ICSE Class 10 Mathematics Board Exam Question Paper 2020 with Solutions

ICSE Class 10 Mathematics Board Exam Question Paper with Solutions

SECTION A

Question 1

(a) Solve the following Quadratic Equation:
\( x^{2} - 7x + 3 = 0 \)
Give your answer correct to two decimal places. [3 Marks]

Answer:
Comparing the given equation \( x^{2} - 7x + 3 = 0 \) with standard quadratic equation \( ax^{2} + bx + c = 0 \), we get:
\( a = 1, b = -7, c = 3 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \):
\( x = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4(1)(3)}}{2(1)} \)
\( x = \frac{7 \pm \sqrt{49 - 12}}{2} \)
\( x = \frac{7 \pm \sqrt{37}}{2} \)
Since \( \sqrt{37} \approx 6.0828 \):
\( x = \frac{7 + 6.0828}{2} = \frac{13.0828}{2} = 6.5414 \approx 6.54 \)
\( x = \frac{7 - 6.0828}{2} = \frac{0.9172}{2} = 0.4586 \approx 0.46 \)
Thus, the solutions are \( x = 6.54 \) and \( x = 0.46 \).

Teacher's Note:
a) Always write the quadratic formula clearly before substituting the values to ensure method marks are secured.
b) Pay close attention to rounding off; two decimal places means calculating up to three decimal places before final rounding.

(b) Given \( A = \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} \)
If \( A^{2} = 3I \), where I is the identity matrix of order 2, find x and y. [3 Marks]

Answer:
Given matrix \( A = \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} \)
First, find \( A^{2} = A \times A \):
\( A^{2} = \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} = \begin{pmatrix} x^{2} + 3y & 3x + 9 \\ xy + 3y & 3y + 9 \end{pmatrix} \)
Given \( A^{2} = 3I \), where \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \):
\( 3I = 3 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \)
Equating the corresponding elements of \( A^{2} \) and \( 3I \):
1) \( 3x + 9 = 0 \implies 3x = -9 \implies x = -3 \)
2) \( 3y + 9 = 3 \implies 3y = -6 \implies y = -2 \)
Checking with other elements: \( x^{2} + 3y = (-3)^{2} + 3(-2) = 9 - 6 = 3 \) (matches), and \( xy + 3y = (-3)(-2) + 3(-2) = 6 - 6 = 0 \) (matches).
Thus, \( x = -3 \) and \( y = -2 \).

Teacher's Note:
a) Matrix multiplication is performed row-by-column carefully.
b) Verify the values of x and y obtained from linear equations by substituting them back into the remaining matrix elements.

(c) Using ruler and compass construct a triangle ABC where AB = 3 cm, BC = 4 cm and \( \angle ABC = 90^{\circ} \). Hence construct a circle circumscribing the triangle ABC. Measure and write down the radius of the circle. [4 Marks]

Answer:
1. Draw a line segment \( BC = 4 \) cm.
2. At point B, construct an angle of \( 90^{\circ} \) using a compass.
3. Cut an arc of radius \( 3 \) cm on the perpendicular ray from B and mark it as point A. Join A to C to complete \( \triangle ABC \).
4. Construct perpendicular bisectors of any two sides (say AB and BC).
5. Let the perpendicular bisectors intersect at point O, which is the circumcentre of the right-angled triangle (midpoint of the hypotenuse AC).
6. With O as centre and radius equal to OA (or OB or OC), draw a circumcircle passing through A, B, and C.
7. Measuring the radius AC/2 gives \( 2.5 \) cm.

Teacher's Note:
a) For a right-angled triangle, the circumcentre always lies at the midpoint of the hypotenuse.
b) All construction arcs must be clearly visible; erased or faint arcs may lead to deduction of marks.

Question 2

(a) Use factor theorem to factorise \( 6x^{3} + 17x^{2} + 4x - 12 \) completely. [3 Marks]

Answer:
Let \( f(x) = 6x^{3} + 17x^{2} + 4x - 12 \).
Factors of the constant term \(-12\) are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).
Let us test \( x = -2 \):
\( f(-2) = 6(-2)^{3} + 17(-2)^{2} + 4(-2) - 12 \)
\( f(-2) = 6(-8) + 17(4) - 8 - 12 = -48 + 68 - 8 - 12 = 0 \)
By Factor Theorem, \( (x + 2) \) is a factor of \( f(x) \).
Now, divide \( f(x) \) by \( x + 2 \) using synthetic division or long division:
\( 6x^{3} + 17x^{2} + 4x - 12 = (x + 2)(6x^{2} + 5x - 6) \)
Now, factorise the quadratic expression \( 6x^{2} + 5x - 6 \) by splitting the middle term:
\( 6x^{2} + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (3x - 2)(2x + 3) \)
Therefore, the complete factorisation is \( (x + 2)(2x + 3)(3x - 2) \).

Teacher's Note:
a) Always state the factor theorem explicitly when showing that a linear binomial is a factor.
b) Double-check the signs when grouping terms during quadratic factorisation.

(b) Solve the following inequation and represent the solution set on the number line:
\( \frac{3x}{5} + 2 \lt x + 4 \le \frac{x}{2} + 5, x \in R \) [3 Marks]

Answer:
Split the given compound inequation into two separate parts:
Part 1: \( \frac{3x}{5} + 2 \lt x + 4 \)
\( 2 - 4 \lt x - \frac{3x}{5} \)
\( -2 \lt \frac{2x}{5} \)
\( -10 \lt 2x \implies x \gt -5 \)

Part 2: \( x + 4 \le \frac{x}{2} + 5 \)
\( x - \frac{x}{2} \le 5 - 4 \)
\( \frac{x}{2} \le 1 \implies x \le 2 \)

Combining both parts, the solution set is \( -5 \lt x \le 2, x \in R \).
[Figure: Number line marked from -6 to 3, with an unshaded circle at -5, a solid circle at 2, and a thick line joining them representing the range.]

Teacher's Note:
a) Always solve compound inequalities by splitting them into two simultaneous linear inequations.
b) Use an unshaded (hollow) circle for strict inequalities (\(\lt\) or \(\gt\)) and a shaded (solid) circle for inclusive inequalities (\(\le\) or \(\ge\)).

(c) Draw a Histogram for the given data, using a graph paper:
Estimate the mode from the graph [4 Marks]

Answer:
1. Take Weekly Wages along the x-axis (with a kink/origin break starting at 3000) and No. of People along the y-axis.
2. Draw rectangles for each class interval corresponding to their frequencies.
3. The modal class is 5000-6000 with the highest frequency of 18.
4. To estimate the mode, draw cross-lines from the top corners of the highest rectangle to the adjacent rectangles' top corners. From their intersection point, drop a perpendicular to the x-axis.
5. The estimated mode from the graph is approximately Rs. 5,350 (acceptable range: Rs. 5,300 to Rs. 5,400).

Teacher's Note:
a) A kink (jagged line) must be used on the x-axis when the scale does not start from zero.
b) Ensure accurate diagonal line intersections on the highest bar to read the mode correctly.

Question 3

(a) In the figure given below, O is the center of the circle and AB is a diameter.
If AC = BD and \( \angle AOC = 72^{\circ} \) Find:
(i) \( \angle ABC \)
(ii) \( \angle BAD \)
(iii) \( \angle ABD \) [3 Marks]

[Figure: Circle with centre O and diameter AB. Chords AC, BC, AD, BD form triangles inside the circle. \( \angle AOC = 72^{\circ} \), AC = BD.]

Answer:
(i) \( \angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 72^{\circ} = 36^{\circ} \) (Angle subtended by an arc at the centre is double the angle subtended at the remaining part of the circle).
(ii) Since chords AC = BD, the corresponding arcs are equal, hence subtend equal angles. Also, \( \angle ADB = 90^{\circ} \) (Angle in a semicircle).
In \( \triangle ABD \), \( \angle BAD + \angle ABD + \angle ADB = 180^{\circ} \).
Since AC = BD, arc AC = arc BD, which implies \( \angle ABC = \angle BAD = 36^{\circ} \).
(iii) In right-angled \( \triangle ABD \) (\( \angle ADB = 90^{\circ} \)), \( \angle ABD = 90^{\circ} - \angle BAD = 90^{\circ} - 36^{\circ} = 54^{\circ} \).

Teacher's Note:
a) Recall standard circle theorems: angle in a semicircle is a right angle, and equal chords subtend equal angles.
b) Clearly state the geometric theorem used for each step to secure full credit.

(b) Prove that:
\( \frac{\sin A}{1 + \cot A} - \frac{\cos A}{1 + \tan A} = \sin A - \cos A \) [3 Marks]

Answer:
Consider the LHS:
\( \text{LHS} = \frac{\sin A}{1 + \cot A} - \frac{\cos A}{1 + \tan A} \)
Express \( \cot A \) as \( \frac{\cos A}{\sin A} \) and \( \tan A \) as \( \frac{\sin A}{\cos A} \):
\( = \frac{\sin A}{1 + \frac{\cos A}{\sin A}} - \frac{\cos A}{1 + \frac{\sin A}{\cos A}} \)
\( = \frac{\sin A}{\frac{\sin A + \cos A}{\sin A}} - \frac{\cos A}{\frac{\cos A + \sin A}{\cos A}} \)
\( = \frac{\sin^{2} A}{\sin A + \cos A} - \frac{\cos^{2} A}{\cos A + \sin A} \)
Since denominators are the same:\( = \frac{\sin^{2} A - \cos^{2} A}{\sin A + \cos A} \)
Using algebraic identity \( a^{2} - b^{2} = (a - b)(a + b) \):
\( = \frac{(\sin A - \cos A)(\sin A + \cos A)}{\sin A + \cos A} \)
\( = \sin A - \cos A = \text{RHS} \). Hence proved.

Teacher's Note:
a) Converting all trigonometric ratios into sine and cosine is a reliable strategy for proving identities.
b) Show all algebraic expansion and cancellation steps clearly.

(c) In what ratio is the line joining P(5, 3) and Q(-5, 3) divided by the y-axis? Also find coordinates of the point of intersection. [4 Marks]

Answer:
Let the y-axis divide the line segment PQ in the ratio \( k : 1 \) at point R.
Since the point lies on the y-axis, its x-coordinate is \( 0 \).
Using the section formula for the x-coordinate:
\( x = \frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}} \)
\( 0 = \frac{k(-5) + 1(5)}{k + 1} \)
\( -5k + 5 = 0 \implies 5k = 5 \implies k = 1 \)
Thus, the ratio is \( 1 : 1 \) (divided internally).
Now, find the y-coordinate of the point of intersection R:
\( y = \frac{1(3) + 1(3)}{1 + 1} = \frac{3 + 3}{2} = \frac{6}{2} = 3 \)
Therefore, the coordinates of the point of intersection are \( (0, 3) \).

Teacher's Note:
a) Any point on the y-axis has an x-coordinate equal to zero, which is the key condition to set up the equation.
b) A ratio of \( 1 : 1 \) indicates that the y-axis is the midpoint of the line segment PQ in this case.

Question 4

(a) A solid spherical ball of radius 6 cm is melted and recast into 64 identical spherical marbles. Find the radius of each marble. [3 Marks]

Answer:
Radius of the large spherical ball (\( R \)) = \( 6 \) cm.
Volume of the large sphere = \( \frac{4}{3} \pi R^{3} = \frac{4}{3} \pi (6)^{3} \).
Let the radius of each small marble be \( r \).
Volume of 64 small marbles = \( 64 \times \frac{4}{3} \pi r^{3} \).
Since volume remains unchanged when melted and recast:
\( 64 \times \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (6)^{3} \)
\( 64 r^{3} = 6^{3} = 216 \)
\( r^{3} = \frac{216}{64} = \frac{27}{8} \)
\( r = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} = 1.5 \) cm.
Thus, the radius of each marble is \( 1.5 \) cm.

Teacher's Note:
a) In mensuration problems involving melting and recasting, equate the volume of the original body to the total volume of the new bodies.
b) Avoid expanding \( \pi \) and large cubes prematurely; simplification before cube root extraction makes calculation easier.

(b) Each of the letters of the word AUTHORIZES is written on identical circular discs and put in a bag. They are well shuffled. If a disc is drawn at random from the bag, what is the probability that the letter is:
(i) a vowel
(ii) one of the first 9 letters of the English alphabet which appears in the given word
(iii) one of the last 9 letters of the English alphabet which appears in the given word [3 Marks]

Answer:
The word AUTHORIZES has 10 letters: A, U, T, H, O, R, I, Z, E, S (all unique). Total possible outcomes \( n(S) = 10 \).
(i) Vowels in AUTHORIZES are A, U, O, I, E (5 vowels).
Probability = \( \frac{5}{10} = \frac{1}{2} \).
(ii) First 9 letters of the English alphabet are A, B, C, D, E, F, G, H, I. Those appearing in AUTHORIZES are A, E, H, I (4 letters).
Probability = \( \frac{4}{10} = \frac{2}{5} \).
(iii) Last 9 letters of the English alphabet are R, S, T, U, V, W, X, Y, Z. Those appearing in AUTHORIZES are R, S, T, U, Z (5 letters).
Probability = \( \frac{5}{10} = \frac{1}{2} \).

Teacher's Note:
a) Count the total number of distinct letters in the given word carefully before finding probabilities.
b) Always reduce fractions to their simplest form.

(c) Mr. Bedi visits the market and buys the following articles:
Medicines costing Rs. 950, GST @ 5%
A pair of shoes costing Rs. 3000, GST @ 18%
A Laptop bag costing Rs. 1000 with a discount of 30%, GST @ 18%
(i) Calculate the total amount of GST paid.
(ii) The total bill amount including GST paid by Mr. Bedi [4 Marks]

Answer:
1. Medicines: Marked Price = Rs. 950, GST @ 5%
GST on medicines = \( 950 \times \frac{5}{100} = \text{Rs. } 47.50 \)

2. Pair of Shoes: Marked Price = Rs. 3000, GST @ 18%
GST on shoes = \( 3000 \times \frac{18}{100} = \text{Rs. } 540 \)

3. Laptop Bag: Marked Price = Rs. 1000, Discount = 30%
Selling Price (taxable value) = \( 1000 - (30\% \text{ of } 1000) = 1000 - 300 = \text{Rs. } 700 \)
GST on laptop bag @ 18% = \( 700 \times \frac{18}{100} = \text{Rs. } 126 \)

(i) Total amount of GST paid = \( 47.50 + 540 + 126 = \text{Rs. } 713.50 \)
(ii) Total bill amount including GST = Sum of selling prices + Total GST
= \( (950 + 3000 + 700) + 713.50 = 4650 + 713.50 = \text{Rs. } 5,363.50 \)

Teacher's Note:
a) GST is always calculated on the taxable value (Selling Price after discount, if any), not on the printed marked price.
b) Double-check addition of decimal amounts for GST calculations.

SECTION B

Attempt any four questions from this section

Question 5

(a) A company with 500 shares of nominal value 120 declares an annual dividend of 15%. Calculate:
(i) the total amount of dividend paid by the company.
(ii) annual income of Mr. Sharma who holds 80 shares of the company
If the return percent of Mr. Sharma from his shares is 10%, find the market value of each share. [3 Marks]

Answer:
Given: Nominal Value (NV) of 1 share = Rs. 120, Total shares of company = 500, Dividend % = 15%.
(i) Total nominal value of company shares = \( 500 \times 120 = \text{Rs. } 60,000 \)
Total dividend paid by company = \( 15\% \text{ of } 60,000 = \frac{15}{100} \times 60,000 = \text{Rs. } 9,000 \).
(ii) Mr. Sharma holds 80 shares.
Annual income of Mr. Sharma = \( 80 \times (\text{Dividend on 1 share}) = 80 \times (15\% \text{ of } 120) = 80 \times 18 = \text{Rs. } 1,440 \).
(iii) Given Mr. Sharma's return percent = 10%.
Return % = \( \frac{\text{Annual Income}}{\text{Total Investment}} \times 100 \)
\( 10 = \frac{1440}{\text{Investment}} \times 100 \implies \text{Investment} = \text{Rs. } 14,400 \)
Market Value (MV) of 1 share = \( \frac{\text{Total Investment}}{\text{Number of shares held}} = \frac{14400}{80} = \text{Rs. } 180 \).

Teacher's Note:
a) Dividend is always calculated on the Nominal (Face) Value, while return percent is calculated on the Market Value (Investment).
b) Clearly distinguish between company-level calculations and individual shareholder calculations.

(b) The mean of the following data is 16. Calculate the value of f. [3 Marks]

Answer:
Construct a frequency distribution table:
- For \( x_{1} = 5, f_{1} = 3 \implies f_{1}x_{1} = 15 \)
- For \( x_{2} = 10, f_{2} = 7 \implies f_{2}x_{2} = 70 \)
- For \( x_{3} = 15, f_{3} = f \implies f_{3}x_{3} = 15f \)
- For \( x_{4} = 20, f_{4} = 9 \implies f_{4}x_{4} = 180 \)
- For \( x_{5} = 25, f_{5} = 6 \implies f_{5}x_{5} = 150 \)

Sum of frequencies \( \sum f_{i} = 3 + 7 + f + 9 + 6 = 25 + f \)
Sum of products \( \sum f_{i}x_{i} = 15 + 70 + 15f + 180 + 150 = 415 + 15f \)
Given Mean \( \bar{x} = 16 \):
\( \bar{x} = \frac{\sum f_{i}x_{i}}{\sum f_{i}} \)
\( 16 = \frac{415 + 15f}{25 + f} \)
\( 16(25 + f) = 415 + 15f \)
\( 400 + 16f = 415 + 15f \)
\( 16f - 15f = 415 - 400 \implies f = 15 \).

Teacher's Note:
a) Set up a clear table for calculating \( f_{i}x_{i} \) to avoid algebraic errors.
b) Cross-multiply carefully and group variable terms on one side.

(c) The 4th, 6th and the last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio, and number of terms of the series. [4 Marks]

Answer:
Let the first term of the G.P. be \( a \) and the common ratio be \( r \).
The nth term of a G.P. is given by \( T_{n} = ar^{n-1} \).
Given:
1) \( T_{4} = ar^{3} = 10 \) ---(Equation 1)
2) \( T_{6} = ar^{5} = 40 \) ---(Equation 2)
3) Last term \( T_{n} = ar^{n-1} = 640 \) ---(Equation 3)

Dividing Equation 2 by Equation 1:
\( \frac{ar^{5}}{ar^{3}} = \frac{40}{10} \)
\( r^{2} = 4 \implies r = \pm 2 \)
Since the common ratio is positive, \( r = 2 \).
Substitute \( r = 2 \) in Equation 1:
\( a(2)^{3} = 10 \implies 8a = 10 \implies a = \frac{10}{8} = \frac{5}{4} = 1.25 \)
Now, find the number of terms \( n \) using Equation 3:
\( ar^{n-1} = 640 \)
\( \frac{5}{4}(2)^{n-1} = 640 \)
\( 2^{n-1} = 640 \times \frac{4}{5} = 128 \times 4 = 512 \)
Since \( 512 = 2^{9} \):
\( 2^{n-1} = 2^{9} \implies n - 1 = 9 \implies n = 10 \).
Thus, first term \( a = 1.25 \), common ratio \( r = 2 \), and number of terms \( n = 10 \).

Teacher's Note:
a) Use the ratio of two terms in a G.P. to eliminate 'a' and find the common ratio directly.
b) Express the final term value as a power of the common ratio to solve for n.

Question 6

(a) If \( A = \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix} \) and \( B = \begin{pmatrix} -4 & 2 \\ 1 & 0 \end{pmatrix} \), find \( A^{2} - 2AB + B^{2} \) [3 Marks]

Answer:
Note: \( A^{2} - 2AB + B^{2} = (A - B)^{2} \).
First, find \( A - B \):
\( A - B = \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix} - \begin{pmatrix} -4 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 3 - (-4) & 0 - 2 \\ 5 - 1 & 1 - 0 \end{pmatrix} = \begin{pmatrix} 7 & -2 \\ 4 & 1 \end{pmatrix} \)
Now, find \( (A - B)^{2} = (A - B)(A - B) \):
\( = \begin{pmatrix} 7 & -2 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 7 & -2 \\ 4 & 1 \end{pmatrix} \)
\( = \begin{pmatrix} (7)(7) + (-2)(4) & (7)(-2) + (-2)(1) \\ (4)(7) + (1)(4) & (4)(-2) + (1)(1) \end{pmatrix} \)
\( = \begin{pmatrix} 49 - 8 & -14 - 2 \\ 28 + 4 & -8 + 1 \end{pmatrix} = \begin{pmatrix} 41 & -16 \\ 32 & -7 \end{pmatrix} \).

Teacher's Note:
a) Using the algebraic identity \( (A - B)^{2} \) saves calculation steps compared to finding \( A^{2} \), \( 2AB \), and \( B^{2} \) separately.
b) Check each matrix multiplication step carefully for sign errors.

(b) In the given figure AB = 9 cm, PA = 7.5 cm and PC = 5 cm. Chords AD and BC intersect at P.
(i) Prove that \( \triangle PAB \sim \triangle PCD \)
(ii) Find the length of CD
(iii) Find area of \( \triangle PAB \) : area of \( \triangle PCD \) [3 Marks]

[Figure: Circle with intersecting chords AD and BC at point P. AB = 9 cm, PA = 7.5 cm, PC = 5 cm.]

Answer:
(i) In \( \triangle PAB \) and \( \triangle PCD \):
- \( \angle APB = \angle DPC \) (Vertically opposite angles)
- \( \angle PAB = \angle PDC \) (Angles in the same segment subtended by arc BC)
Therefore, by AA similarity criterion, \( \triangle PAB \sim \triangle PCD \).

(ii) By intersecting chords theorem: \( PA \times PD = PC \times PB \).
Wait, looking at the figure, chords are AD and BC intersecting at P. So \( PA \times PD = PC \times PB \), but PB is not directly given. Instead, using similarity of triangles \( \triangle PAB \) and \( \triangle PCD \):
Corresponding sides ratio: \( \frac{PA}{PD} = \frac{PB}{PC} = \frac{AB}{CD} \)... Actually, let's use the intersecting chords property properly: \( PA \times PD = PC \times PB \) or use triangle similarity. Since \( AB = 9 \), let's use similarity ratios: Note that \( \angle PBA = \angle PCD \) (angles in same segment).
From similarity \( \triangle PAB \sim \triangle PCD \):
\( \frac{PA}{PC} = \frac{AB}{CD} \implies \frac{7.5}{5} = \frac{9}{CD} \)
\( CD = \frac{9 \times 5}{7.5} = \frac{45}{7.5} = 6 \) cm.

(iii) Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\( \frac{\text{Area}(\triangle PAB)}{\text{Area}(\triangle PCD)} = \left(\frac{AB}{CD}\right)^{2} = \left(\frac{9}{6}\right)^{2} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4} \) or \( 9 : 4 \).

Teacher's Note:
a) Establishing triangle similarity correctly ensures the proper correspondence of side lengths.
b) The ratio of the areas of two similar triangles equals the square of the ratio of any pair of corresponding sides.

(c) From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be \( 45^{\circ} \) and \( 60^{\circ} \) respectively. If the height of the tower is 20 m.
Find:
(i) the height of the cliff
(ii) the distance between the cliff and the tower [4 Marks]

Answer:
Let AB be the cliff of height \( h \) meters, and CD be the tower of height \( 20 \) m. Let the distance between the cliff and the tower be \( x \) meters.
Draw a horizontal line from C meeting AB at E. Then \( BE = CD = 20 \) m, and \( AE = h - 20 \).
Distance \( EC = x \).
Angle of depression of the top of the tower from the cliff is \( 45^{\circ} \), so \( \angle ACE = 45^{\circ} \).
In right-angled \( \triangle AEC \):
\( \tan 45^{\circ} = \frac{AE}{EC} \implies 1 = \frac{h - 20}{x} \implies x = h - 20 \) ---(Equation 1)
Angle of depression of the bottom of the tower from the cliff is \( 60^{\circ} \), so \( \angle ADB = 60^{\circ} \).
In right-angled \( \triangle ABD \):
\( \tan 60^{\circ} = \frac{AB}{BD} \implies \sqrt{3} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}} \) ---(Equation 2)
Equating (1) and (2):
\( h - 20 = \frac{h}{\sqrt{3}} \)
\( \sqrt{3}h - 20\sqrt{3} = h \)
\( \sqrt{3}h - h = 20\sqrt{3} \)
\( h(\sqrt{3} - 1) = 20\sqrt{3} \)
\( h = \frac{20\sqrt{3}}{\sqrt{3} - 1} = \frac{20(1.732)}{1.732 - 1} = \frac{34.64}{0.732} \approx 47.32 \) m.
(i) Height of the cliff \( h \approx 47.32 \) m (or \( 20 + 10\sqrt{3}(\sqrt{3}+1) = 47.32 \) m).
(ii) Distance between cliff and tower \( x = h - 20 = 47.32 - 20 = 27.32 \) m.

Teacher's Note:
a) Draw a clear schematic diagram showing angles of depression as alternate interior angles of elevation.
b) Rationalise denominators carefully when substituting numerical values for square roots like \( \sqrt{3} \).

Question 7

(a) Find the value of 'p' if the lines \( 5x - 3y + 2 = 0 \) and \( 6x - py + 7 = 0 \) are perpendicular to each other. Hence find the equation of a line passing through \( (-2, -1) \) and parallel to \( 6x - py + 7 = 0 \). [3 Marks]

Answer:
1. Slope of line 1 (\( 5x - 3y + 2 = 0 \)):
\( 3y = 5x + 2 \implies y = \frac{5}{3}x + \frac{2}{3} \). Slope \( m_{1} = \frac{5}{3} \).
2. Slope of line 2 (\( 6x - py + 7 = 0 \)):
\( py = 6x + 7 \implies y = \frac{6}{p}x + \frac{7}{p} \). Slope \( m_{2} = \frac{6}{p} \).
Since the two lines are perpendicular, \( m_{1} \times m_{2} = -1 \):
\( \frac{5}{3} \times \frac{6}{p} = -1 \)
\( \frac{10}{p} = -1 \implies p = -10 \).

3. Substitute \( p = -10 \) into the equation of the parallel line: \( 6x - (-10)y + 7 = 0 \implies 6x + 10y + 7 = 0 \).
The required line is parallel to this line, so its slope is \( m = -\frac{6}{10} = -\frac{3}{5} \).
Using point-slope form for the line passing through \( (-2, -1) \):
\( y - y_{1} = m(x - x_{1}) \)
\( y - (-1) = -\frac{3}{5}(x - (-2)) \)
\( y + 1 = -\frac{3}{5}(x + 2) \)
\( 5(y + 1) = -3(x + 2) \)
\( 5y + 5 = -3x - 6 \)
\( 3x + 5y + 11 = 0 \).

Teacher's Note:
a) Two lines are perpendicular if the product of their slopes is \(-1\), and parallel if their slopes are equal.
b) Parallel lines share the exact same coefficients of x and y in their general equation, differing only in the constant term.

(b) Using properties of proportion, find \( x : y \), given:
\( \frac{x^{2} + 2x}{2x + 4} = \frac{y^{2} + 3y}{3y + 9} \) [3 Marks]

Answer:
Given equation:
\( \frac{x^{2} + 2x}{2x + 4} = \frac{y^{2} + 3y}{3y + 9} \)
Factorise the numerators and denominators:
\( \frac{x(x + 2)}{2(x + 2)} = \frac{y(y + 3)}{3(y + 3)} \)
Cancelling common terms \( (x + 2) \) and \( (y + 3) \):
\( \frac{x}{2} = \frac{y}{3} \)
Rearranging using alternation property:
\( \frac{x}{y} = \frac{2}{3} \)
Therefore, \( x : y = 2 : 3 \).

Teacher's Note:
a) Always simplify algebraic fractions by factorisation before applying properties of proportion.
b) Clearly state the final ratio in colon notation as requested.

(c) In the given figure TP and TQ are two tangents to the circle with center O, touching at A and C respectively. If \( \angle BCQ = 55^{\circ} \) and \( \angle BAP = 60^{\circ} \), find:
(i) \( \angle OBA \) and \( \angle OBC \)
(ii) \( \angle AOC \)
(iii) \( \angle ATC \) [4 Marks]

[Figure: Circle with centre O, tangents TP and TQ touching at A and C. Chords AB and BC inside. \( \angle BCQ = 55^{\circ} \), \( \angle BAP = 60^{\circ} \).]

Answer:
(i) By the alternate segment theorem, the angle between tangent TP and chord AB is equal to angle in the alternate segment: \( \angle ACB = \angle BAP = 60^{\circ} \).
Similarly, angle between tangent TQ and chord BC: \( \angle BAC = \angle BCQ = 55^{\circ} \).
In \( \triangle ABC \), \( \angle ABC = 180^{\circ} - (60^{\circ} + 55^{\circ}) = 180^{\circ} - 115^{\circ} = 65^{\circ} \).
In \( \triangle OAB \), OA = OB (radii), so \( \angle OAB = \angle OBA \).
Also, radius OA is perpendicular to tangent TP, so \( \angle OAP = 90^{\circ} \).
\( \angle OAB = 90^{\circ} - \angle BAP = 90^{\circ} - 60^{\circ} = 30^{\circ} \).
Thus, \( \angle OBA = 30^{\circ} \).
Similarly, in \( \triangle OCB \), OC = OB (radii), and \( \angle OCT = 90^{\circ} \):
\( \angle OCB = 90^{\circ} - \angle BCQ = 90^{\circ} - 55^{\circ} = 35^{\circ} \).
Thus, \( \angle OBC = 35^{\circ} \).
(ii) In \( \triangle ABC \), \( \angle AOC = 2 \times \angle ABC = 2 \times 65^{\circ} = 130^{\circ} \) (Angle at centre is twice the angle at circumference).
(iii) In quadrilateral OATC, the sum of angles is \( 360^{\circ} \):
\( \angle ATC + \angle OAT + \angle OCT + \angle AOC = 360^{\circ} \)
\( \angle ATC + 90^{\circ} + 90^{\circ} + 130^{\circ} = 360^{\circ} \)
\( \angle ATC + 310^{\circ} = 360^{\circ} \implies \angle ATC = 50^{\circ} \).

Teacher's Note:
a) Use alternate segment theorem and radius-tangent perpendicularity property efficiently in circle geometry problems.
b) The sum of angles in the quadrilateral formed by two radii and two tangents from an external point is always \( 360^{\circ} \), with opposite angles summing to \( 180^{\circ} \).

Question 8

(a) What must be added to the polynomial \( 2x^{3} - 3x^{2} - 8x \), so that it leaves a remainder 10 when divided by \( 2x + 1 \)? [3 Marks]

Answer:
Let the polynomial be \( f(x) = 2x^{3} - 3x^{2} - 8x \).
Let the expression to be added be \( k \).
New polynomial \( g(x) = 2x^{3} - 3x^{2} - 8x + k \).
When \( g(x) \) is divided by \( 2x + 1 \), the divisor root is \( x = -\frac{1}{2} \).
According to the remainder theorem, remainder \( R = g\left(-\frac{1}{2}\right) \):
Given remainder \( = 10 \).
\( 2\left(-\frac{1}{2}\right)^{3} - 3\left(-\frac{1}{2}\right)^{2} - 8\left(-\frac{1}{2}\right) + k = 10 \)
\( 2\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) - 8\left(-\frac{1}{2}\right) + k = 10 \)
\( -\frac{2}{8} - \frac{3}{4} + 4 + k = 10 \)
\( -\frac{1}{4} - \frac{3}{4} + 4 + k = 10 \)
\( -\frac{4}{4} + 4 + k = 10 \)
\( -1 + 4 + k = 10 \)
\( 3 + k = 10 \implies k = 7 \).
Therefore, \( 7 \) must be added.

Teacher's Note:
a) Use the Remainder Theorem by equating the value of the polynomial at the zero of the divisor to the given remainder.
b) Be careful with fraction arithmetic and negative signs when substituting fractional values.

(b) Mr. Sonu has a recurring deposit account and deposits Rs. 750 per month for 2 years. If he gets Rs. 19,125 at the time of maturity, find the rate of interest. [3 Marks]

Answer:
Given: Monthly deposit \( P = \text{Rs. } 750 \), Time \( n = 2 \text{ years} = 24 \text{ months} \), Maturity value \( MV = \text{Rs. } 19,125 \).
Total money deposited \( = P \times n = 750 \times 24 = \text{Rs. } 18,000 \).
Interest \( I = MV - \text{Total Deposit} = 19,125 - 18,000 = \text{Rs. } 1,125 \).
Formula for interest in a recurring deposit account:
\( I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( 1125 = 750 \times \frac{24(24 + 1)}{24} \times \frac{r}{100} \)
\( 1125 = 750 \times \frac{25}{2} \times \frac{r}{100} \)
\( 1125 = 750 \times \frac{25r}{200} = \frac{750 \times r}{8} \)
\( 750r = 1125 \times 8 \)
\( 750r = 9000 \implies r = \frac{9000}{750} = 12\% \).
Thus, the rate of interest is \( 12\% \) per annum.

Teacher's Note:
a) Always calculate total interest first by subtracting total deposits from the maturity value.
b) Ensure the number of months n is used correctly in the denominator formula \( \frac{n(n+1)}{2 \times 12} \).

(c) Use graph paper for this question. Take 1 cm = 1 unit on both x and y axes.
(i) Plot the following points on your graph sheets: A(-4, 0), B(-3, 2), C(0, 4), D(4, 1) and E(7, 3).
(ii) Reflect the points B, C, D and E on the x-axis and name them as B', C', D' and E' respectively.
(iii) Join the points A, B, C, D, E, E', D', C', B' and A in order.
(iv) Name the closed figure formed. [4 Marks]

Answer:
(i) Given points plotted: A(-4, 0), B(-3, 2), C(0, 4), D(4, 1), E(7, 3).
(ii) Coordinates of reflected points on the x-axis (\( (x, y) \to (x, -y) \)):
- B' = (-3, -2)
- C' = (0, -4)
- D' = (4, -1)
- E' = (7, -3)
(iii) Joining A, B, C, D, E, E', D', C', B', A in order forms a closed polygon.
(iv) The closed figure formed is a decagon (specifically, a symmetric polygon with 10 vertices resembling an arrow or kite-like symmetric shape across the x-axis).

Teacher's Note:
a) Reflection across the x-axis changes the sign of the y-coordinate while keeping the x-coordinate unchanged.
b) Ensure all points are plotted and joined in the exact order specified in the question.

Question 9

(a) 40 students enter for a game of shot-put competition. The distance thrown (in meters) is recorded below: [6 Marks]

Use a graph paper to draw an ogive for the above distribution. Use a scale of 2 cm = 1 m on one axis and 2 cm = 5 students on the other axis. Hence using your graph find:
(i) the median
(ii) Upper Quartile
(iii) number of students who cover a distance which is above \( 16\frac{1}{2} \) m.

Answer:
Construct cumulative frequency table:
- 12-13: f = 3, c.f. = 3
- 13-14: f = 9, c.f. = 12
- 14-15: f = 12, c.f. = 24
- 15-16: f = 9, c.f. = 33
- 16-17: f = 4, c.f. = 37
- 17-18: f = 2, c.f. = 39
- 18-19: f = 1, c.f. = 40
Total frequency \( N = 40 \).
Plot upper limits against cumulative frequencies: (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39), (19, 40), and join with a smooth freehand curve starting from lower limit of first class (12, 0).
(i) Median = Value at \( \frac{N}{2} = \frac{40}{2} = 20 \)th term. From graph, median \( \approx 14.6 \) m.
(ii) Upper Quartile (\( Q_{3} \)) = Value at \( \frac{3N}{4} = \frac{3 \times 40}{4} = 30 \)th term. From graph, \( Q_{3} \approx 15.75 \) m.
(iii) Number of students above \( 16\frac{1}{2} \) m (\( 16.5 \) m):
Locate \( 16.5 \) on the x-axis, find its corresponding cumulative frequency on the ogive curve (\( \approx 35.5 \)), and subtract from total N: \( 40 - 35.5 = 4.5 \to \) approximately \( 4 \) to \( 5 \) students.

Teacher's Note:
a) An ogive is always plotted using upper class boundaries against cumulative frequencies.
b) For "above" values, subtract the cumulative frequency corresponding to that x-value from the total number of observations (N).

(a) [Note: Question numbering in paper repeats (a) here] If \( x = \frac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}} \), prove that \( x^{2} - 4ax + 1 = 0 \) [4 Marks]

Answer:
Given equation: \( x = \frac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}} \)
Using componendo and dividendo:
\( \frac{x + 1}{x - 1} = \frac{(\sqrt{2a + 1} + \sqrt{2a - 1}) + (\sqrt{2a + 1} - \sqrt{2a - 1})}{(\sqrt{2a + 1} + \sqrt{2a - 1}) - (\sqrt{2a + 1} - \sqrt{2a - 1})} \)
\( \frac{x + 1}{x - 1} = \frac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}} = \frac{\sqrt{2a + 1}}{\sqrt{2a - 1}} \)
Squaring both sides:
\( \left(\frac{x + 1}{x - 1}\right)^{2} = \frac{2a + 1}{2a - 1} \)
\( \frac{x^{2} + 2x + 1}{x^{2} - 2x + 1} = \frac{2a + 1}{2a - 1} \)
Again applying componendo and dividendo on both sides:
\( \frac{(x^{2} + 2x + 1) + (x^{2} - 2x + 1)}{(x^{2} + 2x + 1) - (x^{2} - 2x + 1)} = \frac{(2a + 1) + (2a - 1)}{(2a + 1) - (2a - 1)} \)
\( \frac{2x^{2} + 2}{4x} = \frac{4a}{2} \)
\( \frac{2(x^{2} + 1)}{4x} = 2a \)
\( \frac{x^{2} + 1}{2x} = 2a \)
\( x^{2} + 1 = 4ax \)
\( x^{2} - 4ax + 1 = 0 \). Hence proved.

Teacher's Note:
a) Applying Componendo and Dividendo successively is the most efficient method for solving surd equations of this type.
b) Ensure proper expansion of algebraic squares before applying properties of proportion a second time.

Question 10

(a) If the 6th term of an A.P. is equal to four times its first term and the sum of first six terms is 75, find the first term and the common difference. [3 Marks]

Answer:
Let the first term be \( a \) and common difference be \( d \).
Given 1: 6th term \( T_{6} = 4a \)
\( a + 5d = 4a \implies 5d = 3a \implies a = \frac{5}{3}d \) ---(Equation 1)
Given 2: Sum of first 6 terms \( S_{6} = 75 \)
\( S_{n} = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{6} = \frac{6}{2}[2a + 5d] = 75 \)
\( 3(2a + 5d) = 75 \implies 2a + 5d = 25 \) ---(Equation 2)
Substitute \( 5d = 3a \) into Equation 2:
\( 2a + 3a = 25 \implies 5a = 25 \implies a = 5 \).
Now find \( d \) using \( 5d = 3a \):
\( 5d = 3(5) = 15 \implies d = 3 \).
Therefore, first term \( a = 5 \) and common difference \( d = 3 \).

Teacher's Note:
a) Form simultaneous linear equations using the nth term formula and sum formula for AP.
b) Substitution method simplifies solving equations when one variable can be expressed directly in terms of the other.

(b) The difference of two natural numbers is 7 and their product is 450. Find the numbers. [3 Marks]

Answer:
Let the two natural numbers be \( x \) and \( x + 7 \) (or \( x \) and \( y \) such that \( y - x = 7 \)).
Let smaller number be \( x \), then larger number is \( x + 7 \).
Given product is 450:
\( x(x + 7) = 450 \)
\( x^{2} + 7x - 450 = 0 \)
Factorise the quadratic equation by splitting the middle term (\( 25 \times -18 = -450 \), \( 25 - 18 = 7 \)):
\( x^{2} + 25x - 18x - 450 = 0 \)
\( x(x + 25) - 18(x + 25) = 0 \)
\( (x - 18)(x + 25) = 0 \)
\( x = 18 \) or \( x = -25 \).
Since numbers are natural numbers, \( x = 18 \).
The smaller number is \( 18 \) and the larger number is \( 18 + 7 = 25 \).
Therefore, the numbers are 18 and 25.

Teacher's Note:
a) Representing two numbers with a difference of 7 as x and x + 7 reduces the problem to a single-variable quadratic equation.
b) Reject negative roots since the question specifies natural numbers.

(c) Use ruler and compass for this question. Construct a circle of radius 4.5 cm. Draw a chord AB = 6 cm.
(i) Find the locus of points equidistant from A and B. Mark the point where it meets the circle as D.
(ii) Join AD and find the locus of points which are equidistant from AD and AB. Mark the point where it meets the circle as C.
(iii) Join CD. Measure and write down the length of side CD of the quadrilateral ABCD. [3 Marks]

Answer:
1. Draw a circle with centre O and radius \( 4.5 \) cm.
2. Draw a chord \( AB = 6 \) cm.
(i) The locus of points equidistant from A and B is the perpendicular bisector of chord AB. Draw the perpendicular bisector of AB, which passes through centre O and meets the circle at point D.
(ii) Join AD. The locus of points equidistant from AD and AB is the angle bisector of \( \angle DAB \). Draw the angle bisector of \( \angle DAB \) meeting the circle at point C.
(iii) Join CD to complete quadrilateral ABCD. Measuring with a ruler, the length of side CD is approximately \( 4.2 \) cm (acceptable range: \( 4.1 \) cm to \( 4.3 \) cm).

Teacher's Note:
a) Locus of points equidistant from two points is their perpendicular bisector, and from two intersecting lines is their angle bisector.
b) Ensure all construction lines and intersection points are clearly labelled.

Question 11

(a) A model of a high rise building is made to a scale of 1 : 50.
(i) If the height of the model is 0.8 m, find the height of the actual building.
(ii) If the floor area of a flat in the building is \( 20\text{ m}^{2} \), find the floor area of that in the model. [3 Marks]

Answer:
Given scale factor \( k = 1 : 50 \implies \frac{\text{Dimension of model}}{\text{Dimension of actual}} = \frac{1}{50} \).
(i) Height of model = \( 0.8 \) m.
\( \frac{\text{Height of model}}{\text{Height of actual}} = \frac{1}{50} \)
\( \frac{0.8}{\text{Height of actual}} = \frac{1}{50} \implies \text{Height of actual} = 0.8 \times 50 = 40 \) m.
(ii) Ratio of areas is the square of the linear scale factor: \( \frac{\text{Area of model}}{\text{Area of actual}} = \left(\frac{1}{50}\right)^{2} = \frac{1}{2500} \).
Given actual floor area = \( 20\text{ m}^{2} \).
\( \frac{\text{Area of model}}{20} = \frac{1}{2500} \)
Area of model = \( \frac{20}{2500} = \frac{2}{250} = \frac{1}{125} = 0.008\text{ m}^{2} \) (or \( 80\text{ cm}^{2} \)).

Teacher's Note:
a) Linear dimensions scale directly, while areas scale as the square of the scale factor and volumes as the cube.
b) State units clearly in the final answer.

(b) From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Taking \( \pi = \frac{22}{7} \), find the volume of the remaining solid. [3 Marks]

[Figure: Cylinder of height 28 cm and diameter 6 cm with two conical cavities of height 10.5 cm hollowed out from both ends.]

Answer:
Radius of cylinder \( r = \frac{6}{2} = 3 \) cm, Height of cylinder \( H = 28 \) cm.
Radius of each cone \( r = 3 \) cm, Height of each cone \( h = 10.5 = \frac{21}{2} \) cm.
Volume of cylinder \( V_{\text{cyl}} = \pi r^{2} H = \frac{22}{7} \times (3)^{2} \times 28 = \frac{22}{7} \times 9 \times 28 = 22 \times 9 \times 4 = 792\text{ cm}^{3} \).
Volume of one cone \( V_{\text{cone}} = \frac{1}{3} \pi r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (3)^{2} \times 10.5 = \frac{1}{3} \times \frac{22}{7} \times 9 \times \frac{21}{2} = 22 \times \frac{9}{3} \times \frac{3}{2} = 99\text{ cm}^{3} \).
Volume of two cones = \( 2 \times 99 = 198\text{ cm}^{3} \).
Volume of remaining solid = Volume of cylinder - Volume of two cones
\( = 792 - 198 = 594\text{ cm}^{3} \).

Teacher's Note:
a) When cavities are hollowed out from a solid, subtract the volumes of the cavities from the total volume of the solid body.
b) Using fractional values for heights like \( 10.5 = \frac{21}{2} \) simplifies manual calculations with \( \pi = \frac{22}{7} \).

(c) Prove the identity:
\( \left(\frac{1 - \tan \theta}{1 - \cot \theta}\right)^{2} = \tan^{2} \theta \) [4 Marks]

Answer:
Consider the LHS:
\( \text{LHS} = \left(\frac{1 - \tan \theta}{1 - \cot \theta}\right)^{2} \)
Substitute \( \cot \theta = \frac{1}{\tan \theta} \):
\( = \left(\frac{1 - \tan \theta}{1 - \frac{1}{\tan \theta}}\right)^{2} \)
\( = \left(\frac{1 - \tan \theta}{\frac{\tan \theta - 1}{\tan \theta}}\right)^{2} \)
Rewrite the denominator and simplify:
\( = \left(\frac{1 - \tan \theta}{-(1 - \tan \theta)} \times \tan \theta\right)^{2} \)
\( = \left(\frac{-1 \times \tan \theta}{1}\right)^{2} \)
\( = (-\tan \theta)^{2} = \tan^{2} \theta = \text{RHS} \). Hence proved.

Teacher's Note:
a) Expressing cotangent in terms of tangent is the quickest way to simplify this identity.
b) Pay close attention to negative signs when cancelling terms like \( (1 - \tan \theta) \) and \( (\tan \theta - 1) \).