Section A [64 marks] Answer all the questions in this section.

Here's the solution to your assignment:

1. (a) Simplify as far as possible $\frac{(2x-5)-(x-3)}{x^2-4}$.

Step 1: Simplify the numerator. $(2x-5)-(x-3) = 2x - 5 - x + 3 = x - 2$

Step 2: Factor the denominator using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$). $x^2 - 4 = x^2 - 2^2 = (x-2)(x+2)$

Step 3: Substitute the simplified numerator and factored denominator back into the expression and simplify. $\frac{x-2}{(x-2)(x+2)} = \frac{1}{x+2}$ The simplified expression is $\boxed{\frac{1}{x+2}}$.

1. (b) Given that $p = \frac{q-t}{r+t}$

(i) Calculate the value of $p$ when $q=3$, $t=-5$ and $r=2$.

Step 1: Substitute the given values into the formula for $p$. $p = \frac{3 - (-5)}{2 + (-5)}$

Step 2: Perform the arithmetic operations. $p = \frac{3 + 5}{2 - 5} = \frac{8}{-3}$ The value of $p$ is $\boxed{-\frac{8}{3}}$.

(ii) Express $t$ in terms of $p$, $q$ and $r$.

Step 1: Multiply both sides of the equation by $(r+t)$ to eliminate the denominator. $p(r+t) = q-t$ $pr + pt = q-t$

Step 2: Gather all terms containing $t$ on one side and all other terms on the other side. $pt + t = q - pr$

Step 3: Factor out $t$ from the terms on the left side. $t(p+1) = q - pr$

Step 4: Divide by $(p+1)$ to isolate $t$. $t = \frac{q - pr}{p+1}$ The expression for $t$ is $\boxed{\frac{q - pr}{p+1}}$.

1. (c) Solve the simultaneous equations: $4x - 2y = 5$ (Equation 1) $x + y = -\frac{1}{4}$ (Equation 2)

Step 1: Multiply Equation 2 by 2 to make the coefficients of $y$ opposites. $2(x + y) = 2\left(-\frac{1}{4}\right)$ $2x + 2y = -\frac{1}{2} \text{ (Equation 3)}$

Step 2: Add Equation 1 and Equation 3 to eliminate $y$. $(4x - 2y) + (2x + 2y) = 5 + \left(-\frac{1}{2}\right)$ $6x = 5 - \frac{1}{2}$ $6x = \frac{10}{2} - \frac{1}{2}$ $6x = \frac{9}{2}$

Step 3: Solve for $x$. $x = \frac{9}{2 \times 6} = \frac{9}{12} = \frac{3}{4}$

Step 4: Substitute the value of $x$ into Equation 2 to solve for $y$. $\frac{3}{4} + y = -\frac{1}{4}$ $y = -\frac{1}{4} - \frac{3}{4}$ $y = -\frac{4}{4}$ $y = -1$ The solution is $\boxed{x = \frac{3}{4}, y = -1}$.

2. (a) Given that $m = 3 \times 10^2$ and $n = 5 \times 10^{-4}$, express in standard form

(i) $mn$

Step 1: Multiply $m$ and $n$. $mn = (3 \times 10^2) \times (5 \times 10^{-4})$ $mn = (3 \times 5) \times (10^2 \times 10^{-4})$ $mn = 15 \times 10^{2-4}$ $mn = 15 \times 10^{-2}$

Step 2: Convert to standard form (a number between 1 and 10 multiplied by a power of 10). $mn = (1.5 \times 10^1) \times 10^{-2}$ $mn = 1.5 \times 10^{1-2}$ $mn = 1.5 \times 10^{-1}$ The product $mn$ in standard form is $\boxed{1.5 \times 10^{-1}}$.

(ii) $\frac{m}{n}$

Step 1: Divide $m$ by $n$. $\frac{m}{n} = \frac{3 \times 10^2}{5 \times 10^{-4}}$ $\frac{m}{n} = \frac{3}{5} \times \frac{10^2}{10^{-4}}$ $\frac{m}{n} = 0.6 \times 10^{2 - (-4)}$ $\frac{m}{n} = 0.6 \times 10^{2+4}$ $\frac{m}{n} = 0.6 \times 10^6$

Step 2: Convert to standard form. $\frac{m}{n} = (6 \times 10^{-1}) \times 10^6$ $\frac{m}{n} = 6 \times 10^{-1+6}$ $\frac{m}{n} = 6 \times 10^5$ The quotient $\frac{m}{n}$ in standard form is $\boxed{6 \times 10^5}$.

2. (b) Jenny knits jerseys for her aunt. In one week, she knits 8 jerseys which are sold at $750.00 each. She is paid a basic weekly wage of$500.00 plus a commission of $2\frac{1}{2}\%$ per jersey. Calculate

(i) her commission in one week,

Step 1: Calculate the total sales value of the jerseys. $\text{Total sales} = 8 \text{ jerseys} \times \$750.00/\text{jersey} = \$6000.00$

Step 2: Calculate the commission. The commission rate is $2\frac{1}{2}\% = 2.5\% = 0.025$. $\text{Commission} = 0.025 \times \$6000.00 = \$150.00$ Her commission in one week is \boxed{\150.00}$.

(ii) her total weekly earnings.

Step 1: Add her basic weekly wage and her commission. $\text{Total earnings} = \text{Basic wage} + \text{Commission}$ $\text{Total earnings} = \$500.00 + \$150.00 = \$650.00$ Her total weekly earnings are \boxed{\650.00}$.

2. (c) A bank exchanges South African rands for Zimbabwean dollars at the rate of R1 to Z$6.25. Calculate

(i) the amount received in rands for Z$1 250.00,

Step 1: Use the exchange rate to convert Zimbabwean dollars to South African rands. $\text{Amount in Rands} = \frac{\text{Amount in Z\$}}{\text{Exchange rate}}$ $\text{Amount in Rands} = \frac{Z\$1250.00}{Z\$6.25/\text{R1}}$

Step 2: Perform the calculation. $\text{Amount in Rands} = 200$ The amount received in rands is $\boxed{\text{R}200.00}$.

(ii) the amount received in dollars for R80.

Step 1: Use the exchange rate to convert South African rands to Zimbabwean dollars. $\text{Amount in Z\$} = \text{Amount in Rands} \times \text{Exchange rate}$ $\text{Amount in Z\$} = \text{R}80 \times Z\$6.25/\text{R1}$

Step 2: Perform the calculation. $\text{Amount in Z\$} = 500$ The amount received in dollars is \boxed{\text{Z}\500.00}$.