ICSE Class 10 Mathematics Board Exam Question Paper 2019 with Solutions

ICSE Class 10 Mathematics Board Exam Question Paper with Solutions

SECTION A (40 Marks)

Attempt all questions from this Section.

Question 1

(a) Solve the following in equation and write down the solution set:
\( 11x - 4 \lt 15x + 4 \le 13x + 14, \, x \in W \)
Represent the solution on a real number line. [3 Marks]

Answer:
The given inequality is \( 11x - 4 \lt 15x + 4 \le 13x + 14 \).
Splitting into two parts:
Part 1: \( 11x - 4 \lt 15x + 4 \)
\( \Rightarrow 11x - 15x \lt 4 + 4 \)
\( \Rightarrow -4x \lt 8 \)
\( \Rightarrow x \gt -2 \)
Part 2: \( 15x + 4 \le 13x + 14 \)
\( \Rightarrow 15x - 13x \le 14 - 4 \)
\( \Rightarrow 2x \le 10 \)
\( \Rightarrow x \le 5 \)
Combining both parts, we get \( -2 \lt x \le 5 \).
Since \( x \in W \) (Whole numbers), the solution set is \( \{0, 1, 2, 3, 4, 5\} \).

[Figure: Number line showing solid dots at 0, 1, 2, 3, 4, 5 and an empty circle at -2 with an arrow towards 5]

Teacher's Note:
a) Separate the simultaneous inequation into two independent inequalities and solve them individually before finding their intersection.
b) Pay close attention to the replacement set (\( W \) means whole numbers starting from 0, not integers).

(b) A man invests Rs. 4500 in shares of a company which is paying 7.5% dividend. If Rs. 100 shares are available at a discount of 10%. Find:
(i) Number of shares he purchases.
(ii) His annual income. [3 Marks]

Answer:
1. Face Value (FV) = Rs. 100, Discount = \( 10\% \) of Rs. 100 = Rs. 10.
Market Value (MV) = FV - Discount = \( 100 - 10 = \) Rs. 90.
Number of shares = \( \frac{\text{Investment}}{\text{Market Value}} = \frac{4500}{90} = 50 \).\br />2. Annual Income = Number of shares \(\times\) Dividend percentage \(\times\) Face Value
\( = 50 \times \frac{7.5}{100} \times 100 = \) Rs. 375.

Teacher's Note:
a) Market value is calculated by subtracting the discount from the face value.
b) Dividend is always calculated on the total face value of the shares held, never on the market value or investment amount.

(c) In class of 40 students, marks obtained by the students a class test (out of 10) are given below,

Calculate the following for the given distribution:
(i) Median
(ii) Mode [4 Marks]

Answer:
Total frequency \( n = 40 \) (even).
1. Cumulative frequency table:
Marks (1): f = 1, cf = 1
Marks (2): f = 2, cf = 3
Marks (3): f = 3, cf = 6
Marks (4): f = 3, cf = 9
Marks (5): f = 6, cf = 15
Marks (6): f = 10, cf = 25
Marks (7): f = 5, cf = 30
Marks (8): f = 4, cf = 34
Marks (9): f = 3, cf = 37
Marks (10): f = 3, cf = 40
Median = Mean of \( (\frac{n}{2})^{\text{th}} \) and \( (\frac{n}{2} + 1)^{\text{th}} \) observations
\( = \text{Mean of } 20^{\text{th}} \text{ and } 21^{\text{th}} \text{ observations} \)
From the cumulative frequency table, both the \( 20^{\text{th}} \) and \( 21^{\text{th}} \) observations lie in the cumulative frequency 25, which corresponds to the mark 6.
Median = \( \frac{6 + 6}{2} = 6 \).
2. Mode is the observation with the highest frequency.
Highest frequency is 10, which corresponds to the mark 6.
Mode = 6.

Teacher's Note:
a) For discrete frequency distributions, always write out the cumulative frequency to locate the median position correctly.
b) Mode can be read directly by identifying the maximum frequency in a frequency distribution table.

Question 2

(a) Using the factor theorem, show that \((x - 2)\) is a factor of \(x^3 + x^2 - 4x - 4\), Hence, factorise the polynomial completely. [3 Marks]

Answer:
Let \(P(x) = x^3 + x^2 - 4x - 4\).
To show \((x - 2)\) is a factor, we evaluate \(P(2)\):
\(P(2) = (2)^3 + (2)^2 - 4(2) - 4\)
\(P(2) = 8 + 4 - 8 - 4 = 0\)
Since \(P(2) = 0\), by Factor Theorem, \((x - 2)\) is a factor of \(P(x)\).
Now, factorising the polynomial completely:
\(x^3 + x^2 - 4x - 4 = x^2(x + 1) - 4(x + 1)\)
\(= (x^2 - 4)(x + 1)\)
\(= (x - 2)(x + 2)(x + 1)\).

Teacher's Note:
a) Always state the Factor Theorem explicitly when showing that a binomial is a factor.
b) Grouping terms by taking common factors is the most efficient way to factorise cubic polynomials once the linear factor is identified.

(b) Prove that: \((\csc \theta - \sin \theta)(\sec \theta - \cos \theta)(\tan \theta + \cot \theta) = 1\) [3 Marks]

Answer:
L.H.S. = \((\csc \theta - \sin \theta)(\sec \theta - \cos \theta)(\tan \theta + \cot \theta)\)
\(= (\frac{1}{\sin \theta} - \sin \theta)(\frac{1}{\cos \theta} - \cos \theta)(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta})\)
\(= (\frac{1 - \sin^2 \theta}{\sin \theta})(\frac{1 - \cos^2 \theta}{\cos \theta})(\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta})\)
\(= (\frac{\cos^2 \theta}{\sin \theta})(\frac{\sin^2 \theta}{\cos \theta})(\frac{1}{\sin \theta \cos \theta})\)
\(= \frac{\cos^2 \theta \sin^2 \theta}{\sin^2 \theta \cos^2 \theta} = 1 =\) R.H.S.

Teacher's Note:
a) Converting all trigonometric ratios into sine and cosine is a reliable standard strategy for proving identities.
b) Remember to use the fundamental identity \(\sin^2 \theta + \cos^2 \theta = 1\) at appropriate steps to simplify expressions.

(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively, Find the:
(i) First term
(ii) Common difference
(iii) Sum of the first 20 terms [4 Marks]

Answer:
Let the first term be \(a\) and the common difference be \(d\).
1. Given \(a_4 = 8 \Rightarrow a + 3d = 8\) ---(1)
Given \(a_6 = 14 \Rightarrow a + 5d = 14\) ---(2)
Subtracting (1) from (2):
\(2d = 6 \Rightarrow d = 3\).
Substituting \(d = 3\) in equation (1):
\(a + 3(3) = 8 \Rightarrow a + 9 = 8 \Rightarrow a = -1\).
First term = \(-1\), Common difference = 3.
2. Sum of the first 20 terms (\(S_{20}\)):
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(S_{20} = \frac{20}{2}[2(-1) + (20 - 1)3]\)
\(S_{20} = 10[-2 + 19(3)]\)
\(S_{20} = 10[-2 + 57] = 10 \times 55 = 550\).

Teacher's Note:
a) Form simultaneous linear equations using the formula \(a_n = a + (n - 1)d\) for given terms.
b) Substitute the calculated values of \(a\) and \(d\) carefully into the sum formula with correct order of operations.

Question 3

(a) Simplify:
\(\sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix}\) [3 Marks]

Answer:
\(= \begin{bmatrix} \sin^2 A & -\sin A \cos A \\ \sin A \cos A & \sin^2 A \end{bmatrix} + \begin{bmatrix} \cos^2 A & \sin A \cos A \\ -\sin A \cos A & \cos^2 A \end{bmatrix}\)
\(= \begin{bmatrix} \sin^2 A + \cos^2 A & -\sin A \cos A + \sin A \cos A \\ \sin A \cos A - \sin A \cos A & \sin^2 A + \cos^2 A \end{bmatrix}\)
Since \(\sin^2 A + \cos^2 A = 1\):
\(= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Teacher's Note:
a) Multiply each element of the matrix by the scalar outside before performing matrix addition.
b) Simplify trigonometric identities inside the matrix elements to obtain the final identity matrix.

(b) M and N are two points on the X axis and Y axis respectively. P(3, 2) divides the line segment MN in the ratio 2 : 3. Find:
(i) The coordinates of M and N
(ii) Slope of the line MN. [3 Marks]

Answer:
1. Let M be \((a, 0)\) on the X-axis and N be \((0, b)\) on the Y-axis.
Using the section formula for internal division in ratio \(m : n = 2 : 3\) at point \(P(3, 2)\):
\((\frac{2(0) + 3(a)}{2 + 3}, \frac{2(b) + 3(0)}{2 + 3}) = (3, 2)\)
\((\frac{3a}{5}, \frac{2b}{5}) = (3, 2)\)
\( \frac{3a}{5} = 3 \Rightarrow 3a = 15 \Rightarrow a = 5 \)
\( \frac{2b}{5} = 2 \Rightarrow 2b = 10 \Rightarrow b = 5 \)
Coordinates of M are \((5, 0)\) and N are \((0, 5)\).
2. Slope of line MN passing through \((5, 0)\) and \((0, 5)\):
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 0}{0 - 5} = \frac{5}{-5} = -1\).

Teacher's Note:
a) Points on the X-axis have a y-coordinate of zero, and points on the Y-axis have an x-coordinate of zero.
b) Apply the section formula correctly by matching respective x and y coordinates.

(c) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32cm. Find the:
(i) Radius of the cylinder
(ii) Curved surface area of the cylinder
Take \(\pi = 3.1\) [4 Marks]

Answer:
1. Radius of sphere (\(r\)) = 6 cm.
Volume of sphere = \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi \) cm\({}^3\).
Height of cylinder (\(h\)) = 32 cm. Let radius of cylinder be \(r_1\).
Volume of cylinder = \(\pi r_1^2 h = \pi r_1^2 (32)\).
Since the volume remains constant when melted:
\(32\pi r_1^2 = 288\pi\)
\(r_1^2 = \frac{288}{32} = 9 \Rightarrow r_1 = 3\) cm.
Radius of the cylinder is 3 cm.
2. Curved Surface Area (CSA) of the cylinder = \(2\pi r_1 h\)
\(CSA = 2 \times 3.1 \times 3 \times 32\)
\(CSA = 595.2\) cm\({}^2\).

Teacher's Note:
a) Equate the volume of the original solid to the volume of the newly formed solid to find unknown dimensions.
b) Use the exact value of \(\pi\) specified in the question (in this case, \(\pi = 3.1\)) for all decimal calculations.

Question 4

(a) The following numbers, \(K + 3\), \(K + 2\), \(3K - 7\) and \(2K - 3\) are in proportion. Find K. [3 Marks]

Answer:
Since \(K + 3, K + 2, 3K - 7, 2K - 3\) are in proportion:
\(\frac{K + 3}{K + 2} = \frac{3K - 7}{2K - 3}\)
\((K + 3)(2K - 3) = (3K - 7)(K + 2)\)
\(2K^2 - 3K + 6K - 9 = 3K^2 + 6K - 7K - 14\)
\(2K^2 + 3K - 9 = 3K^2 - K - 14\)
\(K^2 - 4K - 5 = 0\)
\(K^2 - 5K + K - 5 = 0\)
\(K(K - 5) + 1(K - 5) = 0\)
\((K - 5)(K + 1) = 0\)
\(K = 5\) or \(K = -1\).

Teacher's Note:
a) Four quantities \(a, b, c, d\) in proportion satisfy the relation \(a : b :: c : d\), which means \(\frac{a}{b} = \frac{c}{d}\).
b) Solve the resulting quadratic equation carefully by factorisation or formula method.

(b) Solve for x the quadratic equation \(x^2 - 4x - 8 = 0\). Give your answer correct to three significant figures. [3 Marks]

Answer:
Given equation: \(x^2 - 4x - 8 = 0\).
Comparing with \(ax^2 + bx + c = 0\), we get \(a = 1, b = -4, c = -8\).
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(b^2 - 4ac = (-4)^2 - 4(1)(-8) = 16 + 32 = 48\)
\(x = \frac{-(-4) \pm \sqrt{48}}{2(1)}\)
\(x = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}\)
Given \(\sqrt{3} \approx 1.732\):
\(x = 2 \pm 2(1.732) = 2 \pm 3.464\)
\(x = 2 + 3.464 = 5.464\) or \(x = 2 - 3.464 = -1.464\)
Correct to three significant figures, \(x = 5.46\) or \(x = -1.46\).

Teacher's Note:
a) Always substitute values into the quadratic formula with correct signs, especially negative coefficients.
b) Round off the final answers strictly to the required number of significant figures as requested in the question.

(c) Use ruler and compass only for answering this question. Draw a circle of radius 4 cm. Mark the centre as O. Mark a point P outside the circle at a distance of 7 cm from the centre, Construct two tangents to the circle from the external point P. Measure and write down the length of any one tangent.

Answer:
Steps of construction:
1. Draw a circle with centre O and radius 4 cm.
2. Mark a point P at a distance of 7 cm from O and join OP.
3. Draw the perpendicular bisector of OP to find its midpoint, let it be X.
4. With X as centre and XO as radius, draw a circle (or arcs) to intersect the given circle at points S and T.
5. Join PS and PT. These are the required tangents.
6. Measure length of PS, which is approximately 5.74 cm.

Teacher's Note:
a) Leave all construction arcs clearly visible and do not erase them.
b) The length of tangents drawn from an external point to a circle are equal in length.

SECTION B (40 Marks)

Attempt any four questions from this Section.

Question 5

(a) There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. [3 Marks]
Find the probability that the number on the disc is:
(i) An odd number
(ii) Divisible by 2 and 3 both.
(iii) A number less than 16.

Answer:
Total number of possible outcomes \(n(S) = 25\).
1. Let A be the event of getting an odd number.
\(A = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25\}\), \(n(A) = 13\).
Probability = \(\frac{n(A)}{n(S)} = \frac{13}{25}\).
2. Let B be the event of getting a number divisible by both 2 and 3 (i.e., divisible by 6).
\(B = \{6, 12, 18, 24\}\), \(n(B) = 4\).
Probability = \(\frac{n(B)}{n(S)} = \frac{4}{25}\).
3. Let C be the event of getting a number less than 16.
\(C = \{1, 2, 3, \dots, 15\}\), \(n(C) = 15\).
Probability = \(\frac{n(C)}{n(S)} = \frac{15}{25} = \frac{3}{5}\).

Teacher's Note:
a) Divisibility by both 2 and 3 requires finding the least common multiple, which is 6.
b) Always reduce probability fractions to their simplest form.

(b) Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives Rs. 441 as interest at the time of maturity. Find the amount Rekha deposited each month. [3 Marks]

Answer:
Let monthly deposit be \(P\).
Time period \(n = 20\) months.
Rate \(r = 9\%\) per annum.
Interest \(I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}\)
\(441 = P \times \frac{20(20 + 1)}{24} \times \frac{9}{100}\)
\(441 = P \times \frac{20 \times 21}{24} \times \frac{9}{100}\)
\(441 = P \times \frac{420}{24} \times \frac{9}{100} = P \times 17.5 \times 0.09 = P \times 1.575\)
\(P = \frac{441}{1.575} = 280\).
Monthly deposit = Rs. 280.

Teacher's Note:
a) Ensure the time period in months is correctly applied in the denominator (\(2 \times 12 = 24\)).
b) Double-check decimal arithmetic when isolating the principal amount \(P\).

(c) Use a graph sheet for this question. Take 1 cm = 1 unit along both x and y axis.
(i) Plot the following points: A(0, 5), B(3, 0), C(1, 0), and D(1, -5)
(ii) Reflect the points B, C, and D on the y axis and name them as B’, C’ and D’ respectively.
(iii) Write down the coordinates of B’, C’ and D’.
(iv) Join the points A, B, C, D, D’, C’, B’, A in order and give a name to the closed figure ABCDD’C’B’. [4 Marks]

Answer:
1. Plotting points A(0, 5), B(3, 0), C(1, 0), and D(1, -5) on the graph sheet.
2. Reflection across the y-axis changes the sign of the x-coordinate (\((x, y) \rightarrow (-x, y)\)).
3. Coordinates of reflected points:
B'( -3, 0)
C'( -1, 0)
D'( -1, -5)
4. Joining points A, B, C, D, D', C', B', A in order forms a symmetrical arrow-like polygon, which is identified as an Arrowhead or Delta shape / Kite.

Teacher's Note:
a) Recall the coordinate transformation rules for reflections across axes: reflection in the y-axis negates the x-coordinate.
b) Label all plotted and reflected points clearly on the graph sheet.

Question 6

(a) In the given figure \(\angle PQR = \angle PST = 90^{\circ}\), PQ = 5 cm and PS = 2 cm.
(i) Prove that \(\triangle PQR \sim \triangle PST\)
(ii) Find Area of \(\triangle PQR : \text{Area of quadrilateral SRQT}\). [3 Marks]

[Figure: Triangle PQR with perpendicular line segment ST on PQ such that angle PQR = angle PST = 90 degrees, PQ = 5 cm, PS = 2 cm]

Answer:
1. In \(\triangle PQR\) and \(\triangle PST\):
\(\angle PQR = \angle PST = 90^{\circ}\) (Given)
\(\angle RPQ = \angle SPT\) (Common angle)
Therefore, \(\triangle PQR \sim \triangle PST\) (by AA similarity).
2. Since the triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides:
\(\frac{\text{ar}(\triangle PQR)}{\text{ar}(\triangle PST)} = (\frac{PQ}{PS})^2 = (\frac{5}{2})^2 = \frac{25}{4}\).
Let \(\text{ar}(\triangle PQR) = 25k\) and \(\text{ar}(\triangle PST) = 4k\).
\(\text{Area of quadrilateral SRQT} = \text{ar}(\triangle PQR) - \text{ar}(\triangle PST) = 25k - 4k = 21k\).
Ratio of Area of \(\triangle PQR\) to Area of quadrilateral SRQT = \(25 : 21\).

Teacher's Note:
a) Prove triangle similarity using two pairs of equal corresponding angles.
b) The ratio of areas of two similar triangles is equal to the square of the ratio of any pair of corresponding sides.

(b) The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find:
(i) ‘n’ the number of terms of the G.P.
(ii) Sum of the n terms. [3 Marks]

Answer:
Given first term \(a = 3\), last term \(l = a_n = 96\), common ratio \(r = 2\).
1. Formula for \(n^{\text{th}}\) term of G.P.: \(a_n = a r^{n-1}\)
\(96 = 3(2)^{n-1}\)
\(2^{n-1} = \frac{96}{3} = 32\)
\(2^{n-1} = 2^5\)
\(n - 1 = 5 \Rightarrow n = 6\).
Number of terms \(n = 6$.
2. Sum of \(n\) terms (\(S_n\)) = \(\frac{a(r^n - 1)}{r - 1}\)
\(S_6 = \frac{3(2^6 - 1)}{2 - 1} = \frac{3(64 - 1)}{1} = 3 \times 63 = 189\).

Teacher's Note:
a) Express both sides of the exponential equation with the same base to solve for \(n\).
b) Use the appropriate sum formula for G.P. when \(r \gt 1\).

(c) A hemispherical and a conical hole is scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows: The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm. Give your answer correct to the nearest whole number Take \(\pi = \frac{22}{7}\) [4 Marks]

[Figure: Solid cylinder of height 7 cm and radius 3 cm with a conical cavity at the top and a hemispherical cavity at the bottom, both having radius 3 cm and cone height 3 cm]

Answer:
Radius \(r = 3\) cm, height of cylinder \(h = 7\) cm, height of cone \(H = 3\) cm.
1. Volume of cylinder (\(V_1\)) = \(\pi r^2 h = \frac{22}{7} \times (3)^2 \times 7 = \frac{22}{7} \times 9 \times 7 = 198\) cm\({}^3\).
2. Volume of cone (\(V_2\)) = \(\frac{1}{3}\pi r^2 H = \frac{1}{3} \times \frac{22}{7} \times (3)^2 \times 3 = \frac{22}{7} \times 9 = \frac{198}{7} \approx 28.29\) cm\({}^3\).
3. Volume of hemisphere (\(V_3\)) = \(\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (3)^3 = \frac{2}{3} \times \frac{22}{7} \times 27 = \frac{396}{7} \approx 56.57\) cm\({}^3\).
4. Volume of remaining solid = \(V_1 - V_2 - V_3\)
\(= 198 - \frac{198}{7} - \frac{396}{7} = 198 - \frac{594}{7} = 198 - 84.857 = 113.143\) cm\({}^3\).
Correct to the nearest whole number, the volume is 113 cm\({}^3\).

Teacher's Note:
a) Scooping out cavities means subtracting their individual volumes from the total volume of the cylinder.
b) Pay close attention to rounding instructions (nearest whole number) at the end of the calculation.

Question 7

(a) In the given figure AC is a tangent to the circle with centre O. If \(\angle ADB = 55^{\circ}\), find x and y, Give reasons for your answer. [3 Marks]

[Figure: Circle with centre O, chord AB, point D on circumference, tangent AC at A, angles labeled x and y at centre and vertices]

Answer:
1. In \(\triangle ABD\), \(\angle BAD = 90^{\circ}\) (Angle in a semicircle is a right angle).
Given \(\angle ADB = 55^{\circ}\).
In \(\triangle ABD\), sum of angles = \(180^{\circ}\):
\(\angle ABD = 180^{\circ} - (90^{\circ} + 55^{\circ}) = 180^{\circ} - 145^{\circ} = 35^{\circ}\).
2. Angle subtended by arc AE at the centre (\(\angle AOE\), labeled \(y\)) is double the angle subtended at the remaining part of the circle (\(\angle ABD\)):
\(y = 2 \times \angle ABD = 2 \times 35^{\circ} = 70^{\circ}\).
3. Radius OA is perpendicular to tangent AC at point A, so \(\angle OAC = 90^{\circ}\).
Also, \(\angle OAC = \angle AOC + x\) or from right triangle geometry, since \(\angle AOC = y = 70^{\circ}\):
\(x = 90^{\circ} - 35^{\circ} = 55^{\circ}\) (or using angle sum in \(\triangle AOC\)).
Thus, \(x = 55^{\circ}\) and \(y = 70^{\circ}\).

Teacher's Note:
a) Use standard circle theorems: angle in a semicircle is \(90^{\circ}\), and the angle at the centre is twice the angle at the circumference.
b) The radius drawn to the point of contact is perpendicular to the tangent.

(b) The model of a building is constructed with scale factor 1 : 30. [3 Marks]
(i) If the height of the model is 80 cm, find the actual height of the building in meters.
(ii) If the actual volume of a tank at the top of the building is \(27 m^3\), find the volume of the tank on the top of the model.

Answer:
1. Scale factor \(k = \frac{1}{30}\).
Height of model = 80 cm.
Actual height = Height of model \(\times 30 = 80 \times 30 = 2400\) cm = 24 meters.
2. Ratio of volumes is the cube of the scale factor: \(\frac{\text{Volume of model}}{\text{Actual Volume}} = (\frac{1}{30})^3 = \frac{1}{27000}\).
Actual volume = \(27 \text{ m}^3 = 27 \times 10^6 \text{ cm}^3\) (or using \(1 \text{ m}^3 = 10^6 \text{ cm}^3\)).
Volume of model tank = \(\frac{\text{Actual Volume}}{27000} = \frac{27 \text{ m}^3}{27000} = \frac{1}{1000} \text{ m}^3\).
Since \(1 \text{ m}^3 = 1000 \text{ litres}\):
Volume of model tank = \(\frac{1}{1000} \times 1000 \text{ litres} = 1 \text{ litre}\).

Teacher's Note:
a) Remember that for similar figures, length scales linearly, area scales as the square, and volume scales as the cube of the scale factor.
b) Convert units correctly between cubic centimeters and cubic meters or liters.

(c) Given, \(\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} M = 6I\), where M is a matrix and I is unit matrix of order 2 x 2. [4 Marks]
(i) State the order of matrix M.
(ii) Find the matrix M

Answer:
1. The given matrix has order \(2 \times 2\) and \(6I\) has order \(2 \times 2\). For matrix multiplication to be valid and result in a \(2 \times 2\) matrix, matrix M must be of order \(2 \times 2\).
2. Let \(M = \begin{bmatrix} p & q \\ r & s \end{bmatrix}\).
Given equation is \(\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} p & q \\ r & s \end{bmatrix} = 6 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 4p + 2r & 4q + 2s \\ -p + r & -q + s \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}\)
Equating corresponding elements:
\(4p + 2r = 6\) ---(1)
\(-p + r = 0 \Rightarrow r = p\) ---(2)
Substitute \(r = p\) in (1):
\(4p + 2p = 6 \Rightarrow 6p = 6 \Rightarrow p = 1\), hence \(r = 1$.
Also:
\(4q + 2s = 0\) ---(3)
\(-q + s = 6 \Rightarrow s = 6 + q\) ---(4)
Substitute \(s = 6 + q\) in (3):
\(4q + 2(6 + q) = 0\)
\(4q + 12 + 2q = 0 \Rightarrow 6q = -12 \Rightarrow q = -2\).
Then \(s = 6 + (-2) = 4\).
Therefore, matrix \(M = \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}\).

Teacher's Note:
a) Determine the order of an unknown matrix by checking the dimensions of conformable matrices in multiplication.
b) Form simultaneous linear equations by equating corresponding elements of equal matrices.

Question 8

(a) The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. Find the first term and the common difference. [3 Marks]

Answer:
Let the three consecutive terms of the A.P. be \((a - d)\), \(a\), and \((a + d)\).
1. Sum of the first three terms = 42:
\((a - d) + a + (a + d) = 42\)
\(3a = 42 \Rightarrow a = 14\).
2. Product of the first and third term = 52:
\((a - d)(a + d) = 52\)
\(a^2 - d^2 = 52\)
Substitute \(a = 14\):
\((14)^2 - d^2 = 52\)
\(196 - d^2 = 52\)
\(d^2 = 196 - 52 = 144\)
\(d = \pm 12\).
First term \(a = 14\) and common difference \(d = 12\) (or \(-12\)).

Teacher's Note:
a) For three terms in an A.P., assuming them as \((a - d), a, (a + d)\) simplifies calculations significantly.
b) Remember that a square root has both positive and negative values unless restricted by problem context.

(b) The vertices of a \(\triangle ABC\) are A(3, 8), B(-1, 2) and C(6, -6), Find:
(i) Slope of BC.
(ii) Equation of a line perpendicular to BC and passing through A. [3 Marks]

Answer:
1. Slope of BC passing through \(B(-1, 2)\) and \(C(6, -6)\):
\(m_{BC} = \frac{-6 - 2}{6 - (-1)} = \frac{-8}{7}\).
2. Let the line perpendicular to BC be AE. Its slope \(m_{AE}\) is the negative reciprocal of \(m_{BC}\):
\(m_{AE} = -\frac{1}{m_{BC}} = -\frac{1}{-8/7} = \frac{7}{8}\).
Equation of line AE passing through \(A(3, 8)\) with slope \(\frac{7}{8}\):
\(y - y_1 = m(x - x_1)\)
\(y - 8 = \frac{7}{8}(x - 3)\)
\(8(y - 8) = 7(x - 3)\)
\(8y - 64 = 7x - 21\)
\(7x - 8y + 43 = 0\).

Teacher's Note:
a) Perpendicular lines have slopes whose product is \(-1\) (\(m_1 m_2 = -1\)).
b) Use the point-slope form to find the equation of a straight line given a point and its slope.

(c) Using ruler and a compass only construct a semi-circle with diameter BC = 7cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Measure \(\angle ADC\) and write it down. [4 Marks]

Answer:
Steps of construction:
1. Draw line segment \(BC = 7\) cm and construct a semi-circle on BC as diameter.
2. Draw the perpendicular bisector of BC to intersect the semi-circle at point A. Join AB and AC.
3. Point A is equidistant from B and C.
4. Draw the angle bisector of \(\angle ABC\) to meet the semi-circle at point D. Join AD and CD.
5. Point D is equidistant from AB and BC.
6. Measure \(\angle ADC\), which is equal to \(135^{\circ}\).

Teacher's Note:
a) The perpendicular bisector of a chord passes through the centre and bisects the arc, locating point A equidistant from B and C.
b) The angle bisector of an angle of a cyclic quadrilateral helps locate equidistant points on the circumference.

Question 9

(a) The data on the number of patients attending a hospital in a month are given below. Find the average (mean) number of patients attending the hospital in a month by using the shortcut method. Take the assumed mean as 45. Give your answer correct to 2 decimal places. [3 Marks]

Answer:
Assumed mean \(A = 45\).
Frequency table with mid-points (\(x\)), deviation \(d = x - A\), and \(fd\):
- Class 10-20: Mid-point \(x = 15\), \(f = 5\), \(d = 15 - 45 = -30\), \(fd = -150\)
- Class 20-30: Mid-point \(x = 25\), \(f = 2\), \(d = 25 - 45 = -20\), \(fd = -40\)
- Class 30-40: Mid-point \(x = 35\), \(f = 7\), \(d = 35 - 45 = -10\), \(fd = -70\)
- Class 40-50: Mid-point \(x = 45\), \(f = 9\), \(d = 45 - 45 = 0\), \(fd = 0\)
- Class 50-60: Mid-point \(x = 55\), \(f = 2\), \(d = 55 - 45 = 10\), \(fd = 20\)
- Class 60-70: Mid-point \(x = 65\), \(f = 5\), \(d = 65 - 45 = 20\), \(fd = 100\)
Totals: \(\sum f = 30\), \(\sum fd = -140\).
Mean (\(\bar{x}\)) = \(A + \frac{\sum fd}{\sum f} = 45 + \frac{-140}{30} = 45 - 4.67 = 40.33\) (daily average).
Total patients in a month (assuming 30 days) = \(40.33 \times 30 = 1209.9\) (or approx 1210).

Teacher's Note:
a) The assumed mean method uses deviations \(d = x - A\) to simplify calculations.
b) Ensure all decimal answers are rounded to the specified number of decimal places.

(b) Using properties of proportion solve for x, given. \(\frac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4\) [3 Marks]

Answer:
Given equation: \(\frac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = \frac{4}{1}\).
Applying Componendo and Dividendo:
\(\frac{(\sqrt{5x} + \sqrt{2x - 6}) + (\sqrt{5x} - \sqrt{2x - 6})}{(\sqrt{5x} + \sqrt{2x - 6}) - (\sqrt{5x} - \sqrt{2x - 6})} = \frac{4 + 1}{4 - 1}\)
\(\frac{2\sqrt{5x}}{2\sqrt{2x - 6}} = \frac{5}{3}\)
\(\frac{\sqrt{5x}}{\sqrt{2x - 6}} = \frac{5}{3}\)
Squaring both sides:
\(\frac{5x}{2x - 6} = \frac{25}{9}\)
Cross-multiplying:
\(9(5x) = 25(2x - 6)\)
\(45x = 50x - 150\)
\(50x - 45x = 150\)
\(5x = 150 \Rightarrow x = 30\).

Teacher's Note:
a) Componendo and Dividendo (\(\frac{a+b}{a-b} = \frac{c+d}{c-d}\)) is the quickest method to solve equations involving square roots in proportions.
b) Always verify the calculated value of \(x\) by substituting it back into the original equation to check for extraneous roots.

(c) Sachin invests Rs. 8500 in 10%, Rs. 100 shares at Rs. 170 He sells the shares when the price of each share rises by Rs. 30. He invests the proceeds in 12% Rs. 100 shares at Rs. 125, Find:
(i) The sale proceeds
(ii) The number of Rs. 125 shares he buys.
(iii) The change in his annual income. [4 Marks]

Answer:
1. Initial investment = Rs. 8500, Market Value (MV) = Rs. 170.
Number of shares purchased initially = \(\frac{8500}{170} = 50\) shares.
Selling price of each share = \(170 + 30 =\) Rs. 200.
(i) Sale proceeds = \(50 \times 200 =\) Rs. 10,000.
(ii) New shares bought at MV = Rs. 125.
Number of new shares = \(\frac{10000}{125} = 80\) shares.
(iii) Initial annual income:
Face Value (FV) of 50 shares = \(50 \times 100 =\) Rs. 5000.
Initial dividend = \(10\% \text{ of } 5000 =\) Rs. 500.
New annual income:
Face Value of 80 shares = \(80 \times 100 =\) Rs. 8000.
New dividend = \(12\% \text{ of } 8000 = 0.12 \times 8000 =\) Rs. 960.
Change in annual income = \(960 - 500 =\) Rs. 460 (increase).

Teacher's Note:
a) Sale proceeds represent the total money obtained after selling all shares at the new market price.
b) Income is always calculated on the total face value using the dividend percentage.

Question 10

(a) Use graph paper for this question. The marks obtained by 120 students in an English test are given below. [6 Marks]

Draw the ogive and hence, estimate:
(i) The median marks.
(ii) The number of students who did not pass the test if the pass percentage was 50.
(iii) The upper quartile marks.

Answer:
Cumulative frequency (c.f.) table:
- 0-10: 5
- 10-20: \(5 + 9 = 14$
- 20-30: \(14 + 16 = 30$
- 30-40: \(30 + 22 = 52$
- 40-50: \(52 + 26 = 78$
- 50-60: \(78 + 18 = 96$
- 60-70: \(96 + 11 = 107$
- 70-80: \(107 + 6 = 113$
- 80-90: \(113 + 4 = 117$
- 90-100: \(117 + 3 = 120$
Total number of students \(n = 120\).
1. Median = term at \(\frac{n}{2} = \frac{120}{2} = 60^{\text{th}}\) position.
From the ogive graph, corresponding to cumulative frequency 60, the median marks = 43.
2. Pass percentage is 50%, which corresponds to passing marks of 50 out of 100.
Number of students who did not pass (scoring less than 50) can be read from the ogive at mark 50, which is 78 students.
3. Upper quartile (\(Q_3\)) = term at \(\frac{3n}{4} = \frac{3 \times 120}{4} = 90^{\text{th}}\) position.
From the ogive graph, corresponding to cumulative frequency 90, the upper quartile marks = 58 (or approx from graph reading around 57-58).

Teacher's Note:
a) Plot cumulative frequencies against the upper class limits of each class interval to draw the less-than ogive.
b) Locate quartiles and median on the cumulative frequency axis (y-axis) and drop perpendiculars to read values from the x-axis.

(b) A man observes the angle of elevation of the top of the lower to be \(45^{\circ}\). He walks towards it in a horizontal line through its base. On covering 20 m the angle of elevation changes to \(60^{\circ}\). Find the height of the tower correct to 2 significant figures. [4 Marks]

Answer:
Let PQ be the height of the tower = \(h\).
Let the initial distance from point S to the base be \(SP\). After walking 20 m to point R, \(RP = x\), so \(SP = 20 + x\).
1. In right-angled \(\triangle QSP\):
\(\tan 45^{\circ} = \frac{h}{20 + x} \Rightarrow 1 = \frac{h}{20 + x} \Rightarrow h = 20 + x \Rightarrow x = h - 20\).
2. In right-angled \(\triangle PQR\):
\(\tan 60^{\circ} = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}\).
Equating both expressions for \(x\):
\(h - 20 = \frac{h}{\sqrt{3}}\)
\(h - \frac{h}{\sqrt{3}} = 20\)
\(h(\frac{\sqrt{3} - 1}{\sqrt{3}}) = 20\)
\(h = \frac{20\sqrt{3}}{\sqrt{3} - 1}\)
Rationalising the denominator:
\(h = \frac{20\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(3 + \sqrt{3})}{3 - 1} = \frac{20(3 + 1.732)}{2} = 10(4.732) = 47.32\) m.
Correct to 2 significant figures, height \(h = 47\) m.

Teacher's Note:
a) Set up simultaneous trigonometric equations using tangent ratios for both observer positions.
b) Substitute standard values for square roots and round off correctly to significant figures.

Question 11

(a) Using the Remainder Theorem find the remainders obtained when \(x^3 + (kx + 8)x + k\) is divided by \(x + 1\) and \(x - 2\). Hence find k if the sum of the two remainders is 1. [3 Marks]

Answer:
Let \(P(x) = x^3 + kx^2 + 8x + k\).
1. Remainder when divided by \((x + 1)\) is \(P(-1)\):
\(P(-1) = (-1)^3 + k(-1)^2 + 8(-1) + k\)
\(= -1 + k - 8 + k = 2k - 9\).
2. Remainder when divided by \((x - 2)\) is \(P(2)\):
\(P(2) = (2)^3 + k(2)^2 + 8(2) + k\)
\(= 8 + 4k + 16 + k = 5k + 24\).
3. Given that the sum of the two remainders is 1:
\(P(-1) + P(2) = 1$
\((2k - 9) + (5k + 24) = 1$
\(7k + 15 = 1$
\(7k = 1 - 15 = -14$
\(k = -2\).

Teacher's Note:
a) Remainder Theorem states that the remainder of division of polynomial \(P(x)\) by \((x - a)\) is equal to \(P(a)\).
b) Simplify polynomial expressions carefully before applying given linear conditions.

(b) The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers. [3 Marks]

Answer:
Let the two consecutive natural numbers which are multiples of 3 be \(3x\) and \(3x + 3\).
According to the question, their product is 810:
\(3x(3x + 3) = 810$
\(9x^2 + 9x = 810$
Dividing by 9:
\(x^2 + x = 90$
\(x^2 + x - 90 = 0$
\(x^2 + 10x - 9x - 90 = 0$
\(x(x + 10) - 9(x + 10) = 0$
\((x + 10)(x - 9) = 0$
\(x = 9\) or \(x = -10\).
Since natural numbers must be positive, \(x = 9\).
First number = \(3(9) = 27\).
Second number = \(3(9) + 3 = 30\).
The required numbers are 27 and 30 (or \(-27\) and \(-30\) if integers).

Teacher's Note:
a) Represent consecutive multiples of any number \(m\) as \(mx\) and \(mx + m\).
b) Check the domain requirement (natural numbers) to reject inadmissible negative roots.

(c) In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side \(BC \parallel AE\). If \(\angle BAC = 50^{\circ}\), find giving reasons:
(i) \(\angle ACB\)
(ii) \(\angle EDC\)
(iii) \(\angle BEC\)
Hence prove that BE is also a diameter. [4 Marks]

[Figure: Cyclic pentagon ABCDE inscribed in a circle with diameter AC, parallel lines BC and AE, angle BAC = 50 degrees]

Answer:
1. \(\angle ABC = 90^{\circ}\) (Angle in a semicircle).
In \(\triangle ABC\), \(\angle BAC = 50^{\circ}\):
\(\angle ACB = 180^{\circ} - (90^{\circ} + 50^{\circ}) = 40^{\circ}\).
2. Since \(BC \parallel AE\) and AC is a transversal, alternate interior angles are equal:
\(\angle EAC = \angle ACB = 40^{\circ}\).
In \(\triangle AEC\), \(\angle AEC = 90^{\circ}\) (Angle in a semicircle on diameter AC).
\(\angle ACE = 180^{\circ} - (90^{\circ} + 40^{\circ}) = 50^{\circ}\).
Since AEDC is a cyclic quadrilateral, opposite angles sum to \(180^{\circ}\):
\(\angle EAC + \angle EDC = 180^{\circ}\)
\(40^{\circ} + \angle EDC = 180^{\circ} \Rightarrow \angle EDC = 140^{\circ}\).
3. Arc BC subtends \(\angle BAC\) and \(\angle BEC\) in the same segment:
\(\angle BEC = \angle BAC = 50^{\circ}\).
4. Since \(\angle BAE = \angle BAC + \angle EAC = 50^{\circ} + 40^{\circ} = 90^{\circ}\) and the angle subtended by BE at the circumference is \(90^{\circ}\), BE must be a diameter of the circle.

Teacher's Note:
a) Any angle subtended by a diameter at the circumference is a right angle (\(90^{\circ}\)).
b) Opposite angles of a cyclic quadrilateral are supplementary (sum to \(180^{\circ}\)).