This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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here are the solutions to the questions. Question (b): If each block costs GHC 1.80 and the cost of labour is one-third the cost of a block, find the total cost of laying the blocks. To find the total cost of laying the blocks, the number of blocks to be laid must be known. Since this information is not provided in the question, we will calculate the cost per block laid. Step 1: Calculate the cost of labour per block. Cost of labour per block = (1)/(3) × Cost of one block Cost of labour per block = (1)/(3) × GHC 1.80 Cost of labour per block = GHC 0.60 Step 2: Calculate the total cost per block laid. Total cost per block laid = Cost of one block + Cost of labour per block Total cost per block laid = GHC 1.80 + GHC 0.60 Total cost per block laid = GHC 2.40 Assuming the question asks for the cost per block laid, the total cost is GHC 2.40. Geometry Question: In the diagram, MNPQ is a circle with centre O. MNQ = y^, NEP = x^, NMP = (100-x)^ and NFP = (5x)^. Find the values of x and y. Step 1: Express the intercepted arcs in terms of x and y using the Inscribed Angle Theorem. The measure of an inscribed angle is half the measure of its intercepted arc. For MNQ: MNQ = y^ arc MQ = 2 × MNQ = 2y^ For NMP: NMP = (100-x)^ arc NP = 2 × NMP = 2(100-x)^ Step 2: Use the angle formed by intersecting chords inside the circle. The angle formed by two chords intersecting inside a circle is half the sum of the measures of the intercepted arcs. NFP = (1)/(2) (arc NP + arc MQ) Given NFP = (5x)^: 5x = (1)/(2) (2(100-x) + 2y) 5x = (100-x) + y 5x = 100 - x + y 6x - y = 100 (Equation 1) Step 3: Use the angle formed by two secants intersecting outside the circle. The angle formed by two secants intersecting outside a circle is half the difference of the measures of the intercepted arcs. NEP = (1)/(2) (arc MQ - arc NP) Given NEP = x^: x = (1)/(2) (2y - 2(100-x)) x = y - (100-x) x = y - 100 + x 0 = y - 100 y = 100 Step 4: Solve the system of equations for x and y. Substitute y=100 into Equation 1: 6x - 100 = 100 6x = 100 + 100 6x = 200 x = (200)/(6) x = (100)/(3) x ≈ 33.33 The values are x = (100)/(3) and y = 100. The values of x and y are x = (100)/(3) or 33.3^ and y = 100^. Drop the next question.