PROGRAMME: DIPLOMA EXAMINATIONS - 2026 SUBJECT: BUSINESS MATHEMATICS & STATISTICS INSTRUCTIONS: Answer All Questions in Section A and Three (3) Questions in Section B.

Question

Asked on July 9, 2026Mathematics

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.

Accepted Answer · 2026-07-09T18:34:05.089Z

Here are the solutions to the questions from the exam paper: 1a. Simplify: 1a.i. (3)/(y+3) + (2)/(y+5) Step 1: Find a common denominator, which is (y+3)(y+5). (3)/(y+3) + (2)/(y+5) = (3(y+5))/((y+3)(y+5)) + (2(y+3))/((y+3)(y+5)) Step 2: Combine the numerators over the common denominator. = (3(y+5) + 2(y+3))/((y+3)(y+5)) = (3y + 15 + 2y + 6)/((y+3)(y+5)) = (5y + 21)/((y+3)(y+5)) 1a.ii. 5 - (x-1)/(7x) Step 1: Write 5 as a fraction with denominator 7x. 5 - (x-1)/(7x) = (5 × 7x)/(7x) - (x-1)/(7x) = (35x)/(7x) - (x-1)/(7x) Step 2: Combine the numerators over the common denominator. = (35x - (x-1))/(7x) = (35x - x + 1)/(7x) = (34x + 1)/(7x) 1a.iii. (x^2+2x-3)/(x^2+5x-6) Step 1: Factorize the numerator x^2+2x-3. We need two numbers that multiply to -3 and add to 2. These are 3 and -1. x^2+2x-3 = (x+3)(x-1) Step 2: Factorize the denominator x^2+5x-6. We need two numbers that multiply to -6 and add to 5. These are 6 and -1. x^2+5x-6 = (x+6)(x-1) Step 3: Substitute the factored forms back into the expression and cancel common factors. ((x+3)(x-1))/((x+6)(x-1)) = (x+3)/(x+6) 1b. Consider the matrices M = 3 & -2 \\ 2 & 4 and N = 4 & 6 \\ -1 & 5 . Find: 1b.i. -3M^T Step 1: Find the transpose of M, denoted as M^T. This involves swapping rows and columns. M^T = 3 & 2 \\ -2 & 4 Step 2: Multiply M^T by -3. -3M^T = -3 3 & 2 \\ -2 & 4 = -3 × 3 & -3 × 2 \\ -3 × -2 & -3 × 4 = -9 & -6 \\ 6 & -12 1b.ii. N^T - M Step 1: Find the transpose of N, denoted as N^T. N^T = 4 & -1 \\ 6 & 5 Step 2: Subtract matrix M from N^T. N^T - M = 4 & -1 \\ 6 & 5 - 3 & -2 \\ 2 & 4 = 4-3 & -1-(-2) \\ 6-2 & 5-4 = 1 & -1+2 \\ 4 & 1 = 1 & 1 \\ 4 & 1 1c. Calculations and Factorisation: 1c.i. Use a calculator to compute 20C_5. Step 1: Use the combination formula nC_r = (n!)/(r!(n-r)!) or a calculator's combination function. 20C_5 = (20!)/(5!(20-5)!) = (20!)/(5!15!) 20C_5 = (20 × 19 × 18 × 17 × 16)/(5 × 4 × 3 × 2 × 1) 20C_5 = 19 × 3 × 17 × 16 20C_5 = 15504 1c.ii. Factorise completely 2t^2 - 72. Step 1: Find the greatest common factor (GCF) of 2t^2 and 72, which is 2. 2t^2 - 72 = 2(t^2 - 36) Step 2: Recognize that t^2 - 36 is a difference of squares, a^2 - b^2 = (a-b)(a+b), where a=t and b=6. 2(t^2 - 6^2) = 2(t-6)(t+6) = 2(t-6)(t+6) 1d. Equations: 1d.i. Solve the simultaneous equations: 3x - 2y = 5 (1) 4x + 5y = -24 (2) Step 1: Multiply equation (1) by 5 and equation (2) by 2 to eliminate y. (1) × 5 15x - 10y = 25 (3) (2) × 2 8x + 10y = -48 (4) Step 2: Add equation (3) and equation (4). (15x - 10y) + (8x + 10y) = 25 + (-48) 23x = -23 Step 3: Solve for x. x = (-23)/(23) x = -1 Step 4: Substitute x = -1 into equation (1) to find y. 3(-1) - 2y = 5 -3 - 2y = 5 -2y = 5 + 3 -2y = 8 y = (8)/(-2) y = -4 The solution is x = -1, y = -4. 1d.ii. Use the quadratic formula to solve 5x^2 = 20x - 4. (Give your answer correct to 1 d.p) Step 1: Rewrite the equation in standard quadratic form ax^2 + bx + c = 0. 5x^2 - 20x + 4 = 0 Step 2: Identify the coefficients a, b, c. a = 5, b = -20, c = 4 Step 3: Apply the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a. x = -(-20) ± sqrt((-20)^2 - 4(5)(4))2(5) x = 20 ± sqrt(400 - 80)10 x = 20 ± sqrt(320)10 Step 4: Calculate the two solutions and round to 1 decimal place. sqrt(320) ≈ 17.8885 x_1 = (20 + 17.8885)/(10) = (37.8885)/(10) ≈ 3.78885 ≈ 3.8 x_2 = (20 - 17.8885)/(10) = (2.1115)/(10) ≈ 0.21115 ≈ 0.2 The solutions are x ≈ 3.8 or x ≈ 0.2. 2a. Consider the AP 5, 2, -1, -4, 2a.i. Write down the next two terms. Step 1: Find the common difference (d). d = 2 - 5 = -3 d = -1 - 2 = -3 The common difference is -3. Step 2: Add the common difference to the last given term to find the next two terms. The last given term is -4. Next term: -4 + (-3) = -7 Second next term: -7 + (-3) = -10 The next two terms are -7, -10. 2a.ii. Find a formula for the nth term. Step 1: Identify the first term (a) and common difference (d). a = 5 d = -3 Step 2: Use the formula for the nth term of an AP: a_n = a + (n-1)d. a_n = 5 + (n-1)(-3) a_n = 5 - 3n + 3 a_n = 8 - 3n 2a.iii. Use the formula in (ii) to find the 81st term. Step 1: Substitute n=81 into the formula a_n = 8 - 3n. a_81 = 8 - 3(81) a_81 = 8 - 243 a_81 = -235 2a.iv. Find the sum of the first 40 terms. Step 1: Use the formula for the sum of n terms of an AP: S_n = (n)/(2)[2a + (n-1)d]. Here, n=40, a=5, d=-3. S_40 = (40)/(2)[2(5) + (40-1)(-3)] S_40 = 20[10 + (39)(-3)] S_40 = 20[10 - 117] S_40 = 20[-107] S_40 = -2140 2b. Find the value of: 2b.i. 1+2+4+8++16384. Step 1: Identify this as a Geometric Progression (GP). The first term is a = 1. The common ratio is r = (2)/(1) = 2. The last term is a_n = 16384. Step 2: Find the number of terms (n) using the formula for the nth term of a GP: a_n = ar^n-1. 16384 = 1 × 2^n-1 2^n-1 = 16384 Since 2^10 = 1024, 2^14 = 16384. n-1 = 14 n = 15 Step 3: Use the formula for the sum of n terms of a GP: S_n = (a(r^n - 1))/(r-1). S_15 = 1(2^15 - 1)2-1 S_15 = (32768 - 1)/(1) S_15 = 32767 2b.ii. _n=1^40 4n. Step 1: This is the sum of an arithmetic series where the terms are 4(1), 4(2), , 4(40). This can be written as 4 × (1+2+3++40). The sum of the first N natural numbers is (N(N+1))/(2). Here N=40. S = 4 × (40(40+1))/(2) S = 4 × (40 × 41)/(2) S = 4 × (1640)/(2) S = 4 × 820 S = 3280 That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-07-09T18:34:05.089Z

Here are the solutions to the questions from the exam paper: 1a. Simplify: 1a.i. (3)/(y+3) + (2)/(y+5) Step 1: Find a common denominator, which is (y+3)(y+5). (3)/(y+3) + (2)/(y+5) = (3(y+5))/((y+3)(y+5)) + (2(y+3))/((y+3)(y+5)) Step 2: Combine the numerators over the common denominator. = (3(y+5) + 2(y+3))/((y+3)(y+5)) = (3y + 15 + 2y + 6)/((y+3)(y+5)) = (5y + 21)/((y+3)(y+5)) 1a.ii. 5 - (x-1)/(7x) Step 1: Write 5 as a fraction with denominator 7x. 5 - (x-1)/(7x) = (5 × 7x)/(7x) - (x-1)/(7x) = (35x)/(7x) - (x-1)/(7x) Step 2: Combine the numerators over the common denominator. = (35x - (x-1))/(7x) = (35x - x + 1)/(7x) = (34x + 1)/(7x) 1a.iii. (x^2+2x-3)/(x^2+5x-6) Step 1: Factorize the numerator x^2+2x-3. We need two numbers that multiply to -3 and add to 2. These are 3 and -1. x^2+2x-3 = (x+3)(x-1) Step 2: Factorize the denominator x^2+5x-6. We need two numbers that multiply to -6 and add to 5. These are 6 and -1. x^2+5x-6 = (x+6)(x-1) Step 3: Substitute the factored forms back into the expression and cancel common factors. ((x+3)(x-1))/((x+6)(x-1)) = (x+3)/(x+6) 1b. Consider the matrices M = 3 & -2 \\ 2 & 4 and N = 4 & 6 \\ -1 & 5 . Find: 1b.i. -3M^T Step 1: Find the transpose of M, denoted as M^T. This involves swapping rows and columns. M^T = 3 & 2 \\ -2 & 4 Step 2: Multiply M^T by -3. -3M^T = -3 3 & 2 \\ -2 & 4 = -3 × 3 & -3 × 2 \\ -3 × -2 & -3 × 4 = -9 & -6 \\ 6 & -12 1b.ii. N^T - M Step 1: Find the transpose of N, denoted as N^T. N^T = 4 & -1 \\ 6 & 5 Step 2: Subtract matrix M from N^T. N^T - M = 4 & -1 \\ 6 & 5 - 3 & -2 \\ 2 & 4 = 4-3 & -1-(-2) \\ 6-2 & 5-4 = 1 & -1+2 \\ 4 & 1 = 1 & 1 \\ 4 & 1 1c. Calculations and Factorisation: 1c.i. Use a calculator to compute 20C_5. Step 1: Use the combination formula nC_r = (n!)/(r!(n-r)!) or a calculator's combination function. 20C_5 = (20!)/(5!(20-5)!) = (20!)/(5!15!) 20C_5 = (20 × 19 × 18 × 17 × 16)/(5 × 4 × 3 × 2 × 1) 20C_5 = 19 × 3 × 17 × 16 20C_5 = 15504 1c.ii. Factorise completely 2t^2 - 72. Step 1: Find the greatest common factor (GCF) of 2t^2 and 72, which is 2. 2t^2 - 72 = 2(t^2 - 36) Step 2: Recognize that t^2 - 36 is a difference of squares, a^2 - b^2 = (a-b)(a+b), where a=t and b=6. 2(t^2 - 6^2) = 2(t-6)(t+6) = 2(t-6)(t+6) 1d. Equations: 1d.i. Solve the simultaneous equations: 3x - 2y = 5 (1) 4x + 5y = -24 (2) Step 1: Multiply equation (1) by 5 and equation (2) by 2 to eliminate y. (1) × 5 15x - 10y = 25 (3) (2) × 2 8x + 10y = -48 (4) Step 2: Add equation (3) and equation (4). (15x - 10y) + (8x + 10y) = 25 + (-48) 23x = -23 Step 3: Solve for x. x = (-23)/(23) x = -1 Step 4: Substitute x = -1 into equation (1) to find y. 3(-1) - 2y = 5 -3 - 2y = 5 -2y = 5 + 3 -2y = 8 y = (8)/(-2) y = -4 The solution is x = -1, y = -4. 1d.ii. Use the quadratic formula to solve 5x^2 = 20x - 4. (Give your answer correct to 1 d.p) Step 1: Rewrite the equation in standard quadratic form ax^2 + bx + c = 0. 5x^2 - 20x + 4 = 0 Step 2: Identify the coefficients a, b, c. a = 5, b = -20, c = 4 Step 3: Apply the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a. x = -(-20) ± sqrt((-20)^2 - 4(5)(4))2(5) x = 20 ± sqrt(400 - 80)10 x = 20 ± sqrt(320)10 Step 4: Calculate the two solutions and round to 1 decimal place. sqrt(320) ≈ 17.8885 x_1 = (20 + 17.8885)/(10) = (37.8885)/(10) ≈ 3.78885 ≈ 3.8 x_2 = (20 - 17.8885)/(10) = (2.1115)/(10) ≈ 0.21115 ≈ 0.2 The solutions are x ≈ 3.8 or x ≈ 0.2. 2a. Consider the AP 5, 2, -1, -4, 2a.i. Write down the next two terms. Step 1: Find the common difference (d). d = 2 - 5 = -3 d = -1 - 2 = -3 The common difference is -3. Step 2: Add the common difference to the last given term to find the next two terms. The last given term is -4. Next term: -4 + (-3) = -7 Second next term: -7 + (-3) = -10 The next two terms are -7, -10. 2a.ii. Find a formula for the nth term. Step 1: Identify the first term (a) and common difference (d). a = 5 d = -3 Step 2: Use the formula for the nth term of an AP: a_n = a + (n-1)d. a_n = 5 + (n-1)(-3) a_n = 5 - 3n + 3 a_n = 8 - 3n 2a.iii. Use the formula in (ii) to find the 81st term. Step 1: Substitute n=81 into the formula a_n = 8 - 3n. a_81 = 8 - 3(81) a_81 = 8 - 243 a_81 = -235 2a.iv. Find the sum of the first 40 terms. Step 1: Use the formula for the sum of n terms of an AP: S_n = (n)/(2)[2a + (n-1)d]. Here, n=40, a=5, d=-3. S_40 = (40)/(2)[2(5) + (40-1)(-3)] S_40 = 20[10 + (39)(-3)] S_40 = 20[10 - 117] S_40 = 20[-107] S_40 = -2140 2b. Find the value of: 2b.i. 1+2+4+8++16384. Step 1: Identify this as a Geometric Progression (GP). The first term is a = 1. The common ratio is r = (2)/(1) = 2. The last term is a_n = 16384. Step 2: Find the number of terms (n) using the formula for the nth term of a GP: a_n = ar^n-1. 16384 = 1 × 2^n-1 2^n-1 = 16384 Since 2^10 = 1024, 2^14 = 16384. n-1 = 14 n = 15 Step 3: Use the formula for the sum of n terms of a GP: S_n = (a(r^n - 1))/(r-1). S_15 = 1(2^15 - 1)2-1 S_15 = (32768 - 1)/(1) S_15 = 32767 2b.ii. _n=1^40 4n. Step 1: This is the sum of an arithmetic series where the terms are 4(1), 4(2), , 4(40). This can be written as 4 × (1+2+3++40). The sum of the first N natural numbers is (N(N+1))/(2). Here N=40. S = 4 × (40(40+1))/(2) S = 4 × (40 × 41)/(2) S = 4 × (1640)/(2) S = 4 × 820 S = 3280 That's 2 down. 3 left today — send the next one.

PROGRAMME: DIPLOMA EXAMINATIONS - 2026 SUBJECT: BUSINESS MATHEMATICS & STATISTICS INSTRUCTIONS: Answer All Questions in Section A and Three (3) Questions in Section B.

Need help with your own homework?

More Mathematics Questions

PROGRAMME: DIPLOMA EXAMINATIONS - 2026 SUBJECT: BUSINESS MATHEMATICS & STATISTICS INSTRUCTIONS: Answer All Questions in Section A and Three (3) Questions in Section B.

Need help with your own homework?

More Mathematics Questions