ICSE Class 10 Mathematics Board Exam Question Paper 2014 with Solutions

ICSE Class 10 Mathematics Board Exam Question Paper with Solutions

SECTION A (40 Marks)

Question 1

(a) Ranbir borrows Rs. 20,000 at 12% per annum compound interest. If he repays Rs. 8400 at the end of the first year and Rs. 9680 at the end of the second year, find the amount of loan outstanding at the beginning of the third year. [3 Marks]

Answer:
Principal for the 1st year \( P_1 = \text{Rs. } 20000 \)
Interest for the 1st year \( I_1 = \frac{20000 \times 12 \times 1}{100} = \text{Rs. } 2400 \)
Amount at the end of the 1st year \( = 20000 + 2400 = \text{Rs. } 22400 \)
Amount repaid at the end of the 1st year \( = \text{Rs. } 8400 \)
Principal for the 2nd year \( P_2 = 22400 - 8400 = \text{Rs. } 14000 \)
Interest for the 2nd year \( I_2 = \frac{14000 \times 12 \times 1}{100} = \text{Rs. } 1680 \)
Amount at the end of the 2nd year \( = 14000 + 1680 = \text{Rs. } 15680 \)
Amount repaid at the end of the 2nd year \( = \text{Rs. } 9680 \)
Loan outstanding at the beginning of the 3rd year \( = 15680 - 9680 = \text{Rs. } 6000 \).

Teacher's Note:
a) Always subtract the repayment from the total amount at the end of each year to find the principal for the succeeding year.
b) Students often make calculation errors in finding annual interest; ensure simple interest formula is correctly applied year by year for compound interest repayments.

(b) Find the values of \( x \), which satisfy the inequation \(-2\frac{5}{6} \lt -\frac{1}{2} - \frac{2x}{3} \le 2\), \(x \in W\). Graph the solution set on the number line. [3 Marks]

Answer:
Given inequation: \(-\frac{17}{6} \lt -\frac{1}{2} - \frac{2x}{3} \le 2\)
Multiplying throughout by 6 (LCM of 6, 2, 3):
\(-17 \lt -3 - 4x \le 12\)
Splitting into two parts:
Part 1: \(-17 \lt -3 - 4x \implies 4x \lt -3 + 17 \implies 4x \lt 14 \implies x \lt \frac{7}{2} \implies x \lt 3.5\)
Part 2: \(-3 - 4x \le 12 \implies -4x \le 12 + 3 \implies -4x \le 15 \implies x \ge -\frac{15}{4} \implies x \ge -3.75\)
Combining both: \(-3.75 \le x \lt 3.5\)
Since \( x \in W \) (Whole Numbers), the solution set is \( \{0, 1, 2, 3\} \).
[Figure: A number line showing marked points at 0, 1, 2, and 3 with solid dots representing the whole number solutions.]

Teacher's Note:
a) Remember to reverse the inequality sign if you divide or multiply both sides by a negative number.
b) Pay close attention to the replacement set (\(W\) in this case) before writing the final solution set.

(c) A die has 6 faces marked by the given numbers as shown below:
1, 2, 3, -1, -2, -3
The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than \(-3\).
(iii) the smallest integer. [4 Marks]

Answer:
Total possible outcomes \( n(S) = 6 \) (faces are \( 1, 2, 3, -1, -2, -3 \)).
(i) Positive integers are \( 1, 2, 3 \); number of favorable outcomes \( = 3 \).
Probability \( = \frac{3}{6} = \frac{1}{2} \).
(ii) Integers greater than \(-3\) are \(-2, -1, 1, 2, 3 \); number of favorable outcomes \( = 5 \).
Probability \( = \frac{5}{6} \).
(iii) The smallest integer is \(-3\); number of favorable outcomes \( = 1 \).
Probability \( = \frac{1}{6} \).

Teacher's Note:
a) Ensure negative integers are properly evaluated when comparing sizes (e.g., \(-2\) is greater than \(-3\)).
b) Always express probabilities in their simplest fractional form.

Question 2

(a) Find \( x, y \) if \(\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}\). [3 Marks]

Answer:
\(\begin{bmatrix} (-2)(-1) + (0)(2x) \\ (3)(-1) + (1)(2x) \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}\)
\(\begin{bmatrix} 2 \\ -3 + 2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}\)
\(\begin{bmatrix} 2 - 6 \\ -3 + 2x + 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}\)
\(\begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}\)
Comparing corresponding elements:
\(2y = -4 \implies y = -2\)
\(2x = 6 \implies x = 3\)
Therefore, \( x = 3, y = -2 \).

Teacher's Note:
a) Perform matrix multiplication carefully row by column before adding matrices.
b) Equate corresponding elements from both sides of the matrix equation to find the unknowns.

(b) Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs. 800 per month for \(1\frac{1}{2}\) years. If he received Rs. 15,084 at the time of maturity, find the rate of interest per annum. [3 Marks]

Answer:
Monthly deposit \( P = \text{Rs. } 800 \)
Time \( n = 1\frac{1}{2} \text{ years} = 18 \text{ months} \)
Total money deposited \( = P \times n = 800 \times 18 = \text{Rs. } 14,400 \)
Maturity Value \( (MV) = \text{Rs. } 15,084 \)
Total Interest \( (I) = MV - \text{Deposit} = 15,084 - 14,400 = \text{Rs. } 684 \)
Equivalent principal for 1 month \( P_{eq} = P \times \frac{n(n+1)}{2} = 800 \times \frac{18 \times 19}{2} = 800 \times 171 = 1,36,800 \)
\( I = \frac{P_{eq} \times r \times 1}{100 \times 12} \)
\( 684 = \frac{136800 \times r}{1200} \)
\( 684 = 114 \times r \)
\( r = \frac{684}{114} = 6\% \)
Rate of interest is \( 6\% \) per annum.

style="margin-top:0;"Teacher's Note:
a) Convert the time period from years to total months right at the beginning since recurring deposits operate on monthly installments.
b) Verify the formula for equivalent principal: \( \frac{P \times n(n+1)}{2} \).

(c) Calculate the ratio in which the line joining \( A(-4, 2) \) and \( B(3, 6) \) is divided by point \( P(x, 3) \). Also find (i) \( x \) (ii) Length of AP. [4 Marks]

Answer:
Let the point \( P(x, 3) \) divide the line segment \( AB \) in the ratio \( k : 1 \).
Using the section formula for the y-coordinate:
\( y = \frac{k y_2 + 1 y_1}{k + 1} \)
\( 3 = \frac{k(6) + 1(2)}{k + 1} \)
\( 3(k + 1) = 6k + 2 \)
\( 3k + 3 = 6k + 2 \)
\( 3 - 2 = 6k - 3k \)
\( 3k = 1 \implies k = \frac{1}{3} \)
Thus, the ratio is \( 1 : 3 \).
(i) Finding \( x \):
\( x = \frac{k x_2 + 1 x_1}{k + 1} = \frac{\frac{1}{3}(3) + 1(-4)}{\frac{1}{3} + 1} = \frac{1 - 4}{\frac{4}{3}} = \frac{-3}{\frac{4}{3}} = -\frac{9}{4} = -2.25 \)
So, \( P \) is \((-2.25, 3)\).
(ii) Length of AP:
\( AP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-2.25 - (-4))^2 + (3 - 2)^2} \)
\( AP = \sqrt{(1.75)^2 + (1)^2} = \sqrt{3.0625 + 1} = \sqrt{4.0625} \approx 2.015 \text{ units} \).

Teacher's Note:
a) Always use the coordinate whose value is completely known (here y = 3) to find the ratio \( k : 1 \) first.
b) Distance formula should be applied carefully using exact fraction values for higher precision before rounding.

Question 3

(a) Without using trigonometric tables, evaluate
\(\sin^2 34^{\circ} + \sin^2 56^{\circ} + 2 \tan 18^{\circ} \tan 72^{\circ} - \cot^2 30^{\circ}\). [3 Marks]

Answer:
Given expression: \(\sin^2 34^{\circ} + \sin^2 56^{\circ} + 2 \tan 18^{\circ} \tan 72^{\circ} - \cot^2 30^{\circ}\)
\(=\sin^2 34^{\circ} + \sin^2 (90^{\circ} - 34^{\circ}) + 2 \tan 18^{\circ} \tan (90^{\circ} - 18^{\circ}) - (\sqrt{3})^2\)
\(=\sin^2 34^{\circ} + \cos^2 34^{\circ} + 2 \tan 18^{\circ} \cot 18^{\circ} - 3\)
\( = 1 + 2(1) - 3 \)
\( = 1 + 2 - 3 = 0 \).

Teacher's Note:
a) Use complementary angle relations: \(\sin(90^{\circ} - \theta) = \cos \theta\) and \(\tan(90^{\circ} - \theta) = \cot \theta\).
b) Remember standard trigonometric values like \(\cot 30^{\circ} = \sqrt{3}\).

(b) Using the Remainder and Factor Theorem, factorise the following polynomial:
\(x^3 + 10x^2 - 37x + 26\). [3 Marks]

Answer:
Let \( f(x) = x^3 + 10x^2 - 37x + 26 \).
Factors of the constant term 26 are \( \pm 1, \pm 2, \pm 13, \pm 26 \).
Testing \( x = 1 \):
\( f(1) = (1)^3 + 10(1)^2 - 37(1) + 26 = 1 + 10 - 37 + 26 = 37 - 37 = 0 \).
Since \( f(1) = 0 \), by Factor Theorem, \( (x - 1) \) is a factor of \( f(x) \).
Now, dividing \( f(x) \) by \( (x - 1) \) or using synthetic division:
\( f(x) = (x - 1)(x^2 + 11x - 26) \)
Factoring the quadratic part \( x^2 + 11x - 26 \):
\( x^2 + 13x - 2x - 26 = x(x + 13) - 2(x + 13) = (x - 2)(x + 13) \)
Thus, the fully factorised form is \( (x - 1)(x - 2)(x + 13) \).

Teacher's Note:
a) Always test small integers like \( \pm 1, \pm 2 \) first as they are usually the roots in board problems.
b) Complete the factorisation of the resulting quadratic expression fully.

(c) In the figure given below, ABCD is a rectangle. \( AB = 14\text{cm}, BC = 7\text{cm} \). From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle. (Take \( \pi = 22/7 \)) [4 Marks]

[Figure: Rectangle ABCD with AB = 14 cm and BC = 7 cm. A quarter circle BFEC with radius equal to BC (7 cm) is attached to BC. A semicircle DGE with diameter DE is removed from the bottom side.]

Answer:
Area of rectangle \( ABCD = \text{Length} \times \text{Breadth} = 14 \times 7 = 98\text{ cm}^2 \).
Radius of quarter circle BFEC \( r_1 = BC = 7\text{ cm} \).
Area of quarter circle \( = \frac{1}{4} \pi r_1^2 = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = \frac{77}{2} = 38.5\text{ cm}^2 \).
Note: Looking at standard ICSE figures for this problem, semicircle DGE has diameter \( DE = 7\text{ cm} \) (or the remaining part of length), so radius \( r_2 = \frac{7}{2}\text{ cm} \).
Area of semicircle \( = \frac{1}{2} \pi r_2^2 = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{4} = 19.25\text{ cm}^2 \).
Wait, let us re-read carefully: "a quarter circle BFEC and a semicircle DGE are removed."
Area of remaining piece \( = \text{Area of rectangle} - \text{Area of semicircle} - \text{Area of quarter circle} \)
\( = 98 - 19.25 - 38.5 = 98 - 57.75 = 40.25\text{ cm}^2 \).

Teacher's Note:
a) Carefully identify whether the regions are added or removed from the base geometric figure.
b) Double check the radii of the circular arcs from the given dimensions of the rectangle.

Question 4

(a) The numbers 6, 8, 10, 12, 13, and \( x \) are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of \( x \). [3 Marks]

Answer:
Given numbers: 6, 8, 10, 12, 13, \( x \) (already in ascending order, assuming \( x \ge 13 \)).
Number of observations \( n = 6 \) (even).
Median \( = \frac{(\frac{n}{2})^{\text{th}} \text{ term} + (\frac{n}{2} + 1)^{\text{th}} \text{ term}}{2} = \frac{3^{\text{rd}} \text{ term} + 4^{\text{th}} \text{ term}}{2} = \frac{10 + 12}{2} = \frac{22}{2} = 11 \).
Mean \( = \frac{6 + 8 + 10 + 12 + 13 + x}{6} = \frac{49 + x}{6} \).
Given that Mean = Median:
\( \frac{49 + x}{6} = 11 \)
\( 49 + x = 66 \)
\( x = 66 - 49 = 17 \).

Teacher's Note:
a) For an even number of observations, the median is the average of the two middle terms.
b) Equate the mean expression directly to the calculated median to solve for the unknown variable \( x \).

(b) In the figure, \(\angle DBC = 58^{\circ}\). BD is a diameter of the circle. Calculate:
(i) \(\angle BDC\)
(ii) \(\angle BEC\)
(iii) \(\angle BAC\) [3 Marks]

Answer:
(i) Since BD is the diameter, angle in a semicircle \(\angle BCD = 90^{\circ}\).
In right-angled triangle \(\triangle BCD\), \(\angle BDC = 180^{\circ} - (90^{\circ} + 58^{\circ}) = 180^{\circ} - 148^{\circ} = 32^{\circ}\).
(ii) Angles in the same segment subtended by chord BC: \(\angle BEC = \angle BDC = 32^{\circ}\).
(iii) Angles in the same segment subtended by chord BC: \(\angle BAC = \angle BDC = 32^{\circ}\) (or \(\angle BEC = \angle BAC\)).

Teacher's Note:
a) Recall that the angle subtended by a diameter at any point on the circle is a right angle (\(90^{\circ}\)).
b) Angles in the same segment of a circle are equal.

(c) Use graph paper to answer the following questions. (Take \( 2\text{cm} = 1\text{ unit} \) on both axis)
(i) Plot the points \( A(-4, 2) \) and \( B(2, 4) \).
(ii) \( A' \) is the image of \( A \) when reflected in the y-axis. Plot it on the graph paper and write the coordinates of \( A' \).
(iii) \( B' \) is the image of \( B \) when reflected in the line \( AA' \). Write the coordinates of \( B' \).
(iv) Write the geometric name of the figure \( ABA'B' \).
(v) Name a line of symmetry of the figure formed. [4 Marks]

Answer:
(i) Plot points \( A(-4, 2) \) and \( B(2, 4) \).
(ii) Reflection of \( A(-4, 2) \) in the y-axis changes the sign of the x-coordinate: \( A'(4, 2) \).
(iii) The line \( AA' \) is the line \( y = 2 \) (since both \( A \) and \( A' \) have y-coordinate 2).
Point \( B(2, 4) \) is at a perpendicular distance of \( 4 - 2 = 2 \) units above the line \( y = 2 \).
Therefore, reflection of \( B(2, 4) \) in the line \( y = 2 \) will be 2 units below the line \( y = 2 \), giving \( B'(2, 0) \).
(iv) The figure \( ABA'B' \) has vertices \( A(-4, 2), B(2, 4), A'(4, 2), B'(2, 0) \). This is a rhombus (or kite / parallelogram).
(v) The line of symmetry is the line \( AA' \) (or the y-axis).
[Figure: Graph plot showing points A, B, A', B' and the quadrilateral formed.]

Teacher's Note:
a) Reflection in the y-axis negates the x-coordinate: \((x, y) \to (-x, y)\).
b) Reflection across a horizontal line keeps the x-coordinate same while reflecting the y-coordinate equidistant across the line.

SECTION B (40 Marks)
Attempt any four questions from this Section

Question 5

(a) A shopkeeper bought a washing machine at a discount of \( 20\% \) from a wholesaler, the printed price of the washing machine being Rs. 18,000. The shopkeeper sells it to a consumer at a discount of \( 10\% \) on the printed price. If the rate of sales tax is \( 8\% \), find:
(i) the VAT paid by the shopkeeper.
(ii) the total amount that the consumer pays for the washing machine. [3 Marks]

Answer:
Printed Price (PP) \( = \text{Rs. } 18,000 \)
1. Cost Price for the shopkeeper (CP) \( = 18,000 - 20\% \text{ of } 18,000 = 18,000 - 3,600 = \text{Rs. } 14,400 \)
2. Selling Price by the shopkeeper (SP) \( = 18,000 - 10\% \text{ of } 18,000 = 18,000 - 1,800 = \text{Rs. } 16,200 \)
Rate of Sales Tax \( = 8\% \)
(i) VAT paid by the shopkeeper \( = \text{Tax on SP} - \text{Tax on CP} = 8\% \text{ of } (16,200 - 14,400) = 8\% \text{ of } 1,800 = \frac{8}{100} \times 1800 = \text{Rs. } 144 \).
(ii) Total amount paid by the consumer \( = \text{SP} + \text{Sales Tax on SP} = 16,200 + 8\% \text{ of } 16,200 = 16,200 + 1,296 = \text{Rs. } 17,496 \).

Teacher's Note:
a) VAT is calculated as tax on output selling price minus tax on input cost price, or simply \( 8\% \) of the profit margin (\(SP - CP\)).
b) The total amount paid by the consumer is the selling price plus the sales tax charged on it.

(b) If \(\frac{x^2 + y^2}{x^2 - y^2} = \frac{17}{8}\), then find the value of:
(i) \( x : y \)
(ii) \(\frac{x^3 + y^3}{x^3 - y^3}\) [3 Marks]

Answer:
Given: \(\frac{x^2 + y^2}{x^2 - y^2} = \frac{17}{8}\)
Applying Componendo and Dividendo:
\(\frac{(x^2 + y^2) + (x^2 - y^2)}{(x^2 + y^2) - (x^2 - y^2)} = \frac{17 + 8}{17 - 8}\)
\(\frac{2x^2}{2y^2} = \frac{25}{9}\)
\(\frac{x^2}{y^2} = \frac{25}{9}\)
(i) Taking square root on both sides: \(\frac{x}{y} = \frac{5}{3}\), so \( x : y = 5 : 3 \).
(ii) Let \( x = 5k \) and \( y = 3k \).
\(\frac{x^3 + y^3}{x^3 - y^3} = \frac{(5k)^3 + (3k)^3}{(5k)^3 - (3k)^3} = \frac{125k^3 + 27k^3}{125k^3 - 27k^3} = \frac{152k^3}{98k^3} = \frac{152}{98} = \frac{76}{49}\).

Teacher's Note:
a) Componendo and Dividendo is the fastest method to simplify expressions of the form \(\frac{a+b}{a-b}\).
b) Substitute proportional parts after finding \(x/y\) to evaluate higher-power ratios.

(c) In \(\triangle ABC\), \(\angle ABC = \angle DAC\). \( AB = 8\text{cm}, AC = 4\text{cm}, AD = 5\text{cm} \).
(i) Prove that \(\triangle ACD\) is similar to \(\triangle BCA\)
(ii) Find \( BC \) and \( CD \)
(iii) Find area of \(\triangle ACD : \text{area of } \triangle ABC\) [4 Marks]

[Figure: Triangle ABC with a line dividing inside showing point D on BC, such that \(\angle ABC = \angle DAC\).]

Answer:
(i) In \(\triangle ACD\) and \(\triangle BCA\):
\(\angle DAC = \angle ABC\) (given)
\(\angle ACD = \angle BCA\) (common angle)
Therefore, by AA similarity criterion, \(\triangle ACD \sim \triangle BCA\).
(ii) Since \(\triangle ACD \sim \triangle BCA\), the ratio of corresponding sides is equal:
\(\frac{AC}{BC} = \frac{CD}{CA} = \frac{AD}{BA}\)
Given \( AB = 8\text{cm}, AC = 4\text{cm}, AD = 5\text{cm} \):
From \(\frac{AC}{BC} = \frac{AD}{BA} \implies \frac{4}{BC} = \frac{5}{8} \implies 5BC = 32 \implies BC = \frac{32}{5} = 6.4\text{ cm}\).
From \(\frac{CD}{CA} = \frac{AD}{BA} \implies \frac{CD}{4} = \frac{5}{8} \implies 8CD = 20 \implies CD = \frac{20}{8} = 2.5\text{ cm}\).
(iii) Ratio of areas of similar triangles is equal to the square of the ratio of their corresponding sides:
\(\frac{\text{Area}(\triangle ACD)}{\text{Area}(\triangle ABC)} = \left(\frac{AC}{AB}\right)^2 = \left(\frac{4}{8}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\), so \( 1 : 4 \).

Teacher's Note:
a) Write the vertices of similar triangles in correct corresponding order to easily set up side ratios.
b) The ratio of areas of two similar triangles equals the square of the ratio of any pair of corresponding sides.

Question 6

(a) Find the value of 'a' for which the following points \( A(a, 3), B(2, 1) \) and \( C(5, a) \) are collinear. Hence find the equation of the line. [3 Marks]

Answer:
For points \( A(a, 3), B(2, 1), C(5, a) \) to be collinear, the slope of \( AB \) must equal the slope of \( BC \).
Slope of \( AB = \frac{1 - 3}{2 - a} = \frac{-2}{2 - a} \)
Slope of \( BC = \frac{a - 1}{5 - 2} = \frac{a - 1}{3} \)
Equating slopes:
\(\frac{-2}{2 - a} = \frac{a - 1}{3}\)
\(-6 = (2 - a)(a - 1)\)
\(-6 = 2a - 2 - a^2 + a\)
\(-6 = 3a - 2 - a^2\)
\(a^2 - 3a - 4 = 0\)
\((a - 4)(a + 1) = 0\)
So, \( a = 4 \) or \( a = -1 \).
Case 1: If \( a = 4 \), points are \( A(4, 3), B(2, 1), C(5, 4) \).
Equation of line AB: \( y - 1 = \frac{3 - 1}{4 - 2}(x - 2) \implies y - 1 = 1(x - 2) \implies x - y - 1 = 0 \).
Case 2: If \( a = -1 \), points are \( A(-1, 3), B(2, 1), C(5, -1) \).
Equation of line AB: \( y - 1 = \frac{1 - 3}{2 - (-1)}(x - 2) \implies y - 1 = \frac{-2}{3}(x - 2) \implies 3y - 3 = -2x + 4 \implies 2x + 3y - 7 = 0 \).

Teacher's Note:
a) Collinearity can be tested by equating slopes or by setting the area of the triangle formed by the three points to zero.
b) Both values of \( a \) are valid, yielding two distinct lines.

(b) Salman invests a sum of money in Rs. 50 shares, paying \( 15\% \) dividend quoted at \( 20\% \) premium. If his annual dividend is Rs. 600, calculate:
(i) the number of shares he bought.
(ii) his total investment.
(iii) the rate of return on his investment. [3 Marks]

Answer:
Nominal Value (NV) of 1 share \( = \text{Rs. } 50 \)
Dividend rate \( = 15\% \)
Market Value (MV) at \( 20\% \) premium \( = 50 + 20\% \text{ of } 50 = 50 + 10 = \text{Rs. } 60 \)
Annual Dividend per share \( = 15\% \text{ of NV} = \frac{15}{100} \times 50 = \text{Rs. } 7.50 \)
Total Annual Dividend \( = \text{Rs. } 600 \)
(i) Number of shares \( n = \frac{\text{Total Dividend}}{\text{Dividend per share}} = \frac{600}{7.50} = 80 \text{ shares} \).
(ii) Total Investment \( = n \times \text{MV} = 80 \times 60 = \text{Rs. } 4,800 \).
(iii) Rate of return \( = \frac{\text{Total Dividend}}{\text{Total Investment}} \times 100 = \frac{600}{4800} \times 100 = \frac{1}{8} \times 100 = 12.5\% \).

Teacher's Note:
a) Dividend is always calculated on the Nominal (Face) Value, while investment is calculated on the Market Value.
b) Rate of return is the actual percentage yield on the money invested, not on the face value.

(c) The surface area of a solid metallic sphere is \( 2464\text{ cm}^2 \). It is melted and recast into solid right circular cones of radius \( 3.5\text{cm} \) and height \( 7\text{cm} \). Calculate:
(i) the radius of the sphere.
(ii) the number of cones recast. (Take \( \pi = 22/7 \)) [4 Marks]

Answer:
Surface area of sphere \( = 4\pi R^2 = 2464\text{ cm}^2 \)
\( 4 \times \frac{22}{7} \times R^2 = 2464 \)
\( \frac{88}{7} R^2 = 2464 \)
\( R^2 = \frac{2464 \times 7}{88} = 28 \times 7 = 196 \)
(i) Radius of the sphere \( R = \sqrt{196} = 14\text{ cm} \).
(ii) Volume of the sphere \( V_s = \frac{4}{3} \pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (14)^3 = \frac{4 \times 22 \times 2744}{21} \)
Volume of one cone \( V_c = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 7 = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} \times 7 = \frac{22 \times 343}{12} \)
Number of cones \( n = \frac{\text{Volume of sphere}}{\text{Volume of one cone}} = \frac{\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14}{\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 7} \)
\( n = \frac{4 \times 14 \times 14 \times 14}{\frac{49}{4} \times 7} = \frac{4 \times 14 \times 14 \times 14 \times 4}{49 \times 7} = \frac{4 \times 2 \times 2 \times 14 \times 4}{1} = 4 \times 2 \times 2 \times 14 \times 4 = 1792 \dots \) Wait, let us simplify properly:
\( V_s = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 = \frac{4 \times 22 \times 2 \times 14 \times 14}{3} = \frac{34496}{3}\text{ cm}^3 \).
\( V_c = \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 7 = \frac{22 \times 49}{12} = \frac{1078}{12} = \frac{539}{6}\text{ cm}^3 \).
Number of cones \( = \frac{34496 / 3}{539 / 6} = \frac{34496}{3} \times \frac{6}{539} = \frac{34496 \times 2}{539} = \frac{68992}{539} = 128 \).

Teacher's Note:
a) When a solid is recast into another, equate their volumes to find the number of items produced.
b) Keep fractions unsimplified during intermediate steps to avoid rounding errors.

Question 7

(a) Calculate the mean of the distribution given below using the short cut method. [3 Marks]

Answer:

Mean \( \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} = 45.5 + \frac{170}{80} = 45.5 + 2.125 = 47.625 \).

Teacher's Note:
a) For the assumed mean method, choose an intermediate mid-value as \( A \) to simplify calculations.
b) Ensure class intervals are continuous before finding mid-values (they are continuous here).

(b) In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. \( CD = 7.8\text{cm}, PD = 5\text{cm}, PB = 4\text{cm} \). Find:
(i) AB.
(ii) the length of tangent PT. [3 Marks]

Answer:
Given: \( CD = 7.8\text{cm}, PD = 5\text{cm}, PB = 4\text{cm} \).
Length of chord intersecting outside at P: \( PC = PD + CD = 5 + 7.8 = 12.8\text{ cm} \).
Let radius of the circle be \( r \), so diameter \( AB = 2r \).
Let \( PA = x \). Then \( PB = 4 \implies AB = PA - PB \) (or considering secant property).
By secant-secant theorem (or chord intersection theorem for external points):
\( PA \times PB = PC \times PD \)
Let \( PA = x \), so \( x \times 4 = 12.8 \times 5 \)
\( 4x = 64 \implies x = 16\text{ cm} \).
Since \( PA = 16\text{ cm} \) and \( PB = 4\text{ cm} \):
(i) \( AB = PA - PB = 16 - 4 = 12\text{ cm} \).
(ii) Tangent length \( PT^2 = PA \times PB \) (Tangent-secant theorem) or \( PT^2 = PC \times PD \)
\( PT^2 = 16 \times 4 = 64 \implies PT = 8\text{ cm} \).

Teacher's Note:
a) Use the intersecting secant theorem: \( PA \times PB = PC \times PD \) where P is the external point.
b) Tangent-secant theorem states that \( PT^2 = PA \times PB \).

(c) Let \( A = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}, B = \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} \) and \( C = \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix} \).
Find \( A^2 + AC - 5B \). [4 Marks]

Answer:
1. Calculate \( A^2 = A \times A = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} (2)(2) + (1)(0) & (2)(1) + (1)(-2) \\ (0)(2) + (-2)(0) & (0)(1) + (-2)(-2) \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \)
2. Calculate \( AC = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix} \)
\( AC = \begin{bmatrix} (2)(-3) + (1)(-1) & (2)(2) + (1)(4) \\ (0)(-3) + (-2)(-1) & (0)(2) + (-2)(4) \end{bmatrix} = \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} \)
3. Calculate \( 5B = 5 \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} = \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix} \)
Now, evaluate \( A^2 + AC - 5B \):
\( \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} - \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix} \)
\( = \begin{bmatrix} 4 - 7 - 20 & 0 + 8 - 5 \\ 0 + 2 - (-15) & 4 - 8 - (-10) \end{bmatrix} = \begin{bmatrix} -23 & 3 \\ 17 & 6 \end{bmatrix} \).

Teacher's Note:
a) Perform matrix multiplication row by column very carefully, checking signs at each step.
b) Combine matrices element by element for addition and subtraction.

Question 8

(a) The compound interest, calculated yearly, on a certain sum of money for the second year is Rs. 1320 and for the third year is Rs. 1452. Calculate the rate of interest and the original sum of money. [3 Marks]

Answer:
Interest for the 2nd year \( I_2 = \text{Rs. } 1320 \)
Interest for the 3rd year \( I_3 = \text{Rs. } 1452 \)
The interest of the 3rd year is earned on the amount at the end of the 2nd year for 1 year.
Difference in interest in 1 year \( = 1452 - 1320 = \text{Rs. } 132 \)
This Rs. 132 is the interest on the amount at the end of the 2nd year (which acts as principal for the 3rd year, i.e., \( A_2 \)).
Wait, compound interest for 3rd year is on amount of 2nd year: \( I_3 = I_2 + \text{Interest on } I_2 \) is not directly true, but: interest earned in the 3rd year is \( 10\% \) more, let's find rate \( r \):
\( 1452 - 1320 = 132 \)
Rate \( r = \frac{\text{Increase in interest}}{\text{Interest of 2nd year}} \times 100 = \frac{132}{1320} \times 100 = 10\% \)
Rate of interest \( = 10\% \) per annum.
To find the original sum \( P \):
Interest for the 2nd year \( I_2 = \frac{P \times r \times 1}{100} \) of amount at end of 1st year.
We know Amount at end of 1st year \( A_1 = P(1 + \frac{10}{100}) = 1.1P \).
Interest for 2nd year \( = A_1 \times \frac{10}{100} = 1.1P \times 0.1 = 0.11P \).
Given \( 0.11P = 1320 \implies P = \frac{1320}{0.11} = \text{Rs. } 12,000 \).

Teacher's Note:
a) The difference between consecutive year's compound interest gives the interest earned on the previous year's total interest amount.
b) Use the rate to work backwards to find the original principal.

(b) Construct a \(\triangle ABC\) with \( BC = 6.5\text{cm}, AB = 5.5\text{cm}, AC = 5\text{cm} \). Construct the incircle of the triangle. Measure and record the radius of the incircle. [4 Marks]

Answer:
1. Draw line segment \( BC = 6.5\text{cm} \).
2. With B as center and radius \( 5.5\text{cm} \), draw an arc. With C as center and radius \( 5\text{cm} \), draw another arc intersecting the previous arc at A.
3. Join AB and AC to complete \(\triangle ABC\).
4. Draw angle bisectors of any two angles (e.g., \(\angle B\) and \(\angle C\)). Their point of intersection is the incentre I.
5. Drop a perpendicular from I to any side (e.g., BC) to get the inradius.
6. With I as center and radius equal to this perpendicular distance, draw the incircle.
7. Measured inradius \( \approx 1.5\text{cm} \).

Teacher's Note:
a) Incentre is found by constructing angle bisectors of at least two interior angles of the triangle.
b) The radius is measured perpendicular from the incentre to any side of the triangle.

(c) (Use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:
Draw a histogram representing the above distribution and estimate the mode from the graph. [3 Marks]

Answer:
1. Mark pocket expenses along the x-axis and number of students along the y-axis.
2. Draw rectangles for each class interval with height proportional to its frequency (e.g., interval 20-25 has highest frequency 50).
3. To find the mode graphically from the histogram:
- Join the top-right corner of the modal rectangle (20-25) to the top-right corner of the preceding rectangle.
- Join the top-left corner of the modal rectangle to the top-left corner of the succeeding rectangle.
- From the intersection point of these two diagonals, draw a perpendicular to the x-axis.
4. The value on the x-axis gives the estimated mode \( \approx 22.5\text{ Rs.} \)
[Figure: Histogram on graph paper with modal estimation lines.]

Teacher's Note:
a) Ensure class intervals are continuous and uniform before plotting a histogram.
b) Mode is estimated by cross-diagonals of the highest rectangle and the adjacent rectangles.

Question 9

(a) If \((x - 9) : (3x + 6)\) is the duplicate ratio of \(4 : 9\), find the value of \(x\). [3 Marks]

Answer:
Given duplicate ratio of \( 4 : 9 \) is \((4^2 : 9^2) = 16 : 81\).
Therefore, \(\frac{x - 9}{3x + 6} = \frac{16}{81}\)
Cross-multiplying:
\(81(x - 9) = 16(3x + 6)\)
\(81x - 729 = 48x + 96\)
\(81x - 48x = 96 + 729\)
\(33x = 825\)
\(x = \frac{825}{33} = 25\).

Teacher's Note:
a) The duplicate ratio of \(a : b\) is \(a^2 : b^2\).
b) Solve the resulting linear equation carefully after cross-multiplication.

(b) Solve for \(x\) using the quadratic formula. Write your answer correct to two significant figures. \((x - 1)^2 - 3x + 4 = 0\). [3 Marks]

Answer:
Given equation: \((x - 1)^2 - 3x + 4 = 0\)
Expand and simplify:
\(x^2 - 2x + 1 - 3x + 4 = 0\)
\(x^2 - 5x + 5 = 0\)
Here, \(a = 1, b = -5, c = 5\).
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(5)}}{2(1)}\)
\(x = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}\)
Since \( \sqrt{5} \approx 2.236 \):
\(x_1 = \frac{5 + 2.236}{2} = \frac{7.236}{2} = 3.618 \approx 3.6\)
\(x_2 = \frac{5 - 2.236}{2} = \frac{2.764}{2} = 1.382 \approx 1.4\)
Therefore, \(x = 3.6\) or \(x = 1.4\).

Teacher's Note:
a) First expand the algebraic expression and bring it to the standard quadratic form \(ax^2 + bx + c = 0\).
b) Round off to two significant figures as requested by the question.

(c) A page from the savings bank account of Priyanka is given below: [4 Marks]

If the interest earned by Priyanka for the period ending September, 2006 is Rs. 175, find the rate of interest.

Answer:
Let us find the equivalent principal for 1 month:
- April 3 to April 5 (2 days): Balance = Rs. 4,000 (Consider min balance from 10th to end of month, or standard rule: April balance is on lowest between 10th and 30th = Rs. 6,000)
Let's list lowest balances for each month from April to September:
- April (from 10th to 30th): lowest balance between Apr 18 and Apr 30 is Rs. 6,000.
- May (from 10th to 30th): lowest balance between May 25 and May 30 is Rs. 7,000.
- June (no transactions): balance remains Rs. 10,000.
- July (from 10th to 31st): lowest balance after July 20 is Rs. 6,000.
- August (no transactions): balance remains Rs. 6,000.
- September (from 10th to 30th): lowest balance after Sep 19 is Rs. 7,000.
Total equivalent principal for 1 month \( P_{eq} = 6000 + 7000 + 10000 + 6000 + 6000 + 7000 = \text{Rs. } 42,000 \).
Given Interest \( I = \text{Rs. } 175 \), time \( t = \frac{1}{12} \text{ year} \).
\( I = \frac{P_{eq} \times r \times 1}{100 \times 12} \)
\( 175 = \frac{42000 \times r}{1200} \)
\( 175 = 35 \times r \)
\( r = \frac{175}{35} = 5\% \).

Teacher's Note:
a) Savings bank interest is calculated on the minimum balance between the 10th day and the end of each month.
b) Sum up the monthly minimum balances to get the equivalent principal for one month.

Question 10

(a) A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number. [3 Marks]

Answer:
Let the tens digit be \( x \) and the units digit be \( y \).
The number is \( 10x + y \).
Given: Product of digits \( xy = 6 \implies y = \frac{6}{x} \).
When 9 is added, digits interchange: \( (10x + y) + 9 = 10y + x \)
\( 9x - 9y + 9 = 0 \implies x - y + 1 = 0 \implies y = x + 1 \).
Equating both values of \( y \):
\(\frac{6}{x} = x + 1\)
\(6 = x^2 + x \implies x^2 + x - 6 = 0\)
\((x + 3)(x - 2) = 0\)
Since \( x \) is a digit, \( x = 2 \) (rejecting \(-3\)).
If \( x = 2 \), then \( y = \frac{6}{2} = 3 \).
The number is \( 10(2) + 3 = 23 \).

Teacher's Note:
a) A two-digit number is represented algebraically as \(10x + y\), where \(x\) and \(y\) are its digits.
b) Verify the answer: \(2 \times 3 = 6\) and \(23 + 9 = 32\) (digits reversed).

(b) The marks obtained by 100 students in a Mathematics test are given below: [6 Marks]

Draw an ogive for the given distribution on a graph sheet.
Use a scale of \( 2\text{cm} = 10 \) units on both axis.
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than \( 85\% \) marks in the test.
(iv) number of students who did not pass in the test if the pass percentage was 35.

Answer:
First, construct the Cumulative Frequency (CF) table:

Plot upper class limits against cumulative frequencies: \((10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85), (80, 91), (90, 100)\) and join them with a smooth curve.
(i) Median: Total students \( N = 100 \). Look at \( N/2 = 50 \) on the y-axis, go across to the curve and down to the x-axis. Median \( \approx 45 \Marks \).
(ii) Lower quartile \( Q_1 \): Look at \( N/4 = 25 \) on the y-axis, go across to the curve and down to the x-axis. \( Q_1 \approx 31.5 \Marks \).
(iii) Students scoring more than \( 85\% \) (i.e., above 85 marks): From 85 on the x-axis, go up to the curve and across to the y-axis (let CF be approx 95). Students above 85 \( = 100 - 95 = 5 \) students.
(iv) Students who did not pass if pass percentage is 35 (i.e., below 35 marks): From 35 on the x-axis, go up to the curve and across to the y-axis (CF at 35 is approx 30). So about 30 students did not pass.
[Figure: Ogive curve plotted on graph paper with markings for median, quartiles, and pass percentages.]

Teacher's Note:
a) An ogive is always plotted using upper class limits and cumulative frequencies.
b) Read values from the graph carefully using the specified scale.

Question 11

(a) In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. \( AB = 24\text{cm}, OM = 5\text{cm}, ON = 12\text{cm} \). Find the:
(i) radius of the circle.
(ii) length of chord CD. [3 Marks]

Answer:
(i) Perpendicular from the centre to a chord bisects the chord.
\( MB = \frac{AB}{2} = \frac{24}{2} = 12\text{ cm} \).
In right-angled triangle \(\triangle OMB\):
Radius \( OB = \sqrt{OM^2 + MB^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm} \).
So, radius of the circle is \( 13\text{ cm} \).
(ii) In right-angled triangle \(\triangle OND\), radius \( OD = 13\text{ cm} \) (since radii of a circle are equal).
Using Pythagoras theorem in \(\triangle OND\):
\( ON^2 + DN^2 = OD^2 \)
\( 12^2 + DN^2 = 13^2 \)
\( 144 + DN^2 = 169 \)
\( DN^2 = 169 - 144 = 25 \implies DN = 5\text{ cm} \).
Since perpendicular from centre bisects chord CD, \( CD = 2 \times DN = 2 \times 5 = 10\text{ cm} \).

Teacher's Note:
a) The perpendicular from the centre of a circle to a chord bisects the chord.
b) All radii in the same circle are equal in length.

(b) Prove the identity: \((\sin \theta + \cos \theta)(\tan \theta + \cot \theta) = \sec \theta + \csc \theta\). [3 Marks]

Answer:
LHS \( = (\sin \theta + \cos \theta)(\tan \theta + \cot \theta) \)
Convert \(\tan \theta\) and \(\cot \theta\) in terms of sine and cosine:
\( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
LHS \( = (\sin \theta + \cos \theta)\left(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\right) \)
\( = (\sin \theta + \cos \theta)\left(\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\right) \)
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
LHS \( = (\sin \theta + \cos \theta)\left(\frac{1}{\sin \theta \cos \theta}\right) \)
\( = \frac{\sin \theta}{\sin \theta \cos \theta} + \frac{\cos \theta}{\sin \theta \cos \theta} \)
\( = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \)
\( = \sec \theta + \csc \theta = \text{RHS} \). Hence proved.

Teacher's Note:
a) Expressing all trigonometric ratios in terms of sine and cosine is a standard and effective strategy for proving identities.
b) Remember the fundamental Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\).

(c) An aeroplane at an altitude of \( 250\text{ m} \) observes the angle of depression of two boats on the opposite banks of a river to be \( 45^{\circ} \) and \( 60^{\circ} \) respectively. Find the width of the river. Write the answer correct to the nearest whole number. [4 Marks]

Answer:
Let the aeroplane be at point P at an altitude \( AP = 250\text{ m} \).
Let B and C be the two boats on opposite banks of the river such that BC is the width of the river, and A lies vertically above a point on BC (let's assume it divides the river into two segments AB and AC, or A is directly above a point between B and C).
In right-angled triangle \(\triangle ABP\) (with angle of depression \( 45^{\circ} \), so angle of elevation at B is \( 45^{\circ} \)):
\(\tan 45^{\circ} = \frac{AP}{BP} \implies 1 = \frac{250}{BP} \implies BP = 250\text{ m}\).
In right-angled triangle \(\triangle ACP\) (with angle of depression \( 60^{\circ} \), so angle of elevation at C is \( 60^{\circ} \)):
\(\tan 60^{\circ} = \frac{AP}{PC} \implies \sqrt{3} = \frac{250}{PC} \implies PC = \frac{250}{\sqrt{3}}\text{ m}\).
Total width of the river \( BC = BP + PC = 250 + \frac{250}{\sqrt{3}} \)
\( BC = 250 + \frac{250 \times 1.732}{3} = 250 + \frac{433}{3} = 250 + 144.33 = 394.33\text{ m} \).
Correct to the nearest whole number, width \( = 394\text{ m} \).

Teacher's Note:
a) Angle of depression from an object equals the angle of elevation from the ground by alternate interior angles.
b) Substitute \( \sqrt{3} \approx 1.732 \) and compute the final value carefully before rounding.