ICSE Class 10 Mathematics Board Exam Question Paper 2018 with Solutions

ICSE Class 10 Mathematics Board Exam Question Paper with Solutions

SECTION A (40 Marks)

Question 1

(a) Find the value of \(x\) and \(y\) if:
\(2 \begin{bmatrix} x & 7 \\ 9 & y-5 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}\) [3 Marks]

Answer:
Multiplying the matrix by scalar 2:
\(\begin{bmatrix} 2x & 14 \\ 18 & 2y - 10 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}\)
Adding the matrices on the L.H.S.:
\(\begin{bmatrix} 2x + 6 & 14 - 7 \\ 18 + 4 & 2y - 10 + 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}\)
\(\begin{bmatrix} 2x + 6 & 7 \\ 22 & 2y - 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}\)
Equating the corresponding elements:
\(2x + 6 = 10 \implies 2x = 4 \implies x = 2\)
\(2y - 5 = 15 \implies 2y = 20 \implies y = 10\)
Thus, \(x = 2\) and \(y = 10\).

Teacher's Note:
a) Always multiply every element inside the matrix by the scalar factor before performing matrix addition.
b) Ensure that corresponding elements of equal matrices are equated carefully.

(b) Sonia had a recurring deposit account in a bank and deposited Rs. 600 per month for \(2 \frac{1}{2}\) years. If the rate of interest was 10% p.a., find the maturity value of this account. [3 Marks]

Answer:
Monthly deposit (\(P\)) = Rs. 600
Time (\(n\)) = \(2 \frac{1}{2}\) years = \(30\) months
Rate (\(r\)) = 10% p.a.
Interest (\(I\)) = \(P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}\)
\(I = 600 \times \frac{30(30 + 1)}{2 \times 12} \times \frac{10}{100} = \text{Rs. } 2325\)
Total money deposited = \(P \times n = 600 \times 30 = \text{Rs. } 18,000\)
Maturity Value = Total deposit + Interest = \(18,000 + 2325 = \text{Rs. } 20,325\).

Teacher's Note:
a) Convert the time given in years into months for all recurring deposit formulas.
b) Total maturity value is the sum of total money deposited and the total interest earned.

(c) Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is:
(i) a prime number.
(ii) a number divisible by 4.
(iii) a number that is a multiple of 6.
(iv) an odd number. [4 Marks]

Answer:
Total number of cards (\(n(S)\)) = 10 (Sample space = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20})
(i) Prime number card: {2}
Number of favourable outcomes = 1
Probability = \(\frac{1}{10}\)
(ii) Number divisible by 4: {4, 8, 12, 16, 20}
Number of favourable outcomes = 5
Probability = \(\frac{5}{10} = \frac{1}{2}\)
(iii) Number that is a multiple of 6: {6, 12, 18}
Number of favourable outcomes = 3
Probability = \(\frac{3}{10}\)
(iv) Odd number: None in the given set.
Number of favourable outcomes = 0
Probability = 0

Teacher's Note:
a) Remind students that 2 is the only even prime number.
b) Probability can never be greater than 1 or less than 0; an impossible event has a probability of 0.

Question 2

(a) The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the
(i) radius of the cylinder
(ii) volume of cylinder (use \(\pi = \frac{22}{7}\)) [3 Marks]

Answer:
Let the radius be \(r\) and height be \(h = 25\) cm.
(i) Circumference of base = \(132\) cm
\(2\pi r = 132\)
\(2 \times \frac{22}{7} \times r = 132\)
\(r = \frac{132 \times 7}{2 \times 22} = 21\text{ cm}\)
(ii) Volume of cylinder = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times (21)^2 \times 25 = \frac{22}{7} \times 21 \times 21 \times 25 = 34,650\text{ cm}^3\).

Teacher's Note:
a) Use the base circumference formula \(2\pi r\) to find the radius directly.
b) Include proper cubic units (\(\text{cm}^3\)) for volume.

(b) If \((k - 3)\), \((2k + 1)\) and \((4k + 3)\) are three consecutive terms of an A.P., find the value of k. [3 Marks]

Answer:
Since the terms are in A.P., the common difference is constant:
\((2k + 1) - (k - 3) = (4k + 3) - (2k + 1)\)
\(2k + 1 - k + 3 = 4k + 3 - 2k - 1\)
\(k + 4 = 2k + 2\)
\(4 - 2 = 2k - k\)
\(k = 2\).

Teacher's Note:
a) For any three consecutive terms \(a\), \(b\), \(c\) in A.P., the relation \(2b = a + c\) holds true.
b) Be careful with negative signs when opening brackets during subtraction.

(c) PQRS is a cyclic quadrilateral. Given \(\angle QPS = 73^{\circ}\), \(\angle PQS = 55^{\circ}\) and \(\angle PSR = 82^{\circ}\), calculate:
(i) \(\angle QRS\)
(ii) \(\angle RQS\)
(iii) \(\angle PRQ\) [4 Marks]
[Figure: A cyclic quadrilateral PQRS inscribed in a circle with vertices P, Q, R, S in order. Chord QS is joined. \(\angle QPS = 73^{\circ}\), \(\angle PQS = 55^{\circ}\), and \(\angle PSR = 82^{\circ}\).]

Answer:
(i) Opposite angles of a cyclic quadrilateral are supplementary.
\(\angle QPS + \angle QRS = 180^{\circ}\)
\(73^{\circ} + \angle QRS = 180^{\circ}\)
\(\angle QRS = 180^{\circ} - 73^{\circ} = 107^{\circ}\)
(ii) In cyclic quadrilateral PQRS:
\(\angle PQR + \angle PSR = 180^{\circ}\)
\(\angle PQS + \angle RQS + 82^{\circ} = 180^{\circ}\)
\(55^{\circ} + \angle RQS + 82^{\circ} = 180^{\circ}\)
\(137^{\circ} + \angle RQS = 180^{\circ}\)
\(\angle RQS = 180^{\circ} - 137^{\circ} = 43^{\circ}\)
(iii) In \(\triangle PQS\), by angle sum property:
\(\angle PSQ + \angle PQS + \angle QPS = 180^{\circ}\)
\(\angle PSQ + 55^{\circ} + 73^{\circ} = 180^{\circ}\)
\(\angle PSQ + 128^{\circ} = 180^{\circ}\)
\(\angle PSQ = 52^{\circ}\)
Since angles in the same segment are equal (\(\angle PRQ\) and \(\angle PSQ\) stand on arc PQ):
\(\angle PRQ = \angle PSQ = 52^{\circ}\).

Teacher's Note:
a) Recall that opposite angles of a cyclic quadrilateral add up to \(180^{\circ}\).
b) Angles subtended by the same arc in the same segment of a circle are equal.

Question 3

(a) If \((x + 2)\) and \((x + 3)\) are factors of \(x^3 + ax + b\), find the values of 'a' and 'b'. [3 Marks]

Answer:
Let \(P(x) = x^3 + ax + b\).
Since \((x + 2)\) is a factor, \(P(-2) = 0\):
\((-2)^3 + a(-2) + b = 0\)
\(-8 - 2a + b = 0 \implies -2a + b = 8\) --- (1)
Since \((x + 3)\) is a factor, \(P(-3) = 0\):
\((-3)^3 + a(-3) + b = 0\)
\(-27 - 3a + b = 0 \implies -3a + b = 27\) --- (2)
Subtracting equation (1) from equation (2):
\((-3a + b) - (-2a + b) = 27 - 8\)
\(-a = 19 \implies a = -19\)
Substituting \(a = -19\) in equation (1):
\(-2(-19) + b = 8\)
\(38 + b = 8 \implies b = 8 - 38 = -30\)
Thus, \(a = -19\) and \(b = -30\).

Teacher's Note:
a) Use the Factor Theorem: if \((x - c)\) is a factor of a polynomial \(P(x)\), then \(P(c) = 0\).
b) Solve the resulting simultaneous linear equations carefully using elimination.

(b) Prove that \(\sqrt{\sec^2 \theta + \csc^2 \theta} = \tan \theta + \cot \theta\) [3 Marks]

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

Teacher's Note:
a) Express secant and cosecant in terms of sine and cosine as the first step.
b) Use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify the numerator.

(c) Using a graph paper draw a histogram for the given distribution showing the number of runs scored by 50 batsmen. Estimate the mode of the data: [4 Marks]

Answer:
[Figure: Histogram plotted with Runs scored on X-axis and Number of batsmen on Y-axis. The modal class is 4000-5000 with frequency 18. Cross-lines drawn from the top corners of the modal rectangle to the adjacent rectangles intersect to determine the mode at 4600.]
Estimated Mode = 4600.

Teacher's Note:
a) Ensure class intervals are continuous before plotting the histogram.
b) The mode is located graphically by drawing intersecting lines from the top corners of the highest rectangle to the adjacent rectangles.

Question 4

(a) Solve the following inequation, write down the solution set and represent it on the real number line:
\(-2 + 10x \le 13x + 10 \lt 24 + 10x\), \(x \in \mathbb{Z}\) [3 Marks]

Answer:
Splitting the given inequation into two parts:
Part 1: \(-2 + 10x \le 13x + 10\)
\(-2 - 10 \le 13x - 10x\)
\(-12 \le 3x \implies x \ge -4\)
Part 2: \(13x + 10 \lt 24 + 10x\)
\(13x - 10x \lt 24 - 10\)
\(3x \lt 14 \implies x \lt \frac{14}{3}\ (\text{i.e., } x \lt 4.67)\)
Combining both parts: \(-4 \le x \lt 4.67\)
Since \(x \in \mathbb{Z}\), the solution set is \(\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}\).
[Figure: Real number line showing solid dots at integers from -4 to 4.]

Teacher's Note:
a) Always split a double inequation into two separate inequalities and solve them independently.
b) Pay close attention to the replacement set (\(\mathbb{Z}\) in this case) when writing the final solution set.

(b) If the straight lines \(3x - 5y = 7\) and \(4x + ay + 9 = 0\) are perpendicular to one another, find the value of a. [3 Marks]

Answer:
For the first line \(3x - 5y = 7\):
\(5y = 3x - 7 \implies y = \frac{3}{5}x - \frac{7}{5}\)
Slope (\(m_1\)) = \(\frac{3}{5}\)
For the second line \(4x + ay + 9 = 0\):
\(ay = -4x - 9 \implies y = -\frac{4}{a}x - \frac{9}{a}\)
Slope (\(m_2\)) = \(-\frac{4}{a}\)
Since the lines are perpendicular, \(m_1 \times m_2 = -1\):
\(\frac{3}{5} \times \left(-\frac{4}{a}\right) = -1\)
\(-\frac{12}{5a} = -1 \implies 5a = 12 \implies a = \frac{12}{5} = 2.4\).

Teacher's Note:
a) Two lines are perpendicular if and only if the product of their slopes is \(-1\).
b) Convert linear equations into slope-intercept form (\(y = mx + c\)) to easily identify their slopes.

(c) Solve \(x^2 + 7x = 7\) and give your answer correct to two decimal places. [4 Marks]

Answer:
Given equation: \(x^2 + 7x - 7 = 0\)
Comparing with \(ax^2 + bx + c = 0\), we get \(a = 1, b = 7, c = -7\).
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-7)}}{2(1)}\)
\(x = \frac{-7 \pm \sqrt{49 + 28}}{2} = \frac{-7 \pm \sqrt{77}}{2}\)
Since \(\sqrt{77} \approx 8.7749\):
\(x = \frac{-7 + 8.7749}{2} = \frac{1.7749}{2} \approx 0.89\)
\(x = \frac{-7 - 8.7749}{2} = \frac{-15.7749}{2} \approx -7.89\)
Thus, \(x = 0.89\) or \(x = -7.89\).

Teacher's Note:
a) Rearrange the equation into standard quadratic form before applying the formula.
b) Calculate square roots to at least three decimal places to ensure accuracy when rounding off to two decimal places.

SECTION B (40 Marks)

Question 5

(a) The \(4^{\text{th}}\) term of a G.P. is 16 and the \(7^{\text{th}}\) term is 128. Find the first term and common ratio of the series. [3 Marks]

Answer:
Let the first term be \(a\) and common ratio be \(r\).
\(T_4 = ar^{4-1} = ar^3 = 16\) --- (1)
\(T_7 = ar^{7-1} = ar^6 = 128\) --- (2)
Dividing equation (2) by equation (1):
\(\frac{ar^6}{ar^3} = \frac{128}{16}\)
\(r^3 = 8 \implies r = 2\)
Substituting \(r = 2\) in equation (1):
\(a(2)^3 = 16 \implies 8a = 16 \implies a = 2\)
Thus, the first term is 2 and the common ratio is 2.

Teacher's Note:
a) The \(n^{\text{th}}\) term of a geometric progression is given by \(T_n = ar^{n-1}\).
b) Dividing one term equation by another eliminates the first term \(a\), making it easy to solve for \(r\).

(b) A man invests Rs. 22,500 in Rs. 50 shares available at 10% discount. If the dividend paid by the company is 12%, calculate:
(i) The number of shares purchased
(ii) The annual dividend received.
(iii) The rate of return he gets on his investment. Give your answer correct to the nearest whole number. [3 Marks]

Answer:
Face Value (F.V.) = Rs. 50
Discount = \(10\%\text{ of } 50 = \text{Rs. } 5\)
Market Value (M.V.) = F.V. - Discount = \(50 - 5 = \text{Rs. } 45\)
Total Investment = Rs. 22,500
(i) Number of shares purchased = \(\frac{\text{Total Investment}}{\text{Market Value}} = \frac{22500}{45} = 500\)
(ii) Annual Dividend = \(\text{Number of shares} \times \text{Dividend } \% \times \text{Face Value}\)
\(\text{Annual Dividend} = 500 \times \frac{12}{100} \times 50 = \text{Rs. } 3000\)
(iii) Rate of Return = \(\frac{\text{Annual Dividend}}{\text{Total Investment}} \times 100\)
\(\text{Rate of Return} = \frac{3000}{22500} \times 100 = 13.33\%\)
Correct to the nearest whole number, the rate of return is \(13\%\).

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, regardless of the purchase price.

(c) Use graph paper for this question (Take 2 cm = 1 unit along both x and y axis). ABCD is a quadrilateral whose vertices are A(2, 2), B(2, -2), C(0, -1) and D(0, 1).
(i) Reflect quadrilateral ABCD on the y-axis and name it as A'B'CD.
(ii) Write down the coordinates of A' and B'.
(iii) Name two points which are invariant under the above reflection.
(iv) Name the polygon A'B'CD. [4 Marks]

Answer:
[Figure: Cartesian plane with points A(2, 2), B(2, -2), C(0, -1), D(0, 1) and their reflections across the y-axis forming quadrilateral A'B'CD with A'(-2, 2) and B'(-2, -2).]
(i) The reflection of quadrilateral ABCD across the y-axis gives A'B'CD.
(ii) Coordinates of A' are \((-2, 2)\) and B' are \((-2, -2)\).
(iii) Points C \((0, -1)\) and D \((0, 1)\) lie on the y-axis and are invariant.
(iv) Polygon A'B'CD is a trapezium.

Teacher's Note:
a) Reflection in the y-axis changes the sign of the x-coordinate while leaving the y-coordinate unchanged \((x, y) \to (-x, y)\).
b) Points lying on the line of reflection remain invariant.

Question 6

(a) Using properties of proportion, solve for x. Given that x is positive:
\(\frac{2x + \sqrt{4x^2 - 1}}{2x - \sqrt{4x^2 - 1}} = 4\) [3 Marks]

Answer:
Applying Componendo and Dividendo:
\(\frac{(2x + \sqrt{4x^2 - 1}) + (2x - \sqrt{4x^2 - 1})}{(2x + \sqrt{4x^2 - 1}) - (2x - \sqrt{4x^2 - 1})} = \frac{4 + 1}{4 - 1}\)
\(\frac{4x}{2\sqrt{4x^2 - 1}} = \frac{5}{3}\)
\(\frac{2x}{\sqrt{4x^2 - 1}} = \frac{5}{3}\)
Squaring both sides:
\(\frac{4x^2}{4x^2 - 1} = \frac{25}{9}\)
\(36x^2 = 25(4x^2 - 1)\)
\(36x^2 = 100x^2 - 25\)
\(64x^2 = 25 \implies x^2 = \frac{25}{64}\)
\(x = \pm \frac{5}{8}\)
Since \(x\) is positive, \(x = \frac{5}{8}\).

Teacher's Note:
a) Componendo and Dividendo states that if \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\).
b) Always check the given condition (here \(x\) is positive) to reject inadmissible roots.

(b) If \(A = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\), \(B = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix}\) and \(C = \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix}\), find \(AC + B^2 - 10C\). [3 Marks]

Answer:
1. Calculating \(AC\):
\(AC = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} (2)(1) + (3)(-1) & (2)(0) + (3)(4) \\ (5)(1) + (7)(-1) & (5)(0) + (7)(4) \end{bmatrix} = \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix}\)
2. Calculating \(B^2\):
\(B^2 = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} (0)(0) + (4)(-1) & (0)(4) + (4)(7) \\ (-1)(0) + (7)(-1) & (-1)(4) + (7)(7) \end{bmatrix} = \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix}\)
3. Calculating \(10C\):
\(10C = 10 \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix}\)
4. Evaluating \(AC + B^2 - 10C\):
\(\begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix} + \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix} = \begin{bmatrix} -1 - 4 - 10 & 12 + 28 - 0 \\ -2 - 7 - (-10) & 28 + 45 - 40 \end{bmatrix} = \begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}\).

Teacher's Note:
a) Perform matrix multiplication row by column carefully.
b) Double-check signs when adding and subtracting resulting matrices.

(c) Prove that \((1 + \cot \theta - \csc \theta)(1 + \tan \theta + \sec \theta) = 2\) [4 Marks]

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

Teacher's Note:
a) Convert all trigonometric ratios into sine and cosine terms.
b) Group terms as \([(\sin \theta + \cos \theta) - 1][(\sin \theta + \cos \theta) + 1]\) to apply the algebraic identity \((a-b)(a+b) = a^2 - b^2\).

Question 7

(a) Find the value of k for which the following equation has equal roots.
\(x^2 + 4kx + (k^2 - k + 2) = 0\) [3 Marks]

Answer:
Given quadratic equation: \(x^2 + 4kx + (k^2 - k + 2) = 0\)
Here, \(a = 1\), \(b = 4k\), \(c = k^2 - k + 2\).
For equal roots, the discriminant must be zero (\(b^2 - 4ac = 0\)):
\((4k)^2 - 4(1)(k^2 - k + 2) = 0\)
\(16k^2 - 4k^2 + 4k - 8 = 0\)
\(12k^2 + 4k - 8 = 0\)
Dividing by 4:
\(3k^2 + k - 2 = 0\)
\(3k^2 + 3k - 2k - 2 = 0\)
\(3k(k + 1) - 2(k + 1) = 0\)
\((k + 1)(3k - 2) = 0\)
\(k + 1 = 0 \implies k = -1\)
\(3k - 2 = 0 \implies k = \frac{2}{3}\)
Thus, \(k = -1\) or \(k = \frac{2}{3}\).

Teacher's Note:
a) A quadratic equation has equal roots when its discriminant \(b^2 - 4ac = 0\).
b) Factorize the resulting quadratic in terms of \(k\) carefully to find all possible values.

(b) On a map drawn to a scale of \(1 : 50,000\), a rectangular plot of land ABCD has the following dimensions. \(AB = 6\) cm; \(BC = 8\) cm and all angles are right angles. Find:
(i) the actual length of the diagonal distance AC of the plot in km.
(ii) the actual area of the plot in sq. km. [3 Marks]

Answer:
Scale: \(1\text{ cm} = 50,000\text{ cm} = 0.5\text{ km}\)
(i) In rectangle ABCD, by Pythagoras theorem, diagonal \(AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ cm}\)
Actual length of diagonal \(AC = 10 \times 0.5\text{ km} = 5\text{ km}\)
(ii) Area of rectangle on map = \(AB \times BC = 6 \times 8 = 48\text{ cm}^2\)
Since \(1\text{ cm} = 0.5\text{ km}\), \(1\text{ cm}^2 = (0.5)^2\text{ km}^2 = 0.25\text{ km}^2\)
Actual area of the plot = \(48 \times 0.25 = 12\text{ km}^2\).

Teacher's Note:
a) Convert map scale units from centimeters to kilometers before calculating actual dimensions.
b) For area scales, square the linear scale conversion factor.

(c) A(2, 5), B(-1, 2) and C(5, 8) are the vertices of a triangle ABC, 'M' is a point on AB such that \(AM : MB = 1 : 2\). Find the co-ordinates of 'M'. Hence find the equation of the line passing through the points C and M. [4 Marks]

Answer:
Using the section formula for point M dividing AB in the ratio \(m_1 : m_2 = 1 : 2\):
\(x_M = \frac{1(-1) + 2(2)}{1 + 2} = \frac{-1 + 4}{3} = \frac{3}{3} = 1\)
\(y_M = \frac{1(2) + 2(5)}{1 + 2} = \frac{2 + 10}{3} = \frac{12}{3} = 4\)
Coordinates of M are \((1, 4)\).
Now, finding the equation of the line passing through C\((5, 8)\) and M\((1, 4)\):
Slope (\(m\)) = \(\frac{8 - 4}{5 - 1} = \frac{4}{4} = 1\)
Using point-slope form with point M\((1, 4)\):
\(y - 4 = 1(x - 1)\)
\(y - 4 = x - 1 \implies y = x + 3\).

Teacher's Note:
a) Use the section formula to find internal division coordinates.
b) Two-point form or point-slope form can be used to find the equation of a straight line.

Question 8

(a) Rs. 7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received Rs. 100 more. Find the original number of children. [3 Marks]

Answer:
Let the original number of children be \(x\).
Money received by each child originally = \(\frac{7500}{x}\)
If there are \(x - 20\) children, money received by each = \(\frac{7500}{x - 20}\)
According to the question:
\(\frac{7500}{x - 20} - \frac{7500}{x} = 100\)
\(7500\left(\frac{1}{x - 20} - \frac{1}{x}\right) = 100\)
\(\frac{x - (x - 20)}{x(x - 20)} = \frac{100}{7500} = \frac{1}{75}\)
\(\frac{20}{x^2 - 20x} = \frac{1}{75}\)
\(x^2 - 20x = 1500\)
\(x^2 - 20x - 1500 = 0\)
\((x - 50)(x + 30) = 0\)
\(x = 50\) or \(x = -30\)
Since the number of children cannot be negative, \(x = 50\).
Thus, the original number of children is 50.

Teacher's Note:
a) Set up the algebraic equation carefully by equating the difference in per capita share to 100.
b) Reject negative roots in word problems involving physical quantities like people or objects.

(b) If the mean of the following distribution is 24, find the value of 'a'. [3 Marks]

Answer:
Constructing the frequency distribution table:
- 0-10: Midpoint (\(x_1\)) = 5, \(f_1 = 7\), \(f_1x_1 = 35\)
- 10-20: Midpoint (\(x_2\)) = 15, \(f_2 = a\), \(f_2x_2 = 15a\)
- 20-30: Midpoint (\(x_3\)) = 25, \(f_3 = 8\), \(f_3x_3 = 200\)
- 30-40: Midpoint (\(x_4\)) = 35, \(f_4 = 10\), \(f_4x_4 = 350\)
- 40-50: Midpoint (\(x_5\)) = 45, \(f_5 = 5\), \(f_5x_5 = 225\)
Total frequency \(\sum f = 7 + a + 8 + 10 + 5 = 30 + a\)
Total \(\sum fx = 35 + 15a + 200 + 350 + 225 = 810 + 15a\)
Given Mean = 24:
\(\text{Mean} = \frac{\sum fx}{\sum f}\)
\(\frac{810 + 15a}{30 + a} = 24\)
\(810 + 15a = 24(30 + a)\)
\(810 + 15a = 720 + 24a\)
\(810 - 720 = 24a - 15a\)
\(90 = 9a \implies a = 10\).

Teacher's Note:
a) Calculate class midpoints accurately for continuous grouped data.\n b) Use the direct mean formula \(\bar{x} = \frac{\sum fx}{\sum f}\) and solve linearly for the unknown frequency.

(c) Using ruler and compass only, construct a \(\triangle ABC\) such that \(BC = 5\) cm and \(AB = 6.5\) cm and \(\angle ABC = 120^{\circ}\)
(i) Construct a circum-circle of \(\triangle ABC\)
(ii) Construct a cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. [4 Marks]

Answer:
[Figure: Geometric construction showing \(\triangle ABC\) with \(BC = 5\text{ cm}\), \(\angle B = 120^{\circ}\), \(AB = 6.5\text{ cm}\), perpendicular bisectors intersecting at circumcenter O, circumcircle, and point D on the circle equidistant from AB and BC.]
Steps of construction:
1. Draw line segment \(BC = 5\) cm.
2. At B, construct an angle of \(120^{\circ}\) and cut an arc of radius \(6.5\) cm to locate point A. Join AC.
3. Draw perpendicular bisectors of sides AB and BC to intersect at circumcenter O.
4. With center O and radius OB, draw the circumcircle of \(\triangle ABC$.
5. Draw the angle bisector of \(\angle ABC\); it intersects the circumcircle at point D.
6. Join AD and DC to form the cyclic quadrilateral ABCD.

Teacher's Note:
a) Construct the \(120^{\circ}\) angle precisely using a compass (60 degree plus 60 degree).
b) The circumcenter is found by the intersection of perpendicular bisectors of any two sides of the triangle.

Question 9

(a) Priyanka has a recurring deposit account of Rs. 1000 per month at 10% per annum. If she gets Rs. 5550 as interest at the time of maturity, find the total time for which the account was held. [3 Marks]

Answer:
\(P = \text{Rs. } 1000\)
\(r = 10\%\)
\(I = \text{Rs. } 5550\)
\(I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}\)
\(5550 = 1000 \times \frac{n(n + 1)}{24} \times \frac{10}{100}\)
\(5550 = \frac{1000 \times 10 \times n(n + 1)}{2400} = \frac{25}{6}n(n + 1)\)
\(n(n + 1) = \frac{5550 \times 6}{25} = 222 \times 6 = 1332\)
\(n^2 + n - 1332 = 0\)
\((n + 37)(n - 36) = 0\)
\(n = 36\) or \(n = -37\)
Since time cannot be negative, \(n = 36\) months (i.e., 3 years).

Teacher's Note:
a) Substitute all known values into the recurring deposit interest formula and simplify carefully.
b) Solve the resulting quadratic equation in terms of \(n\) (months) and reject negative values.

(b) In \(\triangle PQR\), MN is parallel to QR and \(\frac{PM}{MQ} = \frac{2}{3}\)
(i) Find \(\frac{MN}{QR}\)
(ii) Prove that \(\triangle OMN\) and \(\triangle ORQ\) are similar.
(iii) Find, Area of \(\triangle OMN : \text{Area of } \triangle ORQ\) [3 Marks]
[Figure: Triangle PQR with line segment MN parallel to QR intersecting PQ and PR at M and N. diagonals MQ and NR intersect at O inside the triangle.]

Answer:
(i) Given \(\frac{PM}{MQ} = \frac{2}{3}\), so \(\frac{PM}{PQ} = \frac{2}{2 + 3} = \frac{2}{5}\).
In \(\triangle PMN\) and \(\triangle PQR\), since \(MN \parallel QR\):
\(\triangle PMN \sim \triangle PQR\) (AA similarity)
Therefore, \(\frac{MN}{QR} = \frac{PM}{PQ} = \frac{2}{5}\).
(ii) In \(\triangle OMN\) and \(\triangle ORQ\):
\(\angle OMN = \angle OQR\) (alternate interior angles since \(MN \parallel QR\))
\(\angle ONM = \angle ORQ\) (alternate interior angles)
Thus, \(\triangle OMN \sim \triangle ORQ\) by AA similarity.
(iii) The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\(\frac{\text{Area of } \triangle OMN}{\text{Area of } \triangle ORQ} = \left(\frac{MN}{QR}\right)^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25}\).

Teacher's Note:
a) Use Basic Proportionality Theorem / similarity properties for parallel lines inside triangles.
b) Remember that the ratio of areas of similar triangles equals the square of the ratio of corresponding sides.

(c) The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and cone are each of 4 cm. Find the volume of the solid. [4 Marks]
[Figure: A solid figure showing a central cylinder of height 4 cm, capped by a cone of height 4 cm on top and a hemisphere of radius 7 cm at the bottom. Common radius is 7 cm.]

Answer:
Radius (\(r\)) = 7 cm
Height of cone (\(h_1\)) = 4 cm
Height of cylinder (\(h_2\)) = 4 cm
Radius of hemisphere (\(r\)) = 7 cm
1. Volume of cone = \(\frac{1}{3}\pi r^2 h_1 = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 4 = \frac{22 \times 7 \times 4}{3} = \frac{616}{3}\text{ cm}^3\)
2. Volume of cylinder = \(\pi r^2 h_2 = \frac{22}{7} \times 7^2 \times 4 = 22 \times 7 \times 4 = 616\text{ cm}^3\)
3. Volume of hemisphere = \(\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2 \times 22 \times 7 \times 7}{3} = \frac{2156}{3}\text{ cm}^3\)
Total Volume = Volume of cone + Volume of cylinder + Volume of hemisphere
Total Volume = \(\frac{616}{3} + 616 + \frac{2156}{3} = 616 + \frac{2772}{3} = 616 + 924 = 1540\text{ cm}^3\).

Teacher's Note:
a) Break down composite solids into standard geometric shapes (cone, cylinder, hemisphere) and sum their volumes.
b) Maintain fractional forms during intermediate steps to avoid rounding errors.

Question 10

(a) Use Remainder theorem to factorize the following polynomial: \(2x^3 + 3x^2 - 9x - 10\). [3 Marks]

Answer:
Let \(P(x) = 2x^3 + 3x^2 - 9x - 10\).
Testing factors of constant term \(-10\):
\(P(2) = 2(2)^3 + 3(2)^2 - 9(2) - 10 = 2(8) + 3(4) - 18 - 10 = 16 + 12 - 18 - 10 = 0\)
Since \(P(2) = 0\), by Factor Theorem, \((x - 2)\) is a factor of \(P(x)\).
Dividing \(P(x)\) by \((x - 2)\):
\(P(x) = (x - 2)(2x^2 + 7x + 5)\)
Factoring the quadratic part \(2x^2 + 7x + 5\):
\(2x^2 + 5x + 2x + 5 = x(2x + 5) + 1(2x + 5) = (2x + 5)(x + 1)\)
Thus, completely factorized form is \((x - 2)(2x + 5)(x + 1)\).

Teacher's Note:
a) Use trial and error with factors of the constant term to find the first linear factor using the Remainder Theorem.
b) Perform synthetic division or long division to find the remaining quadratic factor.

(b) In the figure given below 'O' is the centre of the circle. If QR = OP and \(\angle ORP = 20^{\circ}\). Find the value of 'x' giving reasons. [3 Marks]
[Figure: Circle with center O. Points P, Q, R on circumference and tangent/secant lines forming triangles with center O. \(\angle ORP = 20^{\circ}\), \(QR = OP\), and central angle marked as \(x^{\circ}\).]

Answer:
Given \(OP = QR\). Since radii of the circle are \(OP = OQ = OR\), we have \(OQ = OR = QR\), which means \(\triangle OQR\) is an equilateral triangle.
Therefore, \(\angle QOR = 60^{\circ}\), but using given angles: in \(\triangle ORP\), \(OR = OP\) (radii), so \(\angle OPR = \angle ORP = 20^{\circ}\).
Extending and analyzing angles:
In \(\triangle ORP\), exterior angle \(\angle QOR = \angle ORP + \angle OPR = 20^{\circ} + 20^{\circ} = 40^{\circ}\).
In \(\triangle OPQ\), \(OP = OQ\) (radii), so \(\angle OQP = \angle OPQ\).
Using triangle properties and straight line angles, \(x^{\circ} = 60^{\circ}\).

Teacher's Note:
a) Radii of the same circle are always equal, forming isosceles triangles.
b) The exterior angle of a triangle equals the sum of the two opposite interior angles.

(c) The angle of elevation from a point P of the top of a tower QR, 50 m high is \(60^{\circ}\) and that of the tower PT from a point Q is \(30^{\circ}\). Find the height of the tower PT, correct to the nearest metre. [4 Marks]
[Figure: Two towers QR and PT situated on horizontal ground. Point P is at the base of tower PT, point Q is at the base of tower QR. Angles of elevation from P to R is \(60^{\circ}\) and from Q to T is \(30^{\circ}\).]

Answer:
Height of tower QR = 50 m.
In right-angled \(\triangle PQR\):
\(\tan 60^{\circ} = \frac{QR}{PQ}\)
\(\sqrt{3} = \frac{50}{PQ} \implies PQ = \frac{50}{\sqrt{3}}\text{ m}\)
In right-angled \(\triangle PQT\), let height of tower PT be \(h\):
\(\tan 30^{\circ} = \frac{PT}{PQ}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{\frac{50}{\sqrt{3}}}\)
\(h = \frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}} = \frac{50}{3} = 16.67\text{ m}\)
Correct to the nearest metre, the height of tower PT is 17 m.

Teacher's Note:
a) Use standard trigonometric ratios (\(\tan \theta\)) for right-angled triangles involving heights and distances.
b) Express intermediate lengths in exact radical form before final calculation.

Question 11

(a) The \(4^{\text{th}}\) term of an A.P. is 22 and \(15^{\text{th}}\) term is 66. Find the first term and the common difference. Hence find the sum of the series to 8 terms. [4 Marks]

Answer:
Let the first term be \(a\) and common difference be \(d\).
\(T_4 = a + 3d = 22\) --- (1)
\(T_{15} = a + 14d = 66\) --- (2)
Subtracting equation (1) from equation (2):
\(11d = 44 \implies d = 4\)
Substituting \(d = 4\) in equation (1):
\(a + 3(4) = 22 \implies a + 12 = 22 \implies a = 10\)
First term = 10, Common difference = 4.
Sum of first 8 terms (\(S_8\)):
\(S_8 = \frac{n}{2}[2a + (n - 1)d] = \frac{8}{2}[2(10) + (8 - 1)4] = 4[20 + 28] = 4 \times 48 = 192\).

Teacher's Note:
a) Set up linear equations using the formula \(T_n = a + (n-1)d\).
b) Use the standard arithmetic progression sum formula for \(S_n\).

(b) Use Graph paper for this question.
A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded: [6 Marks]

Taking 2 cm = height of 10 cm along one axis and 2 cm = 10 boys along the other axis draw an ogive of the above distribution. Use the graph to estimate the following:
(i) the median
(ii) lower Quartile
(iii) if above 158 cm is considered as the tall boys of the class. Find the number of boys in the class who are tall.

Answer:
Constructing the cumulative frequency table:
- 135-140: f = 4, c.f. = 4
- 140-145: f = 8, c.f. = 12
- 145-150: f = 20, c.f. = 32
- 150-155: f = 14, c.f. = 46
- 155-160: f = 7, c.f. = 53
- 160-165: f = 6, c.f. = 59
- 165-170: f = 1, c.f. = 60
[Figure: Ogive (cumulative frequency curve) plotted with upper class boundaries on X-axis and cumulative frequencies on Y-axis.]
Total frequency \(N = 60\).
(i) Median = Value at \(\frac{N}{2}^{\text{th}}\) term = \(30^{\text{th}}\) term.
From graph, Median = 152 cm.
(ii) Lower Quartile (\(Q_1\)) = Value at \(\frac{N}{4}^{\text{th}}\) term = \(15^{\text{th}}\) term.
From graph, \(Q_1 = 148\) cm.
(iii) Number of boys with height above 158 cm:
At height 158 cm on the X-axis, the corresponding cumulative frequency on the ogive is 48.
Number of tall boys = \(60 - 48 = 12\).

Teacher's Note:
a) Always plot cumulative frequencies against upper class limits when drawing an ogive.
b) Median corresponds to \(\frac{N}{2}\) and lower quartile to \(\frac{N}{4}\) on the cumulative frequency axis.