T24 511 • ICSE CLASS X
MATHEMATICS (THEORY)
YEAR 2024
ICSE Class 10 Mathematics 2024 Solved Paper
Official CISCE Board Examination Question Paper (Code T24 511) complete with authentic stepwise marking schemes, vector geometric diagrams, and chapter mappings.
Q1 (15 MCQs)
Q2 (Matrices, Cylinders, GP)
Q3 (Trig, Circles, Reflection)
Q4 (Banking, Mean, Proportion)
Q5 (Histogram, Heights)
Q6 (Quad Eq, AP, Similarity)
Q7 (Circles, Inequation, Centroid)
Q8 (Factorization, Probability, Sphere)
Q9 (Tangent-Secant, Shares, Ogive)
Q10 (GST, Straight Line, Loci)
Question 1
Compulsory Objective Section
[15 Marks]
Choose the correct answers to the questions from the given options. (Write the correct answers only).
(i) Goods & Services Tax (GST)
Chapter 1
For an Intra-state sale, the CGST paid by a dealer to the Central government is ₹120. If the marked price of the article is ₹2000, the rate of GST is:
View Stepwise Solution & Rubric
Correct Option: (c) 12%
For an intra-state sale, CGST = SGST = 1/2 of Total GST.
Given CGST = ₹120 ⇒ Total GST = 2 × 120 = ₹240.
Rate of GST = (Total GST / Marked Price) × 100% = (240 / 2000) × 100% = 12%. 1 Mark
For an intra-state sale, CGST = SGST = 1/2 of Total GST.
Given CGST = ₹120 ⇒ Total GST = 2 × 120 = ₹240.
Rate of GST = (Total GST / Marked Price) × 100% = (240 / 2000) × 100% = 12%. 1 Mark
(ii) Factorisation (Remainder Theorem)
Chapter 6
What must be subtracted from the polynomial x3 + x2 − 2x + 1, so that the result is exactly divisible by (x − 3)?
View Stepwise Solution & Rubric
Correct Option: (d) 31
Let f(x) = x3 + x2 − 2x + 1.
By the Remainder Theorem, the remainder when f(x) is divided by (x − 3) is f(3):
f(3) = (3)3 + (3)2 − 2(3) + 1 = 27 + 9 − 6 + 1 = 31.
Therefore, subtracting the remainder 31 makes the polynomial exactly divisible by (x − 3). 1 Mark
Let f(x) = x3 + x2 − 2x + 1.
By the Remainder Theorem, the remainder when f(x) is divided by (x − 3) is f(3):
f(3) = (3)3 + (3)2 − 2(3) + 1 = 27 + 9 − 6 + 1 = 31.
Therefore, subtracting the remainder 31 makes the polynomial exactly divisible by (x − 3). 1 Mark
(iii) Quadratic Equations (Nature of Roots)
Chapter 5
The roots of the quadratic equation px2 − qx + r = 0 are real and equal if:
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Correct Option: (b) q2 = 4pr
For a quadratic equation Ax2 + Bx + C = 0, roots are real and equal when discriminant D = B2 − 4AC = 0.
Here A = p, B = −q, C = r.
⇒ (−q)2 − 4(p)(r) = 0 ⇒ q2 = 4pr. 1 Mark
For a quadratic equation Ax2 + Bx + C = 0, roots are real and equal when discriminant D = B2 − 4AC = 0.
Here A = p, B = −q, C = r.
⇒ (−q)2 − 4(p)(r) = 0 ⇒ q2 = 4pr. 1 Mark
(iv) Matrices (Multiplication)
Chapter 7
If matrix A = [[2, 2], [0, 2]] and A2 = [[4, x], [0, 4]], then the value of x is:
View Stepwise Solution & Rubric
Correct Option: (c) 8
A2 = A × A = [[2, 2], [0, 2]] × [[2, 2], [0, 2]]
Row 1 × Col 2 = (2)(2) + (2)(2) = 4 + 4 = 8.
Comparing with the (1, 2) element of A2: x = 8. 1 Mark
A2 = A × A = [[2, 2], [0, 2]] × [[2, 2], [0, 2]]
Row 1 × Col 2 = (2)(2) + (2)(2) = 4 + 4 = 8.
Comparing with the (1, 2) element of A2: x = 8. 1 Mark
(v) Measures of Central Tendency (Median)
Chapter 22
The median of the following observations arranged in ascending order is 64. Find the value of x:
27, 31, 46, 52, x, x + 4, 71, 79, 85, 90
27, 31, 46, 52, x, x + 4, 71, 79, 85, 90
View Stepwise Solution & Rubric
Correct Option: (c) 62
Number of observations n = 10 (even).
Median = ½ [ (n/2)th observation + (n/2 + 1)th observation ]
Median = ½ [ 5th + 6th ] = ½ [ x + (x + 4) ] = ½ (2x + 4) = x + 2.
Given median = 64 ⇒ x + 2 = 64 ⇒ x = 62. 1 Mark
Number of observations n = 10 (even).
Median = ½ [ (n/2)th observation + (n/2 + 1)th observation ]
Median = ½ [ 5th + 6th ] = ½ [ x + (x + 4) ] = ½ (2x + 4) = x + 2.
Given median = 64 ⇒ x + 2 = 64 ⇒ x = 62. 1 Mark
(vi) Equation of a Straight Line (Collinearity)
Chapter 13
Points A(x, y), B(3, −2) and C(4, −5) are collinear. The value of y in terms of x is:
View Stepwise Solution & Rubric
Correct Option: (d) 7 − 3x
Since points A, B, C are collinear, Slope of AB = Slope of BC.
Slope of BC = (−5 − (−2)) / (4 − 3) = (−3) / 1 = −3.
Slope of AB = (y − (−2)) / (x − 3) = (y + 2) / (x − 3).
⇒ (y + 2) / (x − 3) = −3 ⇒ y + 2 = −3(x − 3) = −3x + 9 ⇒ y = 7 − 3x. 1 Mark
Since points A, B, C are collinear, Slope of AB = Slope of BC.
Slope of BC = (−5 − (−2)) / (4 − 3) = (−3) / 1 = −3.
Slope of AB = (y − (−2)) / (x − 3) = (y + 2) / (x − 3).
⇒ (y + 2) / (x − 3) = −3 ⇒ y + 2 = −3(x − 3) = −3x + 9 ⇒ y = 7 − 3x. 1 Mark
(vii) Ratio and Proportion (Direct Variation)
Chapter 10
The given table shows the distance covered and the time taken by a train moving at a uniform speed along a straight track.
The values of x and y are:
| Distance (in m) | 60 | 90 | y |
|---|---|---|---|
| Time (in sec) | 2 | x | 5 |
View Stepwise Solution & Rubric
Correct Option: (d) x = 3, y = 150
At uniform speed, Distance / Time = constant = 60 / 2 = 30 m/s.
⇒ 90 / x = 30 ⇒ x = 3.
⇒ y / 5 = 30 ⇒ y = 150. 1 Mark
At uniform speed, Distance / Time = constant = 60 / 2 = 30 m/s.
⇒ 90 / x = 30 ⇒ x = 3.
⇒ y / 5 = 30 ⇒ y = 150. 1 Mark
(viii) Arithmetic Progression (A.P.)
Chapter 8
The 7th term of the given Arithmetic Progression (A.P.): 1/a, (1/a + 1), (1/a + 2) … is:
View Stepwise Solution & Rubric
Correct Option: (a) (1/a + 6)
First term T1 = 1/a.
Common difference d = (1/a + 1) − 1/a = 1.
7th term T7 = a1 + (7 − 1)d = 1/a + 6. 1 Mark
First term T1 = 1/a.
Common difference d = (1/a + 1) − 1/a = 1.
7th term T7 = a1 + (7 − 1)d = 1/a + 6. 1 Mark
(ix) Shares and Dividends
Chapter 3
The sum invested to purchase 15 shares of a company of nominal value ₹75 available at a discount of 20% is:
View Stepwise Solution & Rubric
Correct Option: (d) ₹900
Nominal Value (NV) = ₹75. Discount = 20% of 75 = ₹15.
Market Value (MV) per share = 75 − 15 = ₹60.
Sum invested = Number of shares × MV = 15 × 60 = ₹900. 1 Mark
Nominal Value (NV) = ₹75. Discount = 20% of 75 = ₹15.
Market Value (MV) per share = 75 − 15 = ₹60.
Sum invested = Number of shares × MV = 15 × 60 = ₹900. 1 Mark
(x) Loci
Chapter 15
The circumcentre of a triangle is the point which is:
View Stepwise Solution & Rubric
Correct Option: (b) at equal distance from the three vertices of the triangle.
The circumcentre is the point of concurrence of perpendicular bisectors of the sides. By definition of the perpendicular bisector locus, every point on it is equidistant from the ends of the segment. Hence, the circumcentre is equidistant from all three vertices (and serves as the centre of the circumscribed circle). 1 Mark
The circumcentre is the point of concurrence of perpendicular bisectors of the sides. By definition of the perpendicular bisector locus, every point on it is equidistant from the ends of the segment. Hence, the circumcentre is equidistant from all three vertices (and serves as the centre of the circumscribed circle). 1 Mark
(xi) Trigonometrical Identities
Chapter 20
Statement 1: sin2 θ + cos2 θ = 1
Statement 2: cosec2 θ + cot2 θ = 1
Which of the following is valid?
Statement 2: cosec2 θ + cot2 θ = 1
Which of the following is valid?
View Stepwise Solution & Rubric
Correct Option: (a) only 1
Statement 1 is the fundamental identity: sin2 θ + cos2 θ = 1 (Valid).
Statement 2 is invalid because the standard identity is cosec2 θ − cot2 θ = 1 (not addition). 1 Mark
Statement 1 is the fundamental identity: sin2 θ + cos2 θ = 1 (Valid).
Statement 2 is invalid because the standard identity is cosec2 θ − cot2 θ = 1 (not addition). 1 Mark
(xii) Tangents and Intersecting Chords
Chapter 17
In the given diagram, PS and PT are tangents to the circle. SQ ∥ PT and ∠SPT = 80°. The value of ∠QST is:
Figure: Circle with Tangents PS, PT and Parallel Chord SQ ∥ PT
View Stepwise Solution & Rubric
Correct Option: (d) 50°
1. In ΔPST, PS = PT (tangents drawn from external point P are equal in length).
⇒ ∠PST = ∠PTS = (180° − 80°) / 2 = 100° / 2 = 50°.
2. Since chord SQ ∥ tangent PT and ST is transversal:
Alternate interior angles are equal ⇒ ∠QST = ∠PTS = 50°. 1 Mark
1. In ΔPST, PS = PT (tangents drawn from external point P are equal in length).
⇒ ∠PST = ∠PTS = (180° − 80°) / 2 = 100° / 2 = 50°.
2. Since chord SQ ∥ tangent PT and ST is transversal:
Alternate interior angles are equal ⇒ ∠QST = ∠PTS = 50°. 1 Mark
(xiii) Probability (Assertion-Reasoning)
Chapter 24
Assertion (A): A die is thrown once and the probability of getting an even number is 2/3.
Reason (R): The sample space for even numbers on a die is {2, 4, 6}.
Reason (R): The sample space for even numbers on a die is {2, 4, 6}.
View Stepwise Solution & Rubric
Correct Option: (b) A is false, R is true.
Total outcomes when a die is thrown: S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6.
Favorable outcomes for even numbers: E = {2, 4, 6} ⇒ n(E) = 3 (Reason R is true).
P(even) = 3/6 = 1/2 ≠ 2/3 (Assertion A is false). 1 Mark
Total outcomes when a die is thrown: S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6.
Favorable outcomes for even numbers: E = {2, 4, 6} ⇒ n(E) = 3 (Reason R is true).
P(even) = 3/6 = 1/2 ≠ 2/3 (Assertion A is false). 1 Mark
(xiv) Mensuration (Cylinders)
Chapter 19
A rectangular sheet of paper of size 11 cm × 7 cm is first rotated about the side 11 cm and then about the side 7 cm to form a cylinder, as shown in the diagram. The ratio of their curved surface areas is:
Figure: Sheet of Paper rotated/rolled into two cylinders
View Stepwise Solution & Rubric
Correct Option: (a) 1 : 1
Whether the sheet is folded/rolled or rotated:
- Curved Surface Area of cylinder formed by wrapping a sheet = Area of rectangular sheet = 11 × 7 = 77 cm2.
- When rotated about 11 cm side: height h1 = 11, radius r1 = 7 ⇒ CSA1 = 2πr1h1 = 2π(7)(11) = 154π.
- When rotated about 7 cm side: height h2 = 7, radius r2 = 11 ⇒ CSA2 = 2πr2h2 = 2π(11)(7) = 154π.
In either case, the ratio of their curved surface areas is 1 : 1. 1 Mark
Whether the sheet is folded/rolled or rotated:
- Curved Surface Area of cylinder formed by wrapping a sheet = Area of rectangular sheet = 11 × 7 = 77 cm2.
- When rotated about 11 cm side: height h1 = 11, radius r1 = 7 ⇒ CSA1 = 2πr1h1 = 2π(7)(11) = 154π.
- When rotated about 7 cm side: height h2 = 7, radius r2 = 11 ⇒ CSA2 = 2πr2h2 = 2π(11)(7) = 154π.
In either case, the ratio of their curved surface areas is 1 : 1. 1 Mark
(xv) Similarity of Triangles
Chapter 14
In the given diagram, ΔABC ∼ ΔPQR. If AD and PS are bisectors of ∠BAC and ∠QPR respectively, then:
Figure: Similar Triangles ΔABC ∼ ΔPQR with Angle Bisectors AD and PS
View Stepwise Solution & Rubric
Correct Option: (b) ΔABD ∼ ΔPQS
Given ΔABC ∼ ΔPQR ⇒ ∠B = ∠Q and ∠BAC = ∠QPR.
Since AD and PS bisect ∠BAC and ∠QPR:
∠BAD = ½ ∠BAC = ½ ∠QPR = ∠QPS.
In ΔABD and ΔPQS:
- ∠B = ∠Q
- ∠BAD = ∠QPS
By AA similarity criterion, ΔABD ∼ ΔPQS. 1 Mark
Given ΔABC ∼ ΔPQR ⇒ ∠B = ∠Q and ∠BAC = ∠QPR.
Since AD and PS bisect ∠BAC and ∠QPR:
∠BAD = ½ ∠BAC = ½ ∠QPR = ∠QPS.
In ΔABD and ΔPQS:
- ∠B = ∠Q
- ∠BAD = ∠QPS
By AA similarity criterion, ΔABD ∼ ΔPQS. 1 Mark
Question 2
Compulsory Subjective
[12 Marks]
(i) Matrices • Matrix Multiplication & Equality
4 Marks
A = [[x, 0], [1, 1]], B = [[4, 0], [y, 1]] and C = [[4, 0], [x, 1]].
Find the values of x and y, if AB = C.
Find the values of x and y, if AB = C.
View Stepwise Solution & Rubric
Step 1: Compute matrix product AB
AB = [[x, 0], [1, 1]] × [[4, 0], [y, 1]]
Row 1 × Col 1 = x(4) + 0(y) = 4x.
Row 1 × Col 2 = x(0) + 0(1) = 0.
Row 2 × Col 1 = 1(4) + 1(y) = 4 + y.
Row 2 × Col 2 = 1(0) + 1(1) = 1.
⇒ AB = [[4x, 0], [4 + y, 1]]. 2 Marks
Step 2: Equate AB to C
[[4x, 0], [4 + y, 1]] = [[4, 0], [x, 1]]
Equating corresponding elements:
4x = 4 ⇒ x = 1. 1 Mark
4 + y = x ⇒ 4 + y = 1 ⇒ y = −3. 1 Mark
AB = [[x, 0], [1, 1]] × [[4, 0], [y, 1]]
Row 1 × Col 1 = x(4) + 0(y) = 4x.
Row 1 × Col 2 = x(0) + 0(1) = 0.
Row 2 × Col 1 = 1(4) + 1(y) = 4 + y.
Row 2 × Col 2 = 1(0) + 1(1) = 1.
⇒ AB = [[4x, 0], [4 + y, 1]]. 2 Marks
Step 2: Equate AB to C
[[4x, 0], [4 + y, 1]] = [[4, 0], [x, 1]]
Equating corresponding elements:
4x = 4 ⇒ x = 1. 1 Mark
4 + y = x ⇒ 4 + y = 1 ⇒ y = −3. 1 Mark
(ii) Mensuration • Cylinder Cut into Identical Halves
4 Marks
A solid metallic cylinder is cut into two identical halves along its height (as shown in the diagram). The diameter of the cylinder is 7 cm and the height is 10 cm. Find:
(a) The total surface area (both the halves).
(b) The total cost of painting the two halves at the rate of ₹30 per cm2. (Use π = 22/7)
(a) The total surface area (both the halves).
(b) The total cost of painting the two halves at the rate of ₹30 per cm2. (Use π = 22/7)
Figure: Solid Cylinder Cut Vertically into Two Identical Halves
View Stepwise Solution & Rubric
Diameter d = 7 cm ⇒ Radius r = 3.5 cm = 7/2 cm. Height h = 10 cm.
Part (a) Total Surface Area of BOTH halves:
When cut vertically along the diameter, the cylinder's original surface remains intact, and two identical new flat rectangular cross-sections are exposed.
Original Total Surface Area of cylinder = 2πrh + 2πr2 = 2πr(h + r)
= 2 × (22/7) × (7/2) × (10 + 3.5) = 22 × 13.5 = 297 cm2. 1.5 Marks
Area of two rectangular cut faces = 2 × (diameter × height) = 2 × (7 × 10) = 140 cm2. 1 Mark
Total surface area (both halves) = 297 + 140 = 437 cm2. 0.5 Mark
Part (b) Total Cost of Painting:
Cost = Total Area × Rate = 437 cm2 × ₹30/cm2 = ₹13,110. 1 Mark
Part (a) Total Surface Area of BOTH halves:
When cut vertically along the diameter, the cylinder's original surface remains intact, and two identical new flat rectangular cross-sections are exposed.
Original Total Surface Area of cylinder = 2πrh + 2πr2 = 2πr(h + r)
= 2 × (22/7) × (7/2) × (10 + 3.5) = 22 × 13.5 = 297 cm2. 1.5 Marks
Area of two rectangular cut faces = 2 × (diameter × height) = 2 × (7 × 10) = 140 cm2. 1 Mark
Total surface area (both halves) = 297 + 140 = 437 cm2. 0.5 Mark
Part (b) Total Cost of Painting:
Cost = Total Area × Rate = 437 cm2 × ₹30/cm2 = ₹13,110. 1 Mark
(iii) Geometric Progression (G.P.)
4 Marks
15, 30, 60, 120… are in G.P. (Geometric Progression).
(a) Find the nth term of this G.P. in terms of n.
(b) How many terms of the above G.P. will give the sum 945?
(a) Find the nth term of this G.P. in terms of n.
(b) How many terms of the above G.P. will give the sum 945?
View Stepwise Solution & Rubric
First term a = 15, common ratio r = 30/15 = 2.
Part (a) nth term:
Tn = a × rn−1 = 15 × 2n−1.
Tn = 15(2n−1). 1.5 Marks
Part (b) Number of terms for sum = 945:
Sum of n terms of G.P. Sn = [a(rn − 1)] / (r − 1).
945 = [15(2n − 1)] / (2 − 1) = 15(2n − 1). 1 Mark
2n − 1 = 945/15 = 63.
2n = 64 = 26 ⇒ n = 6 terms. 1.5 Marks
Part (a) nth term:
Tn = a × rn−1 = 15 × 2n−1.
Tn = 15(2n−1). 1.5 Marks
Part (b) Number of terms for sum = 945:
Sum of n terms of G.P. Sn = [a(rn − 1)] / (r − 1).
945 = [15(2n − 1)] / (2 − 1) = 15(2n − 1). 1 Mark
2n − 1 = 945/15 = 63.
2n = 64 = 26 ⇒ n = 6 terms. 1.5 Marks
Question 3
Compulsory Subjective
[13 Marks]
(i) Trigonometrical Identities • Algebraic Factoring
4 Marks
Factorize: sin3θ + cos3θ.
Hence, prove the following identity:
Hence, prove the following identity:
[ (sin3θ + cos3θ) / (sinθ + cosθ) ] + sinθ cosθ = 1
View Stepwise Solution & Rubric
Figure: Reflection on Coordinate Plane Forming Square ABA'B'
Using algebraic identity a3 + b3 = (a + b)(a2 − ab + b2):
sin3θ + cos3θ = (sinθ + cosθ)(sin2θ − sinθ cosθ + cos2θ)
Since sin2θ + cos2θ = 1:
sin3θ + cos3θ = (sinθ + cosθ)(1 − sinθ cosθ). 2 Marks
Step 2: Prove the given identity
L.H.S. = [ (sin3θ + cos3θ) / (sinθ + cosθ) ] + sinθ cosθ
Substitute the factorized expression into L.H.S.:
L.H.S. = [ (sinθ + cosθ)(1 − sinθ cosθ) / (sinθ + cosθ) ] + sinθ cosθ 1 Mark
Cancelling (sinθ + cosθ):
= (1 − sinθ cosθ) + sinθ cosθ
= 1 = R.H.S. (Hence proved). 1 Mark
(ii) Circles • Tangents & Cyclic Quadrilateral
4 Marks
In the given diagram, O is the centre of the circle. PR and PT are two tangents drawn from external point P touching the circle at Q and S respectively. MN is a diameter of the circle. Given ∠PQM = 42° and ∠PSM = 25°.
Find: (a) ∠OQM (b) ∠QNS (c) ∠QOS (d) ∠QMS.
Find: (a) ∠OQM (b) ∠QNS (c) ∠QOS (d) ∠QMS.
Figure: Circle with Diameter MN, Tangents PR, PT and Cyclic Quadrilateral MQNS
View Stepwise Solution & Rubric
(a) Find ∠OQM:
Radius OQ ⊥ Tangent PR at point of contact Q ⇒ ∠OQP = 90°.
∠OQM = ∠OQP − ∠PQM = 90° − 42° = 48°. 1 Mark
(b) Find ∠QNS:
In ΔOQM, OM = OQ (radii) ⇒ ∠OMQ = ∠OQM = 48°.
Radius OS ⊥ Tangent PT at S ⇒ ∠OSP = 90°.
∠OSM = 90° − ∠PSM = 90° − 25° = 65°.
In ΔOMS, OM = OS (radii) ⇒ ∠OMS = ∠OSM = 65°.
⇒ ∠QMS = ∠OMQ + ∠OMS = 48° + 65° = 113°.
Quadrilateral MQNS is cyclic ⇒ opposite angles sum to 180°:
∠QNS = 180° − ∠QMS = 180° − 113° = 67°. 1 Mark
(c) Find ∠QOS:
Angle subtended by arc QS at the centre is twice the angle subtended at the circumference:
∠QOS = 2 × ∠QNS = 2 × 67° = 134°. 1 Mark
(d) Find ∠QMS:
∠QMS = ∠OMQ + ∠OMS = 48° + 65° = 113°. 1 Mark
Radius OQ ⊥ Tangent PR at point of contact Q ⇒ ∠OQP = 90°.
∠OQM = ∠OQP − ∠PQM = 90° − 42° = 48°. 1 Mark
(b) Find ∠QNS:
In ΔOQM, OM = OQ (radii) ⇒ ∠OMQ = ∠OQM = 48°.
Radius OS ⊥ Tangent PT at S ⇒ ∠OSP = 90°.
∠OSM = 90° − ∠PSM = 90° − 25° = 65°.
In ΔOMS, OM = OS (radii) ⇒ ∠OMS = ∠OSM = 65°.
⇒ ∠QMS = ∠OMQ + ∠OMS = 48° + 65° = 113°.
Quadrilateral MQNS is cyclic ⇒ opposite angles sum to 180°:
∠QNS = 180° − ∠QMS = 180° − 113° = 67°. 1 Mark
(c) Find ∠QOS:
Angle subtended by arc QS at the centre is twice the angle subtended at the circumference:
∠QOS = 2 × ∠QNS = 2 × 67° = 134°. 1 Mark
(d) Find ∠QMS:
∠QMS = ∠OMQ + ∠OMS = 48° + 65° = 113°. 1 Mark
(iii) Reflection • Coordinate Geometry on Graph
5 Marks
Use graph sheet for this question. Take 2 cm = 1 unit along the axes.
(a) Plot A(0, 3), B(2, 1) and C(4, −1).
(b) Reflect point B and C in y-axis and name their images as B' and C' respectively. Plot and write coordinates of the points B' and C'.
(c) Reflect point A in the line BB' and name its image as A'.
(d) Plot and write coordinates of point A'.
(e) Join the points ABA'B' and give the geometrical name of the closed figure so formed.
(a) Plot A(0, 3), B(2, 1) and C(4, −1).
(b) Reflect point B and C in y-axis and name their images as B' and C' respectively. Plot and write coordinates of the points B' and C'.
(c) Reflect point A in the line BB' and name its image as A'.
(d) Plot and write coordinates of point A'.
(e) Join the points ABA'B' and give the geometrical name of the closed figure so formed.
View Stepwise Solution & Rubric
(a) Correct plotting of A(0, 3), B(2, 1) and C(4, −1). 1 Mark
(b) Reflections in y-axis:
Rule: Ry-axis(x, y) = (−x, y).
- B' is image of B(2, 1) ⇒ B'(−2, 1).
- C' is image of C(4, −1) ⇒ C'(−4, −1). 1 Mark
(c) & (d) Reflection of A(0, 3) in line BB':
Line BB' connects (−2, 1) and (2, 1) ⇒ equation of line is y = 1.
Point A(0, 3) is 2 units above the line y = 1 (since 3 − 1 = 2).
Its reflection A' lies 2 units below the line y = 1 ⇒ y-coordinate = 1 − 2 = −1.
Coordinates of image: A'(0, −1). 1 Mark
(e) Geometrical Name of closed figure ABA'B':
Diagonals are AA' (along y-axis, length = 4 units) and BB' (along line y = 1, length = 4 units).
The diagonals are equal (4 units each), bisect each other at right angles at (0, 1).
Side lengths: AB = BA' = A'B' = B'A = √(22 + 22) = √8 = 2√2 units.
The figure is a Square (also accepted: Rhombus). 2 Marks
(b) Reflections in y-axis:
Rule: Ry-axis(x, y) = (−x, y).
- B' is image of B(2, 1) ⇒ B'(−2, 1).
- C' is image of C(4, −1) ⇒ C'(−4, −1). 1 Mark
(c) & (d) Reflection of A(0, 3) in line BB':
Line BB' connects (−2, 1) and (2, 1) ⇒ equation of line is y = 1.
Point A(0, 3) is 2 units above the line y = 1 (since 3 − 1 = 2).
Its reflection A' lies 2 units below the line y = 1 ⇒ y-coordinate = 1 − 2 = −1.
Coordinates of image: A'(0, −1). 1 Mark
(e) Geometrical Name of closed figure ABA'B':
Diagonals are AA' (along y-axis, length = 4 units) and BB' (along line y = 1, length = 4 units).
The diagonals are equal (4 units each), bisect each other at right angles at (0, 1).
Side lengths: AB = BA' = A'B' = B'A = √(22 + 22) = √8 = 2√2 units.
The figure is a Square (also accepted: Rhombus). 2 Marks
Question 4
Section B • 10 Marks
[3 + 3 + 4 = 10 Marks]
(i) Banking • Recurring Deposit (RD) Account
3 Marks
Suresh has a recurring deposit account in a bank. He deposits ₹2000 per month and the bank pays interest at the rate of 8% per annum. If he gets ₹1040 as interest at the time of maturity, find in years total time for which the account was held.
View Stepwise Solution & Rubric
Monthly deposit P = ₹2000, Rate R = 8% p.a., Interest I = ₹1040.
Let the number of months be n.
Formula: I = P × [ n(n + 1) / (2 × 12) ] × (R / 100) 1 Mark
1040 = 2000 × [ n(n + 1) / 24 ] × (8 / 100)
1040 = (160 / 24) × n(n + 1) = (20 / 3) × n(n + 1)
n(n + 1) = (1040 × 3) / 20 = 52 × 3 = 156. 1 Mark
n2 + n − 156 = 0 ⇒ (n + 13)(n − 12) = 0
Since months cannot be negative, n = 12 months.
Time in years = 12/12 = 1 year. 1 Mark
Let the number of months be n.
Formula: I = P × [ n(n + 1) / (2 × 12) ] × (R / 100) 1 Mark
1040 = 2000 × [ n(n + 1) / 24 ] × (8 / 100)
1040 = (160 / 24) × n(n + 1) = (20 / 3) × n(n + 1)
n(n + 1) = (1040 × 3) / 20 = 52 × 3 = 156. 1 Mark
n2 + n − 156 = 0 ⇒ (n + 13)(n − 12) = 0
Since months cannot be negative, n = 12 months.
Time in years = 12/12 = 1 year. 1 Mark
(ii) Measures of Central Tendency • Mean by Step-Deviation Method
3 Marks
The following table gives the duration of movies in minutes.
Using step-deviation method, find the mean duration of the movies.
| Duration (in minutes) | 100–110 | 110–120 | 120–130 | 130–140 | 140–150 | 150–160 |
|---|---|---|---|---|---|---|
| No. of movies | 5 | 10 | 17 | 8 | 6 | 4 |
View Stepwise Solution & Rubric
Let Assumed Mean A = 125, Class width i = 10.
Σfu = −10 − 10 + 0 + 8 + 12 + 12 = 12. 1.5 Marks
Mean = A + [ (Σfu / Σf) × i ]
Mean = 125 + [ (12 / 50) × 10 ] = 125 + 2.4 = 127.4 minutes. 1.5 Marks
| Class | Mid-val (x) | f | u = (x−125)/10 | fu |
|---|---|---|---|---|
| 100–110 | 105 | 5 | −2 | −10 |
| 110–120 | 115 | 10 | −1 | −10 |
| 120–130 | 125 (A) | 17 | 0 | 0 |
| 130–140 | 135 | 8 | 1 | 8 |
| 140–150 | 145 | 6 | 2 | 12 |
| 150–160 | 155 | 4 | 3 | 12 |
| Total | Σf = 50 | — | Σfu = 12 | |
Mean = A + [ (Σfu / Σf) × i ]
Mean = 125 + [ (12 / 50) × 10 ] = 125 + 2.4 = 127.4 minutes. 1.5 Marks
(iii) Ratio and Proportion • Componendo and Dividendo
4 Marks
If (a + b)3 / (a − b)3 = 64/27.
(a) Find (a + b) / (a − b).
(b) Hence using properties of proportion, find a : b.
(a) Find (a + b) / (a − b).
(b) Hence using properties of proportion, find a : b.
View Stepwise Solution & Rubric
Part (a) Find (a + b) / (a − b):
Taking cube roots on both sides:
3√[ (a + b)3 / (a − b)3 ] = 3√(64/27)
⇒ (a + b) / (a − b) = 4/3. 2 Marks
Part (b) Find a : b using Componendo and Dividendo:
Applying componendo and dividendo:
[ (a + b) + (a − b) ] / [ (a + b) − (a − b) ] = (4 + 3) / (4 − 3) 1 Mark
2a / 2b = 7/1 ⇒ a / b = 7/1.
⇒ a : b = 7 : 1. 1 Mark
Taking cube roots on both sides:
3√[ (a + b)3 / (a − b)3 ] = 3√(64/27)
⇒ (a + b) / (a − b) = 4/3. 2 Marks
Part (b) Find a : b using Componendo and Dividendo:
Applying componendo and dividendo:
[ (a + b) + (a − b) ] / [ (a + b) − (a − b) ] = (4 + 3) / (4 − 3) 1 Mark
2a / 2b = 7/1 ⇒ a / b = 7/1.
⇒ a : b = 7 : 1. 1 Mark
Question 5
Section B • 10 Marks
[5 + 5 = 10 Marks]
(i) Ogives and Histograms • Finding Mode from Histogram
5 Marks
The given graph with a histogram represents the number of plants of different heights grown in a school campus. Study the graph carefully and answer the following questions:
(a) Make a frequency table with respect to the class boundaries and their corresponding frequencies.
(b) State the modal class.
(c) Identify and note down the mode of the distribution.
(d) Find the number of plants whose height range is between 80 cm to 90 cm.
(a) Make a frequency table with respect to the class boundaries and their corresponding frequencies.
(b) State the modal class.
(c) Identify and note down the mode of the distribution.
(d) Find the number of plants whose height range is between 80 cm to 90 cm.
Figure: Histogram of Plant Heights and Modal Value Determination
View Stepwise Solution & Rubric
Part (a) Frequency Table:
2 Marks
Part (b) Modal class:
The class with the highest frequency (12) is 60 − 70 cm. 1 Mark
Part (c) Mode of the distribution:
From the intersection of the diagonal lines in the modal rectangle dropped vertically to the horizontal axis:
Mode = 64 cm. 1 Mark
Part (d) Number of plants with height range 80 cm to 90 cm:
Reading from the bar for 80–90 cm: 3 plants. 1 Mark
| Class Boundaries (cm) | Frequency (Plants) |
|---|---|
| 30 − 40 | 4 |
| 40 − 50 | 2 |
| 50 − 60 | 8 |
| 60 − 70 | 12 (Maximum) |
| 70 − 80 | 6 |
| 80 − 90 | 3 |
| 90 − 100 | 4 |
Part (b) Modal class:
The class with the highest frequency (12) is 60 − 70 cm. 1 Mark
Part (c) Mode of the distribution:
From the intersection of the diagonal lines in the modal rectangle dropped vertically to the horizontal axis:
Mode = 64 cm. 1 Mark
Part (d) Number of plants with height range 80 cm to 90 cm:
Reading from the bar for 80–90 cm: 3 plants. 1 Mark
(ii) Heights and Distances • Tree Viewed from Opposite Sides
5 Marks
The angle of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are 52° and 45° respectively. Find the distance AB, to the nearest metre.
Figure: Tree CD with Observers A and B on Opposite Sides
View Stepwise Solution & Rubric
Let CD = 100 m represent the tree, where C is the top and D is the foot on the ground.
Points A and B are on opposite sides along line AB, with ∠CAD = 52° and ∠CBD = 45°.
Step 1: In right ΔCDA
tan 52° = CD / AD = 100 / AD.
AD = 100 / tan 52° = 100 / 1.2799 ≈ 78.13 m. 2 Marks
Step 2: In right ΔCDB
tan 45° = CD / BD = 100 / BD.
Since tan 45° = 1 ⇒ BD = 100 / 1 = 100 m. 1.5 Marks
Step 3: Calculate total distance AB
AB = AD + BD = 78.13 + 100 = 178.13 m.
Rounding to the nearest metre: AB = 178 m. 1.5 Marks
Points A and B are on opposite sides along line AB, with ∠CAD = 52° and ∠CBD = 45°.
Step 1: In right ΔCDA
tan 52° = CD / AD = 100 / AD.
AD = 100 / tan 52° = 100 / 1.2799 ≈ 78.13 m. 2 Marks
Step 2: In right ΔCDB
tan 45° = CD / BD = 100 / BD.
Since tan 45° = 1 ⇒ BD = 100 / 1 = 100 m. 1.5 Marks
Step 3: Calculate total distance AB
AB = AD + BD = 78.13 + 100 = 178.13 m.
Rounding to the nearest metre: AB = 178 m. 1.5 Marks
Question 6
Section B • 10 Marks
[3 + 3 + 4 = 10 Marks]
(i) Quadratic Equations • Formula Method to 3 Sig Figs
3 Marks
Solve the following quadratic equation for x and give your answer correct to three significant figures:
2x2 − 10x + 5 = 0 (Use mathematical tables if necessary).
2x2 − 10x + 5 = 0 (Use mathematical tables if necessary).
View Stepwise Solution & Rubric
Comparing with ax2 + bx + c = 0: a = 2, b = −10, c = 5.
Discriminant D = b2 − 4ac = (−10)2 − 4(2)(5) = 100 − 40 = 60.
√60 = 2√15 = 2 × 3.74166 = 7.7460. 1 Mark
x = [ −b ± √D ] / (2a) = [ 10 ± 7.746 ] / 4
x1 = (10 + 7.746) / 4 = 17.746 / 4 = 4.4365 ≈ 4.44 (3 sig fig). 1 Mark
x2 = (10 − 7.746) / 4 = 2.254 / 4 = 0.5635 ≈ 0.564 (3 sig fig). 1 Mark
Answers: x = 4.44, 0.564.
Discriminant D = b2 − 4ac = (−10)2 − 4(2)(5) = 100 − 40 = 60.
√60 = 2√15 = 2 × 3.74166 = 7.7460. 1 Mark
x = [ −b ± √D ] / (2a) = [ 10 ± 7.746 ] / 4
x1 = (10 + 7.746) / 4 = 17.746 / 4 = 4.4365 ≈ 4.44 (3 sig fig). 1 Mark
x2 = (10 − 7.746) / 4 = 2.254 / 4 = 0.5635 ≈ 0.564 (3 sig fig). 1 Mark
Answers: x = 4.44, 0.564.
(ii) Arithmetic Progression (A.P.) • nth Term Formula
3 Marks
The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 − n).
Find: (a) its first term and common difference (b) sum of its first 25 terms.
Find: (a) its first term and common difference (b) sum of its first 25 terms.
View Stepwise Solution & Rubric
Given Tn = 6(7 − n) = 42 − 6n.
Part (a) First term and Common Difference:
For n = 1: First term a = T1 = 42 − 6(1) = 36.
For n = 2: T2 = 42 − 6(2) = 30.
Common difference d = T2 − T1 = 30 − 36 = −6. 1.5 Marks
Part (b) Sum of first 25 terms:
Sn = (n / 2) [ 2a + (n − 1)d ]
S25 = (25 / 2) [ 2(36) + (24)(−6) ] = (25 / 2) [ 72 − 144 ]
= (25 / 2) × (−72) = 25 × (−36) = −900. 1.5 Marks
Part (a) First term and Common Difference:
For n = 1: First term a = T1 = 42 − 6(1) = 36.
For n = 2: T2 = 42 − 6(2) = 30.
Common difference d = T2 − T1 = 30 − 36 = −6. 1.5 Marks
Part (b) Sum of first 25 terms:
Sn = (n / 2) [ 2a + (n − 1)d ]
S25 = (25 / 2) [ 2(36) + (24)(−6) ] = (25 / 2) [ 72 − 144 ]
= (25 / 2) × (−72) = 25 × (−36) = −900. 1.5 Marks
(iii) Similarity of Triangles • Intersecting Chords in Right Triangles
4 Marks
In the given diagram ΔADB and ΔACB are two right angled triangles with ∠ADB = ∠BCA = 90°. If AB = 10 cm, AD = 6 cm, BC = 2.4 cm and DP = 4.5 cm.
(a) Prove that ΔAPD ∼ ΔBPC.
(b) Find the length of BD and PB.
(c) Hence, find the length of PA.
(d) Find area ΔAPD : area ΔBPC.
(a) Prove that ΔAPD ∼ ΔBPC.
(b) Find the length of BD and PB.
(c) Hence, find the length of PA.
(d) Find area ΔAPD : area ΔBPC.
Figure: Right Triangles ΔADB and ΔACB with Chords Intersecting at P
View Stepwise Solution & Rubric
(a) Prove ΔAPD ∼ ΔBPC:
In ΔAPD and ΔBPC:
- ∠ADP = ∠BCP = 90° (Given)
- ∠APD = ∠BPC (Vertically opposite angles)
By AA similarity criterion, ΔAPD ∼ ΔBPC. 1 Mark
(b) Find length of BD and PB:
In right ΔADB, by Pythagoras Theorem:
BD2 = AB2 − AD2 = 102 − 62 = 100 − 36 = 64 ⇒ BD = 8 cm.
Given DP = 4.5 cm ⇒ PB = BD − DP = 8 − 4.5 = 3.5 cm. 1 Mark
(c) Find length of PA:
Since ΔAPD ∼ ΔBPC, the ratio of corresponding sides is equal:
PA / PB = AD / BC
PA / 3.5 = 6 / 2.4 = 2.5
PA = 2.5 × 3.5 = 8.75 cm. 1 Mark
(d) Area ratio ΔAPD : ΔBPC:
Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
Area(ΔAPD) / Area(ΔBPC) = (AD / BC)2 = (6 / 2.4)2 = (2.5)2 = 6.25 = 25/4.
⇒ Area ΔAPD : Area ΔBPC = 25 : 4. 1 Mark
In ΔAPD and ΔBPC:
- ∠ADP = ∠BCP = 90° (Given)
- ∠APD = ∠BPC (Vertically opposite angles)
By AA similarity criterion, ΔAPD ∼ ΔBPC. 1 Mark
(b) Find length of BD and PB:
In right ΔADB, by Pythagoras Theorem:
BD2 = AB2 − AD2 = 102 − 62 = 100 − 36 = 64 ⇒ BD = 8 cm.
Given DP = 4.5 cm ⇒ PB = BD − DP = 8 − 4.5 = 3.5 cm. 1 Mark
(c) Find length of PA:
Since ΔAPD ∼ ΔBPC, the ratio of corresponding sides is equal:
PA / PB = AD / BC
PA / 3.5 = 6 / 2.4 = 2.5
PA = 2.5 × 3.5 = 8.75 cm. 1 Mark
(d) Area ratio ΔAPD : ΔBPC:
Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
Area(ΔAPD) / Area(ΔBPC) = (AD / BC)2 = (6 / 2.4)2 = (2.5)2 = 6.25 = 25/4.
⇒ Area ΔAPD : Area ΔBPC = 25 : 4. 1 Mark
Question 7
Section B • 10 Marks
[3 + 3 + 4 = 10 Marks]
(i) Circles • Tangent and Inscribed Isosceles Triangle
3 Marks
In the given diagram, an isosceles ΔABC is inscribed in a circle with centre O. PQ is a tangent to the circle at C. OM is perpendicular to chord AC and ∠COM = 65°.
Find: (a) ∠ABC (b) ∠BAC (c) ∠BCQ.
Find: (a) ∠ABC (b) ∠BAC (c) ∠BCQ.
Figure: Inscribed Isosceles ΔABC with Perpendicular Bisector OM and Tangent PQ
View Stepwise Solution & Rubric
In ΔAOC, OA = OC (radii of the circle).
Since OM ⊥ AC, ΔOMA ≅ ΔOMC (RHS criterion), so ∠AOM = ∠COM = 65°.
Central angle ∠AOC = 65° + 65° = 130°.
(a) Find ∠ABC:
The angle subtended by an arc at the circumference is half the angle subtended by it at the centre:
∠ABC = ½ ∠AOC = ½ × 130° = 65°. 1 Mark
(b) Find ∠BAC:
Since ΔABC is isosceles with AB = AC:
∠ACB = ∠ABC = 65° (angles opposite to equal sides).
In ΔABC: ∠BAC = 180° − (∠ABC + ∠ACB) = 180° − (65° + 65°) = 50°. 1 Mark
(c) Find ∠BCQ:
By the Alternate Segment Theorem, the angle between chord BC and tangent CQ is equal to the angle subtended by chord BC in the alternate segment:
∠BCQ = ∠BAC = 50°. 1 Mark
Since OM ⊥ AC, ΔOMA ≅ ΔOMC (RHS criterion), so ∠AOM = ∠COM = 65°.
Central angle ∠AOC = 65° + 65° = 130°.
(a) Find ∠ABC:
The angle subtended by an arc at the circumference is half the angle subtended by it at the centre:
∠ABC = ½ ∠AOC = ½ × 130° = 65°. 1 Mark
(b) Find ∠BAC:
Since ΔABC is isosceles with AB = AC:
∠ACB = ∠ABC = 65° (angles opposite to equal sides).
In ΔABC: ∠BAC = 180° − (∠ABC + ∠ACB) = 180° − (65° + 65°) = 50°. 1 Mark
(c) Find ∠BCQ:
By the Alternate Segment Theorem, the angle between chord BC and tangent CQ is equal to the angle subtended by chord BC in the alternate segment:
∠BCQ = ∠BAC = 50°. 1 Mark
(ii) Linear Inequations • Integer Solution Set & Number Line
3 Marks
Solve the following inequation, write down the solution set and represent it on the real number line:
−3 + x ≤ 7x/2 + 2 < 8 + 2x, x ∈ I
−3 + x ≤ 7x/2 + 2 < 8 + 2x, x ∈ I
View Stepwise Solution & Rubric
Figure: Solution Set Represented as Dark Solid Dots on Real Number Line (x ∈ I)
Part 1: −3 + x ≤ 7x/2 + 2
Multiply by 2: −6 + 2x ≤ 7x + 4
−6 − 4 ≤ 7x − 2x ⇒ −10 ≤ 5x ⇒ x ≥ −2. 1 Mark
Part 2: 7x/2 + 2 < 8 + 2x
Multiply by 2: 7x + 4 < 16 + 4x
7x − 4x < 16 − 4 ⇒ 3x < 12 ⇒ x < 4. 1 Mark
Combining both parts: −2 ≤ x < 4.
Since x ∈ I (set of integers):
Solution Set = {−2, −1, 0, 1, 2, 3}. 0.5 Mark
Correct representation with darkened solid dots on −2, −1, 0, 1, 2, 3 on number line. 0.5 Mark
(iii) Coordinate Geometry • Section Formula, Centroid & Line Equation
4 Marks
In the given diagram, ABC is a triangle, where B(4, −4) and C(−4, −2). D is a point on AC.
(a) Write down the coordinates of A and D.
(b) Find the coordinates of the centroid of ΔABC.
(c) If D divides AC in the ratio k : 1, find the value of k.
(d) Find the equation of the line BD.
(a) Write down the coordinates of A and D.
(b) Find the coordinates of the centroid of ΔABC.
(c) If D divides AC in the ratio k : 1, find the value of k.
(d) Find the equation of the line BD.
Figure: Cartesian Diagram Showing Triangle ABC, Point D and Line BD
View Stepwise Solution & Rubric
(a) Coordinates of A and D:
From the graph, point A lies on the y-axis at 6 ⇒ A(0, 6).
Point D lies on the x-axis at −3 ⇒ D(−3, 0). 1 Mark
(b) Centroid of ΔABC:
Formula: G = [ (x1 + x2 + x3)/3, (y1 + y2 + y3)/3 ]
xG = (0 + 4 + (−4)) / 3 = 0 / 3 = 0.
yG = (6 + (−4) + (−2)) / 3 = 0 / 3 = 0.
⇒ Centroid G = (0, 0) (the origin). 1 Mark
(c) Value of k (ratio k : 1):
D divides AC joining A(0, 6) and C(−4, −2) in ratio k : 1.
x-coordinate: xD = [ k(−4) + 1(0) ] / (k + 1) = −4k / (k + 1).
Given xD = −3 ⇒ −4k / (k + 1) = −3 ⇒ −4k = −3k − 3 ⇒ −k = −3 ⇒ k = 3. (Ratio is 3 : 1). 1 Mark
(d) Equation of line BD:
Passing through B(4, −4) and D(−3, 0):
Slope m = [ 0 − (−4) ] / [ −3 − 4 ] = 4/−7 = −4/7.
Using point-slope form with D(−3, 0):
y − 0 = −4/7 (x + 3) ⇒ 7y = −4x − 12 ⇒ 4x + 7y + 12 = 0. 1 Mark
From the graph, point A lies on the y-axis at 6 ⇒ A(0, 6).
Point D lies on the x-axis at −3 ⇒ D(−3, 0). 1 Mark
(b) Centroid of ΔABC:
Formula: G = [ (x1 + x2 + x3)/3, (y1 + y2 + y3)/3 ]
xG = (0 + 4 + (−4)) / 3 = 0 / 3 = 0.
yG = (6 + (−4) + (−2)) / 3 = 0 / 3 = 0.
⇒ Centroid G = (0, 0) (the origin). 1 Mark
(c) Value of k (ratio k : 1):
D divides AC joining A(0, 6) and C(−4, −2) in ratio k : 1.
x-coordinate: xD = [ k(−4) + 1(0) ] / (k + 1) = −4k / (k + 1).
Given xD = −3 ⇒ −4k / (k + 1) = −3 ⇒ −4k = −3k − 3 ⇒ −k = −3 ⇒ k = 3. (Ratio is 3 : 1). 1 Mark
(d) Equation of line BD:
Passing through B(4, −4) and D(−3, 0):
Slope m = [ 0 − (−4) ] / [ −3 − 4 ] = 4/−7 = −4/7.
Using point-slope form with D(−3, 0):
y − 0 = −4/7 (x + 3) ⇒ 7y = −4x − 12 ⇒ 4x + 7y + 12 = 0. 1 Mark
Question 8
Section B • 10 Marks
[3 + 3 + 4 = 10 Marks]
(i) Factorisation • Factor Theorem & Complete Factorisation
3 Marks
The polynomial 3x3 + 8x2 − 15x + k has (x − 1) as a factor. Find the value of k. Hence factorize the resulting polynomial completely.
View Stepwise Solution & Rubric
Let f(x) = 3x3 + 8x2 − 15x + k.
Since (x − 1) is a factor, f(1) = 0:
3(1)3 + 8(1)2 − 15(1) + k = 0
3 + 8 − 15 + k = 0 ⇒ −4 + k = 0 ⇒ k = 4. 1 Mark
Polynomial is 3x3 + 8x2 − 15x + 4.
Dividing f(x) by (x − 1):
3x3 + 8x2 − 15x + 4 = (x − 1)(3x2 + 11x − 4). 1 Mark
Factoring the quadratic quotient: 3x2 + 12x − x − 4 = 3x(x + 4) − 1(x + 4) = (3x − 1)(x + 4).
Complete factorisation: (x − 1)(3x − 1)(x + 4). 1 Mark
Since (x − 1) is a factor, f(1) = 0:
3(1)3 + 8(1)2 − 15(1) + k = 0
3 + 8 − 15 + k = 0 ⇒ −4 + k = 0 ⇒ k = 4. 1 Mark
Polynomial is 3x3 + 8x2 − 15x + 4.
Dividing f(x) by (x − 1):
3x3 + 8x2 − 15x + 4 = (x − 1)(3x2 + 11x − 4). 1 Mark
Factoring the quadratic quotient: 3x2 + 12x − x − 4 = 3x(x + 4) − 1(x + 4) = (3x − 1)(x + 4).
Complete factorisation: (x − 1)(3x − 1)(x + 4). 1 Mark
(ii) Probability • Word Letter Cards
3 Marks
The following letters A, D, M, N, O, S, U, Y of the English alphabet are written on separate cards and put in a box. The cards are well shuffled and one card is drawn at random. What is the probability that the card drawn is a letter of the word:
(a) MONDAY?
(b) which does not appear in MONDAY?
(c) which appears both in SUNDAY and MONDAY?
(a) MONDAY?
(b) which does not appear in MONDAY?
(c) which appears both in SUNDAY and MONDAY?
View Stepwise Solution & Rubric
Sample space S = {A, D, M, N, O, S, U, Y} ⇒ Total outcomes n(S) = 8.
Part (a) Letter of the word MONDAY:
Letters of MONDAY present in sample space: {A, D, M, N, O, Y} ⇒ 6 cards.
P(letter of MONDAY) = 6/8 = 3/4. 1 Mark
Part (b) Letter which does NOT appear in MONDAY:
Remaining letters in S: {S, U} ⇒ 2 cards.
P(not in MONDAY) = 2/8 = 1/4. 1 Mark
Part (c) Letter appearing in BOTH SUNDAY and MONDAY:
Common letters between SUNDAY and MONDAY: {A, D, N, Y}.
All 4 letters are present in S: {A, D, N, Y} ⇒ 4 cards.
P(in both words) = 4/8 = 1/2. 1 Mark
Part (a) Letter of the word MONDAY:
Letters of MONDAY present in sample space: {A, D, M, N, O, Y} ⇒ 6 cards.
P(letter of MONDAY) = 6/8 = 3/4. 1 Mark
Part (b) Letter which does NOT appear in MONDAY:
Remaining letters in S: {S, U} ⇒ 2 cards.
P(not in MONDAY) = 2/8 = 1/4. 1 Mark
Part (c) Letter appearing in BOTH SUNDAY and MONDAY:
Common letters between SUNDAY and MONDAY: {A, D, N, Y}.
All 4 letters are present in S: {A, D, N, Y} ⇒ 4 cards.
P(in both words) = 4/8 = 1/2. 1 Mark
(iii) Mensuration • Sphere Capacity Transferred to Cylinder
4 Marks
Oil is stored in a spherical vessel occupying 3/4 of its full capacity. Radius of this spherical vessel is 28 cm. This oil is then poured into a cylindrical vessel with a radius of 21 cm. Find the height of the oil in the cylindrical vessel (correct to the nearest cm). (Take π = 22/7)
View Stepwise Solution & Rubric
Radius of sphere R = 28 cm. Radius of cylinder r = 21 cm.
Full capacity of spherical vessel = 4/3 π R3.
Volume of oil = 3/4 × (4/3 π R3) = π R3. 1.5 Marks
Volume of oil in cylindrical vessel = π r2 h.
Equating volumes:
π r2 h = π R3 ⇒ r2 h = R3 1 Mark
(21)2 × h = (28)3
h = (28 × 28 × 28) / (21 × 21) = (4 × 4 × 28) / (3 × 3) = 448/9 ≈ 49.78 cm.
Correct to the nearest cm: h = 50 cm. 1.5 Marks
Full capacity of spherical vessel = 4/3 π R3.
Volume of oil = 3/4 × (4/3 π R3) = π R3. 1.5 Marks
Volume of oil in cylindrical vessel = π r2 h.
Equating volumes:
π r2 h = π R3 ⇒ r2 h = R3 1 Mark
(21)2 × h = (28)3
h = (28 × 28 × 28) / (21 × 21) = (4 × 4 × 28) / (3 × 3) = 448/9 ≈ 49.78 cm.
Correct to the nearest cm: h = 50 cm. 1.5 Marks
Question 9
Section B • 10 Marks
[3 + 3 + 4 = 10 Marks]
(i) Tangents and Intersecting Chords • Tangent-Secant Theorem
3 Marks
The figure shows a circle of radius 9 cm with O as the centre. The diameter AB produced meets the tangent PQ at P. If PA = 24 cm, find the length of tangent PQ.
View Stepwise Solution & Rubric
Figure: Cumulative Frequency Curve (Ogive) Showing Median Age and Age > 52 Years
Figure: Circle with Diameter AB Produced to Tangent PQ
Given PA = 24 cm.
Then PB = PA − AB = 24 − 18 = 6 cm. 1 Mark
By the Tangent-Secant Theorem (when a tangent and a secant are drawn to a circle from an external point):
PQ2 = PA × PB 1 Mark
PQ2 = 24 × 6 = 144
PQ = √144 = 12 cm. 1 Mark
(Alternative method using right ΔOQP: OP = 9 + 6 = 15 cm; PQ = √(152 − 92) = √144 = 12 cm also earns full marks).
(ii) Shares and Dividends • Investment at Premium
3 Marks
Mr. Gupta invested ₹33,000 in buying ₹100 shares of a company at 10% premium. The dividend declared by the company is 12%. Find:
(a) the number of shares purchased by him.
(b) his annual dividend.
(a) the number of shares purchased by him.
(b) his annual dividend.
View Stepwise Solution & Rubric
Nominal Value (NV) = ₹100.
Premium = 10% of 100 = ₹10.
Market Value (MV) per share = 100 + 10 = ₹110.
Part (a) Number of shares:
Number of shares = Total Investment / MV per share = 33000 / 110 = 300 shares. 1.5 Marks
Part (b) Annual dividend:
Annual Dividend = Number of shares × (Dividend % × NV)
= 300 × (12% of 100) = 300 × 12 = ₹3,600. 1.5 Marks
Premium = 10% of 100 = ₹10.
Market Value (MV) per share = 100 + 10 = ₹110.
Part (a) Number of shares:
Number of shares = Total Investment / MV per share = 33000 / 110 = 300 shares. 1.5 Marks
Part (b) Annual dividend:
Annual Dividend = Number of shares × (Dividend % × NV)
= 300 × (12% of 100) = 300 × 12 = ₹3,600. 1.5 Marks
(iii) Ogives and Histograms • Cumulative Frequency Curve
4 Marks
A life insurance agent found the following data for distribution of ages of 100 policy holders:
On a graph sheet draw an ogive using the given data. Take 2 cm = 5 years along one axis and 2 cm = 10 policy holders along the other axis. Use your graph to find:
(a) The median age.
(b) Number of policy holders whose age is above 52 years.
| Age (years) | 20–25 | 25–30 | 30–35 | 35–40 | 40–45 | 45–50 | 50–55 | 55–60 |
|---|---|---|---|---|---|---|---|---|
| Frequency | 2 | 4 | 12 | 20 | 28 | 22 | 8 | 4 |
| Cumulative Freq | 2 | 6 | 18 | 38 | 66 | 88 | 96 | 100 |
(a) The median age.
(b) Number of policy holders whose age is above 52 years.
View Stepwise Solution & Rubric
Points plotted at upper limits:
(20, 0), (25, 2), (30, 6), (35, 18), (40, 38), (45, 66), (50, 88), (55, 96), (60, 100). Smooth S-curve drawn. 2 Marks
Part (a) Median age:
Total N = 100.
Median rank = N/2 = 50th policy holder.
Locating 50 on y-axis and reading across to the curve and down to x-axis:
Median age ≈ 42.1 years (Acceptable board range: 41.5 to 42.5 years). 1 Mark
Part (b) Number of policy holders whose age is above 52 years:
Locate 52 years on the horizontal axis and read up to the ogive curve: cumulative frequency ≈ 91.
Number of policy holders above 52 years = 100 − 91 = 9 policy holders (Acceptable range: 8 to 10). 1 Mark
(20, 0), (25, 2), (30, 6), (35, 18), (40, 38), (45, 66), (50, 88), (55, 96), (60, 100). Smooth S-curve drawn. 2 Marks
Part (a) Median age:
Total N = 100.
Median rank = N/2 = 50th policy holder.
Locating 50 on y-axis and reading across to the curve and down to x-axis:
Median age ≈ 42.1 years (Acceptable board range: 41.5 to 42.5 years). 1 Mark
Part (b) Number of policy holders whose age is above 52 years:
Locate 52 years on the horizontal axis and read up to the ogive curve: cumulative frequency ≈ 91.
Number of policy holders above 52 years = 100 − 91 = 9 policy holders (Acceptable range: 8 to 10). 1 Mark
Question 10
Section B • 10 Marks
[3 + 3 + 4 = 10 Marks]
(i) Goods and Services Tax (GST) • Sweet Mart Bill Calculation
3 Marks
Rohan bought the following eatables for his friends:
Calculate: (a) Total GST paid (b) Total bill amount including GST.
| S. No. | Item | Price | Quantity | Rate of GST |
|---|---|---|---|---|
| 1 | Laddu | ₹500 per kg | 2 kg | 5% |
| 2 | Pastries | ₹100 per piece | 12 pieces | 18% |
View Stepwise Solution & Rubric
Item 1 (Laddu):
Cost = 2 kg × ₹500 = ₹1,000.
GST on Laddu = 5% of 1000 = ₹50. 1 Mark
Item 2 (Pastries):
Cost = 12 × ₹100 = ₹1,200.
GST on Pastries = 18% of 1200 = ₹216. 1 Mark
(a) Total GST paid:
Total GST = 50 + 216 = ₹266. 0.5 Mark
(b) Total bill amount including GST:
Total basic cost = 1000 + 1200 = ₹2,200.
Total bill amount = 2200 + 266 = ₹2,466. 0.5 Mark
Cost = 2 kg × ₹500 = ₹1,000.
GST on Laddu = 5% of 1000 = ₹50. 1 Mark
Item 2 (Pastries):
Cost = 12 × ₹100 = ₹1,200.
GST on Pastries = 18% of 1200 = ₹216. 1 Mark
(a) Total GST paid:
Total GST = 50 + 216 = ₹266. 0.5 Mark
(b) Total bill amount including GST:
Total basic cost = 1000 + 1200 = ₹2,200.
Total bill amount = 2200 + 266 = ₹2,466. 0.5 Mark
Figure: Ruler & Compass Construction of Right ΔABC, Circumcircle and Loci
(ii) Equation of a Straight Line • Perpendicular & Parallel Lines
3 Marks
(a) If the lines kx − y + 4 = 0 and 2y = 6x + 7 are perpendicular to each other, find the value of k.
(b) Find the equation of a line parallel to 2y = 6x + 7 and passing through (−1, 1).
(b) Find the equation of a line parallel to 2y = 6x + 7 and passing through (−1, 1).
View Stepwise Solution & Rubric
Part (a) Find k:
Line 1: kx − y + 4 = 0 ⇒ y = kx + 4 ⇒ slope m1 = k.
Line 2: 2y = 6x + 7 ⇒ y = 3x + 3.5 ⇒ slope m2 = 3.
Since the lines are perpendicular, m1 × m2 = −1:
k × 3 = −1 ⇒ k = −1/3. 1.5 Marks
Part (b) Equation of parallel line:
Parallel lines have identical slopes ⇒ m = 3.
Passing through (x1, y1) = (−1, 1):
y − y1 = m(x − x1)
y − 1 = 3(x − (−1)) = 3(x + 1) = 3x + 3
y = 3x + 4 (or 3x − y + 4 = 0). 1.5 Marks
Line 1: kx − y + 4 = 0 ⇒ y = kx + 4 ⇒ slope m1 = k.
Line 2: 2y = 6x + 7 ⇒ y = 3x + 3.5 ⇒ slope m2 = 3.
Since the lines are perpendicular, m1 × m2 = −1:
k × 3 = −1 ⇒ k = −1/3. 1.5 Marks
Part (b) Equation of parallel line:
Parallel lines have identical slopes ⇒ m = 3.
Passing through (x1, y1) = (−1, 1):
y − y1 = m(x − x1)
y − 1 = 3(x − (−1)) = 3(x + 1) = 3x + 3
y = 3x + 4 (or 3x − y + 4 = 0). 1.5 Marks
(iii) Constructions and Loci • Ruler & Compass Geometry
4 Marks
Use ruler and compass to answer this question. Construct ∠ABC = 90°, where AB = 6 cm, BC = 8 cm.
(a) Construct the locus of points equidistant from B and C.
(b) Construct the locus of points equidistant from A and B.
(c) Mark the point which satisfies both conditions (a) and (b) as O. Construct the locus of points keeping a fixed distance OA from the fixed point O.
(d) Construct the locus of points which are equidistant from BA and BC.
(a) Construct the locus of points equidistant from B and C.
(b) Construct the locus of points equidistant from A and B.
(c) Mark the point which satisfies both conditions (a) and (b) as O. Construct the locus of points keeping a fixed distance OA from the fixed point O.
(d) Construct the locus of points which are equidistant from BA and BC.
View Stepwise Solution & Rubric
Construction Steps & Marking Rubric:
1. Draw line segment BC = 8 cm. At B, construct ∠B = 90° using compass arcs and cut off BA = 6 cm. Join AC to complete right ΔABC (Hypotenuse AC = √(62 + 82) = 10 cm). 1 Mark
2. Locus (a): Draw the perpendicular bisector of segment BC (points equidistant from B and C). 1 Mark
3. Locus (b): Draw the perpendicular bisector of segment AB (points equidistant from A and B). 0.5 Mark
4. Locus (c): The intersection of the two perpendicular bisectors is point O. In a right triangle, O lies at the midpoint of hypotenuse AC (OA = OB = OC = 5 cm). Draw a circle with centre O and radius OA = 5 cm. This circumcircle passes through A, B, and C. 1 Mark
5. Locus (d): Construct the angle bisector of ∠ABC using compass arcs. The ray equidistant from arms BA and BC bisects the 90° angle into two 45° angles. 0.5 Mark
1. Draw line segment BC = 8 cm. At B, construct ∠B = 90° using compass arcs and cut off BA = 6 cm. Join AC to complete right ΔABC (Hypotenuse AC = √(62 + 82) = 10 cm). 1 Mark
2. Locus (a): Draw the perpendicular bisector of segment BC (points equidistant from B and C). 1 Mark
3. Locus (b): Draw the perpendicular bisector of segment AB (points equidistant from A and B). 0.5 Mark
4. Locus (c): The intersection of the two perpendicular bisectors is point O. In a right triangle, O lies at the midpoint of hypotenuse AC (OA = OB = OC = 5 cm). Draw a circle with centre O and radius OA = 5 cm. This circumcircle passes through A, B, and C. 1 Mark
5. Locus (d): Construct the angle bisector of ∠ABC using compass arcs. The ray equidistant from arms BA and BC bisects the 90° angle into two 45° angles. 0.5 Mark
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