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📐 Mathematics Standard — Class X

Algebra · Geometry · Trigonometry · Statistics · Coordinate Geometry

80 Marks 15 Chapters 3hr Paper MCQ + LAQ
5
Algebra Chapters
3
Geometry Chapters
2
Trigonometry Ch.
2
Statistics Ch.
🔢
Algebra
Polynomials, Equations, AP
📐
Geometry
Triangles, Circles
📏
Trigonometry
Ratios, Heights & Distances
📍
Coordinate Geometry
Distance, Section, Area
📊
Statistics
Mean, Median, Mode, Prob.

🏅 Badges

🔢 Algebra
Polynomials · Pair of Linear Equations · Quadratic Equations · Arithmetic Progressions

📖 Core Theory

Ch 2: Polynomials

A polynomial in x of degree n: p(x) = an·x^n + ... + a0. Number of zeroes = degree. Relationship between zeroes and coefficients for quadratic ax^2 + bx + c: sum of zeroes = -b/a; product of zeroes = c/a.

Linear Polynomial
ax + b. One zero: x = -b/a. Graph is a straight line cutting x-axis once.
Quadratic Polynomial
ax^2 + bx + c. At most two zeroes. Graph is parabola: opens up if a > 0, down if a < 0.
Cubic Polynomial
ax^3 + bx^2 + cx + d. At most three zeroes. Sum of zeroes = -b/a; product = -d/a.
Division Algorithm
p(x) = g(x) × q(x) + r(x). Dividend = Divisor × Quotient + Remainder. Degree of remainder < degree of divisor.

🔑 Key Formulae — Polynomials and Quadratics

For ax^2 + bx + c: Sum of zeroes = -b/a ; Product of zeroes = c/a
Quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / 2a
Discriminant D = b^2 - 4ac: D > 0 (2 real roots), D = 0 (equal roots), D < 0 (no real roots)
For cubic: Sum of all zeroes = -b/a ; Sum of products of pairs = c/a ; Product = -d/a

Ch 3: Pair of Linear Equations in Two Variables

Two linear equations in x and y. Solutions: unique (lines intersect), infinitely many (lines coincide), no solution (lines parallel). Methods: substitution, elimination, cross-multiplication, graphical.

🔑 Key Facts — Linear Equations

a1x + b1y + c1 = 0 ; a2x + b2y + c2 = 0
Unique solution (consistent): a1/a2 ≠ b1/b2
Infinitely many (dependent): a1/a2 = b1/b2 = c1/c2
No solution (inconsistent): a1/a2 = b1/b2 ≠ c1/c2

Ch 5: Arithmetic Progressions

AP: sequence with constant common difference d. First term a, common difference d, nth term = a + (n-1)d. Sum of n terms = n/2 × [2a + (n-1)d] = n/2 × (first + last).

nth Term
an = a + (n-1)d where a = first term, d = common difference, n = term number.
Sum of n Terms
Sn = n/2 × [2a + (n-1)d] = n/2 × (a + l) where l = last term.
Common Difference
d = a2 - a1 = a3 - a2. Can be positive (increasing AP), negative (decreasing), or zero (constant).
Mean of AP
Arithmetic mean of any two terms equidistant from middle = middle term. a, b, c in AP → b = (a+c)/2.

✏️ Worked Examples

Example 1 — Quadratic Equation (Board 2023, 3M)

Q: Solve: 2x^2 - 5x + 3 = 0 using the quadratic formula.

1
a = 2, b = -5, c = 3. D = b^2 - 4ac = 25 - 24 = 1. sqrt(D) = 1.
2
x = (5 ± 1)/4. So x = 6/4 = 3/2 or x = 4/4 = 1.
✅ x = 3/2 or x = 1.

Example 2 — AP (Board 2022, 3M)

Q: The 17th term of an AP is 7 more than its 10th term. Find the common difference.

1
a17 = a + 16d and a10 = a + 9d. Given: a17 = a10 + 7.
2
a + 16d = a + 9d + 7 → 7d = 7 → d = 1.
✅ Common difference d = 1.

📝 Practice Questions — Algebra

MCQ 1 The HCF of 96 and 404 is:
(A) 4 (B) 8 (C) 12 (D) 2
Ans: (A) 4 — 96 = 2⁵ × 3; 404 = 2² × 101. HCF = 2² = 4.
MCQ 2 If the zeroes of the polynomial x² − 5x + k are reciprocals of each other, then k =
(A) 5 (B) −5 (C) 1 (D) −1
Ans: (C) 1 — Product of zeroes = k/1 = 1 ⟹ k = 1.
MCQ 3 The pair of equations x + 2y = 5 and 3x + 6y = 15 has:
(A) unique solution (B) no solution (C) infinitely many solutions (D) exactly two solutions
Ans: (C) — a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/3 ⟹ coincident lines.
MCQ 4 The nth term of an AP is 3n − 4. The common difference is:
(A) 3 (B) −4 (C) 7 (D) 1
Ans: (A) 3
A-R 1 A: √2 is irrational.
R: If √2 = p/q (in lowest terms), then squaring gives 2q² = p², so p² is even ⟹ p is even. Let p = 2m. Then 2q² = 4m² ⟹ q² = 2m² ⟹ q is even. This contradicts p/q being in lowest terms. Therefore √2 is irrational.
Ans: (A)
A-R 2 A: The quadratic equation x² − 4x + 5 = 0 has no real roots.
R: Discriminant D = b² − 4ac = 16 − 20 = −4 < 0. When D < 0, the equation has no real roots.
Ans: (A)

Case — Arithmetic Progressions: A student saves ₹5 in the first week, ₹8 in the second week, ₹11 in the third week, and continues this pattern.

Q1 (1M) Identify the first term (a) and common difference (d).
Ans: a = 5, d = 3.
Q2 (1M) What will be the savings in the 20th week?
Ans: a₂₀ = a + 19d = 5 + 57 = ₹62
Q3 (2M) Find total savings in 52 weeks (one year).
Ans: S₅₂ = 52/2 × (2×5 + 51×3) = 26 × (10 + 153) = 26 × 163 = ₹4238
Q4 (1M) In which week does savings exceed ₹50?
Ans: aₙ > 50 ⟹ 5 + (n−1)×3 > 50 ⟹ 3n > 48 ⟹ n > 16. So in the 17th week.
SA (3M) Find the zeroes of the polynomial p(x) = 6x² − 3 − 7x and verify the relationship between zeroes and coefficients.
Ans: 6x² − 7x − 3 = 0. Factoring: (3x + 1)(2x − 3) = 0. Zeroes: α = −1/3, β = 3/2. Sum = −1/3 + 3/2 = −2/6 + 9/6 = 7/6 = −b/a ✓. Product = (−1/3)(3/2) = −1/2 = c/a = −3/6 ✓.
LA (5M) Solve: 2x + 3y = 11 and 2x − 4y = −24. Also find the value of m for which y = mx + 3.
Ans: Subtracting: 7y = 35 ⟹ y = 5. Substituting: 2x + 15 = 11 ⟹ 2x = −4 ⟹ x = −2. Solution: x = −2, y = 5. For y = mx + 3: Substituting (−2, 5): 5 = m(−2) + 3 ⟹ 2 = −2m ⟹ m = −1.
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📐 Geometry
Triangles · Circles · Constructions · Surface Area · Volume

📖 Core Theory

Ch 6: Triangles

Similar triangles have equal corresponding angles and proportional corresponding sides. Criteria: AA, SAS, SSS similarity. Area ratio = (side ratio)^2. Pythagoras theorem: for right triangle, hypotenuse^2 = sum of squares of other two sides.

AA Similarity
If two angles of one triangle equal two angles of another, triangles are similar.
BPT (Thales' Theorem)
If a line parallel to one side of a triangle intersects the other two sides, it divides them in the same ratio. AD/DB = AE/EC.
Pythagoras Theorem
In right triangle: AC^2 = AB^2 + BC^2. Converse: if AC^2 = AB^2 + BC^2, angle B = 90 degrees.
Area of Similar Triangles
Ratio of areas = (ratio of corresponding sides)^2 = (ratio of medians)^2 = (ratio of perimeters)^2.

Ch 10: Circles

Tangent
Line touching circle at exactly one point. Tangent is perpendicular to radius at point of contact.
Two Tangents
From external point, two tangents are equal in length. PA = PB if P is external point, A and B are points of contact.
Chord-Tangent Angle
Angle between tangent and chord at point of contact = angle in alternate segment (Tangent-Chord Angle theorem).
Angle in Semicircle
Angle subtended by diameter at the circumference = 90 degrees (angle in semicircle).

🔑 Key Formulae — Mensuration

Sphere: Volume = (4/3) × pi × r^3 ; Surface area = 4 × pi × r^2
Cone: Volume = (1/3) × pi × r^2 × h ; Curved SA = pi × r × l ; l = sqrt(r^2 + h^2)
Cylinder: Volume = pi × r^2 × h ; Curved SA = 2 × pi × r × h ; Total SA = 2 × pi × r × (r + h)
Frustum: Volume = (pi × h/3) × (R^2 + r^2 + Rr) ; l = sqrt(h^2 + (R-r)^2)

✏️ Worked Examples

Example 1 — BPT (Board 2023, 3M)

Q: In triangle ABC, DE || BC. AD = 3cm, DB = 5cm, AE = 2.4cm. Find EC.

1
By BPT: AD/DB = AE/EC. So 3/5 = 2.4/EC.
2
EC = 2.4 × 5/3 = 12/3 = 4cm.
✅ EC = 4cm.

📝 Practice Questions — Geometry

MCQ 1 If △ABC ∼ △DEF and their areas are 64 cm² and 121 cm² respectively, then AB/DE =
(A) 8/11 (B) 11/8 (C) 64/121 (D) 4/√11
Ans: (A) 8/11 — Ratio of areas = (ratio of corresponding sides)².
MCQ 2 If a tangent PA is drawn from external point P to a circle with centre O, then OA is:
(A) parallel to PA (B) equal to PA (C) perpendicular to PA (D) half of PA
Ans: (C) perpendicular to PA
MCQ 3 The curved surface area of a cylinder with radius 7 cm and height 10 cm is:
(A) 440 cm² (B) 154 cm² (C) 308 cm² (D) 220 cm²
Ans: (A) 440 cm² — CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².
A-R 1 A: The tangents drawn from an external point to a circle are equal in length.
R: By RHS congruence: in △OAP and △OBP (A, B are tangent points), OA = OB (radii), OP = OP (common), ∠OAP = ∠OBP = 90°. So PA = PB.
Ans: (A)

Case — Similar Triangles and Heights: A building casts a shadow 40m long when the sun's altitude is 30°. At the same time, a flagpole of height h casts a shadow of 10m.

Q1 (2M) Find the height of the building.
Ans: tan 30° = height/shadow. Height = 40 × tan 30° = 40 × (1/√3) = 40/√3 = 40√3/3 ≈ 23.1m.
Q2 (1M) Using similar triangles, find h (height of flagpole).
Ans: Building height / shadow = h / 10. h = (40/√3) × (10/40) = 10/√3 = 10√3/3 ≈ 5.77m.
Q3 (1M) What is the ratio of the heights of the building to the flagpole?
Ans: 40/10 = 4:1
SA (3M) Prove that the tangent to a circle is perpendicular to the radius at the point of tangency.
Ans: Let O be the centre, P the point of tangency, and PT the tangent. Take any other point Q on PT. Since PT is a tangent, it touches the circle only at P, so Q lies outside the circle. Thus OQ > OP (radius). Since this is true for every point Q on PT other than P, the shortest distance from O to the line PT is OP. Therefore OP ⊥ PT.
LA (5M) A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, surmounted by a cone of height 60 cm. Find the mass of the pole, given that 1 cm³ of iron has mass 8g. (π = 3.14)
Ans: r = 12 cm. Cylinder: V = πr²h = 3.14 × 144 × 220 = 99,475.2 cm³. Cone: V = (1/3)πr²h = (1/3) × 3.14 × 144 × 60 = 9,043.2 cm³. Total volume = 99,475.2 + 9,043.2 = 1,08,518.4 cm³. Mass = 108518.4 × 8 = 8,68,147.2 g = 868.15 kg.
📏 Trigonometry
Trig Ratios · Identities · Heights and Distances

📖 Core Theory

Ch 8 & 9: Trigonometry

Trigonometric ratios relate angles to sides of a right triangle. For angle A in right triangle: sin A = opposite/hypotenuse; cos A = adjacent/hypotenuse; tan A = opposite/adjacent. Identities hold for all values of angle.

Standard Values
sin 30° = 1/2, sin 45° = 1/sqrt(2), sin 60° = sqrt(3)/2, sin 90° = 1. cos: reverse order. tan 30° = 1/sqrt(3), tan 45° = 1, tan 60° = sqrt(3).
Reciprocal Ratios
cosec A = 1/sin A; sec A = 1/cos A; cot A = 1/tan A = cos A/sin A.
Pythagorean Identities
sin^2 A + cos^2 A = 1; 1 + tan^2 A = sec^2 A; 1 + cot^2 A = cosec^2 A.
Angle of Elevation/Depression
Elevation: angle above horizontal to object. Depression: angle below horizontal. Drawn from the observer's eye level.

🔑 Key Identities — Trigonometry

sin^2 A + cos^2 A = 1 (fundamental identity)
1 + tan^2 A = sec^2 A (divide above by cos^2 A)
1 + cot^2 A = cosec^2 A (divide by sin^2 A)
tan A = sin A / cos A ; cot A = cos A / sin A

✏️ Worked Examples

Example 1 — Height and Distance (Board 2023, 3M)

Q: The angle of elevation of the top of a tower from a point on the ground 30m away is 60°. Find the height of the tower.

1
Let height = h. tan 60° = h/30.
2
sqrt(3) = h/30 → h = 30sqrt(3) = 30 × 1.732 = 51.96m ≈ 52m.
✅ Height of tower = 30 × sqrt(3) m ≈ 51.96m.

📝 Practice Questions — Trigonometry

MCQ 1 The value of (sin²60° + cos²60°) is:
(A) 0 (B) 1/2 (C) 1 (D) 2
Ans: (C) 1 — Pythagorean identity.
MCQ 2 If tan θ = 1, then θ =
(A) 30° (B) 45° (C) 60° (D) 90°
Ans: (B) 45°
MCQ 3 A ladder 10 m long reaches a wall at height 6 m. The distance of the foot of the ladder from the wall is:
(A) 4 m (B) 6 m (C) 8 m (D) 10 m
Ans: (C) 8 m — 10² − 6² = 100 − 36 = 64 ⟹ √64 = 8m.
A-R 1 A: The angle of depression from the top of a building to an object on the ground equals the angle of elevation of the top of the building from the object.
R: The angle of depression (measured below horizontal from top) and the angle of elevation (measured above horizontal from ground) are alternate interior angles formed by the horizontal line and the line of sight as transversal — they are equal.
Ans: (A)

Case — Heights and Distances: From the top of a building 30 m high, the angle of elevation of a tower is 30°. The angle of depression of the foot of the tower from the top of the building is 60°.

Q1 (2M) Find the horizontal distance between the building and the tower.
Ans: Let d = horizontal distance, h = height of building = 30m. tan 60° = 30/d ⟹ √3 = 30/d ⟹ d = 30/√3 = 10√3 m.
Q2 (2M) Find the height of the tower.
Ans: Let extra height above building = x. tan 30° = x/d = x/(10√3). So x = 10√3 × (1/√3) = 10m. Height of tower = 30 + 10 = 40 m.
Q3 (1M) What is the ratio of heights of building to tower?
Ans: 30:40 = 3:4
SA (3M) Prove: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A.
Ans: LHS = sin²A + 2 + cosec²A + cos²A + 2 + sec²A = (sin²A + cos²A) + 4 + cosec²A + sec²A = 1 + 4 + (1 + cot²A) + (1 + tan²A) = 7 + tan²A + cot²A = RHS ✓
LA (5M) The angle of elevation of the top of a tower from two points P and Q at distances of a and b metres from the base of the tower (on the same side) are 30° and 60° respectively. Prove that the height of the tower is √(ab) metres.
Ans: Let height = h. From P: tan 30° = h/a ⟹ h = a/√3. From Q: tan 60° = h/b ⟹ h = b√3. Multiplying: h² = (a/√3)(b√3) = ab. Therefore h = √(ab). ∎
📍 Coordinate Geometry
Distance Formula · Section Formula · Area of Triangle

📖 Core Theory

Ch 7: Coordinate Geometry

Points in a plane are located using x and y coordinates. Distance between two points, dividing a line segment in a given ratio, and finding the area of a triangle from coordinates are key topics.

Distance Formula
d = sqrt[(x2-x1)^2 + (y2-y1)^2]. Distance of point (x,y) from origin = sqrt(x^2 + y^2).
Section Formula
Point dividing (x1,y1) and (x2,y2) in ratio m:n internally: x = (mx2 + nx1)/(m+n); y = (my2 + ny1)/(m+n).
Midpoint Formula
Special case of section formula with m=n=1: M = ((x1+x2)/2, (y1+y2)/2).
Area of Triangle
Area = (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|. If area = 0, points are collinear.

🔑 Key Formulae — Coordinate Geometry

Distance: d = sqrt[(x2-x1)^2 + (y2-y1)^2]
Section formula (internal): P = [(mx2+nx1)/(m+n) , (my2+ny1)/(m+n)]
Midpoint: M = [(x1+x2)/2 , (y1+y2)/2]
Area of triangle: (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|

✏️ Worked Examples

Example 1 — Section Formula (Board 2022, 3M)

Q: Find the point which divides the line segment joining A(4, -3) and B(8, 5) in ratio 3:1 internally.

1
m = 3, n = 1. x = (3×8 + 1×4)/(3+1) = (24+4)/4 = 28/4 = 7.
2
y = (3×5 + 1×(-3))/(3+1) = (15-3)/4 = 12/4 = 3.
✅ Point = (7, 3).

📝 Practice Questions — Coordinate Geometry

MCQ 1 The distance between points A(2, 3) and B(−1, −1) is:
(A) 3 (B) 4 (C) 5 (D) 7
Ans: (C) 5 — √[(2−(−1))²+(3−(−1))²] = √[9+16] = 5.
MCQ 2 The mid-point of segment joining (2, −3) and (−6, 5) is:
(A) (−2, 1) (B) (2, −1) (C) (−4, 8) (D) (4, −2)
Ans: (A) (−2, 1)
A-R 1 A: The point (0, 4) lies on the y-axis.
R: Any point with x-coordinate = 0 lies on the y-axis, since the y-axis is the set of all points (0, y) in the coordinate plane.
Ans: (A)

Case — Section Formula: Three friends A, B, C live at points (2, 3), (−4, −6), and (5, 12) on a coordinate map. A park is to be built at the centroid of their triangle.

Q1 (1M) Find the location of the park (centroid).
Ans: Centroid = ((2−4+5)/3, (3−6+12)/3) = (3/3, 9/3) = (1, 3)
Q2 (2M) B wants to meet A. Find the distance BA.
Ans: BA = √[(2−(−4))²+(3−(−6))²] = √[36+81] = √117 = 3√13 units.
Q3 (2M) Find the point that divides BC in ratio 2:1 internally.
Ans: x = (2×5 + 1×(−4))/(2+1) = (10−4)/3 = 2. y = (2×12 + 1×(−6))/3 = (24−6)/3 = 6. Point = (2, 6).
SA (3M) Find the area of triangle with vertices A(1, 2), B(3, −4), C(5, 6).
Ans: Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| = ½|1(−4−6) + 3(6−2) + 5(2−(−4))| = ½|−10 + 12 + 30| = ½ × 32 = 16 sq. units
📊 Statistics and Probability
Mean · Median · Mode · Probability

📖 Core Theory

Ch 14: Statistics

For grouped data: Mean by direct, assumed mean, or step-deviation method. Median: l + [(n/2 - cf)/f] × h. Mode: l + [(f1-f0)/(2f1-f0-f2)] × h. Modal class = class with highest frequency.

Direct Mean
Mean = Sum(fi × xi) / Sum(fi). fi = frequency; xi = class midpoint (class mark).
Assumed Mean
a = assumed mean; di = xi - a. Mean = a + Sum(fi × di)/Sum(fi).
Median
Median = l + [(n/2 - cf)/f] × h. l = lower limit of median class; cf = cumulative freq before median class.
Mode
Mode = l + [(f1-f0)/(2f1-f0-f2)] × h. f1 = modal class freq; f0 = preceding class freq; f2 = following class freq.

🔑 Key Formulae — Probability

P(event) = Number of favourable outcomes / Total number of outcomes
0 ≤ P(A) ≤ 1 for any event A
P(A) + P(not A) = 1
For a die: P(any face) = 1/6; For a coin: P(head) = 1/2

✏️ Worked Examples

Example 1 — Mean of Grouped Data (Board 2023, 3M)

Q: Find mean for: Class: 10-20, 20-30, 30-40; Frequency: 5, 8, 7.

1
Midpoints: 15, 25, 35. f×x: 5×15=75, 8×25=200, 7×35=245. Sum(fi)=20, Sum(fi×xi)=520.
2
Mean = 520/20 = 26.
✅ Mean = 26.

📝 Practice Questions — Statistics and Probability

MCQ 1 The empirical relationship between mean, median and mode is:
(A) Mode = 3 Median − 2 Mean (B) Mode = 2 Mean − 3 Median (C) Mean = 3 Mode − 2 Median (D) Median = 2 Mean − Mode
Ans: (A)
MCQ 2 A card is drawn from a deck of 52 cards. Probability of getting a king is:
(A) 1/52 (B) 1/13 (C) 4/52 (D) Both B and C
Ans: (D) — 4 kings in 52 cards = 4/52 = 1/13.
A-R 1 A: The sum of probabilities of all elementary events of a sample space is 1.
R: In a sample space, all elementary events are mutually exclusive and exhaustive — one of them MUST occur, so their probabilities add to 1.
Ans: (A)

Case — Statistics: Marks scored by 30 students in a maths test are grouped: 0-10: 2 students, 10-20: 4, 20-30: 8, 30-40: 10, 40-50: 6.

Q1 (1M) Identify the modal class.
Ans: 30–40 (highest frequency = 10).
Q2 (2M) Find the mode.
Ans: Mode = l + (f₁−f₀)/(2f₁−f₀−f₂) × h = 30 + (10−8)/(20−8−6) × 10 = 30 + 2/6 × 10 = 30 + 3.33 = 33.33
Q3 (2M) Calculate the mean using assumed mean method (A = 25).
Ans: Class midpoints: 5,15,25,35,45. Deviations (uᵢ = xᵢ−25): −20,−10,0,10,20. Σfᵢuᵢ = 2(−20)+4(−10)+8(0)+10(10)+6(20) = −40−40+0+100+120 = 140. Mean = A + Σfᵢuᵢ/n = 25 + 140/30 = 25 + 4.67 = 29.67
SA (3M) A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both balls are red.
Ans: P(1st red) = 5/8. After drawing red, 4 red and 3 blue remain. P(2nd red | 1st red) = 4/7. P(both red) = 5/8 × 4/7 = 20/56 = 5/14.
LA (5M) The following table gives the ages of 100 residents of a colony. Find the mean and median age.
Age: 0-10: 5; 10-20: 8; 20-30: 20; 30-40: 30; 40-50: 22; 50-60: 15.
Ans (Mean): Midpoints: 5, 15, 25, 35, 45, 55. Σfᵢxᵢ = 5×5+8×15+20×25+30×35+22×45+15×55 = 25+120+500+1050+990+825 = 3510. Mean = 3510/100 = 35.1 years. Ans (Median): n/2 = 50. CF: 5, 13, 33, 63, 85, 100. 50th value falls in class 30-40 (CF crosses 50 at 63). Median = l + (n/2−CF)/f × h = 30 + (50−33)/30 × 10 = 30 + 17/30 × 10 = 30 + 5.67 = 35.67 years.
📝 Practice Problems
Mixed Problems · Board-Style Questions

📖 Important Board Questions

Problem 1 — Volume (Board 2023, 4M)

Q: A solid sphere of radius 3cm is melted and recast into small spheres of radius 0.3cm. Find the number of small spheres.

1
Volume of large sphere = (4/3) × pi × 3^3 = 36pi cm^3.
2
Volume of small sphere = (4/3) × pi × 0.3^3 = (4/3) × pi × 0.027 = 0.036pi cm^3.
3
Number = 36pi / 0.036pi = 36/0.036 = 1000.
✅ Number of small spheres = 1000.
📜 PYQ Bank CBSE 2022–2024
2024 3 Marks Quadratic Equations
Find the discriminant of the quadratic equation 3x² − 5x + 2 = 0 and hence find its roots.
D = b² − 4ac = (−5)² − 4(3)(2) = 25 − 24 = 1

Since D > 0, two distinct real roots.
x = (5 ± √1) / 6
x₁ = (5 + 1)/6 = 1  |  x₂ = (5 − 1)/6 = 2/3
2023 5 Marks Arithmetic Progressions
The sum of first 16 terms of an AP is 112 and the sum of its next 14 terms is 518. Find the first term and common difference.
S₁₆ = 112 → (16/2)[2a + 15d] = 112 → 2a + 15d = 14  …(i)

Sum of next 14 terms = S₃₀ − S₁₆ = 518
S₃₀ = 518 + 112 = 630 → (30/2)[2a + 29d] = 630 → 2a + 29d = 42  …(ii)

(ii) − (i): 14d = 28 → d = 2
2a + 30 = 42 → a = 6 → a = 6
2024 5 Marks Circles
Prove that the tangents drawn from an external point to a circle are equal in length.
Given: PA and PB are tangents from external point P to circle with centre O.
To Prove: PA = PB

Proof:
In △OAP and △OBP:
OA = OB (radii) | OP = OP (common) | ∠OAP = ∠OBP = 90° (radius ⊥ tangent)
By RHS congruence: △OAP ≅ △OBP
∴ PA = PB   (CPCT)   QED
2023 4 Marks Heights & Distances
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.
Let height of building = 7 m, distance between building and tower = d, height of tower = h.

Angle of depression to foot = 45°: tan 45° = 7/d → d = 7 m

Angle of elevation to top = 60°: tan 60° = (h − 7)/d
√3 = (h − 7)/7 → h − 7 = 7√3
h = 7(1 + √3) m = 7(1 + 1.732) ≈ 19.12 m
2024 3 Marks Probability
Two dice are thrown simultaneously. Find the probability that the sum of the two numbers appearing on the top is: (i) 7 (ii) a prime number.
Total outcomes = 6 × 6 = 36

(i) Sum = 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes
P(sum = 7) = 6/36 = 1/6

(ii) Prime sums (2,3,5,7,11): 2→1, 3→2, 5→4, 7→6, 11→2 = 15 outcomes
P(prime) = 15/36 = 5/12
⚠️ Mistake Bank Common Errors

HCF × LCM ≠ Always Product of All Numbers

The theorem applies to exactly TWO numbers only.

❌ WRONG: "HCF(a,b,c) × LCM(a,b,c) = a × b × c for three numbers."
✅ CORRECT: HCF × LCM = Product is valid ONLY for two numbers: HCF(a,b) × LCM(a,b) = a × b. For three numbers, use prime factorisation to find HCF and LCM separately.

Discriminant Sign for Nature of Roots

Students mix up the three cases of discriminant.

❌ WRONG: "D = 0 means the quadratic has no real roots."
✅ CORRECT: D > 0 → two distinct real roots | D = 0 → two equal real roots | D < 0 → no real roots. D = 0 means exactly one repeated root (equal roots), not no roots.

AP Sum Formula — Wrong n

Students use n as last term index incorrectly.

❌ WRONG: Using Sn = n/2(a + an) when the nth term isn't known, or using Sn = n/2[2a + (n−1)d] with wrong n.
✅ CORRECT: Two forms: Sn = n/2[2a + (n−1)d] (when d is known) | Sn = n/2(a + l) (when last term l is known). Identify which information is given before choosing the formula.

Tangent-Radius: Forgetting the Right Angle

Students forget that tangent is perpendicular to radius at point of contact.

❌ WRONG: Drawing tangent at an angle to the radius in diagrams, or not using ∠OAP = 90° in proofs.
✅ CORRECT: The radius to the point of tangency is ALWAYS perpendicular to the tangent (∠ = 90°). This is the key theorem that enables all circle-tangent proofs. Mark it in every diagram.

Probability: Events Not Equally Likely

Students apply the basic formula without checking equal likelihood.

❌ WRONG: P(Head) = 1/2 for a biased coin because there are 2 outcomes (H or T).
✅ CORRECT: P(E) = Favourable/Total only works when all outcomes are EQUALLY LIKELY. For a biased coin, P(H) ≠ 1/2. Always check whether the outcomes are equally likely before applying the formula.

Similar Triangles — Ratio of Areas

Students write ratio of areas = ratio of sides instead of ratio of squares.

❌ WRONG: "If sides of similar triangles are in ratio 2:3, then areas are also in ratio 2:3."
✅ CORRECT: Ratio of areas of similar triangles = (Ratio of corresponding sides)². If sides = 2:3, then areas = 4:9. Similarly, ratio of perimeters = ratio of sides (not squared).

Heights & Distances: Angle of Depression vs Elevation

Students confuse which angle to use for which calculation.

❌ WRONG: Using the same triangle/formula regardless of whether the angle is of elevation or depression.
✅ CORRECT: Angle of elevation = angle measured UPWARD from horizontal. Angle of depression = angle measured DOWNWARD from horizontal. Draw a clear diagram first. Alternate angles make angle of depression = angle of elevation (parallel lines).

Mean vs Mode vs Median — Formula Confusion

Students use mode formula for mean or misidentify modal class.

❌ WRONG: "Modal class is the class with the highest frequency; mode = lower boundary + (f1−f0)/(2f1−f0−f2) × h applied to grouped data mean."
✅ CORRECT: Mode formula for grouped data: Mode = l + [(f1−f0)/(2f1−f0−f2)] × h (l = lower limit of modal class). Mean uses direct/assumed mean/step-deviation method. Median uses l + [(n/2−cf)/f] × h. Don't mix these up.
🎯 Examiner's Eye Scoring Tips

🎯 Tip 1 — Show ALL Steps in Proofs

In geometry proofs (Similar Triangles, Circles), every step earns marks. Write: "Given, To Prove, Construction, Proof" as separate headings. State every reason in brackets: "by AA similarity" / "by CPCT" / "by Pythagoras." Skipping steps loses step marks even if the conclusion is correct.

🎯 Tip 2 — Algebra: Write Formula Before Substituting

For every formula-based question (Quadratic formula, AP sum, Volume of sphere), write the formula first, then substitute values. This earns a formula mark + step marks + answer mark. Even a computation error won't cost the formula and step marks.

🎯 Tip 3 — Draw Diagrams for Heights & Distances

A correctly labelled diagram (with the given angle, height, and horizontal distance marked) earns 1 mark before any calculation. Many students skip this and lose easy marks. Draw → label → set up equation → solve is the winning sequence.

🎯 Tip 4 — Statistics: Complete the Frequency Table First

For Mean, Median, and Mode questions from grouped data, always construct the complete table (class, frequency, midpoint, cumulative frequency as needed) before computing. An incomplete table with correct computation still scores partial marks. A complete table with a small calculation error scores more than no table with the right answer.

🎯 Tip 5 — Probability: List Sample Space for Small Problems

For dice, cards, and coin problems, write out the complete sample space or at least the favourable outcomes. "Sample space n(S) = 36" shows you know the denominator. Then list favourable events. Examiners check whether you've correctly identified the total.

🎯 Tip 6 — Quadratic: Check Roots Satisfy the Equation

After finding roots, substitute back into the original equation to verify. This is worth 0.5 marks in verification sub-questions and also catches sign errors in your working. Takes 30 seconds and prevents mark loss from silly mistakes.

🎯 Tip 7 — Circles: Mark the 90° Angle First

In any circle question involving tangents, mark the 90° angle (radius ⊥ tangent) in the diagram before anything else. This single fact drives 80% of circle proofs. If you forget to use it, you'll circle back (pun intended) after getting stuck.

🎯 Tip 8 — Time Management: Section D Last

Attempt Section A (MCQs), B (2-marks), C (3-marks) first in that order. Tackle Section D (5-marks) and Section E (case-based) last. Don't spend 15 minutes on one 5-mark proof early on and run out of time for easier marks later. Attempted > unattempted always.

🔥 High-Probability Questions Exam Forecast
🔥🔥🔥 VERY HIGH Circles — Tangent Proofs (5 Marks)

The proof "tangents from external point are equal in length" appears almost every year in Section D. It's a memorisation question with a fixed structure. Know: Given, To Prove, Construction (join OA, OB, OP), Proof (RHS congruence), Conclusion (CPCT). Practice writing it in under 5 minutes.

🔥🔥🔥 VERY HIGH Arithmetic Progressions — Numericals (5 Marks)

AP word problems (sum of n terms, finding a and d, sum of first n natural numbers) appear in every paper. Master both sum formulas: Sn = n/2[2a + (n−1)d] and Sn = n/2(a+l). The two-equation system type (given S₁₆ and sum of next 14) is very frequent.

🔥🔥🔥 VERY HIGH Heights & Distances — Two-Observer Problems (5 Marks)

Problems with angle of elevation AND angle of depression in the same figure appear every year. Set up two right triangles, write two equations, solve simultaneously. The "7 m building looking at a tower" type is a perennial favourite. Draw the diagram first.

🔥🔥 HIGH Statistics — Mean, Median, Mode (4–5 Marks)

Computing mean by step-deviation, median by ogive or formula, and mode from grouped data each appear in every session. Know all three formulas perfectly. Case-based questions often ask for all three measures from the same data table.

🔥🔥 HIGH Quadratic Equations — Word Problems (5 Marks)

Number problems, age problems, and speed-distance problems that reduce to a quadratic equation appear in Section D every year. Form the equation → use factorisation or quadratic formula → check both roots in context (negative values for age/length are rejected).

🔥🔥 HIGH Surface Areas & Volumes — Combination Solids (5 Marks)

Questions combining cone + cylinder, hemisphere + cone, or frustum problems appear in Section D. Know all formulas. For combination problems: total surface area carefully removes the joined face. These often come with real-life context (tent, toy, capsule).

🔥 MODERATE-HIGH Probability — Two Dice / Cards (3 Marks)

Two-dice problems (sum = 7, prime, perfect square) and card problems (face cards, red queens, numbered cards) appear in Section C. Know: total outcomes for two dice = 36, for a deck = 52. List favourable outcomes systematically.

🔥 MODERATE-HIGH Real Numbers — HCF/LCM + Proof of Irrationality (3 Marks)

HCF and LCM using prime factorisation (often for 3 numbers), and proofs that √2, √3, or √5 is irrational (by contradiction) appear regularly. The irrationality proof is completely fixed — memorise the proof structure for any prime square root.

⚡ Last-Minute Revision Quick Recall
📌 Real Numbers & Polynomials
  • HCF × LCM = Product (for TWO numbers only)
  • Irrationality proof: assume p/q in lowest terms → contradiction via prime factorisation
  • Sum of zeros of quadratic: α + β = −b/a | Product: αβ = c/a
  • Sum of zeros of cubic: α+β+γ = −b/a | αβ+βγ+γα = c/a | αβγ = −d/a
📌 Algebra (Equations & AP)
  • Quadratic formula: x = [−b ± √(b²−4ac)] / 2a | D = b²−4ac
  • D > 0 → 2 distinct real roots | D = 0 → equal roots | D < 0 → no real roots
  • AP: aₙ = a + (n−1)d | Sₙ = n/2[2a + (n−1)d] = n/2(a + l)
  • Pair of linear equations: substitution, elimination, cross-multiplication
📌 Geometry
  • Similar △s: AA, SAS, SSS similarity criteria
  • Basic Proportionality Theorem (Thales): DE ∥ BC → AD/DB = AE/EC
  • Ratio of areas of similar △s = square of ratio of corresponding sides
  • Tangent ⊥ radius at point of contact (∠OAP = 90°)
  • Tangents from external point are equal: PA = PB
  • Angle in semicircle = 90° | Angle subtended at centre = 2 × angle at circumference
📌 Trigonometry & Statistics
  • sin²θ + cos²θ = 1 | 1 + tan²θ = sec²θ | 1 + cot²θ = cosec²θ
  • sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1, sin 60° = √3/2
  • Mode (grouped) = l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h
  • Median = l + [(n/2 − cf)/f] × h
  • Mean (step deviation) = a + (Σfᵢuᵢ/Σfᵢ) × h
  • P(E) = Favourable outcomes / Total outcomes (equally likely events only)
🎮 Exam Simulator
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