ICSE Class 10 Mathematics Sample Paper 2026 with Solutions

SECTION A (40 Marks)

(Attempt all questions from this Section.)

Question 1

Choose the correct answers to the questions from the given options.
(Do not copy the question, write the correct answers only.) [15]

(i) \((x - 2)\) and \((x + 2)\) are the factors of \(x^3 + x^2 - 4x - 4\). The third factor of the given polynomial is: [1 Mark]
(a) \((x - 1)\)
(b) \((x - 4)\)
(c) \((x + 1)\)
(d) \((x + 4)\)

Answer: (c) \((x + 1)\)

Dividing the given cubic polynomial by the product \((x - 2)(x + 2) = x^2 - 4\), we get \((x + 1)\).

Teacher's Note:
a) The product of two linear factors of a polynomial is always a factor of that polynomial.
b) Long division or synthetic division can be used to find the remaining linear factor.

(ii) Radha deposited ₹400 per month in a recurring deposit account for 18 months. The qualifying sum of money for the calculation of interest is: [1 Mark]
(a) ₹ 3,600
(b) ₹ 7,200
(c) ₹ 68,400
(d) ₹ 1,36,800

Answer: (c) ₹ 68,400

Equivalent principal for 1 month = \(P \times \frac{n(n+1)}{2} = 400 \times \frac{18 \times 19}{2} = \text{₹ } 68,400\).

Teacher's Note:
a) The qualifying sum of money is the equivalent principal for one month calculated using the formula \(\frac{Pn(n+1)}{2}\).
b) Students often confuse the total deposited amount with the equivalent principal.

(iii) In the figure given below, AC is a diameter of the circle. AP = 3 cm and PB = 4 cm and \(QP \perp AB\). If the area of \(\Delta APQ\) is \(18\text{ cm}^2\), then the area of shaded portion QPBC is: [1 Mark]
(a) \(32\text{ cm}^2$
(b) \(49\text{ cm}^2$
(c) \(80\text{ cm}^2$
(d) \(98\text{ cm}^2$

[Figure: A circle with diameter AC and centre O. Chord AB has point P such that AP = 3 cm and PB = 4 cm. Line segment QP is perpendicular to AB at P, connecting to point Q on the circumference. \(\Delta APQ\) is shaded grey.]

Answer: (c) \(80\text{ cm}^2$

Using intersecting chords theorem and proportionality in similar triangles, the area of the shaded portion works out to \(80\text{ cm}^2\).

Teacher's Note:
a) Apply properties of circles, perpendicular from the center or chord properties along with triangle area ratios.
b) Verify dimensional units carefully while calculating area.

(iv) In the given diagram, the radius of the circle with centre O is 3 cm. PA and PB are the tangents to the circle which are at right angle to each other. The length of OP is: [1 Mark]
(a) \(\frac{3}{\sqrt{2}}\text{ cm}\)
(b) \(3\text{ cm}\)
(c) \(3\sqrt{2}\text{ cm}\)
(d) \(6\sqrt{2}\text{ cm}\)

[Figure: A circle with centre O and radius 3 cm. Tangents PA and PB meet at external point P at right angles. Dashed line connects O to B.]

Answer: (c) \(3\sqrt{2}\text{ cm}\)

Since tangents are perpendicular to radius, \(OA \perp PA\) and \(OB \perp PB\). OAPB is a square of side 3 cm. The length of diagonal \(OP = 3\sqrt{2}\text{ cm}\).

Teacher's Note:
a) Tangents drawn from an external point to a circle subtend equal angles at the center and form a square with radii when perpendicular.
b) Always remember that the diagonal of a square of side \(a\) is \(a\sqrt{2}\).

(v) Assertion (A): If \(\sec\theta + \tan\theta = a\) and \(\sec\theta - \tan\theta = b\) then \(ab = 1\)
Reason (R): \(\sec^2\theta - \tan^2\theta = 1\) [1 Mark]
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true and (R) is the correct explanation of (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Answer: (c) Both (A) and (R) are true and (R) is the correct explanation of (A).

\(ab = (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = \sec^2\theta - \tan^2\theta = 1\).

Teacher's Note:
a) Fundamental trigonometric identities are direct tools for such proofs.
b) Check the algebraic identity \((x+y)(x-y) = x^2 - y^2\) applied here.

(vi) A solid sphere is cut into two identical hemispheres.
Assertion (A): The total volume of two hemispheres is equal to the volume of the original sphere.
Reason (R): The total surface area of two hemispheres together is equal to the surface area of the original sphere. [1 Mark]
(a) (A) is true, (R) is false.
(b) (A) is false, (R) is true.
(c) Both (A) and (R) are true and (R) is the correct explanation of (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Answer: (a) (A) is true, (R) is false.

Volume remains conserved (\(V = 2 \times \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3\)), but total surface area increases due to two new circular flat faces (\(2 \times 3\pi r^2 = 6\pi r^2\) vs \(4\pi r^2\)).

Teacher's Note:
a) Volume is additive and conserved upon cutting, whereas surface area increases due to exposed cross-sectional areas.
b) Read carefully whether curved surface area or total surface area is referenced.

(vii) Given that the sum of the squares of the first seven natural numbers is 140, then their mean is: [1 Mark]
(a) \(20$
(b) \(70$
(c) \(280$
(d) \(980$

Answer: (a) \(20$

Mean = \(\frac{\text{Sum of observations}}{\text{Number of observations}} = \frac{140}{7} = 20\).

Teacher's Note:
a) Mean is simply the sum divided by the count of items.
b) Do not confuse sum of squares with the arithmetic mean of the numbers themselves.

(viii) A bag contains 3 red and 2 blue marbles. A marble is drawn at random. The probability of drawing a black marble is: [1 Mark]
(a) \(0$
(b) \(\frac{1}{5}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{3}{5}\)

Answer: (a) \(0$

Since there are no black marbles in the bag, the event is impossible, so its probability is 0.

Teacher's Note:
a) The probability of an impossible event is always 0.
b) Verify the favourable outcomes against the total sample space carefully.

(ix) If matrix \(A = \begin{bmatrix}-1 & 2\end{bmatrix}\) and matrix \(B = \begin{bmatrix}3 \\ 4\end{bmatrix}\), then matrix AB is equal to: [1 Mark]
(a) \([-3]\)
(b) \([8]\)
(c) \([5]\)
(d) \(\begin{bmatrix}-1 & 2 \\ 3 & 4\end{bmatrix}\)

Answer: (c) \([5]\)

\(AB = \begin{bmatrix}-1 & 2\end{bmatrix} \begin{bmatrix}3 \\ 4\end{bmatrix} = [(-1)(3) + (2)(4)] = [-3 + 8] = [5]\).

Teacher's Note:
a) Matrix multiplication requires the number of columns in the first matrix to equal the rows in the second.
b) The resulting matrix order will be \((1 \times 2) \times (2 \times 1) = 1 \times 1\).

(x) A mixture of paint is prepared by mixing 2 parts of red pigments with 5 parts of the base. Using the given information in the following table, find the values of a, b & c to get the required mixture of paint. [1 Mark]

(a) \(a = 10, b = 10, c = 10\)
(b) \(a = 5, b = 2, c = 5$
(c) \(a = 10, b = 5, c = 10$
(d) \(a = 10, b = 5, c = 15\)

Answer: (d) \(a = 10, b = 5, c = 15$

Ratio is constant: \(\frac{2}{5} = \frac{4}{a} \Rightarrow a = 10\); \(\frac{2}{5} = \frac{b}{12.5} \Rightarrow b = 5\); \(\frac{2}{5} = \frac{6}{c} \Rightarrow c = 15\).

Teacher's Note:
a) Direct proportion implies that the ratio of the two quantities remains constant throughout.
b) Cross-multiplication is the quickest way to solve for unknown variables in proportional tables.

(xi) An article which is marked at ₹ 1,200 is available at a discount of 20% and the rate of GST is 18%. The amount of SGST is: [1 Mark]
(a) ₹ 216.00
(b) ₹ 172.80
(c) ₹ 108.00
(d) ₹ 86.40

Answer: (d) ₹ 86.40

Selling Price = \(80\% \text{ of ₹ } 1200 = \text{₹ } 960\). Total GST = \(18\% \text{ of ₹ } 960 = \text{₹ } 172.80\). SGST = \(\frac{1}{2} \times 172.80 = \text{₹ } 86.40\).

Teacher's Note:
a) GST is always calculated on the selling price (marked price after discount).
b) SGST is half of the total GST amount.

(xii) The sum of money required to buy 50, ₹ 40 shares at ₹ 38.50 is: [1 Mark]
(a) ₹ 1,920
(b) ₹ 1,924
(c) ₹ 1,925
(d) ₹ 1,952

Answer: (c) ₹ 1925

Investment = Number of shares \(\times\) Market price = \(50 \times \text{₹ } 38.50 = \text{₹ } 1,925\).

Teacher's Note:
a) Investment is always calculated using the market price, regardless of the nominal (face) value.
b) Multiplication can be simplified as \(50 \times \frac{77}{2} = 25 \times 77 = 1925\).

(xiii) The roots of quadratic equation \(x^2 - 1 = 0\) are: [1 Mark]
(a) \(0, 0$
(b) \(1, 1$
(c) \(-1, -1$
(d) \(+1, -1\)

Answer: (d) \(+1, -1$

\(x^2 - 1 = 0 \Rightarrow (x - 1)(x + 1) = 0 \Rightarrow x = 1$ or \(x = -1\).

Teacher's Note:
a) Difference of squares can be factored directly as \((a-b)(a+b) = 0\).
b) Every quadratic equation of degree 2 has two distinct or equal roots.

(xiv) Which of the following equations represents a line equally inclined to the axes? [1 Mark]
(a) \(2x - 3y + 7 = 0$
(b) \(x - y = 7$
(c) \(x = 7$
(d) \(y = -7\)

Answer: (b) \(x - y = 7$

A line equally inclined to the axes makes an angle of \(45^{\circ}\) or \(135^{\circ}\), meaning its slope \(m = \pm 1\). For \(x - y = 7\), \(y = x - 7\), so \(m = 1\).

Teacher's Note:
a) Lines equally inclined to the coordinate axes have slopes equal to \(1\) or \(-1\).
b) Rearrange equations into slope-intercept form (\(y = mx + c\)) to easily identify the slope.

(xv) Given, \(x + 2 \le \frac{x}{3} + 3\) and x is a prime number. The solution set for x is: [1 Mark]
(a) \(\emptyset$
(b) \(\{0\}$
(c) \(\{1\}$
(d) \(\{0, 1\}\)

Answer: (a) \(\emptyset$

Solving \(x + 2 \le \frac{x}{3} + 3 \Rightarrow \frac{2x}{3} \le 1 \Rightarrow x \le \frac{3}{2} = 1.5\). Since there are no prime numbers less than or equal to \(1.5\) (prime numbers start from 2), the solution set is empty.

Teacher's Note:
a) Pay close attention to number definitions like prime, natural, or integer numbers.
b) The number 1 is neither prime nor composite, and 2 is the smallest prime number.

Question 2

(i) While factorizing a given polynomial, using remainder & factor theorem, a student finds that \((2x + 1)\) is a factor of \(2x^3 + 7x^2 + 2x - 3\).
(a) Is the student's solution correct stating that \((2x + 1)\) is a factor of the given polynomial?
(b) Give a valid reason for your answer.
Also, factorize the given polynomial completely. [4 Marks]

Answer:
Let \(f(x) = 2x^3 + 7x^2 + 2x - 3\).
(a) No, the student's statement is incorrect.
(b) Checking for \((2x + 1)\), put \(x = -\frac{1}{2}\):
\(f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 + 7\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) - 3 = 2\left(-\frac{1}{8}\right) + 7\left(\frac{1}{4}\right) - 1 - 3 = -\frac{1}{4} + \frac{7}{4} - 4 = \frac{6}{4} - 4 = -\frac{5}{2} \neq 0\).
Since the remainder is not zero, \((2x + 1)\) is not a factor.
Checking for \((2x - 1)\), put \(x = \frac{1}{2}\):
\(f\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) + 7\left(\frac{1}{4}\right) + 2\left(\frac{1}{2}\right) - 3 = \frac{1}{4} + \frac{7}{4} + 1 - 3 = 2 + 1 - 3 = 0\).
Thus, \((2x - 1)\) is a factor.
Dividing \(f(x)\) by \((2x - 1)\), quotient is \(x^2 + 4x + 3 = (x + 3)(x + 1)\).
Complete factorization: \(f(x) = (2x - 1)(x + 3)(x + 1)\).

Teacher's Note:
a) Always substitute the root corresponding to the linear factor into the polynomial to test factorability.
b) Division or synthetic substitution must be executed carefully to avoid sign errors.

(ii) P is a point on the x-axis which divides the line joining A (-6, 2) and B (9, -4). Find:
(a) the ratio in which P divides the line segment AB.
(b) the coordinates of the point P.
(c) equation of a line parallel to AB and passing through (-3, -2). [4 Marks]

Answer:
(a) Since P lies on the x-axis, its y-coordinate is 0. Let the ratio be \(k:1\).
\(y = \frac{k(y_2) + 1(y_1)}{k+1} \Rightarrow 0 = \frac{k(-4) + 1(2)}{k+1} \Rightarrow -4k + 2 = 0 \Rightarrow k = \frac{2}{4} = \frac{1}{2}\).
Ratio \(k:1 = 1:2\).
(b) Coordinates of P: \(x = \frac{\frac{1}{2}(9) + 1(-6)}{\frac{1}{2} + 1} = \frac{\frac{9}{2} - 6}{\frac{3}{2}} = \frac{-3/2}{3/2} = -1\).
Point P is \((-1, 0)\).
(c) Slope of AB (\(m_{AB}\)) = \(\frac{-4 - 2}{9 - (-6)} = \frac{-6}{15} = -\frac{2}{5}\).
Equation of line parallel to AB passing through \((-3, -2)\):
\(y - (-2) = -\frac{2}{5}(x - (-3)) \Rightarrow 5(y + 2) = -2(x + 3) \Rightarrow 5y + 10 = -2x - 6 \Rightarrow 2x + 5y = -16\).

Teacher's Note:
a) A point on the x-axis always has a y-coordinate of zero \((x, 0)\).
b) Parallel lines share the exact same slope.

(iii) In the given figure, AC is the diameter of the circle with centre O. CD is parallel to BE. \(\angle AOB = 80^{\circ}\) and \(\angle ACE = 20^{\circ}\). Calculate:
(a) \(\angle BEC$
(b) \(\angle BCD$
(c) \(\angle CED\) [4 Marks]

[Figure: Circle with diameter AC and centre O. Chord BE and CD are parallel. Angle \(\angle AOB = 80^{\circ}\) at center and \(\angle ACE = 20^{\circ}\) at circumference.]

Answer:
(a) \(\angle BOC = 180^{\circ} - \angle AOB = 180^{\circ} - 80^{\circ} = 100^{\circ}\).
\(\angle BEC = \frac{1}{2} \angle BOC = \frac{1}{2} \times 100^{\circ} = 50^{\circ}\) (Angle at circumference is half the angle at centre subtended by the same arc).
(b) \(\angle BCD = \angle BCA + \angle ACE + \angle ECD\). Since \(CD \parallel BE\), alternate interior angles apply. Here \(\angle BCE = \angle CED\). Also \(\angle BCA = 90^{\circ}\) (angle in a semicircle).
\(\angle BCD = 40^{\circ} + 20^{\circ} + 50^{\circ} = 110^{\circ}\).
(c) \(\angle CED = 180^{\circ} - 110^{\circ} - 50^{\circ} = 20^{\circ}\).

Teacher's Note:
a) Angle in a semicircle is always a right angle (\(90^{\circ}\)).
b) Parallel lines property (alternate interior angles) is extremely useful in cyclic quadrilateral and circle geometry problems.

Question 3

(i) \(-11, -7, -3, \dots, 49, 53\) are the terms of a progression.
Answer the following:
(a) What is the type of progression?
(b) How many terms are there in all?
(c) Calculate the value of middle most term. [4 Marks]

Answer:
(a) Arithmetic Progression (A.P.), because the difference between consecutive terms is constant (\(-7 - (-11) = 4\)).
(b) Here, \(a = -11\), \(d = 4\), and \(T_n = 53\).
\(T_n = a + (n - 1)d \Rightarrow 53 = -11 + (n - 1)4 \Rightarrow 64 = 4(n - 1) \Rightarrow n - 1 = 16 \Rightarrow n = 17\).
There are 17 terms in all.
(c) Middle term = \(\left(\frac{17 + 1}{2}\right)^{\text{th}}\) term = \(9^{\text{th}}\) term.
\(T_9 = a + 8d = -11 + 8(4) = -11 + 32 = 21\).

Teacher's Note:
a) Always verify if the common difference is uniform to confirm an A.P.
b) For an odd number of terms \(n\), the middle term position is given by \(\frac{n+1}{2}\).

(ii) In the diagram given below, a tilted right circular cylindrical vessel with base diameter 7 cm contains a liquid. When placed vertically, the height of the liquid in the vessel is the mean of two heights shown in the diagram. Find the area of wet surface, when the cylinder is placed vertically on a horizontal surface. \(\left(\text{Use }\pi = \frac{22}{7}\right)\). [4 Marks]

[Figure: Tilted cylinder with base diameter 7 cm, minimum height 1 cm and maximum height 6 cm marked on the side.]

Answer:
Radius \(r = \frac{7}{2}\text{ cm}\).
Height of liquid when placed vertically \(h = \frac{1 + 6}{2} = \frac{7}{2}\text{ cm}\).
Area of wet surface = Curved surface area + Base area
\(= 2\pi rh + \pi r^2 = \pi r(2h + r)\)
\(= \frac{22}{7} \times \frac{7}{2} \left(2 \times \frac{7}{2} + \frac{7}{2}\right) = 11 \left(7 + \frac{7}{2}\right) = 11 \left(\frac{21}{2}\right) = \frac{231}{2} = 115.5\text{ cm}^2\).

Teacher's Note:
a) The wet surface of a cylinder standing vertically includes the circular base and the curved surface up to the liquid height.
b) Pay close attention to whether diameter or radius is given in mensuration problems.

(iii) Use a ruler and compass to answer this question.
(a) Construct a circle of radius 4.5 cm and draw a chord AB of length 6.5 cm.
(b) At A, construct \(\angle CAB = 75^{\circ}\), where C lies on the circumference of the circle.
(c) Construct the locus of all points equidistant from A and B.
(d) Construct the locus of all points equidistant from CA and BA.
(e) Mark the point of intersection of the two loci as P. Measure and write down the length of CP. [5 Marks]

[Figure: Construction diagram showing circle, chord AB, angle at A, perpendicular bisector of AB, angle bisector of \(\angle CAB\), intersecting at P inside the circle.]

Answer:
(a) - (d) Constructions completed using ruler and compass as per standard geometric steps.
(e) Length of \(CP = 4.9\text{ cm}\) (within acceptable tolerance limits \(\pm 0.1\text{ cm}\)).

Teacher's Note:
a) Locus of points equidistant from two points is the perpendicular bisector of the line segment joining them.
b) Locus of points equidistant from two intersecting lines is the angle bisector between them.

SECTION B (40 Marks)

(Attempt any four questions from this Section.)

Question 4

(i) Ms. Kaur invested ₹ 8,000 in buying ₹ 100 shares of a company paying 6% dividend at ₹ 80. After a year, she sold these shares at ₹ 75 each and invested the proceeds including the dividend received during the first year in buying ₹ 20 shares, paying 15% dividend at ₹ 27 each. Find the:
(a) dividend received by her during the first year.
(b) number of shares purchased by her using the total proceeds. [3 Marks]

Answer:
(a) Number of initial shares = \(\frac{\text{Investment}}{\text{Market Price}} = \frac{8000}{80} = 100\text{ shares}\).
Nominal value = \(100 \times \text{₹ } 100 = \text{₹ } 10,000\).
Annual Dividend = \(6\% \text{ of ₹ } 10,000 = \text{₹ } 600\).
(b) Sale proceeds from 100 shares at ₹ 75 each = \(100 \times 75 = \text{₹ } 7,500\).
Total proceeds including dividend = \(\text{₹ } 7,500 + \text{₹ } 600 = \text{₹ } 8,100\).\br />Number of new shares purchased at ₹ 27 each = \(\frac{8100}{27} = 300\text{ shares}\).

Teacher's Note:
a) Dividend is always calculated on the total nominal (face) value of the shares held.
b) Total proceeds for reinvestment comprise the sale amount plus any dividends earned.

(ii) Solve the following inequation, write the solution set, and represent it on the real number line.
\(5x - 21 < \frac{5x}{7} - 6 \le -3\frac{3}{7} + x, x \in R\). [3 Marks]

Answer:
Splitting the inequation into two parts:
Part 1: \(5x - 21 < \frac{5x}{7} - 6\)
\(\Rightarrow 5x - \frac{5x}{7} < -6 + 21 \Rightarrow \frac{30x}{7} < 15 \Rightarrow 30x < 105 \Rightarrow x < \frac{105}{30} = 3.5\ (\text{or } \frac{7}{2})\).
Part 2: \(\frac{5x}{7} - 6 \le -\frac{24}{7} + x\)
\(\Rightarrow \frac{5x}{7} - x \le -\frac{24}{7} + 6 \Rightarrow -\frac{2x}{7} \le \frac{18}{7} \Rightarrow -2x \le 18 \Rightarrow x \ge -9\).
Solution set: \(\{x : -9 \le x < 3.5, x \in R\}\).
[Figure: Real number line showing shaded region from -9 (inclusive) to 3.5 (exclusive).]

Teacher's Note:
a) Remember that dividing or multiplying an inequality by a negative number reverses the inequality sign.
b) Use a solid circle for inclusive inequalities (\(\le\), \(\ge\)) and a hollow circle for strict inequalities (\(<\), \(>\)).

(iii) Prove the following trigonometry identity:
\((\sin\theta + \cos\theta)(\csc\theta - \sec\theta) = \csc\theta\sec\theta - 2\tan\theta\). [4 Marks]

Answer:
\(\text{LHS} = (\sin\theta + \cos\theta)\left(\frac{1}{\sin\theta} - \frac{1}{\cos\theta}\right)\)
\(= (\sin\theta + \cos\theta)\left(\frac{\cos\theta - \sin\theta}{\sin\theta\cos\theta}\right)\)
\(= \frac{\cos^2\theta - \sin^2\theta}{\sin\theta\cos\theta} = \frac{\cos^2\theta}{\sin\theta\cos\theta} - \frac{\sin^2\theta}{\sin\theta\cos\theta} = \frac{\cos\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta}\) -- wait, expanding numerator: \(\frac{\cos^2\theta - \sin^2\theta}{\sin\theta\cos\theta}\) is not directly giving rhs unless expanded properly.
Let us re-evaluate standard expansion: \((s+c)(\frac{1}{s}-\frac{1}{c}) = \frac{s}{s} - \frac{s}{c} + \frac{c}{s} - \frac{c}{c} = 1 - \tan\theta + \cot\theta - 1 = \cot\theta - \tan\theta\)...
Checking key steps:
\(\text{LHS} = (\sin\theta + \cos\theta)(\csc\theta - \sec\theta) = (\sin\theta + \cos\theta)\left(\frac{\cos\theta - \sin\theta}{\sin\theta\cos\theta}\right) = \frac{\cos^2\theta - \sin^2\theta}{\sin\theta\cos\theta}\)
\(= \frac{1 - 2\sin^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} - \frac{2\sin^2\theta}{\sin\theta\cos\theta} = \csc\theta\sec\theta - 2\tan\theta = \text{RHS}\).

Teacher's Note:
a) Convert secant and cosecant into sine and cosine terms to simplify complex trigonometric expressions.
b) Split fractions into separate terms to match the required RHS expression step-by-step.

Question 5

(i) In the given figure (not drawn to scale) chords AD and BC intersect at P, where AB = 9 cm, PB = 3 cm and PD = 2 cm.
(a) Prove that \(\Delta APB \sim \Delta CPD$.
(b) Find the length of CD.
(c) Find \(\text{area }\Delta APB : \text{area }\Delta CPD\). [3 Marks]

[Figure: Circle with intersecting chords AD and BC at P. AB = 9 cm, PB = 3 cm, PD = 2 cm marked.]

Answer:
(a) In \(\Delta APB\) and \(\Delta CPD\):
\(\angle APB = \angle CPD\) (Vertically opposite angles)
\(\angle PAB = \angle PCD\) (Angles in the same segment subtended by arc BD)
Therefore, \(\Delta APB \sim \Delta CPD\) (AA axiom).
(b) By intersecting chords theorem: \(AP \times PD = BP \times PC\).
Also using similarity ratio: \(\frac{AB}{CD} = \frac{PB}{PD}\) is incorrect naming, corresponding sides are \(\frac{AP}{CP} = \frac{PB}{PD} = \frac{AB}{CD}\).
Given \(AB = 9\), \(PB = 3\), \(PD = 2\). From chord theorem \(PA \times 2 = 3 \times PC\)...
Wait, using ratio of similar triangles: \(\frac{AB}{CD} = \frac{PB}{PD}\) -- let us use standard corresponding sides: \(\frac{AB}{CD} = \frac{PB}{PD}\) is not correct unless matched by vertices. Vertices correspondence: \(A \leftrightarrow C, B \leftrightarrow D, P \leftrightarrow P\). So \(\frac{AB}{CD} = \frac{AP}{CP} = \frac{BP}{DP}\).
Thus \(\frac{9}{CD} = \frac{3}{2} \Rightarrow CD = \frac{9 \times 2}{3} = 6\text{ cm}\).
(c) Ratio of areas = \((\text{Ratio of corresponding sides})^2 = \left(\frac{PB}{PD}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 9 : 4\).

Teacher's Note:
a) Correct vertex matching is essential when writing similarity statements for triangles.
b) The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

(ii) Mr. Sam has a recurring deposit account and deposits ₹ 600 per month for 2 years. If he gets ₹ 15,600 at the time of maturity, find the rate of interest earned by him. [3 Marks]

Answer:
\(P = \text{₹ } 600\), \(n = 2\text{ years} = 24\text{ months}\), Maturity Value \((\text{MV}) = \text{₹ } 15,600\).
Total deposit = \(P \times n = 600 \times 24 = \text{₹ } 14,400\).
Interest \(I = \text{MV} - \text{Total deposit} = 15,600 - 14,400 = \text{₹ } 1,200\).
\(I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \Rightarrow 1200 = 600 \times \frac{24 \times 25}{24} \times \frac{r}{100}\)
\(1200 = 600 \times \frac{25}{2} \times \frac{r}{100} = 150r \Rightarrow r = \frac{1200}{150} = 8\%\).

Teacher's Note:
a) Interest is calculated as Maturity Value minus total money deposited.
b) Always convert the time period into months (\(n\) in months) for the recurring deposit interest formula.

(iii) Using step-deviation method, find mean for the following frequency distribution: [4 Marks]

Answer:

Assumed mean \(A = 52.5\), class size \(i = 15\).
Mean \(\bar{x} = A + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times i = 52.5 + \left(\frac{-12}{30}\right) \times 15 = 52.5 - 6 = 46.50\).

Teacher's Note:
a) Step-deviation method simplifies calculations by reducing large numbers into smaller integers using \(u_i = \frac{x_i - A}{i}\).
b) Ensure class intervals are continuous and uniform before applying the method.

Question 6

(i) Find the coordinates of the centroid P of the \(\Delta ABC\), whose vertices are \(A(-1, 3), B(3, -1)\) and \(C(0, 0)\). Hence, find the equation of a line passing through P and parallel to AB. [3 Marks]

Answer:
Centroid P = \(\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) = \left(\frac{-1 + 3 + 0}{3}, \frac{3 + (-1) + 0}{3}\right) = \left(\frac{2}{3}, \frac{2}{3}\right)\).
Slope of AB (\(m_{AB}\)) = \(\frac{-1 - 3}{3 - (-1)} = \frac{-4}{4} = -1\).
Since the required line is parallel to AB, its slope \(m = -1\).
Equation of line passing through \(P\left(\frac{2}{3}, \frac{2}{3}\right)\) with slope \(-1\):
\(y - \frac{2}{3} = -1\left(x - \frac{2}{3}\right) \Rightarrow y - \frac{2}{3} = -x + \frac{2}{3} \Rightarrow x + y = \frac{4}{3} \Rightarrow 3x + 3y = 4\).

Teacher's Note:
a) The centroid formula averages the x-coordinates and y-coordinates of the three triangle vertices.
b) Parallel lines share identical slopes.

(ii) In the given figure, the parallelogram ABCD circumscribe a circle, touching circle at P, Q, R and S.
(a) Prove that: AB = BC
(b) What special name can be given to the parallelogram ABCD? [3 Marks]

[Figure: Parallelogram ABCD circumscribing a circle, touching at P, Q, R, S on sides AB, BC, CD, DA respectively.]

Answer:
(a) Tangents drawn from an external point to a circle are equal in length.
\(AP = AS, BP = BQ, DR = DS, CR = CQ\).
Adding these: \((AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ) \Rightarrow AB + CD = AD + BC\).
Since ABCD is a parallelogram, \(AB = CD\) and \(AD = BC\), so \(2AB = 2BC \Rightarrow AB = BC\).
(b) Since adjacent sides of the parallelogram are equal (\(AB = BC\)), the parallelogram is a Rhombus.

Teacher's Note:
a) Tangent segments from a common external point are always equal.
b) A parallelogram with all sides equal is a rhombus.

(iii) The following bill shows the GST rate and the marked price of articles: [4 Marks]

Rajdhani Departmental Store

Find the total amount to be paid (including GST) for the above bill.

Answer:

Grand Total (Discounted Price + Total GST) = \(\text{₹ } 1836 + \text{₹ } 200.30 = \text{₹ } 2036.30\).

Teacher's Note:
a) Calculate the selling price for each item by subtracting the discount from the marked price.
b) GST is computed on each item's individual selling price and then summed up.

Question 7

(i) Draw the necessary diagram for this question.
A man on the top of a lighthouse observes the angle of depression of two ships on the opposite sides of the lighthouse as \(30^{\circ}\) and \(50^{\circ}\) respectively. If the height of the lighthouse is 80m, find the distance between the two ships. Give your answer correct to the nearest meter. (Use Mathematical Tables for this Question) [5 Marks]

[Figure: Lighthouse of height 80m with two ships on opposite sides at angles of depression \(30^{\circ}\) and \(50^{\circ}\).]

Answer:
Let height of lighthouse \(AB = 80\text{ m}\). Let the two ships be C and D on opposite sides.
In right \(\Delta ABC\): \(\tan 30^{\circ} = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{80}{BC} \Rightarrow BC = 80\sqrt{3} = 80 \times 1.7321 = 138.568\text{ m}\).
In right \(\Delta ABD\): \(\tan 50^{\circ} = \frac{AB}{BD} \Rightarrow BD = \frac{80}{\tan 50^{\circ}} = \frac{80}{1.1918} = 67.125\text{ m}\) (or using \(\cot 50^{\circ} = 0.8391\), \(BD = 80 \times 0.8391 = 67.128\text{ m}\)).
Total distance between ships \(CD = BC + BD = 138.568 + 67.128 = 205.696\text{ m} \approx 206\text{ m}\).

Teacher's Note:
a) Angle of elevation equals the angle of depression due to alternate interior angles.
b) Always round off the final answer to the requested precision (nearest meter).

(ii) The marks of 200 students in a test were recorded as follows: [5 Marks]

Using a graph sheet draw ogive for the given data and use it to find the:
(a) median.
(b) number of students who obtained more than 65% marks.
(c) number of students who did not pass, if the pass percentage was 35.

Answer:
Cumulative frequency table construction:
\(0-10: 5\), \(0-20: 12\), \(0-30: 23\), \(0-40: 43\), \(0-50: 83\), \(0-60: 135\), \(0-70: 171\), \(0-80: 186\), \(0-90: 195\), \(0-100: 200\).
(a) Median = Value at \(\frac{N}{2} = \frac{200}{2} = 100^{\text{th}}\) term \(\approx 53 \pm 1\).
(b) Students with more than 65% marks = Total students (\(200\)) - Cumulative frequency at 65% (\(\approx 154\)) = \(46 \pm 2\).
(c) Students who did not pass (below 35%) corresponds to cumulative frequency at 35 \(\approx 31 \pm 2\).

Teacher's Note:
a) An ogive is plotted using upper class boundaries against cumulative frequencies.
b) For "more than" type questions, subtract the cumulative frequency from the total frequency \(N\).

Question 8

(i) A box containing cards numbered between 0 and 100 are shuffled and a card is picked at random. Find the probability of getting a card which is:
(a) divisible by 6.
(b) not divisible by 6. [3 Marks]

Answer:
Total numbers between 0 and 100 (excluding 0 and 100, or inclusive? Standard interpretation for 1 to 99 is 99 cards; if inclusive of 1 to 99, \(N = 99\)).
Multiples of 6 between 1 and 99: \(\{6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96\}\) (Total 16 cards).
(a) \(P(\text{divisible by } 6) = \frac{16}{99}\).
(b) \(P(\text{not divisible by } 6) = 1 - \frac{16}{99} = \frac{83}{99}\).

Teacher's Note:
a) Probability equals favourable outcomes divided by total possible outcomes.
b) The sum of probabilities of complementary events is always equal to 1.

(ii) If \(x, y\) and \(z\) are in continued proportion, prove that:
\(\frac{x}{y^2 \cdot z^2} + \frac{y}{z^2 \cdot x^2} + \frac{z}{x^2 \cdot y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}\). [3 Marks]

Answer:
Since \(x, y, z\) are in continued proportion, \(\frac{x}{y} = \frac{y}{z} \Rightarrow y^2 = xz\).
\(\text{LHS} = \frac{x}{y^2z^2} + \frac{y}{z^2x^2} + \frac{z}{x^2y^2} = \frac{x^3 + y^3 + z^3}{x^2y^2z^2}\)
Substitute \(y^2 = xz\): denominator becomes \(x^2(xz)z^2 = x^3z^3\).
\(\text{LHS} = \frac{x^3 + y^3 + z^3}{x^3z^3} = \frac{x^3}{x^3z^3} + \frac{y^3}{x^3z^3} + \frac{z^3}{x^3z^3} = \frac{1}{z^3} + \frac{y^3}{(xz)^3} + \frac{1}{x^3} = \frac{1}{z^3} + \frac{y^3}{(y^2)^3} + \frac{1}{x^3} = \frac{1}{z^3} + \frac{y^3}{y^6} + \frac{1}{x^3} = \frac{1}{z^3} + \frac{1}{y^3} + \frac{1}{x^3} = \text{RHS}\).

Teacher's Note:
a) Continued proportion means \(y^2 = xz\) or \(y = \sqrt{xz}\).
b) Simplifying algebraic fractions requires careful substitution of proportional relationships.

(iii) A manufacturing company prepares spherical ball bearings, each of radius 7 mm and mass 4 gm. These ball bearings are packed into boxes. Each box can have a maximum of \(2156\text{ cm}^3\) of ball bearings. Find the:
(a) maximum number of ball bearings that each box can have.
(b) mass of each box of ball bearings in kg.
\(\left(\text{Use }\pi = \frac{22}{7}\right)\). [4 Marks]

Answer:
Radius of each ball bearing \(r = 7\text{ mm} = 0.7\text{ cm} = \frac{7}{10}\text{ cm}\).
Volume of one spherical ball bearing = \(\frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{10}\right)^3\text{ cm}^3\).
(a) Maximum number of ball bearings = \(\frac{\text{Volume of box}}{\text{Volume of 1 ball bearing}} = \frac{2156}{\frac{4}{3} \times \frac{22}{7} \times \frac{7}{10} \times \frac{7}{10} \times \frac{7}{10}}\)
\(= \frac{2156 \times 3 \times 7 \times 10 \times 10 \times 10}{4 \times 22 \times 7 \times 7 \times 7} = 1500\text{ bearings}\).
(b) Total mass in grams = \(1500 \times 4\text{ gm} = 6000\text{ gm} = 6\text{ kg}\).

Teacher'sNote:
a) Consistent units are vital: convert millimeters to centimeters before calculating volume in cubic centimeters.
b) Convert grams to kilograms by dividing by 1000.

Question 9

(i) Study the graph given below and answer the following:
(a) Number of batsmen who scored 500 to 700 runs
(b) Modal class interval
(c) The value of mode [3 Marks]

[Figure: Histogram showing number of batsmen on y-axis and number of runs on x-axis with scale: x-axis 2cm = 100 runs, y-axis 2cm = 02 batsmen.]

Answer:
(a) Number of batsmen who scored 500 to 700 runs = Frequency of \((500-600)\) + Frequency of \((600-700) = 3 + 2 = 5\).
(b) Modal class interval = \(400 - 500\) (highest bar in the histogram).
(c) The value of mode \(= 430\) runs (determined graphically from the highest rectangle).

Teacher's Note:
a) Mode represents the most frequent observation and corresponds to the highest bar in a histogram.
b) To find the exact mode graphically, draw cross-lines from the top corners of the modal rectangle to adjacent rectangles.

(ii) An Arithmetic Progression (A.P.) has 3 as its first term. The sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference of the A.P. [3 Marks]

Answer:
\(a = 3\). Given \(S_8 = 2S_5\).
\(\frac{8}{2}[2(3) + (8 - 1)d] = 2 \times \frac{5}{2}[2(3) + (5 - 1)d]\)
\(\Rightarrow 4[6 + 7d] = 5[6 + 4d]\)
\(\Rightarrow 24 + 28d = 30 + 20d\)
\(\Rightarrow 28d - 20d = 30 - 24 \Rightarrow 8d = 6 \Rightarrow d = \frac{6}{8} = \frac{3}{4}\ (\text{or } 0.75)\).

Teacher's Note:
a) Use the sum formula for an A.P.: \(S_n = \frac{n}{2}[2a + (n-1)d]\).
b) Formulate equations carefully according to the given conditional statements.

(iii) The roots of equation \((q - r)x^2 + (r - p)x + (p - q) = 0\) are equal. Prove that: \(2q = p + r\), that is, \(p, q \& r\) are in A.P. [4 Marks]

Answer:
Let coefficients be \(A = q - r\), \(B = r - p\), \(C = p - q\).
Since roots are equal, discriminant \(B^2 - 4AC = 0 \Rightarrow (r - p)^2 = 4(q - r)(p - q)\).
Expanding both sides:
\(r^2 + p^2 - 2pr = 4(qp - q^2 - rp + rq) = 4qp - 4q^2 - 4rp + 4rq\)
\(r^2 + p^2 + 2pr = 4qp - 4q^2 + 4rq \Rightarrow (p + r)^2 = 4q(p + r) - 4q^2\)
Let \((p + r) = y\):
\(y^2 - 4qy + 4q^2 = 0 \Rightarrow (y - 2q)^2 = 0 \Rightarrow y - 2q = 0 \Rightarrow y = 2q \Rightarrow p + r = 2q\).
This proves that \(p, q, r\) are in Arithmetic Progression.

Teacher's Note:
a) Equal roots of a quadratic equation imply that the discriminant (\(b^2 - 4ac\)) is equal to zero.
b) Algebraic manipulation and substitution help simplify symmetrical quadratic relations.

Question 10

(i) The sum of the squares of three consecutive even numbers is 596. Find the numbers. [3 Marks]

Answer:
Let the three consecutive even numbers be \((x - 2), x, (x + 2)\).
\((x - 2)^2 + x^2 + (x + 2)^2 = 596\)
\(\Rightarrow x^2 - 4x + 4 + x^2 + x^2 + 4x + 4 = 596\)
\(\Rightarrow 3x^2 + 8 = 596 \Rightarrow 3x^2 = 588 \Rightarrow x^2 = 196 \Rightarrow x = 14\).
The numbers are \(14 - 2 = 12\), \(14\), and \(14 + 2 = 16\).
Required numbers are \(12, 14\text{ and } 16\).

Teacher's Note:
a) Representing consecutive even numbers as \((x-2, x, x+2)\) simplifies algebraic expansion.
b) Discard negative roots if the context specifies positive integers.

(ii) Given matrix, \(X = \begin{bmatrix}1 & 1 \\ 8 & 3\end{bmatrix}\) and \(I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\), prove that \(X^2 = 4X + 5I\). [3 Marks]

Answer:
\(X^2 = \begin{bmatrix}1 & 1 \\ 8 & 3\end{bmatrix} \begin{bmatrix}1 & 1 \\ 8 & 3\end{bmatrix} = \begin{bmatrix}(1)(1) + (1)(8) & (1)(1) + (1)(3) \\ (8)(1) + (3)(8) & (8)(1) + (3)(3)\end{bmatrix}\)
\(= \begin{bmatrix}1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9\end{bmatrix} = \begin{bmatrix}9 & 4 \\ 32 & 17\end{bmatrix}\).
Now, \(4X + 5I = 4\begin{bmatrix}1 & 1 \\ 8 & 3\end{bmatrix} + 5\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}4 & 4 \\ 32 & 12\end{bmatrix} + \begin{bmatrix}5 & 0 \\ 0 & 5\end{bmatrix} = \begin{bmatrix}4+5 & 4+0 \\ 32+0 & 12+5\end{bmatrix} = \begin{bmatrix}9 & 4 \\ 32 & 17\end{bmatrix}\).
Since \(LHS = RHS\), \(X^2 = 4X + 5I\) is proved.

Teacher's Note:
a) Matrix multiplication is performed row-by-column.
b) Scalar multiplication multiplies every element inside the matrix by the given scalar.

(iii) Use a graph sheet for this question. Take 1 cm = 1 unit along both the x and y axis. Plot ABCDE, where A (4, 0), B (4, 2), C (2, 2), D (2, 4) and E (0, 4).
(a) Reflect the points A, B, C and D on the y-axis and name them as F, G, H and I respectively.
(b) Join the points A, B, C, D, E, I, H, G and F in order. Reflect the figure ABCDEIHGF on the x-axis and name it as AMNPQRSTF.
(c) Give the geometrical name of the closed figure AEFQ. [4 Marks]

[Figure: Graph showing plotted coordinates, reflections across y-axis and x-axis forming a symmetrical polygon, with closed figure AEFQ highlighted as a square.]

Answer:
(a) Reflections on the y-axis \((x, y) \rightarrow (-x, y)\):
\(F(-4, 0), G(-4, 2), H(-2, 2), I(-2, 4)\).
(b) Reflected figure plotted across x-axis \((x, y) \rightarrow (x, -y)\) to form AMNPQRSTF.
(c) Geometrical name of the closed figure AEFQ: Square.

Teacher's Note:
a) Reflection in the y-axis changes the sign of the x-coordinate (\((-x, y)\)).
b) Reflection in the x-axis changes the sign of the y-coordinate (\((x, -y)\)).