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Mathematics Exam Solutions - Part 2

Question 7: Logic and Venn Diagrams

Consider the statements below:

  • P: All good Literature students are Arts students
  • Q: Some Science students offer Literature

(a) Draw a Venn diagram to illustrate the statements P and Q

The Venn diagram would show:

  • A large circle representing "Arts students"
  • A smaller circle inside Arts representing "Good Literature students" (showing P: all good lit students are arts students)
  • Another circle representing "Science students"
  • An overlapping region between Science and Literature circles (showing Q: some science students offer literature)

(b) From your diagram, indicate whether the following conclusions from P and Q are valid:

(α) Mercy is a science student ⇒ she does not offer Literature Invalid - Statement Q says some science students DO offer literature, so we cannot conclude that being a science student means not offering literature.

(β) James is an Art student ⇒ he is a good Literature student Invalid - Statement P says good literature students are arts students, but this doesn't mean all arts students are good literature students.

(γ) Angela is a literature student otherwise, she is not an Art student Invalid - This conclusion doesn't follow logically from the given statements.

c) A fire station C is on a bearing of 042° from the western station at A and 023° from the eastern station at B

(i) Sketch a triangular diagram to illustrate the information above [A triangular diagram would show stations A, B, and C with the given bearings marked]

(ii) Calculate how far is the fire station from the western station at A and from the eastern station at B

Using the given bearings and trigonometry:

  • From A: bearing 042° to C
  • From B: bearing 023° to C
  • Need additional information (like distance AB) to solve completely

Question 8: Business Mathematics and Coordinate Geometry

(a) Mrs. Donkor paid GH₵594.00 for three crates of ideal milk. If the marked price for a crate is GH₵270.00, find:

(i) The discount allowed Cost per crate = 594 ÷ 3 = GH₵198.00 Discount per crate = 270 - 198 = GH₵72.00 Total discount = 72 × 3 = GH₵216.00

(ii) How much should Mrs. Donkor sell a tin of ideal milk to make a profit of 20%? Cost per crate = GH₵198.00 For 20% profit: Selling price = 198 × 1.20 = GH₵237.60 per crate

(b) Points B and C are on the horizontal plane such that |BC| = 30 m, |AB| = 26 m and the area of ABCD is 990m²

(i) Calculate the angles of depression of: (α) B from D (β) A from D

Area of quadrilateral ABCD = 990 m² Using the given measurements and coordinate geometry principles:

Let's set up coordinates with B at origin and C at (30, 0) Area = 990 m² |AB| = 26 m

Using area formula and the constraint |AB| = 26: Height of parallelogram = 990 ÷ 30 = 33 m

For angle of depression from D to B: tan(angle) = height/horizontal distance Angle of depression = tan⁻¹(33/horizontal component)

(ii) Determine the perimeter of ABCD Given |BC| = 30 m and |AB| = 26 m If ABCD is a parallelogram: |AD| = |BC| = 30 m and |CD| = |AB| = 26 m Perimeter = 2(26 + 30) = 112 m

Question 9: Geometric Construction and Logarithms

(a) Using a ruler and a pair of compasses only, construct: (i) Triangle QRT with |QR| = 8 cm, |RT| = 6 cm and |QT| = 4.5cm (ii) A quadrilateral QRSP which has a common base QR with △QRT such that QTP is a straight line. PQ is parallel to SR, |QP| = 9 cm and |RS| = 4.5 cm (iii) What is QRS, P?

[These would require geometric construction steps with compass and ruler]

(b) Given that f(x) = x² - 2log₃(x + 5) - log₃(2x - 13)

(i) Find the value of x for which f(x) does not exist For f(x) to exist, we need:

  • x + 5 > 0 ⇒ x > -5
  • 2x - 13 > 0 ⇒ x > 6.5

Therefore, f(x) does not exist for x ≤ 6.5

(ii) Solve: f(x) = 1 x² - 2log₃(x + 5) - log₃(2x - 13) = 1 x² - log₃[(x + 5)²] - log₃(2x - 13) = 1 x² - log₃[(x + 5)²(2x - 13)] = 1 x² = 1 + log₃[(x + 5)²(2x - 13)]

This requires numerical methods or further algebraic manipulation to solve.

Question 10: Circle Geometry and Variation

(a) In the diagram below, A, B, C and D are points on the circumference of a circle. Given that XY is a tangent to the circle at point A, ∠ZAB = 90° and ∠AYB = 60°

(i) ∠CAX Since XY is tangent at A and ∠ZAB = 90°: Using the tangent-chord angle theorem: ∠CAX = ∠ABC (alternate segment theorem)

(ii) ∠ABY Given ∠AYB = 60° and using circle theorems: ∠ABY can be found using the fact that XY is tangent and the given angles.

(b) If (y, x) is partly constant and partly varies as x. Given that y = 8 when x = 3 and y = 12 when x = 5, find:

Let y = a + bx where a is constant and bx is the variable part.

From the given conditions: 8 = a + 3b ... (1) 12 = a + 5b ... (2)

Subtracting (1) from (2): 4 = 2b b = 2

Substituting back: a = 8 - 3(2) = 2

Therefore: y = 2 + 2x

(i) A relation connecting y and x y = 2 + 2x

(ii) The value of x when y = 20 20 = 2 + 2x 18 = 2x x = 9

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