T26 511
This paper consists of 16 printed pages.
MATHEMATICS ICSE Class 10 Examination 2026 Solved Paper
Time allowed: Three hours (Answers must be written on separate answer sheet)
- You will not be allowed to write during first 15 minutes. This time is to be spent in reading the question paper.
- The time given at the head of this Paper is the time allowed for writing the answers.
- Attempt all questions from Section A and any four questions from Section B.
- All working, including rough work, must be clearly shown on the same sheet as the rest of the answer.
- Omission of essential working will result in loss of marks. Intended marks are given in brackets [ ].
- Mathematical tables and graph papers are to be provided by the school.
Question 1 • Multiple Choice Questions
[15 × 1 = 15 Marks]
(i)
Ch 10: Ratio & Proportion
(x + 3), 1, (3x − 7) and −5 are in proportion. The value of x is:
View Stepwise Solution & Rubric
Correct Option: (a) −1
Method Since quantities are in proportion: (x + 3) / 1 = (3x − 7) / (−5)
⇒ −5(x + 3) = 1(3x − 7)
⇒ −5x − 15 = 3x − 7 ⇒ −8x = 8 ⇒ x = −1.
Method Since quantities are in proportion: (x + 3) / 1 = (3x − 7) / (−5)
⇒ −5(x + 3) = 1(3x − 7)
⇒ −5x − 15 = 3x − 7 ⇒ −8x = 8 ⇒ x = −1.
(ii)
Ch 1: GST
The marked price of a refrigerator is ₹12,000 and GST paid by the customer is ₹2,160. The rate of GST is:
View Stepwise Solution & Rubric
Correct Option: (c) 18%
Method GST Rate = (GST Paid / Marked Price) × 100 = (2,160 / 12,000) × 100 = 18%.
Method GST Rate = (GST Paid / Marked Price) × 100 = (2,160 / 12,000) × 100 = 18%.
(iii)
Ch 22: Statistics
Rakhi’s mobile number has the following integers: 1, 6, 9, 8, 9, 1, 7, 8, 9. The mode of the above given data is:
View Stepwise Solution & Rubric
Correct Option: (d) 9
Method Frequency count: 1 occurs 2 times, 6 occurs 1 time, 7 occurs 1 time, 8 occurs 2 times, 9 occurs 3 times. Mode = 9.
Method Frequency count: 1 occurs 2 times, 6 occurs 1 time, 7 occurs 1 time, 8 occurs 2 times, 9 occurs 3 times. Mode = 9.
(iv)
Ch 2: Banking
A and B opened a recurring deposit account in a bank which is paying simple interest at 9% per annum. A deposited ₹1,500 for one year and B deposited ₹1,200 for 15 months. The amount invested by:
View Stepwise Solution & Rubric
Correct Option: (d) Both A and B are same (₹18,000)
Method Amount invested = Monthly deposit × number of months.
A’s principal = 1,500 × 12 = ₹18,000. B’s principal = 1,200 × 15 = ₹18,000.
Method Amount invested = Monthly deposit × number of months.
A’s principal = 1,500 × 12 = ₹18,000. B’s principal = 1,200 × 15 = ₹18,000.
(v)
Ch 13: Straight Line
Find the equation of a line whose y-intercept is 6 and is parallel to x-axis.
View Stepwise Solution & Rubric
Correct Option: (a) y = 6
Method A line parallel to x-axis has slope m = 0. Its equation in slope-intercept form is y = mx + c = 0(x) + 6 ⇒ y = 6.
Method A line parallel to x-axis has slope m = 0. Its equation in slope-intercept form is y = mx + c = 0(x) + 6 ⇒ y = 6.
(vi)
Ch 3: Shares & Dividends
Asha buys ₹20 shares of a company which pays 9% dividend at such a price that she gets a return of 12% on her investment. At what price did she buy each share?
View Stepwise Solution & Rubric
Correct Option: (b) ₹15
Method Nominal Value FV = ₹20. Dividend per share = 9% of 20 = ₹1.80.
Return % = (Dividend / Market Value) × 100 ⇒ 12 = (1.80 / MV) × 100 ⇒ MV = 180 / 12 = ₹15.
Method Nominal Value FV = ₹20. Dividend per share = 9% of 20 = ₹1.80.
Return % = (Dividend / Market Value) × 100 ⇒ 12 = (1.80 / MV) × 100 ⇒ MV = 180 / 12 = ₹15.
(vii)
Ch 19: Mensuration
The total surface area of a solid sphere (S1) and a solid hemisphere (S2), as shown in the diagram, are equal. The ratio of radii R and r is:
Sphere S₁ (radius R) and Solid Hemisphere S₂ (radius r)
View Stepwise Solution & Rubric
Correct Option: (c) √3 : 2
Method TSA of solid sphere S1 = 4πR². TSA of solid hemisphere S2 = 3πr².
Given 4πR² = 3πr² ⇒ R² / r² = 3 / 4 ⇒ R / r = √3 / 2.
Method TSA of solid sphere S1 = 4πR². TSA of solid hemisphere S2 = 3πr².
Given 4πR² = 3πr² ⇒ R² / r² = 3 / 4 ⇒ R / r = √3 / 2.
(viii)
Ch 16: Circles
In the given diagram, O is the centre of the circle and ABCD is a cyclic quadrilateral. If ∠CDE = 65°, then the value of x is:
Cyclic Quad ABCD with ray ADE extended to E and ∠CDE = 65°
View Stepwise Solution & Rubric
Correct Option: (d) 130°
Method Exterior angle of cyclic quad equals opposite interior angle: ∠B = ∠CDE = 65°.
Angle subtended by chord AC at centre is double the angle at circumference: x = 2 × ∠B = 2 × 65° = 130°.
Method Exterior angle of cyclic quad equals opposite interior angle: ∠B = ∠CDE = 65°.
Angle subtended by chord AC at centre is double the angle at circumference: x = 2 × ∠B = 2 × 65° = 130°.
(ix)
Ch 5: Quadratic Equations
The nature of roots of quadratic equation 3x² − 6x − 3 = 0 are:
View Stepwise Solution & Rubric
Correct Option: (c) real, distinct and irrational
Method Discriminant D = b² − 4ac = (−6)² − 4(3)(−3) = 36 + 36 = 72 > 0.
Since 72 is positive but not a perfect square, roots are real, distinct and irrational.
Method Discriminant D = b² − 4ac = (−6)² − 4(3)(−3) = 36 + 36 = 72 > 0.
Since 72 is positive but not a perfect square, roots are real, distinct and irrational.
(x)
Ch 24: Probability
Assertion (A): If a die is rolled, the probability of getting a number greater than 6 is 1/6.
Reason (R): There are six possible outcomes when rolling a die, {1, 2, 3, 4, 5, 6}.
Reason (R): There are six possible outcomes when rolling a die, {1, 2, 3, 4, 5, 6}.
View Stepwise Solution & Rubric
Correct Option: (b) (A) is false and (R) is true.
Method Numbers greater than 6 on a die is 0. Probability = 0/6 = 0 ≠ 1/6. Hence Assertion is false, Reason is true.
Method Numbers greater than 6 on a die is 0. Probability = 0/6 = 0 ≠ 1/6. Hence Assertion is false, Reason is true.
(xi)
Ch 14: Similarity
If the areas of two similar triangles are in the ratio 9 : 64, then the ratio of their corresponding altitudes is:
View Stepwise Solution & Rubric
Correct Option: (a) 3 : 8
Method Ratio of areas of similar triangles = (ratio of corresponding altitudes)².
Altitudes ratio = √(9/64) = 3 : 8.
Method Ratio of areas of similar triangles = (ratio of corresponding altitudes)².
Altitudes ratio = √(9/64) = 3 : 8.
(xii)
Ch 6: Factorisation
What must be added to x³ + 7x² + 3x + 2 so that the result is completely divisible by (x + 2)?
View Stepwise Solution & Rubric
Correct Option: (b) −16
Method Let k be added. By factor theorem, P(−2) = 0.
(−2)³ + 7(−2)² + 3(−2) + 2 + k = 0 ⇒ −8 + 28 − 6 + 2 + k = 0 ⇒ 16 + k = 0 ⇒ k = −16.
Method Let k be added. By factor theorem, P(−2) = 0.
(−2)³ + 7(−2)² + 3(−2) + 2 + k = 0 ⇒ −8 + 28 − 6 + 2 + k = 0 ⇒ 16 + k = 0 ⇒ k = −16.
(xiii)
Ch 12: Section & Midpoint
In the given diagram, ΔAOB is a right-angled triangle and C is the mid-point of AB. The coordinates of the point which is equidistant from the three vertices of ΔAOB is:
Right-angled ΔAOB with midpoint C of hypotenuse AB
View Stepwise Solution & Rubric
Correct Option: (a) (x, y)
Method In a right-angled triangle, the circumcentre (equidistant from all three vertices) is the midpoint of the hypotenuse C.
Midpoint C = ((0 + 2x)/2, (2y + 0)/2) = (x, y).
Method In a right-angled triangle, the circumcentre (equidistant from all three vertices) is the midpoint of the hypotenuse C.
Midpoint C = ((0 + 2x)/2, (2y + 0)/2) = (x, y).
(xiv)
Ch 7: Matrices
Given matrix A = [[2, 3], [1, 2]] and matrix B = [2, −4]. Product AB is a matrix of order:
View Stepwise Solution & Rubric
Correct Option: (d) product AB is not possible
Method Matrix A has order 2 × 2 (2 columns). Matrix B has order 1 × 2 (1 row).
Since columns of A (2) ≠ rows of B (1), product AB is not possible.
Method Matrix A has order 2 × 2 (2 columns). Matrix B has order 1 × 2 (1 row).
Since columns of A (2) ≠ rows of B (1), product AB is not possible.
(xv)
Ch 9: GP
Assertion (A): The 9th term of a Geometric Progression (G.P.) 6, −12, 24, −48... is a positive term.
Reason (R): The value of (−2)8 is always positive.
Reason (R): The value of (−2)8 is always positive.
View Stepwise Solution & Rubric
Correct Option: (c) Both (A) and (R) are true and (R) is the correct explanation of (A).
Method First term a = 6, common ratio r = −2. T9 = a × r8 = 6 × (−2)8.
Since (−2)8 is positive, T9 is positive. Reason correctly explains Assertion.
Method First term a = 6, common ratio r = −2. T9 = a × r8 = 6 × (−2)8.
Since (−2)8 is positive, T9 is positive. Reason correctly explains Assertion.
Question 2
[4 + 4 + 4 = 12 Marks]
(i) [4 Marks]
Ch 8: Arithmetic Progression
The fourth and seventh terms of an Arithmetic Progression (A.P.), are 60 and 114 respectively. Find the:
(a) first term and common difference.
(b) sum of its first 10 terms.
(a) first term and common difference.
(b) sum of its first 10 terms.
View Stepwise Solution & Rubric
Part (a) [2M] T4 = a + 3d = 60 …(1); T7 = a + 6d = 114 …(2)
Subtracting (1) from (2): 3d = 54 ⇒ d = 18 [1M]
Substitute d = 18 in (1): a + 3(18) = 60 ⇒ a + 54 = 60 ⇒ a = 6 [1M]
Part (b) [2M] Sum of 10 terms S10 = (n/2)[2a + (n−1)d]
= (10/2)[2(6) + 9(18)] = 5[12 + 162] = 5 × 174 = 870 [2M].
Subtracting (1) from (2): 3d = 54 ⇒ d = 18 [1M]
Substitute d = 18 in (1): a + 3(18) = 60 ⇒ a + 54 = 60 ⇒ a = 6 [1M]
Part (b) [2M] Sum of 10 terms S10 = (n/2)[2a + (n−1)d]
= (10/2)[2(6) + 9(18)] = 5[12 + 162] = 5 × 174 = 870 [2M].
(ii) [4 Marks]
Ch 7: Matrices
Given, A = [[3, 1], [5, 3]] and B = [[−1, a], [3, −5]] and product AB = [[b, 7], [4, 5]]. Find the values of ‘a’ and ‘b’.
View Stepwise Solution & Rubric
Matrix Multiplication [2M]
AB = [[3(−1) + 1(3), 3(a) + 1(−5)], [5(−1) + 3(3), 5(a) + 3(−5)]]
= [[0, 3a − 5], [4, 5a − 15]]
Equating Elements [2M]
Equating (1, 1): b = 0 [1M]
Equating (1, 2): 3a − 5 = 7 ⇒ 3a = 12 ⇒ a = 4 [1M].
AB = [[3(−1) + 1(3), 3(a) + 1(−5)], [5(−1) + 3(3), 5(a) + 3(−5)]]
= [[0, 3a − 5], [4, 5a − 15]]
Equating Elements [2M]
Equating (1, 1): b = 0 [1M]
Equating (1, 2): 3a − 5 = 7 ⇒ 3a = 12 ⇒ a = 4 [1M].
(iii) [4 Marks]
Ch 17: Circle Tangents
In the given diagram, O is the centre of the circle and the tangent DE touches the circle at B. If, ∠ADB = 32°. Find the values of x and y.
Circle with diameter AC, tangent DBE touching at B, and ∠ADB = 32°
View Stepwise Solution & Rubric
Finding x [2M]
Radius OB ⊥ tangent DE ⇒ ∠OBD = 90°.
In ΔOBD: ∠BOD = 180° − (90° + 32°) = 58°.
Since D-C-O-A is a straight line, ∠AOB = 180° − 58° = 122°.
In isosceles ΔOAB (OA = OB): x = ∠OAB = (180° − 122°) / 2 = 29° [2M].
Finding y [2M]
∠CBA = 90° (angle in semicircle).
In ΔABC: ∠BCA = 90° − 29° = 61°.
By Alternate Segment Theorem, angle between tangent BE and chord BA equals ∠BCA:
y = 61° [2M].
Radius OB ⊥ tangent DE ⇒ ∠OBD = 90°.
In ΔOBD: ∠BOD = 180° − (90° + 32°) = 58°.
Since D-C-O-A is a straight line, ∠AOB = 180° − 58° = 122°.
In isosceles ΔOAB (OA = OB): x = ∠OAB = (180° − 122°) / 2 = 29° [2M].
Finding y [2M]
∠CBA = 90° (angle in semicircle).
In ΔABC: ∠BCA = 90° − 29° = 61°.
By Alternate Segment Theorem, angle between tangent BE and chord BA equals ∠BCA:
y = 61° [2M].
Question 3
[4 + 4 + 5 = 13 Marks]
(i) [4 Marks]
Ch 6: Factorisation
The polynomial kx³ + 3x² − 11x − 6 when divided by (x + 1), leaves a remainder of 6.
(a) Find the value of k.
(b) Using the value of k factorise completely the polynomial kx³ + 3x² − 11x − 6.
(a) Find the value of k.
(b) Using the value of k factorise completely the polynomial kx³ + 3x² − 11x − 6.
View Stepwise Solution & Rubric
Part (a) [2M] By Remainder Theorem, P(−1) = 6:
k(−1)³ + 3(−1)² − 11(−1) − 6 = 6 ⇒ −k + 3 + 11 − 6 = 6 ⇒ −k + 8 = 6 ⇒ k = 2 [2M].
Part (b) [2M] Polynomial P(x) = 2x³ + 3x² − 11x − 6.
For x = 2: P(2) = 2(8) + 3(4) − 11(2) − 6 = 16 + 12 − 22 − 6 = 0 ⇒ (x − 2) is a factor.
Dividing P(x) by (x − 2): 2x³ + 3x² − 11x − 6 = (x − 2)(2x² + 7x + 3)
Factorising quadratic: 2x² + 6x + x + 3 = (2x + 1)(x + 3).
Complete factors = (x − 2)(2x + 1)(x + 3) [2M].
k(−1)³ + 3(−1)² − 11(−1) − 6 = 6 ⇒ −k + 3 + 11 − 6 = 6 ⇒ −k + 8 = 6 ⇒ k = 2 [2M].
Part (b) [2M] Polynomial P(x) = 2x³ + 3x² − 11x − 6.
For x = 2: P(2) = 2(8) + 3(4) − 11(2) − 6 = 16 + 12 − 22 − 6 = 0 ⇒ (x − 2) is a factor.
Dividing P(x) by (x − 2): 2x³ + 3x² − 11x − 6 = (x − 2)(2x² + 7x + 3)
Factorising quadratic: 2x² + 6x + x + 3 = (2x + 1)(x + 3).
Complete factors = (x − 2)(2x + 1)(x + 3) [2M].
(ii) [4 Marks]
Ch 19: Mensuration
An eye drop bottle is prepared consisting of a hemisphere, a cylinder and a conical cap, as shown in the given diagram. Height of the cylindrical and conical parts are each, equal to the diameter (7 cm). Find the:
(a) minimum height of the cylindrical box required to pack this bottle.
(b) volume of the liquid medicine (shaded part: cylinder + hemisphere) in the bottle. Give your answer to the nearest whole number. (Use π = 22/7)
(a) minimum height of the cylindrical box required to pack this bottle.
(b) volume of the liquid medicine (shaded part: cylinder + hemisphere) in the bottle. Give your answer to the nearest whole number. (Use π = 22/7)
Eye Drop Bottle: Conical Cap + Cylinder + Inverted Hemispherical Base (d = 7 cm)
View Stepwise Solution & Rubric
Part (a) [1M] Radius r = 7/2 = 3.5 cm.
Height of bottle = height of cone + height of cylinder + radius of hemisphere
= 7 + 7 + 3.5 = 17.5 cm [1M].
Part (b) [3M] Liquid fills cylinder + hemisphere.
Volume of cylinder = πr²h = (22/7) × (3.5)² × 7 = 269.5 cm³ [1M]
Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × (3.5)³ = 89.83 cm³ [1M]
Total volume = 269.5 + 89.83 = 359.33 ≈ 359 cm³ [1M].
Height of bottle = height of cone + height of cylinder + radius of hemisphere
= 7 + 7 + 3.5 = 17.5 cm [1M].
Part (b) [3M] Liquid fills cylinder + hemisphere.
Volume of cylinder = πr²h = (22/7) × (3.5)² × 7 = 269.5 cm³ [1M]
Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × (3.5)³ = 89.83 cm³ [1M]
Total volume = 269.5 + 89.83 = 359.33 ≈ 359 cm³ [1M].
(iii) [5 Marks]
Ch 15 & 18: Loci & Constructions
Use ruler and compass for the following construction:
(a) construct an equilateral triangle ABC of side 5 cm.
(b) construct the circumcircle of ΔABC.
(c) construct the locus of points which are equidistant from AB and BC. Mark the point where the circumcircle and locus meet, as D.
(d) give the geometrical name of quadrilateral ABCD.
(a) construct an equilateral triangle ABC of side 5 cm.
(b) construct the circumcircle of ΔABC.
(c) construct the locus of points which are equidistant from AB and BC. Mark the point where the circumcircle and locus meet, as D.
(d) give the geometrical name of quadrilateral ABCD.
View Stepwise Solution & Rubric
Steps of Construction
(a) Draw BC = 5 cm. With B and C as centres and radius 5 cm, draw arcs intersecting at A. Join AB and AC. [1.5M]
(b) Draw perpendicular bisectors of any two sides (e.g. AB and BC) meeting at circumcentre O. With radius OA, draw circumcircle. [1.5M]
(c) The locus of points equidistant from AB and BC is the angle bisector of ∠ABC. Construct the angle bisector of ∠B to meet circumcircle at D. [1M]
(d) Since AB = BC and AD = CD, quadrilateral ABCD is a Kite (or cyclic kite). [1M]
(a) Draw BC = 5 cm. With B and C as centres and radius 5 cm, draw arcs intersecting at A. Join AB and AC. [1.5M]
(b) Draw perpendicular bisectors of any two sides (e.g. AB and BC) meeting at circumcentre O. With radius OA, draw circumcircle. [1.5M]
(c) The locus of points equidistant from AB and BC is the angle bisector of ∠ABC. Construct the angle bisector of ∠B to meet circumcircle at D. [1M]
(d) Since AB = BC and AD = CD, quadrilateral ABCD is a Kite (or cyclic kite). [1M]
Question 4
[3 + 3 + 4 = 10 Marks]
(i) [3 Marks]
Ch 20: Trigonometric Identities
Prove that: (sec θ − cos θ)(cosec θ − sin θ) = sin θ cos θ
View Stepwise Solution & Rubric
Proof LHS = (sec θ − cos θ)(cosec θ − sin θ)
= (1/cos θ − cos θ)(1/sin θ − sin θ) [1M]
= [(1 − cos²θ) / cos θ] × [(1 − sin²θ) / sin θ]
= (sin²θ / cos θ) × (cos²θ / sin θ) [1M]
= sin θ cos θ = RHS. Hence proved. [1M]
= (1/cos θ − cos θ)(1/sin θ − sin θ) [1M]
= [(1 − cos²θ) / cos θ] × [(1 − sin²θ) / sin θ]
= (sin²θ / cos θ) × (cos²θ / sin θ) [1M]
= sin θ cos θ = RHS. Hence proved. [1M]
(ii) [3 Marks]
Ch 1: GST
The cost price of a TV set is ₹20,000. The shopkeeper marked it for ₹24,000. He sells it to a customer at a discount of 10% on the marked price. If the sale is intra-state and the rate of GST is 12%, find the:
(a) discounted price of the TV set.
(b) amount paid by the customer to clear the bill.
(a) discounted price of the TV set.
(b) amount paid by the customer to clear the bill.
View Stepwise Solution & Rubric
Part (a) [1.5M] Discount = 10% of 24,000 = ₹2,400.
Discounted price (Selling Price) = 24,000 − 2,400 = ₹21,600.
Part (b) [1.5M] GST = 12% of 21,600 = ₹2,592.
Amount paid by customer = 21,600 + 2,592 = ₹24,192.
Discounted price (Selling Price) = 24,000 − 2,400 = ₹21,600.
Part (b) [1.5M] GST = 12% of 21,600 = ₹2,592.
Amount paid by customer = 21,600 + 2,592 = ₹24,192.
(iii) [4 Marks]
Ch 14: Similarity
In the given diagram, DE || BC and AD : DB = 2 : 3.
(a) Prove that: ΔADE ∼ ΔABC and hence find DE : BC
(b) Prove: ΔDFE ∼ ΔCFB
(c) Given, area of ΔDFE = 16 square units, find the area of ΔCFB.
(a) Prove that: ΔADE ∼ ΔABC and hence find DE : BC
(b) Prove: ΔDFE ∼ ΔCFB
(c) Given, area of ΔDFE = 16 square units, find the area of ΔCFB.
ΔABC with DE ∥ BC, AD : DB = 2 : 3, and diagonals intersecting at F
View Stepwise Solution & Rubric
Part (a) [1.5M] In ΔADE and ΔABC: ∠ADE = ∠ABC (corr. ∠s), ∠A = ∠A (common).
By AA criterion, ΔADE ∼ ΔABC.
DE / BC = AD / AB = 2 / (2 + 3) = 2 : 5.
Part (b) [1M] In ΔDFE and ΔCFB: ∠FDE = ∠FCB (alt. int. ∠s), ∠DFE = ∠CFB (vert. opp. ∠s).
By AA criterion, ΔDFE ∼ ΔCFB.
Part (c) [1.5M] Area(ΔDFE) / Area(ΔCFB) = (DE / BC)² = (2/5)² = 4/25.
16 / Area(ΔCFB) = 4/25 ⇒ Area(ΔCFB) = (16 × 25) / 4 = 100 square units.
By AA criterion, ΔADE ∼ ΔABC.
DE / BC = AD / AB = 2 / (2 + 3) = 2 : 5.
Part (b) [1M] In ΔDFE and ΔCFB: ∠FDE = ∠FCB (alt. int. ∠s), ∠DFE = ∠CFB (vert. opp. ∠s).
By AA criterion, ΔDFE ∼ ΔCFB.
Part (c) [1.5M] Area(ΔDFE) / Area(ΔCFB) = (DE / BC)² = (2/5)² = 4/25.
16 / Area(ΔCFB) = 4/25 ⇒ Area(ΔCFB) = (16 × 25) / 4 = 100 square units.
Question 5
[3 + 3 + 4 = 10 Marks]
(i) [3 Marks]
Ch 23: Ogives & Histograms
The histogram drawn on the graph represents the number of students of different heights (in cm).
Using the graph, answer the following:
(a) the number of students whose height is 150 cm and above.
(b) the modal height.
(c) the total number of students.
Using the graph, answer the following:
(a) the number of students whose height is 150 cm and above.
(b) the modal height.
(c) the total number of students.
Histogram with intersecting modal diagonals determining Mode = 137 cm
View Stepwise Solution & Rubric
(a) [1M] Heights 150 cm and above = bars for (150−160) + (160−170) = 9 + 4 = 13 students.
(b) [1M] The modal class is 130−140 cm. The x-coordinate of the intersection of diagonals gives Modal Height = 136 cm (acceptable: 135.5–136.5 cm).
(c) [1M] Total students = 7 + 2 + 9 + 14 + 12 + 9 + 4 = 57 students.
(b) [1M] The modal class is 130−140 cm. The x-coordinate of the intersection of diagonals gives Modal Height = 136 cm (acceptable: 135.5–136.5 cm).
(c) [1M] Total students = 7 + 2 + 9 + 14 + 12 + 9 + 4 = 57 students.
(ii) [3 Marks]
Ch 12: Section Formula
A(−10, −2) and B(2, 10) are two end points of a line segment. If AB intersects the x-axis at P, find the:
(a) ratio in which ‘P’ divides AB.
(b) coordinates of point P.
(a) ratio in which ‘P’ divides AB.
(b) coordinates of point P.
View Stepwise Solution & Rubric
Part (a) [1.5M] Since P lies on the x-axis, its y-coordinate is 0: P(x, 0).
Let P divide AB in ratio k : 1. By section formula for y:
0 = [k(10) + 1(−2)] / (k + 1) ⇒ 10k − 2 = 0 ⇒ 10k = 2 ⇒ k = 1/5.
Ratio = 1 : 5.
Part (b) [1.5M] x = [1(2) + 5(−10)] / (1 + 5) = (2 − 50) / 6 = −48 / 6 = −8.
Coordinates of P = (−8, 0).
Let P divide AB in ratio k : 1. By section formula for y:
0 = [k(10) + 1(−2)] / (k + 1) ⇒ 10k − 2 = 0 ⇒ 10k = 2 ⇒ k = 1/5.
Ratio = 1 : 5.
Part (b) [1.5M] x = [1(2) + 5(−10)] / (1 + 5) = (2 − 50) / 6 = −48 / 6 = −8.
Coordinates of P = (−8, 0).
(iii) [4 Marks]
Ch 5: Quadratic Equations
Solve the quadratic equation (x − 2)² − 5x − 3 = 0 and give your answer correct to 3 significant figures. (Use Mathematical Tables for this question if necessary.)
View Stepwise Solution & Rubric
Simplification [1M] (x − 2)² − 5x − 3 = 0 ⇒ x² − 4x + 4 − 5x − 3 = 0 ⇒ x² − 9x + 1 = 0.
Formula [2M] a = 1, b = −9, c = 1. Discriminant D = (−9)² − 4(1)(1) = 81 − 4 = 77.
√77 = 8.775.
x = [9 ± 8.775] / 2.
Roots to 3 Significant Figures [1M]
x1 = (9 + 8.775) / 2 = 17.775 / 2 = 8.8875 ≈ 8.89.
x2 = (9 − 8.775) / 2 = 0.225 / 2 = 0.1125 ≈ 0.113.
Formula [2M] a = 1, b = −9, c = 1. Discriminant D = (−9)² − 4(1)(1) = 81 − 4 = 77.
√77 = 8.775.
x = [9 ± 8.775] / 2.
Roots to 3 Significant Figures [1M]
x1 = (9 + 8.775) / 2 = 17.775 / 2 = 8.8875 ≈ 8.89.
x2 = (9 − 8.775) / 2 = 0.225 / 2 = 0.1125 ≈ 0.113.
Question 6
[3 + 3 + 4 = 10 Marks]
(i) [3 Marks]
Ch 3: Shares & Dividends
Kabir bought 120 shares of a company with nominal value ₹100, available at a premium of ₹25. Find:
(a) the money invested by Kabir in buying these shares.
(b) the rate of dividend, if he received ₹1,080 as dividend from these shares after one year.
(c) his rate of return.
(a) the money invested by Kabir in buying these shares.
(b) the rate of dividend, if he received ₹1,080 as dividend from these shares after one year.
(c) his rate of return.
View Stepwise Solution & Rubric
(a) [1M] Market Value per share = 100 + 25 = ₹125.
Investment = 120 × 125 = ₹15,000.
(b) [1M] Total Face Value = 120 × 100 = ₹12,000.
Dividend % = (Annual Dividend / Total Face Value) × 100 = (1,080 / 12,000) × 100 = 9%.
(c) [1M] Return % = (Annual Income / Investment) × 100 = (1,080 / 15,000) × 100 = 7.2%.
Investment = 120 × 125 = ₹15,000.
(b) [1M] Total Face Value = 120 × 100 = ₹12,000.
Dividend % = (Annual Dividend / Total Face Value) × 100 = (1,080 / 12,000) × 100 = 9%.
(c) [1M] Return % = (Annual Income / Investment) × 100 = (1,080 / 15,000) × 100 = 7.2%.
(ii) [3 Marks]
Ch 22: Mean (Step-Deviation)
Find the mean of the following frequency distribution using step-deviation method. Take assumed mean A = 28.
| Class Interval | 0–8 | 8–16 | 16–24 | 24–32 | 32–40 | 40–48 |
| Frequency (f) | 10 | 20 | 14 | 16 | 18 | 22 |
View Stepwise Solution & Rubric
Table [2M] Class width i = 8, Assumed Mean A = 28.
• 0–8: x=4, u=(4−28)/8=−3, f=10, fu=−30
• 8–16: x=12, u=−2, f=20, fu=−40
• 16–24: x=20, u=−1, f=14, fu=−14
• 24–32: x=28, u=0, f=16, fu=0
• 32–40: x=36, u=+1, f=18, fu=+18
• 40–48: x=44, u=+2, f=22, fu=+44
∑f = 100, ∑fu = −84 + 62 = −22.
Mean Calculation [1M]
Mean = A + [(∑fu / ∑f) × i] = 28 + [(−22 / 100) × 8] = 28 − 1.76 = 26.24.
• 0–8: x=4, u=(4−28)/8=−3, f=10, fu=−30
• 8–16: x=12, u=−2, f=20, fu=−40
• 16–24: x=20, u=−1, f=14, fu=−14
• 24–32: x=28, u=0, f=16, fu=0
• 32–40: x=36, u=+1, f=18, fu=+18
• 40–48: x=44, u=+2, f=22, fu=+44
∑f = 100, ∑fu = −84 + 62 = −22.
Mean Calculation [1M]
Mean = A + [(∑fu / ∑f) × i] = 28 + [(−22 / 100) × 8] = 28 − 1.76 = 26.24.
(iii) [4 Marks]
Ch 5: Quadratic Word Problem
The difference of two natural numbers is 5 and sum of their reciprocals is 3/10. Find the two numbers.
View Stepwise Solution & Rubric
Forming Equation [2M] Let the natural numbers be x and (x + 5).
1/x + 1/(x + 5) = 3/10 ⇒ (2x + 5) / (x² + 5x) = 3/10
10(2x + 5) = 3(x² + 5x) ⇒ 20x + 50 = 3x² + 15x ⇒ 3x² − 5x − 50 = 0.
Solving [2M] 3x² − 15x + 10x − 50 = 0 ⇒ 3x(x − 5) + 10(x − 5) = 0
(x − 5)(3x + 10) = 0 ⇒ x = 5 (since x ∈ N, x ≠ −10/3).
First number = 5, Second number = 5 + 5 = 10.
The two numbers are 5 and 10.
1/x + 1/(x + 5) = 3/10 ⇒ (2x + 5) / (x² + 5x) = 3/10
10(2x + 5) = 3(x² + 5x) ⇒ 20x + 50 = 3x² + 15x ⇒ 3x² − 5x − 50 = 0.
Solving [2M] 3x² − 15x + 10x − 50 = 0 ⇒ 3x(x − 5) + 10(x − 5) = 0
(x − 5)(3x + 10) = 0 ⇒ x = 5 (since x ∈ N, x ≠ −10/3).
First number = 5, Second number = 5 + 5 = 10.
The two numbers are 5 and 10.
Question 7
[5 + 5 = 10 Marks]
(i) [5 Marks]
Ch 21: Heights & Distances
A flagpole is erected at the top of a building. The angle of elevation of the top and foot of the flagpole from a point 100 m away, on the same level as that of the foot of the building, are 33° and 31° respectively. Find the height of the flagpole. Give your answer correct to the nearest metre. (Use Mathematical Tables for this question.)
Vertical flagpole CD on building BC with elevation angles 31° and 33° from 100 m away
View Stepwise Solution & Rubric
Building Height [2M] Let BC = building height, CD = flagpole height.
In right ΔABC: tan 31° = BC / 100 ⇒ BC = 100 × tan 31° = 100 × 0.6009 = 60.09 m.
Total Height [2M] In right ΔABD: tan 33° = BD / 100 ⇒ BD = 100 × tan 33° = 100 × 0.6494 = 64.94 m.
Flagpole Height [1M] Flagpole height CD = BD − BC = 64.94 − 60.09 = 4.85 m.
Correct to nearest metre = 5 metres.
In right ΔABC: tan 31° = BC / 100 ⇒ BC = 100 × tan 31° = 100 × 0.6009 = 60.09 m.
Total Height [2M] In right ΔABD: tan 33° = BD / 100 ⇒ BD = 100 × tan 33° = 100 × 0.6494 = 64.94 m.
Flagpole Height [1M] Flagpole height CD = BD − BC = 64.94 − 60.09 = 4.85 m.
Correct to nearest metre = 5 metres.
(ii) [5 Marks]
Ch 23: Ogive
Using a graph paper, draw an ogive for the following distribution which shows a record of weight in kilograms of 100 students.
Use your ogive to estimate the following:
(a) the median weight of the students.
(b) percentage of students whose weight is 60 kg or more.
(c) the weight above which 20% of the students lie.
| Weight (kg) | 35–40 | 40–45 | 45–50 | 50–55 | 55–60 | 60–65 | 65–70 | 70–75 |
| Students (f) | 4 | 6 | 10 | 24 | 26 | 17 | 8 | 5 |
(a) the median weight of the students.
(b) percentage of students whose weight is 60 kg or more.
(c) the weight above which 20% of the students lie.
View Stepwise Solution & Rubric
Ogive (Cumulative Frequency) curve with Median ≈ 56.5 kg and ≥60 kg estimate = 30%
Graph Plotting [2M] Smooth S-curve passing through points from (35,0).
(a) [1M] Median N/2 = 100/2 = 50th student. From graph, Median = 56 kg (±0.5 kg).
(b) [0.5M] Students with weight ≤ 60 kg = 70. Students ≥ 60 kg = 100 − 70 = 30%.
(c) [0.5M] Weight above which 20% lie corresponds to the 80th student: Weight ≈ 63 kg.
Question 8
[3 + 3 + 4 = 10 Marks]
(i) [3 Marks]
Ch 2: Banking
Rohit and Vinay both opened a recurring deposit account in a bank for 2 years at 8% simple interest. Vinay deposited ₹300 per month. On maturity, Rohit’s interest was ₹800 more than Vinay’s interest. Find the:
(a) interest earned by Vinay.
(b) sum deposited by Rohit every month.
(a) interest earned by Vinay.
(b) sum deposited by Rohit every month.
View Stepwise Solution & Rubric
Parallelogram OABC plotted on grid with x-axis reflection ODEC
Interest I = P × [n(n+1)/24] × (r/100) = P × [24 × 25 / 24] × (8/100) = 2P.
(a) [1.5M] Vinay: P = ₹300 ⇒ Vinay’s interest = 2 × 300 = ₹600.
(b) [1.5M] Rohit’s interest = 600 + 800 = ₹1,400.
2 × PRohit = 1,400 ⇒ PRohit = ₹700 per month.
(ii) [3 Marks]
Ch 9: Geometric Progression
The fourth term of a Geometric Progression (G.P.) is 16 and its seventh term is 128. Find its:
(a) common ratio
(b) first term
(a) common ratio
(b) first term
View Stepwise Solution & Rubric
(a) [1.5M] T4 = ar³ = 16; T7 = ar6 = 128.
T7 / T4 = ar6 / ar³ = r³ = 128 / 16 = 8 ⇒ common ratio r = 2.
(b) [1.5M] Substitute r = 2 in ar³ = 16: a(8) = 16 ⇒ first term a = 2.
T7 / T4 = ar6 / ar³ = r³ = 128 / 16 = 8 ⇒ common ratio r = 2.
(b) [1.5M] Substitute r = 2 in ar³ = 16: a(8) = 16 ⇒ first term a = 2.
(iii) [4 Marks]
Ch 11: Reflection
Use graph sheet for this question. Take 2 cm = 1 unit along both x and y axis.
Graphically represent parallelogram OABC, where O(0, 0), A(2, 3), B(5, 3) and C(3, 0).
Reflect OABC:
(a) on the x-axis and name its image as ODEC.
(b) through the origin and name its image as OIJH.
(c) on the y-axis and name its image as OFGH.
Graphically represent parallelogram OABC, where O(0, 0), A(2, 3), B(5, 3) and C(3, 0).
Reflect OABC:
(a) on the x-axis and name its image as ODEC.
(b) through the origin and name its image as OIJH.
(c) on the y-axis and name its image as OFGH.
View Stepwise Solution & Rubric
Plotting [1M] Parallelogram O(0,0), A(2,3), B(5,3), C(3,0).
(a) [1M] Reflection in x-axis: (x, y) → (x, −y):
O(0, 0), D(2, −3), E(5, −3), C(3, 0).
(b) [1M] Reflection in origin: (x, y) → (−x, −y):
O(0, 0), I(−2, −3), J(−5, −3), H(−3, 0).
(c) [1M] Reflection in y-axis: (x, y) → (−x, y):
O(0, 0), F(−2, 3), G(−5, 3), H(−3, 0).
(a) [1M] Reflection in x-axis: (x, y) → (x, −y):
O(0, 0), D(2, −3), E(5, −3), C(3, 0).
(b) [1M] Reflection in origin: (x, y) → (−x, −y):
O(0, 0), I(−2, −3), J(−5, −3), H(−3, 0).
(c) [1M] Reflection in y-axis: (x, y) → (−x, y):
O(0, 0), F(−2, 3), G(−5, 3), H(−3, 0).
Question 9
[3 + 3 + 4 = 10 Marks]
(i) [3 Marks]
Ch 4: Linear Inequations
Solve the following inequation, write the solution set and represent it on the real number line:
−1 < (2x − 3)/3 − x/5 ≤ 1, x ∈ R
−1 < (2x − 3)/3 − x/5 ≤ 1, x ∈ R
View Stepwise Solution & Rubric
Solution set on the real number line: open circle at −1 and solid dot at 1
−1 < (7x − 15) / 15 ≤ 1
Multiply by 15: −15 < 7x − 15 ≤ 15 ⇒ 0 < 7x ≤ 30 ⇒ 0 < x ≤ 30/7.
Solution Set & Graph [1M]
Solution set = {x : x ∈ R, 0 < x ≤ 4²⁄‚}.
(ii) [3 Marks]
Ch 13: Straight Line
Use the following graph and answer the given questions:
(a) Write the co-ordinates of points A, B and C.
(b) Find the equation of a line passing through the mid-point of AC and parallel to AB.
(a) Write the co-ordinates of points A, B and C.
(b) Find the equation of a line passing through the mid-point of AC and parallel to AB.
ΔABC on grid with midpoint M(5, 5) of AC and parallel line 6x − 5y = 5
View Stepwise Solution & Rubric
(a) [1M] Coordinates from graph: A(4, 8), B(−1, 2), C(6, 2).
(b) [2M] Midpoint of AC = M((4+6)/2, (8+2)/2) = M(5, 5) [1M]
Slope of AB = (8 − 2) / (4 − (−1)) = 6/5.
Since required line is parallel to AB, its slope m = 6/5.
Equation through (5, 5): y − 5 = (6/5)(x − 5) ⇒ 5y − 25 = 6x − 30 ⇒ 6x − 5y − 5 = 0 [1M].
(b) [2M] Midpoint of AC = M((4+6)/2, (8+2)/2) = M(5, 5) [1M]
Slope of AB = (8 − 2) / (4 − (−1)) = 6/5.
Since required line is parallel to AB, its slope m = 6/5.
Equation through (5, 5): y − 5 = (6/5)(x − 5) ⇒ 5y − 25 = 6x − 30 ⇒ 6x − 5y − 5 = 0 [1M].
(iii) [4 Marks]
Ch 19: Mensuration
A solid wooden toy is prepared by joining a cone, a cylinder and a sphere, as shown in the given diagram. The radius of each of the three solids is 7 cm and heights of each of the cone and the cylinder is 24 cm. Find:
(a) the total surface area of the given solid.
(b) the cost of painting the total surface at the rate of ₹0.50 per cm². (Use π = 22/7)
(a) the total surface area of the given solid.
(b) the cost of painting the total surface at the rate of ₹0.50 per cm². (Use π = 22/7)
Wooden toy composed of Sphere (r=7 cm) + Cylinder (h=24 cm) + Cone (h=24 cm, r=7 cm)
View Stepwise Solution & Rubric
Surface Areas [2.5M] Radius r = 7 cm, height h = 24 cm.
1. Slant height of cone l = √(r² + h²) = √(7² + 24²) = √625 = 25 cm.
CSA of cone = πrl = (22/7) × 7 × 25 = 550 cm².
2. CSA of cylinder = 2πrh = 2 × (22/7) × 7 × 24 = 1,056 cm².
3. Exposed surface of sphere = 4πr² − πr² (circular contact) = 3πr² = 3 × (22/7) × 49 = 462 cm².
(a) Total Surface Area [0.5M]
TSA = 550 + 1,056 + 462 = 2,068 cm².
(b) Cost [1M]
Cost = 2,068 × ₹0.50 = ₹1,034.
1. Slant height of cone l = √(r² + h²) = √(7² + 24²) = √625 = 25 cm.
CSA of cone = πrl = (22/7) × 7 × 25 = 550 cm².
2. CSA of cylinder = 2πrh = 2 × (22/7) × 7 × 24 = 1,056 cm².
3. Exposed surface of sphere = 4πr² − πr² (circular contact) = 3πr² = 3 × (22/7) × 49 = 462 cm².
(a) Total Surface Area [0.5M]
TSA = 550 + 1,056 + 462 = 2,068 cm².
(b) Cost [1M]
Cost = 2,068 × ₹0.50 = ₹1,034.
Question 10
[3 + 3 + 4 = 10 Marks]
(i) [3 Marks]
Ch 10: Ratio & Proportion
If x = 5ab / (a − b), a ≠ b,
(a) Find: x / a
(b) Using properties of proportion, find: (x + a) / (x − a)
(a) Find: x / a
(b) Using properties of proportion, find: (x + a) / (x − a)
View Stepwise Solution & Rubric
(a) [1M] x / a = 5b / (a − b).
(b) [2M] By Componendo and Dividendo:
(x + a) / (x − a) = [5b + (a − b)] / [5b − (a − b)]
= (5b + a − b) / (5b − a + b) = (a + 4b) / (6b − a).
(b) [2M] By Componendo and Dividendo:
(x + a) / (x − a) = [5b + (a − b)] / [5b − (a − b)]
= (5b + a − b) / (5b − a + b) = (a + 4b) / (6b − a).
(ii) [3 Marks]
Ch 24: Probability
A survey was conducted on 300 families having 2 children each. The results obtained are given below:
If one family is selected at random, find the probability that it will have:
(a) one girl child
(b) one or more girl children
(c) no boy child
| Number of girl children | 2 | 1 | 0 | Total |
| Number of families | 95 | 165 | 40 | 300 |
(a) one girl child
(b) one or more girl children
(c) no boy child
View Stepwise Solution & Rubric
(a) [1M] P(1 girl child) = 165 / 300 = 11 / 20 = 0.55.
(b) [1M] Families with 1 or 2 girls = 165 + 95 = 260.
P(≥ 1 girl) = 260 / 300 = 13 / 15 ≈ 0.867.
(c) [1M] In a 2-child family, “no boy” means 2 girls.
P(no boy child) = 95 / 300 = 19 / 60 ≈ 0.317.
(b) [1M] Families with 1 or 2 girls = 165 + 95 = 260.
P(≥ 1 girl) = 260 / 300 = 13 / 15 ≈ 0.867.
(c) [1M] In a 2-child family, “no boy” means 2 girls.
P(no boy child) = 95 / 300 = 19 / 60 ≈ 0.317.
(iii) [4 Marks]
Ch 16 & 17: Circles
In the given figure ‘O’ is the centre of the circle. PQ is a tangent to the circle at B and AB = AC. If ∠CBQ = 40°, find the unknown angles x, y, z and w.
Circle with tangent PQ at B, AB = AC, diameter BOE, and ∠CBQ = 40°
View Stepwise Solution & Rubric
Angle x [1M] By Alternate Segment Theorem, angle between tangent BQ and chord BC equals angle in opposite alternate segment: x = ∠BAC = 40°.
Angle w [1M] In ΔABC, AB = AC ⇒ ∠ABC = ∠ACB = (180° − 40°) / 2 = 70°.
By Alternate Segment Theorem, angle between tangent BP and chord AB equals ∠ACB: w = 70°.
Angle y [1M] Angle at centre subtended by chord BC is double the angle at circumference: y = ∠BOC = 2 × x = 2 × 40° = 80°.
Angle z [1M] Angles in the same segment subtended by chord AC are equal: z = ∠ADC = ∠ABC = 70°.
Angle w [1M] In ΔABC, AB = AC ⇒ ∠ABC = ∠ACB = (180° − 40°) / 2 = 70°.
By Alternate Segment Theorem, angle between tangent BP and chord AB equals ∠ACB: w = 70°.
Angle y [1M] Angle at centre subtended by chord BC is double the angle at circumference: y = ∠BOC = 2 × x = 2 × 40° = 80°.
Angle z [1M] Angles in the same segment subtended by chord AC are equal: z = ∠ADC = ∠ABC = 70°.