a)

We want to ensure we won't go over the limit, so it is better to overestimate. If we underestimated, we could potentially think we have enough money when we don't.

To overestimate our total, we round each price up.

b)

We round up all our values to 1 decimal place:

Now we calculate the total of these rounded numbers:

\[ \var{pounds(precround(p[0]+0.05,1))} + \var{pounds(precround(p[1]+0.05,1))} + \var{pounds(precround(ice_cream,1))} = \var{pounds(total_rounded_up)} \]

c)

As the estimated total, £3.20, is higher than £3, we may not have enough money to purchase the Neapolitan ice cream.

We need to find the original price paid by Kate. This value represents 100%.

By the time Trevor bought the PC, the price had decreased by 27%.

Trevor therefore paid 73% of the price Kate paid.

We use the unitary method to find the original price. We know the price paid by Trevor

\[\var{100-percentage}\text{%} = \var{newprice} \text{.}\]

Divide both sides by 73 to get

\[\begin{align} 1\text{%} &= \var{newprice} \div \var{100-percentage} \\&= \var{newprice/(100-percentage)} \text{.} \end{align}\]

Multiply both sides by 100 to get

\[\begin{align} 100\text{%} &= \var{newprice/(100-percentage)} \times 100 \\&= \var{newprice/(100-percentage)*100} \\&= \var{oldprice}\text{.} \end{align}\]

This is the original price paid by Kate before the 27% decrease.

We can check our answer with a different method.

\[\begin{align} \var{100-percentage}\text{% of } \var{oldprice} &= \var{(100-percentage)/100} \times \var{oldprice} \\&=  \var{(100-percentage)/100*oldprice} \\&= \var{precround((100-percentage)/100*oldprice, 2)}  \text{.} \end{align}\]

We are told that Henrietta gets paid a total of $£\var{payment}$ at the end of her summer job and that she works at her job for $\var{weeks}$ weeks.

To calculate the amount of money Henrietta gets paid per week, we divide the total amount of money that she earns at the end of her job by how many weeks that she works for.

\[£\displaystyle\frac{\var{payment}}{\var{weeks}} = £\var{{payment/weeks}}.\]

Therefore Henrietta gets paid $£\var{{payment/weeks}}$/week.

Note that in compound measures, a forward slash symbol / is often used instead of the word 'per'. So $£\var{{payment/weeks}}$/week means the same as $£\var{{payment/weeks}}$ per week.

This is simple interest, which means the amount added each year is a percentage of the original amount. The amount we add is fixed for all 6 years.

First, we work out the amount of interest for one year:

\begin{align}
\var{perc2} \text{% of } \var{money} &= \frac{\var{perc2}}{100} \times \var{money} \\
&= \var{perc2/100} \times \var{money} \\
&= £\var{dpformat(perc2/100*money,2)} \text{.} 
\end{align}

The money has been in the account for 6 years, so we multiply $£\var{dpformat(perc2/100*money,2)}$ by $\var{years}$.

\[ £\var{dpformat(perc2/100*money,2)} \times \var{years} = £\var{dpformat(perc2/100*money*years,2)} \text{.} \]

Adding this to the original balance:

\[ £\var{money} + £\var{dpformat(perc2/100*money*years,2)}  = £\var{dpformat(perc2/100*money*years + money,2)} \text{.} \]

This is the amount we would get if we withdrew the whole savings balance today.