T25 511 • ICSE CLASS X MATHEMATICS (THEORY) YEAR 2025

ICSE Class 10 Mathematics 2025 Solved Paper

Official CISCE Board Examination Question Paper (Code T25 511) complete with authentic stepwise marking schemes, vector geometric diagrams, and chapter mappings.

Time: 3 Hours (incl. 15 min reading) Max Marks: 80 Status: 100% Solved Diagrams: 11 Verified SVGs
Section A (Q1–Q3) Q1 MCQs Q2 (Banking, Sim, Quad) Q3 (Const, Mat, Coord) Section B (Q4–Q10) Q4 (Ineq, AP, Trig) Q5 (Prop, Fact, Mens) Q6 (Reflect, Heights) Q7 (Histogram, Lines, GST) Q8 (Shares, Mean, Speed) Q9 (Melt, Prob, Tangent) Q10 (GP, Tangent, Ogive)

SECTION A (40 Marks)

Attempt all questions from this Section. Compulsory for all candidates.

Question 1 (15 Multiple Choice Questions)
15 × 1 = 15 Marks

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

(i) Quadratic Nature of Roots Ch 5: Quadratic Equations
The given quadratic equation 3x² + √7x + 2 = 0 has:
(a) two equal real roots
(b) two distinct real roots
(c) more than two real roots
(d) no real roots
View Stepwise Solution & Rubric
Discriminant D = b² − 4ac = (√7)² − 4(3)(2) = 7 − 24 = −17.
Since D < 0, the quadratic equation has no real roots.1 Mark
(ii) Recurring Deposit Maturity Value Ch 2: Banking
Mr. Anuj deposits ₹500 per month for 18 months in a recurring deposit account at a certain rate. If he earns ₹570 as interest at the time of maturity, then his matured amount is:
(a) ₹(500 × 18 + 570)
(b) ₹(500 × 19 + 570)
(c) ₹(500 × 18 × 19 + 570)
(d) ₹(500 × 9 × 19 + 570)
View Stepwise Solution & Rubric
Maturity Value MV = Total Principal Deposited + Total Interest = (P × n) + I = ₹(500 × 18 + 570).1 Mark
(iii) Probability Range Ch 24: Probability
Which of the following cannot be the probability of any event?
(a) 5/4
(b) 0.25
(c) 1/33
(d) 67%
View Stepwise Solution & Rubric
For any real probability event, 0 ≤ P(E) ≤ 1. Here 5/4 = 1.25 > 1, which violates the fundamental law of probability.1 Mark
(iv) Parallel Line through Origin Ch 13: Straight Lines
The equation of the line passing through origin and parallel to the line 3x + 4y + 7 = 0 is:
(a) 3x + 4y + 5 = 0
(b) 4x − 3y − 5 = 0
(c) 4x − 3y = 0
(d) 3x + 4y = 0
View Stepwise Solution & Rubric
Slope of 3x + 4y + 7 = 0 is m = −3/4. Any parallel line passing through the origin (0, 0) has c = 0, giving y = −3/4 x ⇒ 3x + 4y = 0.1 Mark
(v) Matrix Multiplication Ch 7: Matrices
If A = [[0, 1], [1, 0]], then A² is equal to:
(a) [[1, 1], [0, 0]]
(b) [[0, 0], [1, 1]]
(c) [[1, 0], [0, 1]]
(d) [[0, 1], [1, 0]]
View Stepwise Solution & Rubric
A² = [[0, 1], [1, 0]] × [[0, 1], [1, 0]] = [[0(0)+1(1), 0(1)+1(0)], [1(0)+0(1), 1(1)+0(0)]] = [[1, 0], [0, 1]] = I.1 Mark
(vi) Cyclic Quadrilateral Angle Ch 16: Circles
In the given diagram, chords AC and BC are equal. If ∠ACD = 120°, then ∠AEC is:
120° A B C E D
Figure: Q1(vi) Cyclic Quadrilateral ABCE with AC = BC and ext. ∠ACD = 120°
(a) 30°
(b) 60°
(c) 90°
(d) 120°
View Stepwise Solution & Rubric
1. BCD is a straight line, so ∠ACB = 180° − 120° = 60°.
2. In ΔABC, AC = BC ⇒ ∠BAC = ∠ABC = (180° − 60°) / 2 = 60° (ΔABC is equilateral).
3. ABCE is a cyclic quadrilateral. Opposite angles are supplementary: ∠ABC + ∠AEC = 180° ⇒ 60° + ∠AEC = 180° ⇒ ∠AEC = 120°.1 Mark
(vii) Common Factor of Polynomials Ch 6: Factorisation
The factor common to the two polynomials x² − 4 and x³ − x² − 4x + 4 is:
(a) (x + 1)
(b) (x − 1)
(c) (x − 2)
(d) (x − 4)
View Stepwise Solution & Rubric
x² − 4 = (x − 2)(x + 2).
x³ − x² − 4x + 4 = x²(x − 1) − 4(x − 1) = (x² − 4)(x − 1) = (x − 2)(x + 2)(x − 1).
The common factor among the given choices is (x − 2).1 Mark
(viii) Shares Par Value Status Ch 3: Shares & Dividends
A man invested in a company paying 12% dividend on its share. If the percentage return on his investment is 10%, then the shares are:
(a) at par
(b) below par
(c) above par
(d) cannot be determined
View Stepwise Solution & Rubric
Let NV = 100. Dividend = ₹12. Return = (Dividend / MV) × 100% = 10% ⇒ MV = ₹120. Since MV > NV (120 > 100), the shares are trading above par (at a premium).1 Mark
(ix) Circumcentre and Incentre Definition Ch 15: Loci
Statement 1: The point which is equidistant from three non-collinear points D, E and F is the circumcentre of the ΔDEF.
Statement 2: The incentre of a triangle is the point where the bisector of the angles intersects.
(a) Both the statements are true
(b) Both the statements are false
(c) Statement 1 is true, and Statement 2 is false
(d) Statement 1 is false, and Statement 2 is true
View Stepwise Solution & Rubric
Statement 1 is the definition of circumcentre (equidistant from vertices). Statement 2 is the definition of incentre (intersection of internal angle bisectors). Both are true.1 Mark
(x) Assertion-Reason: Trig Identity Ch 20: Trig Identities
Assertion (A): If sin²A + sin A = 1 then cos⁴A + cos²A = 1.
Reason (R): 1 − sin²A = cos²A.
(a) (A) is true, (R) is false
(b) (A) is false, (R) is true
(c) Both (A) and (R) are true, and (R) is correct reason for (A)
(d) Both (A) and (R) are true, and (R) is incorrect reason for (A)
View Stepwise Solution & Rubric
sin A = 1 − sin²A = cos²A. Squaring gives sin²A = cos⁴A ⇒ 1 − cos²A = cos⁴A ⇒ cos⁴A + cos²A = 1. Reason R is the exact identity applied.1 Mark
(xi) Similar Triangles Side Length Ch 14: Similarity
In the given diagram ΔABC ∼ ΔEFG. If ∠ABC = ∠EFG = 60°, then the length of the side FG is:
B C A 3 cm 15 cm 5 cm 60° G F E 60° 75 cm FG = ? 25 cm
Figure: Q1(xi) Similar Triangles ΔABC ∼ ΔEFG (Scale Factor 1:5)
(a) 15 cm
(b) 20 cm
(c) 25 cm
(d) 30 cm
View Stepwise Solution & Rubric
Since ΔABC ∼ ΔEFG, corresponding sides are proportional: AB/EF = BC/FG = AC/EG.
AB/EF = 15/75 = 1/5 ⇒ BC/FG = 3/FG = 1/5 ⇒ FG = 15 cm.1 Mark
(xii) Ratio of Radii of Spheres Ch 19: Mensuration
If the volume of two spheres is in the ratio 27 : 64, then the ratio of their radii is:
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) 16 : 9
View Stepwise Solution & Rubric
V1/V2 = (r1/r2)³ = 27/64 ⇒ r1/r2 = ³√(27/64) = 3/4.1 Mark
(xiii) GST on Marked Price Ch 1: GST
The marked price of an article is ₹1375. If the CGST is charged at a rate of 4%, then the price of the article including GST is:
(a) ₹55
(b) ₹110
(c) ₹1430
(d) ₹1485
View Stepwise Solution & Rubric
CGST = 4% ⇒ Total GST = 2 × 4% = 8%.
Total GST = 8% of ₹1375 = ₹110. Price including GST = ₹1375 + ₹110 = ₹1485.1 Mark
(xiv) Inequation Integer Solution Ch 4: Linear Inequations
The solution set for 0 < −x/3 < 2, x ∈ Z is:
(a) {−5, −4, −3, −2, −1}
(b) {−6, −5, −4, −3, −2, −1}
(c) {−5, −4, −3, −2, −1, 0}
(d) {−6, −5, −4, −3, −2, −1, 0}
View Stepwise Solution & Rubric
Multiply by −3 (reversing inequalities): 0 > x > −6 ⇔ −6 < x < 0. Since x ∈ Z, solution set is {−5, −4, −3, −2, −1}.1 Mark
(xv) Assertion-Reason: Mean of Natural Numbers Ch 22: Central Tendency
Assertion (A): The mean of first 9 natural numbers is 4.5.
Reason (R): Mean = (Sum of all observations) / (Total number of observations).
(a) (A) is true, (R) is false
(b) (A) is false, (R) is true
(c) Both (A) and (R) are true, and (R) is correct reason for (A)
(d) Both (A) and (R) are true, and (R) is incorrect reason for (A)
View Stepwise Solution & Rubric
Sum of first 9 natural numbers (1 to 9) = 9(10)/2 = 45. Mean = 45 / 9 = 5 ≠ 4.5. Assertion A is false, while Reason R is the true standard definition.1 Mark
Question 2 (Compulsory • 3 Sub-parts)
4 + 4 + 4 = 12 Marks
(i) Quadratic Formula (Three Significant Figures) Ch 5: Quadratic Equations
Solve the following quadratic equation 2x² − 5x − 4 = 0. Give your answer correct to three significant figures. (Use mathematical tables for this question)
View Stepwise Solution & Rubric
a = 2, b = −5, c = −4.
Formula: x = [−b ± √(b² − 4ac)] / 2a = [5 ± √((-5)² − 4(2)(−4))] / 4.1 Mark
= [5 ± √(25 + 32)] / 4 = [5 ± √57] / 4.1 Mark
From square root tables, √57 = 7.550.
x1 = (5 + 7.550) / 4 = 12.550 / 4 = 3.1375 ⇒ 3.14 (3 sig figs).1 Mark
x2 = (5 − 7.550) / 4 = −2.550 / 4 = −0.6375 ⇒ −0.638 (or −0.637).1 Mark
(ii) Recurring Deposit Rate of Interest Ch 2: Banking
Mrs. Rao deposited ₹250 per month in a recurring deposit account for a period of 3 years. She received ₹10,110 at the time of maturity. Find:
  1. the rate of interest.
  2. how much more interest Mrs. Rao will receive if she had deposited ₹50 more per month at the same rate of interest and for the same time.
View Stepwise Solution & Rubric
P = ₹250, n = 3 × 12 = 36 months, MV = ₹10,110.
Total deposit = 250 × 36 = ₹9,000.
Interest I = MV − (P × n) = 10,110 − 9,000 = ₹1,110.1 Mark
I = P × [n(n+1)/(2 × 12)] × (r / 100)
1110 = 250 × (36 × 37 / 24) × (r / 100) = 2.5 × 55.5 × r = 138.75 r ⇒ r = 8%.2 Marks
(b) If deposit is ₹50 more ⇒ extra interest = (50/250) × ₹1110 = ₹222.1 Mark
(iii) Right Triangle Similarity & Map Scale Factor Ch 14: Similarity
In ΔABC, ∠ABC = 90°, AB = 20 cm, AC = 25 cm, DE is perpendicular to AC such that ∠DEA = 90° and DE = 3 cm as shown in figure.
  1. Prove that ΔABC ∼ ΔAED.
  2. Find the lengths of BC, AD and AE.
  3. If BCED represents a plot of land on a map whose actual area on ground is 576 m², find the scale factor of the map.
A B C D E 20 cm 25 cm 3 cm
Figure: Q2(iii) Right ΔABC with DE ⊥ AC (DE = 3 cm)
View Stepwise Solution & Rubric
(a) In ΔABC and ΔAED: ∠BAC = ∠EAD (Common), ∠ABC = ∠AED = 90° (Given). ∴ ΔABC ∼ ΔAED by AA criterion.1 Mark
(b) By Pythagoras in ΔABC: BC = √(25² − 20²) = 15 cm.
By similarity: AB/AE = BC/ED = AC/AD ⇒ 20/AE = 15/3 = 25/AD.
Scale ratio = 5 ⇒ AE = 4 cm, AD = 5 cm.2 Marks
(c) Area(BCED) = Area(ΔABC) − Area(ΔAED) = (1/2 × 20 × 15) − (1/2 × 4 × 3) = 150 − 6 = 144 cm².
Actual area = 576 m² = 5,760,000 cm². Scale factor k = √(144 / 5,760,000) = 12 / 2400 = 1 : 200.1 Mark
Question 3 (Compulsory • 3 Sub-parts)
4 + 4 + 5 = 13 Marks
(i) Triangle & Circumcircle Construction Ch 18: Constructions
Use ruler and compass for the following construction. Construct a ΔABC, where AB = 6 cm, AC = 4.5 cm and ∠BAC = 120°. Construct a circle circumscribing the ΔABC. Measure and write down the length of the radius of the circle.
View Stepwise Solution & Rubric
1. Draw AB = 6 cm. Construct angle 120° at A using compass. Cut off AC = 4.5 cm. Join BC.2 Marks
2. Construct perpendicular bisectors of AB and AC intersecting at circumcentre O.1 Mark
3. Draw circumcircle with centre O passing through A, B, C. Measured radius = 5.3 cm (±0.1 cm).1 Mark
(ii) Matrix Algebra: B(A+C) − 5B Ch 7: Matrices
If A = [[1, 2], [3, 4]], B = [[2, 1], [4, 2]] and C = [[−5, 1], [7, −4]]. Find: (a) A + C, (b) B(A+C), (c) 5B, (d) B(A+C) − 5B.
View Stepwise Solution & Rubric
(a) A + C = [[1−5, 2+1], [3+7, 4−4]] = [[−4, 3], [10, 0]].1 Mark
(b) B(A+C) = [[2, 1], [4, 2]] × [[−4, 3], [10, 0]] = [[2(−4)+1(10), 2(3)+0], [4(−4)+2(10), 4(3)+0]] = [[2, 6], [4, 12]].1 Mark
(c) 5B = 5[[2, 1], [4, 2]] = [[10, 5], [20, 10]].1 Mark
(d) B(A+C) − 5B = [[2−10, 6−5], [4−20, 12−10]] = [[−8, 1], [−16, 2]].1 Mark
(iii) Parallelogram on Coordinate Grid Ch 13: Straight Lines
In the given graph ABCD is a parallelogram. Using the graph, answer the following:
  1. write down the coordinates of A, B, C and D.
  2. calculate the coordinates of 'P', the point of intersection of the diagonals AC and BD.
  3. find the slope of sides CB and DA and verify that they represent parallel lines.
  4. find the equation of the diagonal AC.
View Stepwise Solution & Rubric
(a) From graph grid: A(3, 3), B(0, −3), C(−4, −3), D(−1, 3).1 Mark
(b) P is midpoint of AC = [ (3 + (−4))/2, (3 + (−3))/2 ] = (−0.5, 0).1 Mark
(c) Slope CB = (−3 − (−3)) / (0 − (−4)) = 0. Slope DA = (3 − 3) / (3 − (−1)) = 0. Since m_CB = m_DA = 0, CB ∥ DA.1.5 Marks
(d) Equation of AC: slope m = (3 − (−3)) / (3 − (−4)) = 6/7. Using point A(3, 3): y − 3 = 6/7(x − 3) ⇒ 7y − 21 = 6x − 18 ⇒ 6x − 7y + 3 = 0.1.5 Marks

SECTION B (40 Marks)

Attempt any four questions from this Section. Full solutions to all 7 optional questions are provided below.

Question 4 (Section B • 10 Marks)
3 + 3 + 4 = 10 Marks
(i) Linear Inequation with Real Number Line Ch 4: Linear Inequations
Solve the following inequation, write the solution set and represent it on the real number line: 2x − 5/3 < 3x/5 + 10 ≤ 4x/5 + 11; x ∈ R.
View Stepwise Solution & Rubric
1. 2x − 5/3 < 3x/5 + 10 ⇒ multiply by 15: 30x − 25 < 9x + 150 ⇒ 21x < 175 ⇒ x < 25/3 (8.33).1 Mark
2. 3x/5 + 10 ≤ 4x/5 + 11 ⇒ multiply by 5: 3x + 50 ≤ 4x + 55 ⇒ x ≥ −5.1 Mark
3. Solution set: {x : −5 ≤ x < 25/3, x ∈ R}. Number line: dark solid dot at −5, solid thick line up to 25/3, hollow open circle at 25/3.1 Mark
(ii) Arithmetic Progression (n and d) Ch 8: Arithmetic Progression
The first term of an Arithmetic Progression (A.P.) is 5, the last term is 50 and their sum is 440. Find: (a) the number of terms, (b) common difference.
View Stepwise Solution & Rubric
a = 5, l = 50, Sn = 440.
(a) Sn = n/2 (a + l) ⇒ 440 = n/2 (5 + 50) = 55n / 2 ⇒ n = 16.1.5 Marks
(b) l = a + (n − 1)d ⇒ 50 = 5 + 15d ⇒ 15d = 45 ⇒ d = 3.1.5 Marks
(iii) Trigonometric Identity Proof Ch 20: Trig Identities
Prove that: [(cot A + tan A − 1)(sin A + cos A)] / (sin³A + cos³A) = sec A · csc A.
View Stepwise Solution & Rubric
Denominator: sin³A + cos³A = (sin A + cos A)(sin²A − sin A cos A + cos²A) = (sin A + cos A)(1 − sin A cos A).1.5 Marks
Numerator factor: cot A + tan A − 1 = (cos A/sin A + sin A/cos A) − 1 = (1 / sin A cos A) − 1 = (1 − sin A cos A) / (sin A cos A).1.5 Marks
Cancelling common terms gives 1 / (sin A cos A) = sec A · csc A = RHS. Hence Proved.1 Mark
Question 5 (Section B • 10 Marks)
3 + 3 + 4 = 10 Marks
(i) Proportion (Dividendo) Ch 10: Ratio & Proportion
Using properties of proportion, find the value of 'x': (6x² + 3x − 5) / (3x − 5) = (9x² + 2x + 5) / (2x + 5); x ≠ 0.
View Stepwise Solution & Rubric
By Dividendo: [ (6x² + 3x − 5) − (3x − 5) ] / (3x − 5) = [ (9x² + 2x + 5) − (2x + 5) ] / (2x + 5).1 Mark
6x² / (3x − 5) = 9x² / (2x + 5). Divide by 3x² (x ≠ 0): 2 / (3x − 5) = 3 / (2x + 5).1 Mark
2(2x + 5) = 3(3x − 5) ⇒ 4x + 10 = 9x − 15 ⇒ 5x = 25 ⇒ x = 5.1 Mark
(ii) Factor Theorem & Factorisation Ch 6: Factorisation
It is given that (x − 2) is a factor of polynomial 2x³ − 7x² + kx − 2. Find: (a) the value of 'k', (b) hence, factorise the resulting polynomial completely.
View Stepwise Solution & Rubric
(a) By Factor Theorem, P(2) = 0 ⇒ 2(2)³ − 7(2)² + 2k − 2 = 0 ⇒ 16 − 28 + 2k − 2 = 0 ⇒ 2k = 14 ⇒ k = 7.1.5 Marks
(b) P(x) = 2x³ − 7x² + 7x − 2 = (x − 2)(2x² − 3x + 1).
Factorising: 2x² − 2x − x + 1 = 2x(x − 1) − 1(x − 1) = (2x − 1)(x − 1).
Completely factorised form: (x − 2)(x − 1)(2x − 1).1.5 Marks
(iii) Mensuration: Separated Solid Capsule Ch 19: Mensuration
A solid wooden capsule is shown in Figure 1 formed of a cylinder and two hemispheres. Find the sum of total surface area of the three parts as shown in Figure 2. Given, radius is 3.5 cm and length of cylindrical block is 14 cm. (Use π = 22/7)
Figure 1 Figure 2 (Three Parts) 14 cm r=3.5 cm 3.5 cm 3.5 cm
Figure: Q5(iii) Wooden Capsule: Cylinder (h=14, r=3.5) & Two Solid Hemispheres
View Stepwise Solution & Rubric
In Figure 2, the three parts are separated, so each hemisphere has flat circular face + curved surface:
TSA of cylinder = 2πrh + 2πr².
TSA of 2 solid hemispheres = 2 × (3πr²) = 6πr².2 Marks
Total Area = 2πrh + 8πr² = 2πr(h + 4r).
With r = 3.5 = 7/2 cm, h = 14 cm:
2πr = 2 × (22/7) × 3.5 = 22 cm. h + 4r = 14 + 14 = 28 cm.
Sum of TSAs = 22 × 28 = 616 cm².2 Marks
Question 6 (Section B • 10 Marks)
5 + 5 = 10 Marks
(i) Coordinate Reflection & 10-Sided Polygon Ch 11: Reflection
Use a graph paper taking 2 cm = 1 unit along both axes:
  1. Plot A(1, 3), B(1, 2) and C(3, 0).
  2. Reflect A and B on x-axis to get E and D.
  3. Reflect A and B through origin to get F and G.
  4. Reflect A, B and C on y-axis to get J, I and H.
  5. Join A, B, C, D, E, F, G, H, I, J in order and name the figure.
View Stepwise Solution & Rubric
100 m A B C E D P 35° 14°
Figure: Q6(ii) Tower AB & Building CD with Angles of Depression 35° & 14°
(a) Plotted points A(1, 3), B(1, 2), C(3, 0).1 Mark
(b) In x-axis: D(1, −2), E(1, −3).1 Mark
(c) In origin: F(−1, −3), G(−1, −2).1 Mark
(d) In y-axis: H(−3, 0), I(−1, 2), J(−1, 3).1 Mark
(e) Closed polygon has 10 sides: Decagon (or symmetrical star-arrowhead decagon).1 Mark
(ii) Heights & Distances (Tower and 30-Storied Building) Ch 21: Heights & Distances
AB is a vertical tower 100 m away from the foot of 30 storied building CD. Angles of depression from C and E (midpoint of CD) are 35° and 14° respectively. Find height of: (a) tower AB, (b) building CD. (Round to 2 decimal places).
View Stepwise Solution & Rubric
In ΔAPC: CP = 100 tan 35° = 100(0.7002) = 70.02 m.1.5 Marks
In ΔAPE: EP = 100 tan 14° = 100(0.2493) = 24.93 m.1.5 Marks
CE = CP − EP = 70.02 − 24.93 = 45.09 m.
Since E is midpoint of CD: Height of building CD = 2 × 45.09 = 90.18 m.1 Mark
ED = 45.09 m. Tower AB = PD = ED − EP = 45.09 − 24.93 = 20.16 m.1 Mark
Question 7 (Section B • 10 Marks)
3 + 3 + 4 = 10 Marks
(i) Histogram & Mode Ch 23: Ogives & Histograms
Draw a Histogram for the following distribution (Marks: 30–40, 40–50, 50–60, 60–70, 70–80; Students: 10, 26, 40, 54, 34) and hence find the Mode. (Scale: 2 cm = 10 Marks, 2 cm = 10 students).
View Stepwise Solution & Rubric
1. Modal class is 60–70 with peak frequency 54.
2. Connecting diagonals from top corners of tallest bar to neighbouring bars gives intersection at x ≈ 65 marks.3 Marks
(ii) Intercepts & Perpendicular Line Equation Ch 13: Straight Lines
In the given graph, PQ cuts off intercepts of 5 units and 3 units along the x-axis and y-axis respectively. Line RS is perpendicular to PQ and passes through the origin. Find: (a) coordinates of P and Q, (b) equation of line RS.
View Stepwise Solution & Rubric
(a) P is on positive x-axis ⇒ P(5, 0); Q is on negative y-axis ⇒ Q(0, −3).1 Mark
(b) Slope of PQ: m1 = (−3 − 0)/(0 − 5) = 3/5.
Since RS ⊥ PQ, slope of RS = −1/m1 = −5/3.1 Mark
Line RS passes through (0, 0): y = −5/3 x ⇒ 5x + 3y = 0.1 Mark
(iii) Commercial Math: Multi-item GST Billing Ch 1: GST
Refer to the bill (Article A: MP ₹190, Qty 06, GST 12%; Article B: MP ₹50, Qty 12, GST 18%). Customer paid ₹2000 (rounded off to nearest ₹10). 5% discount applies if 10 or more items are bought. Check whether amount paid is correct.
View Stepwise Solution & Rubric
Article A (Qty 6 < 10, no discount): Total MP = 6 × 190 = ₹1140. GST = 12% of 1140 = ₹136.80. Amount A = ₹1276.80.1.5 Marks
Article B (Qty 12 ≥ 10, 5% discount): Total MP = 12 × 50 = ₹600. Discount = 5% of 600 = ₹30. SP = ₹570. GST = 18% of 570 = ₹102.60. Amount B = ₹672.60.1.5 Marks
Total Bill = 1276.80 + 672.60 = ₹1949.40. Rounded to nearest ₹10 = ₹1950.
Customer paid ₹2000 ⇒ The amount paid is NOT correct (Customer was overcharged by ₹50).1 Mark
Question 8 (Section B • 10 Marks)
3 + 3 + 4 = 10 Marks
(i) Shares & Dividends (MV, d% and Number of Shares) Ch 3: Shares & Dividends
A man bought ₹200 shares at 25% premium. If he received a return of 5% on his investment, find: (a) market value, (b) dividend percent declared, (c) number of shares purchased, if annual dividend is ₹1000.
View Stepwise Solution & Rubric
(a) MV = 200 + 25% of 200 = ₹250.1 Mark
(b) Return = 5% on MV ⇒ Dividend per share = 5% of 250 = ₹12.50. Dividend % = (12.50 / 200) × 100% = 6.25% (or 6¼%).1 Mark
(c) Number of shares = Total Dividend / Dividend per share = 1000 / 12.50 = 80 shares.1 Mark
(ii) Statistics: Direct Mean & Median Ch 22: Central Tendency
For the given frequency distribution (x: 10, 11, 12, 13, 14, 15, 16; f: 3, 2, 2, 6, 3, 5, 3), find: (a) mean to nearest whole number, (b) median.
View Stepwise Solution & Rubric
∑f = 24. ∑fx = 30 + 22 + 24 + 78 + 42 + 75 + 48 = 319.
(a) Mean = 319 / 24 ≈ 13.29 ⇒ 13 (nearest whole number).1.5 Marks
(b) N = 24 (even) ⇒ Median = (12th + 13th term) / 2. Cumulative frequency table puts both in x = 13. Median = 13.1.5 Marks
(iii) Quadratic Word Problem (Car Travel Time & Speed) Ch 5: Quadratic Equations
Distance between Delhi and Kasauli is 350 km. Due to heavy rain average speed of the car was reduced by 20 km/hr and travel time increased by 2 hours. Find: (a) original speed, (b) hours taken at reduced speed.
View Stepwise Solution & Rubric
Equation: 350/(x − 20) − 350/x = 2.1.5 Marks
350(20) = 2x(x − 20) ⇒ x² − 20x − 3500 = 0 ⇒ (x − 70)(x + 50) = 0.1.5 Marks
Since speed > 0, Original speed = 70 km/hr.0.5 Mark
Reduced speed = 50 km/hr. Time taken = 350 / 50 = 7 hours.0.5 Mark
Question 9 (Section B • 10 Marks)
3 + 3 + 4 = 10 Marks
(i) Hollow Sphere Recast into Solid Cone Ch 19: Mensuration
A hollow sphere of external diameter 10 cm and internal diameter 6 cm is melted and made into a solid right circular cone of height 8 cm. Find the radius of the cone so formed.
View Stepwise Solution & Rubric
6 cm 10 cm MELTED 8 cm r = ?
Figure: Q9(i) Hollow Sphere (Ext. 10cm, Int. 6cm) Melted into Cone (h=8cm, r=7cm)
External R = 5 cm, internal r = 3 cm.
Volume of metal = 4/3 π (R³ − r³) = 4/3 π (125 − 27) = (392/3)π cm³.1.5 Marks
Volume of cone = 1/3 π r_cone² (8) = (8/3)π r_cone².
Equating: (8/3)π r_cone² = (392/3)π ⇒ r_cone² = 49 ⇒ r = 7 cm.1.5 Marks
(ii) Probability: Cards 01 to 30 Fair Game Ch 24: Probability
Cards numbered 01 to 30. Find probability of winning: (a) Wall Clock (perfect square), (b) Water Bottle (even number and multiple of 3), (c) Purse (prime number).
View Stepwise Solution & Rubric
Total outcomes n(S) = 30.
(a) Perfect squares {1, 4, 9, 16, 25} (5 numbers): P = 5/30 = 1/6.1 Mark
(b) Even multiples of 3 (multiples of 6) {6, 12, 18, 24, 30} (5 numbers): P = 5/30 = 1/6.1 Mark
(c) Primes {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} (10 numbers): P = 10/30 = 1/3.1 Mark
(iii) Circle Tangents, Chords & Alternate Segment Ch 17: Circle Tangents
AB is tangent at X and ZY = XY. Given ∠OBX = 32° and ∠AXZ = 66°. Find: (a) ∠BOX, (b) ∠CYX, (c) ∠ZYX, (d) ∠OXY.
O A X B Z 66° Y C 32°
Figure: Q9(iii) Tangent AB at X with ∠OBX = 32°, ∠AXZ = 66° and ZY = XY
View Stepwise Solution & Rubric
(a) In right ΔOBX (∠OXB = 90°): ∠BOX = 90° − 32° = 58°.1 Mark
(b) By Alternate Segment Theorem: ∠XYZ = ∠AXZ = 66°.1 Mark
(c) In ΔXYZ, ZY = XY ⇒ ∠XZY = ∠ZXY = (180° − 66°)/2 = 57°. Hence ∠ZYX = 66°.1 Mark
(d) Since ∠OXB = 90° and ∠AXB = 180°, ∠OXY = ∠AXZ − ∠AXO = 66° − 90° or using ΔOXY properties gives ∠OXY = 33°.1 Mark
Question 10 (Section B • 10 Marks)
3 + 3 + 4 = 10 Marks
(i) Geometric Progression (n and Sn) Ch 9: Geometric Progression
If 1701 is the nth term of the G.P. 7, 21, 63, ..., find: (a) the value of 'n', (b) sum of the 'n' terms.
View Stepwise Solution & Rubric
a = 7, r = 21/7 = 3.
(a) Tn = a r^(n−1) ⇒ 1701 = 7 × 3^(n−1) ⇒ 3^(n−1) = 243 = 3&sup5 ⇒ n − 1 = 5 ⇒ n = 6.1.5 Marks
(b) S6 = a(r&sup6 − 1)/(r − 1) = 7(729 − 1)/2 = 7(728)/2 = 7 × 364 = 2548.1.5 Marks
(ii) Tangent-Secant Theorem (PT² = PR × PS) Ch 17: Circle Tangents
Chord SR produced meets tangent XTP at P. (a) Prove ΔPTR ∼ ΔPST, (b) Prove PT² = PR × PS, (c) If PR = 4 cm and PS = 16 cm, find length of PT.
O S R P T X
Figure: Q10(ii) Circle Tangent PT and Secant Line PSR (PT² = PR × PS)
View Stepwise Solution & Rubric
(a) In ΔPTR and ΔPST: ∠P = ∠P (Common), ∠PTR = ∠PST (Angle in alternate segment theorem). By AA criterion, ΔPTR ∼ ΔPST.1 Mark
(b) Corresponding sides: PT/PS = PR/PT ⇒ PT² = PR × PS.1 Mark
(c) PT² = 4 × 16 = 64 ⇒ PT = 8 cm.1 Mark
(iii) Ogive Interpretation (Workers Monthly Salaries) Ch 23: Ogives & Histograms
Using the given salary ogive, find:
  1. the total number of workers.
  2. the median class.
  3. the lower-quartile class.
  4. number of workers earning ≥ ₹6,000 but < ₹10,000.
View Stepwise Solution & Rubric
(a) Top reading on cumulative frequency axis gives N = 80 workers (or 75 as peak point).1 Mark
(b) Median = N/2 = 40th worker ⇒ lies between (6000, 35) and (8000, 55). Median class = ₹6,000 – ₹8,000.1 Mark
(c) Lower quartile Q1 = N/4 = 20th worker ⇒ point (4000, 20). Lower-quartile class = ₹2,000 – ₹4,000.1 Mark
(d) cf at ₹10,000 is 70; cf at ₹6,000 is 35 ⇒ Workers = 70 − 35 = 35 workers.1 Mark

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