📊 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(3, −5) in the origin is:
(a) (3, 5)
(b) (−3, −5)
(c) (−3, 5)
(d) (−5, 3)

✅ Answer: (c) (−3, 5)
📖 Explanation: Under reflection in the origin, (x, y) → (−x, −y). So, (3, −5) → (−3, 5).

(ii) 📍 The coordinates of the point dividing the line segment joining A(1, 3) and B(4, 6) in the ratio 1 : 2 internally are:
(a) (2, 4)
(b) (3, 5)
(c) (2, 3)
(d) (1, 2)

✅ Answer: (a) (2, 4)
📖 Explanation: P = ((1×4 + 2×1)/(1+2), (1×6 + 2×3)/(1+2)) = ((4+2)/3, (6+6)/3) = (6/3, 12/3) = (2, 4).

(iii) 📈 The slope of the line 2x + 3y − 6 = 0 is:
(a) 2/3
(b) −2/3
(c) 3/2
(d) −3/2

✅ Answer: (b) −2/3
📖 Explanation: 2x + 3y − 6 = 0 ⇒ 3y = −2x + 6 ⇒ y = (−2/3)x + 2. Slope = −2/3.

(iv) ⭕ In the given figure, O is the centre of the circle. If ∠AOB = 100°, then ∠ACB is:

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

(a) 100°
(b) 50°
(c) 80°
(d) 25°

✅ Answer: (b) 50°
📖 Explanation: Angle at the centre = 2 × Angle at the circumference. ∠ACB = 100°/2 = 50°.

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

✅ Answer: (b) 1
📖 Explanation: From a point on the circle, exactly one tangent can be drawn.

(vi) 🧊 The volume of a sphere of radius 3 cm is:
(a) 36π cm³
(b) 27π cm³
(c) 9π cm³
(d) 18π cm³

✅ Answer: (a) 36π cm³
📖 Explanation: Volume = (4/3)πr³ = (4/3)π(3)³ = (4/3)π × 27 = 36π cm³.

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

✅ Answer: (a) 12/13
📖 Explanation: sin²θ = 1 − cos²θ = 1 − 25/169 = 144/169. sin θ = 12/13.

(viii) 🔢 The value of sin²30° + cos²30° is:
(a) 0
(b) 1
(c) 1/2
(d) √3/2

✅ Answer: (b) 1
📖 Explanation: sin²θ + cos²θ = 1 (fundamental identity).

(ix) 🗼 The angle of depression of a point on the ground from the top of a tower is 30°. If the height of the tower is 30 m, the distance of the point from the foot of the tower is:
(a) 30√3 m
(b) 30/√3 m
(c) 60 m
(d) 15 m

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

(x) 📊 The mean of 10, 20, 30, 40, 50 is:
(a) 25
(b) 30
(c) 35
(d) 40

✅ Answer: (b) 30
📖 Explanation: Mean = (10+20+30+40+50)/5 = 150/5 = 30.

(xi) 📉 The mode of the data: 2, 3, 3, 5, 7, 3, 2 is:
(a) 2
(b) 3
(c) 5
(d) 7

✅ Answer: (b) 3
📖 Explanation: 3 appears most frequently (3 times).

(xii) 🎲 A card is drawn from a well-shuffled deck of 52 cards. The probability of getting a king is:
(a) 1/13
(b) 1/26
(c) 1/52
(d) 4/13

✅ Answer: (a) 1/13
📖 Explanation: Number of kings = 4. P(king) = 4/52 = 1/13.

(xiii) 🔤 A bag contains 3 red, 4 green, and 5 blue balls. One ball is drawn at random. The probability that it is NOT green is:
(a) 1/3
(b) 2/3
(c) 4/12
(d) 5/12

✅ Answer: (b) 2/3
📖 Explanation: Total = 12. P(not green) = 1 − P(green) = 1 − 4/12 = 8/12 = 2/3.

(xiv) ⭕ In the given figure, PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then ∠OAB is:

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

(a) 30°
(b) 60°
(c) 90°
(d) 120°

✅ Answer: (a) 30°
📖 Explanation: ∠AOB = 180° − 60° = 120°. In triangle OAB, OA = OB (radii), so ∠OAB = ∠OBA = (180° − 120°)/2 = 30°.

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

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

❓ Question 2 [12 Marks]

(Solve the following)

(i) 📍 Find the coordinates of the point which divides the line segment joining A(−1, 7) and B(4, −3) in the ratio 2 : 3 internally. [4]

✅ Answer:
Using section formula:
P = ((m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂))
Here, m₁ = 2, m₂ = 3, (x₁, y₁) = (−1, 7), (x₂, y₂) = (4, −3)
P = ((2×4 + 3×(−1))/(2+3), (2×(−3) + 3×7)/(2+3))
P = ((8−3)/5, (−6+21)/5) = (5/5, 15/5) = (1, 3)
✔️ Coordinates of P = (1, 3)

(ii) 🔺 Prove that: (1 + cot²A) sin²A = 1. [4]

✅ Answer:
LHS = (1 + cot²A) sin²A
= cosec²A × sin²A [Since 1 + cot²A = cosec²A]
= (1/sin²A) × sin²A
= 1 = RHS
✔️ Hence proved.

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

✅ Answer:

N/2 = 15. Median class = 20–30 (cf before = 10, f = 10, l = 20, h = 10)
Median = l + [(N/2 − cf)/f] × h
= 20 + [(15 − 10)/10] × 10
= 20 + (5/10) × 10
= 20 + 5 = ✔️ 25

❓ Question 3 [12 Marks]

(Solve the following)

(i) 🧊 A solid cylinder of radius 7 cm and height 10 cm is melted and recast into a sphere. Find the radius of the sphere. [4]

✅ Answer:
Volume of cylinder = Volume of sphere
πr²h = (4/3)πR³
π(7)²(10) = (4/3)πR³
49 × 10 = (4/3)R³
490 = (4/3)R³
R³ = (490 × 3)/4 = 1470/4 = 367.5
R = ∛367.5 ≈ ✔️ 7.16 cm

(ii) ⭕ Construct a circle of radius 4 cm. Draw a chord PQ of length 6 cm. Construct the perpendicular bisector of PQ and verify that it passes through the centre. [4]

✅ Answer:
📝 Steps of Construction:

1. Draw a circle with centre O and radius 4 cm.

2. Mark a point P on the circle.

3. With P as centre and radius 6 cm, draw an arc intersecting the circle at Q.

4. Join PQ to get chord PQ.

5. With P and Q as centres, draw arcs of radius more than half of PQ on both sides of PQ, intersecting at M and N.

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

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

(iii) 🗼 A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall. [4]

✅ Answer:

        |\
        | \
    8 m |  \ 10 m
        |   \
        |____\
          x

In right triangle:
x² + 8² = 10²
x² + 64 = 100
x² = 36
x = 6 m
✔️ Distance of the foot of the ladder from the wall = 6 m

📗 SECTION B (40 Marks)

(Attempt any four questions from this Section.)

❓ Question 4 [10 Marks]

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

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

(ii) 📈 Find the equation of a line passing through the point (−1, 2) and parallel to the line 3x + 4y = 7. [3]

✅ Answer:
Given line: 3x + 4y = 7 ⇒ 4y = −3x + 7 ⇒ y = (−3/4)x + 7/4
Slope of given line = −3/4
Slope of parallel line = −3/4
Equation of required line through (−1, 2):
y − 2 = (−3/4)(x + 1)
4(y − 2) = −3(x + 1)
4y − 8 = −3x − 3
3x + 4y − 5 = 0
📝 Equation of the line: 3x + 4y − 5 = 0

(iii) 🎲 A bag contains 5 white balls, 3 black balls, and 2 red balls. Two balls are drawn at random without replacement. Find the probability that:
(a) both are white
(b) one is white and one is black [4]

✅ Answer:
Total balls = 10
(a) P(both white) = (5/10) × (4/9) = 20/90 = 2/9
(b) P(one white, one black) = P(WB) + P(BW) = (5/10)(3/9) + (3/10)(5/9) = 15/90 + 15/90 = 30/90 = 1/3

❓ Question 5 [10 Marks]

(i) 🧊 A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights. [3]

✅ Answer:
Volume of cone = Volume of hemisphere
(1/3)πr²h = (2/3)πr³
h = 2r
Ratio of height of cone to radius = 2 : 1
✔️ h : r = 2 : 1

(ii) 🔺 Prove that: (sin A + cos A)² = 1 + 2 sin A cos A. [3]

✅ Answer:
LHS = (sin A + cos A)²
= sin²A + 2sinAcosA + cos²A
= (sin²A + cos²A) + 2sinAcosA
= 1 + 2sinAcosA = RHS
✔️ Hence proved.

(iii) 📊 The mean of the following data is 25. Find the value of f. [4]

✅ Answer:

Mean = Σfx / Σf = 25
(810 + 25f)/(30 + f) = 25
810 + 25f = 750 + 25f
810 = 750 (Contradiction)
📝 Note: This indicates the data needs adjustment. Let’s solve correctly:
810 + 25f = 25(30 + f)
810 + 25f = 750 + 25f
810 = 750
This is inconsistent. The correct approach:
If mean = 25, then total = 25(30+f) = 750 + 25f
But Σfx = 810 + 25f
So, 810 + 25f = 750 + 25f ⇒ 810 = 750 (impossible)
✔️ The given data is inconsistent. Please verify the frequencies.

❓ Question 6 [10 Marks]

(i) 📍 Find the value of p if the point (p, −3) lies on the line 3x − 2y = 12. [3]

✅ Answer:
Substitute (p, −3) in 3x − 2y = 12:
3p − 2(−3) = 12
3p + 6 = 12
3p = 6
✔️ p = 2

(ii) ⭕ Construct a tangent to a circle of radius 5 cm from a point P at a distance of 8 cm from the centre. [3]

✅ Answer:
📝 Steps of Construction:

1. Draw a circle with centre O and radius 5 cm.

2. Mark a point P at a distance of 8 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 building 60 m high, the angles of depression of the top and bottom of a tower are 30° and 60° respectively. Find the height of the tower. [4]

✅ Answer:

    A (top of building)
    |\
    | \
    |  \
    |   \
    |    \
    |     \
    |      \
    B-------C (top of tower)
    |       |
    |       |
    D-------E (bottom of tower)

Let AB = 60 m, CE = h (height of tower), DE = BC = x
From A, angle of depression to C = 30°, to E = 60°
In triangle ADE: tan 60° = 60/x ⇒ x = 60/√3 = 20√3 m
In triangle ABC: tan 30° = (60 − h)/x ⇒ 1/√3 = (60 − h)/(20√3)
20 = 60 − h
h = 40 m
✔️ Height of the tower = 40 m

❓ Question 7 [10 Marks]

(i) 📐 Show that the points A(1, 1), B(3, 3), and C(5, 5) are collinear. [3]

✅ Answer:
Slope of AB = (3−1)/(3−1) = 2/2 = 1
Slope of BC = (5−3)/(5−3) = 2/2 = 1
Since slope of AB = slope of BC, the points are ✔️ collinear.

(ii) 🧊 A solid sphere of radius 4.2 cm is melted and recast into a cylinder of radius 2.1 cm. Find the height of the cylinder. [3]

✅ Answer:
Volume of sphere = Volume of cylinder
(4/3)πR³ = πr²h
(4/3)(4.2)³ = (2.1)²h
(4/3)(74.088) = 4.41h
98.784 = 4.41h
h = 98.784/4.41 = ✔️ 22.4 cm

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

✅ Answer:

Mean = Σfx / Σf = 1330/50 = ✔️ 26.6

❓ Question 8 [10 Marks]

(i) 🔺 Prove that: tan θ + cot θ = sec θ cosec θ. [3]

✅ Answer:
LHS = tan θ + cot θ
= (sin θ/cos θ) + (cos θ/sin θ)
= (sin²θ + cos²θ)/(sin θ cos θ)
= 1/(sin θ cos θ)
= (1/cos θ)(1/sin θ)
= sec θ cosec θ = RHS
✔️ Hence proved.

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

✅ Answer:
Slope m = (−1−3)/(4+2) = −4/6 = −2/3
Equation: y − 3 = (−2/3)(x + 2)
3(y − 3) = −2(x + 2)
3y − 9 = −2x − 4
2x + 3y − 5 = 0
📝 Equation: 2x + 3y − 5 = 0

(iii) 🎲 A die is thrown twice. Find the probability that:
(a) both throws show even numbers
(b) the sum of the numbers is 8 [4]

✅ Answer:
Total outcomes = 36
(a) Even numbers: 2, 4, 6 (3 outcomes each)
P(both even) = (3/6) × (3/6) = 9/36 = 1/4
(b) Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes
P(sum = 8) = 5/36

❓ Question 9 [10 Marks]

(i) 📍 Find the coordinates of the point on the y-axis which is equidistant from A(3, 2) and B(−1, 4). [3]

✅ Answer:
Let the point on y-axis be P(0, y).
PA = PB
(0−3)² + (y−2)² = (0+1)² + (y−4)²
9 + y² − 4y + 4 = 1 + y² − 8y + 16
y² − 4y + 13 = y² − 8y + 17
−4y + 13 = −8y + 17
4y = 4
y = 1
✔️ Point is (0, 1)

(ii) 🧊 The curved surface area of a cone is 220 cm² and its slant height is 10 cm. Find its radius. [3]

✅ Answer:
CSA of cone = πrl
220 = (22/7) × r × 10
220 = 220r/7
r = 7 cm
✔️ Radius = 7 cm

(iii) ⭕ In the given figure, O is the centre of the circle and AB is a chord. If OA = 13 cm and AB = 24 cm, find the distance of the chord from the centre.

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

✅ Answer:
AM = AB/2 = 12 cm (perpendicular from centre bisects chord)
In right triangle OMA:
OM² = OA² − AM² = 13² − 12² = 169 − 144 = 25
OM = 5 cm
✔️ Distance of the chord from the centre = 5 cm

❓ Question 10 [10 Marks]

(i) 🔢 Solve for θ: tan θ = √3 (for 0° < θ < 90°). [3]

✅ Answer:
tan θ = √3 ⇒ tan θ = tan 60°
✔️ θ = 60°

(ii) 📊 The median of 5 numbers arranged in ascending order is 20. If the numbers are 12, 15, x, 25, 30, find x. [3]

✅ Answer:
Median (5 numbers) = 3rd number = x
Given median = 20
✔️ x = 20

(iii) 🗼 A man observes the top of a tower from two points A and B on the ground, which are 20 m apart. The angles of elevation from A and B are 30° and 60° respectively. Find the height of the tower. [4]

✅ Answer:

        T (top of tower)
        |\
        | \
      h |  \
        |   \
        |    \
        |     \
        A------B------C
        |--20--|

Let TC = h, BC = x, AC = x + 20
From A: tan 30° = h/(x + 20) ⇒ h = (x + 20)/√3
From B: tan 60° = h/x ⇒ h = x√3
Equating: x√3 = (x + 20)/√3
3x = x + 20
2x = 20 ⇒ x = 10
h = 10√3 = 10 × 1.732 = 17.32 m
✔️ Height of the tower = 10√3 m (≈ 17.32 m)

🎓 END OF QUESTION PAPER

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