ICSE Class 10 Mathematics Sample Paper 2024 with Solutions

SECTION A

Question 1
Choose the correct answers to the questions from the given options. [15]
(Do not copy the question, write the correct answers only.)

(i) If \( A = \begin{bmatrix} -1 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} \) [1 Mark]
Which of the following operation is possible?
(a) A - B
(b) A + B
(c) AB
(d) BA

Answer: (d) BA

Order of matrix \( A \) is \( 1 \times 2 \) and order of matrix \( B \) is \( 2 \times 2 \). For matrix multiplication \( BA \), the number of columns in \( B \) (which is 2) is equal to the number of rows in \( A \) (which is 2), so the product \( BA \) is defined.

Teacher's Note:
a) Two matrices can be added or subtracted only if they are of the exact same order, which is not true for \( A \) and \( B \).
b) For matrix multiplication \( AB \), the inner dimensions (columns of \( A \) and rows of \( B \)) must match, i.e., \( 2 \neq 2 \) for \( AB \) since \( 2 \times 2 \) and \( 1 \times 2 \); hence only \( BA \) (\( 2 \times 2 \) multiplied by \( 2 \times 1 \)) is possible.

(ii) If \( x^{2} + kx + 6 = (x - 2)(x - 3) \) for all values of x, then the value of k is: [1 Mark]
(a) -5
(b) -3
(c) -2
(d) 5

Answer: (a) -5

Expanding the RHS: \( (x - 2)(x - 3) = x^{2} - 3x - 2x + 6 = x^{2} - 5x + 6 \). Comparing with \( x^{2} + kx + 6 \), we get \( k = -5 \).

Teacher's Note:
a) Expand the product of linear factors and equate coefficients of like powers of \( x \).
b) Students often miss the negative sign when combining the middle terms.

(iii) A retailer purchased an item for Rs. 1500 from a wholesaler and sells it to a customer at 10% profit. The sales are intra-state and the rate of GST is 10%. The amount of GST paid by the customer: [1 Mark]
(a) Rs. 15
(b) Rs. 30
(c) Rs. 150
(d) Rs. 165

Answer: (d) Rs. 165

Selling price for the retailer = \( \text{Rs. } 1500 + 10\% \text{ of } 1500 = \text{Rs. } 1500 + \text{Rs. } 150 = \text{Rs. } 1650 \). Total GST paid by the customer = \( 10\% \text{ of } \text{Rs. } 1650 = \text{Rs. } 165 \).

Teacher's Note:
a) The customer pays GST on the final selling price of the item, which includes the retailer's profit.
b) Do not calculate GST only on the wholesaler's cost price.

(iv) If the roots of equation \( x^{2} - 6x + k = 0 \) are real and distinct, then value of k is: [1 Mark]
(a) \( \gt -9 \)
(b) \( \gt -6 \)
(c) \( \lt 6 \)
(d) \( \lt 9 \)

Answer: (c) \( \lt 6 \)

For real and distinct roots, discriminant \( D \gt 0 \). Here, \( D = b^{2} - 4ac = (-6)^{2} - 4(1)(k) = 36 - 4k \). Therefore, \( 36 - 4k \gt 0 \implies 4k \lt 36 \implies k \lt 9 \).

Teacher's Note:
a) Remember the conditions for roots: real and distinct means \( D \gt 0 \), real and equal means \( D = 0 \), and not real means \( D \lt 0 \).
b) Watch out for inequality sign reversal when dividing by a negative number.

(v) Which of the following is/are an Arithmetic Progression (A.P.)? [1 Mark]
1. \( 1, 4, 9, 16, \dots \)
2. \( \sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}, \dots \)
3. \( 8, 6, 4, 2, \dots \)
(a) only 1.
(b) only 2.
(c) only 2. and 3.
(d) all 1., 2. and 3.

Answer: (c) only 2. and 3.

Sequence 1 has differences \( 3, 5, 7 \) (not constant). Sequence 2 has a common difference of \( \sqrt{3} \). Sequence 3 has a common difference of \( -2 \).

Teacher's Note:
a) An A.P. requires a constant common difference between consecutive terms.
b) Sequence 1 is the sequence of squares of natural numbers, which forms a quadratic progression.

(vi) The table shows the values of x and y, where x is proportional to y. [1 Mark]

What are the values of M and N?
(a) M = 4, N = 9
(b) M = 9, N = 3
(c) M = 9, N = 4
(d) M = 12, N = 0

Answer: (b) M = 9, N = 3

Since \( x \) is proportional to \( y \), \( \frac{x}{y} = k \). Using the given pair \( x = 12 \) and \( y = 18 \), \( k = \frac{12}{18} = \frac{2}{3} \). For \( x = 6 \), \( \frac{6}{M} = \frac{2}{3} \implies M = 9 \). For \( y = 6 \), \( \frac{N}{6} = \frac{2}{3} \implies N = 4 \).

[Note: Calculating N: \( \frac{N}{6} = \frac{2}{3} \implies N = 4 \). Hence option (c) is the correct pair.]

Answer: (c) M = 9, N = 4

Teacher's Note:
a) Direct proportionality implies constant ratio \( \frac{x}{y} \).
b) Check the calculated values by substituting them back into the ratio.

(vii) In the given diagram, \( \Delta ABC \sim \Delta PQR \) and \( \frac{AD}{PS} = \frac{3}{8} \). The value of AB : PQ is: [1 Mark]
(a) 8 : 3
(b) 3 : 5
(c) 3 : 8
(d) 5 : 8

[Figure: Two similar triangles ABC and PQR with corresponding altitudes AD and PS respectively, where AD is perpendicular to BC and PS is perpendicular to QR.]

Answer: (c) 3 : 8

The ratio of corresponding sides of similar triangles is equal to the ratio of their corresponding altitudes. Therefore, \( \frac{AB}{PQ} = \frac{AD}{PS} = \frac{3}{8} \).

Teacher's Note:
a) Recall that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides, but the ratio of altitudes is equal to the ratio of corresponding sides.
b) Maintain the correct order of vertices.

(viii) A right angle triangle shaped piece of hard board is rotated completely about its hypotenuse, as shown in the diagram. The solid so formed is always: [1 Mark]
1. a single cone
2. a double cone
Which of the statement is valid?
(a) only 1.
(b) only 2.
(c) both 1. and 2.
(d) neither 1. nor 2.

[Figure: A right-angled triangle with a curved arrow showing complete rotation about its hypotenuse, forming two joined circular cones sharing a common base.]

Answer: (b) only 2.

Rotating a right-angled triangle about its hypotenuse produces two cones joined base-to-base, which forms a double cone.

Teacher's Note:
a) Visualising solid shapes formed by revolving plane figures helps in mensuration problems.
b) The common base of the two cones is the perpendicular dropped from the opposite vertex onto the hypotenuse.

(ix) Event A: The sun will rise from east tomorrow.
Event B: It will rain on Monday.
Event C: February month has 29 days in a leap year.
Which of the above event(s) has probability equal to 1? [1 Mark]
(a) all events A, B and C
(b) both events A and B
(c) both events B and C
(d) both events A and C

Answer: (d) both events A and C

Event A is a sure event (probability 1). Event C is also a sure event since a leap year always has 29 days in February (probability 1). Event B is an uncertain event.

Teacher's Note:
a) A sure or certain event always has a probability of 1.
b) An impossible event has a probability of 0.

(x) The three vertices of a scalene triangle are always equidistant from a fixed point. The point is: [1 Mark]
(a) Orthocentre of the triangle.
(b) Incentre of the triangle.
(c) Circumcentre of the triangle.
(d) Centroid of the triangle.

Answer: (c) Circumcentre of the triangle.

The circumcentre of a triangle is equidistant from all three vertices, as it is the centre of the circumcircle passing through the vertices.

Teacher's Note:
a) Incentre is equidistant from the sides of the triangle.
b) Circumcentre is equidistant from the vertices of the triangle.

(xi) In a circle with radius R, the shortest distance between two parallel tangents is equal to: [1 Mark]
(a) R
(b) 2R
(c) \( 2\pi R \)
(d) \( \pi R \)

Answer: (b) 2R

The shortest distance between two parallel tangents of a circle is the length of the segment joining their points of contact passing through the centre, which is equal to the diameter (\( 2R \)).

Teacher's Note:
a) Parallel tangents to a circle can only exist at the opposite ends of a diameter.
b) The distance between them is the diameter of the circle.

(xii) An observer at point E, which is at a certain distance from the lamp post AB, finds the angle of elevation of top of lamp post from positions C, D and E as \( \alpha \), \( \beta \) and \( \gamma \). It is given that B, C, D and E are along a straight line.
Which of the following condition is satisfied? [1 Mark]
(a) \( \tan\alpha \gt \tan\beta \)
(b) \( \tan\beta \lt \tan\gamma \)
(c) \( \tan\gamma \gt \tan\alpha \)
(d) \( \tan\alpha \lt \tan\beta \)

[Figure: A vertical lamp post AB with points C, D, E on a horizontal line away from B. Angles of elevation from C, D, E are \( \alpha \), \( \beta \), \( \gamma \) respectively.]

Answer: (a) \( \tan\alpha \gt \tan\beta \)

As an observer moves away from the base of a tower, the angle of elevation decreases. Since point C is closer to B than point D, \( \alpha \gt \beta \), which implies \( \tan\alpha \gt \tan\beta \) (since tangent is an increasing function in \( (0^{\circ}, 90^{\circ}) \)).

Teacher's Note:
a) The closer the observer is to the vertical object, the greater the angle of elevation.
b) The angles decrease in magnitude as the distance from the base increases: \( \alpha \gt \beta \gt \gamma \).

(xiii) 1. Shares of company A, paying 12%, Rs. 100 shares are at Rs. 80.
2. Shares of company B, paying 12%, Rs. 100 shares at Rs. 100.
3. Shares of company C, paying 12%, Rs. 100 shares are at Rs. 120.
Shares of which company are at premium? [1 Mark]
(a) Company A
(b) Company B
(c) Company C
(d) Company A and C

Answer: (c) Company C

Shares are said to be at a premium when their market value is greater than their nominal (face) value. Here, Company C has a market value of Rs. 120 against a face value of Rs. 100.

Teacher's Note:
a) Market value \( \gt \) Face value means shares are at a premium.
b) Market value \( \lt \) Face value means shares are at a discount.

(xiv) Which of the following equation represent a line passing through origin? [1 Mark]
(a) \( 3x - 2y + 5 = 0 \)
(b) \( 2x - 3y = 0 \)
(c) \( x = 5 \)
(d) \( y = -6 \)

Answer: (b) \( 2x - 3y = 0 \)

A line passes through the origin if its equation is satisfied by the point \( (0, 0) \). Substituting \( x = 0 \) and \( y = 0 \) into \( 2x - 3y = 0 \) gives \( 0 = 0 \).

Teacher's Note:
a) Equations of lines passing through the origin have no constant term (i.e., of the form \( y = mx \)).
b) Always test the point \( (0, 0) \) to verify if a line passes through the origin.

(xv) For the given 25 variables: \( x_{1}, x_{2}, x_{3}, \dots, x_{25} \)
Assertion (A): To find median of the given data, the variate needs to be arranged in ascending or descending order.
Reason (R): The median is the central most term of the arranged data. [1 Mark]
(a) A is true, R is false
(b) A is false, R is true
(c) both A and R are true
(d) both A and R are false

Answer: (c) both A and R are true

Arranging data in order of magnitude is a necessary first step to identify the middle observation, which defines the median.

Teacher's Note:
a) Never calculate the median of raw data without sorting it first.
b) Reason (R) correctly explains why the sorting in Assertion (A) is required.

Question 2

(i) Shown below is a horizontal water tank composed of a cylinder and two hemispheres. The tank is filled up to a height of 7 m. Find the surface area of the tank in contact with water. Use \( \pi = \frac{22}{7} \). [4 Marks]

[Figure: A horizontal cylindrical tank of length 34 m with hemispherical ends. Total height/diameter is 14 m, and water fills it up to a height of 7 m.]

Answer:
1. Radius of the cylinder and hemispheres, \( r = \frac{14}{2} = 7 \text{ m} \).
2. Total length of the tank \( = 34 \text{ m} \).
3. Length of the cylindrical part \( h = 34 - 7 - 7 = 20 \text{ m} \).
4. Since the tank is filled up to a height of 7 m (which is the radius, meaning it is half-filled horizontally), the inner wetted surface consists of:
- Half of the curved surface area of the cylinder.
- The curved surface area of one full sphere (since two half-hemispheres filled to half form one full sphere, or lower half of both hemispheres makes a full spherical surface in contact with water).
5. Wetted curved surface area of cylinder \( = \frac{1}{2} \times (2\pi rh) = \pi rh = \frac{22}{7} \times 7 \times 20 = 440 \text{ m}^{2} \).
6. Wetted surface area of two hemispheres \( = 4\pi r^{2} \) (lower half of two hemispheres equals full sphere surface area) \( = 4 \times \frac{22}{7} \times 7^{2} = 616 \text{ m}^{2} \).
7. Total surface area in contact with water \( = 440 + 616 = 1056 \text{ m}^{2} \).

Teacher's Note:
a) Carefully determine which parts of the solid are in contact with water when it is half-filled horizontally.
b) Note that two hemispheres filled up to the middle radius combined form a complete sphere's curved surface area.

(ii) In a recurring deposit account for 2 years, the total amount deposited by a person is Rs. 9600. If the interest earned by him is one-twelfth of his total deposit, then find:
(a) the interest he earns.
(b) his monthly deposit.
(c) the rate of interest. [4 Marks]

Answer:
1. Time period \( n = 2 \text{ years} = 24 \text{ months} \).
2. Total amount deposited \( = \text{Rs. } 9600 \).
(a) Interest earned \( I = \frac{1}{12} \times \text{Total deposit} = \frac{1}{12} \times 9600 = \text{Rs. } 800 \).
(b) Monthly deposit \( P = \frac{\text{Total deposit}}{n} = \frac{9600}{24} = \text{Rs. } 400 \).
(c) Using the interest formula for recurring deposit:
\( I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100} \)
\( 800 = 400 \times \frac{24 \times 25}{24} \times \frac{R}{100} \)
\( 800 = 400 \times \frac{25}{2} \times \frac{R}{100} = 50 R \)
\( R = \frac{800}{50} = 16\% \).

Teacher's Note:
a) Ensure time \( n \) is always converted to months for recurring deposit formula.
b) Verify the relation between total deposit, monthly deposit, and number of months.

(iii) Find:
(a) \( (\sin\theta + \csc\theta)^{2} \)
(b) \( (\cos\theta + \sec\theta)^{2} \)
Using the above results prove the following trigonometry identity.
\( (\sin\theta + \csc\theta)^{2} + (\cos\theta + \sec\theta)^{2} = 7 + \tan^{2}\theta + \cot^{2}\theta \) [4 Marks]

Answer:
(a) \( (\sin\theta + \csc\theta)^{2} = \sin^{2}\theta + \csc^{2}\theta + 2\sin\theta\csc\theta = \sin^{2}\theta + \csc^{2}\theta + 2 \) (since \( \csc\theta = \frac{1}{\sin\theta} \)).
(b) \( (\cos\theta + \sec\theta)^{2} = \cos^{2}\theta + \sec^{2}\theta + 2\cos\theta\sec\theta = \cos^{2}\theta + \sec^{2}\theta + 2 \) (since \( \sec\theta = \frac{1}{\cos\theta} \)).
Adding both results:
LHS \( = (\sin^{2}\theta + \csc^{2}\theta + 2) + (\cos^{2}\theta + \sec^{2}\theta + 2) \)
\( = (\sin^{2}\theta + \cos^{2}\theta) + \csc^{2}\theta + \sec^{2}\theta + 4 \)
\( = 1 + (1 + \cot^{2}\theta) + (1 + \tan^{2}\theta) + 4 \) (using \( \csc^{2}\theta = 1 + \cot^{2}\theta \) and \( \sec^{2}\theta = 1 + \tan^{2}\theta \))
\( = 1 + 1 + \cot^{2}\theta + 1 + \tan^{2}\theta + 4 = 7 + \tan^{2}\theta + \cot^{2}\theta = \) RHS.

Teacher's Note:
a) Use algebraic identity \( (a + b)^{2} = a^{2} + b^{2} + 2ab \) and reciprocal trigonometric relations.
b) Substitute fundamental Pythagorean identities \( \sin^{2}\theta + \cos^{2}\theta = 1 \), \( 1 + \cot^{2}\theta = \csc^{2}\theta \), and \( 1 + \tan^{2}\theta = \sec^{2}\theta \).

Question 3

(i) If a, b and c are in continued proportion, then prove that: [4 Marks]
\( \frac{3a^{2} + 5ab + 7b^{2}}{3b^{2} + 5bc + 7c^{2}} = \frac{a}{c} \)

Answer:
1. Since \( a, b, c \) are in continued proportion, \( \frac{a}{b} = \frac{b}{c} = k \), which implies \( b = ck \) and \( a = bk = ck^{2} \).
2. Substitute \( a = ck^{2} \) and \( b = ck \) into the LHS:
Numerator \( = 3(ck^{2})^{2} + 5(ck^{2})(ck) + 7(ck)^{2} \)
\( = 3c^{2}k^{4} + 5c^{2}k^{3} + 7c^{2}k^{2} = c^{2}k^{2}(3k^{2} + 5k + 7) \)
Denominator \( = 3(ck)^{2} + 5(ck)(c) + 7c^{2} \)
\( = 3c^{2}k^{2} + 5c^{2}k + 7c^{2} = c^{2}(3k^{2} + 5k + 7) \)
3. Ratio \( = \frac{c^{2}k^{2}(3k^{2} + 5k + 7)}{c^{2}(3k^{2} + 5k + 7)} = k^{2} \).
4. RHS \( = \frac{a}{c} = \frac{ck^{2}}{c} = k^{2} \).
Hence, LHS = RHS.

Teacher's Note:
a) Express both \( a \) and \( b \) in terms of \( c \) and common ratio \( k \).
b) Factor out common terms from numerator and denominator to simplify.

(ii) In the given diagram, O is the centre of circle circumscribing the \( \Delta ABC \). CD is perpendicular to chord AB. \( \angle OAC = 32^{\circ} \). Find each of the unknown angles x, y and z. [4 Marks]

[Figure: A circle with centre O circumscribing triangle ABC. Chord AB has perpendicular CD from C to AB. Angle OAC is \( 32^{\circ} \), angle at O is \( x^{\circ} \), angle B is \( y^{\circ} \), and angle at C formed by radii/chords is \( z^{\circ} \).]

Answer:
1. In \( \Delta OAC \), \( OA = OC \) (radii of the same circle). Therefore, \( \angle OCA = \angle OAC = 32^{\circ} \).
2. Angle \( \angle AOC = 180^{\circ} - (32^{\circ} + 32^{\circ}) = 180^{\circ} - 64^{\circ} = 116^{\circ} \). Thus, \( x = 116 \).
3. The angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle. Therefore, \( \angle B = \frac{1}{2}\angle AOC = \frac{116^{\circ}}{2} = 58^{\circ} \). Thus, \( y = 58 \).
4. In right-angled triangle \( CDB \) (where \( CD \perp AB \)), \( \angle BCD = 90^{\circ} - \angle B = 90^{\circ} - 58^{\circ} = 32^{\circ} \). Thus, \( z = 32 \).

Teacher's Note:
a) Use properties of isosceles triangles formed by radii.
b) Apply the theorem relating the angle at the centre to the angle at the circumference.

(iii) Study the graph and answer each of the following:
(a) Name the curve plotted
(b) Total number of students
(c) The median marks
(d) Number of students scoring between 50 and 80 marks [5 Marks]

[Figure: An ogive graph showing cumulative frequency on the y-axis (0 to 45) and marks on the x-axis (30 to 100), with plotted points A(40,3), B(50,12), C(60,24), D(70,33), E(80,37), F(90,39), G(100,40).]

Answer:
(a) Cumulative frequency curve or Ogive (specifically, less than cumulative frequency curve).
(b) Total number of students = 40 (highest cumulative frequency at 100 marks).
(c) Median marks: Total frequency \( N = 40 \). \( \frac{N}{2} = \frac{40}{2} = 20 \). Locating 20 on the y-axis and finding the corresponding x-value on the curve gives median marks \( = 57 \).
(d) Number of students scoring between 50 and 80 marks = Cumulative frequency at 80 marks minus cumulative frequency at 50 marks \( = 37 - 12 = 25 \).

Teacher's Note:
a) An ogive is always a smooth cumulative frequency curve.
b) Always show dotted lines from the axes on the graph to justify the readings for median and cumulative frequencies.

SECTION B

(Attempt any four questions from this Section.)

Question 4

(i) If \( A = \begin{bmatrix} 4 & -4 \\ -4 & 4 \end{bmatrix} \), find \( A^{2} \). If \( A^{2} = p A \), then find the value of p. [3 Marks]

Answer:
1. \( A^{2} = A \times A = \begin{bmatrix} 4 & -4 \\ -4 & 4 \end{bmatrix} \begin{bmatrix} 4 & -4 \\ -4 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} (4)(4) + (-4)(-4) & (4)(-4) + (-4)(4) \\ (-4)(4) + (4)(-4) & (-4)(-4) + (4)(4) \end{bmatrix} \)
\( = \begin{bmatrix} 16 + 16 & -16 - 16 \\ -16 - 16 & 16 + 16 \end{bmatrix} = \begin{bmatrix} 32 & -32 \\ -32 & 32 \end{bmatrix} \)
2. Given \( A^{2} = pA \):
\( \begin{bmatrix} 32 & -32 \\ -32 & 32 \end{bmatrix} = p \begin{bmatrix} 4 & -4 \\ -4 & 4 \end{bmatrix} = \begin{bmatrix} 4p & -4p \\ -4p & 4p \end{bmatrix} \)
3. Comparing corresponding elements: \( 4p = 32 \implies p = 8 \).

Teacher's Note:
a) Follow row-by-column matrix multiplication carefully.
b) Equate any corresponding non-zero element to find the scalar multiplier \( p \).

(ii) Solve the given equation \( x^{2} - 4x - 2 = 0 \) and express your answer correct to two places of decimal.
(You may use mathematical tables for this question). [3 Marks]

Answer:
1. Comparing \( x^{2} - 4x - 2 = 0 \) with standard quadratic equation \( ax^{2} + bx + c = 0 \), we get \( a = 1, b = -4, c = -2 \).
2. Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \):
\( x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(1)(-2)}}{2(1)} \)
\( x = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2} \)
3. Since \( \sqrt{24} = 2\sqrt{6} \approx 2(2.449) = 4.898 \):
\( x = \frac{4 \pm 4.898}{2} = 2 \pm 2.449 \)
4. Roots are:
\( x = 2 + 2.449 = 4.449 \approx 4.45 \)
\( x = 2 - 2.449 = -0.449 \approx -0.45 \).

Teacher's Note:
a) Always write the formula clearly before substituting values.
b) Round off to two decimal places only in the final step.

(iii) In the given diagram, \( \Delta ABC \) is right angled at B. BDFE is a rectangle.
AD = 6 cm, CE = 4 cm and BC = 12 cm
(a) prove that \( \Delta ADF \sim \Delta FEC \)
(b) prove that \( \Delta ADF \sim \Delta ABC \)
(c) find the length of FE
(d) find area \( \Delta ADF \) : area \( \Delta ABC \) [4 Marks]

[Figure: Right-angled triangle ABC with B at right angle. Rectangle BDFE inscribed with D on BC and E on BC/AC, F on AC.]

Answer:
(a) In \( \Delta ADF \) and \( \Delta FEC \):
- Since BDFE is a rectangle, \( \angle FDA = \angle C \) (corresponding angles, as \( FD \parallel BE \) or \( FD \parallel AB \)).
- Also, \( \angle FAD = \angle CFE \) (corresponding angles).
- By AA similarity, \( \Delta ADF \sim \Delta FEC \).
(b) In \( \Delta ADF \) and \( \Delta ABC \):
- \( \angle ADF = \angle B = 90^{\circ} \) (since BDFE is a rectangle).
- \( \angle FAD = \angle CAB \) (common angle).
- By AA similarity, \( \Delta ADF \sim \Delta ABC \).
(c) Let \( FE = x \). Since \( \Delta ADF \sim \Delta FEC \), corresponding sides are proportional:
\( \frac{AD}{FE} = \frac{DF}{EC} \)
Also, since BDFE is a rectangle, \( DF = BE = BC - CE = 12 - 4 = 8 \text{ cm} \).
Thus, \( \frac{6}{x} = \frac{8}{4} \implies 8x = 24 \implies x = 3 \text{ cm} \). Length of \( FE = 3 \text{ cm} \).
(d) Ratio of areas of similar triangles is equal to the square of the ratio of their corresponding sides:
\( \frac{\text{Area}(\Delta ADF)}{\text{Area}(\Delta ABC)} = \left(\frac{AD}{AB}\right)^{2} \)
Here, \( AB = BD + DA \). Since \( BD = FE = 3 \text{ cm} \) (opposite sides of rectangle), \( AB = 3 + 6 = 9 \text{ cm} \).
Therefore, ratio \( = \left(\frac{6}{9}\right)^{2} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9} \) or \( 4 : 9 \).

Teacher's Note:
a) Use properties of rectangles (opposite sides equal and parallel) to establish angle equalities.
b) Remember that the ratio of areas of two similar triangles equals the square of the ratio of any pair of corresponding sides.

Question 5

(i) Shown below is a table illustrating the monthly income distribution of a company with 100 employees. [3 Marks]

Using step-deviation method, find the mean monthly income of an employee.

Answer:

Class width \( h = 4 \), Assumed mean \( A = 10 \).
Mean \( \bar{x} = A + \left(\frac{\sum f_{i}u_{i}}{\sum f_{i}}\right) \times h \)
\( \bar{x} = 10 + \left(\frac{-81}{100}\right) \times 4 = 10 - 3.24 = 6.76 \)
Since income is in units of Rs. 10,000, mean monthly income \( = 6.76 \times 10,000 = \text{Rs. } 67,600 \).

Teacher's Note:
a) Clearly state the assumed mean and class width in the step-deviation method table.
b) Do not forget to multiply the final coded mean by the class size and scale factor.

(ii) The following bill shows the GST rate and the marked price of articles: [3 Marks]

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

Answer:
1. For LED TV set:
- Marked Price \( = \text{Rs. } 12000 \)
- GST amount \( = 28\% \text{ of } 12000 = \frac{28}{100} \times 12000 = \text{Rs. } 3360 \)
- Total price for LED TV \( = 12000 + 3360 = \text{Rs. } 15360 \).
2. For MP4 player:
- Marked Price \( = \text{Rs. } 5000 \)
- GST amount \( = 18\% \text{ of } 5000 = \frac{18}{100} \times 5000 = \text{Rs. } 900 \)
- Total price for MP4 player \( = 5000 + 900 = \text{Rs. } 5900 \).
3. Total amount to be paid (including GST) \( = 15360 + 5900 = \text{Rs. } 21,260 \).

Teacher's Note:
a) Calculate GST separately for each item according to its specified tax rate.
b) Add the individual total amounts (Price + GST) to get the final bill amount.

(iii) In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If \( \angle PQB = 55^{\circ} \).
(a) find the value of the angles x, y and z.
(b) prove that RB is parallel to PQ. [4 Marks]

[Figure: A circle with centre O and tangent AB at B. Chord PQ and secant/chord PR. Angle PQB is \( 55^{\circ} \), angle at R is \( x^{\circ} \), angle at centre or subtended is \( y^{\circ} \), angle at S is \( z^{\circ} \).]

Answer:
(a) - Angle in the alternate segment theorem: The angle between tangent AB and chord BQ equals the angle subtended by chord BQ in the alternate segment, so \( x = \angle PQB = 55^{\circ} \).
- Considering chord PB subtending angle \( x \) at R and angle at centre O being \( y \): \( y = 2x = 2 \times 55^{\circ} = 110^{\circ} \).
- In triangle formed by radius and chords or cyclic properties, \( z \) (angle related to the remaining segments) can be found. Since OB is radius and AB is tangent, \( \angle OBA = 90^{\circ} \). Using cyclic quadrilateral or angle properties, \( z = 35^{\circ} \).
(b) To prove \( RB \parallel PQ \):
- We observe alternate interior angles or corresponding angles. Since \( \angle RBR' \) or angle between RB and PQ matches alternate segment angles, alternate interior angles are equal, proving \( RB \parallel PQ \).

Teacher's Note:
a) Apply the alternate segment theorem relating tangents and chords.
b) Use the central angle theorem to relate circumference and centre angles.

Question 6

(i) There are three positive numbers in a Geometric Progression (G.P.) such that:
(a) their product is 3375
(b) the result of the product of first and second number added to the product of second and third number is 750.
Find the numbers. [3 Marks]

Answer:
1. Let the three numbers in G.P. be \( \frac{a}{r}, a, ar \).
2. Given product \( = 3375 \):
\( \left(\frac{a}{r}\right)(a)(ar) = 3375 \implies a^{3} = 3375 \implies a = 15 \).
3. Given condition: product of first and second number added to product of second and third number is 750:
\( \left(\frac{a}{r}\right)(a) + (a)(ar) = 750 \)
\( \frac{a^{2}}{r} + a^{2}r = 750 \)
4. Substitute \( a = 15 \):
\( \frac{225}{r} + 225r = 750 \)
Divide by 225:
\( \frac{1}{r} + r = \frac{750}{225} = \frac{10}{3} \)
\( 3 + 3r^{2} = 10r \implies 3r^{2} - 10r + 3 = 0 \)
5. Factorize the quadratic equation:
\( 3r^{2} - 9r - r + 3 = 0 \implies 3r(r - 3) - 1(r - 3) = 0 \implies (3r - 1)(r - 3) = 0 \)
Thus, \( r = 3 \) or \( r = \frac{1}{3} \).
6. The numbers are: for \( a = 15, r = 3 \), the numbers are \( \frac{15}{3}, 15, 15(3) \), i.e., \( 5, 15, 45 \).

Teacher's Note:
a) For three numbers in G.P., always assume them as \( \frac{a}{r}, a, ar \) to easily find \( a \) from their product.
b) Both values of \( r \) yield the same set of numbers in reverse order.

(ii) The table given below shows the ages of members of a society. [3 Marks]

Use graph sheet for this question.
Take 2cm = 10 years along one axis and 2cm=10 members along the other axis.
(a) Draw a histogram representing the above distribution.
(b) Hence find the modal age of the members.

Answer:
(a) Histogram is drawn by taking class intervals on the x-axis and frequencies on the y-axis with the specified scale.
(b) Modal age is determined graphically by drawing cross lines from the top corners of the highest rectangle (class 55 - 65) to the adjacent rectangles. The x-coordinate of the intersection point gives the mode, which is approximately \( 59 \text{ years} \).

Teacher's Note:
a) Ensure class intervals are continuous before plotting a histogram.
b) State the exact scale used on both axes clearly on the graph sheet.

(iii) A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 7 m each and the total height of the tent is 14 m. Find the:
(a) quantity of air contained inside the tent.
(b) radius of a sphere whose volume is equal to the quantity of air inside the tent.
Use \( \pi = \frac{22}{7} \) [4 Marks]

Answer:
1. Radius of cylinder \( r = 7 \text{ m} \), height of cylinder \( h = 7 \text{ m} \).
2. Total height of the tent \( = 14 \text{ m} \). Height of the conical part \( H = 14 - 7 = 7 \text{ m} \).
(a) Quantity of air inside the tent = Volume of cylinder + Volume of cone
- Volume of cylinder \( = \pi r^{2}h = \frac{22}{7} \times 7^{2} \times 7 = \frac{22}{7} \times 49 \times 7 = 1078 \text{ m}^{3} \).
- Volume of cone \( = \frac{1}{3}\pi r^{2}H = \frac{1}{3} \times \frac{22}{7} \times 7^{2} \times 7 = \frac{1078}{3} = 359.33 \text{ m}^{3} \).
- Total volume \( = 1078 + 359.33 = 1437.33 \text{ m}^{3} \).
(b) Let the radius of the sphere be \( R_{s} \). Volume of sphere \( = \frac{4}{3}\pi R_{s}^{3} \).
\( \frac{4}{3} \times \frac{22}{7} \times R_{s}^{3} = \frac{4312}{3} \) (where \( 1437.33 = \frac{4312}{3} \))
\( \frac{88}{21} \times R_{s}^{3} = \frac{4312}{3} \)
\( R_{s}^{3} = \frac{4312}{3} \times \frac{21}{88} = 4312 \times \frac{7}{88} = 49 \times 7 = 343 \)
\( R_{s} = \sqrt[3]{343} = 7 \text{ m} \).

Teacher's Note:
a) Calculate the height of the cone by subtracting the cylindrical height from the total height.
b) Equate the total volume directly to the formula for the volume of a sphere to solve for its radius.

Question 7

(i) The line segment joining A(2,-3) and B(-3, 2) is intercepted by the x-axis at the point M and the y axis at the point N. PQ is perpendicular to AB produced at R and meets the y-axis at a distance of 6 units from the origin O, as shown in the diagram, at S. Find the:
(a) coordinates of M and N
(b) coordinates of S
(c) slope of AB.
(d) equation of line PQ. [5 Marks]

[Figure: Cartesian plane showing line segment AB with points A(2,-3) and B(-3,2), intercepts M and N, and perpendicular line PQ meeting y-axis at S(0,6).]

Answer:
1. Equation of line AB using two-point formula \( \frac{y - y_{1}}{x - x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \):
\( \frac{y - (-3)}{x - 2} = \frac{2 - (-3)}{-3 - 2} \implies \frac{y + 3}{x - 2} = \frac{5}{-5} = -1 \)
\( y + 3 = -x + 2 \implies x + y + 1 = 0 \).
(a) Coordinates of M (x-intercept, where \( y = 0 \)): \( x + 0 + 1 = 0 \implies x = -1 \), so \( M(-1, 0) \).
Coordinates of N (y-intercept, where \( x = 0 \)): \( 0 + y + 1 = 0 \implies y = -1 \), so \( N(0, -1) \).
(b) Coordinates of S: Since S is on the y-axis at a distance of 6 units from the origin, \( S(0, 6) \).
(c) Slope of AB \( = -1 \) (from equation \( y = -x - 1 \)).
(d) Equation of line PQ: PQ is perpendicular to AB, so slope of PQ \( m_{PQ} = -\frac{1}{\text{slope of AB}} = -\frac{1}{-1} = 1 \).
Line PQ passes through \( S(0, 6) \). Using slope-point form:\
\( y - 6 = 1(x - 0) \implies x - y + 6 = 0 \).

Teacher's Note:
a) Find the equation of the line first to easily determine intercepts and slope.
b) Perpendicular lines have slopes whose product is -1.

(ii) The angle of depression of two ships A and B on opposite sides of a light house of height 100m are respectively \( 42^{\circ} \) and \( 54^{\circ} \). The line joining the two ships passes through the foot of the lighthouse.
(a) Find the distance between the two ships A and B.
(b) Give your final answer correct to the nearest whole number.
(Use mathematical tables for this question) [5 Marks]

[Figure: A lighthouse of height 100 m with angles of depression of ships A and B on opposite sides as \( 42^{\circ} \) and \( 54^{\circ} \).]

Answer:
1. Let the height of lighthouse CD = 100 m.
2. In right-angled triangle CAD:
\( \tan 42^{\circ} = \frac{CD}{AC} \implies AC = \frac{100}{\tan 42^{\circ}} = \frac{100}{0.9004} \approx 111.06 \text{ m} \).
3. In right-angled triangle CBD:
\( \tan 54^{\circ} = \frac{CD}{BC} \implies BC = \frac{100}{\tan 54^{\circ}} = \frac{100}{1.3764} \approx 72.65 \text{ m} \).
4. Total distance between the two ships AB \( = AC + BC = 111.06 + 72.65 = 183.71 \text{ m} \).
5. Correct to the nearest whole number, the distance is \( 184 \text{ m} \).

Teacher's Note:
a) Angle of depression equals the alternate angle of elevation from the ships.
b) Calculate the two base distances separately and add them since the ships are on opposite sides.

Question 8

(i) Solve the following inequation write the solution set and represent it on the real number line.
\( 3 - 2x \ge x + \frac{1 - x}{3} \gt \frac{2x}{5}, x \in R \) [3 Marks]

Answer:
1. Split into two inequalities:
Part 1: \( 3 - 2x \ge x + \frac{1 - x}{3} \)
Multiply by 3: \( 9 - 6x \ge 3x + 1 - x \)
\( 9 - 6x \ge 2x + 1 \implies 8x \le 8 \implies x \le 1 \).
Part 2: \( x + \frac{1 - x}{3} \gt \frac{2x}{5} \)
Multiply by 15: \( 15x + 5(1 - x) \gt 6x \)
\( 15x + 5 - 5x \gt 6x \implies 10x + 5 \gt 6x \implies 4x \gt -5 \implies x \gt -1.25 \).
2. Combined solution set: \( -1.25 \lt x \le 1 \).
3. Representation on real number line: A line with a hollow circle at -1.25, a dark solid circle at 1, and a thick line segment joining them.

Teacher's Note:
a) Solve compound inequalities by splitting them into two separate linear inequalities.
b) Use open circles for strict inequalities (\(\lt, \gt\)) and closed circles for inclusive inequalities (\(\le, \ge\)) on the number line.

(ii) ABCD is a cyclic quadrilateral in which BC = CD and EF is a tangent at A. \( \angle CBD = 43^{\circ} \) and \( \angle ADB = 62^{\circ} \). Find:
(a) \( \angle ADC \)
(b) \( \angle ABD \)
(c) \( \angle FAD \) [3 Marks]

[Figure: Cyclic quadrilateral ABCD with tangent EF at A. Angle CBD = \( 43^{\circ} \), Angle ADB = \( 62^{\circ} \), and BC = CD.]

Answer:
(a) Since \( BC = CD \), the chords are equal, which means their subtended angles are equal: \( \angle CAD = \angle CBD = 43^{\circ} \).
- Also, \( \angle CAB = \angle ADB = 43^{\circ} \) or using angles in the same segment.
- \( \angle ADC = 180^{\circ} - \angle ABC \) (opposite angles of cyclic quadrilateral are supplementary).
- Here \( \angle ABC = \angle CBD + \angle ABD \). Since \( \angle ADB = \angle ACB = 62^{\circ} \), we calculate: \( \angle ADC = \angle ADB + \angle BDC = 62^{\circ} + 62^{\circ} = 124^{\circ} \).
(b) In \( \Delta BDC \), \( BC = CD \implies \angle CBD = \angle CDB = 43^{\circ} \).
- \( \angle ABD = \angle ACD \) (angles in same segment) \( = 43^{\circ} \).
(c) By alternate segment theorem, the angle between tangent EF and chord AD (\( \angle FAD \)) is equal to the angle in the alternate segment, \( \angle ABD \) or \( \angle ACD \), so \( \angle FAD = \angle ABD = 43^{\circ} \).

Teacher's Note:
a) Equal chords subtend equal angles at the circumference.
b) Apply the alternate segment theorem for angles involving tangents.

(iii) A (a, b), B(-4, 3) and C(8,-6) are the vertices of a \( \Delta ABC \). Point D is on BC such that BD : DC is 2 : 1 and M (6, 0) is mid point of AD. Find:
(a) coordinates of point D.
(b) coordinates of point A.
(c) equation of a line passing through M and parallel to line BC. [4 Marks]

Answer:
(a) Coordinates of D using section formula for \( B(-4, 3) \) and \( C(8, -6) \) in ratio \( m : n = 2 : 1 \):
\( D = \left(\frac{2(8) + 1(-4)}{2 + 1}, \frac{2(-6) + 1(3)}{2 + 1}\right) = \left(\frac{16 - 4}{3}, \frac{-12 + 3}{3}\right) = \left(\frac{12}{3}, \frac{-9}{3}\right) = (4, -3) \).
(b) Let \( A(a, b) \). M(6, 0) is the mid-point of AD where \( D(4, -3) \):
\( \left(\frac{a + 4}{2}, \frac{b - 3}{2}\right) = (6, 0) \)
\( \frac{a + 4}{2} = 6 \implies a + 4 = 12 \implies a = 8 \)
\( \frac{b - 3}{2} = 0 \implies b - 3 = 0 \implies b = 3 \)
Thus, coordinates of \( A(8, 3) \).
(c) Slope of line BC using \( B(-4, 3) \) and \( C(8, -6) \):
\( m = \frac{-6 - 3}{8 - (-4)} = \frac{-9}{12} = -\frac{3}{4} \).
Since the required line is parallel to BC, its slope is also \( -\frac{3}{4} \).
Equation of line passing through \( M(6, 0) \) with slope \( -\frac{3}{4} \):
\( y - 0 = -\frac{3}{4}(x - 6) \)
\( 4y = -3x + 18 \implies 3x + 4y - 18 = 0 \).

Teacher's Note:
a) Use the section formula for internal division to find coordinates of D.
b) Parallel lines share the exact same slope.

Question 9

(i) Using componendo and dividend, find the value of x, when:
\( \frac{x^{3} + 3x}{3x^{2} + 1} = \frac{14}{13} \) [3 Marks]

Answer:
1. Given equation: \( \frac{x^{3} + 3x}{3x^{2} + 1} = \frac{14}{13} \).
2. Applying Componendo and Dividendo (\( \frac{a + b}{a - b} = \frac{c + d}{c - d} \)):
\( \frac{(x^{3} + 3x) + (3x^{2} + 1)}{(x^{3} + 3x) - (3x^{2} + 1)} = \frac{14 + 13}{14 - 13} \)
\( \frac{x^{3} + 3x^{2} + 3x + 1}{x^{3} - 3x^{2} + 3x - 1} = \frac{27}{1} \)
3. Expressing numerator and denominator as cubes:
\( \frac{(x + 1)^{3}}{(x - 1)^{3}} = 27 \)
4. Taking cube root on both sides:
\( \frac{x + 1}{x - 1} = 3 \)
\( x + 1 = 3(x - 1) \)
\( x + 1 = 3x - 3 \)
\( 2x = 4 \implies x = 2 \).

Teacher's Note:
a) Recognize standard algebraic expansions like \( (x + 1)^{3} = x^{3} + 3x^{2} + 3x + 1 \) after applying componendo and dividendo.
b) Take the cube root of both sides to simplify the equation into a linear one.

(ii) The total expense of a trip for certain number of people is Rs. 18000. If three more people join them, then the share of each reduces by Rs. 3000. Taking x to be the original number of people, form a quadratic equation in x and solve it to find the value of x. [3 Marks]

Answer:
1. Original number of people = \( x \).
2. Original share of each person = \( \frac{18000}{x} \).
3. New number of people = \( x + 3 \).
4. New share of each person = \( \frac{18000}{x + 3} \).
5. Given condition: \( \frac{18000}{x} - \frac{18000}{x + 3} = 3000 \).
6. Dividing by 3000:
\( \frac{6}{x} - \frac{6}{x + 3} = 1 \)
\( 6\left(\frac{x + 3 - x}{x(x + 3)}\right) = 1 \implies \frac{18}{x^{2} + 3x} = 1 \)
\( x^{2} + 3x - 18 = 0 \)
7. Factorizing the quadratic equation:
\( (x + 6)(x - 3) = 0 \)
\( x = -6 \) or \( x = 3 \).
8. Since the number of people cannot be negative, \( x = 3 \).

Teacher's Note:
a) Formulate equations by setting up expressions for individual shares before and after the change.
b) Reject negative values of \( x \) since number of people must be a positive integer.

(iii) Using ruler and compass only construct \( \angle ABC = 60^{\circ} \), AB = 6 cm and BC = 5 cm.
(a) construct the locus of points equidistant from AB and BC.
(b) construct the locus of points equidistant from A and B.
(c) Mark the point which satisfies both the conditions (a) and (b) as P.
Hence, construct a circle with centre P and passing through A and B. [4 Marks]

Answer:
1. Construct line segment \( AB = 6 \text{ cm} \). At B, construct an angle of \( 60^{\circ} \) and cut off \( BC = 5 \text{ cm} \), then join AC.
(a) The locus of points equidistant from AB and BC is the angle bisector of \( \angle ABC \).
(b) The locus of points equidistant from A and B is the perpendicular bisector of line segment AB.
(c) Point P is marked at the intersection of the angle bisector of \( \angle ABC \) and the perpendicular bisector of AB.
- With centre P and radius equal to PA (or PB), construct a circle passing through A and B.

Teacher's Note:
a) Use only a ruler and compass for all constructions; construction arcs must be clearly visible.
b) The circumcentre of a triangle is equidistant from its vertices.

Question 10

(i) Using remainder and factor theorem, factorize completely, the given polynomial: [3 Marks]
\( 2x^{3} - 9x^{2} + 7x + 6 \)

Answer:
1. Let \( f(x) = 2x^{3} - 9x^{2} + 7x + 6 \).
2. Factors of the constant term 6 are \( \pm 1, \pm 2, \pm 3, \pm 6 \).
Testing \( x = 2 \):
\( f(2) = 2(2)^{3} - 9(2)^{2} + 7(2) + 6 = 2(8) - 9(4) + 14 + 6 = 16 - 36 + 14 + 6 = 36 - 36 = 0 \).
Since \( f(2) = 0 \), by the factor theorem, \( (x - 2) \) is a factor of \( f(x) \).
3. Dividing \( f(x) \) by \( (x - 2) \) using synthetic division or long division:
\( 2x^{3} - 9x^{2} + 7x + 6 = (x - 2)(2x^{2} - 5x - 3) \).
4. Factorizing the quadratic expression \( 2x^{2} - 5x - 3 \):
\( 2x^{2} - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3) \).
5. Complete factorization: \( (x - 2)(2x + 1)(x - 3) \).

Teacher's Note:
a) Use trial and error with factors of the constant term to find the first root.
b) Fully factorize the resulting quadratic quotient to obtain all linear factors.

(ii) Each of the letter of the word “HOUSEWARMING” is written on cards and put in a bag. If a card is drawn at random from the bag after shuffling, what is the probability that the letter on the card is:
(a) a vowel
(b) one of the letters of the word SEWING.
(c) not a letter from the word WEAR. [3 Marks]

Answer:
1. Word given: "HOUSEWARMING". Total number of letters \( n = 12 \) (H, O, U, S, E, W, A, R, M, I, N, G - all distinct letters: H-1, O-1, U-1, S-1, E-1, W-1, A-1, R-1, M-2, I-1, N-1, G-1). Let's list all 12 letters: H, O, U, S, E, W, A, R, M, I, N, G (Note: M appears twice: HOUSEWARMING has M twice, total 12 letters).
(a) Vowels in "HOUSEWARMING" are O, U, E, A, I (5 vowels).
- Probability of getting a vowel \( = \frac{5}{12} \).
(b) Letters of the word "SEWING" present in "HOUSEWARMING": S, E, W, I, N, G (6 letters).
- Probability \( = \frac{6}{12} = \frac{1}{2} \).
(c) Letters from the word "WEAR": W, E, A, R (4 letters).
- Letters NOT from the word "WEAR": total letters minus letters in WEAR \( = 12 - 4 = 8 \).
- Probability \( = \frac{8}{12} = \frac{2}{3} \).

Teacher's Note:
a) Count the total number of letters carefully, accounting for any repetitions if present.
b) For complementary probability like "not a letter", subtract the favorable count from the total sample space.

(iii) Use graph sheet for this question. Take 2 cm = 1 unit along the axes.
(a) Plot A (1, 2), B(1, 1) and C (2, 1)
(b) Reflect A, B and C about y-axis and name them as A\( ' \), B\( ' \) and C\( ' \).
(c) Reflect A, B, C, A\( ' \), B\( ' \) and C\( ' \) about x-axis and name them as A\( '' \), B\( '' \), C\( '' \), A\( ''' \), B\( ''' \) and C\( ''' \) respectively.
(d) Join A, B, C, C\( '' \), B\( '' \), A\( '' \), A\( ''' \), B\( ''' \), C\( ''' \), C\( ' \), B\( ' \), A\( ' \) and A to form a closed figure. [4 Marks]

Answer:
1. Original points plotted: \( A(1, 2), B(1, 1), C(2, 1) \).
2. Reflection about y-axis (\( (x, y) \to (-x, y) \)):
- \( A'(-1, 2) \)
- \( B'(-1, 1) \)
- \( C'(-2, 1) \)
3. Reflection about x-axis (\( (x, y) \to (x, -y) \)):
- Reflections of original points \( A, B, C \) give: \( A''(1, -2), B''(1, -1), C''(2, -1) \).
- Reflections of reflected points \( A', B', C' \) give: \( A'''(-1, -2), B'''(-1, -1), C'''(-2, -1) \).
4. Joining the points in the specified sequence forms a symmetric closed geometric figure across both axes.

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