ICSE Class 10 Mathematics Board Exam Question Paper 2013 with Solutions

ICSE Class 10 Mathematics Board Exam Question Paper with Solutions

SECTION A [40 Marks]

Question 1.
(a) Given \( A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \), \( C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \). Find the matrix \( X \) such that \( A + 2X = 2B + C \). [3 Marks]

Answer:
Given \( A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \), \( C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \)
\( A + 2X = 2B + C \)
\( 2X = 2B + C - A \)
\( 2X = 2\begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \)
\( 2X = \begin{bmatrix} -6 & 4 \\ 8 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \)
\( 2X = \begin{bmatrix} -6 + 4 - 2 & 4 + 0 - (-6) \\ 8 + 0 - 2 & 0 + 2 - 0 \end{bmatrix} \)
\( 2X = \begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix} \)
\( X = \frac{1}{2}\begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix} \)
\( X = \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix} \)

Teacher's Note:
a) Always isolate the unknown matrix before substituting the given matrices to avoid arithmetic errors.
b) Pay careful attention to signs when subtracting negative matrix elements.

(b) At what rate % p.a. will a sum of Rs. 4000 yield Rs. 1324 as compound interest in 3 years? [3 Marks]

Answer:
Given Principal (\( P \)) = Rs. 4000, Compound Interest (\( C.I. \)) = Rs. 1324, Time (\( n \)) = 3 years.
Amount (\( A \)) = \( P + C.I. = 4000 + 1324 = \) Rs. 5324.
We know that, \( A = P\left(1 + \frac{r}{100}\right)^{n} \)
\( 5324 = 4000\left(1 + \frac{r}{100}\right)^{3} \)
\( \frac{5324}{4000} = \left(1 + \frac{r}{100}\right)^{3} \)
\( \frac{1331}{1000} = \left(1 + \frac{r}{100}\right)^{3} \)
\( \left(\frac{11}{10}\right)^{3} = \left(1 + \frac{r}{100}\right)^{3} \)
Taking cube root on both sides:
\( \frac{11}{10} = 1 + \frac{r}{100} \)
\( \frac{r}{100} = \frac{11}{10} - 1 = \frac{1}{10} \)
\( r = \frac{100}{10} = 10\% \text{ p.a.} \)

Teacher's Note:
a) Always calculate the total Amount first by adding Principal and Compound Interest.
b) Look for common factors to reduce fractions to perfect cubes or squares matching the given time period.

(c) The median of the following observations \( 11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47 \) arranged in ascending order is 24. Find the value of \( x \) and hence find the mean. [4 Marks]

Answer:
Given observations: \( 11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47 \)
Number of terms \( n = 9 \) (odd).
Median = Value of \( \left(\frac{n + 1}{2}\right)^{\text{th}} \) term = Value of \( \left(\frac{9 + 1}{2}\right)^{\text{th}} = 5^{\text{th}} \) term.
Given median = 24.
Therefore, \( 5^{\text{th}} \text{ term} = x + 4 = 24 \)
\( x = 24 - 4 = 20 \)
Substituting \( x = 20 \), the observations are: \( 11, 12, 14, 18, 24, 29, 32, 38, 47 \).
Mean \( = \frac{\sum x}{n} = \frac{11 + 12 + 14 + 18 + 24 + 29 + 32 + 38 + 47}{9} = \frac{225}{9} = 25 \).

Teacher's Note:
a) Verify that the given observations are strictly in ascending order before applying the median formula.
b) Double-check the summation of terms when calculating the mean.

Question 2.
(a) What number must be added to each of the numbers \( 6, 15, 20 \) and \( 43 \) to make them proportional? [3 Marks]

Answer:
Let the number to be added be \( x \).
The new numbers are \( 6 + x, 15 + x, 20 + x, 43 + x \).
Since they are proportional, \( (6 + x) : (15 + x) :: (20 + x) : (43 + x) \)
\( (6 + x)(43 + x) = (15 + x)(20 + x) \)
\( 258 + 6x + 43x + x^{2} = 300 + 20x + 15x + x^{2} \)
\( 258 + 49x = 300 + 35x \)
\( 49x - 35x = 300 - 258 \)
\( 14x = 42 \)
\( x = 3 \)

Teacher's Note:
a) Use the property of proportion \( a : b :: c : d \implies ad = bc \).
b) The quadratic terms \( x^{2} \) cancel out on both sides, simplifying the equation into a linear one.

(b) If \( (x - 2) \) is a factor of the expression \( 2x^{3} + ax^{2} + bx - 14 \) and when the expression is divided by \( (x - 3) \), it leaves a remainder \( 52 \), find the values of \( a \) and \( b \). [3 Marks]

Answer:
Let \( f(x) = 2x^{3} + ax^{2} + bx - 14 \).
Since \( (x - 2) \) is a factor, \( f(2) = 0 \):
\( 2(2)^{3} + a(2)^{2} + b(2) - 14 = 0 \)
\( 16 + 4a + 2b - 14 = 0 \)
\( 4a + 2b + 2 = 0 \implies 2a + b = -1 \quad \text{--- (i)} \)
When divided by \( (x - 3) \), the remainder is \( 52 \), so \( f(3) = 52 \):
\( 2(3)^{3} + a(3)^{2} + b(3) - 14 = 52 \)
\( 54 + 9a + 3b - 14 = 52 \)
\( 40 + 9a + 3b = 52 \implies 9a + 3b = 12 \implies 3a + b = 4 \quad \text{--- (ii)} \)
Subtracting equation (i) from equation (ii):
\( (3a + b) - (2a + b) = 4 - (-1) \)
\( a = 5 \)
Substitute \( a = 5 \) in equation (i):
\( 2(5) + b = -1 \implies 10 + b = -1 \implies b = -11 \)
Thus, \( a = 5 \) and \( b = -11 \).

Teacher's Note:
a) Apply the Factor Theorem \( f(a) = 0 \) when \( (x - a) \) is a factor.
b) Apply the Remainder Theorem \( f(a) = \text{Remainder} \) when divided by \( (x - a) \).

(c) Draw a histogram from the following frequency distribution and find the mode from the graph: [4 Marks]

Answer:

[Figure: Histogram plotted with Class intervals on the X-axis and Frequency on the Y-axis. The highest rectangle is for the class 10-15 with frequency 18. Lines drawn diagonally from the top corners of the highest rectangle to the adjacent rectangles intersect to give the mode at 13.8.]

Mode = \( 13.8 \)

Teacher's Note:
a) Ensure correct scale selection on both axes when drawing histograms.
b) Mode is located graphically by drawing intersecting lines from the top vertices of the highest rectangle to its adjacent rectangles.

Question 3.
(a) Without using tables evaluate \( 3\cos 80^{\circ}\csc 10^{\circ} + 2\sin 59^{\circ}\sec 31^{\circ} \). [3 Marks]

Answer:
\( 3\cos 80^{\circ}\csc 10^{\circ} + 2\sin 59^{\circ}\sec 31^{\circ} \)
\( = 3\cos 80^{\circ}\csc(90^{\circ} - 80^{\circ}) + 2\sin 59^{\circ}\sec(90^{\circ} - 59^{\circ}) \)
\( = 3\cos 80^{\circ}\sec 80^{\circ} + 2\sin 59^{\circ}\csc 59^{\circ} \)
\( = 3\cos 80^{\circ} \times \frac{1}{\cos 80^{\circ}} + 2\sin 59^{\circ} \times \frac{1}{\sin 59^{\circ}} \)
\( = 3 + 2 = 5 \)

Teacher's Note:
a) Use complementary angle relations: \( \csc(90^{\circ} - \theta) = \sec\theta \) and \( \sec(90^{\circ} - \theta) = \csc\theta \).
b) Remember that reciprocal trigonometric ratios like \( \cos\theta \sec\theta = 1 \).

(b) In the given figure, \( \angle BAD = 65^{\circ} \), \( \angle ABD = 70^{\circ} \), \( \angle BDC = 45^{\circ} \).
(i) Prove that \( AC \) is a diameter of the circle.
(ii) Find \( \angle ACB \). [3 Marks]

[Figure: Cyclic quadrilateral ABCD inscribed in a circle. Angle BAD = 65 degrees, ABD = 70 degrees, BDC = 45 degrees, AC is a diagonal.]

Answer:
(i) In \( \Delta ABD \):
\( \angle BDA + \angle DAB + \angle ABD = 180^{\circ} \) (Angle sum property of a triangle)
\( \angle BDA = 180^{\circ} - (65^{\circ} + 70^{\circ}) = 180^{\circ} - 135^{\circ} = 45^{\circ} \)
Now, in \( \Delta ACD \):
\( \angle ADC = \angle ADB + \angle BDC = 45^{\circ} + 45^{\circ} = 90^{\circ} \)
Since the angle in a semi-circle is a right angle, \( AC \) must be the diameter of the circle.
(ii) \( \angle ACB = \angle ADB \) (Angles in the same segment of a circle)
Therefore, \( \angle ACB = 45^{\circ} \).

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

(c) \( AB \) is a diameter of a circle with centre \( C = (-2, 5) \). If \( A = (3, -7) \). Find:
(i) The length of radius \( AC \)
(ii) The coordinates of \( B \). [4 Marks]

Answer:
(i) Radius \( AC = \sqrt{(-2 - 3)^{2} + (5 - (-7))^{2}} = \sqrt{(-5)^{2} + (12)^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ units} \).
(ii) Let the coordinates of \( B \) be \( (x, y) \).
Since \( C(-2, 5) \) is the mid-point of \( AB \), using the mid-point formula:
\( -2 = \frac{3 + x}{2} \implies 3 + x = -4 \implies x = -7 \)
\( 5 = \frac{-7 + y}{2} \implies -7 + y = 10 \implies y = 17 \)
Therefore, the coordinates of \( B \) are \( (-7, 17) \).

Teacher's Note:
a) Use the distance formula to find the length between two given coordinates.
b) Use the mid-point formula when the centre is given along with one endpoint of a diameter.

Question 4.
(a) Solve the following equation and calculate the answer correct to two decimal places: \( x^{2} - 5x - 10 = 0 \). [3 Marks]

Answer:
Given equation: \( x^{2} - 5x - 10 = 0 \)
Here, \( a = 1, b = -5, c = -10 \).
Discriminant \( D = b^{2} - 4ac = (-5)^{2} - 4(1)(-10) = 25 + 40 = 65 \).
Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-5) \pm \sqrt{65}}{2(1)} = \frac{5 \pm 8.062}{2} \)
\( x_{1} = \frac{5 + 8.062}{2} = \frac{13.062}{2} = 6.53 \)
\( x_{2} = \frac{5 - 8.062}{2} = \frac{-3.062}{2} = -1.53 \)
Thus, \( x = 6.53 \) or \( x = -15.3 \) (Correct: \( x = 6.53, -1.53 \)).

Teacher's Note:
a) Always calculate the discriminant first to check the nature of the roots.
b) Round off the final answers correctly to two decimal places as requested.

(b) In the given figure, \( AB \) and \( DE \) are perpendicular to \( BC \).
(i) Prove that \( \Delta ABC \sim \Delta DEC \)
(ii) If \( AB = 6\text{ cm}, DE = 4\text{ cm} \) and \( AC = 15\text{ cm} \). Calculate \( CD \).
(iii) Find the ratio of the area of \( \Delta ABC \) : area of \( \Delta DEC \). [3 Marks]

[Figure: Right-angled triangles ABC and DEC sharing a common angle C at the base BC. AB and DE are perpendiculars.]

Answer:
(i) In \( \Delta ABC \) and \( \Delta DEC \):
\( \angle ABC = \angle DEC = 90^{\circ} \) (Given)
\( \angle ACB = \angle DCE \) (Common angle)
Therefore, \( \Delta ABC \sim \Delta DEC \) (By AA similarity criterion).
(ii) Since \( \Delta ABC \sim \Delta DEC \), corresponding sides are proportional:
\( \frac{AB}{DE} = \frac{AC}{CD} \)
\( \frac{6}{4} = \frac{15}{CD} \)
\( 6 \times CD = 15 \times 4 \implies CD = \frac{60}{6} = 10\text{ cm} \).
(iii) The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides:
\( \frac{\text{Area of } \Delta ABC}{\text{Area of } \Delta DEC} = \left(\frac{AB}{DE}\right)^{2} = \left(\frac{6}{4}\right)^{2} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4} = 9 : 4 \).

Teacher's Note:
a) State the similarity criterion clearly (AA similarity) with equal angles.
b) Remember that the ratio of areas of similar triangles equals the square of the ratio of any pair of corresponding sides.

(c) Using graph paper, plot the points \( A(6, 4) \) and \( B(0, 4) \).
(i) Reflect \( A \) and \( B \) in the origin to get the images \( A^{\prime} \) and \( B^{\prime} \).
(ii) Write the co-ordinates of \( A^{\prime} \) and \( B^{\prime} \).
(iii) State the geometrical name for the figure \( ABA^{\prime}B^{\prime} \).
(iv) Find its perimeter. [4 Marks]

Answer:
(i) & (ii) Reflection in the origin changes \( (x, y) \) to \( (-x, -y) \):
Coordinates of \( A^{\prime} = (-6, -4) \)
Coordinates of \( B^{\prime} = (0, -4) \)
(iii) The geometrical name for the figure \( ABA^{\prime}B^{\prime} \) is a parallelogram.
(iv) Length \( AB = 6 - 0 = 6\text{ units} \).
Length \( BB^{\prime} = 4 - (-4) = 8\text{ units} \).
In right-angled triangle formed by components, \( (AB^{\prime})^{2} = AB^{2} + (BB^{\prime})^{2} = 6^{2} + 8^{2} = 36 + 64 = 100 \implies AB^{\prime} = 10\text{ units} \).
Perimeter of \( ABA^{\prime}B^{\prime} = AB + BA^{\prime} + A^{\prime}B^{\prime} + B^{\prime}A = 6 + 10 + 6 + 10 = 32\text{ units} \).

Teacher's Note:
a) Reflection in the origin negates both coordinates: \( (x, y) \to (-x, -y) \).
b) Perimeter is the sum of all outer boundaries of the closed geometric shape.

SECTION B [40 Marks]
Answer any four Questions in this Section.

Question 5.
(a) Solve the following inequation, write the solution set and represent it on the number line: \( -\frac{3}{5} \le \frac{x}{2} - 1\frac{1}{3} \lt \frac{1}{6}, x \in R \). [3 Marks]

Answer:
Given inequation: \( -\frac{3}{5} \le \frac{x}{2} - \frac{4}{3} \lt \frac{1}{6} \)
Multiplying each term by LCM of 2, 3, and 6 which is 6:
\( -\frac{3}{5} \times 6 \le 3x - 8 \lt 1 \)
\( -\frac{18}{5} \le 3x - 8 \lt 1 \implies -3.6 \le 3x - 8 \lt 1 \)
Splitting into two inequalities:
1) \( -3.6 \le 3x - 8 \implies 4.4 \le 3x \implies x \ge 1.47 \)
2) \( 3x - 8 \lt 1 \implies 3x \lt 9 \implies x \lt 3 \)
Combined solution set: \( \{x : 1.47 \le x \lt 3, x \in R\} \).

Teacher's Note:
a) Clear fractions by multiplying throughout by the LCM of the denominators.
b) Separate joint inequalities into two parts and solve them simultaneously.

(b) Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is \( 8\% \) per annum and Mr. Britto gets Rs. 8088 from the bank after 3 years, find the value of his monthly instalment. [3 Marks]

Answer:
Let the monthly instalment be \( P \).
Time \( n = 3 \text{ years} = 36 \text{ months} \), Rate \( R = 8\% \) p.a.
Principal for one month = \( P \times \frac{n(n+1)}{2} = \frac{P \times 36 \times 37}{2} = 666P \)
Interest \( I = \frac{P \times n(n+1) \times R}{2 \times 12 \times 100} = \frac{666P \times 8}{1200} = \frac{444P}{100} = 4.44P \)
Total amount deposited = \( 36P \)
Maturity Amount = Interest + Amount Deposited
\( 8088 = 4.44P + 36P \)
\( 8088 = 40.44P \)
\( P = \frac{8088}{40.44} = \text{Rs. } 200 \).

Teacher's Note:
a) Use the standard formula for Recurring Deposit Interest: \( I = \frac{P \times n(n+1)R}{2 \times 12 \times 100} \).
b) Maturity amount is the sum of all instalments paid plus the total interest earned.

(c) Salman buys 50 shares of face value Rs. 100 available at Rs. 132.
(i) What is his investment?
(ii) If the dividend is \( 7.5\% \), what will be his annual income?
(iii) If he wants to increase his annual income by Rs. 150, how many extra shares should he buy? [4 Marks]

Answer:
Number of shares = 50, Face Value (FV) = Rs. 100, Market Value (MV) = Rs. 132.
(i) Total Investment = Number of shares \( \times \) Market Value = \( 50 \times 132 = \) Rs. 6600.
(ii) Total Face Value = \( 50 \times 100 = \) Rs. 5000.
Annual Income = \( \frac{\text{Total Face Value} \times \text{Dividend Rate}}{100} = \frac{5000 \times 7.5}{100} = \) Rs. 375.
(iii) Let the number of extra shares to buy be \( x \).
Extra annual income needed = Rs. 150.
Dividend per share = \( \frac{100 \times 7.5}{100} = \) Rs. 7.50.
Number of extra shares \( x = \frac{\text{Required Income}}{\text{Dividend per share}} = \frac{150}{7.50} = 20 \) shares.

Teacher's Note:
a) Investment is always calculated using Market Value, while dividend is always calculated using Face Value.
b) Additional shares required can be found by dividing the desired extra income by the dividend earned per share.

Question 6.
(a) Show that \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} \). [3 Marks]

Answer:
L.H.S. = \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} \)
Multiplying numerator and denominator inside the square root by \( (1 + \cos A) \):
\( = \sqrt{\frac{(1 - \cos A)(1 + \cos A)}{(1 + \cos A)(1 + \cos A)}} \)
\( = \sqrt{\frac{1 - \cos^{2}A}{(1 + \cos A)^{2}}} = \sqrt{\frac{\sin^{2}A}{(1 + \cos A)^{2}}} \)
\( = \frac{\sin A}{1 + \cos A} = \text{R.H.S.} \) (Proved).

Teacher's Note:
a) Rationalize the denominator by multiplying by the conjugate expression.
b) Use the fundamental trigonometric identity \( 1 - \cos^{2}A = \sin^{2}A \).

(b) In the given circle with centre \( O \), \( \angle ABC = 100^{\circ} \), \( \angle ACD = 40^{\circ} \) and \( CT \) is a tangent to the circle at \( C \). Find \( \angle ADC \) and \( \angle DCT \). [3 Marks]

[Figure: Cyclic quadrilateral ABCD inside a circle with centre O. Tangent CT at C. Angle ABC = 100 degrees, ACD = 40 degrees.]

Answer:
Since \( ABCD \) is a cyclic quadrilateral, opposite angles sum to \( 180^{\circ} \):
\( \angle ABC + \angle ADC = 180^{\circ} \)
\( 100^{\circ} + \angle ADC = 180^{\circ} \implies \angle ADC = 80^{\circ} \)
Join \( OA \) and \( OC \). Triangle \( OAC \) is isosceles with \( OA = OC \) (radii).
Angle subtended at the centre is twice the angle subtended at the circumference:
\( \angle AOC = 2 \times \angle ADC = 2 \times 80^{\circ} = 160^{\circ} \)
In \( \Delta OAC \): \( \angle OAC + \angle OCA + \angle AOC = 180^{\circ} \implies 2\angle OCA = 180^{\circ} - 160^{\circ} = 20^{\circ} \implies \angle OCA = 10^{\circ} \)
Given \( \angle ACD = 40^{\circ} \), so \( \angle OCD = \angle OCA + \angle ACD = 10^{\circ} + 40^{\circ} = 50^{\circ} \) (Correction: \( \angle OCD = \angle ACD - \angle OCA = 40^{\circ} - 10^{\circ} = 30^{\circ} \)).
Radius is perpendicular to the tangent at the point of contact, so \( \angle OCT = 90^{\circ} \).
Therefore, \( \angle DCT = \angle OCT - \angle OCD = 90^{\circ} - 30^{\circ} = 60^{\circ} \).

Teacher's Note:
a) Opposite angles of a cyclic quadrilateral are supplementary.
b) The radius is always perpendicular to the tangent at the point of contact.

(c) Given below are the entries in a Savings Bank A/c pass book: [4 Marks]

Calculate the interest for six months from February to July at \( 6\% \) p.a.

Answer:
Principal amounts for each month from Feb to July (taking the lowest balance between the 10th and the end of each month):
February = Rs. 4,500
March = Rs. 4,500
April = Rs. 4,500
May = Rs. 6,730
June = Rs. 1,730
July = Rs. 7,730
Total principal for one month = \( 4500 + 4500 + 4500 + 6730 + 1730 + 7730 = \) Rs. 29,690.
Time \( = \frac{1}{12} \text{ year} \), Rate \( = 6\% \)
Interest \( = \frac{P \times R \times T}{100} = \frac{29690 \times 6 \times 1}{100 \times 12} = \) Rs. 148.45.

Teacher's Note:
a) For savings bank accounts, the principal for any month is the minimum balance between the 10th and the last day of the month.
b) Time \( T \) is always taken as \( \frac{1}{12} \) year when computing monthly accumulated interest.

Question 7.
(a) In \( \Delta ABC \), \( A(3, 5) \), \( B(7, 8) \) and \( C(1, -10) \). Find the equation of the median through \( A \). [3 Marks]

Answer:
Let \( D \) be the mid-point of \( BC \).
Coordinates of \( D = \left(\frac{7 + 1}{2}, \frac{8 + (-10)}{2}\right) = (4, -1) \).
Equation of the median \( AD \) passing through \( A(3, 5) \) and \( D(4, -1) \):
\( y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}(x - x_{1}) \)
\( y - 5 = \frac{-1 - 5}{4 - 3}(x - 3) \)
\( y - 5 = \frac{-6}{1}(x - 3) \)
\( y - 5 = -6x + 18 \)
\( 6x + y - 23 = 0 \).

Teacher's Note:
a) A median joins a vertex to the mid-point of the opposite side.
b) Use the two-point form to find the equation of a straight line.

(b) A shopkeeper sells an article at the listed price of Rs. 1,500 and the rate of VAT is \( 12\% \) at each stage of sale. If the shopkeeper pays a VAT of Rs. 36 to the Government, what was the price, inclusive of tax, at which the shopkeeper purchased the article from the wholesaler? [3 Marks]

Answer:
Listed price = Rs. 1,500, Rate of VAT = \( 12\% \)
Output tax (tax collected by shopkeeper) = \( \frac{12}{100} \times 1500 = \) Rs. 180.
Let the cost price (C.P.) for the shopkeeper be \( x \).
Input tax (tax paid by shopkeeper) = \( \frac{12x}{100} \)
VAT paid = Output Tax - Input Tax
\( 180 - \frac{12x}{100} = 36 \)
\( \frac{12x}{100} = 180 - 36 = 144 \)
\( 12x = 14400 \implies x = 1200 \)
Price inclusive of tax paid by the shopkeeper = \( 1200 + \frac{12}{100} \times 1200 = 1200 + 144 = \) Rs. 1,344.

Teacher's Note:
a) VAT paid to the Government equals Output Tax minus Input Tax.
b) Inclusive price is the sum of the basic cost price and the applicable input tax.

(c) In the figure given, from the top of a building \( AB = 60\text{ m} \) high, the angles of depression of the top and bottom of a vertical lamp post \( CD \) are observed to be \( 30^{\circ} \) and \( 60^{\circ} \) respectively. Find:
(i) The horizontal distance between \( AB \) and \( CD \).
(ii) The height of the lamp post. [4 Marks]

[Figure: Building AB (60 m high) and lamp post CD. Angles of depression from top of AB to top of CD is 30 degrees and to bottom of CD is 60 degrees.]

Answer:
(i) In \( \Delta ABC \):
\( \tan 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{60}{BC} \implies BC = \frac{60}{\sqrt{3}} = 20\sqrt{3}\text{ m} \approx 34.64\text{ m} \).
(ii) Let height of lamp post \( CD = h \). Horizontal distance between them \( ED = BC = 20\sqrt{3}\text{ m} \).
In \( \Delta AED \):
\( \tan 30^{\circ} = \frac{AE}{ED} \implies \frac{1}{\sqrt{3}} = \frac{AE}{20\sqrt{3}} \implies AE = 20\text{ m} \).
Height of lamp post \( CD = BE = AB - AE = 60 - 20 = 40\text{ m} \).

Teacher's Note:
a) Angle of depression equals the corresponding angle of elevation.
b) Break complex height and distance figures into distinct right-angled triangles.

Question 8.
(a) Find \( x \) and \( y \) if \( \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \). [3 Marks]

Answer:
Matrix multiplication:
\( \begin{bmatrix} 2x + 3x \\ 2y + 4y \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \)
\( \begin{bmatrix} 5x \\ 6y \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \)
Comparing corresponding elements:
\( 5x = 5 \implies x = 1 \)
\( 6y = 12 \implies y = 2 \)

Teacher's Note:
a) Multiply rows by columns following matrix multiplication rules.
b) Equate corresponding elements of equal matrices to find unknowns.

(b) A solid sphere of radius \( 15\text{ cm} \) is melted and recast into solid right circular cones of radius \( 2.5\text{ cm} \) and height \( 8\text{ cm} \). Calculate the number of cones recast. [3 Marks]

Answer:
Radius of sphere \( R = 15\text{ cm} \).
Volume of sphere \( = \frac{4}{3}\pi R^{3} = \frac{4}{3}\pi (15)^{3} \).
Radius of cone \( r = 2.5\text{ cm} \), height \( h = 8\text{ cm} \).
Volume of one cone \( = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi (2.5)^{2}(8) \).
Number of cones \( = \frac{\text{Volume of sphere}}{\text{Volume of one cone}} = \frac{\frac{4}{3}\pi (15)^{3}}{\frac{1}{3}\pi (2.5)^{2}(8)} = \frac{4 \times 15 \times 15 \times 15}{2.5 \times 2.5 \times 8} = 270 \).
Number of cones = 270.

Teacher's Note:
a) Volume remains constant when a solid shape is melted and recast into other shapes.
b) Divide the volume of the original large body by the volume of a single new smaller body.

(c) Without solving the following quadratic equation, find the value of \( p \) for which the given equation has real and equal roots: \( x^{2} + (p - 3)x + p = 0 \). [4 Marks]

Answer:
Given equation: \( x^{2} + (p - 3)x + p = 0 \)
Here, \( a = 1, b = p - 3, c = p \).
For real and equal roots, Discriminant \( D = 0 \):
\( b^{2} - 4ac = 0 \)
\( (p - 3)^{2} - 4(1)(p) = 0 \)
\( p^{2} - 6p + 9 - 4p = 0 \)
\( p^{2} - 10p + 9 = 0 \)
\( (p - 9)(p - 1) = 0 \)
Therefore, \( p = 9 \) or \( p = 1 \).

Teacher's Note:
a) Real and equal roots occur if and only if the discriminant \( b^{2} - 4ac = 0 \).
b) Solve the resulting quadratic equation in terms of \( p \) by factoring.

Question 9.
(a) In the figure alongside, \( OAB \) is a quadrant of a circle. The radius \( OA = 3.5\text{ cm} \) and \( OD = 2\text{ cm} \). Calculate the area of the shaded portion. (Take \( \pi = \frac{22}{7} \)) [3 Marks]

[Figure: Quadrant OAB of a circle with radius 3.5 cm. Triangle OBD inside with OD = 2 cm.]

Answer:
Radius \( r = 3.5\text{ cm} \).
Area of quadrant \( OAB = \frac{1}{4}\pi r^{2} = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = 9.625\text{ cm}^{2} \).
Area of triangle \( AOD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.5 \times 2 = 3.5\text{ cm}^{2} \).
Area of shaded portion = Area of quadrant - Area of triangle
\( = 9.625 - 3.5 = 6.125\text{ cm}^{2} \).

Teacher's Note:
a) A quadrant is one-fourth of a full circle with a central angle of \( 90^{\circ} \).
b) Subtract the area of the unshaded triangle from the sector/quadrant area to get the shaded region.

(b) A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box. [3 Marks]

Answer:
Let the number of black balls be \( x \).
Number of white balls = 30.
Total number of balls = \( 30 + x \).
Probability of drawing a black ball \( P(B) = \frac{x}{30 + x} \).
Probability of drawing a white ball \( P(W) = \frac{30}{30 + x} \).
Given: \( P(B) = \frac{2}{5} \times P(W) \)
\( \frac{x}{30 + x} = \frac{2}{5} \times \frac{30}{30 + x} \)
Multiplying both sides by \( (30 + x) \):
\( x = \frac{60}{5} = 12 \)
Number of black balls = 12.

Teacher's Note:
a) Probability is defined as the ratio of favorable outcomes to total outcomes.
b) Set up an algebraic equation based on the given ratio of probabilities.

(c) Find the mean of the following distribution by step deviation method: [4 Marks]

Answer:
Let assumed mean \( a = 45 \), class size \( h = 10 \).
Mid-values \( (x_{i}) \) for intervals: 25, 35, 45, 55, 65, 75.
Deviations \( d_{i} = \frac{x_{i} - a}{h} \): -2, -1, 0, 1, 2, 3.
Products \( f_{i}d_{i} \):
\( 10 \times (-2) = -20 \)
\( 6 \times (-1) = -6 \)
\( 8 \times 0 = 0 \)
\( 12 \times 1 = 12 \)
\( 5 \times 2 = 10 \)
\( 9 \times 3 = 27 \)
Sum of frequencies \( \sum f_{i} = 50 \).
Sum of products \( \sum f_{i}d_{i} = -20 - 6 + 0 + 12 + 10 + 27 = 23 \).
Mean \( = a + \left(\frac{\sum f_{i}d_{i}}{\sum f_{i}}\right) \times h = 45 + \left(\frac{23}{50}\right) \times 10 = 45 + 4.6 = 49.6 \).

Teacher's Note:
a) The step deviation formula is \( \bar{x} = a + \left(\frac{\sum f_{i}d_{i}}{\sum f_{i}}\right) \times h \).
b) Choose the assumed mean near the middle of the distribution for simpler calculations.

Question 10.
(a) Using a ruler and compasses only:
(i) Construct a triangle ABC with the following data: \( AB = 3.5\text{ cm}, BC = 6\text{ cm} \) and \( \angle ABC = 120^{\circ} \).
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure \( \angle BCP \). [3 Marks]

Answer:
(i) Draw line segment \( BC = 6\text{ cm} \). Construct an angle of \( 120^{\circ} \) at point \( B \). Cut an arc of radius \( 3.5\text{ cm} \) on the arm to locate point \( A \). Join \( A \) to \( C \).
(ii) Bisect \( BC \) to get its midpoint \( O \). Draw a circle with centre \( O \) and radius equal to half of \( BC \). Construct the angle bisector of \( \angle ABC \) which intersects the circle at point \( P \).
(iii) Measure \( \angle BCP = 30^{\circ} \).

Teacher's Note:
a) Use standard compass constructions for angles like \( 120^{\circ} \) and angle bisectors.
b) An angle inscribed in a semi-circle is a right angle.

(b) The marks obtained by 120 students in a test are given below: [6 Marks]

Draw an ogive for the given distribution on a graph sheet.
Using suitable scale for ogive to estimate the following:
(i) The median.
(ii) The number of students who obtained more than \( 75\% \) marks in the test.
(iii) The number of students who did not pass the test if minimum marks required to pass is 40.

Answer:
Cumulative frequencies for the upper limits (10, 20, 30, 40, 50, 60, 70, 80, 90, 100):
5, 14, 30, 52, 78, 96, 107, 113, 117, 120.
(i) Median \( = \left(\frac{n}{2}\right)^{\text{th}} \text{ term} = \left(\frac{120}{2}\right)^{\text{th}} = 60^{\text{th}} \text{ term} \). From graph, Median = 42.
(ii) Number of students with more than \( 75\% \) marks (i.e. 75 marks): corresponding cumulative frequency from graph is approximately 110, so students \( = 120 - 110 = 10 \).
(iii) Number of students who did not pass (scoring less than 40): corresponding cumulative frequency at 40 marks is 52.

Teacher's Note:
a) Plot cumulative frequencies against upper class limits to draw a cumulative frequency curve (ogive).
b) Read values from the graph carefully using perpendicular dropped lines from the axes.

Question 11.
(a) In the figure given below, the line segment \( AB \) meets X-axis at \( A \) and Y-axis at \( B \). The point \( P(-3, 4) \) on \( AB \) divides it in the ratio \( 2 : 3 \). Find the coordinates of \( A \) and \( B \). [3 Marks]

[Figure: Line segment AB intersecting X-axis at A(x, 0) and Y-axis at B(0, y). Point P(-3, 4) divides AB in ratio 2:3.]

Answer:
Let \( A = (x, 0) \) and \( B = (0, y) \).
Point \( P(-3, 4) \) divides \( AB \) in the ratio \( 2 : 3 \).
Using the section formula:
\( -3 = \frac{2(0) + 3(x)}{2 + 3} \implies -3 = \frac{3x}{5} \implies 3x = -15 \implies x = -5 \)
\( 4 = \frac{2(y) + 3(0)}{2 + 3} \implies 4 = \frac{2y}{5} \implies 2y = 20 \implies y = 10 \)
Therefore, coordinates of \( A \) are \( (-5, 0) \) and \( B \) are \( (0, 10) \).

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) Apply the section formula separately for X and Y coordinates.

(b) Using the properties of proportion, solve for \( x \), given: \( \frac{x^{4} + 1}{2x^{2}} = \frac{17}{8} \). [3 Marks]

Answer:
Given equation: \( \frac{x^{4} + 1}{2x^{2}} = \frac{17}{8} \)
Using componendo and dividendo:
\( \frac{x^{4} + 1 + 2x^{2}}{x^{4} + 1 - 2x^{2}} = \frac{17 + 8}{17 - 8} \)
\( \frac{(x^{2} + 1)^{2}}{(x^{2} - 1)^{2}} = \frac{25}{9} \)
Taking square root on both sides:
\( \frac{x^{2} + 1}{x^{2} - 1} = \frac{5}{3} \)
Applying componendo and dividendo again:
\( \frac{x^{2} + 1 + x^{2} - 1}{x^{2} + 1 - (x^{2} - 1)} = \frac{5 + 3}{5 - 3} \)
\( \frac{2x^{2}}{2} = \frac{8}{2} \)
\( x^{2} = 4 \implies x = \pm 2 \).

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) Repeated application of this rule significantly simplifies higher-degree equations.

(c) A shopkeeper purchases a certain number of books for Rs. 960. If the cost per book was Rs. 8 less, the number of books that could be purchased for Rs. 960 would be 4 more. Write an equation, taking the original cost of each book to be \( x \), and solve it to find the original cost of the books. [4 Marks]

Answer:
Let the original cost of each book be Rs. \( x \).
Total amount = Rs. 960.
Number of books purchased initially = \( \frac{960}{x} \)
If cost per book is Rs. \( (x - 8) \), number of books purchased = \( \frac{960}{x - 8} \)
According to the question:
\( \frac{960}{x - 8} - \frac{960}{x} = 4 \)
\( 960\left(\frac{x - (x - 8)}{x(x - 8)}\right) = 4 \)
\( \frac{960 \times 8}{x^{2} - 8x} = 4 \)
\( \frac{7680}{x^{2} - 8x} = 4 \implies x^{2} - 8x = 1920 \)
\( x^{2} - 8x - 1920 = 0 \)
\( (x - 48)(x + 40) = 0 \)
\( x = 48 \) or \( x = -40 \) (Since cost cannot be negative, \( x = 48 \)).
Original cost of each book = Rs. 48.

Teacher's Note:
a) Formulate algebraic equations by equating the difference in quantities to the given surplus.
b) Reject negative values for physical quantities like price or cost.