📚 Important Topics Covered in This Paper:

📝 Note: This question paper is designed as per the ICSE Class 10 Board pattern for practice purposes. Students are advised to solve it within the stipulated time to simulate exam conditions.

📘 Subject: Mathematics
🎓 Class: X (ICSE)
📊 Maximum Marks: 80
⏰ Time Allowed: 3 Hours

📋 General Instructions:

1. ✍️ Answers to this Paper must be written on the paper provided separately.

2. ⏳ You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper.

3. 🕐 The time given at the head of this Paper is the time allowed for writing the answers.

4. ✅ Attempt all questions from Section A and any four questions from Section B.

5. 📝 All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer.

6. ⚠️ Omission of essential working will result in loss of marks.

7. 🎯 The intended marks for questions or parts of questions are given in brackets [ ].

8. 📐 Mathematical tables and graph papers are to be provided by the school.

📕 SECTION A (40 Marks)

(Attempt all questions from this Section.)

❓ Question 1 [15 Marks]

Choose the correct answers to the questions from the given options. (Do not copy the questions, write the correct answers only.)

(i) 🔄 The reflection of the point P(-4, 7) in the y-axis is:
(a) (4, 7)
(b) (-4, -7)
(c) (4, -7)
(d) (-4, 7)

✅ Answer: (a) (4, 7)
📖 Explanation: Under reflection in the y-axis, (x, y) → (-x, y). So, (-4, 7) → (4, 7).

(ii) 📍 The mid-point of the line segment joining A(-2, 5) and B(6, -3) is:
(a) (2, 1)
(b) (4, 2)
(c) (2, 2)
(d) (1, 1)

✅ Answer: (a) (2, 1)
📖 Explanation: Mid-point = ((-2+6)/2, (5+(-3))/2) = (4/2, 2/2) = (2, 1).

(iii) 📈 The slope of a line parallel to the line 3x − 4y + 7 = 0 is:
(a) 3/4
(b) −3/4
(c) 4/3
(d) −4/3

✅ Answer: (a) 3/4
📖 Explanation: 3x − 4y + 7 = 0 ⇒ 4y = 3x + 7 ⇒ y = (3/4)x + 7/4. Slope = 3/4. Parallel lines have equal slopes.

(iv) ⭕ In a circle, the angle subtended by an arc at the centre is 80°. The angle subtended by the same arc at any point on the remaining part of the circle is:
(a) 160°
(b) 100°
(c) 40°
(d) 80°

✅ Answer: (c) 40°
📖 Explanation: The angle at the centre is twice the angle at the circumference. So, angle at circumference = 80°/2 = 40°.

(v) 📏 The number of tangents that can be drawn from an external point to a circle is:
(a) 0
(b) 1
(c) 2
(d) Infinite

✅ Answer: (c) 2
📖 Explanation: From an external point, exactly two tangents can be drawn to a circle.

(vi) 🧊 If the volume of a cone is 100π cm³ and the volume of a cylinder with the same base and height is:
(a) 100π cm³
(b) 200π cm³
(c) 300π cm³
(d) 400π cm³

✅ Answer: (c) 300π cm³
📖 Explanation: Volume of cylinder = 3 × Volume of cone (for same base and height) = 3 × 100π = 300π cm³.

(vii) ⚽ The total surface area of a solid hemisphere of radius r is:
(a) 2πr²
(b) 3πr²
(c) 4πr²
(d) πr²

✅ Answer: (b) 3πr²
📖 Explanation: TSA of hemisphere = Curved surface area + Area of base = 2πr² + πr² = 3πr².

(viii) 📐 If sin θ = 3/5, then the value of cos θ (for acute θ) is:
(a) 4/5
(b) 5/4
(c) 3/4
(d) 5/3

✅ Answer: (a) 4/5
📖 Explanation: sin²θ + cos²θ = 1 ⇒ (3/5)² + cos²θ = 1 ⇒ 9/25 + cos²θ = 1 ⇒ cos²θ = 16/25 ⇒ cos θ = 4/5.

(ix) 🔢 The value of (1 − cos²θ) cosec²θ is:
(a) 0
(b) 1
(c) −1
(d) 2

✅ Answer: (b) 1
📖 Explanation: 1 − cos²θ = sin²θ. So, sin²θ × cosec²θ = sin²θ × (1/sin²θ) = 1.

(x) 🗼 The angle of elevation of the top of a tower from a point 20 m away from its base is 45°. The height of the tower is:
(a) 10 m
(b) 20 m
(c) 20√3 m
(d) 40 m

✅ Answer: (b) 20 m
📖 Explanation: tan 45° = height/distance ⇒ 1 = h/20 ⇒ h = 20 m.

(xi) 📊 The mean of the first five natural numbers is:
(a) 2
(b) 3
(c) 4
(d) 5

✅ Answer: (b) 3
📖 Explanation: Mean = (1+2+3+4+5)/5 = 15/5 = 3.

(xii) 📉 The median of the data: 3, 5, 7, 9, 11, 13 is:
(a) 7
(b) 8
(c) 9
(d) 10

✅ Answer: (b) 8
📖 Explanation: The data has 6 values. Median = average of 3rd and 4th values = (7+9)/2 = 8.

(xiii) 🎲 A die is thrown once. The probability of getting a prime number is:
(a) 1/6
(b) 1/3
(c) 1/2
(d) 2/3

✅ Answer: (c) 1/2
📖 Explanation: Prime numbers on a die: 2, 3, 5 (3 outcomes). Total outcomes = 6. Probability = 3/6 = 1/2.

(xiv) 🔤 If a letter is chosen at random from the word “MATHEMATICS”, the probability that it is a vowel is:
(a) 4/11
(b) 5/11
(c) 3/11
(d) 6/11

✅ Answer: (a) 4/11
📖 Explanation: Vowels in “MATHEMATICS”: A, E, A, I = 4 vowels. Total letters = 11. Probability = 4/11.

(xv) 🎯 The point (0, 0) lies on the line:
(a) x = 0 only
(b) y = 0 only
(c) both x = 0 and y = 0
(d) neither x = 0 nor y = 0

✅ Answer: (c) both x = 0 and y = 0
📖 Explanation: The origin (0, 0) satisfies both x = 0 (y-axis) and y = 0 (x-axis).

❓ Question 2 [12 Marks]

(Solve the following)

(i) 📍 Find the coordinates of the point which divides the line segment joining A(2, −3) and B(−4, 6) in the ratio 2 : 1 internally. [4]

✅ Answer:
Using section formula:
P = ((m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂))
Here, m₁ = 2, m₂ = 1, (x₁, y₁) = (2, −3), (x₂, y₂) = (−4, 6)
P = ((2×(−4) + 1×2)/(2+1), (2×6 + 1×(−3))/(2+1))
P = ((−8+2)/3, (12−3)/3) = (−6/3, 9/3) = (−2, 3)
📍 Coordinates of P = (−2, 3)

(ii) 🔺 Prove that: (1 + tan²A)(1 − sin²A) = 1. [4]

✅ Answer:
LHS = (1 + tan²A)(1 − sin²A)
= sec²A × cos²A [Since 1 + tan²A = sec²A, and 1 − sin²A = cos²A]
= (1/cos²A) × cos²A
= 1 = RHS
✔️ Hence proved.

(iii) 📊 Find the mean of the following frequency distribution: [4]

✅ Answer:

Mean = Σfx / Σf = 440/20 = 📊 22

❓ Question 3 [12 Marks]

(Solve the following)

(i) ⚽ A solid metallic sphere of radius 6 cm is melted and recast into a cone of height 24 cm. Find the radius of the base of the cone. [4]

✅ Answer:
Volume of sphere = Volume of cone
(4/3)πr³ = (1/3)πR²h
(4/3) × π × 6³ = (1/3) × π × R² × 24
(4/3) × 216 = (1/3) × R² × 24
288 = 8R²
R² = 36
R = 6 cm
📏 Radius of the cone = 6 cm

(ii) ⭕ Construct a circle of radius 3.5 cm. Draw a chord AB of length 5 cm. Construct the perpendicular bisector of AB and verify that it passes through the centre of the circle. [4]

✅ Answer:
📝 Steps of Construction:

1. Draw a circle with centre O and radius 3.5 cm.

2. Mark a point A on the circle.

3. With A as centre and radius 5 cm, draw an arc intersecting the circle at B.

4. Join AB to get chord AB.

5. With A and B as centres, draw arcs of radius more than half of AB on both sides of AB, intersecting at P and Q.

6. Join PQ. PQ is the perpendicular bisector of AB.

7. Observe that PQ passes through the centre O.
   ✔️ Conclusion: The perpendicular bisector of a chord passes through the centre of the circle. Verified.

(iii) 🗼 A tower stands vertically on the ground. From a point on the ground, 15 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower. [4]

✅ Answer:
Let AB be the tower and C be the point on the ground.
In right triangle ABC:
tan 60° = AB/BC
√3 = h/15
h = 15√3
h = 15 × 1.732 = 25.98 m ≈ 📏 25.98 m
Height of the tower = 15√3 m (or 25.98 m)

📗 SECTION B (40 Marks)

(Attempt any four questions from this Section.)

❓ Question 4 [10 Marks]

(i) 🔄 The reflection of point P(m, n) in the x-axis is P'(m, −n). If P’ is reflected in the origin to get P”(−3, 5), find the values of m and n. [3]

✅ Answer:
Reflection of P(m, n) in x-axis = P'(m, −n)
Reflection of P'(m, −n) in origin = P”(−m, n)
Given P” = (−3, 5)
So, −m = −3 ⇒ m = 3
And n = 5 ⇒ n = 5
✔️ m = 3, n = 5

(ii) 📈 Find the equation of a line passing through the point (2, −3) and perpendicular to the line 3x − 4y + 5 = 0. [3]

✅ Answer:
Given line: 3x − 4y + 5 = 0 ⇒ 4y = 3x + 5 ⇒ y = (3/4)x + 5/4
Slope of given line = 3/4
Slope of perpendicular line = −4/3
Equation of required line through (2, −3):
y − (−3) = (−4/3)(x − 2)
y + 3 = (−4/3)(x − 2)
3(y + 3) = −4(x − 2)
3y + 9 = −4x + 8
4x + 3y + 1 = 0
📝 Equation of the line: 4x + 3y + 1 = 0

(iii) 🎲 A bag contains 5 red balls, 4 blue balls, and 3 green balls. A ball is drawn at random. Find the probability that it is:
(a) red
(b) not blue
(c) either red or green [4]

✅ Answer:
Total balls = 5 + 4 + 3 = 12
(a) P(red) = 5/12
(b) P(not blue) = 1 − P(blue) = 1 − 4/12 = 8/12 = 2/3
(c) P(red or green) = (5+3)/12 = 8/12 = 2/3
✔️ Answers: (a) 5/12 (b) 2/3 (c) 2/3

❓ Question 5 [10 Marks]

(i) 🧊 A cylinder and a cone have the same base radius and same height. Find the ratio of their volumes. [3]

✅ Answer:
Volume of cylinder : Volume of cone
= πr²h : (1/3)πr²h
= 1 : 1/3
= 📊 3 : 1

(ii) 🔺 Prove that: sec²θ − cosec²θ = tan²θ − cot²θ. [3]

✅ Answer:
LHS = sec²θ − cosec²θ
= (1 + tan²θ) − (1 + cot²θ) [Since sec²θ = 1 + tan²θ, cosec²θ = 1 + cot²θ]
= 1 + tan²θ − 1 − cot²θ
= tan²θ − cot²θ = RHS
✔️ Hence proved.

(iii) 📊 The median of the following data is 28. Find the value of x. [4]

✅ Answer:
Total frequency = 5 + 8 + x + 12 + 7 = 32 + x
Median class is 20–30 (given median = 28)
Cumulative frequency before median class = 5 + 8 = 13
Median = l + [(N/2 − cf)/f] × h
28 = 20 + [((32+x)/2 − 13)/x] × 10
8 = [((32+x)/2 − 13)/x] × 10
8x/10 = (32+x)/2 − 13
0.8x = (32+x)/2 − 13
Multiply by 2: 1.6x = 32 + x − 26
1.6x = 6 + x
0.6x = 6
✔️ x = 10

❓ Question 6 [10 Marks]

(i) 📍 Find the value of k if the point (k, 2) lies on the line 2x + 3y = 16. [3]

✅ Answer:
Substitute (k, 2) in 2x + 3y = 16:
2k + 3(2) = 16
2k + 6 = 16
2k = 10
✔️ k = 5

(ii) ⭕ Construct a tangent to a circle of radius 4 cm from a point P on the concentric circle of radius 6 cm. [3]

✅ Answer:
📝 Steps of Construction:

1. Draw two concentric circles with centre O and radii 4 cm and 6 cm.

2. Take a point P on the larger circle.

3. Join OP. Draw the perpendicular bisector of OP, intersecting OP at M.

4. With M as centre and radius MO, draw a circle intersecting the smaller circle at points A and B.

5. Join PA and PB. PA and PB are the required tangents.
   ✔️ Verification: ∠OAP = ∠OBP = 90° (angle in semicircle). Hence, PA and PB are tangents.

(iii) 🗼 A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is 60°. When he moves 40 m away from the bank, the angle of elevation becomes 30°. Find the height of the tree and the width of the river. [4]

✅ Answer:
Let the height of the tree be h and the width of the river be x.
From first position: tan 60° = h/x ⇒ h = x√3 … (1)
From second position: tan 30° = h/(x + 40) ⇒ h = (x + 40)/√3 … (2)
From (1) and (2):
x√3 = (x + 40)/√3
3x = x + 40
2x = 40 ⇒ x = 20
From (1): h = 20√3 = 20 × 1.732 = 34.64 m
✔️ Height of tree = 20√3 m (≈ 34.64 m), Width of river = 20 m

❓ Question 7 [10 Marks]

(i) 📐 If A(1, 2), B(4, 6), and C(5, 8) are three points, show that they are collinear. [3]

✅ Answer:
Slope of AB = (6−2)/(4−1) = 4/3
Slope of BC = (8−6)/(5−4) = 2/1 = 2
Since slope of AB ≠ slope of BC, the points are ❌ not collinear.
(📝 Note: This question tests understanding. If collinearity is intended, the points would need adjustment.)

(ii) 🧊 A solid cone of radius 7 cm and height 24 cm is melted and recast into small spheres of radius 3.5 cm. Find the number of spheres formed. [3]

✅ Answer:
Volume of cone = (1/3)πr²h = (1/3)π(7)²(24) = (1/3)π × 49 × 24 = 392π cm³
Volume of one sphere = (4/3)πr³ = (4/3)π(3.5)³ = (4/3)π × 42.875 = (171.5/3)π cm³
Number of spheres = (392π) / ((171.5/3)π) = (392 × 3)/171.5 = 1176/171.5 ≈ 6.86
Since we need whole spheres, ✔️ 6 spheres can be formed completely.

(iii) 📊 Find the mode of the following data: [4]

✅ Answer:
The mode is the value with the highest frequency.
Highest frequency = 8 (for marks = 30)
✔️ Mode = 30

❓ Question 8 [10 Marks]

(i) 🔺 Prove that: (sin θ + cos θ)² + (sin θ − cos θ)² = 2. [3]

✅ Answer:
LHS = (sin θ + cos θ)² + (sin θ − cos θ)²
= (sin²θ + 2sinθcosθ + cos²θ) + (sin²θ − 2sinθcosθ + cos²θ)
= 2sin²θ + 2cos²θ
= 2(sin²θ + cos²θ)
= 2(1) = 2 = RHS
✔️ Hence proved.

(ii) 📈 Find the equation of the line passing through the points A(2, 3) and B(4, 7). [3]

✅ Answer:
Slope m = (7−3)/(4−2) = 4/2 = 2
Equation: y − 3 = 2(x − 2)
y − 3 = 2x − 4
📝 2x − y − 1 = 0 (or y = 2x − 1)

(iii) 🎲 A bag contains 6 white balls and 4 black balls. Two balls are drawn at random without replacement. Find the probability that both balls are white. [4]

✅ Answer:
Total balls = 10
P(first white) = 6/10
P(second white | first white) = 5/9
P(both white) = (6/10) × (5/9) = 30/90 = ✔️ 1/3

❓ Question 9 [10 Marks]

(i) 📍 Find the coordinates of the point on the x-axis which is equidistant from A(2, −5) and B(−2, 9). [3]

✅ Answer:
Let the point on x-axis be P(x, 0).
PA = PB
(x−2)² + (0+5)² = (x+2)² + (0−9)²
(x−2)² + 25 = (x+2)² + 81
x² − 4x + 4 + 25 = x² + 4x + 4 + 81
−4x + 29 = 4x + 85
−8x = 56
x = −7
✔️ Point is (−7, 0)

(ii) 🧊 The circumference of the base of a cylinder is 44 cm and its height is 10 cm. Find its curved surface area. [3]

✅ Answer:
2πr = 44 ⇒ r = 44/(2π) = 44/(2×22/7) = 7 cm
Curved surface area = 2πrh = 44 × 10 = ✔️ 440 cm²

(iii) ⭕ In a circle of radius 10 cm, a chord is at a distance of 6 cm from the centre. Find the length of the chord. [4]

✅ Answer:
Let AB be the chord and O be the centre. Draw OM ⊥ AB.
OM = 6 cm, OA = 10 cm
In right triangle OMA:
AM² = OA² − OM² = 10² − 6² = 100 − 36 = 64
AM = 8 cm
Since perpendicular from centre bisects the chord:
AB = 2 × AM = 2 × 8 = ✔️ 16 cm

❓ Question 10 [10 Marks]

(i) 🔢 Solve for θ: 2 sin²θ = 1 (for 0° < θ < 90°). [3]

✅ Answer:
2 sin²θ = 1 ⇒ sin²θ = 1/2 ⇒ sin θ = 1/√2
Since θ is acute, ✔️ θ = 45°

(ii) 📊 The mean of 5 numbers is 18. If one number is excluded, the mean becomes 16. Find the excluded number. [3]

✅ Answer:
Sum of 5 numbers = 5 × 18 = 90
Sum of 4 numbers = 4 × 16 = 64
Excluded number = 90 − 64 = ✔️ 26

(iii) 🗼 From the top of a cliff 100 m high, the angles of depression of two boats on the same side of the cliff are 30° and 45°. Find the distance between the two boats. [4]

✅ Answer:
Let the cliff be AB with A at the top (height = 100 m). Boats at C and D (C nearer, D farther).
For boat C: tan 45° = 100/BC ⇒ BC = 100 m
For boat D: tan 30° = 100/BD ⇒ BD = 100√3 = 173.2 m
Distance between boats = BD − BC = 100√3 − 100 = 100(√3 − 1) = 100(1.732 − 1) = 100 × 0.732 = ✔️ 73.2 m

🎓 END OF QUESTION PAPER

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