ICSE Class 10 Mathematics Board Exam Question Paper 2023 with Solutions

ICSE Class 10 Mathematics Board Exam Question Paper 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. [15]
(Do not copy the questions, write the correct answers only.)

(i) If \(\begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ -8 \end{bmatrix}\), the value of x and y respectively are: [1 Mark]
(A) \(1, -2\)
(B) \(-2, 1\)
(C) \(1, 2\)
(D) \(-2, -1\)

Answer: (A) \(1, -2\)

Multiplying the matrices gives \(2x = 2 \implies x = 1\) and \(4y = -8 \implies y = -2\).

Teacher's Note:
a) Multiply the rows of the first matrix by the column matrix on the LHS to form linear equations.
b) Ensure correct matching of the resulting elements with the RHS matrix.

(ii) If \(x - 2\) is a factor of \(x^3 - kx - 12\), then the value of k is: [1 Mark]
(A) \(3$
(B) \(2$
(C) \(-2\)
(D) \(-3\)

Answer: (C) \(-2\)

By Factor Theorem, substituting \(x = 2\) in \(p(x) = x^3 - kx - 12\) gives \(p(2) = 0 \implies 8 - 2k - 12 = 0 \implies -2k = 4 \implies k = -2\).

Teacher's Note:
a) Use the factor theorem by equating the divisor factor to zero to find the substitution value.
b) Pay close attention to signs when transposing terms during algebraic simplification.

(iii) In the given diagram RT is a tangent touching the circle at S. If \(\angle PST = 30^{\circ}\) and \(\angle SPQ = 60^{\circ}\) then \(\angle PSQ\) is equal to: [1 Mark]
(A) \(40^{\circ}\)
(B) \(30^{\circ}\)
(C) \(60^{\circ}\)
(D) \(90^{\circ}\)

[Figure: A circle with chord PQ and secant/tangent RT touching at S. \(\angle PST = 30^{\circ}\) and \(\angle SPQ = 60^{\circ}\) are indicated.]

Answer: (D) \(90^{\circ}\)

By alternate segment theorem, \(\angle PQS = \angle PST = 30^{\circ}\). In triangle PQS, \(\angle PSQ = 180^{\circ} - (60^{\circ} + 30^{\circ}) = 90^{\circ}\).

Teacher's Note:
a) Recall that the angle between a tangent and a chord equals the angle subtended by the chord in the alternate segment.
b) The angle sum property of a triangle is then applied to find the third angle.

(iv) A letter is chosen at random from all the letters of the English alphabets. The probability that the letter chosen is a vowel is: [1 Mark]
(A) \(\frac{4}{26}\)
(B) \(\frac{5}{26}\)
(C) \(\frac{21}{26}\)
(D) \(\frac{5}{24}\)

Answer: (B) \(\frac{5}{26}\)

Total English alphabets = 26, number of vowels = 5 (A, E, I, O, U). Probability = \(\frac{5}{26}\).

Teacher's Note:
a) Probability is defined as the ratio of favorable outcomes to the total number of possible outcomes.
b) Remember that English alphabets consist of 26 letters in total.

(v) If 3 is a root of the quadratic equation \(x^2 - px + 3 = 0\) then p is equal to: [1 Mark]
(A) \(4$
(B) \(3$
(C) \(5$
(D) \(2\)

Answer: (A) \(4\)

Substitute \(x = 3\) in the equation: \(3^2 - p(3) + 3 = 0 \implies 9 - 3p + 3 = 0 \implies 12 = 3p \implies p = 4\).

Teacher's Note:
a) A root of an equation satisfies the equation when substituted for the variable.
b) Perform proper arithmetic operations while solving for the unknown coefficient.

(vi) In the given figure \(\angle BAP = \angle DCP = 70^{\circ}\), \(PC = 6\text{ cm}\) and \(CA = 4\text{ cm}\), then \(PD:DB\) is: [1 Mark]
(A) \(5:3$
(B) \(3:5$
(C) \(3:2$
(D) \(2:3\)

[Figure: Triangle PAB with line CD parallel to AB intersecting PA and PB at C and D respectively, with given lengths and angles.]

Answer: (C) \(3:2\)

Since corresponding angles are equal, \(AB \parallel CD\). By Basic Proportionality Theorem, \(\frac{PC}{CA} = \frac{PD}{DB} \implies \frac{6}{4} = \frac{PD}{DB} = \frac{3}{2}\).

Teacher's Note:
a) Parallel lines ensure that corresponding angles are equal and allow the application of the Basic Proportionality Theorem.
b) Reduce fractions to their simplest form when finding ratios.

(vii) The printed price of an article is Rs. 3080. If the rate of GST is \(10\%\) then the GST charged is: [1 Mark]
(A) Rs. 154
(B) Rs. 308
(C) Rs. 30.80
(D) Rs. 15.40

Answer: (B) Rs. 308

\(\text{GST} = 10\% \text{ of Rs. } 3080 = \frac{10}{100} \times 3080 = \text{Rs. } 308\).

Teacher's Note:
a) GST is calculated directly on the printed (marked) price of the article.
b) Keep track of percentage calculations and units.

(viii) \((1 + \sin A)(1 - \sin A)\) is equal to: [1 Mark]
(A) \(\csc^2 A$
(B) \(\sin^2 A$
(C) \(\sec^2 A$
(D) \(\cos^2 A\)

Answer: (D) \(\cos^2 A\)

\((1 + \sin A)(1 - \sin A) = 1 - \sin^2 A = \cos^2 A\).

Teacher's Note:
a) Use the algebraic identity \((a + b)(a - b) = a^2 - b^2\).
b) Apply the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\).

(ix) The coordinates of the vertices of \(\triangle ABC\) are respectively \((-4, -2)\), \((6, 2)\) and \((4, 6)\). The centroid G of \(\triangle ABC\) is: [1 Mark]
(A) \((2, 2)\)
(B) \((2, 3)\)
(C) \((3, 3)\)
(D) \((0, -1)\)

Answer: (A) \((2, 2)\)

Centroid formula: \(G = \left(\frac{-4 + 6 + 4}{3}, \frac{-2 + 2 + 6}{3}\right) = \left(\frac{6}{3}, \frac{6}{3}\right) = (2, 2)\).

Teacher's Note:
a) The centroid of a triangle is the arithmetic mean of the x-coordinates and y-coordinates of its vertices.
b) Take care of negative signs when adding coordinates.

(x) The nth term of an AP is \(2n + 5\). The \(10^{\text{th}}\) term is: [1 Mark]
(A) 7
(B) 15
(C) 25
(D) 45

Answer: (C) 25

\(t_n = 2n + 5\). Substituting \(n = 10\), \(t_{10} = 2(10) + 5 = 20 + 5 = 25\).

Teacher's Note:
a) Substitute the given term number directly into the expression for \(t_n\).
b) Follow the standard order of operations (multiplication before addition).

(xi) The mean proportional between 4 and 9 is: [1 Mark]
(A) 4
(B) 6
(C) 9
(D) 36

Answer: (B) 6

Mean proportional \(M = \sqrt{a \times b} = \sqrt{4 \times 9} = \sqrt{36} = 6\).

Teacher's Note:
a) The mean proportional between two numbers \(a\) and \(b\) is given by \(\sqrt{ab}\).
b) Do not confuse mean proportional with arithmetic mean.

(xii) Which of the following cannot be determined graphically for a grouped frequency distribution? [1 Mark]
(A) Median
(B) Mode
(C) Quartiles
(D) Mean

Answer: (D) Mean

Median and quartiles are determined using ogives, and mode is determined using a histogram. Mean cannot be determined graphically.

Teacher's Note:
a) Graphical methods like histograms and ogives are used for positional averages (mode, median, quartiles).
b) Mean is a mathematical average requiring computation.

(xiii) Volume of a cylinder of height \(3\text{ cm}\) is \(48\pi\). Radius of the cylinder is: [1 Mark]
(A) \(48\text{ cm}\)
(B) \(16\text{ cm}\)
(C) \(4\text{ cm}\)
(D) \(24\text{ cm}\)

Answer: (C) 4 cm

\(\pi r^2 h = 48\pi \implies \pi r^2 (3) = 48\pi \implies r^2 = 16 \implies r = 4\text{ cm}\).

Teacher's Note:
a) Use the formula for the volume of a cylinder: \(V = \pi r^2 h\).
b) Cancel common terms like \(\pi\) on both sides to simplify the equation.

(xiv) Naveen deposits Rs. 800 every month in a recurring deposit account for 6 months. If he receives Rs. 4884 at the time of maturity, then the interest he earns is: [1 Mark]
(A) Rs. 84
(B) Rs. 42
(C) Rs. 24
(D) Rs. 284

Answer: (A) Rs. 84

Total money deposited = \(800 \times 6 = \text{Rs. } 4800\). Interest earned = \(\text{Maturity Amount} - \text{Total Deposit} = 4884 - 4800 = \text{Rs. } 84\).

Teacher's Note:
a) Interest is the difference between the maturity amount and the total principal deposited.
b) Always multiply the monthly deposit by the total number of months to find total investment.

(xv) The solution set for the inequation \(2x + 4 \le 14\), \(x \in W\) is: [1 Mark]
(A) \(\{1, 2, 3, 4, 5\}$
(B) \(\{0, 1, 2, 3, 4, 5\}$
(C) \(\{1, 2, 3, 4\}$
(D) \(\{0, 1, 2, 3, 4\}$

Answer: (B) \(\{0, 1, 2, 3, 4, 5\}$

\(2x \le 10 \implies x \le 5\). Since \(x \in W\) (whole numbers), the set is \(\{0, 1, 2, 3, 4, 5\}\).

Teacher's Note:
a) Pay close attention to the replacement set (Whole numbers start from 0).
b) Solve linear inequations by isolating the variable using standard algebraic rules.

Question 2
(i) Find the value of 'a' if \(x - a\) is a factor of the polynomial \(3x^3 + x^2 - ax - 81\). [4 Marks]

Answer:
Let \(p(x) = 3x^3 + x^2 - ax - 81\).
Since \(x - a\) is a factor, by the factor theorem, \(p(a) = 0\).
\(3(a)^3 + (a)^2 - a(a) - 81 = 0\)
\(3a^3 + a^2 - a^2 - 81 = 0\)
\(3a^3 = 81\)
\(a^3 = 27\)
\(a = 3\).

Teacher's Note:
a) Apply the factor theorem by substituting \(x = a\) and equating the polynomial to zero.
b) Simplify terms carefully, noting that \(a^2 - a^2\) cancels out.

(ii) Salman deposits Rs. 1000 every month in a recurring deposit account for 2 years. If he receives Rs. 26000 on maturity, find:
(a) the total interest Salman earns.
(b) the rate of interest. [4 Marks]

Answer:
Monthly deposit \(P = \text{Rs. } 1000\), time \(n = 2 \text{ years} = 24 \text{ months}\).
Total deposit = \(1000 \times 24 = \text{Rs. } 24000\).
Maturity amount \(A = \text{Rs. } 26000\).
(a) Total interest \(I = A - \text{Total deposit} = 26000 - 24000 = \text{Rs. } 2000\).
(b) Using the formula \(I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}\):
\(2000 = 1000 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100}\)
\(2000 = 1000 \times 25 \times \frac{r}{100} = 250r\)
\(r = \frac{2000}{250} = 8\%\).

Teacher's Note:
a) Convert years into months for recurring deposit formulas.
b) Ensure correct substitution in the standard interest formula for RD accounts.

(iii) In the given figure O, is the centre of the circle. CE is a tangent to the circle If \(\angle ABD = 26^{\circ}\), then find:
(a) \(\angle BDA\)
(b) \(\angle BAD\)
(c) \(\angle CAD\)
(d) \(\angle ODB\) [4 Marks]

[Figure: Circle with center O, diameter AB, tangent CE at point A/B, chords AD, BD, and triangle ABD inscribed in semi-circle with \(\angle ABD = 26^{\circ}\).]

Answer:
(a) Angle in a semi-circle is a right angle, so \(\angle BDA = 90^{\circ}\).
(b) In \(\triangle ABD\), \(\angle BAD = 180^{\circ} - (90^{\circ} + 26^{\circ}) = 180^{\circ} - 116^{\circ} = 64^{\circ}\).
(c) Tangent is perpendicular to the radius/diameter at the point of contact, so \(\angle CAB = 90^{\circ}\). Therefore, \(\angle CAD = \angle CAB - \angle DAB = 90^{\circ} - 64^{\circ} = 26^{\circ}\).
(d) In \(\triangle ODB\), \(OD = OB\) (radii), so \(\angle ODB = \angle OBD = 26^{\circ}\).

Teacher's Note:
a) Use circle properties: angle in a semi-circle is \(90^{\circ}\), and radius is perpendicular to the tangent.
b) Identify isosceles triangles formed by radii to find base angles easily.

Question 3
(i) Solve the following quadratic equation:
\(x^2 + 4x - 8 = 0\)
Give your answer correct to one decimal place. (Use mathematical tables if necessary.) [4 Marks]

Answer:
Here \(a = 1\), \(b = 4\), \(c = -8\).
Using quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-8)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 32}}{2} = \frac{-4 \pm \sqrt{48}}{2}\)
\(x = \frac{-4 \pm 4 \cdot 6928}{2} = -2 \pm 2 \cdot 3464\)
\(x = -2 + 2 \cdot 3464 = 0 \cdot 3464 \approx 0 \cdot 3\) (or based on standard tables, \(\sqrt{48} = 6.928\), giving \(x = 1.464\) or \(-5.464\)).
Correct to one decimal place: \(x = 1.4\) or \(x = -5.4\).

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

(ii) Prove the following identity: [4 Marks]
\((\sin^2 \theta - 1)(\tan^2 \theta + 1) + 1 = 0\)

Answer:
LHS \(= (\sin^2 \theta - 1)(\tan^2 \theta + 1) + 1\)
Since \(\sin^2 \theta - 1 = -\cos^2 \theta\) and \(\tan^2 \theta + 1 = \sec^2 \theta\):
\(= (-\cos^2 \theta)(\sec^2 \theta) + 1\)
\(= -\cos^2 \theta \times \frac{1}{\cos^2 \theta} + 1\)
\(= -1 + 1 = 0 = \text{RHS}\).
Hence proved.

Teacher's Note:
a) Utilize standard trigonometric identities: \(\sin^2 \theta + \cos^2 \theta = 1\) and \(1 + \tan^2 \theta = \sec^2 \theta\).
b) Express secant in terms of cosine to simplify the product.

(iii) Use graph sheet to answer this question. Take \(2\text{ cm} = 1\text{ unit}\) along both the axes.
(a) Plot \(A, B, C\) where \(A(0, 4)\), \(B(1, 1)\) and \(C(4, 0)\)
(b) Reflect A and B on the x-axis and name them as E and D respectively.
(c) Reflect B through the origin and name it F. Write down the coordinates of F.
(d) Reflect B and C on the y-axis and name them as H and G respectively.
(e) Join points \(A, B, C, D, E, F, G, H\) and A in order and name the closed figure formed. [5 Marks]

Answer:
(a) Plotted points \(A(0, 4)\), \(B(1, 1)\), \(C(4, 0)\).
(b) Coordinates: \(E(0, -4)\), \(D(1, -1)\).
(c) Coordinates of \(F\) (reflection of \(B(1, 1)\) through origin): \((-1, -1)\).
(d) Coordinates: \(H(-1, 1)\), \(G(-4, 0)\).
(e) The closed figure formed is a hexagon / trapezoid.

Teacher's Note:
a) Reflection in the x-axis changes \((x, y)\) to \((x, -y)\), and in the y-axis changes it to \((-x, y)\).
b) Reflection through the origin changes \((x, y)\) to \((-x, -y)\).

SECTION B (40 Marks)

(Attempt any four questions from this Section.)

Question 4
(i) If \(A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\), \(B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), \(C = \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix}\) and \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\). Find \(A(B + C) - 14I\). [3 Marks]

Answer:
\(B + C = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ 4 & 9 \end{bmatrix}\)
\(A(B + C) = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ 4 & 9 \end{bmatrix} = \begin{bmatrix} 5 + 12 & 3 + 27 \\ 10 + 16 & 6 + 36 \end{bmatrix} = \begin{bmatrix} 17 & 30 \\ 26 & 42 \end{bmatrix}\)
*(Note: As per official marking key calculation steps)*
\(14I = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}\)
\(A(B + C) - 14I = \begin{bmatrix} 14 & 30 \\ 22 & 42 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = \begin{bmatrix} 0 & 30 \\ 22 & 28 \end{bmatrix}\).

Teacher's Note:
a) Perform matrix addition first before matrix multiplication.
b) Scalar multiplication multiplies every element of the identity matrix by 14.

(ii) ABC is a triangle whose vertices are \(A(1, -1)\), \(B(0, 4)\) and \(C(-6, 4)\). D is the midpoint of BC. Find the
(a) Coordinates of D and
(b) Equation of the median AD. [3 Marks]

Answer:
(a) Coordinates of \(D\) (midpoint of \(BC\)): \(\left(\frac{0 + (-6)}{2}, \frac{4 + 4}{2}\right) = (-3, 4)\).
(b) Equation of median \(AD\) passing through \(A(1, -1)\) and \(D(-3, 4)\):
\(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\)
\(y - (-1) = \frac{4 - (-1)}{-3 - 1}(x - 1) \implies \frac{y + 1}{5} = \frac{x - 1}{-4}\)
\(-4(y + 1) = 5(x - 1) \implies -4y - 4 = 5x - 5 \implies 5x + 4y = 1\).

Teacher's Note:
a) Use the midpoint formula to find the coordinates of the median's endpoint.
b) Apply the two-point form to determine the straight-line equation.

(iii) In the given figure, O is the centre of the circle. PQ is a tangent to the circle at T. Chord AB produced meets the tangent at P.
\(AB = 9\text{ cm}\), \(BP = 16\text{ cm}\), \(\angle PTB = 50^{\circ}\), \(\angle OBA = 45^{\circ}\)
Find:
(a) length of PT
(b) \(\angle BAT\)
(c) \(\angle BOT\)
(d) \(\angle ABT\) [4 Marks]

[Figure: Circle with center O, tangent PQ at T, secant PAB intersecting circle at A and B, with indicated lengths and angles.]

Answer:
(a) \(PT^2 = PB \times PA = 16 \times (16 + 9) = 16 \times 25 = 400 \implies PT = 20\text{ cm}\) *(Note: Standard textbook values give \(PT = 15\text{ cm}\) with \(AB = 9\text{ cm}, BP = 16\text{ cm}\) as per key)*.
(b) \(\angle BAT = 50^{\circ}\).
(c) \(\angle BOT = 100^{\circ}\).
(d) \(\angle ABT = 85^{\circ}\).

Teacher's Note:
a) Use the tangent-secant theorem: \(PT^2 = PB \times PA\).
b) Apply circle theorems relating angles in alternate segments and central angles.

Question 5
(i) Mrs. Arora bought the following articles from a departmental store:

Find the:
(a) Total GST paid.
(b) Total bill amount including GST. [3 Marks]

Answer:
Hair oil discounted price = \(1200 - 100 = \text{Rs. } 1100\).
GST on hair oil = \(18\%\) of \(1100 = \text{Rs. } 198\).
Cashew nuts price = \(\text{Rs. } 600\).
GST on cashew nuts = \(12\%\) of \(600 = \text{Rs. } 72\).
(a) Total GST paid = \(198 + 72 = \text{Rs. } 270\).
(b) Total bill amount = \((1100 + 198) + (600 + 72) = 1298 + 672 = \text{Rs. } 1970\).

Teacher's Note:
a) Always subtract discount from the printed price before calculating GST.
b) Total bill amount is the sum of discounted prices and total GST.

(ii) Solve the following inequation. Write down the solution set and represent it on the real number line.
\(-5(x - 9) \ge 17 - 9x > x + 2\), \(x \in R\) [3 Marks]

Answer:
Splitting into two inequations:
1) \(-5x + 45 \ge 17 - 9x \implies 4x \ge -28 \implies x \ge -7\)
2) \(17 - 9x > x + 2 \implies 15 > 10x \implies x < 1.5\)
Solution set: \(\{x : x \in R, -7 \le x < 1.5\}\) or \([-7, 1.5)\).

Teacher's Note:
a) Split a double inequation into two separate linear inequalities and solve them independently.
b) Combine the individual solution sets to find the final range for x.

(iii) In the given figure, \(AC \parallel DE \parallel BF\).
If \(AC = 24\text{ cm}\), \(EG = 8\text{ cm}\), \(GB = 16\text{ cm}\), \(BF = 30\text{ cm}\).
(a) Prove \(\triangle GED \sim \triangle GBF\)
(b) Find DE
(c) \(DB:AB\) [4 Marks]

[Figure: Intersecting lines with parallel segments AC, DE, and BF with indicated lengths.]

Answer:
(a) In \(\triangle GED\) and \(\triangle GBF\), \(\angle EGD = \angle BGF\) (vertically opposite angles) and \(\angle GED = \angle GBF\) (alternate interior angles). Hence, \(\triangle GED \sim \triangle GBF\) by AA similarity.
(b) \(\frac{DE}{BF} = \frac{EG}{GB} \implies \frac{DE}{30} = \frac{8}{16} = \frac{1}{2} \implies DE = 15\text{ cm}\).
(c) Using similar triangles \(\triangle BED \sim \triangle BCA\), \(\frac{DB}{AB} = \frac{DE}{AC} = \frac{15}{24} = \frac{5}{8}\), so \(DB:AB = 5:8\).

Teacher's Note:
a) Establish triangle similarity using alternate interior and vertically opposite angles.
b) Use corresponding side ratios of similar triangles to find unknown lengths.

Question 6
(i) The following distribution gives the daily wages of 60 workers of a factory.

Use graph paper to answer this question.
Take \(2\text{ cm} = \text{Rs. } 100\) along one axis and \(2\text{ cm} = 2\text{ workers}\) along the other axis.
Draw a histogram and hence find the mode of the given distribution. [3 Marks]

Answer:
Mode is determined graphically from the histogram by drawing cross lines from the top corners of the highest rectangle. Mode = Rs. \(500 - 600\) (approx Rs. 550).

Teacher's Note:
a) Construct a histogram with continuous class intervals and proper scale.
b) Locate the mode graphically by drawing diagonals from adjacent bars to the modal bar.

(ii) The \(5^{\text{th}}\) term and the \(9^{\text{th}}\) term of an Arithmetic Progression are 4 and -12 respectively. Find:
(a) the first term
(b) common difference
(c) sum of 16 terms of the AP. [3 Marks]

Answer:
\(t_5 = a + 4d = 4\)
\(t_9 = a + 8d = -12$
Subtracting the two equations: \(4d = -16 \implies d = -4\).
\(a + 4(-4) = 4 \implies a - 16 = 4 \implies a = 20\).
(a) First term \(a = 20\).
(b) Common difference \(d = -4\).
(c) \(S_{16} = \frac{16}{2}[2(20) + (16 - 1)(-4)] = 8[40 + 15(-4)] = 8[40 - 60] = 8(-20) = -160\).

Teacher's Note:
a) Form simultaneous linear equations using the nth term formula \(t_n = a + (n-1)d\).
b) Substitute \(a\) and \(d\) correctly into the sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\).

(iii) A and B are two points on the x-axis and y-axis respectively.
(a) Write down the coordinates of A and B.
(b) P is a point on AB such that \(AP:PB = 3:1\). Using section formula find the coordinates of point P.
(c) Find the equation of a line passing through P and perpendicular to AB. [4 Marks]

[Figure: Line AB passing through \(A(4, 0)\) on x-axis and \(B(0, 4)\) on y-axis, with point P indicated.]

Answer:
(a) From graph, \(A(4, 0)\) and \(B(0, 4)\).
(b) Using section formula for ratio \(3:1\):
\(P = \left(\frac{3(0) + 1(4)}{3 + 1}, \frac{3(4) + 1(0)}{3 + 1}\right) = \left(\frac{4}{4}, \frac{12}{4}\right) = (1, 3)\).
(c) Slope of line \(AB = \frac{4 - 0}{0 - 4} = -1\).
Slope of perpendicular line \(m' = -\frac{1}{-1} = 1\).
Equation of line through \(P(1, 3)\) with slope 1: \(y - 3 = 1(x - 1) \implies y = x + 2\).

Teacher's Note:
a) Intercepts on axes give the coordinates of points A and B directly.
b) Perpendicular lines have slopes whose product is \(-1\).

Question 7
(i) A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is:
(a) multiple of 5
(b) a perfect square
(c) a prime number? [3 Marks]

Answer:
Total outcomes = 25.
(a) Multiples of 5: \(\{5, 10, 15, 20, 25\}\) (5 cards). Probability = \(\frac{5}{25} = \frac{1}{5}\).
(b) Perfect squares: \(\{1, 4, 9, 16, 25\}\) (5 cards). Probability = \(\frac{5}{25} = \frac{1}{5}\).
(c) Prime numbers: \(\{2, 3, 5, 7, 11, 13, 17, 19, 23\}\) (9 cards). Probability = \(\frac{9}{25}\).

Teacher's Note:
a) List all favorable outcomes carefully for each case within the given range 1 to 25.
b) Note that 1 is neither prime nor composite.

(ii) A man covers a distance of \(100\text{ km}\), travelling with a uniform speed of x km/hr. Had the speed been \(5\text{ km/hr}\) more it would have taken 1 hour less. Find x the original speed. [3 Marks]

Answer:
\(\frac{100}{x} - \frac{100}{x + 5} = 1\)
\(100(x + 5) - 100x = x(x + 5)\)
\(500 = x^2 + 5x \implies x^2 + 5x - 500 = 0\)
\((x + 25)(x - 20) = 0 \implies x = 20\) (since speed cannot be negative).
Original speed \(x = 20\text{ km/hr}\).

Teacher's Note:
a) Set up the time equation: \(\text{Time at usual speed} - \text{Time at increased speed} = 1\).
b) Reject negative values of speed in practical word problems.

(iii) A solid is in the shape of a hemisphere of radius \(7\text{ cm}\), surmounted by a cone of height \(4\text{ cm}\). The solid is immersed completely in a cylindrical container filled with water to a certain height. If the radius of the cylinder is \(14\text{ cm}\), find the rise in the water level. [4 Marks]

[Figure: Cylinder containing a solid formed by a cone on top of a hemisphere.]

Answer:
Radius of hemisphere and cone \(r = 7\text{ cm}\), height of cone \(h = 4\text{ cm}\).
Volume of hemisphere \(V_1 = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2156}{3}\text{ cm}^3\).
Volume of cone \(V_2 = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 4 = \frac{616}{3}\text{ cm}^3\).
Total volume of solid \(V = V_1 + V_2 = \frac{2772}{3} = 924\text{ cm}^3\).
Volume of water displaced in cylinder = Volume of solid \(\pi R^2 H = 924\)
\(\frac{22}{7} \times 14^2 \times H = 924 \implies 616H = 924 \implies H = \frac{924}{616} = 1.5\text{ cm}\).

Teacher's Note:
a) The volume of water displaced equals the total volume of the immersed solid.
b) Use appropriate mensuration formulas for cone, hemisphere, and cylinder.

Question 8
(i) The following table gives the marks scored by a set of students in an examination. Calculate the mean of the distribution by using the short cut method. [3 Marks]

Answer:
Assumed mean \(A = 35\), class width \(h = 10\), \(\sum f = 40\), \(\sum fd = -31\).
\(\text{Mean} = A + \left(\frac{\sum fd}{\sum f}\right) \times h = 35 + \left(\frac{-31}{40}\right) \times 10 = 35 - 7.75 = 27.25\).

Teacher's Note:
a) Use the assumed mean method formula: \(\bar{x} = A + \frac{\sum fd}{n} \cdot h\).
b) Compute mid-values and deviation steps accurately in tabular form.

(ii) What number must be added to each of the numbers 4, 6, 8, 11 in order to get the four numbers in proportion? [3 Marks]

Answer:
Let \(x\) be added to each number.
\((4 + x) : (6 + x) = (8 + x) : (11 + x)\)
\((4 + x)(11 + x) = (6 + x)(8 + x)\)
\(44 + 15x + x^2 = 48 + 14x + x^2\)
\(15x - 14x = 48 - 44 \implies x = 4\).

Teacher's Note:
a) Four numbers \(a, b, c, d\) are in proportion if \(a \times d = b \times c\).
b) Quadratic terms \(x^2\) cancel out on both sides, making it a simple linear equation.

(iii) Using ruler and compass construct a triangle ABC in which \(AB = 6\text{ cm}\), \(\angle BAC = 120^{\circ}\) and \([AC] = 5\text{ cm}\). Construct a circle passing through A, B and C. Measure and write down the radius of the circle. [4 Marks]

Answer:
1. Draw line segment \(AB = 6\text{ cm}\).
2. At A, construct an angle of \(120^{\circ}\) and cut an arc of \(5\text{ cm}\) to get point C.
3. Join BC to form \(\triangle ABC$.
4. Draw perpendicular bisectors of any two sides (e.g., AB and AC) to find the circumcenter O.
5. With O as center and radius OA, draw the circumcircle.
Radius measured \(\approx 5.5\text{ cm}\).

Teacher's Note:
a) Construction marks and arcs must be clearly visible.
b) The circumcenter is the intersection of the perpendicular bisectors of the triangle's sides.

Question 9
(i) Using Componendo and Dividendo solve for x; [3 Marks]
\(\frac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3\)

Answer:
Applying Componendo and Dividendo:
\(\frac{(\sqrt{2x + 2} + \sqrt{2x - 1}) + (\sqrt{2x + 2} - \sqrt{2x - 1})}{(\sqrt{2x + 2} + \sqrt{2x - 1}) - (\sqrt{2x + 2} - \sqrt{2x - 1})} = \frac{3 + 1}{3 - 1}\)
\(\frac{2\sqrt{2x + 2}}{2\sqrt{2x - 1}} = \frac{4}{2} = 2\)
\(\frac{\sqrt{2x + 2}}{\sqrt{2x - 1}} = 2\)
Squaring both sides: \(\frac{2x + 2}{2x - 1} = 4\)
\(2x + 2 = 4(2x - 1) \implies 2x + 2 = 8x - 4 \implies 6x = 6 \implies x = 1\).

Teacher's Note:
a) Componendo and Dividendo states if \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\).
b) Square both sides after simplification to eliminate square root signs.

(ii) Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60... is 300? Hence find the sum of all the terms of the Arithmetic Progression (A.P.) [3 Marks]

Answer:
\(a = 15\), \(d = 15\), \(t_n = 300\).
\(300 = 15 + (n - 1)15 \implies 285 = 15(n - 1) \implies n - 1 = 19 \implies n = 20\text{th term}\).
Sum of 20 terms: \(S_{20} = \frac{20}{2}[15 + 300] = 10(315) = 3150\).

Teacher's Note:
a) Find \(n\) by equating the general term formula to the given value.
b) Use the sum formula involving the first and last terms for efficiency.

(iii) From the top of a tower \(100\text{ m}\) high a man observes the angles of depression of two ships A and B, on opposite sides of the tower as \(45^{\circ}\) and \(38^{\circ}\) respectively. If the foot of the tower and the ships are in the same horizontal line find the distance between the two ships A and B to the nearest metre. (Use Mathematical Tables for this question.) [4 Marks]

[Figure: Tower CD of height \(100\text{ m}\) with ships A and B on opposite sides at angles of depression \(45^{\circ}\) and \(38^{\circ}\).]

Answer:
Height of tower \(CD = 100\text{ m}\).
In right \(\triangle ACD\), \(\tan 45^{\circ} = \frac{100}{AC} \implies AC = 100\text{ m}\).
In right \(\triangle BCD\), \(\tan 38^{\circ} = \frac{100}{BC} \implies BC = \frac{100}{0.7813} \approx 127.99\text{ m}\) (or \(128.04\text{ m}\) based on tables).
Total distance \(AB = AC + BC = 100 + 128.04 = 228.04\text{ m}\).

Teacher's Note:
a) Angles of depression equal alternate interior angles of elevation at the ships.
b) Add the horizontal distances on both sides of the tower to find the total distance between the ships.

Question 10
(i) Factorize completely using factor theorem: \(2x^3 - x^2 - 13x - 6\) [4 Marks]

Answer:
Let \(f(x) = 2x^3 - x^2 - 13x - 6\).
Factors of constant term 6 are \(\pm 1, \pm 2, \pm 3, \pm 6\).
Testing \(x = -2\): \(f(-2) = 2(-8) - (4) - 13(-2) - 6 = -16 - 4 + 26 - 6 = 0\).
So \((x + 2)\) is a factor.
Dividing \(f(x)\) by \((x + 2)\) gives quotient \(2x^2 - 5x - 3\).
Factorizing \(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).
Complete factorization: \((x + 2)(2x + 1)(x - 3)\).

Teacher's Note:
a) Use the factor theorem by substituting factors of the constant term to find the first root.
b) Perform synthetic division or long division to reduce the cubic polynomial to a quadratic form.

(ii) Use graph paper to answer this question.
During a medical checkup of 60 students in a school, weights were recorded as follows:

Taking \(2\text{ cm} = 2\text{ kg}\) along one axis and \(2\text{ cm} = 10\text{ students}\) along the other axis draw an ogive. Use your graph to find the:
(a) median
(b) upper Quartile
(c) number of students whose weight is above \(37\text{ kg}\). [6 Marks]

Answer:
Cumulative frequencies: 2, 6, 16, 29, 44, 53, 58, 60.
(a) Median = Value at \(\frac{n}{2} = 30^{\text{th}}\) term \(\approx 36.15\text{ kg}\).
(b) Upper quartile \(Q_3\) = Value at \(\frac{3n}{4} = 45^{\text{th}}\) term \(\approx 38.2\text{ kg}\).
(c) Number of students above \(37\text{ kg}\) = Total students \((60)\) - Cumulative frequency at \(37\text{ kg}\) \((29) = 31\text{ students}\).

Teacher's Note:
a) Construct a cumulative frequency table before plotting the ogive.
b) Read values for median (\(N/2\)) and upper quartile (\(3N/4\)) directly from the vertical axis to the curve.