PROGRAMME: DIPLOMA EXAMINATIONS - 2026 SUBJECT: BUSINESS MATHEMATICS & STATISTICS INSTRUCTIONS: Answer All…

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

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Here are the solutions to the questions from the exam paper:

1a. Simplify:

1a.i. $\frac{3}{y+3} + \frac{2}{y+5}$

Find a common denominator, which is $(y+3)(y+5)$ .
$\frac{3}{y+3} + \frac{2}{y+5} = \frac{3(y+5)}{(y+3)(y+5)} + \frac{2(y+3)}{(y+3)(y+5)}$
Combine the numerators over the common denominator.
$= \frac{3(y+5) + 2(y+3)}{(y+3)(y+5)}$ $= \frac{3y + 15 + 2y + 6}{(y+3)(y+5)}$ $= \boxed{\frac{5y + 21}{(y+3)(y+5)}}$
1a.ii. $5 - \frac{x-1}{7x}$
Write 5 as a fraction with denominator $7x$.
$5 - \frac{x-1}{7x} = \frac{5 \times 7x}{7x} - \frac{x-1}{7x}$ $= \frac{35x}{7x} - \frac{x-1}{7x}$
Combine the numerators over the common denominator.
$= \frac{35x - (x-1)}{7x}$ $= \frac{35x - x + 1}{7x}$ $= \boxed{\frac{34x + 1}{7x}}$
1a.iii. $\frac{x^2+2x-3}{x^2+5x-6}$
Factorize the numerator $x^2+2x-3$ .
We need two numbers that multiply to -3 and add to 2. These are 3 and -1.
$x^2+2x-3 = (x+3)(x-1)$
Factorize the denominator $x^2+5x-6$ .
We need two numbers that multiply to -6 and add to 5. These are 6 and -1.
$x^2+5x-6 = (x+6)(x-1)$
Substitute the factored forms back into the expression and cancel common factors.
$\frac{(x+3)(x-1)}{(x+6)(x-1)}$ $= \boxed{\frac{x+3}{x+6}}$
1b. Consider the matrices $M = \begin{pmatrix} 3 & -2 \\ 2 & 4 \end{pmatrix}$ and $N = \begin{pmatrix} 4 & 6 \\ -1 & 5 \end{pmatrix}$ . Find:

1b.i. $-3M^T$
Find the transpose of $M$ , denoted as $M^T$ .
This involves swapping rows and columns.
$M^T = \begin{pmatrix} 3 & 2 \\ -2 & 4 \end{pmatrix}$
Multiply $M^T$ by $-3$ .
$-3M^T = -3 \begin{pmatrix} 3 & 2 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} -3 \times 3 & -3 \times 2 \\ -3 \times -2 & -3 \times 4 \end{pmatrix}$ $= \boxed{\begin{pmatrix} -9 & -6 \\ 6 & -12 \end{pmatrix}}$
1b.ii. $N^T - M$
Find the transpose of $N$ , denoted as $N^T$ .
$N^T = \begin{pmatrix} 4 & -1 \\ 6 & 5 \end{pmatrix}$
Subtract matrix $M$ from $N^T$ .
$N^T - M = \begin{pmatrix} 4 & -1 \\ 6 & 5 \end{pmatrix} - \begin{pmatrix} 3 & -2 \\ 2 & 4 \end{pmatrix}$ $= \begin{pmatrix} 4-3 & -1-(-2) \\ 6-2 & 5-4 \end{pmatrix}$ $= \begin{pmatrix} 1 & -1+2 \\ 4 & 1 \end{pmatrix}$ $= \boxed{\begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}}$
1c. Calculations and Factorisation:

1c.i. Use a calculator to compute $20C_5$ .
Use the combination formula $nC_r = \frac{n!}{r!(n-r)!}$ or a calculator's combination function.
$20C_5 = \frac{20!}{5!(20-5)!} = \frac{20!}{5!15!}$ $20C_5 = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1}$ $20C_5 = 19 \times 3 \times 17 \times 16$ $20C_5 = \boxed{15504}$
1c.ii. Factorise completely $2t^2 - 72$ .
Find the greatest common factor (GCF) of $2t^2$ and 72, which is 2.
$2t^2 - 72 = 2(t^2 - 36)$
Recognize that $t^2 - 36$ is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$ , where $a=t$ and $b=6$ .
$2(t^2 - 6^2) = 2(t-6)(t+6)$ $= \boxed{2(t-6)(t+6)}$
1d. Equations:

1d.i. Solve the simultaneous equations:
$3x - 2y = 5 \quad (1)$
$4x + 5y = -24 \quad (2)$
Multiply equation (1) by 5 and equation (2) by 2 to eliminate $y$ .
$(1) \times 5 \implies 15x - 10y = 25 \quad (3)$ $(2) \times 2 \implies 8x + 10y = -48 \quad (4)$
Add equation (3) and equation (4).
$(15x - 10y) + (8x + 10y) = 25 + (-48)$ $23x = -23$
Solve for $x$ .
$x = \frac{-23}{23}$ $x = -1$
Substitute $x = -1$ into equation (1) to find $y$ .
$3(-1) - 2y = 5$ $-3 - 2y = 5$ $-2y = 5 + 3$ $-2y = 8$ $y = \frac{8}{-2}$ $y = -4$
The solution is $\boxed{x = -1, y = -4}$ .

1d.ii. Use the quadratic formula to solve $5x^2 = 20x - 4$ . (Give your answer correct to 1 d.p)
Rewrite the equation in standard quadratic form $ax^2 + bx + c = 0$ .
$5x^2 - 20x + 4 = 0$
Identify the coefficients $a, b, c$.
$a = 5, \quad b = -20, \quad c = 4$
Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
$x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(5)(4)}}{2(5)}$ $x = \frac{20 \pm \sqrt{400 - 80}}{10}$ $x = \frac{20 \pm \sqrt{320}}{10}$
Calculate the two solutions and round to 1 decimal place.
$\sqrt{320} \approx 17.8885$ $x_1 = \frac{20 + 17.8885}{10} = \frac{37.8885}{10} \approx 3.78885 \approx 3.8$ $x_2 = \frac{20 - 17.8885}{10} = \frac{2.1115}{10} \approx 0.21115 \approx 0.2$
The solutions are $\boxed{x \approx 3.8 \text{ or } x \approx 0.2}$ .

2a. Consider the AP $5, 2, -1, -4, \dots$

2a.i. Write down the next two terms.
Find the common difference ( $d$ ).
$d = 2 - 5 = -3$ $d = -1 - 2 = -3$
The common difference is $-3$ .
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 $\boxed{-7, -10}$ .

2a.ii. Find a formula for the nth term.
Identify the first term ( $a$ ) and common difference ( $d$ ).
$a = 5$ $d = -3$
Use the formula for the nth term of an AP
$a_n = a + (n-1)d$ .
$a_n = 5 + (n-1)(-3)$ $a_n = 5 - 3n + 3$ $a_n = \boxed{8 - 3n}$
2a.iii. Use the formula in (ii) to find the 81st term.
Substitute $n=81$ into the formula $a_n = 8 - 3n$ .
$a_{81} = 8 - 3(81)$ $a_{81} = 8 - 243$ $a_{81} = \boxed{-235}$
2a.iv. Find the sum of the first 40 terms.
Use the formula for the sum of $n$ terms of an AP
$S_n = \frac{n}{2}[2a + (n-1)d]$ .
Here, $n=40$ , $a=5$ , $d=-3$ .
$S_{40} = \frac{40}{2}[2(5) + (40-1)(-3)]$ $S_{40} = 20[10 + (39)(-3)]$ $S_{40} = 20[10 - 117]$ $S_{40} = 20[-107]$ $S_{40} = \boxed{-2140}$
2b. Find the value of:

2b.i. $1+2+4+8+\dots+16384$ .
Identify this as a Geometric Progression (GP).
The first term is $a = 1$ .
The common ratio is $r = \frac{2}{1} = 2$ .
The last term is $a_n = 16384$ .
Find the number of terms ( $n$ ) using the formula for the nth term of a GP
$a_n = ar^{n-1}$ .
$16384 = 1 \times 2^{n-1}$ $2^{n-1} = 16384$
Since $2^{10} = 1024$ , $2^{14} = 16384$ .
$n-1 = 14$ $n = 15$
Use the formula for the sum of $n$ terms of a GP
$S_n = \frac{a(r^n - 1)}{r-1}$ .
$S_{15} = \frac{1(2^{15} - 1)}{2-1}$ $S_{15} = \frac{32768 - 1}{1}$ $S_{15} = \boxed{32767}$
2b.ii. $\sum_{n=1}^{40} 4n$ .
This is the sum of an arithmetic series where the terms are $4(1), 4(2), \dots, 4(40)$ .
This can be written as $4 \times (1+2+3+\dots+40)$ .
The sum of the first $N$ natural numbers is $\frac{N(N+1)}{2}$ . Here $N=40$ .
$S = 4 \times \frac{40(40+1)}{2}$ $S = 4 \times \frac{40 \times 41}{2}$ $S = 4 \times \frac{1640}{2}$ $S = 4 \times 820$ $S = \boxed{3280}$

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

Step-by-step solution

Find a common denominator, which is $(y+3)(y+5)$ .

Combine the numerators over the common denominator.

Write 5 as a fraction with denominator $7x$.

Combine the numerators over the common denominator.

Factorize the numerator $x^2+2x-3$ .

Factorize the denominator $x^2+5x-6$ .

Substitute the factored forms back into the expression and cancel common factors.

Find the transpose of $M$ , denoted as $M^T$ .

Multiply $M^T$ by $-3$ .

Find the transpose of $N$ , denoted as $N^T$ .

Subtract matrix $M$ from $N^T$ .

Use the combination formula $nC_r = \frac{n!}{r!(n-r)!}$ or a calculator's combination function.

Find the greatest common factor (GCF) of $2t^2$ and 72, which is 2.

Recognize that $t^2 - 36$ is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$ , where $a=t$ and $b=6$ .

Multiply equation (1) by 5 and equation (2) by 2 to eliminate $y$ .

Add equation (3) and equation (4).

Solve for $x$ .

Substitute $x = -1$ into equation (1) to find $y$ .

Rewrite the equation in standard quadratic form $ax^2 + bx + c = 0$ .

Identify the coefficients $a, b, c$.

Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

Calculate the two solutions and round to 1 decimal place.

Find the common difference ( $d$ ).

Add the common difference to the last given term to find the next two terms.

Identify the first term ( $a$ ) and common difference ( $d$ ).

Use the formula for the nth term of an AP

Substitute $n=81$ into the formula $a_n = 8 - 3n$ .

Use the formula for the sum of $n$ terms of an AP

Identify this as a Geometric Progression (GP).

Find the number of terms ( $n$ ) using the formula for the nth term of a GP

Use the formula for the sum of $n$ terms of a GP

This is the sum of an arithmetic series where the terms are $4(1), 4(2), \dots, 4(40)$ .

Still stuck on this one?

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.

Step-by-step solution

Find a common denominator, which is (y+3)(y+5)(y+3)(y+5)(y+3)(y+5).

Combine the numerators over the common denominator.

Write 5 as a fraction with denominator $7x$.

Combine the numerators over the common denominator.

Factorize the numerator x2+2x−3x^2+2x-3x2+2x−3.

Factorize the denominator x2+5x−6x^2+5x-6x2+5x−6.

Substitute the factored forms back into the expression and cancel common factors.

Find the transpose of MMM, denoted as MTM^TMT.

Multiply MTM^TMT by −3-3−3.

Find the transpose of NNN, denoted as NTN^TNT.

Subtract matrix MMM from NTN^TNT.

Use the combination formula nCr=n!r!(n−r)!nC_r = \frac{n!}{r!(n-r)!}nCr​=r!(n−r)!n!​ or a calculator's combination function.

Find the greatest common factor (GCF) of 2t22t^22t2 and 72, which is 2.

Recognize that t2−36t^2 - 36t2−36 is a difference of squares, a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b), where a=ta=ta=t and b=6b=6b=6.

Multiply equation (1) by 5 and equation (2) by 2 to eliminate yyy.

Add equation (3) and equation (4).

Solve for xxx.

Substitute x=−1x = -1x=−1 into equation (1) to find yyy.

Rewrite the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.

Identify the coefficients $a, b, c$.

Apply the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​.

Calculate the two solutions and round to 1 decimal place.

Find the common difference (ddd).

Add the common difference to the last given term to find the next two terms.

Identify the first term (aaa) and common difference (ddd).

Use the formula for the nth term of an AP

Substitute n=81n=81n=81 into the formula an=8−3na_n = 8 - 3nan​=8−3n.

Use the formula for the sum of nnn terms of an AP

Identify this as a Geometric Progression (GP).

Find the number of terms (nnn) using the formula for the nth term of a GP

Use the formula for the sum of nnn terms of a GP

This is the sum of an arithmetic series where the terms are 4(1),4(2),…,4(40)4(1), 4(2), \dots, 4(40)4(1),4(2),…,4(40).

Still stuck on this one?

More Mathematics Questions

Find a common denominator, which is $(y+3)(y+5)$ .

Factorize the numerator $x^2+2x-3$ .

Factorize the denominator $x^2+5x-6$ .

Find the transpose of $M$ , denoted as $M^T$ .

Multiply $M^T$ by $-3$ .

Find the transpose of $N$ , denoted as $N^T$ .

Subtract matrix $M$ from $N^T$ .

Use the combination formula $nC_r = \frac{n!}{r!(n-r)!}$ or a calculator's combination function.

Find the greatest common factor (GCF) of $2t^2$ and 72, which is 2.

Recognize that $t^2 - 36$ is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$ , where $a=t$ and $b=6$ .

Multiply equation (1) by 5 and equation (2) by 2 to eliminate $y$ .

Solve for $x$ .

Substitute $x = -1$ into equation (1) to find $y$ .

Rewrite the equation in standard quadratic form $ax^2 + bx + c = 0$ .

Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

Find the common difference ( $d$ ).

Identify the first term ( $a$ ) and common difference ( $d$ ).

Substitute $n=81$ into the formula $a_n = 8 - 3n$ .

Use the formula for the sum of $n$ terms of an AP

Find the number of terms ( $n$ ) using the formula for the nth term of a GP

Use the formula for the sum of $n$ terms of a GP

This is the sum of an arithmetic series where the terms are $4(1), 4(2), \dots, 4(40)$ .