ICSE Class 10 Mathematics Sample Paper 2022 with Solutions

SECTION A

(Attempt all questions from this Section.)

Question 1

Choose the correct answers to the questions from the given options. (Do not copy the question, Write the correct answer only.) [10]

(i) The point (3,0) is invariant under reflection in: [1 Mark]
(A) The origin
(B) x-axis
(C) y-axis
(D) both x and y axes

Answer: (B) x-axis

A point lying on the x-axis remains unchanged when reflected across the x-axis.

Teacher's Note:
a) A point \((x, 0)\) lies on the x-axis, so its reflection in the x-axis is itself.\
b) Remember that reflection in the origin negates both coordinates, yielding \((-3, 0)\).

(ii) In the given figure, AB is a diameter of the circle with centre ‘O’. If \(\angle COB = 55^{\circ}\) then the value of x is: [1 Mark]
(A) \(27.5^{\circ}\)
(B) \(55^{\circ}\)
(C) \(110^{\circ}\)
(D) \(125^{\circ}\)

[Figure: Circle with centre O and diameter AB. Points A, B, C lie on the circumference forming triangle AOC with \(\angle COB = 55^{\circ}\) (linear pair with \(\angle AOC\)), and angle at C subtended by arc AB containing x.]

Answer: (A) \(27.5^{\circ}\)

\(\angle AOC = 180^{\circ} - 55^{\circ} = 125^{\circ}\). In \(\triangle AOC\), \(OA = OC\), so \(x = (180^{\circ} - 125^{\circ}) \div 2 = 27.5^{\circ}\).

Teacher's Note:
a) 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.
b) Alternatively, find \(\angle AOB = 125^{\circ}\), so the inscribed angle \(\angle ACB = \frac{1}{2} \times 125^{\circ} = 62.5^{\circ}\), then use triangle angle sum.

(iii) If a rectangular sheet having dimensions \(22\text{ cm} \times 11\text{ cm}\) is rolled along its shorter side to form a cylinder. Then the curved surface area of the cylinder so formed is: [1 Mark]
(A) \(968\text{ cm}^{2}\)
(B) \(424\text{ cm}^{2}\)
(C) \(121\text{ cm}^{2}\)
(D) \(242\text{ cm}^{2}\)

Answer: (D) \(242\text{ cm}^{2}\)

Circumference \(2\pi r = 11\text{ cm}\), height \(h = 22\text{ cm}\). Curved surface area \(= 2\pi rh = 11 \times 22 = 242\text{ cm}^{2}\).

Teacher's Note:
a) When rolling along the shorter side, the shorter side becomes the circumference and the longer side becomes the height.
b) Curved surface area equals base perimeter multiplied by height.

(iv) If the vertices of a triangle are \((1, 3)\), \((2, -4)\) and \((-3, 1)\). Then the co-ordinate of its centroid is: [1 Mark]
(A) \((0, 0)\)
(B) \((0, 1)\)
(C) \((1, 0)\)
(D) \((1, 1)\)

Answer: (A) \((0, 0)\)

Centroid \(x = \frac{1 + 2 + (-3)}{3} = 0\), \(y = \frac{3 + (-4) + 1}{3} = 0\).

Teacher's Note:
a) The formula for centroid is \(\left(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}\right)\).
b) Ensure proper handling of negative signs during addition.

(v) \(\tan \theta \times \sqrt{1 - \sin^{2}\theta}\) is equal to: [1 Mark]
(A) \(\cos \theta\)
(B) \(\sin \theta\)
(C) \(\tan \theta\)
(D) \(\cot \theta\)

Answer: (B) \(\sin \theta\)

\(\sqrt{1 - \sin^{2}\theta} = \cos \theta\). Then \(\frac{\sin \theta}{\cos \theta} \times \cos \theta = \sin \theta\).

Teacher's Note:
a) Use the fundamental trigonometric identity \(\sin^{2}\theta + \cos^{2}\theta = 1\).
b) Express \(\tan \theta\) as \(\frac{\sin \theta}{\cos \theta}\) to simplify expressions.

(vi) The median class for the given distribution is: [1 Mark]
(A) \(1 - 5\)
(B) \(6 - 10\)
(C) \(11 - 15\)
(D) \(11 - 20\)

Answer: (C) \(11 - 15\)

Total frequency \(N = 18\), so \(\frac{N}{2} = 9\). The cumulative frequency just greater than or equal to 9 is 11, corresponding to class \(11 - 15\).

Teacher's Note:
a) The median class is determined by finding the cumulative frequency closest to and greater than \(\frac{N}{2}\).
b) Do not confuse class intervals with frequency values.

(vii) If the lines \(7y = ax + 4\) and \(2y = 3 - x\) are parallel to each other, then the value of ‘a’ is: [1 Mark]
(A) \(-1\)
(B) \(-\frac{7}{2}\)
(C) \(-\frac{2}{7}\)
(D) \(14\)

Answer: (B) \(-\frac{7}{2}\)

Slopes are \(m_{1} = \frac{a}{7}\) and \(m_{2} = -\frac{1}{2}\). For parallel lines, \(m_{1} = m_{2}\), giving \(\frac{a}{7} = -\frac{1}{2}\), so \(a = -\frac{7}{2}\).

Teacher's Note:
a) Parallel lines have equal slopes (\(m_{1} = m_{2}\)).
b) Always rewrite the equation in slope-intercept form \(y = mx + c\) to correctly identify the slope.

(viii) Volume of a cylinder is \(330\text{ cm}^{3}\). The volume of the cone having same radius and height as that of the given cylinder is: [1 Mark]
(A) \(330\text{ cm}^{3}\)
(B) \(165\text{ cm}^{3}\)
(C) \(110\text{ cm}^{3}\)
(D) \(220\text{ cm}^{3}\)

Answer: (C) \(110\text{ cm}^{3}\)

Volume of cone \(= \frac{1}{3} \times \text{Volume of cylinder} = \frac{1}{3} \times 330 = 110\text{ cm}^{3}\).

Teacher's Note:
a) A cone and a cylinder of the same base radius and height are related by the volume formula ratio of \(1 : 3\).
b) Direct application of mensuration formulas saves valuable time.

(ix) In the given graph, the modal class is the class with frequency: [1 Mark]
(A) 72
(B) 21
(C) 48
(D) 36

[Figure: Bar graph / Histogram showing frequencies 21, 48, 72, 36, 48 for class intervals 0-10, 10-20, 20-30, 30-40, 40-50 with maximum frequency 72.]

Answer: (A) 72

The modal class is the class interval with the maximum frequency, which is 72.

Teacher's Note:
a) Mode represents the most frequently occurring observation.
b) Look for the highest bar in the histogram to find the modal class frequency.

(x) If the probability of a player winning a game is 0.56. The probability of his losing this game is: [1 Mark]
(A) 0.56
(B) 1
(C) 0.44
(D) 0

Answer: (C) 0.44

\(P(\text{losing}) = 1 - P(\text{winning}) = 1 - 0.56 = 0.44\).

Teacher's Note:
a) The sum of probabilities of all complementary events is equal to 1.
b) Ensure correct decimal subtraction.

SECTION B

(Attempt any three questions from this Section.)

Question 2

(i) Find the ratio in which the x-axis divides internally the line joining points A (6, -4) and B ( -3, 8). [2 Marks]

Answer:
Let the ratio be \(k : 1\).
Using the section formula, the coordinates of the dividing point are:
\((x, y) = \left(\frac{-3k + 6}{k + 1}, \frac{8k - 4}{k + 1}\right)\)
Since the point lies on the x-axis, its y-coordinate is 0:
\(\frac{8k - 4}{k + 1} = 0\)
\(8k - 4 = 0 \implies k = \frac{4}{8} = \frac{1}{2}\)
Thus, the required ratio is \(1 : 2\).

Teacher's Note:
a) For a point on the x-axis, substitute \(y = 0\); for the y-axis, substitute \(x = 0\).
b) Express the final ratio in simplest form: \(1 : 2\).

(ii) Three rotten apples are accidently mixed with twelve good ones. One apple is picked at random. What is the probability that it is a good one? [2 Marks]

Answer:
Number of good apples \(= 12\)
Number of rotten apples \(= 3\)
Total number of apples \(= 12 + 3 = 15\)
\(P(\text{good apple}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{15} = \frac{4}{5}\)

Teacher's Note:
a) Total outcomes must include both good and defective items.
b) Always reduce fractions to their lowest terms.

(iii) In the given figure , AC is a tangent to circle at point B. \(\triangle EFD\) is an equilateral triangle and \(\angle CBD = 40^{\circ}\). Find: [3 Marks]
(a) \(\angle BFD\)
(b) \(\angle FBD\)
(c) \(\angle ABF\)

[Figure: Circle with inscribed equilateral triangle EFD and tangent AC touching at B. Line chords form various angles with \(\angle CBD = 40^{\circ}\).]

Answer:
(a) By alternate segment theorem, \(\angle BFD = \angle CBD = 40^{\circ}\).
(b) Since \(BFED\) is a cyclic quadrilateral, opposite angles sum to \(180^{\circ}\):
\(\angle FBD + \angle FED = 180^{\circ}\)
Since \(\triangle EFD\) is equilateral, \(\angle FED = 60^{\circ}\):
\(\angle FBD = 180^{\circ} - 60^{\circ} = 120^{\circ}\).
(c) Since \(ABC\) is a straight line, angles on a straight line sum to \(180^{\circ}\):
\(\angle ABF + \angle FBD + \angle CBD = 180^{\circ}\)
\(\angle ABF + 120^{\circ} + 40^{\circ} = 180^{\circ}\)
\(\angle ABF = 180^{\circ} - 160^{\circ} = 20^{\circ}\).

Teacher's Note:
a) Recall the alternate segment theorem: angle between tangent and chord equals angle in alternate segment.
b) Properties of cyclic quadrilaterals and equilateral triangles are crucial for angle-chasing problems.

(iv) A drone camera is used to shoot an object P from two different positions R and S along the same vertical line QRS. The angle of depression of the object P from these two positions are \(35^{\circ}\) and \(60^{\circ}\) respectively as shown in the diagram. If the distance of the object P from point Q is \(50\) metres, then find the distance between R and S correct to the nearest meter. [3 Marks]

[Figure: Right-angled triangles \(\triangle PQR\) and \(\triangle PQS\) sharing base PQ = 50 m, with angles of depression \(35^{\circ}\) at R and \(60^{\circ}\) at S.]

Answer:
In right \(\triangle PQR\):
\(\tan 35^{\circ} = \frac{RQ}{PQ} = \frac{RQ}{50}\)
\(RQ = 50 \times 0.7002 = 35.01\text{ m}\)
In right \(\triangle PQS\):
\(\tan 60^{\circ} = \frac{SQ}{PQ} = \frac{SQ}{50}\)
\(SQ = 50 \times 1.732 = 86.6\text{ m}\)
Distance between R and S is:
\(RS = SQ - RQ = 86.6 - 35.01 = 51.59\text{ m} \approx 52\text{ m}\)
(Note: Using \(\sqrt{3} = 1.73\), \(SQ = 86.5\text{ m}\), giving \(RS = 51.49\text{ m} \approx 51\text{ m}\) as per official key calculations.)

Teacher's Note:
a) Angles of elevation and depression are equal due to alternate interior angles.
b) Subtract the smaller height from the larger height to find the distance between the two observation points.

Question 3

(i) In the given figure, PT is a tangent to the circle at T, chord BA is produced to meet the tangent at P. Perpendicular BC bisects the chord TA at C. If PA = \(9\text{ cm}\) and TB = \(7\text{ cm}\), find the lengths of: [2 Marks]
(a) AB
(b) PT

[Figure: Circle with tangent PT, secant PAB, chord AT bisected perpendicularly by BC at C, with PA = 9 cm and TB = 7 cm.]

Answer:
(a) In \(\triangle ABT\), since \(BC\) is the perpendicular bisector of \(TA\), \(\triangle ACB \cong \triangle BCT\).
Therefore, \(AB = TB = 7\text{ cm}\).
(b) By tangent-secant theorem:
\(PT^{2} = PA \times PB\)
Given \(PB = PA + AB = 9 + 7 = 16\text{ cm}\).
\(PT^{2} = 9 \times 16 = 144\)
\(PT = \sqrt{144} = 12\text{ cm}\).

Teacher's Note:
a) The perpendicular bisector of a chord passes through the centre and bisects the subtended arcs.
b) The tangent-secant theorem states that \(PT^{2} = PA \times PB\).

(ii) How many solid right circular cylinders of radius \(2\text{ cm}\) and height \(3\text{ cm}\) can be made by melting a solid right circular cylinder of diameter \(12\text{ cm}\) and height \(15\text{ cm}\)? [2 Marks]

Answer:
For the bigger cylinder: radius \(R = \frac{12}{2} = 6\text{ cm}\), height \(H = 15\text{ cm}\).
Volume \(V_{1} = \pi R^{2}H = \pi \times 6^{2} \times 15\)
For the smaller cylinder: radius \(r = 2\text{ cm}\), height \(h = 3\text{ cm}\).
Volume \(V_{2} = \pi r^{2}h = \pi \times 2^{2} \times 3\)
Number of cylinders \(= \frac{V_{1}}{V_{2}} = \frac{\pi \times 6 \times 6 \times 15}{\pi \times 2 \times 2 \times 3} = 45\).

Teacher's Note:
a) Number of melted and recast items equals the ratio of the volume of the larger solid to the smaller solid.
b) Do not expand \(\pi\) prematurely; it cancels out in calculations.

(iii) Prove that: [3 Marks]
\(\frac{\cos^{2}A}{\cos A - \sin A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A\)

Answer:
\(LHS = \frac{\cos^{2}A}{\cos A - \sin A} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}}\)
\(= \frac{\cos^{2}A}{\cos A - \sin A} + \frac{\sin^{2}A}{\sin A - \cos A}\)
\(= \frac{\cos^{2}A}{\cos A - \sin A} - \frac{\sin^{2}A}{\cos A - \sin A}\)
\(= \frac{\cos^{2}A - \sin^{2}A}{\cos A - \sin A}\)
\(= \frac{(\cos A + \sin A)(\cos A - \sin A)}{\cos A - \sin A}\)
\(= \cos A + \sin A = RHS\). Hence Proved.

Teacher's Note:
a) Convert all trigonometric ratios into sine and cosine where helpful, or use cotangent definition.\
b) Use algebraic identities like difference of squares to simplify numerators.

(iv) Use graph paper for this question, take \(2\text{ cm} = 10\) marks along one axis and \(2\text{ cm} = 10\) students along the other axis. The following table shows the distribution of marks in a 50 marks test in Mathematics: [3 Marks]

Draw the ogive for the above distribution and hence estimate the median marks.

[Figure: Cumulative frequency table and Ogive graph plotted with points (10, 6), (20, 16), (30, 29), (40, 36), (50, 40) yielding estimated median at 23 marks.]

Answer:

Points plotted: \((10, 6)\), \((20, 16)\), \((30, 29)\), \((40, 36)\), \((50, 40)\).
Median \(= \left(\frac{N}{2}\right)^{\text{th}}\text{ term} = \left(\frac{40}{2}\right)^{\text{th}} = 20^{\text{th}}\text{ term}\).
From the graph, corresponding value is \(23\text{ marks}\).

Teacher's Note:
a) Always plot cumulative frequencies against upper class limits for a less-than ogive.
b) Clearly mark the median position on the cumulative frequency axis and read the corresponding value on the x-axis.

Question 4

(i) Find the equation of the perpendicular dropped from the point P (-1,2) onto the line joining A (1,4) and B (2,3). [2 Marks]

[Figure: Line segment AB with points A(1,4), B(2,3) and point P(-1,2) with perpendicular dropped from P meeting AB at Q.]

Answer:
Slope of line \(AB\), \(m_{1} = \frac{3 - 4}{2 - 1} = \frac{-1}{1} = -1\).
Since the perpendicular line \(PQ\) is perpendicular to \(AB\), its slope \(m_{2} = -\frac{1}{m_{1}} = -\frac{1}{-1} = 1\).
Equation of line passing through \((-1, 2)\) with slope \(1\) is:
\(y - 2 = 1(x - (-1))\)
\(y - 2 = x + 1 \implies y - x = 3\) (or \(x - y + 3 = 0\)).

Teacher's Note:
a) Product of slopes of two perpendicular lines is \(-1\).
b) Use the point-slope form \(y - y_{1} = m(x - x_{1})\) to find the equation.

(ii) Find the mean for the following distribution: [2 Marks]

Answer:
\(\text{Mean } \bar{x} = \frac{\sum f_{i}x_{i}}{\sum f_{i}} = \frac{1160}{20} = 58\).

Teacher's Note:
a) Class mark is the average of the upper and lower class limits: \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\).
b) Double-check column multiplication and summation steps.

(iii) A solid piece of wooden cone is of radius \(OP = 7\text{ cm}\) and height \(OQ = 12\text{ cm}\). A cylinder whose radius and height equal to half of that of the cone is drilled out from this piece of wooden cone. Find the volume of the remaining piece of wood. (Use, \(\pi = \frac{22}{7}\)) [3 Marks]

[Figure: Cone with radius 7 cm and height 12 cm containing an inner coaxial cylinder of radius 3.5 cm and height 6 cm.]

Answer:
Radius of cone \(R = 7\text{ cm}\), Height of cone \(H = 12\text{ cm}\).
Volume of cone \(= \frac{1}{3}\pi R^{2}H = \frac{1}{3} \times \frac{22}{7} \times 7^{2} \times 12 = 616\text{ cm}^{3}\).
Radius of cylinder \(r = \frac{7}{2}\text{ cm}\), Height of cylinder \(h = \frac{12}{2} = 6\text{ cm}\).
Volume of cylinder \(= \pi r^{2}h = \frac{22}{7} \times \left(\frac{7}{2}\right)^{2} \times 6 = \frac{22}{7} \times \frac{49}{4} \times 6 = 231\text{ cm}^{3}\).
Volume of remaining solid \(= 616 - 231 = 385\text{ cm}^{3}\).

Teacher's Note:
a) When a solid is drilled out, subtract the volume of the removed part from the total volume.
b) Substitute dimensions carefully keeping fractions intact before final calculation.

(iv) Use a graph sheet for this question, take \(2\text{cm} = 1\) unit along both x and y axis: [3 Marks]
(a) Plot the points A \((3, 2)\) and B \((5, 0)\). Reflect point A on the y-axis to \(A'\). Write co-ordinates of \(A'\).
(b) Reflect point B on the line \(AA'\) to \(B'\). Write the co-ordinates of \(B'\).
(c) Name the closed figure \(A'B'AB\).

[Figure: Cartesian plane showing triangle/quadrilateral transformations with points A(3,2), B(5,0), A'(-3,2), B'(-5,4).]

Answer:
(a) \(A'(-3, 2)\).
(b) Using reflection formula in line \(y = k\), point \((a, b)\) reflects to \((a, 2k - b)\). Here line \(AA'\) is \(y = 2\), so \(B'(5, 2(2) - 0)\) i.e., \(B'(5, 4)\).
(c) The closed figure \(A'B'AB\) is an arrowhead (or delta / deltoid / kite shape).

Teacher's Note:
a) Reflection in the y-axis negates the x-coordinate: \((x, y) \to (-x, y)\).
b) Verify coordinates by plotting accurately on graph paper.

Question 5

(i) In the given figure, the sides of the quadrilateral PQRS touches the circle at A,B,C and D. If RC = \(4\text{ cm}\), RQ = \(7\text{ cm}\) and PD = \(5\text{ cm}\). Find the length of PQ: [2 Marks]

[Figure: Quadrilateral PQRS circumscribing a circle, touching at points A, B, C, D with RC = 4 cm, RQ = 7 cm, PD = 5 cm.]

Answer:
\(PD = PA = 5\text{ cm}\) (tangents from external point P)
\(RC = RB = 4\text{ cm}\) (tangents from external point R)
Given \(RQ = RB + BQ = 7\text{ cm} \implies 4 + BQ = 7 \implies BQ = 3\text{ cm}\).
Since \(QA = QB = 3\text{ cm}\) (tangents from Q):
\(PQ = PA + AQ = 5 + 3 = 8\text{ cm}\).

Teacher's Note:
a) Tangents drawn from an external point to a circle are equal in length.
b) Break down side lengths using segment addition postulates.

(ii) Prove that: [2 Marks]
\(\frac{\sin^{3}\theta + \cos^{3}\theta}{\sin \theta + \cos \theta} = 1 - \sin \theta \cos \theta\)

Answer:
\(LHS = \frac{(\sin \theta + \cos \theta)(\sin^{2}\theta - \sin \theta \cos \theta + \cos^{2}\theta)}{\sin \theta + \cos \theta}\)
\(= \sin^{2}\theta + \cos^{2}\theta - \sin \theta \cos \theta\)
Since \(\sin^{2}\theta + \cos^{2}\theta = 1\),
\(= 1 - \sin \theta \cos \theta = RHS\). Hence Proved.

Teacher's Note:
a) Use the algebraic expansion formula for sum of cubes: \(a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\).
b) Apply the Pythagorean trigonometric identity \(\sin^{2}\theta + \cos^{2}\theta = 1\).

(iii) In the given diagram, \(OA = OB\), \(\angle OAB = \theta\) and the line AB passes through point P \((-3, 4)\). [3 Marks]
Find:
(a) Slope and inclination (\(\theta\)) of the line AB
(b) Equation of the line AB

[Figure: Line AB passing through P(-3, 4) making angle \(\theta\) with x-axis, where OA = OB with origin O.]

Answer:
(a) Since \(OA = OB\), \(\triangle OAB\) is an isosceles triangle with \(\angle OAB = \angle OBA = \theta\).
The exterior angle at O is \(90^{\circ}\), so \(\theta + \theta = 90^{\circ} \implies 2\theta = 90^{\circ} \implies \theta = 45^{\circ}\).
Slope \(m = \tan 45^{\circ} = 1\).
(b) Equation of line passing through \((-3, 4)\) with slope \(1\):
\(y - 4 = 1(x - (-3))\)
\(y - 4 = x + 3 \implies x - y + 7 = 0\).

Teacher's Note:
a) Base angles of an isosceles triangle are equal.
b) Slope of a line is equal to the tangent of its inclination angle (\(\tan \theta\)).

(iv) Use graph paper for this question. Estimate the mode of the given distribution by plotting a histogram. [Take \(2\text{ cm} = 10\) marks along one axis and \(2\text{ cm} = 5\) students along the other axis] [3 Marks]

[Figure: Histogram showing frequency bars with modal class 50-60 and estimated mode graphically at 56.]

Answer:
Modal class is \(50 - 60\) (highest frequency 20).
Plotting the histogram with the given scale and drawing cross lines from the top corners of the modal rectangle to adjacent rectangles, the perpendicular dropped to the x-axis gives the estimated mode as \(\text{Rs. } 56\).

Teacher's Note:
a) The modal class has the maximum height in a histogram.
b) Mode is determined graphically by intersecting diagonal lines from the top corners of the highest bar and its adjacent bars.

Question 6

(i) A box contains tokens numbered 5 to 16. A token is drawn at random. Find the probability that the token drawn bears a number divisible by: [2 Marks]
(a) 5
(b) Neither by 2 nor by 3

Answer:
Sample space \(= \{5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16\}\), Total tokens \(= 12\).
(a) Favorable outcomes divisible by 5: \(\{5, 10, 15\}\) (3 outcomes).
\(P(\text{divisible by } 5) = \frac{3}{12} = \frac{1}{4}\).
(b) Favorable outcomes neither divisible by 2 nor by 3: \(\{5, 7, 11, 13\}\) (4 outcomes).
\(P(\text{neither by 2 nor by 3}) = \frac{4}{12} = \frac{1}{3}\).

Teacher's Note:
a) Count sample space carefully when starting from numbers other than 1 (from 5 to 16 inclusive gives \(16 - 5 + 1 = 12\) items).
b) List elements explicitly to avoid counting errors.

(ii) Point M (2, b) is the mid-point of the line segment joining points P (a, 7) and Q (6, 5). Find the values of ‘a’ and ‘b’. [2 Marks]

[Figure: Line segment PQ with midpoint M(2, b), P(a, 7), and Q(6, 5).]

Answer:
Using the midpoint formula \(\left(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right)\):
\((2, b) = \left(\frac{a + 6}{2}, \frac{7 + 5}{2}\right)\)
Equating x-coordinates:
\(\frac{a + 6}{2} = 2 \implies a + 6 = 4 \implies a = -2\)
Equating y-coordinates:
\(b = \frac{12}{2} = 6\)
Thus, \(a = -2, b = 6\).

Teacher's Note:
a) The midpoint coordinates are the arithmetic mean of respective endpoints.
b) Solve independent linear equations for each coordinate separately.

(iii) An aeroplane is flying horizontally along a straight line at a height of \(3000\text{ m}\) from the ground at a speed of \(160\text{ m/s}\). Find the time it would take for the angle of elevation of the plane as seen from a particular point on the ground to change from \(60^{\circ}\) to \(45^{\circ}\). Give your answer correct to the nearest second. [3 Marks]

[Figure: Two right-angled triangles sharing perpendicular height 3000 m, with base distances corresponding to angles of elevation \(60^{\circ}\) and \(45^{\circ}\).]

Answer:
Height \(AC = ED = 3000\text{ m}\).
In right \(\triangle ACB\) (\(60^{\circ}\)):
\(\tan 60^{\circ} = \frac{AC}{BC} \implies \sqrt{3} = \frac{3000}{BC} \implies BC = \frac{3000}{\sqrt{3}} = 1000\sqrt{3}\text{ m}\).
In right \(\triangle EDB\) (\(45^{\circ}\)):
\(\tan 45^{\circ} = \frac{ED}{BD} \implies 1 = \frac{3000}{BD} \implies BD = 3000\text{ m}\).
Distance travelled \(AE = CD = BD - BC = 3000 - 1000\sqrt{3}\)
\(= 3000 - 1000(1.732) = 3000 - 1732 = 1268\text{ m}\).
\(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1268}{160} = 7.925\text{ seconds} \approx 8\text{ seconds}\).

Teacher's Note:
a) Distance travelled by the plane equals the difference between the base distances of the two observer angles.
b) Round off the final time to the nearest second as requested.

(iv) Given that the mean of the following frequency distribution is \(30\), find the missing frequency ‘f’ [3 Marks]

Answer:
Using assumed mean / step deviation method:
\(\text{Mean } \bar{x} = A + \left(\frac{\sum f_{i}t_{i}}{\sum f_{i}}\right) \times i\)
\(30 = 35 + \left(\frac{-20}{30 + f}\right) \times 10\)
\(30 - 35 = \frac{-200}{30 + f}\)
\(-5 = \frac{-200}{30 + f}\)
\(5(30 + f) = 200 \implies 30 + f = 40 \implies f = 10\).

Teacher's Note:
a) Step deviation method simplifies calculations for missing frequency problems.
b) Verify the computed missing frequency by substituting back into the mean formula.