ICSE Class 10 Mathematics Board Exam Question Paper 2016 with Solutions

ICSE Class 10 Mathematics Board Exam Question Paper 2016 with Solutions

SECTION A (40 Marks)

Question 1

(a) Using remainder theorem, find the value of k if on dividing \( 2x^{3} + 3x^{2} - kx + 5 \) by \( x - 2 \), leaves a remainder 7. [3 Marks]

Answer:
Let \( f(x) = 2x^{3} + 3x^{2} - kx + 5 \).
Using the Remainder Theorem, when \( f(x) \) is divided by \( x - 2 \), the remainder is \( f(2) \).
Given remainder = \( 7 \), so \( f(2) = 7 \).
\( 2(2)^{3} + 3(2)^{2} - k(2) + 5 = 7 \)
\( 2(8) + 3(4) - 2k + 5 = 7 \)
\( 16 + 12 - 2k + 5 = 7 \)
\( 33 - 2k = 7 \)
\( 2k = 33 - 7 \)
\( 2k = 26 \)
\( k = 13 \)

Teacher's Note:
a) The Remainder Theorem states that the remainder of the division of a polynomial \( f(x) \) by a linear divisor \( x - a \) is equal to \( f(a) \).
b) Ensure careful substitution of positive and negative signs while evaluating polynomial expressions at given values.

(b) Given \( A = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( A^{2} = 9A + mI \). Find m. [4 Marks]

Answer:
Given equation: \( A^{2} = 9A + ml \)
\( \Rightarrow A^{2} - 9A = ml \) ...... (1)
First, find \( A^{2} = A \cdot A \):
\( A^{2} = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \)
\( A^{2} = \begin{bmatrix} (2)(2) + (0)(-1) & (2)(0) + (0)(7) \\ (-1)(2) + (7)(-1) & (-1)(0) + (7)(7) \end{bmatrix} \)
\( A^{2} = \begin{bmatrix} 4 + 0 & 0 + 0 \\ -2 - 7 & 0 + 49 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} \)
Now substitute \( A^{2} \), \( A \), and \( I \) into equation (1):
\( \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} - 9\begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} = m\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} - \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} \)
\( \begin{bmatrix} 4 - 18 & 0 - 0 \\ -9 - (-9) & 49 - 63 \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} \)
\( \begin{bmatrix} -14 & 0 \\ 0 & -14 \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} \)
Comparing corresponding elements:
\( m = -14 \)

Teacher's Note:
a) Matrix multiplication is performed row by column: \( (R_{1} \times C_{1}) \).
b) Scalar multiplication multiplies every element inside the matrix by the scalar factor.

(c) The mean of following numbers is 68. Find the value of x. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence estimate the median. [3 Marks]

Answer:
Given numbers: \( 45, 52, 60, x, 69, 70, 26, 81, 94 \)
Total number of observations \( n = 9 \)
\( \text{Mean} = \frac{\text{Sum of all observations}}{\text{Total number of observations}} \)
\( 68 = \frac{45 + 52 + 60 + x + 69 + 70 + 26 + 81 + 94}{9} \)
\( 68 = \frac{497 + x}{9} \)
\( 497 + x = 68 \times 9 \)
\( 497 + x = 612 \)
\( x = 612 - 497 = 115 \)
The complete set of numbers is: \( 26, 45, 52, 60, 69, 70, 81, 94, 115 \).
Arranging the data in ascending order:
\( 26, 45, 52, 60, 69, 70, 81, 94, 115 \)
Since \( n = 9 \) (odd), the median is the value of the \( \left(\frac{n + 1}{2}\right)^{\text{th}} \) observation.
\( \text{Median} = \left(\frac{9 + 1}{2}\right)^{\text{th}} \text{ observation} = 5^{\text{th}} \text{ observation} \)
\( 5^{\text{th}} \text{ observation} = 69 \).
Hence, the median is 69.

Teacher's Note:
a) Always arrange data in ascending or descending order before finding the median.
b) For an odd number of observations, the median is the middle term directly.

Question 2

(a) The slope of a line joining P(6, k) and Q(1 - 3k, 3) is \( \frac{1}{2} \). Find
(i) k
(ii) Midpoint of PQ, using the value of 'k' found in (i). [3 Marks]

Answer:
(i) Let \( (x_{1}, y_{1}) = (6, k) \) and \( (x_{2}, y_{2}) = (1 - 3k, 3) \).
Slope \( m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \)
\( \frac{1}{2} = \frac{3 - k}{(1 - 3k) - 6} \)
\( \frac{1}{2} = \frac{3 - k}{-3k - 5} \)
Cross-multiplying:
\( -3k - 5 = 2(3 - k) \)
\( -3k - 5 = 6 - 2k \)
\( -5 - 6 = 3k - 2k \)
\( k = -11 \)

(ii) Substituting \( k = -11 \) into the coordinates of P and Q:
\( P(6, k) = P(6, -11) \)
\( Q(1 - 3(-11), 3) = Q(1 + 33, 3) = Q(34, 3) \)
Midpoint of PQ = \( \left(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right) \)
Midpoint = \( \left(\frac{6 + 34}{2}, \frac{-11 + 3}{2}\right) = \left(\frac{40}{2}, \frac{-8}{2}\right) = (20, -4) \).

Teacher's Note:
a) The slope formula for two points is \( m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \).
b) Pay close attention to signs when substituting negative values into coordinates.

(b) Without using trigonometrical tables, evaluate:
\( \operatorname{cosec}^{2} 57^{\circ} - \tan^{2} 33^{\circ} + \cos 44^{\circ} \operatorname{cosec} 46^{\circ} - \sqrt{2}\cos 45^{\circ} - \tan^{2} 60^{\circ} \) [4 Marks]

Answer:
Given expression:
\( \operatorname{cosec}^{2} 57^{\circ} - \tan^{2} 33^{\circ} + \cos 44^{\circ} \operatorname{cosec} 46^{\circ} - \sqrt{2}\cos 45^{\circ} - \tan^{2} 60^{\circ} \)
Using complementary angles (\( \operatorname{cosec}(90^{\circ} - \theta) = \sec \theta \) and \( \operatorname{cosec}(90^{\circ} - \theta) = \sec \theta \)):
\( = \operatorname{cosec}^{2}(90^{\circ} - 33^{\circ}) - \tan^{2} 33^{\circ} + \cos 44^{\circ} \operatorname{cosec}(90^{\circ} - 44^{\circ}) - \sqrt{2}\cos 45^{\circ} - \tan^{2} 60^{\circ} \)
\( = \sec^{2} 33^{\circ} - \tan^{2} 33^{\circ} + \cos 44^{\circ} \sec 44^{\circ} - \sqrt{2}\cos 45^{\circ} - \tan^{2} 60^{\circ} \)
Since \( \sec^{2}\theta - \tan^{2}\theta = 1 \) and \( \sec 44^{\circ} = \frac{1}{\cos 44^{\circ}} \):
\( = 1 + (\cos 44^{\circ} \cdot \frac{1}{\cos 44^{\circ}}) - \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) - (\sqrt{3})^{2} \)
\( = 1 + 1 - 1 - 3 \)
\( = 2 - 1 - 3 = -2 \)

Teacher's Note:
a) Use standard trigonometric identities like \( 1 + \tan^{2}\theta = \sec^{2}\theta \) to simplify terms.
b) Substitute exact standard angle values (\( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \), \( \tan 60^{\circ} = \sqrt{3} \)) carefully.

(c) A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones. [3 Marks]

Answer:
Let the number of cones be \( n \).
Radius of the sphere (\( r_{s} \)) = \( 6 \) cm
Radius of each cone (\( r_{c} \)) = \( 2 \) cm
Height of each cone (\( h \)) = \( 3 \) cm
Volume of the solid sphere = \( n \times (\text{Volume of one metallic cone}) \)
\( \frac{4}{3}\pi r_{s}^{3} = n \left(\frac{1}{3}\pi r_{c}^{2}h\right) \)
\( 4r_{s}^{3} = n \cdot r_{c}^{2}h \)
\( n = \frac{4r_{s}^{3}}{r_{c}^{2}h} \)
\( n = \frac{4(6)^{3}}{(2)^{2}(3)} \)
\( n = \frac{4 \times 216}{4 \times 3} \)
\( n = \frac{216}{3} = 72 \)
Hence, the number of cones is 72.

Teacher's Note:
a) When solid shapes are melted and recast, the total volume remains constant.
b) Cancel common terms like \( \frac{1}{3}\pi \) on both sides before substitution to simplify calculations.

Question 3

(a) Solve the following inequation, write the solution set and represent it on the number line.
\( -3(x - 7) \ge 15 - 7x \gt \frac{x + 1}{3}, x \in \mathbb{R} \) [3 Marks]

Answer:
Split the given compound inequation into two parts:
Part 1: \( -3(x - 7) \ge 15 - 7x \)
\( -3x + 21 \ge 15 - 7x \)
\( -3x + 7x \ge 15 - 21 \)
\( 4x \ge -6 \)
\( x \ge \frac{-6}{4} \)
\( x \ge -1.5 \) ...... (1)

Part 2: \( 15 - 7x \gt \frac{x + 1}{3} \)
\( 3(15 - 7x) \gt x + 1 \)
\( 45 - 21x \gt x + 1 \)
\( 45 - 1 \gt x + 21x \)
\( 44 \gt 22x \)
\( 2 \gt x \text{ i.e., } x \lt 2 \) ...... (2)

Combining (1) and (2):
\( -1.5 \le x \lt 2 \)
The solution set is \( \{x : x \in \mathbb{R}, -1.5 \le x \lt 2\} \).
[Figure: Number line showing a shaded segment from -1.5 (solid circle) to 2 (hollow circle)]

Teacher's Note:
a) Solve compound inequations by separating them into two independent inequalities and finding their intersection.
b) Use a solid circle for "greater than or equal to" (\( \ge \)) and a hollow circle for "strictly greater than/less than" (\(\gt\) or \(\lt\)) on the number line.

(b) In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and \( \angle CBD = 32^{\circ} \). Find:
(i) \( \angle OBD \)
(ii) \( \angle AOB \)
(iii) \( \angle BED \) [4 Marks]

[Figure: Circle with diameter AD, centre O. Cyclic points A, B, C, D on the circle. Chord BC is parallel to AD. Line segments connect O to B, A to B, B to D, B to E, and E to D forming triangles. Angle \( \angle CBD = 32^{\circ} \)]

Answer:
(i) Since \( AD \parallel BC \) and \( BD \) is a transversal, alternate interior angles are equal:
\( \angle ODB = \angle CBD = 32^{\circ} \).
In \( \triangle OBD \), \( OD = OB \) (radii of the same circle).
Therefore, \( \angle OBD = \angle ODB = 32^{\circ} \).

(ii) Since \( AD \parallel BC \) and \( OB \) is a transversal, alternate interior angles are equal:
\( \angle AOB = \angle OBC \).
Now, \( \angle OBC = \angle OBD + \angle DBC = 32^{\circ} + 32^{\circ} = 64^{\circ} \).
Hence, \( \angle AOB = 64^{\circ} \).\br />
(iii) In \( \triangle OAB \), \( OA = OB \) (radii of the same circle), so \( \angle OAB = \angle OBA = x \) (say).
The sum of angles in \( \triangle OAB \) is \( 180^{\circ} \):
\( x + x + \angle AOB = 180^{\circ} \)
\( 2x + 64^{\circ} = 180^{\circ} \)
\( 2x = 116^{\circ} \implies x = 58^{\circ} \)
So, \( \angle OAB = 58^{\circ} \), which means \( \angle DAB = 58^{\circ} \).
Since angles in the same segment are equal, \( \angle BED = \angle DAB = 58^{\circ} \).

Teacher's Note:
a) Use properties of parallel lines, isosceles triangles formed by radii, and angles in the same segment.
b) Clearly state geometric theorems used at each step for full credit.

(c) If \( (3a + 2b) : (5a + 3b) = 18 : 29 \). Find \( a : b \). [3 Marks]

Answer:
Given: \( \frac{3a + 2b}{5a + 3b} = \frac{18}{29} \)
Cross-multiplying:
\( 29(3a + 2b) = 18(5a + 3b) \)
\( 87a + 58b = 90a + 54b \)
Rearranging terms:
\( 58b - 54b = 90a - 87a \)
\( 4b = 3a \)
\( \frac{a}{b} = \frac{4}{3} \)
Hence, \( a : b = 4 : 3 \).

Teacher's Note:
a) Cross-multiplication is the standard method for solving linear ratio equations.
b) Group like terms of \( a \) and \( b \) on opposite sides carefully to avoid sign errors.

Question 4

(a) A game of numbers has cards marked with 11, 12, 13, ... 40. A card is drawn at random. Find the probability that the number on the card drawn is:
(i) A perfect square
(ii) Divisible by 7 [3 Marks]

Answer:
Total number of cards from 11 to 40 is \( 40 - 11 + 1 = 30 \).
Total number of possible outcomes = \( 30 \).

(i) Perfect squares between 11 and 40 are \( 16, 25, 36 \) (total 3 numbers).
Probability of getting a perfect square = \( \frac{3}{30} = \frac{1}{10} \).

(ii) Numbers between 11 and 40 divisible by 7 are \( 14, 21, 28, 35 \) (total 4 numbers).
Probability of getting a number divisible by 7 = \( \frac{4}{30} = \frac{2}{15} \).

Teacher's Note:
a) Total outcomes for numbers from \( a \) to \( b \) inclusive is given by \( b - a + 1 \).
b) Always reduce fractions to their simplest form.

(b) Use graph paper for this question. (Take 2 cm = 1 unit along both x and y axis.) Plot the points \( O(0, 0), A(-4, 4), B(-3, 0) \) and \( C(0, -3) \)
(i) Reflect points A and B on the y-axis and name them A' and B' respectively. Write down their coordinates.
(ii) Name the figure OABCB'A'.
(iii) State the line of symmetry of this figure [4 Marks]

[Figure: Cartesian coordinate plane showing points O(0,0), A(-4,4), B(-3,0), C(0,-3), A'(4,4), and B'(3,0) connected to form a symmetrical arrowhead polygon]

Answer:
(i) Reflection across the y-axis changes the sign of the x-coordinate (\( (x, y) \to (-x, y) \)):
Coordinates of \( A' \) = \( (4, 4) \)
Coordinates of \( B' \) = \( (3, 0) \)

(ii) The figure OABCB'A' is an arrowhead.
(iii) The line of symmetry of this figure is the y-axis.

Teacher's Note:
a) Reflection in the y-axis negates the x-coordinate while leaving the y-coordinate unchanged.
b) A line of symmetry divides a geometric figure into two congruent, mirror-image halves.

(c) Mr. Lalit invested Rs. 5000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to Rs. 5325. Calculate
(i) The rate of interest
(ii) The amount at the end of second year, to the nearest rupee. [3 Marks]

Answer:
(i) For the first year: Principal \( P = \text{Rs. } 5000 \), Amount \( A = \text{Rs. } 5325 \), Time \( T = 1 \) year.
Interest \( I = A - P = 5325 - 5000 = \text{Rs. } 325 \).
Rate \( R = \frac{I \times 100}{P \times T} = \frac{325 \times 100}{5000 \times 1} = \frac{32500}{5000} = 6.5\% \).

(ii) The amount at the end of the first year becomes the principal for the second year: \( P' = \text{Rs. } 5325 \), \( T = 1 \) year, \( R = 6.5\% \).
Interest for the second year \( I' = \frac{5325 \times 6.5 \times 1}{100} = \text{Rs. } 346.125 \).
Amount at the end of the second year = \( 5325 + 346.125 = \text{Rs. } 5671.125 \).
To the nearest rupee, the amount is Rs. 5671.

Teacher's Note:
a) For compound interest compounded annually, the amount at the end of each year acts as the principal for the next year.
b) Round off to the nearest rupee only in the final step of the calculation.

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

Question 5

(a) Solve the quadratic equation \( x^{2} - 3(x + 3) = 0 \); Give your answer correct two significant figures. [3 Marks]

Answer:
Given equation: \( x^{2} - 3(x + 3) = 0 \)
\( x^{2} - 3x - 9 = 0 \)
Comparing with standard quadratic form \( ax^{2} + bx + c = 0 \):
\( a = 1, b = -3, c = -9 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \):
\( x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-9)}}{2(1)} \)
\( x = \frac{3 \pm \sqrt{9 + 36}}{2} \)
\( x = \frac{3 \pm \sqrt{45}}{2} \)
\( x = \frac{3 \pm 3\sqrt{5}}{2} \)
Using \( \sqrt{5} \approx 2.236 \):
\( x = \frac{3 \pm 3(2.236)}{2} = \frac{3 \pm 6.708}{2} \)
Case 1: \( x = \frac{3 + 6.708}{2} = \frac{9.708}{2} = 4.854 \approx 4.9 \) (to two significant figures)
Case 2: \( x = \frac{3 - 6.708}{2} = \frac{-3.708}{2} = -1.854 \approx -1.9 \) (to two significant figures)
Hence, \( x = 4.9 \) or \( x = -1.9 \).

Teacher's Note:
a) Significant figures are counted from the first non-zero digit. For two significant figures, round to one decimal place here.
b) Always simplify surds before decimal evaluation to ensure higher accuracy.

(b) A page from the savings bank account of Mrs. Ravi is given below.

She closed the account on 30^th September, 2006. Calculate the interest Mrs. Ravi earned at the end of 30^th September, 2006 at 4.5% per annum interest. Hence, find the amount she receives on closing the account. [4 Marks]

Answer:
Minimum balances for each month (between the 10^th and last day of the month):
April (10^th to 30^th) = Rs. 8300
May (10^th to 31^st) = Rs. 7600
June (10^th to 30^th) = Rs. 10300
July (10^th to 31^st) = Rs. 10300
August (10^th to 31^st) = Rs. 3900
September (10^th to 30^th) = Rs. 0
Total principal for 1 month = \( 8300 + 7600 + 10300 + 10300 + 3900 + 0 = \text{Rs. } 40400 \).
Rate \( R = 4.5\% \) p.a.
\( \text{Interest} = \frac{P \times R \times T}{100} = \frac{40400 \times 4.5 \times \frac{1}{12}}{100} = \frac{40400 \times 4.5}{1200} = \frac{181800}{1200} = \text{Rs. } 151.50 \).
Closing balance in the account on 30^th September = Rs. 5900.
Total amount received = Balance + Interest = \( 5900 + 151.50 = \text{Rs. } 6051.50 \).

Teacher's Note:
a) In savings bank accounts, interest is calculated on the minimum balance between the 10^th and the last day of each month.
b) Time \( T \) for the total principal sum is always taken as \( \frac{1}{12} \) of a year.

(c) In what time will Rs. 1500 yield Rs. 496.50 as compound interest at 10% per annum compounded annually? [3 Marks]

Answer:
Given principal \( P = \text{Rs. } 1500 \), Compound Interest \( I = \text{Rs. } 496.50 \), Rate \( R = 10\% \) p.a.
Total Amount \( A = P + I = 1500 + 496.50 = \text{Rs. } 1996.50 \).
Using the compound interest formula:
\( A = P\left(1 + \frac{R}{100}\right)^{n} \)
\( 1996.50 = 1500\left(1 + \frac{10}{100}\right)^{n} \)
\( \frac{1996.50}{1500} = \left(1 + \frac{1}{10}\right)^{n} \)
\( 1.331 = (1.1)^{n} \)
Since \( (1.1)^{3} = 1.331 \):
\( (1.1)^{3} = (1.1)^{n} \)
\( n = 3 \) years.

Teacher's Note:
a) Always compute the total amount \( A \) first by adding principal and compound interest.
b) Express the ratio as powers of the base term inside the bracket to easily solve for time \( n \).

Question 6

(a) Construct a regular hexagon of side 5 cm. Hence construct all its lines of symmetry and name them. [3 Marks]

[Figure: Regular hexagon ABCDEF with side 5 cm, showing 3 diagonal lines of symmetry (AD, CF, EB) and 3 perpendicular bisector lines of symmetry passing through center Z]

Answer:
Steps of construction:
1. Draw a line segment \( AF = 5 \) cm using a ruler.
2. With A as the centre and radius equal to 5 cm, draw an arc. With F as the centre and same radius, cut the arc at point Z.
3. Draw a circle of radius 5 cm with centre Z passing through A and F.
4. Mark consecutive points B, C, D, E along the circle taking 5 cm radius steps.
5. Join adjacent vertices to form the regular hexagon ABCDEF.
6. Draw perpendicular bisectors of opposite sides and join opposite vertices (AD, CF, EB, and lines joining midpoints of opposite sides) to obtain all 6 lines of symmetry.

Teacher's Note:
a) A regular hexagon has 6 lines of symmetry: 3 joining opposite vertices and 3 joining midpoints of opposite sides.
b) All construction arcs must be kept clearly visible.

(b) In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.
(i) Prove \( \triangle TPS \sim \triangle TRQ \).
(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.
(iii) Find area of quadrilateral PQRS if area of \( \triangle PTS = 27\text{ cm}^{2} \). [4 Marks]

[Figure: Cyclic quadrilateral PQRS inscribed in a circle. Sides QP and RS are extended to meet at exterior point T, forming smaller triangle TRQ and larger triangle TPS]

Answer:
(i) In \( \triangle TPS \) and \( \triangle TRQ \):
\( \angle PTS = \angle RTQ \) (Common angle at T)
Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle, so \( \angle TSP = \angle RQP \) (Wait, considering exterior angle properties: \( \angle TQR = \angle TSP \)).
Therefore, by AA similarity criterion, \( \triangle TPS \sim \triangle TRQ \).

(ii) Since \( \triangle TPS \sim \triangle TRQ \), corresponding sides are proportional:
\( \frac{SP}{RQ} = \frac{TP}{TR} \)
\( \frac{SP}{4} = \frac{18}{6} \)
\( \frac{SP}{4} = 3 \)
\( SP = 3 \times 4 = 12 \) cm.

(iii) The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides:
\( \frac{\text{Area}(\triangle TPS)}{\text{Area}(\triangle TRQ)} = \left(\frac{SP}{RQ}\right)^{2} \)
\( \frac{27}{\text{Area}(\triangle TRQ)} = \left(\frac{12}{4}\right)^{2} = 3^{2} = 9 \)
\( \text{Area}(\triangle TRQ) = \frac{27}{9} = 3\text{ cm}^{2} \)
Area of quadrilateral PQRS = \( \text{Area}(\triangle TPS) - \text{Area}(\triangle TRQ) = 27 - 3 = 24\text{ cm}^{2} \).

Teacher's Note:
a) The ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides.
b) Area of a cyclic quadrilateral part is found by subtracting the smaller exterior triangle area from the larger triangle area.

(c) Given matrix \( A = \begin{bmatrix} 4\sin 30^{\circ} & \cos 0^{\circ} \\ \cos 0^{\circ} & 4\sin 30^{\circ} \end{bmatrix} \) and \( B = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \). If \( AX = B \)
(i) Write the order of matrix X.
(ii) Find the matrix 'X'. [3 Marks]

Answer:
First, evaluate trigonometric values in matrix A: \( \sin 30^{\circ} = \frac{1}{2}, \cos 0^{\circ} = 1 \).
\( A = \begin{bmatrix} 4(\frac{1}{2}) & 1 \\ 1 & 4(\frac{1}{2}) \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \).

(i) Order of matrix A is \( 2 \times 2 \). Order of matrix B is \( 2 \times 1 \).
Since \( AX = B \), let the order of X be \( m \times n \).
\( (2 \times 2) \times (m \times n) = (2 \times 1) \implies m = 2, n = 1 \).
Thus, the order of matrix X is \( 2 \times 1 \).

(ii) Let \( X = \begin{bmatrix} x \\ y \end{bmatrix} \).
\( \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \)
\( \begin{bmatrix} 2x + y \\ x + 2y \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \)
Equating corresponding elements:
\( 2x + y = 4 \) ...... (1)
\( x + 2y = 5 \) ...... (2)
Multiply equation (1) by 2:
\( 4x + 2y = 8 \) ...... (3)
Subtracting equation (2) from (3):
\( (4x + 2y) - (x + 2y) = 8 - 5 \)
\( 3x = 3 \implies x = 1 \)
Substitute \( x = 1 \) in equation (1):
\( 2(1) + y = 4 \implies y = 2 \)
Therefore, \( X = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \).

Teacher's Note:
a) Matrix multiplication compatibility requires the inner dimensions to match: \( (a \times b) \cdot (b \times c) \).
b) Solve simultaneous linear equations derived from matrix equality using elimination or substitution.

Question 7

(a) An aeroplane at an altitude of 1500 metres, finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are \( 45^{\circ} \) and \( 30^{\circ} \) respectively. Find the distance between the two ships. [4 Marks]

[Figure: Right-angled triangles formed by an aeroplane at height 1500m above point B, with two ships at points C and D on the ground, having angles of depression \( 45^{\circ} \) and \( 30^{\circ} \)]

Answer:
Let AB be the altitude of the aeroplane = \( 1500 \) m. Let C and D be the positions of the two ships.
In right-angled \( \triangle ABC \) (with \( \angle ACB = 45^{\circ} \)):
\( \tan 45^{\circ} = \frac{AB}{BC} \)
\( 1 = \frac{1500}{BC} \implies BC = 1500 \) m.

In right-angled \( \triangle ABD \) (with \( \angle ADB = 30^{\circ} \)):
\( \tan 30^{\circ} = \frac{AB}{BD} \)
\( \frac{1}{\sqrt{3}} = \frac{1500}{BD} \)
\( BD = 1500\sqrt{3} \) m = \( 1500 \times 1.732 = 2598 \) m.
Distance between the two ships \( CD = BD - BC \)
\( CD = 2598 - 1500 = 1098 \) m.

Teacher's Note:
a) Draw a clear diagram showing angles of depression equal to corresponding angles of elevation.
b) Use \( \sqrt{3} \approx 1.732 \) for final numerical evaluation.

(b) The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis).
Scores: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100
No. of shooters: 9, 13, 20, 26, 30, 22, 15, 10, 8, 7
Use your graph to estimate the following:
(i) The median.
(ii) The interquartile range.
(iii) The number of shooters who obtained a score of more than 85%. [6 Marks]

[Figure: Cumulative frequency curve (ogive) plotted with upper class limits on the X-axis and cumulative frequency on the Y-axis, showing markings for median, quartiles, and 85% score]

Answer:
Cumulative Frequency (c.f.) Table:

Total \( n = 160 \).
(i) Median = Value of \( \left(\frac{n}{2}\right)^{\text{th}} \) term = \( \left(\frac{160}{2}\right)^{\text{th}} = 80^{\text{th}} \) term. From the ogive graph, corresponding score is \( 43 \).
(ii) Lower quartile \( Q_{1} = \left(\frac{160}{4}\right)^{\text{th}} = 40^{\text{th}} \) term \( = 28 \).
Upper quartile \( Q_{3} = \left(\frac{3 \times 160}{4}\right)^{\text{th}} = 120^{\text{th}} \) term \( = 60 \).
Interquartile range = \( Q_{3} - Q_{1} = 60 - 28 = 32 \).
(iii) Score of 85% means a score of 85. From 85 on the X-axis, the cumulative frequency on the ogive is approximately 150.
Number of shooters scoring more than 85% = \( 160 - 150 = 10 \).

Teacher's Note:
a) An ogive is always plotted using upper class limits against cumulative frequencies.
b) Ensure clear horizontal and vertical reading lines are drawn on the graph for median and quartiles.

Question 8

(a) If \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \), show that \( \frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3xyz}{abc} \) [3 Marks]

Answer:
Let \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \).
Then, \( x = ak, y = bk, z = ck \).
\( \text{L.H.S.} = \frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} \)
\( = \frac{(ak)^{3}}{a^{3}} + \frac{(bk)^{3}}{b^{3}} + \frac{(ck)^{3}}{c^{3}} \)
\( = \frac{a^{3}k^{3}}{a^{3}} + \frac{b^{3}k^{3}}{b^{3}} + \frac{c^{3}k^{3}}{c^{3}} \)
\( = k^{3} + k^{3} + k^{3} = 3k^{3} \)

\( \text{R.H.S.} = \frac{3xyz}{abc} = \frac{3(ak)(bk)(ck)}{abc} = \frac{3abc \cdot k^{3}}{abc} = 3k^{3} \).
Since \( \text{L.H.S.} = \text{R.H.S.} \), hence proved.

Teacher's Note:
a) Use the k-method for ratio and proportion proofs involving continued equality.
b) Substitute variables systematically and simplify both sides independently.

(b) Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct:
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents. [4 Marks]

[Figure: Line AB = 5cm with point C at 3cm from A. Circle of radius 2.5cm passing through A and C with centre O. Tangents PB and QB drawn from external point B with tangent lengths equal to 3 cm]

Answer:
Steps of construction:
1. Draw line segment \( AB = 5 \) cm. Mark point C on AB such that \( AC = 3 \) cm.
2. Draw the perpendicular bisector of AC to locate the centre O of the circle passing through A and C with radius \( 2.5 \) cm.
3. Draw the circle with centre O and radius \( 2.5 \) cm.
4. Join OB. Draw the perpendicular bisector of OB to find its midpoint M.
5. With M as centre and radius equal to OM, draw a circle intersecting the given circle at P and Q.
6. Join PB and QB. These are the required tangents. Measured length \( PB = QB = 3 \) cm.

Teacher's Note:
a) The centre of a circle passing through two points lies on the perpendicular bisector of the line segment joining them.
b) Tangents drawn from an external point to a circle are equal in length.

(c) A line AB meets X - axis at A and Y - axis at B. P(4, -1) divides AB in the ratio 1 : 2.
(i) Find the coordinates of A and B.
(ii) Find the equation of the line through P and perpendicular to AB. [3 Marks]

[Figure: Coordinate plane showing line AB intersecting X-axis at A and Y-axis at B, with point P(4,-1) dividing AB in ratio 1:2]

Answer:
(i) Since A lies on the X-axis, let its coordinates be \( (x, 0) \). Since B lies on the Y-axis, let its coordinates be \( (0, y) \).
Using the section formula with ratio \( m : n = 1 : 2 \):
\( P(4, -1) = \left(\frac{1(0) + 2(x)}{1 + 2}, \frac{1(y) + 2(0)}{1 + 2}\right) \)
\( (4, -1) = \left(\frac{2x}{3}, \frac{y}{3}\right) \)
Equating coordinates:
\( \frac{2x}{3} = 4 \implies 2x = 12 \implies x = 6 \)
\( \frac{y}{3} = -1 \implies y = -3 \)
Coordinates of A are \( (6, 0) \) and B are \( (0, -3) \).

(ii) Slope of line AB \( m_{1} = \frac{-3 - 0}{0 - 6} = \frac{-3}{-6} = \frac{1}{2} \).
The slope of the line perpendicular to AB is \( m_{2} = -\frac{1}{m_{1}} = -2 \).
Equation of the line passing through \( P(4, -1) \) with slope \( -2 \):
\( y - y_{1} = m(x - x_{1}) \)
\( y - (-1) = -2(x - 4) \)
\( y + 1 = -2x + 8 \)
\( 2x + y = 7 \).

Teacher's Note:
a) Points on the X-axis have a y-coordinate of 0, and points on the Y-axis have an x-coordinate of 0.
b) Perpendicular lines have slopes whose product is \( -1 \) (\( m_{1} \cdot m_{2} = -1 \)).

Question 9

(a) A dealer buys an article at a discount of 30% from the wholesaler, the marked price being Rs. 6000. The dealer sells it to a shopkeeper at a discount of 10% on the marked price. If the rate of VAT is 6%, find
(i) The price paid by the shopkeeper including the tax.
(ii) The VAT paid by the dealer. [3 Marks]

Answer:
Marked Price (MP) = Rs. 6000.
(i) Price paid by the shopkeeper to the dealer (excluding tax) = \( 90\% \text{ of } 6000 = \text{Rs. } 5400 \).
Sales tax paid by shopkeeper (VAT rate 6%) = \( 6\% \text{ of } 5400 = \text{Rs. } 324 \).
Total price paid by the shopkeeper including tax = \( 5400 + 324 = \text{Rs. } 5724 \).

(ii) Price paid by the dealer to the wholesaler = \( 70\% \text{ of } 6000 = \text{Rs. } 4200 \).
Value added by the dealer = \( \text{Selling Price} - \text{Cost Price} = 5400 - 4200 = \text{Rs. } 1200 \).
VAT paid by the dealer = \( 6\% \text{ of Value Added} = 6\% \text{ of } 1200 = \text{Rs. } 72 \).

Teacher's Note:
a) VAT paid by a dealer is always calculated as the tax percentage on the value added (Selling Price - Cost Price).
b) Total price including tax includes the marked/selling price plus the applicable sales tax/VAT.

(b) The given figure represents a kite with a circular and a semicircular motifs stuck on it. The radius of a circle is 2.5 cm and the semicircle is 2 cm. If diagonals AC and BD are of lengths 12 cm and 8 cm respectively, find the area of the:
(i) Shaded part. Give your answer correct to the nearest whole number.
(ii) Unshaded part. [4 Marks]

[Figure: Kite shape with diagonals AC = 12cm and BD = 8cm, containing a full circle of radius 2.5cm and a semicircle of radius 2cm]

Answer:
(i) Area of the shaded part = Area of the circle + Area of the semicircle
Radius of circle \( r_{1} = 2.5 \) cm, Radius of semicircle \( r_{2} = 2 \) cm.
\( \text{Area} = \pi(2.5)^{2} + \frac{\pi(2)^{2}}{2} \)
\( = \pi(6.25 + 2) = \pi(8.25) \)
\( = \frac{22}{7} \times 8.25 = \frac{181.5}{7} \approx 25.928 \text{ cm}^{2} \approx 26\text{ cm}^{2} \) (to nearest whole number).

(ii) Area of the kite = \( \frac{1}{2} \times d_{1} \times d_{2} = \frac{1}{2} \times 12 \times 8 = 48\text{ cm}^{2} \).
Area of the unshaded part = Area of the kite - Area of the shaded part
\( = 48 - 26 = 22\text{ cm}^{2} \).

Teacher's Note:
a) Area of a kite is given by half the product of its diagonals: \( \frac{1}{2}d_{1}d_{2} \).
b) Subtract the total shaded motif area from the kite area to get the unshaded region.

(c) A model of a ship is made to a scale 1 : 300
(i) The length of the model of ship is 2 m. Calculate the lengths of the ship.
(ii) The area of the deck ship is \( 180,000\text{ m}^{2} \). Calculate the area of the deck of the model.
(iii) The volume of the model in \( 6.5\text{ m}^{3} \). Calculate the volume of the ship. [3 Marks]

Answer:
Scale factor \( k = \frac{1}{300} \).
(i) Length of model = \( k \times \text{Length of ship} \)
\( 2 = \frac{1}{300} \times \text{Length of ship} \)
Length of the ship = \( 2 \times 300 = 600 \) m.

(ii) Area of model deck = \( k^{2} \times \text{Area of ship deck} \)
\( \text{Area of model deck} = \left(\frac{1}{300}\right)^{2} \times 180000 = \frac{1}{90000} \times 180000 = 2\text{ m}^{2} \).

(iii) Volume of model = \( k^{3} \times \text{Volume of ship} \)
\( 6.5 = \left(\frac{1}{300}\right)^{3} \times \text{Volume of ship} \)
\( 6.5 = \frac{1}{27000000} \times \text{Volume of ship} \)
Volume of the ship = \( 6.5 \times 27000000 = 175,500,000\text{ m}^{3} \).

Teacher's Note:
a) For scale models, lengths scale by \( k \), areas by \( k^{2} \), and volumes by \( k^{3} \).
b) Maintain consistency in units throughout ratio calculations.

Question 10

(a) Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs. 1200 as interest at the time of maturity, find:
(i) the monthly installment
(ii) the amount of maturity [3 Marks]

Answer:
(i) Interest \( I = \text{Rs. } 1200 \), Time \( n = 2 \text{ years} = 24 \) months, Rate \( r = 6\% \) p.a.
Using the recurring deposit interest formula:
\( I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( 1200 = P \times \frac{24(24 + 1)}{24} \times \frac{6}{100} \)
\( 1200 = P \times 25 \times \frac{6}{100} \)
\( 1200 = P \times \frac{150}{100} = P \times \frac{3}{2} \)
\( P = \frac{1200 \times 2}{3} = \text{Rs. } 800 \).
Monthly installment = Rs. 800.

(ii) Total deposit = \( P \times n = 800 \times 24 = \text{Rs. } 19200 \).
Maturity Amount = Total deposit + Interest = \( 19200 + 1200 = \text{Rs. } 20400 \).

Teacher's Note:
a) The number of months \( n \) for 2 years is \( 24 \).
b) Maturity amount is always the sum of total money deposited and total interest earned.

(b) The histogram below represents the scores obtained by 25 students in a mathematics mental test. Use the data to:
(i) Frame a frequency distribution table
(ii) To calculate mean
(iii) To determine the Modal class [4 Marks]

[Figure: Histogram showing frequency bars for score intervals 0-10 (height 2), 10-20 (height 5), 20-30 (height 8), 30-40 (height 4), and 40-50 (height 6)]

Answer:
(i) Frequency Distribution Table:

(ii) Calculation of Mean:

\( \text{Mean} = \frac{\sum fx}{\sum f} = \frac{695}{25} = 27.8 \).

(iii) The maximum frequency is 8, which corresponds to the class interval \( 20 - 30 \).
Hence, the modal class is \( 20 - 30 \).

Teacher's Note:
a) Mid-values are computed as \( \frac{\text{Lower limit} + \text{Upper limit}}{2} \).
b) The modal class is the class interval having the highest frequency.

(c) A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to cover the total distance. Assuming the uniform speed to be 'x' km/h, form an equation and solve it to evaluate 'x'. [3 Marks]

Answer:
Distance = 240 km. Uniform speed = \( x \) km/h.
Initial time taken = \( \frac{240}{x} \) hours.
Reduced speed = \( (x - 10) \) km/h.
New time taken = \( \frac{240}{x - 10} \) hours.
According to the given condition:
\( \frac{240}{x - 10} - \frac{240}{x} = 2 \)
\( 240\left(\frac{1}{x - 10} - \frac{1}{x}\right) = 2 \)
\( \frac{x - (x - 10)}{x(x - 10)} = \frac{2}{240} \)
\( \frac{10}{x^{2} - 10x} = \frac{1}{120} \)
\( x^{2} - 10x = 1200 \)
\( x^{2} - 10x - 1200 = 0 \)
\( (x - 40)(x + 30) = 0 \)
\( x = 40 \) or \( x = -30 \).
Since speed cannot be negative, the uniform speed \( x = 40 \) km/h.

Teacher's Note:
a) Time is calculated as distance divided by speed (\( t = \frac{D}{S} \)).
b) Reject negative speed values as speed is always a positive quantity.

Question 11

(a) Prove that \( \frac{\cos A}{1 + \sin A} + \tan A = \sec A \) [3 Marks]

Answer:
\( \text{L.H.S.} = \frac{\cos A}{1 + \sin A} + \tan A \)
Rewrite \( \tan A \) as \( \frac{\sin A}{\cos A} \):
\( \text{L.H.S.} = \frac{\cos A}{1 + \sin A} + \frac{\sin A}{\cos A} \)
Taking LCM \( \cos A(1 + \sin A) \):
\( = \frac{\cos^{2} A + \sin A(1 + \sin A)}{\cos A(1 + \sin A)} \)
\( = \frac{\cos^{2} A + \sin A + \sin^{2} A}{\cos A(1 + \sin A)} \)
Since \( \sin^{2} A + \cos^{2} A = 1 \):
\( = \frac{1 + \sin A}{\cos A(1 + \sin A)} \)
Canceling \( (1 + \sin A) \) from numerator and denominator:
\( = \frac{1}{\cos A} = \sec A = \text{R.H.S.} \).
Hence proved.

Teacher's Note:
a) Convert all trigonometric ratios into sine and cosine terms to simplify complex identities.
b) Use fundamental Pythagorean identities like \( \sin^{2}A + \cos^{2}A = 1 \).

(b) Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a \( \triangle ABC \) in which \( BC = 6.5 \) cm, \( \angle ABC = 60^{\circ} \), \( AB = 5 \) cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are a distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY. [4 Marks]

[Figure: Triangle ABC with BC=6.5cm, \( \angle B=60^{\circ} \), AB=5cm, angle bisector of C intersecting circle centered at A with radius 3.5cm at points X and Y]

Answer:
Steps of construction:
1. Draw line segment \( BC = 6.5 \) cm. At vertex B, construct an angle of \( 60^{\circ} \).
2. With centre B and radius 5 cm, cut an arc on the \( 60^{\circ} \) ray to mark vertex A. Join AC to complete \( \triangle ABC \).
3. The locus of points at a distance of \( 3.5 \) cm from A is a circle with centre A and radius \( 3.5 \) cm.
4. The locus of points equidistant from AC and BC is the angle bisector of \( \angle ACB \).
5. Mark the intersection points of the circle and the angle bisector as X and Y. Measured length \( XY = 5 \) cm.

Teacher's Note:
a) Locus problems require combining basic locus theorems (circle for fixed distance from a point, angle bisector for equidistant lines).
b) Ensure all construction arcs and intersection points are clearly labelled.

(c) Ashok invested Rs. 26,400 on 12%, Rs. 25 shares of a company. If he receives a dividend of Rs. 2,475. Find the:
(i) number of shares he bought
(ii) Market value of each share [3 Marks]

Answer:
(i) Total dividend = Rs. 2475.
Dividend on 1 share = \( 12\% \text{ of Rs. } 25 = \frac{12}{100} \times 25 = \text{Rs. } 3 \).
Number of shares bought = \( \frac{\text{Total Dividend}}{\text{Dividend on 1 share}} = \frac{2475}{3} = 825 \) shares.

(ii) Total Investment = Rs. 26,400.
Market value of each share = \( \frac{\text{Total Investment}}{\text{Number of shares bought}} = \frac{26400}{825} = \text{Rs. } 32 \).

Teacher's Note:
a) Dividend is always calculated as a percentage of the nominal (face) value of a share.
b) Market value per share is obtained by dividing total investment by the total number of shares purchased.