ICSE Class 10 Mathematics Board Exam Question Paper 2022 with Solutions

ICSE Board Class 10 Mathematics Board Exam Question Paper with Solutions

SECTION A

Question 1
Choose the correct answers to the questions from the given options. (Do not copy the question, Write the correct answer only.)

(i) The probability of getting a number divisible by 3 in throwing a dice is: [1 Mark]
(A) \(\dfrac{1}{6}\)
(B) \(\dfrac{1}{3}\)
(C) \(\dfrac{1}{2}\)
(D) \(\dfrac{2}{3}\)

Answer: (B) \(\dfrac{1}{3}\)

Favourable outcomes divisible by 3 are 3 and 6, so total favourable outcomes = 2. Total possible outcomes = 6. Probability \(P(E) = \dfrac{2}{6} = \dfrac{1}{3}\).

Teacher's Note:
a) Always write the sample space first before finding favourable outcomes.
b) Remember to simplify the fraction to its lowest terms.

(ii) The volume of a conical tent is \(462\text{ m}^3\) and the area of the base is \(154\text{ m}^2\), the height of the cone is: [1 Mark]
(A) \(15\text{ m}\)
(B) \(12\text{ m}\)
(C) \(9\text{ m}\)
(D) \(24\text{ m}\)

Answer: (C) \(9\text{ m}\)

Volume of a cone \( = \dfrac{1}{3} \times \text{Base Area} \times \text{Height}\). Substituting the values, \(462 = \dfrac{1}{3} \times 154 \times h\), which gives \(h = \dfrac{462 \times 3}{154} = 9\text{ m}\).

Teacher's Note:
a) Base area of a cone is given by \(\pi r^2\), which simplifies calculations when given directly.
b) Do not forget to include the \(\dfrac{1}{3}\) factor for the volume of a cone.

(iii) The median class for the given distribution is: [1 Mark]

(A) 0 - 10
(B) 10 - 20
(C) 20 - 30
(D) 30 - 40

Answer: (C) 20 - 30

Total frequency \(n = 14\). \(\dfrac{n}{2} = 7\). The cumulative frequencies are: 0-10 (2), 10-20 (6), 20-30 (9), 30-40 (14). The 7th term lies in the interval 20 - 30.

Teacher's Note:
a) Always construct a cumulative frequency table to locate the median class.
b) The median class is the first class whose cumulative frequency is greater than or equal to \(\dfrac{n}{2}\).

(iv) If two lines are perpendicular to one another then the relation between their slopes \(m_1\) and \(m_2\) is: [1 Mark]
(A) \(m_1 = m_2\)
(B) \(m_1 = \dfrac{1}{m_2}\)
(C) \(m_1 = -m_2\)
(D) \(m_1 \times m_2 = -1\)

Answer: (D) \(m_1 \times m_2 = -1\)

Two non-vertical lines are perpendicular if and only if the product of their slopes is \(-1\).

Teacher's Note:
a) For parallel lines, the slopes are equal (\(m_1 = m_2\)).
b) For perpendicular lines, the product of slopes is \(-1\), or one is the negative reciprocal of the other.

(v) A lighthouse is \(80\text{ m}\) high. The angle of elevation of its top from a point \(80\text{ m}\) away from its foot along the same horizontal line is: [1 Mark]
(A) \(60^{\circ}\)
(B) \(45^{\circ}\)
(C) \(30^{\circ}\)
(D) \(90^{\circ}\)

Answer: (B) \(45^{\circ}\)

\(\tan \theta = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{80}{80} = 1\). Therefore, \(\theta = 45^{\circ}\).

Teacher's Note:
a) When the height and the distance from the base are equal, the angle of elevation is always \(45^{\circ}\).
b) Ensure trigonometric ratios are correctly applied to right-angled triangles.

(vi) The modal class of a given distribution always corresponds to the: [1 Mark]
(A) interval with highest frequency
(B) Interval with lowest frequency
(C) The first interval
(D) The last interval

Answer: (A) interval with highest frequency

By definition, the modal class is the class interval with the maximum frequency in a frequency distribution.

Teacher's Note:
a) Mode represents the most frequently occurring observation.
b) Do not confuse modal class with median class.

(vii) The coordinates of the point P(\(-3, 5\)) on reflecting on the \(x\)-axis are: [1 Mark]
(A) \((3, 5)\)
(B) \((-3, -5)\)
(C) \((3, -5)\)
(D) \((-3, 5\))

Answer: (B) \((-3, -5)\)

When a point is reflected across the \(x\)-axis, its \(x\)-coordinate remains the same and its \(y\)-coordinate changes sign: \((x, y) \to (x, -y)\).

Teacher's Note:
a) Reflection in the \(x\)-axis changes the sign of the ordinate (\(y\)).
b) Reflection in the \(y\)-axis changes the sign of the abscissa (\(x\)).

(viii) ABCD is a cyclic quadrilateral. If \(\angle BAD = (2x + 5)^{\circ}\) and \(\angle BCD = (x + 10)^{\circ}\) then \(x\) is equal to: [1 Mark]
(A) \(65^{\circ}\)
(B) \(45^{\circ}\)
(C) \(55^{\circ}\)
(D) \(5^{\circ}\)

Answer: (C) \(55^{\circ}\)

Opposite angles of a cyclic quadrilateral are supplementary. Therefore, \((2x + 5) + (x + 10) = 180 \implies 3x + 15 = 180 \implies 3x = 165 \implies x = 55^{\circ}\).

Teacher's Note:
a) Recall the theorem that the sum of opposite angles of a cyclic quadrilateral is \(180^{\circ}\).
b) Solve linear equations carefully without missing constant terms.

(ix) A \((1, 4)\), B \((4, 1)\) and C \((x, 4)\) are the vertices of \(\Delta ABC\). If the centroid of the triangle is G \((4, 3)\) then \(x\) is equal to: [1 Mark]
(A) 2
(B) 1
(C) 7
(D) 4

Answer: (C) 7

The \(x\)-coordinate of the centroid is given by \(\dfrac{x_1 + x_2 + x_3}{3}\). So, \(\dfrac{1 + 4 + x}{3} = 4 \implies 5 + x = 12 \implies x = 7\).

Teacher's Note:
a) Centroid formula for coordinates is \(\left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)\).
b) Equate only the \(x\)-coordinates since \(x\) is the unknown required.

(x) The radius of a roller \(100\text{ cm}\) long is \(14\text{ cm}\). The curved surface area of the roller is (Take \(\pi = \dfrac{22}{7}\)): [1 Mark]
(A) \(13200\text{ cm}^2\)
(B) \(15400\text{ cm}^2\)
(C) \(4400\text{ cm}^2\)
(D) \(8800\text{ cm}^2\)

Answer: (D) \(8800\text{ cm}^2\)

Curved Surface Area of cylinder \( = 2\pi rh = 2 \times \dfrac{22}{7} \times 14 \times 100 = 88 \times 100 = 8800\text{ cm}^2\).

Teacher's Note:
a) Identify the length of the roller as the height (\(h\)) of the cylinder.
b) Always substitute the standard value of \(\pi\) as \(\dfrac{22}{7}\) unless specified otherwise.

SECTION B

Question 2:
(i) Prove that: \(\dfrac{1}{1 + \sin \theta} + \dfrac{1}{1 - \sin \theta} = 2\sec^2 \theta\) [3 Marks]

Answer:
\(\text{LHS} = \dfrac{1}{1 + \sin \theta} + \dfrac{1}{1 - \sin \theta}\)
\(=\dfrac{(1 - \sin \theta) + (1 + \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)}\)
\(=\dfrac{2}{1 - \sin^2 \theta}\)
\(=\dfrac{2}{\cos^2 \theta}\)
\(= 2 \sec^2 \theta = \text{RHS}\)

Teacher's Note:
a) Take the common denominator using algebraic identity \((a + b)(a - b) = a^2 - b^2\).
b) Use the fundamental trigonometric identity \(1 - \sin^2 \theta = \cos^2 \theta\).

(ii) Find \(a\) if A \((2a + 2, 3)\), B \((7, 4)\) and C \((2a + 5, 2)\) are collinear. [3 Marks]

Answer:
For three points to be collinear, the slope of AB must be equal to the slope of BC.
Slope of AB \( = \dfrac{4 - 3}{7 - (2a + 2)} = \dfrac{1}{7 - 2a - 2} = \dfrac{1}{5 - 2a}\)
Slope of BC \( = \dfrac{2 - 4}{2a + 5 - 7} = \dfrac{-2}{2a - 2} = \dfrac{-1}{a - 1}\)
Equating slopes: \(\dfrac{1}{5 - 2a} = \dfrac{-1}{a - 1}\)
\(a - 1 = -(5 - 2a)\)
\(a - 1 = -5 + 2a\)
\(-1 + 5 = 2a - a\)
\(a = 4\)

Teacher's Note:
a) Collinearity means points lie on the same straight line, hence their slopes are equal.
b) Be careful with negative signs and brackets when subtracting coordinate expressions.

(iii) Calculate the mean of the following frequency distribution. [4 Marks]

Answer:

Mean \( = \dfrac{\sum fx}{\sum f} = \dfrac{780}{24} = 32.5\)

Teacher's Note:
a) Class mark \(x\) is calculated as \(\dfrac{\text{Lower Limit} + \text{Upper Limit}}{2}\).
b) Always double-check summation calculations for \(\sum f\) and \(\sum fx\).

(iv) In the given figure O is the center of the circle. PQ and PR are tangent and \(\angle QPR = 70^{\circ}\) calculate
(a) \(\angle QOR\)
(b) \(\angle QSR\) [4 Marks]
[Figure: A circle with center O, external point P with tangents PQ and PR, \(\angle QPR = 70^{\circ}\), and chord QR subtending angle at point S on the major arc]

Answer:
(a) In quadrilateral PQOR, the radius drawn to the tangent is perpendicular to it, so \(\angle OQP = 90^{\circ}\) and \(\angle ORP = 90^{\circ}\).
Sum of angles in quadrilateral \(PQOR = 360^{\circ}\).
\(\angle QOR + 90^{\circ} + 90^{\circ} + 70^{\circ} = 360^{\circ}\)
\(\angle QOR + 250^{\circ} = 360^{\circ}\)
\(\angle QOR = 360^{\circ} - 250^{\circ} = 110^{\circ}\).

(b) \(\angle QSR = \dfrac{1}{2} \angle QOR = \dfrac{1}{2} \times 110^{\circ} = 55^{\circ}\).

Teacher's Note:
a) The angle between a radius and tangent at the point of contact is \(90^{\circ}\).
b) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Question 3:
(i) A bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random. What is the probability that the ball drawn is a red ball? [3 Marks]

Answer:
Number of white balls \(= 5\)
Number of red balls \(= 2\)
Number of black balls \(= 3\)
Total number of balls \(= 5 + 2 + 3 = 10\)
Number of favourable outcomes (red ball) \(= 2\)
Probability \(P(\text{Red}) = \dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}} = \dfrac{2}{10} = \dfrac{1}{5}\)

Teacher's Note:
a) Total outcomes must be the sum of all individual categories.
b) Express the final probability in its simplest fraction form.

(ii) A Solid cone of radius \(5\text{ cm}\) and height \(9\text{ cm}\) is melted and made into small cylinders of radius of \(0.5\text{ cm}\) and height \(1.5\text{ cm}\). Find the number of cylinders so formed. [3 Marks]

Answer:
Volume of cone \( = \dfrac{1}{3}\pi r^2h = \dfrac{1}{3} \times \pi \times (5)^2 \times 9 = \pi \times 25 \times 3 = 75\pi\text{ cm}^3\)
Volume of one small cylinder \( = \pi r^2h = \pi \times (0.5)^2 \times 1.5 = \pi \times 0.25 \times 1.5 = 0.375\pi\text{ cm}^3\)
Let \(n\) be the number of cylinders formed.
\(n \times 0.375\pi = 75\pi\)
\(n = \dfrac{75}{0.375} = 200\)

Teacher's Note:
a) When a solid is melted and recast, the total volume remains conserved.
b) Divide the volume of the larger solid by the volume of one smaller solid to find the total count.

(iii) Two lamp posts AB and CD each of height \(100\text{ m}\) are on either side of the road. P is a point on the road between the two lamp posts. The angles of elevation of the top of the lamp posts from the point P are \(60^{\circ}\) and \(40^{\circ}\). Find the distances PB and PD. [4 Marks]
[Figure: Two vertical poles AB and CD of height 100 m each, separated by road BD, with point P between them. Angles of elevation from P to A is \(40^{\circ}\) and to C is \(60^{\circ}\).]

Answer:
In right-angled triangle ABP:
\(\tan 40^{\circ} = \dfrac{AB}{BP} \implies 0.8391 = \dfrac{100}{BP} \implies BP = \dfrac{100}{0.8391} \approx 119.18\text{ m}\) (using standard \(\tan 40^{\circ} \approx 0.8391\), or \(119\text{ m}\) as per standard approximation guidelines).
In right-angled triangle CPD:
\(\tan 60^{\circ} = \dfrac{CD}{PD} \implies \sqrt{3} = \dfrac{100}{PD} \implies PD = \dfrac{100}{\sqrt{3}} = \dfrac{100\sqrt{3}}{3} = \dfrac{100 \times 1.732}{3} = 57.73\text{ m}\).

Teacher's Note:
a) Separate the composite figure into two individual right-angled triangles.
b) Use proper trigonometric tables or standard values for trigonometric ratios.

(iv) Marks obtained by 100 students in an examination are given below. [4 Marks]

Draw a histogram for the given data using a graph and find the mode.
Take \(2\text{ cm} = 10\) marks along one axis and \(2\text{ cm} = 10\) students along the other axis.

Answer:
1. Construct a histogram with class intervals on the \(x\)-axis and frequencies on the \(y\)-axis according to the given scale.
2. Locate the highest rectangle representing the modal class (30 - 40 with frequency 28).
3. Draw straight lines diagonally from the top corners of the modal rectangle to the top corners of the adjacent rectangles.
4. From the intersection point inside the modal bar, draw a perpendicular down to the \(x\)-axis.
5. Reading from the graph, the estimated mode is approximately \(34\).

Teacher's Note:
a) Ensure axes are properly scaled as requested in the question.
b) Mode is graphically determined using the highest bar in a histogram.

Question 4:
(i) Find a point P which divides internally the line segment joining the points A \((-3, 9)\) and B \((1, -3)\) in the ratio \(1:3\). [3 Marks]

Answer:
Using the section formula for internal division:\br />\(x = \dfrac{mx_2 + nx_1}{m + n} = \dfrac{1(1) + 3(-3)}{1 + 3} = \dfrac{1 - 9}{4} = \dfrac{-8}{4} = -2\)
\(y = \dfrac{my_2 + ny_1}{m + n} = \dfrac{1(-3) + 3(9)}{1 + 3} = \dfrac{-3 + 27}{4} = \dfrac{24}{4} = 6\)
Therefore, the coordinates of point P are \((-2, 6\)).

Teacher's Note:
a) State the section formula clearly before substituting values.
b) Pay close attention to negative signs during multiplication and addition.

(ii) A letter of the word SECONDARY is selected at random. What is the probability that the letter selected is not a vowel? [3 Marks]

Answer:
Total letters in the word SECONDARY \(= 9\) (S, E, C, O, N, D, A, R, Y).
Vowels in the word \(= \text{E, O, A}\) (Total 3 vowels).
Letters that are not vowels \(= 9 - 3 = 6\) (S, C, N, D, R, Y).
Probability of selecting a non-vowel \( = \dfrac{6}{9} = \dfrac{2}{3}\).

Teacher's Note:
a) Count the exact number of distinct or total letters as specified in the word.
b) Subtract the number of vowels from the total count to get the number of non-vowel outcomes.

(iii) Use a graph paper for this question. Take \(2\text{ cm} = 1\) unit along both the axes.
(a) Plot the points A\((0, 4)\), B\((2, 2)\), C\((5, 2)\) and D\((4, 0)\). E \((0, 0)\) is the origin.
(b) Reflect B, C, D on the \(y\)-axis and name them as B\('\), C\('\) and D\('\) respectively.
(c) Join the points A, B, C, D, D\('\), C\('\), B\('\) and A in order and give a geometrical name to the closed figure [4 Marks]

Answer:
(a) Plot the given points on the graph paper.
(b) Reflected coordinates across the \(y\)-axis (\(x \to -x\)):
B\('\) \(= (-2, 2)\)
C\('\) \(= (-5, 2)\)
D\('\) \(= (-4, 0)\)
(c) The closed figure formed by joining A, B, C, D, D\('\), C\('\), B\('\) and A in order is a Hexagon.

Teacher's Note:
a) Reflection in the \(y\)-axis negates the \(x\)-coordinate while keeping the \(y\)-coordinate unchanged.
b) Count the number of boundary vertices to correctly identify the polygon.

(iv) A solid wooden cylinder is of radius \(6\text{ cm}\) and height \(16\text{ cm}\). Two cones each of radius \(2\text{ cm}\) and height \(6\text{ cm}\) are drilled out of the cylinder. Find the volume of the remaining solid. (Take \(\pi = \dfrac{22}{7}\)) [4 Marks]
[Figure: Cylinder of height 16 cm and radius 6 cm with two conical cavities of radius 2 cm and height 6 cm drilled out from each end]

Answer:
Volume of the cylinder \(= \pi r^2 h = \pi \times (6)^2 \times 16 = 36 \times 16\pi = 576\pi\text{ cm}^3\)
Volume of one cone \( = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \pi \times (2)^2 \times 6 = 8\pi\text{ cm}^3\)
Volume of two cones \(= 2 \times 8\pi = 16\pi\text{ cm}^3\)
Volume of remaining solid \( = \text{Volume of cylinder} - \text{Volume of two cones}\)
\(= 576\pi - 16\pi = 560\pi\text{ cm}^3\)
\(= 560 \times \dfrac{22}{7} = 80 \times 22 = 1760\text{ cm}^3\).

Teacher's Note:
a) When objects are drilled out, subtract their volumes from the total volume of the solid.
b) Keep \(\pi\) in symbolic form until the final calculation step to simplify arithmetic.

Question 5:
(i) Two chords AB and CD of a circle intersect externally at E. If EC = \(2\text{ cm}\), EA = \(3\text{ cm}\) and AB = \(5\text{ cm}\), Find the length of CD. [3 Marks]
[Figure: Two intersecting secant lines from external point E meeting circle at A, B and C, D with given segments]

Answer:
Given lengths: \(EA = 3\text{ cm}\), \(AB = 5\text{ cm}\), so \(EB = EA + AB = 3 + 5 = 8\text{ cm}\).
Also, \(EC = 2\text{ cm}\). Let \(CD = x\), so \(ED = 2 + x\).
By the secant-secant theorem (intersecting chords theorem externally):
\(EA \times EB = EC \times ED\)
\(3 \times 8 = 2 \times (2 + x)\)
\(24 = 4 + 2x\)
\(2x = 20 \implies x = 10\text{ cm}\).
Therefore, length of CD is \(10\text{ cm}\).

Teacher's Note:
a) Total length EB is the sum of EA and AB, not just AB.
b) Apply the external intersection theorem correctly: whole secant multiplied by its external segment.

(ii) Line AB is perpendicular to CD. Coordinate of B, C and D are respectively \((4, 0)\), \((0, -1)\) and \((4, 3)\).
Find:
(a) Slope of CD
(b) Equation of AB [3 Marks]
[Figure: Line CD passing through C(0, -1) and D(4, 3), and perpendicular line AB passing through B(4, 0)]

Answer:
(a) Slope of CD \( = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{3 - (-1)}{4 - 0} = \dfrac{4}{4} = 1\).
(b) Since AB is perpendicular to CD, the slope of AB (\(m_2\)) satisfies \(m_1 \times m_2 = -1 \implies 1 \times m_2 = -1 \implies m_2 = -1\).
Line AB passes through B\((4, 0)\) with slope \(-1\).
Using point-slope form: \(y - y_1 = m(x - x_1)\)
\(y - 0 = -1(x - 4)\)
\(y = -x + 4 \implies x + y - 4 = 0\).

Teacher's Note:
a) Slopes of perpendicular lines are negative reciprocals of each other.
b) Use the point-slope form to easily determine the equation of a line given a point and slope.

(iii) Prove that: \(\dfrac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{2\cos^2 \theta} = \sec^2 \theta + \tan^2 \theta\) [4 Marks]

Answer:
\(\text{LHS} = \dfrac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{2\cos^2 \theta}\)
\(=\dfrac{(1 + \sin^2 \theta + 2\sin \theta) + (1 + \sin^2 \theta - 2\sin \theta)}{2\cos^2 \theta}\)
\(=\dfrac{2 + 2\sin^2 \theta}{2\cos^2 \theta}\)
\(=\dfrac{2(1 + \sin^2 \theta)}{2\cos^2 \theta} = \dfrac{1 + \sin^2 \theta}{\cos^2 \theta}\)
\(=\dfrac{1}{\cos^2 \theta} + \dfrac{\sin^2 \theta}{\cos^2 \theta}\)
\(= \sec^2 \theta + \tan^2 \theta = \text{RHS}\)

Teacher's Note:
a) Expand the numerator using algebraic identities \((a \pm b)^2 = a^2 + b^2 \pm 2ab\).
b) Split the fraction into two separate terms to arrive at secant and tangent functions.

(iv) The mean of the following distribution is 50. Find the unknown frequency. [4 Marks]

Answer:

Given Mean \(= 50\)
\(\text{Mean} = \dfrac{\sum fx}{\sum f} \implies 50 = \dfrac{2020 + 30f}{34 + f}\)
\(50(34 + f) = 2020 + 30f\)
\(1700 + 50f = 2020 + 30f\)
\(50f - 30f = 2020 - 1700\)
\(20f = 320 \implies f = 16\)

Teacher's Note:
a) Treat the unknown frequency \(f\) algebraically while summing frequencies and products.
b) Cross-multiply carefully and solve the resulting linear equation for \(f\).

Question 6:
(i) Prove that: \(\dfrac{1 + \tan^2 \theta}{1 + \sec \theta} = \sec \theta\) [3 Marks]

Answer:
\(\text{LHS} = \dfrac{1 + \tan^2 \theta}{1 + \sec \theta}\)
Since \(\tan^2 \theta = \sec^2 \theta - 1\):
\(\text{LHS} = \dfrac{1 + (\sec^2 \theta - 1)}{1 + \sec \theta} = \dfrac{\sec^2 \theta}{1 + \sec \theta}\)
Alternatively, starting from numerator manipulation as in standard scheme:
\(\dfrac{1 + \sec \theta + \tan^2 \theta - 1}{1 + \sec \theta} = \dfrac{\sec \theta + \sec^2 \theta}{1 + \sec \theta} = \dfrac{\sec \theta(1 + \sec \theta)}{1 + \sec \theta} = \sec \theta = \text{RHS}\).

Teacher's Note:
a) Use the identity \(\sec^2 \theta - \tan^2 \theta = 1\).
b) Factor out common terms in the numerator to simplify expressions with secant.

(ii) In the given figure A, B, and D are points on the circle with centre O. Given \(\angle ABC = 62^{\circ}\). Find:
(a) \(\angle ADC
(b) \(\angle CAB\) [3 Marks]
[Figure: Circle with center O, diameter AC, points B and D on circumference, with \(\angle ABC = 62^{\circ}\)]

Answer:
(a) \(\angle ADC = \angle ABC = 62^{\circ}\) (Angles in the same segment subtended by the same arc are equal).
(b) Since AC is the diameter, \(\angle ABC = 90^{\circ}\) or \(\angle ACB = 90^{\circ}\) (Angle in a semi-circle).
In \(\Delta ABC\), the sum of angles is \(180^{\circ}\):
\(\angle CAB + \angle ABC + \angle ACB = 180^{\circ}\)
\(\angle CAB + 62^{\circ} + 90^{\circ} = 180^{\circ}\)
\(\angle CAB + 152^{\circ} = 180^{\circ}\)
\(\angle CAB = 180^{\circ} - 152^{\circ} = 28^{\circ}\).

Teacher's Note:
a) Angles in the same segment of a circle are equal.
b) The angle subtended by a diameter at any point on the circle is a right angle (\(90^{\circ}\)).

(iii) Find the equation of a line parallel to the line \(2x + y - 7 = 0\) and passing through the intersection of the line \(x + y - 4 = 0\) and \(2x - y = 8\). [4 Marks]

Answer:
1. Find the slope of the given line \(2x + y - 7 = 0\), which can be written as \(y = -2x + 7\). Slope \(m = -2\).
2. A line parallel to this line will have the same slope, so \(m = -2\).
3. Solve the simultaneous equations to find the point of intersection:
\(x + y - 4 = 0 \implies x + y = 4\)
\(2x - y = 8\)
Adding the two equations: \(3x = 12 \implies x = 4\).
Substitute \(x = 4\) into \(x + y = 4 \implies 4 + y = 4 \implies y = 0\).
So the point of intersection is \((4, 0)\).
4. Equation of the required line passing through \((4, 0)\) with slope \(m = -2\):
\(y - y_1 = m(x - x_1)\)
\(y - 0 = -2(x - 4)\)
\(y = -2x + 8 \implies 2x + y - 8 = 0\).

Teacher's Note:
a) Parallel lines share identical slopes.
b) Solve simultaneous linear equations by elimination to find the intersecting coordinates.

(iv) Marks obtained by students in an examination are given below. [4 Marks]

Using graph paper, draw an ogive and estimate the median marks.
Take \(2\text{ cm} = 10\) marks along one axis and \(2\text{ cm} = 5\) students along the other axis.

Answer:
1. Construct the cumulative frequency (c.f.) table:

2. Plot the points corresponding to upper class limits and cumulative frequencies: \((20, 3)\), \((30, 11)\), \((40, 25)\), \((50, 34)\), \((60, 38)\), and \((70, 40)\), and join them smoothly to form a less-than ogive.
3. Total frequency \(n = 40\). \(\dfrac{n}{2} = \dfrac{40}{2} = 20\).
4. Locate 20 on the \(y\)-axis, draw a horizontal line to intersect the ogive curve, and from that point drop a perpendicular to the \(x\)-axis.
5. The median value read from the \(x\)-axis is approximately \(36\) marks.

Teacher's Note:
a) Less-than ogives are plotted using upper class boundaries and corresponding cumulative frequencies.
b) Median is obtained by locating \(\dfrac{n}{2}\) on the cumulative frequency axis.