ICSE Class 9 Mathematics Sample Paper with Solutions Set 04

SECTION - A (40 Marks)

(Answer all questions from this Section)

Q. 1. (a) Calculate the amount and the compound interest on Rs. 6000 at 10% p.a. for \( 1\frac{1}{2} \) years, when the interest is compounded half yearly. [3 Marks]

Answer:
Given, Principle \( (P) = \text{Rs. } 6000 \)
Rate \( (R) = 10\% \) p.a. = \( 5\% \) per half-year
Time \( (n) = 1\frac{1}{2} \) years = \( 3 \) half-years
\( A = P\left(1 + \frac{R}{2 \times 100}\right)^{2n} \)
\( A = 6000\left(1 + \frac{10}{2 \times 100}\right)^3 \)
\( A = 6000\left(1 + \frac{5}{100}\right)^3 \)
\( A = 6000 \times (1.05)^3 = \text{Rs. } 6945.75 \)
Compound Interest \( (C.I.) = A - P = 6945.75 - 6000 = \text{Rs. } 945.75 \)

Teacher's Note:
a) When interest is compounded half-yearly, the rate is halved and the time period in years is doubled to find the total number of conversion periods.
b) Students must ensure calculations involving decimals are carried out carefully to avoid rounding errors before the final step.

(b) If \( \frac{\sqrt{11} - \sqrt{7}}{\sqrt{11} + \sqrt{7}} = x - y\sqrt{77} \), find the values \( x \) and \( y \). [3 Marks]

Answer:
L.H.S. = \( \frac{\sqrt{11} - \sqrt{7}}{\sqrt{11} + \sqrt{7}} \)
Rationalising the denominator:
\( \frac{(\sqrt{11} - \sqrt{7})(\sqrt{11} - \sqrt{7})}{(\sqrt{11} + \sqrt{7})(\sqrt{11} - \sqrt{7})} = x - y\sqrt{77} \)
\( \frac{11 - \sqrt{77} - \sqrt{77} + 7}{11 - 7} = x - y\sqrt{77} \)
\( \frac{18 - 2\sqrt{77}}{4} = x - y\sqrt{77} \)
\( \frac{18}{4} - \frac{2}{4}\sqrt{77} = x - y\sqrt{77} \)
\( \frac{9}{2} - \frac{1}{2}\sqrt{77} = x - y\sqrt{77} \)
Comparing both sides, we get:
\( x = \frac{9}{2} \) and \( y = \frac{1}{2} \)

Teacher's Note:
a) Rationalisation of denominators with binomial surds is performed by multiplying the numerator and denominator by the conjugate of the denominator.
b) Compare the rational and irrational parts on both sides of the resulting equation accurately to obtain the final values of \( x \) and \( y \).

(c) Sonu and Monu had adjacent triangular fields with a common boundary of \( 25\text{ m} \). The other two sides of Sonu's field were \( 52\text{ m} \) and \( 63\text{ m} \), while Monu's were \( 114\text{ m} \) and \( 101\text{ m} \). If the cost of fertilization is Rs 20 per sq m, then find the total cost of fertilization for both of Sonu and Monu together. [4 Marks]

Answer:
For Sonu's triangular field \( \Delta ABD \):
Sides \( a = 52\text{ m}, b = 25\text{ m}, c = 63\text{ m} \)
Semi-perimeter \( s = \frac{52 + 25 + 63}{2} = 70\text{ m} \)
Area \( = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{70(18)(45)(7)} = \sqrt{396900} = 630\text{ m}^2 \)
For Monu's triangular field \( \Delta BCD \):
Sides \( a = 114\text{ m}, b = 25\text{ m}, c = 101\text{ m} \)
Semi-perimeter \( s = \frac{114 + 25 + 101}{2} = 120\text{ m} \)
Area \( = \sqrt{120(120 - 114)(120 - 25)(120 - 101)} = \sqrt{120 \times 6 \times 95 \times 19} = \sqrt{1299600} = 1140\text{ m}^2 \)
Total area of both fields \( = 630 + 1140 = 1770\text{ m}^2 \)
Cost of fertilization per sq m = Rs 20
Total cost \( = 1770 \times 20 = \text{Rs. } 35,400 \)

Teacher's Note:
a) Heron's formula is applied independently for both triangles sharing the common side.
b) Ensure proper simplification under the square root to avoid calculation errors.

Q. 2. (a) Calculate the area of fig., ABCDE. Given DX = \( 9\text{ cm} \), DC = \( 5\text{ cm} \), FC = \( 4\text{ cm} \) and XB = \( 6\text{ cm} \). Also F and X are the mid-points of EC and AB respectively. [3 Marks]

[Figure: Triangle ABC with altitude DX perpendicular to AB, point E on AC and point F on EC such that DF is perpendicular to EC. Measurements: DX = 9 cm, DC = 5 cm, FC = 4 cm, XB = 6 cm.]

Answer:
In right-angled \( \Delta DFC \), \( DC^2 = DF^2 + FC^2 \) (by Pythagoras Theorem)
\( 5^2 = DF^2 + 4^2 \)
\( 25 - 16 = DF^2 \implies DF = 3\text{ cm} \)
Area of \( \Delta DEC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (4 + 4) \times 3 = 12\text{ cm}^2 \)
\( FX = DX - DF = 9 - 3 = 6\text{ cm} \)
Area of trapezium \( CEBA = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (4 + 4 + 6 + 6) \times 6 = 60\text{ cm}^2 \)
Total area of figure ABCDE \( = \text{Area of } \Delta DEC + \text{Area of trapezium CEBA} = 12 + 60 = 72\text{ cm}^2 \)

Teacher's Note:
a) Split composite geometric figures into standard triangles and quadrilaterals to compute areas systematically.
b) Pay close attention to segment lengths derived from mid-point properties.

(b) If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal. [4 Marks]

[Figure: A circle with centre O and diameter PQ passing through the point of intersection E of two chords AB and CD, such that \( \angle AEQ = \angle DEQ \).]

Answer:
Given: Chords AB and CD intersect at E inside a circle with centre O. PQ is a diameter through E such that \( \angle AEQ = \angle DEQ \).
To prove: \( AB = CD \).
Construction: Draw perpendiculars \( OL \perp AB \) and \( OM \perp CD \).
Proof: In \( \Delta OLE \) and \( \Delta OME \),
\( \angle LEO = \angle MEO \) (since \( \angle AEQ = \angle DEQ \))
\( \angle OLE = \angle OME = 90^{\circ} \)
\( EO = EO \) (common)
Therefore, \( \Delta OLE \cong \Delta OME \) (by AAS congruence criterion)
Hence, \( OL = OM \) (CPCTC)
Since chords equidistant from the centre are equal, \( AB = CD \).

Teacher's Note:
a) Perpendicular from the centre of a circle to a chord bisects the chord.
b) Chords that are equidistant from the centre of a circle are equal in length.

(c) Find \( x \), if \( (\sqrt[3]{4})^{2x + \frac{1}{2}} = (\sqrt[3]{8})^5 \). [3 Marks]

Answer:
\( (\sqrt[3]{4})^{2x + \frac{1}{2}} = (\sqrt[3]{8})^5 \)
\( (4^{1/3})^{2x + \frac{1}{2}} = (8^{1/3})^5 \)
\( ((2^2)^{1/3})^{2x + \frac{1}{2}} = ((2^3)^{1/3})^5 \)
\( (2^{2/3})^{2x + \frac{1}{2}} = (2^1)^5 \)
\( 2^{\frac{2}{3}\left(2x + \frac{1}{2}\right)} = 2^5 \)
Equating powers of same base 2:
\( \frac{2}{3}\left(2x + \frac{1}{2}\right) = 5 \)
\( 2x + \frac{1}{2} = 5 \times \frac{3}{2} \)
\( 2x + \frac{1}{2} = \frac{15}{2} \)
\( 2x = \frac{15}{2} - \frac{1}{2} = \frac{14}{2} = 7 \)
\( x = \frac{7}{2} \)

Teacher's Note:
a) Express all terms with the same base before equating indices.
b) Apply laws of exponents systematically during simplification.

Q. 3. (a) Given \( \log x = a + b \) and \( \log y = a - b \), find the value of \( \log \frac{10x}{y^2} \) in terms of 'a' and 'b'. [3 Marks]

Answer:
\( \log \frac{10x}{y^2} = \log 10x - \log y^2 \) (using quotient law)
\( = \log 10 + \log x - 2\log y \) (using product and power laws)
Substitute \( \log 10 = 1 \), \( \log x = a + b \), and \( \log y = a - b \):
\( = 1 + (a + b) - 2(a - b) \)
\( = 1 + a + b - 2a + 2b \)
\( = 1 - a + 3b \)

Teacher's Note:
a) Recall the standard logarithmic properties: \( \log(mn) = \log m + \log n \), \( \log(m/n) = \log m - \log n \), and \( \log(m^n) = n\log m \).
b) Be careful with signs when substituting and expanding expressions containing negative coefficients.

(b) The bisector of \( \angle A \) of a \( \Delta ABC \) meets BC at D and BC is produced to E. prove that \( \angle ABC + \angle ACE = 2\angle ADC \). [3 Marks]

[Figure: Triangle ABC with AD as the bisector of \( \angle A \), and side BC produced to E.]

Answer:
Proof:
Let \( \angle BAD = \angle DAC = x \) and \( \angle ABC = y \).
Using the exterior angle property in \( \Delta ACE \):
\( \angle ACE = \angle ABC + \angle BAC = y + 2x \)
Now, consider \( \Delta ABD \):
Exterior angle \( \angle ADC = \angle DAB + \angle ABD = x + y \)
Adding \( \angle ABC \) and \( \angle ACE \):
\( \angle ABC + \angle ACE = y + (y + 2x) = 2y + 2x = 2(x + y) \)
Since \( x + y = \angle ADC \), therefore:
\( \angle ABC + \angle ACE = 2\angle ADC \).

Teacher's Note:
a) The exterior angle of a triangle is equal to the sum of the two interior opposite angles.
b) Clearly define variables for angles to simplify algebraic substitution.

(c) Using ruler and compass only, construct a trapezium ABCD, in which the parallel sides AB and DC are \( 3.3\text{ cm} \) apart; AB = \( 4.5\text{ cm} \), \( \angle A = 120^{\circ} \), BC = \( 3.6\text{ cm} \) and \( \angle B \) is obtuse. [4 Marks]

Answer:
Steps of Construction:
1. Draw line segment AB = \( 4.5\text{ cm} \).
2. At A, construct \( \angle BAS = 120^{\circ} \) and draw perpendicular \( EA \perp AB \).
3. From A, cut an arc of measure \( 3.3\text{ cm} \) on EA to locate point X such that AX = \( 3.3\text{ cm} \).
4. Through X, draw a line QP parallel to AB which cuts AS at D.
5. With B as centre, draw an arc of radius \( 3.6\text{ cm} \) to intersect line QP at C.
6. Join CB.
Thus, ABCD is the required trapezium.

Teacher's Note:
a) Ensure all construction arcs and lines are clearly visible.
b) Parallel lines are constructed by maintaining a constant perpendicular distance.

Q. 4. (a) The following figure shows a right-angled triangle ABC with \( \angle B = 90^{\circ} \), AB = \( 15\text{ cm} \) and AC = \( 25\text{ cm} \). D is a point in side BC and CD = \( 7\text{ cm} \). If \( DE \perp AC \), find the length of DE. [4 Marks]

[Figure: Right-angled triangle ABC with right angle at B, AB = 15 cm, AC = 25 cm, point D on BC with CD = 7 cm, and DE perpendicular to AC.]

Answer:
In right-angled \( \Delta ABC \):
\( AB^2 + BC^2 = AC^2 \) (by Pythagoras theorem)
\( 15^2 + BC^2 = 25^2 \)
\( 225 + BC^2 = 625 \implies BC^2 = 400 \implies BC = 20\text{ cm} \)
Given \( BC = BD + CD \implies 20 = BD + 7 \implies BD = 13\text{ cm} \)
In right-angled \( \Delta ABD \):
\( AD^2 = AB^2 + BD^2 = 15^2 + 13^2 = 225 + 169 = 394 \implies AD = \sqrt{394} \approx 19.85\text{ cm} \)
Let \( CE = x \), then in right-angled \( \Delta CDE \):
\( DE^2 = CD^2 - CE^2 = 7^2 - x^2 = 49 - x^2 \)
In right-angled \( \Delta AED \):
\( DE^2 = AD^2 - AE^2 = 19.85^2 - (25 - x)^2 \)
Equating both expressions for \( DE^2 \):
\( 49 - x^2 = 19.85^2 - (625 - 50x + x^2) \)
Solving for \( x \) yields \( x \approx 5.63\text{ cm} \).
Therefore, \( DE = \sqrt{49 - 5.63^2} \approx 4.2\text{ cm} \).

Teacher's Note:
a) Apply Pythagoras theorem repeatedly in multiple right-angled triangles within the figure.
b) Maintain intermediate precision to avoid accumulation of rounding errors.

(b) Prove that the interior angle of a regular pentagon is three times the exterior angle of a regular decagon. [3 Marks]

Answer:
Each interior angle of a regular polygon of \( n \) sides is given by \( \frac{(2n - 4) \times 90^{\circ}}{n} \).
For a regular pentagon (\( n = 5 \)):
Each interior angle \( = \frac{(2(5) - 4) \times 90^{\circ}}{5} = \frac{6 \times 90^{\circ}}{5} = 108^{\circ} \)
Each exterior angle of a regular polygon of \( n \) sides is given by \( \frac{360^{\circ}}{n} \).
For a regular decagon (\( n = 10 \)):
Each exterior angle \( = \frac{360^{\circ}}{10} = 36^{\circ} \)
Since \( 108^{\circ} = 3 \times 36^{\circ} \), the interior angle of a regular pentagon is three times the exterior angle of a regular decagon.

Teacher's Note:
a) Memorise standard formulae for interior and exterior angles of regular polygons.
b) Clearly state the number of sides \( n \) for each polygon before computation.

(c) If \( \tan \theta + \cot \theta = 3 \), find the value of \( \tan^2 \theta + \cot^2 \theta \). [3 Marks]

Answer:
Given \( \tan \theta + \cot \theta = 3 \)
Squaring both sides:
\( (\tan \theta + \cot \theta)^2 = 3^2 \)
\( \tan^2 \theta + \cot^2 \theta + 2\tan \theta \cot \theta = 9 \)
Since \( \cot \theta = \frac{1}{\tan \theta} \), we have \( \tan \theta \cot \theta = 1 \):
\( \tan^2 \theta + \cot^2 \theta + 2(1) = 9 \)
\( \tan^2 \theta + \cot^2 \theta = 9 - 2 = 7 \)

Teacher's Note:
a) Recognise reciprocal trigonometric identities such as \( \tan \theta \cot \theta = 1 \).
b) Algebraic expansion of binomials applies directly to trigonometric expressions.

SECTION - B (40 Marks)

(Answer any four questions from this Section)

Q. 5. (a) Graphically solve the simultaneous equations: \( x - 2y = 1 \); \( x + y = 4 \). [4 Marks]

Answer:
For equation \( x - 2y = 1 \implies y = \frac{x - 1}{2} \):
When \( x = 1, y = 0 \) ; When \( x = 3, y = 1 \) ; When \( x = 5, y = 2 \). Points: \( (1, 0), (3, 1), (5, 2) \).
For equation \( x + y = 4 \implies y = 4 - x \):
When \( x = 0, y = 4 \) ; When \( x = 2, y = 2 \) ; When \( x = 4, y = 0 \). Points: \( (0, 4), (2, 2), (4, 0) \).
Plotting these points and drawing the lines, the lines intersect at \( (3, 1) \).
Therefore, the solution is \( x = 3 \) and \( y = 1 \).

Teacher's Note:
a) At least three points should be plotted to ensure accuracy when drawing linear graphs.
b) The coordinates of the intersection point represent the simultaneous solution of the equations.

(b) A and B together can do a piece of work in 15 days. If A's one day's work is \( \frac{3}{2} \) times B's one day's work; in how many days can A and B do the work alone? [3 Marks]

Answer:
Let A take \( x \) days and B take \( y \) days to complete the work alone.
A's one day's work \( = \frac{1}{x} \), B's one day's work \( = \frac{1}{y} \).
Given \( \frac{1}{x} = \frac{3}{2} \times \frac{1}{y} \implies y = \frac{3}{2}x \).
Together, their one day's work is \( \frac{1}{x} + \frac{1}{y} = \frac{1}{15} \).
Substitute \( y = \frac{3}{2}x \):
\( \frac{1}{x} + \frac{2}{3x} = \frac{1}{15} \)
\( \frac{3 + 2}{3x} = \frac{1}{15} \implies \frac{5}{3x} = \frac{1}{15} \)
\( 3x = 75 \implies x = 25 \) days.
\( y = \frac{3}{2}(25) = 37.5 \) days.
Hence, A can do the work alone in 25 days and B in 37.5 days.

Teacher's Note:
a) Work problems are solved by equating the sum of individual daily work rates to the combined work rate.
b) Fractions must be simplified carefully while solving for variables.

(c) How many sides does a regular polygon have, each angle of which is of measure \( 108^{\circ} \)? [3 Marks]

Answer:
Let the polygon have \( n \) sides.
Each interior angle \( = \frac{(2n - 4) \times 90^{\circ}}{n} = 108^{\circ} \)
\( (2n - 4) \times 90 = 108n \)
\( 180n - 360 = 108n \)
\( 180n - 108n = 360 \)
\( 72n = 360 \implies n = 5 \)
Hence, the polygon has 5 sides.

Teacher's Note:
a) Equate the interior angle formula directly to the given measure to form a linear equation in \( n \).
b) The number of sides \( n \) must always be a positive integer greater than or equal to 3.

Q. 6. (a) A person invests Rs. 5600 at 14% p.a. compound interest for 2 years. Calculate
i. The interest for the first year.
ii. The amount at the end of the first year.
iii. The interest for the second year corrected to the nearest rupee. [3 Marks]

Answer:
i. Interest for the first year \( = \frac{5600 \times 14 \times 1}{100} = \text{Rs. } 784 \)
ii. Amount at the end of the first year \( = 5600 + 784 = \text{Rs. } 6384 \)
iii. Interest for the second year \( = \frac{6384 \times 14 \times 1}{100} = \text{Rs. } 893.76 \approx \text{Rs. } 894 \) (corrected to the nearest rupee).

Teacher's Note:
a) For compound interest computed annually, the amount at the end of the first year serves as the principal for the second year.
b) Follow rounding instructions carefully when requested to the nearest rupee.

(b) A point P lies on the x-axis and another point Q lies on the y-axis.
i. Write the ordinate of point P.
ii. Write the abscissa of point Q.
iii. If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ. [3 Marks]

Answer:
i. Since point P lies on the x-axis, its ordinate is 0.
ii. Since point Q lies on the y-axis, its abscissa is 0.
iii. Coordinates of P are \( (-12, 0) \) and Q are \( (0, -16) \).
Length of \( PQ = \sqrt{(0 - (-12))^2 + (-16 - 0)^2} = \sqrt{12^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \).

Teacher's Note:
a) Any point on the x-axis has y-coordinate 0, and any point on the y-axis has x-coordinate 0.
b) Apply the distance formula correctly keeping track of negative coordinate signs.

(c) \( \Delta ABC \) is right angled at B. If \( m\angle A = 30^{\circ} \) and BC = \( 8\text{ cm} \). Find the remaining angles and sides. [4 Marks]

[Figure: Right-angled triangle ABC with right angle at B, angle A = 30 degrees, and BC = 8 cm.]

Answer:
In \( \Delta ABC \), \( m\angle B = 90^{\circ} \) and \( m\angle A = 30^{\circ} \):
\( m\angle C = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ} \)
Using trigonometric ratios:
\( \tan 30^{\circ} = \frac{BC}{AB} \implies \frac{1}{\sqrt{3}} = \frac{8}{AB} \implies AB = 8\sqrt{3}\text{ cm} \)
\( \sin 30^{\circ} = \frac{BC}{AC} \implies \frac{1}{2} = \frac{8}{AC} \implies AC = 16\text{ cm} \)

Teacher's Note:
a) Use angle sum property of triangles to find remaining interior angles.
b) Apply standard trigonometric ratios for special angles (\( 30^{\circ}, 60^{\circ} \)) to determine unknown side lengths.

Q. 7. (a) Simplify: \( \frac{3^{n+1}}{3^{n(n-1)}} \div \frac{9^{n+1}}{(3^{n+1})^{n-1}} \) [4 Marks]

Answer:
Expression \( = \frac{3^{n+1}}{3^{n^2-n}} \div \frac{(3^2)^{n+1}}{3^{n^2-1}} \)
\( = \frac{3^{n+1}}{3^{n^2-n}} \div \frac{3^{2n+2}}{3^{n^2-1}} \)
\( = \frac{3^{n+1}}{3^{n^2-n}} \times \frac{3^{n^2-1}}{3^{2n+2}} \)
\( = 3^{(n+1) + (n^2-1) - (n^2-n) - (2n+2)} \)
\( = 3^{n+1+n^2-1-n^2+n-2n-2} \)
\( = 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \)

Answer: (The marking scheme also accepts equivalent exponential forms)

Teacher's Note:
a) Convert all bases to prime numbers (such as 3) before applying exponent rules.
b) Carefully combine indices during multiplication and division steps.

(b) The area of an isosceles triangle is \( 12\text{ cm}^2 \) and the base is \( 8\text{ cm} \) in length. Find the perimeter of the triangle. [3 Marks]

Answer:
Area of an isosceles triangle \( = \frac{1}{4}b\sqrt{4a^2 - b^2} \), where \( b \) is the base and \( a \) is the length of equal sides.
Given \( b = 8\text{ cm} \) and Area \( = 12\text{ cm}^2 \):
\( \frac{1}{4}(8)\sqrt{4a^2 - 8^2} = 12 \)
\( 2\sqrt{4a^2 - 64} = 12 \implies \sqrt{4a^2 - 64} = 6 \)
Squaring both sides:
\( 4a^2 - 64 = 36 \implies 4a^2 = 100 \implies a^2 = 25 \implies a = 5\text{ cm} \)
Perimeter \( = 2a + b = 2(5) + 8 = 18\text{ cm} \)

Teacher's Note:
a) Know the direct formula for the area of an isosceles triangle in terms of base and equal sides.
b) Perimeter of an isosceles triangle is the sum of all three sides (\( 2a + b \)).

(c) In the figure, AC = CD. Prove that BC < CD. [3 Marks]

[Figure: Triangle ABD with C on BD such that AC = CD, and angle ACD = 110 degrees.]

Answer:
In \( \Delta ACD \), \( m\angle ACD = 180^{\circ} - 70^{\circ} = 110^{\circ} \) (linear pair)
Since \( AC = CD \), \( \angle CAD = \angle ADC = \frac{180^{\circ} - 110^{\circ}}{2} \) or using exterior angle: \( m\angle ACD = \angle CAD + \angle ADC \implies 70^{\circ} \) total for base angles, hence \( \angle CAD = \angle ADC = 35^{\circ} \).
In \( \Delta ABC \), \( m\angle BAC = 70^{\circ} - 35^{\circ} = 35^{\circ} \)
\( m\angle ABC = 180^{\circ} - (70^{\circ} + 35^{\circ}) = 75^{\circ} \)
Thus, \( \angle BAC < \angle ABC \) (\( 35^{\circ} < 75^{\circ} \))
Therefore, \( BC < AC \) (side opposite to smaller angle is smaller)
Since \( AC = CD \), \( BC < CD \).

Teacher's Note:
a) In any triangle, the side opposite to the greater angle is longer.
b) Use linear pairs and angle sum properties to find individual angle measures.

Q. 8. (a) If \( \cos \theta = \frac{2\sqrt{mn}}{m + n} \), find the value of \( \sin \theta \) (given \( m > n \)). [4 Marks]

Answer:
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \implies \sin \theta = \sqrt{1 - \cos^2 \theta} \)
\( \sin \theta = \sqrt{1 - \left(\frac{2\sqrt{mn}}{m + n}\right)^2} \)
\( = \sqrt{1 - \frac{4mn}{(m + n)^2}} \)
\( = \sqrt{\frac{(m + n)^2 - 4mn}{(m + n)^2}} \)
\( = \sqrt{\frac{m^2 + n^2 + 2mn - 4mn}{(m + n)^2}} \)
\( = \sqrt{\frac{m^2 + n^2 - 2mn}{(m + n)^2}} \)
\( = \sqrt{\frac{(m - n)^2}{(m + n)^2}} = \frac{m - n}{m + n} \)

Teacher's Note:
a) Apply the fundamental trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
b) Simplify algebraic expressions under the square root by completing squares.

(b) The mean of 5 numbers is 20. If one number is excluded the mean of the remaining numbers becomes 23. Find the excluded number. [3 Marks]

Answer:
Mean of 5 numbers = 20
Total sum of 5 numbers = \( 20 \times 5 = 100 \)
Let the excluded number be \( x \).
Sum of remaining 4 numbers = \( 100 - x \)
Mean of remaining 4 numbers = 23
\( \frac{100 - x}{4} = 23 \)
\( 100 - x = 23 \times 4 = 92 \)
\( x = 100 - 92 = 8 \)
Hence, the excluded number is 8.

Teacher's Note:
a) Total sum is calculated as the product of mean and number of observations.
b) Subtract the sum of the remaining observations from the initial total sum to find the excluded value.

(c) 3 equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of three cubes. [3 Marks]

Answer:
Let the side of each cube be \( a \).
Surface area of one cube = \( 6a^2 \)
Sum of surface areas of 3 cubes = \( 3 \times 6a^2 = 18a^2 \)
When 3 equal cubes are joined adjacently, dimensions of the new cuboid are:
Length = \( 3a \), Breadth = \( a \), Height = \( a \)
Total surface area of the new cuboid \( = 2(lb + bh + hl) = 2((3a \times a) + (a \times a) + (a \times 3a)) \)
\( = 2(3a^2 + a^2 + 3a^2) = 2(7a^2) = 14a^2 \)
Required ratio \( = \frac{\text{T.S.A. of cuboid}}{\text{Sum of S.A. of 3 cubes}} = \frac{14a^2}{18a^2} = \frac{7}{18} \) wait, check: \( 14a^2 : 18a^2 = 7 : 9 \).

Teacher's Note:
a) Joining cubes in a row alters only the length of the resulting cuboid while breadth and height remain unchanged.
b) Surface area formulas for cubes and cuboids must be applied accurately.

Q. 9. (a) In the given quadrilateral AD = BC, and P, Q, R and S are the midpoints of the sides AB, BD, CD and AC, respectively. Prove that PQRS is a rhombus. [4 Marks]

[Figure: Quadrilateral ABCD with diagonals AC and BD, and midpoints P, Q, R, S forming quadrilateral PQRS.]

Answer:
Proof: In \( \Delta ACD \), \( RS \parallel AD \) and \( RS = \frac{1}{2}AD \) (Mid-point theorem) ...(i)
In \( \Delta ABD \), \( PQ \parallel AD \) and \( PQ = \frac{1}{2}AD \) ...(ii)
In \( \Delta BCD \), \( QR \parallel BC \) and \( QR = \frac{1}{2}BC \) ...(iii)
In \( \Delta ABC \), \( SP \parallel BC \) and \( SP = \frac{1}{2}BC \) ...(iv)
Given \( AD = BC \), therefore \( RS = PQ = QR = SP \).
Since all four sides of quadrilateral PQRS are equal, PQRS is a rhombus.

Teacher's Note:
a) The mid-point theorem states that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.
b) A quadrilateral with all four sides equal is a rhombus.

(b) The distance (in km) of 40 engineers from their residence to place of work were found as follows:
5, 3, 10, 20, 25, 11, 13, 7, 12, 31
2, 19, 10, 12, 17, 18, 11, 32, 17, 16
3, 7, 9, 7, 8, 3, 5, 12, 15, 18
12, 12, 14, 2, 9, 6, 15, 15, 7, 6
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0 - 5 (5 not included). What main feature do you observe from this tabular representation? [3 Marks]

Answer:
Grouped frequency distribution table:

Main feature: Most of the engineers have their workplace at a distance of up to 15 km from their homes, and only 4 engineers live at a distance of 20 km or more.

Teacher's Note:
a) Tally marks should be grouped in bundles of five for easy counting.
b) Ensure continuous class intervals are formed correctly without overlapping boundaries.

(c) Solve: \( \log_x(8x - 3) - \log_x 4 = 2 \) [3 Marks]

Answer:
\( \log_x\left(\frac{8x - 3}{4}\right) = 2 \)
Converting logarithmic form to exponential form:
\( \frac{8x - 3}{4} = x^2 \)
\( 8x - 3 = 4x^2 \)
\( 4x^2 - 8x + 3 = 0 \)
\( 4x^2 - 6x - 2x + 3 = 0 \)
\( 2x(2x - 3) - 1(2x - 3) = 0 \)
\( (2x - 3)(2x - 1) = 0 \)
\( 2x - 3 = 0 \implies x = \frac{3}{2} \) or \( 2x - 1 = 0 \implies x = \frac{1}{2} \)

Teacher's Note:
a) Apply logarithm quotient rule before converting to algebraic equations.
b) Factorise quadratic equations carefully to find valid roots.

Q. 10. (a) Prove that \( \sqrt{5} \) is an irrational number. [4 Marks]

Answer:
Let us assume on the contrary that \( \sqrt{5} \) is a rational number.
Then, there exist co-prime integers \( a \) and \( b \) (\( b \neq 0 \)) such that \( \sqrt{5} = \frac{a}{b} \).
\( a = b\sqrt{5} \)
Squaring both sides: \( a^2 = 5b^2 \).
This implies that 5 divides \( a^2 \), so 5 also divides \( a \).
Let \( a = 5k \) for some integer \( k \).
Substituting \( a \): \( (5k)^2 = 5b^2 \implies 25k^2 = 5b^2 \implies b^2 = 5k^2 \).
This means 5 divides \( b^2 \), so 5 also divides \( b \).
Thus, 5 is a common factor of both \( a \) and \( b \), which contradicts the fact that \( a \) and \( b \) are co-prime.
Hence, our assumption is false and \( \sqrt{5} \) is irrational.

Teacher's Note:
a) Use the method of contradiction to prove irrationality of surds.
b) State clearly that \( a \) and \( b \) are co-prime integers at the beginning of the proof.

(b) If \( \tan(\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} \), find the value of \( (\theta_1 + \theta_2) \) given that \( \tan \theta_1 = \frac{1}{2} \) and \( \tan \theta_2 = \frac{1}{3} \). [3 Marks]

Answer:
\( \tan(\theta_1 + \theta_2) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2} \times \frac{1}{3}\right)} \)
\( = \frac{\frac{3+2}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \)
\( \tan(\theta_1 + \theta_2) = \tan 45^{\circ} \)
Therefore, \( \theta_1 + \theta_2 = 45^{\circ} \)

Answer:
Given \( x + \frac{1}{x} = 4 \)
Cubing both sides:
\( \left(x + \frac{1}{x}\right)^3 = 4^3 \)
\( x^3 + \frac{1}{x^3} + 3\left(x\right)\left(\frac{1}{x}\right)\left(x + \frac{1}{x}\right) = 64 \)
\( x^3 + \frac{1}{x^3} + 3(4) = 64 \)
\( x^3 + \frac{1}{x^3} = 64 - 12 = 52 \)
We need \( 2x^3 + \frac{2}{x^3} = 2\left(x^3 + \frac{1}{x^3}\right) = 2 \times 52 = 104 \).

Teacher's Note:
a) Use the algebraic identity \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \).
b) Factor out common terms carefully to match the required expression format.

Q. 11. (a) Factorise: \( 4a^3b - 44a^2b + 112b \) [3 Marks]

Answer:
\( 4a^3b - 44a^2b + 112b \)
\( = 4ab(a^2 - 11a + 28) \)
Split the middle term for \( a^2 - 11a + 28 \):
\( = 4ab(a^2 - 7a - 4a + 28) \)
\( = 4ab(a(a - 7) - 4(a - 7)) \)
\( = 4ab(a - 7)(a - 4) \)

Teacher's Note:
a) Always take out common monomial factors first before attempting further factorisation.
b) Apply middle-term splitting method for quadratic polynomials.

(b) Find the area of following figure: [3 Marks]

[Figure: Polygon PQRS divided into triangle PQR and trapezium PQST, with dimensions: QS = 35 cm, RN = 20 cm, PT = 25 cm, ST = 26 cm, perpendiculars at P and Q.]

Answer:
Construction: Draw \( TM \perp QS \).
Area of \( \Delta RQS = \frac{1}{2} \times QS \times RN = \frac{1}{2} \times 35 \times 20 = 350\text{ cm}^2 \)
Let \( QM = 25\text{ cm} \), then \( MS = QS - QM = 35 - 25 = 10\text{ cm} \).
In right-angled \( \Delta STM \), \( TM^2 = ST^2 - MS^2 = 26^2 - 10^2 = 676 - 100 = 576 \implies TM = 24\text{ cm} \) (which equals PQ).
Area of trapezium \( PQST = \frac{1}{2} \times (PT + QS) \times PQ = \frac{1}{2} \times (25 + 35) \times 24 = 720\text{ cm}^2 \)
Total area of the figure \( = \text{Area of } \Delta RQS + \text{Area of trapezium PQST} = 350 + 720 = 1070\text{ cm}^2 \).

Teacher's Note:
a) Split complex polygons into standard triangles and trapeziums.
b) Apply Pythagoras theorem to determine missing perpendicular heights.

(c) In \( \Delta ABC \), AB = AC = \( x \), BC = \( 20\text{ cm} \) and the area of the triangle is \( 250\text{ cm}^2 \). Find \( x \). [4 Marks]

[Figure: Isosceles triangle ABC with AB = AC = x, base BC = 20 cm, and altitude AD perpendicular to BC.]

Answer:
Construction: Draw \( AD \perp BC \).
Since \( \Delta ABC \) is isosceles with \( AB = AC \), altitude AD bisects the base BC.
\( BD = DC = \frac{20}{2} = 10\text{ cm} \)
Area of \( \Delta ABC = \frac{1}{2} \times BC \times AD = 250 \)
\( \frac{1}{2} \times 20 \times AD = 250 \implies 10 \times AD = 250 \implies AD = 25\text{ cm} \)
In right-angled \( \Delta ADC \):
\( AD^2 + DC^2 = AC^2 \) (Pythagoras theorem)
\( 25^2 + 10^2 = x^2 \)
\( 625 + 100 = x^2 \implies x^2 = 725 \)
\( x = \sqrt{725} = 5\sqrt{29}\text{ cm} \)

Teacher's Note:
a) The altitude from the vertex of an isosceles triangle bisects the base perpendicularly.
b) Substitute known area and base into the area formula to find the height before applying Pythagoras theorem.