CPT$2601$ Assessment $\frac{1}{2024}$

Question $1$

Consider Table $1$ below that displays the average tutorial marks for class of $5$ students. There are a total of $4$ tutorials $($Tut $1$ to Tut$4)$ who's marks appear at the very beginning of the table $($columns C to F$).$ The average's for all the students are meant to lie in cells G$5$:G$9,$ and the outcome $($whether they passed or failed$)$ is to be stated in cells $H5$:$H9.$ The table that lies below it $[$Table $2]$ contains the weightings in percentage as well as the total marks that each assessment is out of$.$ Another single cell can be found to right of table $2($F$14)$ that's been labelled "Vlookup".

Table$1$

The very first entries have been made $($those of Mayetu Neo$),$ and the student got an average of $56\mathrm{.}07\%$ and the final outcome is a pass. The example below shows how to determine the average mark using Mayetu Neo as an example:

Average $=(\frac{\text{T}\text{u}\text{t}1\text{M}\text{a}\text{r}\text{k}}{\text{T}\text{u}\text{t}1\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{r}\text{k}})$ Tut $1$ weighting $+(\frac{\text{T}\text{u}\text{t}2\text{M}\text{a}\text{r}\text{k}}{\text{T}\text{u}\text{t}2\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{r}\text{k}})$ Tut $2$ weighting

$+(\frac{\text{T}\text{u}\text{t}3\text{M}\text{a}\text{r}\text{k}}{\text{T}\text{u}\text{t}3\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{r}\text{k}})$ Tut $3$ weighting $+(\frac{\text{T}\text{u}\text{t}4\text{M}\text{a}\text{r}\text{k}}{\text{T}\text{u}\text{t}\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{r}\text{k}})$ Tut $4$ weighting

$=(\frac{34}{30})5+(\frac{25}{45})20+(\frac{31}{50})35+(\frac{22}{50})40=56\mathrm{.}07\%$

$2$