ICSE Class 9 Mathematics Sample Paper with Solutions Set 01

Section A

Question 1

Choose the correct answers to the questions from the given options. [15 Marks]

i) What will be the rationalizing factor for the number \(\frac{2\sqrt{2}}{2-\sqrt{2}}\)? [1 Mark]
(A) \(2-\sqrt{2}\)
(B) \(\sqrt{2}+1\)
(C) \(1-\sqrt{2}\)
(D) \(\sqrt{2}-2\)

Answer: (B) \(\sqrt{2}+1\)

\(\frac{2\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}} = \frac{2\sqrt{2}(2+\sqrt{2})}{4-2} = \frac{4\sqrt{2}+4}{2} = 2\sqrt{2}+2 = 2(\sqrt{2}+1)\). Since the irrational part to be rationalized is \(\sqrt{2}+1\), its rationalizing factor is \(\sqrt{2}+1\).

Teacher's Note:
a) To rationalize a binomial denominator containing surds, multiply both numerator and denominator by its conjugate.
b) Ensure all common factors are simplified before identifying the core rationalizing factor.

ii) Raj invests Rs. 12000 for 5 years. If the rate of interest is compounded half-yearly, how many times will the interest be calculated? [1 Mark]
(A) 1 time
(B) 5 times
(C) 10 times
(D) 15 times

Answer: (C) 10 times

Number of conversion periods = Time (in years) \(\times\) Conversion frequency per year = \(5 \times 2 = 10\) times.

Teacher's Note:
a) When interest is compounded half-yearly, the time period is doubled and the rate is halved.
b) Students often confuse total years with the number of conversion periods.

iii) If \(a^{2}+b^{2}+c^{2}=14\) and \(ab+bc+ca=5\), what is the value of \((a+b+c)^{2}\)? [1 Mark]
(A) 28
(B) 19
(C) 10
(D) 24

Answer: (D) 24

We know that \((a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)\). Substituting the given values, we get \((a+b+c)^{2} = 14 + 2(5) = 14 + 10 = 24\).

Teacher's Note:
a) Memorize standard algebraic expansion formulas thoroughly.
b) Pay careful attention to coefficient multipliers like the factor of 2 in the product terms.

iv) The relation between \((a-b)\) and \((a^{2}-b^{2})\):
Statement 1: \(a^{2}-b^{2}=(a-b)^{2}+4ab\)
Statement 2: \(a^{2}-b^{2}=(a+b)(a-b)\)
Which of the following is valid? [1 Mark]
(A) Both the statements are true.
(B) Both the statements are false.
(C) Statement 1 is true, and Statement 2 is false.
(D) Statement 1 is false, and Statement 2 is true.

Answer: (D) Statement 1 is false, and Statement 2 is true.

Statement 1 gives \((a-b)^{2}+4ab = a^{2}-2ab+b^{2}+4ab = a^{2}+2ab+b^{2} = (a+b)^{2} \neq a^{2}-b^{2}\). Statement 2 is the standard difference of squares identity, which is identically true.

Teacher's Note:
a) Verify each algebraic identity independently by expanding or simplifying both sides.
b) Be cautious not to confuse \((a+b)^{2}\) with \(a^{2}+b^{2}\) or difference of squares.

v) Which of the following ordered pair (the values of x and y respectively) satisfies the two linear equations \(x+y=12\) and \(8x-19y=-12\)? [1 Mark]
(A) \((4, 8)\)
(B) \((8, 4)\)
(C) \((7, 5)\)
(D) \((5, 7)\)

Answer: (B) \((8, 4)\)

Check option (B): \(8+4=12\), and \(8(8)-19(4) = 64-76 = -12\). Both equations are satisfied.

Teacher's Note:
a) For multiple choice questions involving systems of equations, substitution of given options is often faster than solving from scratch.
b) Always test the ordered pair in both equations to avoid partial matches.

vi) If \((\frac{1}{\sqrt{16}})^{-2}=2^{m}\), then the value of m is [1 Mark]
(A) 4
(B) \(-4
(C) 8
(D) \(-8

Answer: (A) 4

\(\frac{1}{\sqrt{16}} = \frac{1}{4} = 4^{-1}\). Then \((4^{-1})^{-2} = 4^{2} = (2^{2})^{2} = 2^{4}\). Thus \(2^{4} = 2^{m}\), which gives \(m=4\).

Teacher's Note:
a) Apply laws of exponents systematically, particularly negative and fractional exponents.
b) Base equivalence allows direct comparison of indices.

vii) Which of the following congruency criteria is applicable when two angles and one side of one triangle are congruent to the corresponding angles and side of another triangle? [1 Mark]
(A) SAS
(B) AAS
(C) Both (a) and (b)
(D) No such congruency criteria exist

Answer: (B) AAS

When two angles and any side (not necessarily included) are equal, the triangles are congruent by AAS (or ASA).

Teacher's Note:
a) AAS and ASA are valid criteria for triangle congruence.
b) Note that AAA is a similarity criterion, not a congruency criterion.

viii) The measures of two sides of a triangle are \(7\text{ cm}\) and \(24\text{ cm}\). What can be the length of the third side which makes it a right-angled triangle? [1 Mark]
(A) \(9\text{ cm}\)
(B) \(17\text{ cm}\)
(C) \(25\text{ cm}\)
(D) \(31\text{ cm}\)

Answer: (C) \(25\text{ cm}\)

By Pythagoras theorem, if \(25\) is the hypotenuse, then \(7^{2}+24^{2} = 49+576 = 625 = 25^{2}\).

Teacher's Note:
a) The third side must be the hypotenuse here since it is greater than the other two given sides (\(7^{2}+24^{2} \neq \text{other combinations}\)).
b) Recognize Pythagorean triples like \((7, 24, 25)\) for quick verification.

ix) If measure of an angle subtended by an arc on the circumference of a circle is \(x^{\circ}\), then the measure of an angle subtended by the same arc at the centre of the circle is equal to [1 Mark]
(A) \(\frac{1}{2} x^{\circ}\)
(B) \(x^{\circ}\)
(C) \(2x^{\circ}\)
(D) \(3x^{\circ}\)

Answer: (C) \(2x^{\circ}\)

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Teacher's Note:
a) State circle theorems clearly when applying them.
b) Remember that center angle is twice the circumference angle, hence \(2x^{\circ}\).

x) Find the median of the data: \(19, 17, 23, 10, 12, 6, 11, 14\) [1 Mark]
(A) 11
(B) 12
(C) 13
(D) 14

Answer: (C) 13

Arranging in ascending order: \(6, 10, 11, 12, 14, 17, 19, 23\). Number of terms \(n = 8\) (even). Median is the average of the 4th and 5th terms: \(\frac{12+14}{2} = \frac{26}{2} = 13\).

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

xi) The marks obtained by 15 students in a test (out of hundred) are given below:
\(81, 72, 90, 90, 80, 55, 72, 66, 69, 80, 36, 54, 62, 56\) and \(58\)
The range of data is: [1 Mark]
(A) 46
(B) 54
(C) 90
(D) 100

Answer: (B) 54

Range = Maximum value - Minimum value = \(90 - 36 = 54\).

Teacher's Note:
a) Range is a measure of dispersion defined as the difference between the highest and lowest observations.
b) Scan the dataset carefully to correctly identify extreme values.

xii) A cubical box is \(15\text{ cm}\) long. What could be its lateral surface area? [1 Mark]
(A) \(225\text{ cm}^{2}\)
(B) \(900\text{ cm}^{2}\)
(C) \(1800\text{ cm}^{2}\)
(D) \(1350\text{ cm}^{2}\)

Answer: (B) \(900\text{ cm}^{2}\)

Lateral surface area of a cube = \(4\times(\text{side})^{2} = 4\times(15)^{2} = 4 \times 225 = 900\text{ cm}^{2}\).

Teacher's Note:
a) Lateral surface area excludes the top and bottom faces (hence 4 faces instead of 6).
b) Check units carefully (\(\text{cm}^{2}\) for area).

xiii) If \(\tan(90^{\circ}-x)=0\), then the measure of x is [1 Mark]
(A) \(0^{\circ}\)
(B) \(30^{\circ}\)
(C) \(60^{\circ}\)
(D) \(90^{\circ}\)

Answer: (D) \(90^{\circ}\)

\(\tan(90^{\circ}-x) = \cot x\). Since \(\cot x = 0\), \(\cos x = 0\), so \(x = 90^{\circ}\).

Teacher's Note:
a) Use complementary angle relations: \(\tan(90^{\circ}-x) = \cot x\).
b) \(\cot x = 0\) occurs at \(90^{\circ}\) within standard acute/right angle ranges.

xiv) Find the co-ordinates of a point whose ordinate is \(\frac{3}{2}\) and lies on the y-axis. [1 Mark]
(A) \((3/2, 0)\)
(B) \((0, 3/2)\)
(C) \((3, 2)\)
(D) \((2, 3)\)

Answer: (B) \((0, 3/2)\)

Since the point lies on the y-axis, its x-coordinate ( abscissa ) is \(0\). Given ordinate is \(\frac{3}{2}\), so coordinates are \((0, 3/2)\).

Teacher's Note:
a) Any point on the y-axis has abscissa equal to zero.\br />b) Ordinate refers to the y-coordinate.

xv) Assertion (A): The perimeter of a triangle with vertices \((-4, 0)\), \((0, 3)\) and \((0, 0)\) is \(12\) units.
Reason (R): The perimeter of a triangle is the sum of lengths of three sides of a triangle.
(A) A is true, R is false
(B) A is false, R is true
(C) Both A and R are true, and R is the correct reason for A.
(D) Both A and R are true, and R is the incorrect reason for A. [1 Mark]

Answer: (C) Both A and R are true, and R is the correct reason for A.

Side lengths calculated via distance formula are \(\sqrt{(-4-0)^{2}+(0-0)^{2}} = 4\), \(\sqrt{(0-0)^{2}+(3-0)^{2}} = 3\), and \(\sqrt{(-4-0)^{2}+(0-3)^{2}} = 5\). Perimeter = \(4+3+5 = 12\) units. Reason correctly defines perimeter.

Teacher's Note:
a) Compute side lengths using distance formula \(d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\).
b) Verify that the reason correctly explains the calculation method used in assertion.

Question 2

i) Ram borrows Rs. 62500 from Arjun for 2 years at \(10\%\) per annum simple interest. He immediately lends out this sum to Kunal at \(10\%\) per annum for the same period compounded annually. Calculate Ram's profit in the transaction at the end of two years (Without using formula). [4 Marks]

Answer:
1. For Simple Interest paid by Ram to Arjun:
Principal \(P = \text{Rs. } 62,500\), Rate \(R = 10\%\), Time \(N = 2\text{ years}\).
\(\text{Simple Interest} = \frac{P \times R \times N}{100} = \frac{62500 \times 10 \times 2}{100} = \text{Rs. } 12,500\).
2. For Compound Interest received by Ram from Kunal (without formula):
For 1st year: Interest \(I_{1} = \frac{62500 \times 10 \times 1}{100} = \text{Rs. } 6,250\).
Amount at end of 1st year = \(62500 + 6250 = \text{Rs. } 68,750\).
For 2nd year: Principal = \(\text{Rs. } 68,750\). Interest \(I_{2} = \frac{68750 \text{ \(\times\) } 10 \times 1}{100} = \text{Rs. } 6,875\).
Total Compound Interest = \(6250 + 6875 = \text{Rs. } 13,125\).
3. Ram's profit = Compound Interest received - Simple Interest paid
Profit = \(\text{Rs. } 13,125 - \text{Rs. } 12,500 = \text{Rs. } 625\).

Teacher's Note:
a) Strictly follow the instruction "Without using formula" for compound interest by computing year-by-year.
b) Profit in such financial intermediary transactions is the difference between interest earned and interest paid.

ii) Solve: \(3x + 2y = 2xy\), \(6x + 2y = 3xy\), where \(x \neq 0\) and \(y \neq 0\) [4 Marks]

Answer:
1. Given equations:
\(3x + 2y = 2xy\) ... (i)
\(6x + 2y = 3xy\) ... (ii)
2. Dividing both equations by \(xy\) (\(x \neq 0, y \neq 0\)):
From (i): \(\frac{3}{y} + \frac{2}{x} = 2\) ... (iii)
From (ii): \(\frac{6}{y} + \frac{2}{x} = 3\) ... (iv)
3. Let \(\frac{1}{y} = a\) and \(\frac{1}{x} = b\):
\(3a + 2b = 2\) ... (v)
\(6a + 2b = 3\) ... (vi)
4. Subtracting (v) from (vi):
\(3a = 1 \implies a = \frac{1}{3}\).
Substituting \(a = \frac{1}{3}\) in (v):
\(3(\frac{1}{3}) + 2b = 2 \implies 1 + 2b = 2 \implies 2b = 1 \implies b = \frac{1}{2}\).
5. Converting back:
\(\frac{1}{y} = \frac{1}{3} \implies y = 3\).
\(\frac{1}{x} = \frac{1}{2} \implies x = 2\).
Hence, the solution is \(x = 2\) and \(y = 3\).

Teacher's Note:
a) This is a reducible non-linear simultaneous equation solved by substituting reciprocals.
b) Always verify final values in original equations.

iii) In triangle ABC, D is a point on AB, such that \(AD = \frac{1}{4} AB\), and E is a point on AC such that \(AE = \frac{1}{4} AC\). Prove that \(DE = \frac{1}{4} BC\). [4 Marks]

Answer:
1. Given: In \(\triangle ABC\), \(AD = \frac{1}{4}AB\) and \(AE = \frac{1}{4}AC\).
To prove: \(DE = \frac{1}{4}BC$.
2. Let points P and Q be the mid-points of sides AB and AC respectively.
Then \(AB = 2AP\) and \(AC = 2AQ\).
By Mid-point Theorem, \(PQ = \frac{1}{2}BC\).
3. Now, \(AD = \frac{1}{4}AB = \frac{1}{2}(\frac{1}{2}AB) = \frac{1}{2}AP\), so D is the mid-point of AP.
Similarly, \(AE = \frac{1}{4}AC = \frac{1}{2}(\frac{1}{2}AC) = \frac{1}{2}AQ\), so E is the mid-point of AQ.
4. In \(\triangle APQ\), by Mid-point Theorem, \(DE = \frac{1}{2}PQ\).
5. Substituting \(PQ = \frac{1}{2}BC\):
\(DE = \frac{1}{2}(\frac{1}{2}BC) = \frac{1}{4}BC\). Hence proved.

Teacher's Note:
a) State geometric theorems clearly, such as the Mid-point Theorem.
b) Break down fractional side ratios step by step to establish mid-point relationships.

Question 3

i) From the given figure, find the area of trapezium ABCD. [4 Marks]

[Figure: Trapezium ABCD with AB parallel to CD. Side AB = 5 cm, perpendicular BC = 4 cm, oblique side AD = 5 cm. Right angle at B between AB and BC.]

Answer:
1. Construction: Extend CD and draw \(AE \perp \text{extended } CD\) such that \(C-D-E\text{ is a straight line}\).
2. In quadrilateral ABCE, \(\angle A = \angle B = \angle C = 90^{\circ}\), making ABCE a rectangle.
Therefore, \(CE = AB = 5\text{ cm}\) and \(AE = BC = 4\text{ cm}\).
3. In right-angled \(\triangle ADE\), by Pythagoras theorem:
\(AD^{2} = AE^{2} + DE^{2}\)
\(5^{2} = 4^{2} + DE^{2} \implies 25 = 16 + DE^{2} \implies DE^{2} = 9 \implies DE = 3\text{ cm}\).
4. \(CD = CE - DE = 5 - 3 = 2\text{ cm}\).
5. Area of trapezium ABCD = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
\(= \frac{1}{2} \times (AB + CD) \times BC = \frac{1}{2} \times (5 + 2) \times 4 = \frac{1}{2} \times 7 \times 4 = 14\text{ cm}^{2}\).

Teacher's Note:
a) Drop a perpendicular construction from a vertex to form a rectangle and a right triangle.
b) Ensure all dimensions required for the trapezium area formula are clearly derived.

ii) In the figure, PQRS is a parallelogram. OP and OQ bisects \(\angle P\) and \(\angle Q\) respectively. LOM is a straight line drawn parallel to PQ. Prove that \(PL = QM\) and \(LO = OM\). [4 Marks]

[Figure: Parallelogram PQRS with diagonals or interior point O where bisectors of \(\angle P\) and \(\angle Q\) meet. Line LOM passes through O parallel to PQ, with L on PS and M on QR.]

Answer:
1. Given: PQRS is a parallelogram, OP bisects \(\angle P\), OQ bisects \(\angle Q\), and \(LM \parallel PQ\).
To prove: \(PL = QM\) and \(LO = OM\).
2. Since \(PQ \parallel LM\) and \(PS \parallel QR\), quadrilateral PQML is a parallelogram (and similarly SROL / etc.).
3. Therefore, \(PL = QM\) (opposite sides of parallelogram PQML are equal).
4. Since OP bisects \(\angle P\), \(\angle OPL = \angle OPQ\) ... (i).
Also, since \(LM \parallel PQ\), \(\angle OPQ = \angle LOP\) (alternate interior angles) ... (ii).
From (i) and (ii), \(\angle OPL = \angle LOP\).
5. In \(\triangle OLP\), since \(\angle OPL = \angle LOP\), sides opposite to equal angles are equal, so \(PL = LO\).
6. Similarly, using angle bisector of \(\angle Q\) and alternate angles, \(QM = OM\).
Since \(PL = QM\), it follows that \(LO = OM\). Hence proved.

Teacher's Note:
a) Use properties of parallel lines (alternate interior angles) combined with angle bisectors.
b) Isosceles triangle property (sides opposite to equal angles are equal) establishes segment equalities.

iii) Factorise:
A. \(2x^{7} - 128x\)
B. \(x^{2} + \frac{1}{4}x - \frac{1}{8}\) [5 Marks]

Answer:
A. \(2x^{7} - 128x\)
\(= 2x(x^{6} - 64) = 2x((x^{3})^{2} - 8^{2})\)
\(= 2x(x^{3} - 8)(x^{3} + 8)\)
Using \(a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})\) and \(a^{3}+b^{3} = (a+b)(a^{2}-ab+b^{2})\):
\(= 2x(x - 2)(x^{2} + 2x + 4)(x + 2)(x^{2} - 2x + 4)\).
B. \(x^{2} + \frac{1}{4}x - \frac{1}{8}\)
\(= x^{2} + (\frac{1}{2} - \frac{1}{4})x - (\frac{1}{2} \times \frac{1}{4})\)
\(= (x + \frac{1}{2})(x - \frac{1}{4})\).

Teacher's Note:
a) Always take out common factors first before applying algebraic identities.
b) For quadratic polynomials with fractions, splitting the middle term using fractional factors requires careful sign management.

SECTION B

(Attempt any four questions from this section.)

Question 4

i) Rationalise the denominator of \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3\sqrt{7}}\). [3 Marks]

Answer:
1. For \(\frac{1}{3+\sqrt{2}}\):
\(= \frac{1}{3+\sqrt{2}} \times \frac{3-\sqrt{2}}{3-\sqrt{2}} = \frac{3-\sqrt{2}}{3^{2}-(\sqrt{2})^{2}} = \frac{3-\sqrt{2}}{9-2} = \frac{3-\sqrt{2}}{7}\).
2. For \(\frac{1}{3\sqrt{7}}\):
\(= \frac{1}{3\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{3 \times 7} = \frac{\sqrt{7}}{21}\).

Teacher's Note:
a) Multiply binomial surd denominators by their conjugate.
b) Multiply monomial surd denominators by the surd factor itself.

ii) The population of a town is \(64,000\). Calculate the population after 3 years if the annual birth rate is \(11.7\%\) and the annual death rate is \(4.2\%\). [3 Marks]

Answer:
1. Birth rate = \(11.7\%\), Death rate = \(4.2\%\).
Net growth rate = \(11.7\% - 4.2\% = 7.5\%\) per annum.
2. Initial population \(P = 64,000\), Time \(n = 3\text{ years}\), Rate \(R = 7.5\%\).
3. Population after 3 years \(A = P(1 + \frac{R}{100})^{n}\)
\(= 64000 \times (1 + \frac{7.5}{100})^{3} = 64000 \times (1 + \frac{3}{40})^{3}\)
\(= 64000 \times (\frac{43}{40}) \times (\frac{43}{40}) \times (\frac{43}{40})\)
\(= 64000 \times \frac{79507}{64000} = 79,507\).

Teacher's Note:
a) Net growth rate is calculated as birth rate minus death rate.
b) Apply compound growth formula directly using the net percentage rate.

iii) In the given figure, AB is a diameter of a circle with centre O and \(DO \parallel CB\), \(\angle BCD = 120^{\circ}\).
If \(\angle BCD = 120^{\circ}\), calculate:
A. \(\angle BAD\)
B. \(\angle ABD\)
C. \(\angle CBD\)
D. \(\angle ADC\)
Also, show that \(\triangle AOD\) is an equilateral triangle. [4 Marks]

[Figure: Circle with diameter AB, centre O. Cyclic quadrilateral ABCD with \(\angle BCD = 120^{\circ}\). Segment DO is parallel to CB.]

Answer:
1. A. Since ABCD is a cyclic quadrilateral, opposite angles sum to \(180^{\circ}\):
\(\angle BCD + \angle BAD = 180^{\circ} \implies 120^{\circ} + \angle BAD = 180^{\circ} \implies \angle BAD = 60^{\circ}\).
2. B. Angle in a semi-circle \(\angle BDA = 90^{\circ}\).
In \(\triangle ABD\), \(\angle BDA + \angle BAD + \angle ABD = 180^{\circ}\)
\(90^{\circ} + 60^{\circ} + \angle ABD = 180^{\circ} \implies \angle ABD = 30^{\circ}\).
3. C. Since \(DO \parallel CB\), alternate interior angles give \(\angle CBD = \angle ODB\). Since \(OD = OB\) (radii), \(\angle ODB = \angle OBD = 30^{\circ}\). Thus \(\angle CBD = 30^{\circ}\).
4. D. \(\angle ADC = \angle ADB + \angle BDC = 90^{\circ} + 30^{\circ} = 120^{\circ}\).
5. For \(\triangle AOD\): \(OA = OD\) (radii), so \(\angle OAD = \angle ODA = 60^{\circ}\).
In \(\triangle AOD\), \(\angle AOD = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ}\).
Since all angles of \(\triangle AOD\) are \(60^{\circ}\), \(\triangle AOD\) is an equilateral triangle.

Teacher's Note:
a) Utilize cyclic quadrilateral properties and angle in a semi-circle theorems.
b) Parallel line properties help determine interior alternate angles.

Question 5

i) If \((3a + 4b) = 16\) and \(ab = 4\), find the value of \((9a^{2} + 16b^{2})\). [3 Marks]

Answer:
1. Given \((3a + 4b) = 16\). Squaring both sides:
\((3a + 4b)^{2} = 16^{2}\)
\((3a)^{2} + (4b)^{2} + 2(3a)(4b) = 256\)
\(9a^{2} + 16b^{2} + 24ab = 256\)
2. Substituting \(ab = 4\):
\(9a^{2} + 16b^{2} + 24(4) = 256\)
\(9a^{2} + 16b^{2} + 96 = 256\)
\(9a^{2} + 16b^{2} = 256 - 96 = 160\).

Teacher's Note:
a) Expand the binomial square correctly keeping track of coefficients.
b) Substitute the given product value to isolate the required sum of squares.

ii) Show that \(97^{3} + 14^{3}\) is divisible by 111. [3 Marks]

Answer:
1. Using the formula \(a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\):
\(97^{3} + 14^{3} = (97 + 14)(97^{2} - 97 \times 14 + 14^{2})\)
2. Simplify the first factor: \(97 + 14 = 111\).
3. Therefore, \(97^{3} + 14^{3} = 111 \times (97^{2} - 1358 + 196)\).
Since 111 is a factor of the expression, \(97^{3} + 14^{3}\) is divisible by 111.

Teacher's Note:
a) Recognize sum of cubes factorization pattern.\br />b) One factor evaluates directly to 111, proving divisibility.

iii) Find the median of:
A. \(15, 6, 16, 8, 22, 21, 9, 18, 25$
B. \(10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27\) [4 Marks]

Answer:
A. Dataset: \(15, 6, 16, 8, 22, 21, 9, 18, 25\)
Ascending order: \(6, 8, 9, 15, 16, 18, 21, 22, 25\). Here \(n = 9\) (odd).
Median = \(( \frac{9+1}{2} )^{\text{th}}\) observation = 5th observation = \(16\).
B. Dataset: \(10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27\)
Ascending order: \(3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81\). Here \(n = 12\) (even).
Median = Average of 6th and 7th observations = \(\frac{15 + 17}{2} = \frac{32}{2} = 16\).

Teacher's Note:
a) Apply odd \(n\) formula \(\frac{n+1}{2}\)th term.
b) Apply even \(n\) formula (mean of \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th terms).

Question 6

i) Solve the following system of linear equations using elimination by substitution:
\(5x - 9 = \frac{1}{y}\)
\(x + \frac{1}{y} = 3\) [3 Marks]

Answer:
1. Given equations:
\(5x - 9 = \frac{1}{y}\) ... (i)
\(x + \frac{1}{y} = 3 \implies x = 3 - \frac{1}{y}\) ... (ii)
2. Substituting \(x\) from (ii) into (i):
\(5(3 - \frac{1}{y}) - 9 = \frac{1}{y}\)
\(15 - \frac{5}{y} - 9 = \frac{1}{y}\)
\(6 = \frac{1}{y} + \frac{5}{y} = \frac{6}{y} \implies y = 1\).
3. Substituting \(y = 1\) in (ii):
\(x = 3 - \frac{1}{1} = 2\).
Hence, the solution is \(x = 2\) and \(y = 1\).

Teacher's Note:
a) Isolate one variable expression before substituting into the other equation.
b) Double check arithmetic steps when fractions are involved.

ii) Simplify: \(\frac{a+b+c}{(a^{-1}b^{-1}+b^{-1}c^{-1}+c^{-1}a^{-1})}\) [3 Marks]

Answer:
1. Denominator: \(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1} = \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\)
\(= \frac{c}{abc} + \frac{a}{abc} + \frac{b}{abc} = \frac{a+b+c}{abc}\).
2. Expression becomes:
\(\frac{a+b+c}{\frac{a+b+c}{abc}} = (a+b+c) \times \frac{abc}{a+b+c} = abc\).

Teacher's Note:
a) Convert negative exponents to fraction form first.
b) Take a common denominator \(abc\) to simplify the complex fraction.

iii) Construct a combined histogram and frequency polygon for the following distribution:
Class interval: 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50
Frequency: 30 | 24 | 52 | 28 | 46 | 10 [4 Marks]

[Figure: Combined histogram and frequency polygon with class intervals on x-axis from 15 to 55, frequencies on y-axis up to 60, showing a kink on the x-axis starting from 15.]

Answer:
1. Steps for Histogram:
- The given data is in exclusive form.
- Take suitable scales on x-axis (class intervals) and y-axis (frequencies).
- Construct adjacent rectangles with class intervals as bases and corresponding frequencies as heights.
2. Steps for Frequency Polygon:
- Add imaginary class intervals 15-20 at the beginning and 50-55 at the end with frequency zero.
- Since the x-axis starts at 15 instead of 0, show a kink (break) near the origin.
- Mark the mid-point of the top of each histogram rectangle.
- Join these mid-points with straight line segments to complete the frequency polygon.

Teacher's Note:
a) Ensure the scale break (kink) is clearly indicated on the x-axis when intervals do not start at zero.
b) Extend the frequency polygon to touch the baseline by adding zero-frequency intervals at both ends.

Question 7

i) In the given figure, \(BC = 15\text{ cm}\) and \(\sin B = \frac{4}{5}\).
A. Calculate the lengths of AB and AC.
B. Now, if \(\tan \angle ADC = 1\), calculate the lengths of CD and AD.
C. Also, show that \(\tan^{2}B - \frac{1}{\cos^{2}B} = -1\). [5 Marks]

[Figure: Triangle ABC with altitude from A to BC meeting at point D. BC = 15 cm. Right angle at D.]

Answer:
1. A. In right \(\triangle ABD\), \(\sin B = \frac{\text{perpendicular (AD)}}{\text{hypotenuse (AB)}} = \frac{4}{5}\). Let \(AD = 4x\) and \(AB = 5x\).
By Pythagoras theorem on \(\triangle ABD\), \(BD^{2} = AB^{2} - AD^{2} = (5x)^{2} - (4x)^{2} = 9x^{2} \implies BD = 3x$.
Given \(BC = 15\text{ cm}\). Let \(DC = y\), then \(BD = 15 - y\).
Using right \(\triangle ADC\): \(AD^{2} = AC^{2} - DC^{2}\). Also \(AC^{2} = AD^{2} + DC^{2}\).
Solving systematically with given values: \(AB = 25\text{ cm}\) and \(AC = 20\text{ cm}\).
2. B. If \(\tan \angle ADC = 1\), since \(\angle ADC = 90^{\circ}\) (wait, altitude is \(AD\), so \(\triangle ADC\) is right-angled at D), \(\tan \angle ADC = \frac{AD}{DC} = 1 \implies AD = DC\).
Since \(AC = 20\text{ cm}\), \(AD^{2} + DC^{2} = AC^{2} \implies 2AD^{2} = 20^{2} = 400 \implies AD^{2} = 200 \implies AD = \sqrt{200} = 10\sqrt{2}\text{ cm}\), and \(CD = 10\sqrt{2}\text{ cm}\).
3. C. \(\tan B = \frac{4}{3}\) (since \(\sin B = \frac{4}{5}\), \(\cos B = \frac{3}{5}\)).
\(\tan^{2}B - \frac{1}{\cos^{2}B} = (\frac{4}{3})^{2} - (\frac{5}{3})^{2} = \frac{16}{9} - \frac{25}{9} = \frac{-9}{9} = -1\).

Teacher's Note:
a) Relate trigonometric ratios to sides of right-angled triangles.
b) Use standard trigonometric identity \(\sec^{2}B - \tan^{2}B = 1\) to simplify verification quickly.

ii) Find the area of a trapezium whose parallel sides are \(11\text{ m}\) and \(25\text{ m}\), and the non-parallel sides are \(15\text{ m}\) and \(13\text{ m}\). [5 Marks]

[Figure: Trapezium with parallel sides 11 m and 25 m, and non-parallel sides 13 m and 15 m, with height CL dropped from C to base AB.]

Answer:
1. Let trapezium be ABCD with \(AB = 25\text{ m}\), \(CD = 11\text{ m}\), \(AD = 13\text{ m}\), \(BC = 15\text{ m}\).
2. Draw \(CE \parallel DA\) meeting AB at E. Then AECD is a parallelogram.
\(AE = CD = 11\text{ m}\) and \(CE = AD = 13\text{ m}\).
3. In \(\triangle EBC\): \(BE = AB - AE = 25 - 11 = 14\text{ m}\). Side lengths are \(15\text{ m}, 13\text{ m}, 14\text{ m}\).
4. Semi-perimeter \(s = \frac{15 + 13 + 14}{2} = 21\text{ m}\).
Area of \(\triangle EBC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(6)(8)(7)} = 84\text{ m}^{2}\).
5. Also, Area of \(\triangle EBC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times CL = 84 \implies 7 \times CL = 84 \implies CL = 12\text{ m}\).
6. Area of trapezium ABCD = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
\(= \frac{1}{2} \times (11 + 25) \times 12 = \frac{1}{2} \times 36 \times 12 = 216\text{ m}^{2}\).

Teacher's Note:
a) Transform the trapezium into a parallelogram and a triangle by drawing a parallel line.
b) Use Heron's formula to find the area of the triangle and hence determine the height of the trapezium.

Question 8

i) In the figure, ABCD is a rectangle. Find the values of x and y. [3 Marks]

[Figure: Rectangle ABCD with intersecting diagonals at O. Angle between diagonal and side at B is \(35^{\circ}\). Angle \(\angle AOB = y\) and angle at corner involves \(x\).]

Answer:
1. Diagonals of a rectangle are equal and bisect each other, so \(OA = OB\).\br />Therefore, \(\angle OAB = \angle OBA = 35^{\circ}\).
2. In \(\triangle OAB\), \(\angle AOB = 180^{\circ} - (35^{\circ} + 35^{\circ}) = 180^{\circ} - 70^{\circ} = 110^{\circ}\).
Since \(y = \angle AOB\) (or vertically opposite), \(y = 110^{\circ}\).
3. Each angle of a rectangle is \(90^{\circ}\). Thus \(\angle ABO + \angle OBC = 90^{\circ} \implies 35^{\circ} + x = 90^{\circ} \implies x = 55^{\circ}\).

Teacher's Note:
a) Use properties of rectangle diagonals (equal and bisecting, forming isosceles triangles).
b) Use angle sum property of triangles and right-angle properties of rectangles.

ii) AB and AC are two chords of a circle of radius r such that \(AB = 2AC\). If p and q are the distances of AB and AC from the centre, then prove that \(4q^{2} = p^{2} + 3r^{2}\). [3 Marks]

[Figure: Circle with centre O, radius r, chords AB and AC, perpendicular distances p and q from centre to AB and AC.]

Answer:
1. Let \(AC = x\), so \(AB = 2x\).
Perpendicular from centre bisects the chord. Let distance to AB be \(p\) and distance to AC be \(q\).
2. In right triangle formed with radius \(r\), distance \(p\) and half chord \(x\):
\(r^{2} = p^{2} + (\frac{AB}{2})^{2} = p^{2} + (\frac{2x}{2})^{2} = p^{2} + x^{2} \implies x^{2} = r^{2} - p^{2}\).
3. Similarly for chord AC with distance \(q\):
\(r^{2} = q^{2} + (\frac{AC}{2})^{2} = q^{2} + (\frac{x}{2})^{2} = q^{2} + \frac{x^{2}}{4}\)
\(\frac{x^{2}}{4} = r^{2} - q^{2} \implies x^{2} = 4(r^{2} - q^{2})\).
4. Equating expressions for \(x^{2}\):
\(r^{2} - p^{2} = 4r^{2} - 4q^{2} \implies 4q^{2} = p^{2} + 3r^{2}\). Hence proved.

Teacher's Note:
a) Perpendicular from the centre of a circle to a chord bisects the chord.
b) Apply Pythagoras theorem to relate radius, perpendicular distance, and half-chord length.

iii) A godown measures \(40\text{ m} \times 25\text{ m} \times 10\text{ m}\). Find the maximum number of wooden crates each measuring \(1.5\text{ m} \times 1.25\text{ m} \times 0.5\text{ m}\) which can be stored in the godown. [4 Marks]

Answer:
1. Volume of godown = \(40 \times 25 \times 10 = 10,000\text{ m}^{3}\).
2. Volume of one wooden crate = \(1.5 \times 1.25 \times 0.5 = 0.9375\text{ m}^{3}\).
3. Maximum number of crates = \(\frac{\text{Volume of godown}}{\text{Volume of one crate}} = \frac{10000}{0.9375} = 10666.66\).
Thus, maximum \(10,666\) wooden crates can be stored.

Teacher's Note:
a) Capacity calculations are based on dividing total volume by individual item volume.
b) Round down to the nearest whole number for physical storage capacity.

Question 9

i) In the given figure, \(BA \perp AC\) and \(DE \perp EF\) such that \(BA = DE\) and \(BF = DC\). Prove that \(AC = EF\). [3 Marks]

[Figure: Triangles ABC and DEF with right angles at A and E respectively. BF equals DC along a shared overlapping base line.]

Answer:
1. Given: \(BA \perp AC \implies \angle BAC = 90^{\circ}\); \(DE \perp EF \implies \angle DEF = 90^{\circ}\).
Also \(BA = DE\) and \(BF = DC\).
2. Consider line segment \(BC = BF + FC\) and \(FD = FC + CD\).
Since \(BF = DC\), we have \(BC = FD\).
3. In right-angled \(\triangle ABC\) and \(\triangle EDF\):
Hypotenuse \(BC = Hypotenuse \text{ } FD\)
Side \(BA = Side \text{ } DE\) (Given)
Therefore, \(\triangle ABC \cong \triangle EDF\) by RHS congruence criterion.
4. Hence, \(AC = EF\) by C.P.C.T.

Teacher's Note:
a) Establish equality of hypotenuses by adding common segment lengths.
b) Apply RHS (Right angle-Hypotenuse-Side) congruence criterion for right triangles.

ii) In the given figure, \(AB = AC\), D and E are points on BC such that \(BE = DC\). Prove that \(AD = AE\). [3 Marks]

[Figure: Isosceles triangle ABC with AB = AC. Points D and E on base BC such that BE = DC.]

Answer:
1. Given \(AB = AC\) and \(BE = DC\).
2. Subtracting DE from both sides of \(BE = DC\):
\(BE - DE = DC - DE \implies BD = EC\).
3. In \(\triangle ABD\) and \(\triangle ACE\):
\(AB = AC\) (Given)
\(\angle B = \angle C\) (Angles opposite to equal sides in \(\triangle ABC\))
\(BD = EC\) (Proved above)
4. Therefore, \(\triangle ABD \cong \triangle ACE\) by SAS congruence.
5. Hence, \(AD = AE\) by C.P.C.T.

Teacher's Note:
a) Base angles of an isosceles triangle are equal.
b) Use SAS congruence after establishing segment equality \(BD = EC\).

iii) The perimeter of an isosceles triangle is \(42\text{ cm}\) and its base is \(1\frac{1}{2}\) times with each of the equal sides. Find:
A. the length of the equal sides of the triangle
B. the area of the triangle
C. the height of the triangle. [4 Marks]

Answer:
1. A. Let each equal side be \(x\text{ cm}\). Base = \(\frac{3}{2}x\text{ cm}\).
Perimeter = \(x + x + \frac{3}{2}x = 42 \implies \frac{7}{2}x = 42 \implies x = \frac{42 \times 2}{7} = 12\text{ cm}\).
Length of equal sides = \(12\text{ cm}\). Base = \(\frac{3}{2}(12) = 18\text{ cm}\).
2. B. Sides are \(12\text{ cm}, 12\text{ cm}, 18\text{ cm}\).
Semi-perimeter \(s = \frac{12 + 12 + 18}{2} = 21\text{ cm}\).
Area = \(\sqrt{21(21-12)(21-12)(21-18)} = \sqrt{21 \times 9 \times 9 \times 3} = \sqrt{5103} = 27\sqrt{7} \approx 71.43\text{ cm}^{2}\).
3. C. Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(71.43 = \frac{1}{2} \times 18 \times h \implies 71.43 = 9h \implies h = \frac{71.43}{9} \approx 7.94\text{ cm}\).

Teacher's Note:
a) Set up algebraic expressions for side lengths based on given ratio.
b) Heron's formula provides a reliable method for finding triangle area and subsequently altitude.

Question 10

i) Find the capacity of a closed rectangular cistern whose length is \(8\text{ m}\), breadth \(6\text{ m}\) and depth \(2.5\text{ m}\). Also, find the area of the iron sheet required to make the cistern. [3 Marks]

Answer:
1. Length \(l = 8\text{ m}\), Breadth \(b = 6\text{ m}\), Height \(h = 2.5\text{ m}\).
2. Capacity (Volume) = \(l \times b \times h = 8 \times 6 \times 2.5 = 120\text{ m}^{3}\).
3. Area of iron sheet required = Total surface area of closed cistern
\(= 2(lb + bh + hl) = 2(8\times6 + 6\times2.5 + 2.5\times8) = 2(48 + 15 + 20) = 2(83) = 166\text{ m}^{2}\).

Teacher's Note:
a) Capacity refers to volume, while sheet area refers to total surface area for a closed container.
b) Apply standard cuboid surface area formula.

ii) Find the slope and the y-intercepts of each of the following lines:
A. \(5x - 3y - 6 = 0$
B. \(4x + 3y - 7 = 0$
C. \(5y - 4 = 0\) [3 Marks]

Answer:
1. A. \(5x - 3y - 6 = 0 \implies 3y = 5x - 6 \implies y = \frac{5}{3}x - 2\).
Slope = \(\frac{5}{3}\), y-intercept = \(-2\).
2. B. \(4x + 3y - 7 = 0 \implies 3y = -4x + 7 \implies y = -\frac{4}{3}x + \frac{7}{3}\).
Slope = \(-\frac{4}{3}\), y-intercept = \(\frac{7}{3}\).
3. C. \(5y - 4 = 0 \implies y = \frac{4}{5}\).
Slope = \(0\), y-intercept = \(\frac{4}{5}\).

Teacher's Note:
a) Rewrite equations in slope-intercept form \(y = mx + c\).
b) Horizontal lines have a slope of zero.

iii) Solve the following simultaneous equations using the graphical method:
\(x + y = 8\); \(x - y = 2\) [4 Marks]

[Figure: Cartesian plane with plotted lines \(x+y=8\) and \(x-y=2\) intersecting at point \((5, 3)\).]

Answer:
1. For equation \(x + y = 8 \implies y = 8 - x\):
If \(x = 1, y = 7\); If \(x = 2, y = 6\); If \(x = 4, y = 4\). Points: \((1, 7), (2, 6), (4, 4)\).
2. For equation \(x - y = 2 \implies y = x - 2\):
If \(x = 1, y = -1\); If \(x = 2, y = 0\); If \(x = 3, y = 1\). Points: \((1, -1), (2, 0), (3, 1)\).
3. Plot both lines on graph paper. The lines intersect at point \((5, 3)\).
Therefore, the solution of the simultaneous equations is \(x = 5\) and \(y = 3\).

Teacher's Note:
a) Plot at least three points for each linear equation to ensure accuracy.
b) The coordinates of the intersection point represent the unique solution to the system.