ICSE Class 10 Mathematics – Model Paper 3 (With Answers)

🏫 Dhingra Classes, Nashik

📊 Lessons & Marks Weightage (As per ICSE Board Pattern)

📘 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(5, −2) in the x-axis is:
(a) (5, 2)
(b) (−5, −2)
(c) (−5, 2)
(d) (2, 5)

✅ Answer: (a) (5, 2)
📖 Explanation: Under reflection in the x-axis, (x, y) → (x, −y). So, (5, −2) → (5, 2).

(ii) 📍 The midpoint of the line segment joining A(3, −1) and B(−1, 5) is:
(a) (1, 2)
(b) (2, 1)
(c) (1, 3)
(d) (2, 2)

✅ Answer: (a) (1, 2)
📖 Explanation: Midpoint = ((3+(−1))/2, (−1+5)/2) = (2/2, 4/2) = (1, 2).

(iii) 📈 The slope of the line 4x − 2y + 3 = 0 is:
(a) 2
(b) −2
(c) 1/2
(d) −1/2

✅ Answer: (a) 2
📖 Explanation: 4x − 2y + 3 = 0 ⇒ 2y = 4x + 3 ⇒ y = 2x + 3/2. Slope = 2.

(iv) ⭕ In the given figure, O is the centre of the circle. If ∠AOB = 120°, then ∠ACB is:

        A
       / \
      /   \
     /     \
    O-------B
     \     /
      \   /
       \ /
        C

(a) 120°
(b) 60°
(c) 30°
(d) 90°

✅ Answer: (b) 60°
📖 Explanation: Angle at the centre = 2 × Angle at the circumference. ∠ACB = 120°/2 = 60°.

(v) 📐 The length of a tangent from a point P at a distance of 5 cm from the centre of a circle of radius 3 cm is:
(a) 2 cm
(b) 4 cm
(c) 8 cm
(d) √34 cm

✅ Answer: (b) 4 cm
📖 Explanation: Tangent length = √(OP² − r²) = √(5² − 3²) = √(25 − 9) = √16 = 4 cm.

(vi) 🧊 The curved surface area of a cylinder of radius 7 cm and height 10 cm is:
(a) 440 cm²
(b) 220 cm²
(c) 1540 cm²
(d) 770 cm²

✅ Answer: (a) 440 cm²
📖 Explanation: CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².

(vii) 🔺 If sin θ = 5/13, then tan θ (for acute θ) is:
(a) 5/12
(b) 12/5
(c) 13/12
(d) 12/13

✅ Answer: (a) 5/12
📖 Explanation: cos θ = 12/13. tan θ = sin θ/cos θ = (5/13)/(12/13) = 5/12.

(viii) 🔢 The value of sec²45° − tan²45° is:
(a) 0
(b) 1
(c) 2
(d) √2

✅ Answer: (b) 1
📖 Explanation: sec²θ − tan²θ = 1 (fundamental identity).

(ix) 🗼 The angle of elevation of the top of a tower from a point 30 m away from its base is 30°. The height of the tower is:
(a) 10√3 m
(b) 30√3 m
(c) 10 m
(d) 30 m

✅ Answer: (a) 10√3 m
📖 Explanation: tan 30° = h/30 ⇒ 1/√3 = h/30 ⇒ h = 30/√3 = 10√3 m.

(x) 📊 The mean of 5, 10, 15, 20, 25 is:
(a) 10
(b) 15
(c) 20
(d) 25

✅ Answer: (b) 15
📖 Explanation: Mean = (5+10+15+20+25)/5 = 75/5 = 15.

(xi) 📉 The median of the data: 4, 6, 8, 10, 12, 14 is:
(a) 8
(b) 9
(c) 10
(d) 11

✅ Answer: (b) 9
📖 Explanation: Median = average of 3rd and 4th values = (8+10)/2 = 9.

(xii) 🎲 A die is thrown once. The probability of getting a number greater than 4 is:
(a) 1/6
(b) 1/3
(c) 1/2
(d) 2/3

✅ Answer: (b) 1/3
📖 Explanation: Numbers > 4: 5, 6 (2 outcomes). P = 2/6 = 1/3.

(xiii) 🔤 A bag contains 4 red, 5 green, and 6 blue balls. One ball is drawn at random. The probability that it is red or green is:
(a) 4/15
(b) 5/15
(c) 9/15
(d) 6/15

✅ Answer: (c) 9/15
📖 Explanation: P(red or green) = (4+5)/15 = 9/15 = 3/5.

(xiv) ⭕ In the given figure, PA and PB are tangents to the circle with centre O. If ∠AOB = 110°, then ∠APB is:

        A
       / \
      /   \
     /     \
    P-------O
     \     /
      \   /
       \ /
        B

(a) 70°
(b) 110°
(c) 55°
(d) 90°

✅ Answer: (a) 70°
📖 Explanation: ∠APB = 180° − ∠AOB = 180° − 110° = 70°.

(xv) 📈 The equation of the y-axis is:
(a) x = 0
(b) y = 0
(c) x = y
(d) x + y = 0

✅ Answer: (a) x = 0
📖 Explanation: The y-axis consists of all points with x-coordinate 0.

❓ Question 2 [12 Marks]

(Solve the following)

(i) 📍 Find the coordinates of the point which divides the line segment joining A(3, 5) and B(−3, −2) 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) = (3, 5), (x₂, y₂) = (−3, −2)
P = ((2×(−3) + 1×3)/(2+1), (2×(−2) + 1×5)/(2+1))
P = ((−6+3)/3, (−4+5)/3) = (−3/3, 1/3) = (−1, 1/3)
✔️ Coordinates of P = (−1, 1/3)

(ii) 🔺 Prove that: (1 + tan²A) cos²A = 1. [4]

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

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

✅ Answer:

Mean = Σfx / Σf = 2040/40 = 📊 51

❓ Question 3 [12 Marks]

(Solve the following)

(i) 🧊 A solid metallic sphere of radius 9 cm is melted and recast into a cone of radius 9 cm. Find the height of the cone. [4]

✅ Answer:
Volume of sphere = Volume of cone
(4/3)πr³ = (1/3)πR²h
(4/3)π(9)³ = (1/3)π(9)²h
(4/3) × 729 = (1/3) × 81 × h
972 = 27h
h = 36 cm
✔️ Height of the cone = 36 cm

(ii) ⭕ Construct a circle of radius 3 cm. Draw a chord AB of length 4 cm. Construct the perpendicular bisector of AB and verify that it passes through the centre. [4]

✅ Answer:
📝 Steps of Construction:

1. Draw a circle with centre O and radius 3 cm.

2. Mark a point A on the circle.

3. With A as centre and radius 4 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, 25 m away from the foot of the tower, the angle of elevation of the top of the tower is 45°. 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 45° = AB/BC
1 = h/25
h = 25 m
✔️ Height of the tower = 25 m

📗 SECTION B (40 Marks)

(Attempt any four questions from this Section.)

❓ Question 4 [10 Marks]

(i) 🔄 The reflection of point P(x, y) in the origin is P'(−x, −y). If P’ is reflected in the x-axis to get P”(−4, 6), find the values of x and y. [3]

✅ Answer:
Reflection of P(x, y) in origin = P'(−x, −y)
Reflection of P'(−x, −y) in x-axis = P”(−x, y)
Given P” = (−4, 6)
So, −x = −4 ⇒ x = 4
And y = 6 ⇒ y = 6
✔️ x = 4, y = 6

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

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

(iii) 🎲 A bag contains 4 white balls, 6 black balls, and 5 red balls. A ball is drawn at random. Find the probability that it is:
(a) white
(b) not black
(c) either white or red [4]

✅ Answer:
Total balls = 4 + 6 + 5 = 15
(a) P(white) = 4/15
(b) P(not black) = 1 − P(black) = 1 − 6/15 = 9/15 = 3/5
(c) P(white or red) = (4+5)/15 = 9/15 = 3/5
✔️ Answers: (a) 4/15 (b) 3/5 (c) 3/5

❓ Question 5 [10 Marks]

(i) 🧊 A cylinder and a sphere have the same radius. If the volume of the cylinder is equal to the volume of the sphere, find the ratio of the height of the cylinder to its radius. [3]

✅ Answer:
Volume of cylinder = Volume of sphere
πr²h = (4/3)πr³
h = (4/3)r
h/r = 4/3
✔️ h : r = 4 : 3

(ii) 🔺 Prove that: (sec A − tan A)(sec A + tan A) = 1. [3]

✅ Answer:
LHS = (sec A − tan A)(sec A + tan A)
= sec²A − tan²A
= 1 = RHS
✔️ Hence proved.

(iii) 📊 The median of the following data is 35. Find the value of x. [4]

✅ Answer:
Total frequency = 4 + 6 + x + 12 + 8 = 30 + x
Median class is 30–40 (given median = 35)
Cumulative frequency before median class = 4 + 6 + x = 10 + x
Median = l + [(N/2 − cf)/f] × h
35 = 30 + [((30+x)/2 − (10+x))/12] × 10
5 = [((30+x)/2 − 10 − x)/12] × 10
60/10 = (30+x)/2 − 10 − x
6 = (30+x)/2 − 10 − x
16 = (30+x)/2 − x
32 = 30 + x − 2x
32 = 30 − x
x = −2
✔️ Since frequency cannot be negative, the data is inconsistent. Please verify.

❓ Question 6 [10 Marks]

(i) 📍 Find the value of k if the point (−2, k) lies on the line 3x + 4y = 10. [3]

✅ Answer:
Substitute (−2, k) in 3x + 4y = 10:
3(−2) + 4k = 10
−6 + 4k = 10
4k = 16
✔️ k = 4

(ii) ⭕ Construct a tangent to a circle of radius 3.5 cm from a point P at a distance of 7 cm from the centre. [3]

✅ Answer:
📝 Steps of Construction:

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

2. Mark a point P at a distance of 7 cm from O.

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 given 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).

(iii) 🗼 From the top of a cliff 80 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 = 80 m). Boats at C and D (C nearer, D farther).
For boat C: tan 45° = 80/BC ⇒ BC = 80 m
For boat D: tan 30° = 80/BD ⇒ BD = 80√3 = 138.56 m
Distance between boats = BD − BC = 80√3 − 80 = 80(√3 − 1) = 80(1.732 − 1) = 80 × 0.732 = ✔️ 58.56 m

❓ Question 7 [10 Marks]

(i) 📐 Show that the points A(2, 3), B(4, 7), and C(6, 11) are collinear. [3]

✅ Answer:
Slope of AB = (7−3)/(4−2) = 4/2 = 2
Slope of BC = (11−7)/(6−4) = 4/2 = 2
Since slope of AB = slope of BC, the points are ✔️ collinear.

(ii) 🧊 A solid cone of radius 6 cm and height 8 cm is melted and recast into small spheres of radius 1 cm. Find the number of spheres formed. [3]

✅ Answer:
Volume of cone = (1/3)πr²h = (1/3)π(6)²(8) = (1/3)π × 36 × 8 = 96π cm³
Volume of one sphere = (4/3)πr³ = (4/3)π(1)³ = (4/3)π cm³
Number of spheres = 96π / (4/3)π = 96 × 3/4 = 72
✔️ Number of spheres = 72

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

✅ Answer:
The modal class is 20–30 (highest frequency = 15).
Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
l = 20, f₁ = 15, f₀ = 8, f₂ = 10, h = 10
Mode = 20 + [(15 − 8)/(2×15 − 8 − 10)] × 10
= 20 + [7/(30 − 18)] × 10
= 20 + (7/12) × 10
= 20 + 5.83 = ✔️ 25.83

❓ Question 8 [10 Marks]

(i) 🔺 Prove that: (sin θ − cos θ)² = 1 − 2 sin θ cos θ. [3]

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

(ii) 📈 Find the equation of the line passing through the points (−1, 4) and (3, −2). [3]

✅ Answer:
Slope m = (−2−4)/(3+1) = −6/4 = −3/2
Equation: y − 4 = (−3/2)(x + 1)
2(y − 4) = −3(x + 1)
2y − 8 = −3x − 3
3x + 2y − 5 = 0
📝 Equation: 3x + 2y − 5 = 0

(iii) 🎲 A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are red. [4]

✅ Answer:
Total balls = 8
P(first red) = 5/8
P(second red | first red) = 4/7
P(both red) = (5/8) × (4/7) = 20/56 = ✔️ 5/14

❓ Question 9 [10 Marks]

(i) 📍 Find the coordinates of the point on the x-axis which is equidistant from A(4, 3) and B(−2, 5). [3]

✅ Answer:
Let the point on x-axis be P(x, 0).
PA = PB
(x−4)² + (0−3)² = (x+2)² + (0−5)²
(x−4)² + 9 = (x+2)² + 25
x² − 8x + 16 + 9 = x² + 4x + 4 + 25
−8x + 25 = 4x + 29
−12x = 4
x = −1/3
✔️ Point is (−1/3, 0)

(ii) 🧊 The total surface area of a cylinder is 462 cm² and its curved surface area is 308 cm². Find its radius. [3]

✅ Answer:
TSA = CSA + 2πr²
462 = 308 + 2πr²
2πr² = 154
πr² = 77
r² = 77 × 7/22 = 24.5
r = √24.5 ≈ ✔️ 4.95 cm

(iii) ⭕ In the given figure, O is the centre of the circle and AB is a chord. If OA = 10 cm and the distance of the chord from the centre is 6 cm, find the length of the chord.

        O
       /|\
      / | \
     /  |  \
    A---M---B

✅ Answer:
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: cos θ = 1/2 (for 0° < θ < 90°). [3]

✅ Answer:
cos θ = 1/2 ⇒ cos θ = cos 60°
✔️ θ = 60°

(ii) 📊 The mean of 6 numbers is 25. If one number is excluded, the mean becomes 22. Find the excluded number. [3]

✅ Answer:
Sum of 6 numbers = 6 × 25 = 150
Sum of 5 numbers = 5 × 22 = 110
Excluded number = 150 − 110 = ✔️ 40

(iii) 🗼 A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Find the height of the hill. [4]

✅ Answer:

        H (top of hill)
        |\
        | \
        |  \
      h |   \
        |    \
        |     \
        D------E (deck level)
        |      |
     10 |      |
        |      |
        W------B (base of hill)

Let the height of the hill be H, and the horizontal distance from the ship to the hill be x.
From deck level: tan 30° = 10/x ⇒ x = 10√3 m
From deck level: tan 60° = (H − 10)/x ⇒ √3 = (H − 10)/(10√3)
30 = H − 10
H = 40 m
✔️ Height of the hill = 40 m

🎓 END OF QUESTION PAPER

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