 2)/2) = (3, - 2)

Taking the fourth vertex as (x, y), the mid-point of the other diagonal joining (2, - 5) and (x, y) must also be (3, - 2), which gives x = 4 and y = 1.

Marking Scheme

1 mark for the correct option (no part marks).

Reason (R): If a = c, then zeroes of the polynomial ax² + bx + c, a ≠ 0 are reciprocal of each other.

Answer and Solution

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

If the zeroes are reciprocals of each other, their product is 1.

∴ c/a = 1, i.e. a = c ⇒ (- 3)/(2k - 1) = 1 ⇒ 2k - 1 = - 3 ⇒ k = - 1

So Reason (R) is the rule that produces the value stated in Assertion (A).

Marking Scheme

1 mark for the correct option (no part marks).

This section comprises of 5 Very Short Answer (VSA) type questions of 2 marks each.

A hall has a length of 9.75 m, breadth of 6.75 m and height of 5.25 m. What is the length of the longest unmarked ruler that can exactly measure the dimensions of the hall?

OR

21 (B)

There are three bells placed at different swings in a park, which toll at intervals of 5, 6 and 8 minutes, respectively. They all toll together when the park opens for the visitors at 10:00 a.m.. How many more times do they all toll together till the park is closed at 7:00 p.m.?

Solution

21 (A)

Length = 9.75 m = 975 cm, Breadth = 6.75 m = 675 cm, Height = 5.25 m = 525 cm

Length of longest unmarked ruler = HCF (975, 675, 525)

975 = 3 × 5² × 13, 675 = 3³ × 5², 525 = 3 × 5² × 7

∴ HCF (975, 675, 525) = 3 × 5² = 75

Hence, the length of the longest unmarked ruler is 75 cm or 0.75 m.

Marking Scheme

Converting the dimensions to cm - ½

Prime factorisation of 975, 675 and 525 - 1

HCF = 75 cm (0.75 m) - ½

Solution

21 (B)

LCM of 5, 6 and 8 is 120, i.e. all the bells toll together after an interval of 120 minutes or 2 hours.

Thus, all the bells will again toll together at 12:00 noon, 2:00 p.m., 4:00 p.m. and 6:00 p.m., i.e. 4 times.

So, the bells will toll together 4 more times, till the park is closed at 7:00 p.m.

Marking Scheme

LCM of 5, 6 and 8 = 120 minutes (2 hours) - 1

Listing the times and concluding 4 more times - 1

ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

OR

22 (B)

In the given figure, if ∠PQR = ∠QSP, PQ = 6 cm and PS = 3 cm, then find the length of PR.

Not to scale.

For visually impaired candidates: In a right-angled triangle ABC, right-angled at B, a perpendicular BD is drawn to the hypotenuse AC. Prove that the triangle BCD is similar to the triangle ACB.

Solution

22 (A)

Not to scale.

In ∆ AOB and ∆ COD,

∠ OAB = ∠ OCD and ∠ OBA = ∠ ODC (Alternate interior angles)

∴ ∆ AOB ∼ ∆ COD (AA criterion)

⇒ AO/CO = BO/DO or AO/BO = CO/DO

Marking Scheme

Correct figure - 1

Pair of equal angles and the similarity statement - ½

Writing the required proportion - ½

Solution

22 (B)

Not to scale.

In ∆ PRQ and ∆ PQS,

∠ PQR = ∠ QSP (Given); ∠ QPR = ∠ QPS (Common)

∴ ∆ PRQ ∼ ∆ PQS (AA criterion)

⇒ PQ/PS = PR/PQ ⇒ 6/3 = PR/6

∴ PR = 12 cm

Marking Scheme

Similarity of the two triangles with reasons - 1

Writing the proportion - ½

PR = 12 cm - ½

Solution

for visually impaired candidates

Not to scale.

In triangle BCD and triangle ACB,

∠ BDC = ∠ ABC (each 90°); ∠ BCD = ∠ ACB (common)

∴ Triangle BCD is similar to triangle ACB (AA criterion)

Marking Scheme

First pair of equal angles - ½

Second pair of equal angles - ½

Similarity statement with the AA criterion - 1

Answer and Solution

tan θ = 1/√5 ⇒ cot θ = √5

sec²θ = 1 + tan²θ = 1 + (1/√5)² = 6/5

cosec²θ = 1 + cot²θ = 1 + (√5)² = 6

∴ (cosec²θ − sec²θ)/(cosec²θ + sec²θ) = (6 − 6/5)/(6 + 6/5) = 2/3

Marking Scheme

cot θ = √5 - ½

Values of sec²θ and cosec²θ - ½

Substitution and the value 2/3 - 1

Answer and Solution

Radius = 28/2 = 14 cm

Area of sector = ½ × l × r = ½ × 22 × 14

= 154 cm²

Marking Scheme

Radius = 14 cm - ½

Correct formula with substitution - 1

Area = 154 cm² - ½

Answer and Solution

Total number of outcomes = 36

Favourable outcomes = 6 : (1, 2), (1, 3), (1, 5), (2, 1), (3, 1), (5, 1)

P (Prime Product) = 6/36 = 1/6

Marking Scheme

Total number of outcomes = 36 - ½

Listing the 6 favourable outcomes - ½

Probability = 1/6 - 1

This section comprises of 6 Short Answer (SA) type questions of 3 marks each.

Answer and Solution

Let us assume that 2 + 3√3 is a rational number.

Then 2 + 3√3 = a/b , where ‘a’ and ‘b’ are integers and b ≠ 0

⇒ √3 = (a − 2b)/3b

Since ‘a’ and ‘b’ are integers, (a − 2b)/3b is a rational number.

So √3 is a rational number, which contradicts the given fact that √3 is an irrational number.

Hence, our assumption is wrong. Therefore 2 + 3√3 is an irrational number.

Marking Scheme

Assumption and writing 2 + 3√3 = a/b - 1

Obtaining √3 = (a − 2b)/3b - 1

Contradiction and conclusion - 1

Answer and Solution

α + β = 4 and αβ = 5/2

(α + 1/β) × (β + 1/α) = ((αβ + 1)/β) × ((αβ + 1)/α) = ((αβ + 1)²)/αβ

= ((5/2 + 1)²)/(5/2) = 49/10

Marking Scheme

α + β = 4 and αβ = 5/2 - 1

Simplifying the product to ((αβ + 1)²)/αβ - ½

Substituting the values - 1

Final value 49/10 - ½

Ridhi drew a polygon with n sides. The smallest exterior angle is 8° and each subsequent exterior angle is 4° more than the previous exterior angle. Find the number of sides of the polygon that Ridhi had drawn.

OR

28 (B)

Find the sum of integers between 1 and 400 that are multiples of 4 as well as of 5.

Solution

28 (A)

Let the number of sides of the given polygon be ‘n’. The exterior angles of the given polygon are in A.P.

Here a = 8 and d = 4. As the sum of all the exterior angles of a polygon is 360°,

∴ Sn = 360 ⇒ n/2 [2 × 8 + (n - 1) × 4] = 360

⇒ n² + 3n - 180 = 0 ⇒ (n + 15) (n - 12) = 0

∴ n = - 15 or 12. An angle is always positive, ∴ n = 12.

So, the number of sides in the given polygon is 12.

Marking Scheme

Identifying the A.P. with a = 8, d = 4 and Sn = 360 - 1

Forming n² + 3n - 180 = 0 - 1

Solving the equation - ½

Rejecting n = - 15 and concluding n = 12 - ½

Solution

28 (B)

The required A.P. is 20, 40, 60, …, 380. Here a = 20 and d = 20.

Let an = 380 ⇒ 20 + (n - 1) × 20 = 380 ⇒ n = 19

∴ S19 = 19/2 [20 + 380] = 3800

Marking Scheme

Writing the A.P. 20, 40, …, 380 with a = 20, d = 20 - 1

Forming the equation for an - ½

n = 19 - ½

Applying the sum formula - ½

Sum = 3800 - ½

Answer and Solution

Not to scale.

Given, ABCD is a parallelogram circumscribing a circle.

AS = AR --- (i); BS = BP --- (ii); CQ = CP --- (iii); DQ = DR --- (iv)

(Lengths of tangents drawn from an external point to a circle are equal)

Adding (i), (ii), (iii) and (iv), we get

AS + BS + CQ + DQ = AR + BP + CP + DR

⇒ AB + CD = AD + BC ⇒ 2 AB = 2 AD ⇒ AB = AD

Since adjacent sides of parallelogram ABCD are equal, ABCD is a rhombus.

Marking Scheme

Correct figure with the four equal tangent pairs - 1

Adding the four relations - 1

Concluding AB = AD and hence a rhombus - 1

Answer and Solution

Not to scale.

Let the x-axis divide the line segment joining the points A (- 4, - 6) and B (-1, 7) at P (x, 0) in the ratio k : 1.

Coordinates of P = ((k (−1) + 1 (−4))/(k + 1) , (k (7) + 1 (−6))/(k + 1)) = ((− k − 4)/(k + 1) , (7k − 6)/(k + 1))

Since the point P lies on the x-axis,

∴ (7k − 6)/(k + 1) = 0 ⇒ k = 6/7

So, point P divides AB in the ratio 6 : 7.

Coordinates of P are (− 34/13 , 0).

Marking Scheme

Applying the section formula to write the coordinates of P - 1

Equating the y-coordinate to zero - ½

k = 6/7, i.e. the ratio 6 : 7 - ½

Coordinates of P - 1

During a math class, Ms. Isha wrote the expression given below on the board and asked the students to simplify it.

(cos θ)/(1 − sin θ) + (1 − sin θ)/(cos θ)

Jyoti solved it in her note book as follows:

(cos θ)/(1 − sin θ) + (1 − sin θ)/(cos θ)

= (cos²θ + (1 − sin θ)²)/((1 − sin θ) × cos θ)…. (step 1)

= (cos²θ + cos²θ)/((1 − sin θ) × cos θ)…. (step 2)

= 2cos²θ/((1 − sin θ) × cos θ)…. (step 3)

= (2cos θ)/(1 − sin θ)…. (step 4)

Identify the step in which Jyoti has made error(s), if any. Rectify the same and find the correct answer.

OR

31 (B)

If 1/(sin x − cos x) = (cosec x)/√2, then prove that (1/(sin x + cos x))² = sec²x/2.

Solution

31 (A)

Jyoti made an error in step 2. The correct solution is as under:

(cos θ)/(1 − sin θ) + (1 − sin θ)/(cos θ)

= (cos²θ + (1 − sin θ)²)/((1 − sin θ) × cos θ)

= (cos²θ + 1 + sin²θ − 2 sin θ)/((1 − sin θ) × cos θ)

= (2 − 2 sin θ)/((1 − sin θ) × cos θ)

= (2 × (1 − sin θ))/((1 − sin θ) × cos θ) = 2/(cos θ) = 2 sec θ

Marking Scheme

Identifying the error in step 2 - 1

Correct expansion of (1 − sin θ)² and simplification - 1

Reaching (2 − 2 sin θ)/((1 − sin θ) × cos θ) - ½

Final answer 2 sec θ - ½

Solution

31 (B)

Given: 1/(sin x − cos x) = (cosec x)/√2 ⇒ sin x − cos x = √2 sin x

Squaring both sides, we get

2 sin x cos x = 1 − 2 sin²x --- (i)

Now, LHS = (1/(sin x + cos x))² = 1/(1 + 2 sin x cos x)

= 1/(1 + (1 − 2 sin²x)) = 1/(2 cos²x)

= sec²x/2 = RHS

Marking Scheme

Obtaining sin x − cos x = √2 sin x - ½

Squaring and obtaining relation (i) - 1

Writing LHS in terms of 2 sin x cos x - ½

Substituting from (i) - ½

Reaching sec²x/2 - ½

This section comprises of 4 Long Answer (LA) type questions of 5 marks each.

In a rectangular park measuring 50 m × 40 m, the gram panchayat plans to construct a rectangular swimming pool in the middle, surrounded by a uniform-width grass strip throughout the park. Find the dimensions of the swimming pool, if the area of the grassy strip is 1184 m².

Answer and Solution

Not to scale.

ABCD is the rectangular park and PQRS is the swimming pool. Let ‘x’ be the uniform width of the grass strip.

Then, length of the swimming pool, PQ = (50 - 2x) m and breadth of the swimming pool, QR = (40 - 2x) m.

Area of rectangular park - Area of swimming pool = Area of grass strip.

50 × 40 - (50 - 2x) (40 - 2x) = 1184

⇒ x² - 45x + 296 = 0 ⇒ (x - 8) (x - 37) = 0 ⇒ x = 8, 37

Rejecting x = 37, as the length and breadth cannot be negative. ∴ x = 8

Thus, length of the swimming pool = 50 - 16 = 34 m and breadth = 40 - 16 = 24 m.

Marking Scheme

Correct figure - ½

Expressing the dimensions of the pool in terms of x - 1

Forming the equation from the area of the grass strip - 1

Reducing to x² - 45x + 296 = 0 - ½

Solving the quadratic equation - 1

Rejecting x = 37 - ½

Dimensions 34 m × 24 m - ½

Using the above theorem solve the following:

PQRS is a trapezium with PQ ‖ SR. X and Y are points on non-parallel sides PS and QR respectively such that XY ‖ PQ. Show that PX/XS = QY/YR.

Answer and Solution

Basic Proportionality Theorem: correct figure, given, to prove, construction and proof.

Application:

Not to scale.

Join PR, intersecting XY at M.

In ∆ PSR, XM ‖ SR ∴ PX/XS = PM/MR --- (i)

In ∆ PQR, MY ‖ PQ ∴ QY/YR = PM/MR --- (ii)

From (i) and (ii), PX/XS = QY/YR

Marking Scheme

Correct figure, given, to prove and construction - 1

Correct proof of the theorem - 2

Construction: joining PR to meet XY at M - ½

Applying the theorem in ∆ PSR - ½

Applying the theorem in ∆ PQR - ½

Combining (i) and (ii) - ½

A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively, and the total height of the tent is 13.5 m, then find the area of the canvas required for making the tent, keeping a provision of 26 m² of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of ₹ 250 per m².

OR

34 (B)

A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of the hemisphere is 3.5 cm and the total wood used in making the toy is 166 5/6 cm³. Find the height of the conical part. Also, find the cost of painting the hemispherical part of the toy at the rate of ₹ 15 per cm².

Solution

34 (A)

Not to scale.

Height of conical part = 13.5 - 3 = 10.5 m

Slant height (l) = √((14)² + (10.5)²) = 17.5 m

CSA of conical part = πrl = 22/7 × 14 × 17.5 = 770 m²

CSA of cylindrical part = 2πrh = 2 × 22/7 × 14 × 3 = 264 m²

Total area of canvas = CSA of conical part + CSA of cylindrical part + wastage

= 770 + 264 + 26 = 1060 m²

Cost of canvas @ ₹ 250 per m² = 250 × 1060 = ₹ 2,65,000

Marking Scheme

Height of the conical part - ½

Slant height l = 17.5 m - 1

CSA of the conical part - 1

CSA of the cylindrical part - 1

Total canvas = 1060 m² - ½

Cost = ₹ 2,65,000 - 1

Solution

34 (B)

Not to scale.

Let ‘h’ cm be the height of the conical part.

Volume of wood in the toy = Volume of conical part + Volume of hemispherical part

166 5/6 = 1/3 × π × r² × h + 2/3 × π × r³

⇒ 1001/6 = 1/3 × 22/7 × (3.5)² × h + 2/3 × 22/7 × (3.5)³

⇒ h = 6 cm

CSA of hemispherical part = 2 × π × r² = 2 × 22/7 × (3.5)² = 77 cm²

Cost of painting the hemispherical part @ ₹ 15 per cm² = 15 × 77 = ₹ 1155

Marking Scheme

Forming the volume equation and substituting - 2

h = 6 cm - 1

CSA of the hemispherical part = 77 cm² - 1

Cost = ₹ 1155 - 1

Manish is standing on level ground and observes a kite flying 200 m away from him at an angle of elevation of 30°. Mahesh, standing on the roof of a 50 m high building on the opposite side of the kite, observes the same kite at an angle of elevation of 45°. Find the distance between the kite and Mahesh.

OR

35 (B)

The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be 45° and 30°. If the ships are 200 m apart and one ship is exactly behind the other, then find the height of the lighthouse.

Solution

35 (A)

Not to scale.

Let Manish be standing at point ‘A’ and Mahesh at point ‘D’. Distance of the kite from Manish = 200 m.

In rt. angled ∆ AEB, BE/200 = sin 30° = ½ ⇒ BE = 100 m

BR = BE - RE = BE - DC = 100 - 50 = 50 m

In rt. angled ∆ BRD, 50/BD = sin 45° = 1/√2 ⇒ BD = 50√2 m

∴ Distance of the kite from Mahesh is 50√2 m.

Marking Scheme

Correct figure - 1

Using sin 30° in ∆ AEB - 1

BE = 100 m - ½

BR = 50 m - 1

Using sin 45° in ∆ BRD - 1

BD = 50√2 m - ½

Solution

35 (B)

Not to scale.

Let the height of the lighthouse be ‘h’ m.

In rt. angled ∆ ABC, h/BC = tan 45° = 1 ⇒ BC = h --- (i)

In rt. angled ∆ ABD, h/(h + 200) = tan 30° = 1/√3

⇒ h = 200/(√3 − 1) × (√3 + 1)/(√3 + 1) ⇒ h = 100 (√3 + 1)

∴ Height of the lighthouse is 100 (√3 + 1) m.

Marking Scheme

Correct figure - 1

Using tan 45° to get BC = h - 1

Setting up the equation with tan 30° - ½

Simplifying by rationalising - 1

h = 100 (√3 + 1) - ½

Stating the height - 1

This section comprises of 3 case-study-based questions of 4 marks each with sub parts. Each case study question has three sub parts (i), (ii), (iii) of marks 1, 1, 2 respectively.

A student entrepreneur started a company that manufactures sanitizers in two sizes - small and large. The cost of a small bottle of sanitizer is ₹10 and that of a large bottle is ₹15. In June, the company sold 1000 bottles and recorded a total sale of ₹12,750. Seeing the increased demand, the company decided to increase the price of both sanitizer bottles by ₹2 each. In the next month, the company sold 2500 bottles and recorded a total sale of ₹34,250.

Based on the above information, answer the following questions:

(i)Form a linear equation in two variables representing the sale for June.[1]

(ii)Form a linear equation in two variables representing the sale for July.[1]

(iii)(A) How many sanitizer bottles of each type were actually sold in June?[2]

OR

(iii)(B) How many sanitizer bottles of each type were sold in July?[2]

Solution

Let the number of small bottles and large bottles be ‘x’ and ‘y’ respectively.

(i) 10x + 15y = 12750 or 2x + 3y = 2550

(ii) 12x + 17y = 34250

(iii) (A) According to the question,

x + y = 1000 --- (i); 10x + 15y = 12750 --- (ii)

Solving (i) and (ii), we get x = 450 and y = 550.

Number of small and large bottles sold are 450 and 550 respectively.

OR

(iii) (B) According to the question,

x + y = 2500 --- (i); 12x + 17y = 34250 --- (ii)

Solving (i) and (ii), we get x = 1650 and y = 850.

Number of small and large bottles sold are 1650 and 850 respectively.

Marking Scheme

(i) Correct equation - 1

(ii) Correct equation - 1

(iii) (A) Forming the pair of equations - 1; solving to get x = 450, y = 550 - 1

(iii) (B) Forming the pair of equations - 1; solving to get x = 1650, y = 850 - 1

Not to scale.

Arjun, a designer, stood at a point P outside the garden and drew two tangents PA and PB to the circular boundary. The radius of the garden is 9 m, and the distance of point P from the centre O is 18 m. For preparing safety guidelines, Arjun needed the lengths of the tangents. During a demonstration to interns, he asked them to compute the angle formed between the two tangents using the properties of circle.

Based on the above information, answer the following questions:

(i)What is the measure of ∠OAP?[1]

(ii)Calculate the length of the tangent PA.[1]

(iii)(A) Calculate the measure of ∠AOB.[2]

OR

(iii)(B) In the above design, if PA and PB were inclined at 60°, then what would be their length?[2]

Solution

Not to scale.

(i) ∠OAP = 90° (The radius is perpendicular to the tangent at the point of contact)

(ii) In right angled ∆ OAP,

OP² = OA² + AP² ⇒ (18)² = (9)² + AP² ⇒ AP = √243 = 9√3 m

(iii) (A)

Not to scale.

In right angled ∆ OAP, cos θ = 9/18 = ½ = cos 60° ⇒ θ = 60°

∴ ∠ AOB = 2θ = 120°

OR

(iii) (B)

Not to scale.

If the tangents are inclined at 60°, then ∠OPA = ½ × 60° = 30°.

In right angled ∆ OPA, 9/AP = tan 30° = 1/√3 ⇒ AP = 9√3 m

Marking Scheme

(i) ∠OAP = 90° - 1

(ii) Applying Pythagoras theorem - ½; AP = 9√3 m - ½

(iii) (A) Finding θ = 60° - 1; ∠AOB = 2θ - ½; ∠AOB = 120° - ½

(iii) (B) ∠OPA = 30° - ½; using tan 30° - 1; AP = 9√3 m - ½

A school conducted a weekly test for Class X students, before the commencement of the pre-board examination and recorded their scores (out of 50). To analyse performance patterns, the academic coordinator grouped the marks into intervals. The grouped frequency distribution is as below:

The academic coordinator computed the central tendencies to judge overall learning level, the most common performance range and to understand consistency across the batch for planning the remedial sessions.

Based on the above information, answer the following questions:

(i)Identify the class with the most common performance range.[1]

(ii)Find the class interval containing the median.[1]

(iii)(A) What is the average performance of the students?[2]

OR

(iii)(B) Find the mode of the data.[2]

Solution

(i) 30 - 40

(ii)

Since N/2 = 25, ∴ Median class is 30 - 40.

(iii) (A)

Average performance, i.e. mean = 1560/50 = 31.2

OR

(iii) (B) Here, the modal class is 30 - 40, f0 = 12, f1 = 15, f2 = 14.

Mode = 30 + ((15 − 12)/(2 × 15 − 12 − 14)) × 10 = 37.5

Marking Scheme

(i) Most common performance range - 1

(ii) Correct cumulative frequency table and median class - 1

(iii) (A) Correct table - 1; mean = 31.2 - 1

(iii) (B) Identifying the modal class with f0, f1, f2 - ½; substituting in the mode formula - 1; mode = 37.5 - ½