ICSE Class 9 Mathematics Sample Paper with Solutions Set 03

SECTION - A (40 Marks)

(Answer all questions from this Section)

Q. 1. (a) Without using tables, find the value of \(\frac{\sin 30^{\circ} - \sin 90^{\circ} + 2\cos 0^{\circ}}{\tan 30^{\circ} \times \tan 60^{\circ}}\) [3 Marks]

Answer:
1. Substitute standard trigonometric values: \(\sin 30^{\circ} = \frac{1}{2}\), \(\sin 90^{\circ} = 1\), \(\cos 0^{\circ} = 1\), \(\tan 30^{\circ} = \frac{1}{\sqrt{3}}\), and \(\tan 60^{\circ} = \sqrt{3}\).
2. Evaluate the numerator: \(\frac{1}{2} - 1 + 2(1) = \frac{1}{2} - 1 + 2 = \frac{1}{2} + 1 = \frac{3}{2}\).
3. Evaluate the denominator: \(\frac{1}{\sqrt{3}} \times \sqrt{3} = 1\).
4. Calculate the final value: \(\frac{\frac{3}{2}}{1} = 1\frac{1}{2}\).

Teacher's Note:
a) Memorise standard trigonometric ratios for angles \(0^{\circ}\), \(30^{\circ}\), \(45^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\) to evaluate such expressions accurately without tables.
b) Students often make arithmetic sign errors when combining fractions and whole numbers in the numerator.

(b) Evaluate: \(\frac{3 \times 27^{n+1} + 9 \times 3^{n-1}}{8 \times 3^{3n} - 5 \times 27^n}\) [3 Marks]

Answer:
1. Express all terms with base \(3\): \(27 = 3^3\) and \(9 = 3^2\).
2. Rewrite the numerator as \(\frac{3 \times (3^3)^{n+1} + 3^2 \times 3^{n-1}}{8 \times 3^{3n} - 5 \times (3^3)^n} = \frac{3 \times 3^{3n+3} + 3^{n+1}}{8 \times 3^{3n} - 5 \times 3^{3n}}\).
3. Simplify powers using laws of indices: \(\frac{3^{3n+4} + 3^{n+1}}{3^{3n}(8 - 5)}\) or factor out \(3^{n+1}\) properly. Alternatively, simplifying step-by-step gives \(\frac{3^{3n}(3 \times 27 + 9 \times \frac{1}{3})}{3^{3n}(8 - 5)} = \frac{3 \times 27 + 3}{3} = \frac{81 + 3}{3} = \frac{84}{3} = 28\).

Teacher's Note:
a) Convert all composite bases into prime factor bases (like base 3) before applying exponent laws.
b) Take care with fractional exponents and common factors to avoid algebraic expansion mistakes.

(c) If \(x\) and \(y\) are rational numbers and \(\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = x + y\sqrt{3}\), find the value of \(x\) and \(y\). [4 Marks]

Answer:
1. Rationalize the denominator of the LHS by multiplying numerator and denominator by \((2 + \sqrt{3})\):
\(\frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = x + y\sqrt{3}\).
2. Expand numerator and denominator: \(\frac{4 + 3 + 4\sqrt{3}}{4 - 3} = x + y\sqrt{3}\).
3. Simplify: \(7 + 4\sqrt{3} = x + y\sqrt{3}\).
4. Compare rational and irrational parts on both sides to get \(x = 7\) and \(y = 4\).

Teacher's Note:
a) The conjugate of \((a - \sqrt{b})\) is \((a + \sqrt{b})\), which eliminates radicals in the denominator via \((a - b)(a + b) = a^2 - b^2$.
b) Always equate the rational part to the rational part and the coefficient of the surd to the corresponding coefficient on the other side.

Q. 2. (a) In the given figure, AOC is a diameter of a circle with centre O and \(\text{arc } AXB = \frac{1}{2} \text{ arc } BYC\). Find \(\angle BOC\). [3 Marks]

[Figure: A circle with centre O and diameter AOC. Points X on arc AB and Y on arc BC. Triangle BOC is formed with radii OB, OC and chord BC.]

Answer:
1. Given that \(\text{arc } AXB = \frac{1}{2} \text{ arc } BYC\), the angles subtended at the centre are in the same ratio: \(\angle AOB = \frac{1}{2} \angle BOC\).
2. Since AOC is a diameter, it is a straight line, so \(\angle AOB + \angle BOC = 180^{\circ}\).
3. Substitute \(\frac{1}{2} \angle BOC\) for \(\angle AOB\): \(\frac{1}{2} \angle BOC + \angle BOC = 180^{\circ}\), which gives \(\frac{3}{2} \angle BOC = 180^{\circ}\).
4. Solve for \(\angle BOC\): \(\angle BOC = 180^{\circ} \times \frac{2}{3} = 120^{\circ}\).

Teacher's Note:
a) Equal arcs subtend equal angles at the centre, and the angle subtended by an arc is proportional to its length.
b) Recognise that angles on a straight line add up to \(180^{\circ}\).

(b) Find \(xy\), if \(x + y = 6\) and \(x - y = 4\). [3 Marks]

Answer:
1. Use the algebraic identity: \((x + y)^2 = (x - y)^2 + 4xy$.
2. Substitute the given values: \(6^2 = 4^2 + 4xy$.
3. Simplify: \(36 = 16 + 4xy \implies 4xy = 20\).
4. Solve for \(xy\): \(xy = 5\).

Teacher's Note:
a) Alternatively, students can solve for \(x\) and \(y\) independently by adding and subtracting the two linear equations, then finding their product.
b) Using algebraic identities directly saves time and minimises calculation errors.

(c) If \(\frac{\log a}{b - c} = \frac{\log b}{c - a} = \frac{\log c}{a - b}\), prove that \(a^a \cdot b^b \cdot c^c = 1\). [4 Marks]

Answer:
1. Let \(\frac{\log a}{b - c} = \frac{\log b}{c - a} = \frac{\log c}{a - b} = k\).
2. Therefore, \(\log a = k(b - c)\), \(\log b = k(c - a)\), and \(\log c = k(a - b)\).
3. Let \(A = a^a \cdot b^b \cdot c^c\). Taking logarithms on both sides:
\(\log A = a \log a + b \log b + c \log c\).
4. Substitute the expressions for logarithms: \(\log A = a \cdot k(b - c) + b \cdot k(c - a) + c \cdot k(a - b)\).
5. Expand and collect terms: \(\log A = k(ab - ac + bc - ab + ac - bc) = k(0) = 0\).
6. Since \(\log A = 0\), \(\log A = \log 1 \implies A = 1\). Hence, \(a^a \cdot b^b \cdot c^c = 1\).

Teacher's Note:
a) This is a standard ratio and proportion proof combined with logarithmic properties.
b) Ensure all expanded terms cancel out completely to yield zero before converting back using logarithms.

Q. 3. (a) From the given figure, find the angles of the parallelogram ABCD. [3 Marks]

[Figure: Parallelogram ABCD with diagonal AC. Angle D is labeled \(x\) and angle C is labeled \(2x\). Angle DAB has a perpendicular symbol showing a right angle to AC inside triangle ADC.]

Answer:
1. In \(\Delta ADC\), the interior angles are \(\angle D = x\), \(\angle ACD = 2x\), and \(\angle DAC = 90^{\circ}\).
2. The sum of angles in \(\Delta ADC\) is \(180^{\circ}\): \(x + 2x + 90^{\circ} = 180^{\circ} \implies 3x = 90^{\circ} \implies x = 30^{\circ}\).
3. Therefore, \(\angle D = 30^{\circ}\) and \(\angle C = 2(30^{\circ}) = 60^{\circ}\).
4. In parallelogram ABCD, opposite angles are equal: \(\angle B = \angle D = 30^{\circ}\).
5. Adjacent angles are supplementary: \(\angle A = 180^{\circ} - 30^{\circ} = 150^{\circ}\) and \(\angle C = 150^{\circ}\).
6. Thus, the angles of the parallelogram are \(150^{\circ}\), \(30^{\circ}\), \(150^{\circ}\), and \(30^{\circ}\).

Teacher's Note:
a) Use triangle angle sum property for the partitioned triangle and properties of parallelograms (opposite angles equal, adjacent angles supplementary).
b) State reasons clearly for each angle calculation step to secure full marks.

(b) Express \(5.3\overline{47}\) in the form \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\). [3 Marks]

Answer:
1. Let \(x = 5.3474747\dots\) ---(i)
2. Multiply by 10 to shift the non-repeating decimal part: \(10x = 53.474747\dots\) ---(ii)
3. Multiply equation (ii) by 100 (since two digits repeat): \(1000x = 5347.474747\dots\) ---(iii)
4. Subtract equation (ii) from equation (iii):
\(1000x - 10x = 5347.474747\dots - 53.474747\dots \implies 990x = 5294\).
5. Solve for \(x\): \(x = \frac{5294}{990} = \frac{2647}{495}\).

Teacher's Note:
a) Multiplying by appropriate powers of 10 aligns the recurring decimal portions so subtraction eliminates the infinite tail.
b) Always reduce the final fraction to its lowest terms.

(c) The table below classifies the days of the months of June, July and August according to the rainfall received in a locality. [3 Marks]

Answer:
1. To draw a frequency polygon, calculate the class marks for each class interval using the formula \(\text{Class Mark} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.
2. Construct the frequency table with class marks:
\(0 - 10\) (Mark: 5, Freq: 0), \(10 - 20\) (Mark: 15, Freq: 8), \(20 - 30\) (Mark: 25, Freq: 10), \(30 - 40\) (Mark: 35, Freq: 14), \(40 - 50\) (Mark: 45, Freq: 20), \(50 - 60\) (Mark: 55, Freq: 15), \(60 - 70\) (Mark: 65, Freq: 8), \(70 - 80\) (Mark: 75, Freq: 7), \(80 - 90\) (Mark: 85, Freq: 6), \(90 - 100\) (Mark: 95, Freq: 4), \(100 - 110\) (Mark: 105, Freq: 0).
3. Plot the class marks on the x-axis and the number of days (frequencies) on the y-axis, then join consecutive points with straight line segments.

Teacher's Note:
a) Include empty classes with frequency zero at both ends (such as \(0 - 10\) and \(100 - 110\)) to anchor the frequency polygon to the x-axis.
b) Ensure proper labeling of axes with appropriate units.

Q. 4. (a) There are two regular polygons with number of sides equal to \((n - 1)\) and \((n + 2)\). Their external angles differ by \(6^{\circ}\). Find the value of \(n\). [3 Marks]

Answer:
1. The exterior angle of a regular polygon with \(N\) sides is given by \(\frac{360^{\circ}}{N}\).
2. Exterior angle of the first polygon with \((n - 1)\) sides = \(\frac{360^{\circ}}{n - 1}\).
3. Exterior angle of the second polygon with \((n + 2)\) sides = \(\frac{360^{\circ}}{n + 2}\).
4. According to the question, their difference is \(6^{\circ}\): \(\frac{360^{\circ}}{n - 1} - \frac{360^{\circ}}{n + 2} = 6^{\circ}\).
5. Divide by \(6\): \(\frac{60}{n - 1} - \frac{60}{n + 2} = 1 \implies 60\left(\frac{n + 2 - (n - 1)}{(n - 1)(n + 2)}\right) = 1\).
6. Simplify: \(\frac{60 \times 3}{n^2 + n - 2} = 1 \implies 180 = n^2 + n - 2 \implies n^2 + n - 182 = 0\).
7. Factorize the quadratic equation: \((n + 14)(n - 13) = 0\).
8. Since number of sides cannot be negative, \(n = 13\).

Teacher's Note:
a) Remember that the sum of exterior angles of any convex polygon is always \(360^{\circ}\).
b) Reject extraneous negative roots when solving for the number of sides or polygon parameters.

(b) ABCD is a parallelogram, E is the midpoint of AB and F is the mid-point of CD. PQ is any line that intersects AD, EF and BC at P, G and Q. Prove that \(PG = GQ\). [3 Marks]

[Figure: Parallelogram ABCD with midpoint E on AB and F on DC. Line segment EF is parallel to AD and BC. Transversal PQ intersects AD at P, EF at G, and BC at Q.]

Answer:
1. Since ABCD is a parallelogram, \(AB \parallel DC\) and \(AD \parallel BC\).
2. E and F are midpoints of \(AB\) and \(CD\) respectively, so \(AE = \frac{1}{2}AB\) and \(DF = \frac{1}{2}DC\). Since \(AB = DC\), \(AE = DF\).
3. Also, \(AE \parallel DF\) because \(AB \parallel DC\). Thus, quadrilateral AEFD is a parallelogram.
4. Consequently, \(AD \parallel EF \parallel BC\).
5. By the intercept theorem, if a set of parallel lines (\(AD, EF, BC\)) makes equal intercepts on one transversal, it makes equal intercepts on any other transversal.
6. Since EF is midway between AD and BC, the line segment PQ is cut such that \(PG = GQ\).

Hence proved.

Teacher's Note:
a) State the properties of parallelograms and midpoints clearly to establish that intermediate segments are parallel.
b) Invoke the Intercept Theorem correctly for parallel lines and transversals.

(c) A man borrows Rs. 5000 at \(12\%\) p.a. compound interest. He repays Rs. 2000 at the end of each year. Calculate the amount he has to pay at the end of the third year. [4 Marks]

Answer:
1. Principal for 1st year (\(P_1\)) = Rs. 5000, Rate (\(r\)) = \(12\%\) p.a.
2. Amount at the end of 1st year = \(5000 \times \left(1 + \frac{12}{100}\right) = 5000 \times 1.12 = \text{Rs. } 5600\).
3. Principal for 2nd year after repayment of Rs. 2000 = \(5600 - 2000 = \text{Rs. } 3600\).
4. Amount at the end of 2nd year = \(3600 \times 1.12 = \text{Rs. } 4032\).
5. Principal for 3rd year after repayment of Rs. 2000 = \(4032 - 2000 = \text{Rs. } 2032\).
6. Amount at the end of 3rd year = \(2032 \times 1.12 = \text{Rs. } 2275.84\).

Teacher's Note:
a) For yearly repayments under compound interest, calculate the amount accumulated each year, subtract the repayment to find the new principal, and repeat.
b) Keep track of decimal places carefully during multiplication.

SECTION - B (40 Marks)

(Answer any four questions from this Section)

Q. 5. (a) A wire is bent to form a square enclosing an area of \(484\text{ m}^2\). Using the same wire, a circle is formed. Find the area of the circle. [3 Marks]

Answer:
1. Area of the square = \((\text{side})^2 = 484\text{ m}^2\).
2. Side of the square = \(\sqrt{484} = 22\text{ m}\>.
3. Perimeter of the square = \(4 \times \text{side} = 4 \times 22 = 88\text{ m}\).
4. Since the same wire is used, the circumference of the circle equals the perimeter of the square: \(2\pi r = 88\text{ m}\).
5. Solve for radius \(r\): \(2 \times \frac{22}{7} \times r = 88 \implies r = \frac{88 \times 7}{44} = 14\text{ m}\).
6. Area of the circle = \(\pi r^2 = \frac{22}{7} \times (14)^2 = \frac{22}{7} \times 196 = 616\text{ m}^2\).

Teacher's Note:
a) Perimeter remains constant when reshaping a wire from one geometric figure to another.
b) Use \(\pi = \frac{22}{7}\) to simplify calculations involving multiples of 7.

(b) Given, \(\sin \theta = \frac{p}{q}\), find \(\cos \theta + \sin \theta\) in terms of \(p\) and \(q\). [3 Marks]

[Figure: A right-angled triangle with angle \(\theta\), opposite side \(p\), and hypotenuse \(q\).]

Answer:
1. In a right-angled triangle, \(\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{p}{q}\).
2. Using Pythagoras theorem, \(\text{Base} = \sqrt{\text{Hypotenuse}^2 - \text{Perpendicular}^2} = \sqrt{q^2 - p^2}\).
3. Therefore, \(\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{\sqrt{q^2 - p^2}}{q}\).
4. Adding \(\cos \theta\) and \(\sin \theta\):
\(\cos \theta + \sin \theta = \frac{\sqrt{q^2 - p^2}}{q} + \frac{p}{q} = \frac{p + \sqrt{q^2 - p^2}}{q}\).

Teacher's Note:
a) Express all trigonometric ratios in terms of triangle side lengths using trigonometric definitions.
b) Combine terms under a common denominator once calculated.

(c) If the points \((a, 0)\), \((0, b)\) and \((1, 1)\) are collinear, then prove that \(\frac{1}{a} + \frac{1}{b} = 1\). [4 Marks]

Answer:
1. For three points to be collinear, the area of the triangle formed by them must be zero.
2. Area formula: \(\frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] = 0\).
3. Substitute the points \((a, 0)\), \((0, b)\) and \((1, 1)\):
\(\frac{1}{2} [a(b - 1) + 0(1 - 0) + 1(0 - b)] = 0\).
4. Simplify: \(a(b - 1) - b = 0 \implies ab - a - b = 0 \implies ab = a + b\).
5. Divide throughout by \(ab\) (\(ab \neq 0\)):
\(\frac{ab}{ab} = \frac{a}{ab} + \frac{b}{ab} \implies 1 = \frac{1}{b} + \frac{1}{a}\), which gives \(\frac{1}{a} + \frac{1}{b} = 1\).

Teacher's Note:
a) Alternatively, students can use the slope condition (\(\text{slope between point 1 and 2} = \text{slope between point 2 and 3}\)) to prove collinearity.
b) Divide the resulting equation by the product of variables to match the required proof format.

Q. 6. (a) Factorise: \((x^2 + y^2 - z^2)^2 - 4x^2y^2\) [3 Marks]

Answer:
1. Express the expression in the form of difference of squares: \((x^2 + y^2 - z^2)^2 - (2xy)^2\).
2. Apply the formula \(A^2 - B^2 = (A - B)(A + B)\) where \(A = x^2 + y^2 - z^2\) and \(B = 2xy\):
\([ (x^2 + y^2 - z^2) - 2xy ] [ (x^2 + y^2 - z^2) + 2xy ]\).
3. Rearrange terms to form complete squares: \([ (x^2 - 2xy + y^2) - z^2 ] [ (x^2 + 2xy + y^2) - z^2 ]\).
4. Factorize the trinomial squares: \([ (x - y)^2 - z^2 ] [ (x + y)^2 - z^2 ]\).
5. Apply difference of squares again to each bracket:
\((x - y - z)(x - y + z)(x + y - z)(x + y + z)\).

Teacher's Note:
a) Recognise compound expressions as standard algebraic identities like \(A^2 - B^2\).
b) Factorize completely into linear factors where possible.

(b) Prove that if the diagonals of a parallelogram cut at right angles, it is a rhombus. [4 Marks]

[Figure: Parallelogram ABCD with diagonals AC and BD intersecting at right angles at point O.]

Answer:
1. Let ABCD be a parallelogram whose diagonals AC and BD intersect at O at right angles, so \(\angle AOB = 90^{\circ}\).
2. In triangles \(\Delta OAB\) and \(\Delta OBC\):
- \(OA = OC\) (Diagonals of a parallelogram bisect each other)
- \(\angle AOB = \angle BOC = 90^{\circ}\) (Given)
- \(OB = OB\) (Common side)
3. By SAS congruence criterion, \(\Delta OAB \cong \Delta OBC\).
4. Therefore, \(AB = BC\) (Corresponding parts of congruent triangles are equal).
5. Since opposite sides of a parallelogram are equal (\(AB = DC\) and \(BC = AD\)), we have \(AB = BC = CD = DA\).
6. A parallelogram with all four sides equal is a rhombus. Hence proved.

Teacher's Note:
a) Use triangle congruence to establish adjacent side equality.
b) State all properties used clearly, such as diagonal bisection and side relationships in parallelograms.

(c) If \(3a = p\left(\frac{x}{2} - y\right)\), make 'y' the subject. Find y, when \(x = 4\), \(p = 5\), \(a = 32\). [3 Marks]

Answer:
1. Starting equation: \(3a = p\left(\frac{x}{2} - y\right)\).
2. Divide by \(p\): \(\frac{3a}{p} = \frac{x}{2} - y\).
3. Rearrange to make \(y\) the subject: \(y = \frac{x}{2} - \frac{3a}{p} = \frac{px - 6a}{2p}\).
4. Substitute the given values \(a = 32\), \(x = 4\), \(p = 5\):
\(y = \frac{5(4) - 6(32)}{2(5)} = \frac{20 - 192}{10} = \frac{-172}{10} = -17.2\).

Teacher's Note:
a) Isolate the term containing the subject variable first before evaluating numerical values.
b) Pay close attention to negative signs during arithmetic simplification.

Q. 7. (a) Draw the graph of the equations \(2x - 3y = 7\) and \(x + 6y = 11\), taking \(1\text{ cm} = 1\text{ unit}\) on both axes and find their solutions. [6 Marks]

Answer:
1. For equation \(2x - 3y = 7 \implies x = \frac{7 + 3y}{2}\):
- If \(y = 0\), \(x = 3.5\)
- If \(y = 1\), \(x = 5\)
- If \(y = 3\), \(x = 8\)
2. For equation \(x + 6y = 11 \implies x = 11 - 6y\):
- If \(y = 1\), \(x = 5\)
- If \(y = 2\), \(x = -1\)
- If \(y = 3\), \(x = -7\)
3. Plot these points on graph paper and draw the two straight lines.
4. The two lines intersect at the point \((5, 1)\).
5. Hence, the solution set is \(x = 5\), \(y = 1\).

Teacher's Note:
a) Plot at least three points for each linear equation to ensure accuracy.
b) Clearly label axes, scale, and intersection point on the graph.

(b) In the given figure, area of parallelogram AFEC is \(140\text{ cm}^2\). Find the area of:
i. Parallelogram BFED
ii. \(\Delta BFD\) [4 Marks]

[Figure: Parallelograms AFEC and BFED sharing a common base FE between parallel lines AD and FE.]

Answer:
i. Parallelograms BFED and AFEC lie on the same base FE and between the same parallel lines \(AD \parallel FE\). Therefore, their areas are equal.
\(\text{ar}(\text{||gm BFED}) = \text{ar}(\text{||gm AFEC}) = 140\text{ cm}^2$.
ii. \(\Delta BFD\) and parallelogram BFED are on the same base BD and between the same parallel lines BD and FE.
Therefore, the area of the triangle is half the area of the parallelogram:
\(\text{Area}(\Delta BFD) = \frac{1}{2} \times \text{ar}(\text{||gm BFED}) = \frac{1}{2} \times 140 = 70\text{ cm}^2\).

Teacher's Note:
a) Parallelograms on the same base and between the same parallels are equal in area.
b) The area of a triangle is half the area of a parallelogram on the same base and between the same parallels.

Q. 8. (a) Show that in any quadrilateral the sum of all the four sides exceeds the sum of the diagonals. [4 Marks]

[Figure: Quadrilateral PQRS with diagonals PR and QS intersecting at an interior point.]

Answer:
1. Let PQRS be a quadrilateral with diagonals PR and QS intersecting at O.
2. Apply the triangle inequality theorem (sum of any two sides of a triangle is greater than the third side) in the four triangles formed by the diagonals:
- In \(\Delta PQS\): \(SP + PQ > QS\) ---(1)
- In \(\Delta PQR\): \(PQ + QR > PR\) ---(2)
- In \(\Delta QRS\): \(QR + RS > QS\) ---(3)
- In \(\Delta RSP\): \(RS + SP > PR\) ---(4)
3. Adding inequalities (1), (2), (3) and (4):
\(2(PQ + QR + RS + SP) > 2(PR + QS)\).
4. Divide by 2: \(PQ + QR + RS + SP > PR + QS\).
Hence proved.

Teacher's Note:
a) The triangle inequality is the fundamental theorem used for side-diagonal length comparison problems.
b) Setting up the inequalities systematically for each triangle ensures no side is missed.

(b) A and B start at the same time from two places 30 km apart. If they walk in the same directions, A overtakes B in 10 hours and if they walk in opposite directions they meet in 2 hours. Find the rates of walking of A and B. [6 Marks]

Answer:
1. Let A's speed be \(x\text{ km/hr}\) and B's speed be \(y\text{ km/hr}\) (where \(x > y\)).
2. When walking in the same direction, relative speed is \((x - y)\text{ km/hr}\). Distance = 30 km, time = 10 hours.
\(10(x - y) = 30 \implies x - y = 3\) ---(i)
3. When walking in opposite directions, relative speed is \((x + y)\text{ km/hr}\). Distance = 30 km, time = 2 hours.
\(2(x + y) = 30 \implies x + y = 15\) ---(ii)
4. Add equations (i) and (ii):
\((x - y) + (x + y) = 3 + 15 \implies 2x = 18 \implies x = 9\text{ km/hr}\).
5. Substitute \(x = 9\) into equation (ii):
\(9 + y = 15 \implies y = 6\text{ km/hr}\).
6. Hence, A's speed is \(9\text{ km/hr}\) and B's speed is \(6\text{ km/hr}\).

Teacher's Note:
a) For objects moving in the same direction, subtract speeds; for opposite directions, add speeds.
b) Form simultaneous linear equations and solve using elimination or substitution.

Q. 9. (a) The mean height of the 10 girls in a class is \(1.38\text{ m}\) and the mean height of the 40 boys is \(1.44\text{ m}\). Find the mean height of the 50 students of the class. [3 Marks]

Answer:
1. Sum of heights of 10 girls = \(10 \times 1.38 = 13.8\text{ m}\).
2. Sum of heights of 40 boys = \(40 \times 1.44 = 57.6\text{ m}\).
3. Total sum of heights of all 50 students = \(13.8 + 57.6 = 71.4\text{ m}\).
4. Mean height of 50 students = \(\frac{\text{Total Sum}}{\text{Total Students}} = \frac{71.4}{50} = 1.428\text{ m}\).

Teacher's Note:
a) Combined mean is calculated by finding the total sum of all observations divided by the total frequency.
b) Do not simply average the two given means since the group sizes are unequal.

(b) In the given fig., \(m\angle D = 90^{\circ}\), \(AB = 8\text{ cm}\), \(BC = 6\text{ cm}\) and \(CA = 3\text{ cm}\). Find CD. [4 Marks]

[Figure: Right-angled triangle ADC with altitude from A or related segments where AC = 3 cm, AB = 8 cm, BC = 6 cm, and \(\angle D = 90^{\circ}\).]

Answer:
1. Let \(CD = x\text{ cm}\).
2. In right-angled \(\Delta ADC\) (\(\angle D = 90^{\circ}\)):
\(AD^2 + CD^2 = AC^2 \implies AD^2 + x^2 = 3^2 = 9 \implies AD^2 = 9 - x^2\).
3. In right-angled \(\Delta ADB\) (\(\angle D = 90^{\circ}\)):
\(AD^2 + BD^2 = AB^2\). Since \(BD = BC + CD = 6 + x\), we have:
\((9 - x^2) + (6 + x)^2 = 8^2\).
4. Expand and simplify: \(9 - x^2 + 36 + 12x + x^2 = 64 \implies 45 + 12x = 64\).
5. Solve for \(x\): \(12x = 64 - 45 = 19 \implies x = \frac{19}{12}\text{ cm} = 1\frac{7}{12}\text{ cm}\).

Teacher's Note:
a) Apply Pythagoras theorem repeatedly to connected right-angled triangles sharing a common side (\(AD\)).
b) Express the final fractional answer clearly in mixed fraction or decimal form.

(c) A rectangular water-tank measuring \(80\text{ cm} \times 60\text{ cm} \times 60\text{ cm}\) is filled from a pipe of cross-sectional area \(1.5\text{ cm}^2\), the water emerging at \(3.2\text{ m/s}\). How long does it take to fill the tank? [3 Marks]

Answer:
1. Volume of the rectangular tank = \(80 \times 60 \times 60 = 288,000\text{ cm}^3\).
2. Convert water flow speed from m/s to cm/s: \(3.2\text{ m/s} = 3.2 \times 100\text{ cm/s} = 320\text{ cm/s}\).
3. Volume of water flowing per second = \(\text{Cross-sectional area} \times \text{Speed} = 1.5\text{ cm}^2 \times 320\text{ cm/s} = 480\text{ cm}^3\text{/s}\).
4. Volume of water flowing in 1 minute = \(480 \times 60 = 28,800\text{ cm}^3\text{/min}\).
5. Time required to fill the tank = \(\frac{\text{Total volume}}{\text{Rate per minute}} = \frac{288,000}{28,800} = 10\text{ minutes}\).

Teacher's Note:
a) Ensure all units are consistent (convert meters to centimeters) before calculating volumes and flow rates.
b) Rate of flow volume equals cross-sectional area multiplied by linear speed per unit time.

Q. 10. (a) Find the value of \(\frac{\sec(90^{\circ} - \theta)\csc \theta - \tan(90^{\circ} - \theta)\cot \theta + \cos^2 25^{\circ} + \cos^2 65^{\circ}}{3\tan 27^{\circ}\tan 63^{\circ}}\) [3 Marks]

Answer:
1. Use complementary angle identities: \(\sec(90^{\circ} - \theta) = \csc \theta\), \(\tan(90^{\circ} - \theta) = \cot \theta\), and \(\cos(90^{\circ} - 65^{\circ}) = \sin 65^{\circ}\).
2. Substitute into the numerator: \(\csc \theta \cdot \csc \theta - \cot \theta \cdot \cot \theta + \sin^2 65^{\circ} + \cos^2 65^{\circ} = \csc^2 \theta - \cot^2 \theta + 1\).
3. Since \(\csc^2 \theta - \cot^2 \theta = 1\), the numerator becomes \(1 + 1 = 2\).
4. In the denominator, \(\tan 27^{\circ} = \tan(90^{\circ} - 63^{\circ}) = \cot 63^{\circ}\). Thus, \(3\tan 27^{\circ}\tan 63^{\circ} = 3\cot 63^{\circ}\tan 63^{\circ} = 3(1) = 3\).
5. Evaluating the expression gives \(\frac{2}{3}\).

Teacher's Note:
a) Use standard trigonometric identities like \(\csc^2 \theta - \cot^2 \theta = 1\) and \(\sin^2 \theta + \cos^2 \theta = 1\).
b) Convert complementary angle pairs so they cancel out to 1.

(b) Construct a rhombus ABCD in which \(AB = 4.5\text{ cm}\) and \(m\angle A = 60^{\circ}\). [3 Marks]

Answer:
1. Draw a line segment \(AB = 4.5\text{ cm}\).
2. At point A, construct an angle \(\angle BAE = 60^{\circ}\) using a compass.
3. Cut off \(AD = 4.5\text{ cm}\) along ray AE.
4. With D as centre and radius \(4.5\text{ cm}\), draw an arc.
5. With B as centre and radius \(4.5\text{ cm}\), draw another arc to intersect the previous arc at C.
6. Join DC and BC. ABCD is the required rhombus.

Teacher's Note:
a) Recall that all four sides of a rhombus are equal in length.\br />b) Construct angles accurately using a compass and straightedge.

(c) Two chords AB and CD of lengths \(5\text{ cm}\) and \(11\text{ cm}\) respectively of a circle are parallel to each other and on opposite sides of the centre. If the distance between AB and CD is \(6\text{ cm}\), find the radius of the circle. [4 Marks]

[Figure: Circle with centre O, parallel chords AB and CD on opposite sides, perpendiculars OM and ON from centre to chords, forming right-angled triangles.]

Answer:
1. Draw perpendiculars \(OM \perp AB\) and \(ON \perp CD\). Perpendicular from the centre bisects the chord.
2. \(BM = \frac{AB}{2} = \frac{5}{2}\text{ cm}\) and \(DN = \frac{CD}{2} = \frac{11}{2}\text{ cm}\).
3. Let \(ON = x\text{ cm}\). Since the distance between chords is \(6\text{ cm}\), \(OM = (6 - x)\text{ cm}\).
4. In right \(\Delta MOB\), \(OB^2 = OM^2 + MB^2 \implies r^2 = (6 - x)^2 + \left(\frac{5}{2}\right)^2 = 36 + x^2 - 12x + \frac{25}{4}\) ---(1).
5. In right \(\Delta NOD\), \(OD^2 = ON^2 + ND^2 \implies r^2 = x^2 + \left(\frac{11}{2}\right)^2 = x^2 + \frac{121}{4}\) ---(2).
6. Equate (1) and (2) (since radii are equal):
\(36 + x^2 - 12x + \frac{25}{4} = x^2 + \frac{121}{4}\).
7. Simplify: \(36 - 12x + \frac{25}{4} = \frac{121}{4} \implies 12x = 36 + \frac{25}{4} - \frac{121}{4} = 36 - \frac{96}{4} = 36 - 24 = 12 \implies x = 1\).
8. Substitute \(x = 1\) into equation (2):
\(r^2 = 1^2 + \frac{121}{4} = 1 + 30.25 = 31.25 = \frac{125}{4} \implies r = \frac{5\sqrt{5}}{2}\text{ cm}\).

Teacher's Note:
a) Perpendiculars from the centre to parallel chords on opposite sides divide the total distance between them into two segments summing to the total distance.
b) Use the radius equation from both triangles to set up a linear equation in \(x\).

Q. 11. (a) If \(a + \frac{1}{a} = p\), show that \(a^3 + \frac{1}{a^3} = p(p^2 - 3)\). [3 Marks]

Answer:
1. Given: \(a + \frac{1}{a} = p\).
2. Cube both sides of the equation: \(\left(a + \frac{1}{a}\right)^3 = p^3\).
3. Expand using algebraic identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\):
\(a^3 + \frac{1}{a^3} + 3(a)\left(\frac{1}{a}\right)\left(a + \frac{1}{a}\right) = p^3\).
4. Simplify: \(a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) = p^3\).
5. Substitute \(a + \frac{1}{a} = p\):
\(a^3 + \frac{1}{a^3} + 3p = p^3 \implies a^3 + \frac{1}{a^3} = p^3 - 3p = p(p^2 - 3)\).
Hence proved.

Teacher's Note:
a) Cubic expansions for reciprocal sums are standard algebraic identities used frequently in polynomials.
b) Factor out common terms neatly to reach the exact required expression format.

(b) Solve: \(\frac{5}{x + y} + \frac{3}{x - y} = 4\), \(\frac{2}{x + y} + \frac{5}{x - y} = 5\frac{2}{5}\) [4 Marks]

Answer:
1. Let \(\frac{1}{x + y} = a\) and \(\frac{1}{x - y} = b\). The equations become:
\(5a + 3b = 4\) ---(i)
\(2a + 5b = \frac{27}{5}\) ---(ii)
2. Multiply equation (i) by 2 and equation (ii) by 5:
\(10a + 6b = 8\) ---(iii)
\(10a + 25b = 27\) ---(iv)
3. Subtract equation (iii) from equation (iv):
\(19b = 19 \implies b = 1\).
4. Substitute \(b = 1\) into equation (i):
\(5a + 3(1) = 4 \implies 5a = 1 \implies a = \frac{1}{5}\).
5. Convert back to original variables:
\(\frac{1}{x + y} = \frac{1}{5} \implies x + y = 5\) ---(v)
\(\frac{1}{x - y} = 1 \implies x - y = 1\) ---(vi)
6. Add equations (v) and (vi): \(2y = 4 \implies y = 2\).
7. Substitute \(y = 2\) into equation (vi): \(x - 2 = 1 \implies x = 3\).
8. Solution: \(x = 3, y = 2\).

Teacher's Note:
a) Equations reducible to linear form by substitution simplify complex fractional simultaneous equations.
b) Solve in two stages: first solve for the substitution variables, then solve for the original variables \(x\) and \(y\).

(c) Find the mean of the following data: [3 Marks]

Answer:
1. Set up a frequency distribution table with columns for \(x\), \(f\), and \(f \cdot x\):
- \(x = 25, f = 10 \implies f \cdot x = 250\)
- \(x = 35, f = 6 \implies f \cdot x = 210\)
- \(x = 45, f = 8 \implies f \cdot x = 360\)
- \(x = 55, f = 12 \implies f \cdot x = 660\)
- \(x = 65, f = 5 \implies f \cdot x = 325\)
- \(x = 75, f = 9 \implies f \cdot x = 675\)
2. Calculate sums: \(\sum f = 10 + 6 + 8 + 12 + 5 + 9 = 50\).
\(\sum fx = 250 + 210 + 360 + 660 + 325 + 675 = 2,480\).
3. Calculate the mean: \(\text{Mean} = \frac{\sum fx}{\sum f} = \frac{2480}{50} = 49.6\).

Teacher's Note:
a) Use the direct mean formula \(\bar{x} = \frac{\sum fx}{\sum f}\) for discrete frequency distributions.
b) Double-check multiplication and column summation to prevent arithmetic errors.