When inserting numbers into your calculator, make sure that you place brackets correctly.

a) 

We can see from the diagram that the radius of the frisbee is $\var{mccall[2]}$ $\mathrm{cm}$.
Replacing the letter $r$ in the formula for the area of a circle with $\var{mccall[2]}$ gives,

\begin{align}
\mathrm{Area} &= \pi r^2 \\
&= \pi\times(\var{mccall[2]})^2 \\
&= \var{dpformat((mccall[2])^2, 2)}\pi\, \mathrm{cm}^2 \\
&= \var{dpformat(pi (mccall[2])^2, 2)}\, \mathrm{cm}^2 \quad \text{to 2 d.p.}  
\end{align}

b) 

We can see from the diagram that the triangle has two sides with lengths $\var{length_cdp2}$ $\mathrm{cm}$, $\var{length_bdp2}$ $\mathrm{cm}$ and an angle $\var{c_thetadp2}\mathrm{°}$ .
Replacing the letters $a$, $b$ and $C$ in the formula for the area of a triangle with $\var{length_cdp2}$, $\var{length_bdp2}$ and $\var{c_thetadp2}$ respectively gives,

\begin{align}
\mathrm{Area} &= \frac{1}{2}ab\sin{C} \\
&= \frac{1}{2} \times \var{length_cdp2} \times \var{length_bdp2} \times \sin(\var{c_thetadp2}) \\
&= \var{dpformat(0.5*(length_cdp2)*(length_bdp2)*sin(c_thetadp2* pi/180), 5)}\, \mathrm{cm}^2  \\
&= \var{dpformat(0.5*(length_cdp2)*(length_bdp2)*sin(c_thetadp2 * pi/180), 2)}\, \mathrm{cm}^2 \quad \text{to 2 d.p.}
\end{align}

c) 

We can see from the diagram that the radius of the cone is $\var{r}$ $\mathrm{cm}$ and the height is $\var{h}$ $\mathrm{cm}$.
Replacing the letters $r$ and $h$ in the formula for the volume of a cone with $\var{r}$ $\mathrm{cm}$ and $\var{h}$ $\mathrm{cm}$ respectively gives,

\begin{align}
\mathrm{Volume} &= \frac{h}{3} \pi r^2 \\
&= \frac{(\var{h})}{3} \times \pi \times (\var{r})^2 \\
&= \var{dpformat((pi)*(h/3)*(r)^2 , 5)}\, \mathrm{cm}^3 \\
&=\var{dpformat(h/3 * pi * (r)^2, 1)}\, \mathrm{cm}^3 \quad \text{to 1 d.p.} \\
\end{align}

d)

We can see from the diagram that the radius of the tennis ball is $\var{mccall[1]}$ $\mathrm{cm}$.
Replacing the letter $r$ in the formula for the volume of a sphere with $\var{mccall[1]}$ gives,

\begin{align}
\mathrm{Volume} &= \frac{4}{3} \pi r^3 \\
&= \frac{(4)}{(3)} \times \pi \times (\var{mccall[1]})^3 \\
&= \var{dpformat((4/3)*pi*mccall[1]^3, 5)}\,  \mathrm{cm}^3 \\
&= \var{precround(((4/3)* pi) *(mccall[1])^3, 1)}\, \mathrm{cm}^3 \quad \text{to 1 d.p.} \\
\end{align}

e)

We can see from the diagram that the trapezium has two parallel sides with length $\var{trap_length_a}$ $\mathrm{cm}$, $\var{trap_length_b}$ $\mathrm{cm}$ and height $\var{trap_h}$ $\mathrm{cm}$.
Replacing the letters $a$, $b$ and $h$ in the formula for the area of a trapezium with $\var{trap_length_a}$, $\var{trap_length_b}$ and $\var{trap_h}$ respectively gives, 

\begin{align}
\mathrm{Area} &= \frac{1}{2} (a + b) h \\
&= \frac{1}{2} \times (\var{trap_length_a} + \var{trap_length_b}) \times \var{trap_h} \\
&= \var{precround((0.5) (trap_length_a + trap_length_b) trap_h, 1)}\, \mathrm{cm}^2 \quad \text{to 1 d.p.} \\
\end{align}

We're not certain about some of the measurements given in this question - we only know the rounded values. This means that the true value could be lower or higher than the given measurement.

We can find upper and lower bounds for the given measurements. Any values we go on to calculate will also be uncertain and have upper and lower bounds.

To find bounds for a given measurement, we divide the degree of accuracy by 2 and subtract or add this to our estimate to get lower and upper bounds respectively.

For example, $52$ rounded to the nearest integer has a lower bound of $51.5$ and an upper bound of $52.5$.

a)

The distance travelled is given by

\[ d = \text{Average speed} \times \text{Time taken} \]

We find bounds for speed and time first.

Lower bound for speed: $\var{speed} - 0.5 = \var{speed - 0.5} \text{ km/h}$

Upper bound for speed: $\var{speed} + 0.5 = \var{speed + 0.5} \text{ km/h}$

First, note that the speed is given in km/h and we want to find the distance in km. We will convert the given time into hours. 

Lower bound for time taken:

\begin{align} \var{atime} - 0.5 &= \var{atime - 0.5} \text{ min} \\[0.5em]
&= \frac{\var{atime - 0.5}}{60} \text{ h}
\end{align}

Upper bound for time taken:

\begin{align} \var{atime} + 0.5 &= \var{atime + 0.5} \text{ min} \\[0.5em]
&= \frac{\var{atime + 0.5}}{60} \text{ h}
\end{align}

Since we're multiplying the speed and time together, the lower bound for distance is the slowest speed multiplied by the shortest time:

\begin{align}
\text{Lower bound} &= \text{lower bound for speed} \times \text{lower bound for time}\\
&= \var{speed - 0.5} \times \frac{\var{(atime - 0.5)}}{60} \\
&= \var{precround((speed-0.5)*(atime - 0.5)/60, 2 )} \text{ km} \quad \text{(rounded to 2 decimal places).}
\end{align}

The upper bound for distance is the fastest speed multiplied by the longest time:

\begin{align}
\text{Upper bound} &= \text{upper bound for speed} \times \text{upper bound for time} \\
&= \var{speed + 0.5} \times \frac{\var{atime + 0.5}}{60} \\
&= \var{precround((speed+0.5)*(atime + 0.5)/60, 2 )} \text{ km} \quad \text{(rounded to 2 decimal places).}
\end{align}

Hence,

\[\var{precround((speed-0.5)*(atime - 0.5)/60, 2 )} \leq d \lt \var{precround((speed+0.5)*(atime + 0.5)/60, 2 )} \text{.}\]

b)

We're told the speed and the distance travelled, so the time taken is given by

\[ t = \frac{\text{Distance travelled}}{\text{Average speed}} \]

We found upper and lower bounds for George's average speed above.

The distance of the evening run is given to the nearest kilometre, so we can compute bounds as follows:

Lower bound for distance: $\var{distance} - 0.5 = \var{distance - 0.5} \mathrm{km}$

Upper bound for distance: $\var{distance} + 0.5 = \var{distance + 0.5} \mathrm{km}$

The upper bound for the time taken is the longest distance divided by the slowest speed:

\begin{align}
\text{Upper bound} &= \text{upper bound for distance} \div \text{lower bound for speed} \\
&= \var{distance + 0.5} \div \var{speed - 0.5} \\
&= \var{(distance + 0.5)/(speed - 0.5)} \text{ hours.}
\end{align}

We're asked for the answer in minutes, to two decimal places.

\begin{align} 
\var{(distance + 0.5)/(speed - 0.5)} \text{ hours} &= \var{(distance + 0.5)/(speed - 0.5)}\times 60  \text{ min} \\
&= \var{precround((distance + 0.5)/(speed - 0.5)*60, 2)} \text{ minutes} \quad \text{(rounded to 2 decimal places)}
\end{align}

The lower bound for time is the shortest distance divided by the fastest speed:

\begin{align}
\text{Lower bound} &= \text{lower bound for distance} \div \text{upper bound for speed} \\
&= \var{distance - 0.5} \div \var{speed + 0.5} \\
&= \var{(distance - 0.5)/(speed + 0.5)}  \text{ hours.}
\end{align}

Converting into minutes, to two decimal places:

\begin{align} 
\var{(distance - 0.5)/(speed + 0.5)} \text{ hours} &= \var{(distance - 0.5)/(speed + 0.5)}\times 60  \text{ min} \\
&= \var{precround((distance - 0.5)/(speed + 0.5)*60, 2)} \text{ min.}
\end{align}

Therefore, we cannot confidently say George's time was less than 31 minutes as the upper bound for his time, $\var{precround((distance + 0.5)/(speed - 0.5)*60, 2)}$ minutes, is above this threshold.

a)

It is much better to have spare paint than not to have enough of it. So it is better to overestimate the area.

Therefore, we round each measurement up.

b)

We round each of our measurements up to the nearest whole metre:

Length: $\var{length}\,\mathrm{m} \approx \var{l}\,\mathrm{m}$.

Width: $\var{width}\,\mathrm{m} \approx \var{w}\,\mathrm{m}$.

Height: $\var{height}\,\mathrm{m} \approx \var{h}\,\mathrm{m}$.

The total area consists of five areas: two walls of $\var{l}\,\text{m} \times \var{h}\,\text{m}$ (length by height); two walls of $\var{w}\,\text{m} \times \var{h}\,\text{m}$ (width by height); and a ceiling of $\var{l}\,\text{m} \times \var{w}\,\text{m}$ (length by width).

\[ \begin{align}
\var{l}\,\text{m} \times \var{h}\,\text{m} &= \var{l*h}\,\text{m}^2
\\ \var{w}\,\text{m} \times \var{h}\,\text{m} &= \var{w*h}\,\text{m}^2
\\ \var{l}\,\text{m} \times \var{w}\,\text{m} &= \var{l*w}\,\text{m}^2
\end{align}\]

Therefore, the total area Luke needs to paint is

\[ \var{2*l*h} + \var{2*w*h} + \var{l*w} \,\mathrm{m}^2 = \var{rall}\,\mathrm{m}^2 \text{.} \]

c)

The exact number of buckets needed is

\[\var{rall}\,\text{m}^2 \div 15\,\text{m}^2 = \var{rall/15} \text{.}\]

Luke can only buy a whole number of buckets, so he needs to decide between 5 and 6 paint buckets. As it is better to buy more paint than not buy enough, he should buy 6 buckets of yellow paint.

a)

Ellie's training run is $\var{km}$ km long. 

To convert $\var{km}$ km into metres, we multiply $\var{km}$ by $1000$. 

\[\var{km}\times1000= \var{km*1000}\text{ metres.}\]

b)

To convert $\var{km}$ km into miles, we multiply $\var{km}$ by the conversion rate given: $0.62$.

\[\begin{align}
\var{km}\times\frac{5}{8}&= \var{miles}\\
&=\var{dpformat(miles,0)}\text{ miles, rounded to the nearest integer.}
\end{align}\]

c)

The Great Welsh Marathon is $26$ miles long. To convert to km we multiply by the inverse of the conversion rate given in part b):

\[ 26 \times \frac{1}{0.62} = 42\text{ miles, rounded to the nearest integer.} \]

Abid is $5$ft $\var{inches}$ inches tall.

a)

To find Abid's height into cm, we first convert it into inches. 

We can convert feet into inches by multiplying by $12$. So $5$ feet is

\[ 5\times12=60 \text{ inches.}\]

Therefore $5$ft $\var{inches}$ inches is 

\[
60+\var{inches} = \var{60+inches}\text{ inches.}
\]

We can now convert this to cm by multiplying by $2.54$.

\[\var{60+inches}\times2.54=\var{cm}\text{ cm, rounded to the nearest integer.}\]

b)

To convert centimetres into metres, we divide by $100$:

\[\var{cm}\div100=\var{dpformat(cm/100,2)} \text{ metres.}\]

Therefore, Abid is $\var{dpformat(cm/100,2)}$ metres tall. 

a)

To find the average speed of the runner in meters per second (m/s), we divide the distance covered by the runner (in metres) by the time taken for the runner to run this distance (in seconds).

\[
\begin{align}
\text{Average speed} &= \displaystyle\frac{\var{distance}}{\var{seconds}}\\
&= \var{distance/seconds}\\
&= \var{dpformat(distance/seconds,2)}\; \text{m/s} \; (\text{rounded to $2$ decimal places}).
\end{align}
\]

b)

We can convert the average speed of the runner that we calculated in a) in metres per second to kilometres per hour using the following two equivalences:

\[1\text{m} = \displaystyle\frac{1}{1000}\text{km},\]

\[
1 \; \text{second} = \displaystyle\frac{1}{60} \; \text{minutes} = \displaystyle\frac{1}{3600} \; \text{hours}.
\]

We know from a) that the average speed of the runner in m/s was $\var{dpformat(distance/seconds,5)}$ m/s ($5$ d.p), so to convert this speed to km/h we first need to convert metres to kilometres,

\[\var{dpformat(distance/seconds,5)} \; \text{m/s} = \var{dpformat(distance/seconds/1000,5)} \text{km/s} \; (5 \; \text{d.p})\]

Then we convert seconds to hours,

\[1 \; \text{second} = \displaystyle\frac{1}{3600} \; \text{hours}.\]

Now we have

\[\var{dpformat(distance/seconds,5)} \; \text{m/s} = \var{sigformat(distance/seconds/1000,5)} \; \text{kilometres per $\displaystyle\frac{1}{3600}$ hours}.\]

We want a rate per one hour, so we multiply by $3600$ to obtain a measurement in km/h:

\[\begin{align}\var{sigformat(distance/seconds/1000,5)} \; \text{kilometres per $\displaystyle\frac{1}{3600}$ hours} &= \var{siground(distance/seconds/1000,5)*3600} \; \text{km}/\text{h}\\&=\var{dpformat(distance/seconds*3.6, 2)} \; \text{km}/\text{h} \; (\text{rounded to $2$ decimal places}).\end{align}\]

Note that throughout this calculation we have rounded all figures to $5$ decimal places for convenience; when doing calculations which involve long decimals, you should always input the full figure into your calculator to avoid getting an incorrect answer due to rounding.

a)

The price per hour is the total price divide by the number of hours.

\[ \text{Price per hour} = \frac{\var{block_price_per_hour*hours}}{\var{hours}} = £\var{dpformat(block_price_per_hour,2)} \text{ per hour} \]

b)

The price is given in pence per minute. To convert to pounds per minute, divide by $100$:

\[ \var{100*competitor_price_per_minute} \text{ p/minute} = £\var{dpformat(competitor_price_per_minute,2)} \text{ per minute} \]

Then to convert to pounds per hour, multiply by $60$:

\[ £\var{dpformat(competitor_price_per_minute,2)} \text{ per minute} = £\var{dpformat(competitor_price_per_minute*60,2)} \text{ per hour} \]

c)

Emily should choose the method with the lowest cost per hour, which is paying per minute at RoomCo.

a)

The area of a rectangle is calculated using the formula

\[\mathrm{Area} = \mathrm{base} \times \mathrm{height}\text{.}\]

We have a base of $\var{w0}$m and a height $\var{h0}$m, therefore

\begin{align}
\mathrm{Area} &= \mathrm{base} \times \mathrm{height} \\
&= \var{w0} \times \var{h0} \\ &= \var{w0*h0} \\
&= \var{dpformat(w0*h0,1)}\, \mathrm{m}^2 \quad \text{to 1 d.p.} \\
\end{align}

b)

The parallelogram is just a slanted rectangle:

Animation by Picknick.

Therefore, the area of a parallelogram is calculated using the formula

\[\mathrm{Area} = \mathrm{base} \times \mathrm{height}.\]

We have a base $\var{w1}$m and perpendicular height $\var{h1}$m.

\begin{align}
\mathrm{Area} &= \mathrm{base} \times \mathrm{height} \\
&= \var{w1} \times \var{h1} \\
&= \var{dpformat({w1}{h1},1)}\, \mathrm{m}^2 \quad \text{to 1 d.p.} \\
\end{align}

c)

The area of a triangle is calculated using the formula

\[\mathrm{Area} = \frac{\mathrm{base} \times \mathrm{height}}{2}.\]

Note that the triangle is half of a rectangle:

Our triangle has a base $\var{w2}$m and a height $\var{h2}$m, therefore

\begin{align} \mathrm{Area} &= \frac{1}{2} \times \mathrm{base} \times \mathrm{height} \\ 
&= \frac{1}{2} \times \var{w2} \times \var{h2} \\
&= \var{0.5*w2*h2}\, \mathrm{m}^2 \\
&= \var{dpformat(0.5*w2*h2, 1)}\, \mathrm{m}^2 \quad \text{to 1 d.p.} \\
\end{align}

d)

A trapezium can be interpreted as half of a parallelogram, this is shown below:

As we only want the area of one half of this shape, the area is half of

\[\mathrm{area} = (a+b) \times \mathrm{height}\text{,}\]

with ${a} = \var{w5a}$m, ${b} = \var{w5b}$m, and height $\var{h5}$m.

\begin{align}
\mathrm{Area} &= \frac{(a+b)}{2} \times \mathrm{height} \\
&= \frac{(\var{w5a}+\var{w5b})}{2} \times \var{h5} \\
&= \var{(w5a+w5b)*0.5} \times \var{h5} \\
&= \var{(w5a+w5b)*(h5)/2}\, \mathrm{m}^2 \\
&= \var{dpformat((w5a+w5b)*(h5)/2, 1)}\, \mathrm{m}^2 \quad \text{to 1 d.p.}  
\end{align}

a)

We are given the price of a bag of $\var{weight}$ grams of sweets.

To find the price per 100g of sweets we divide the price of a bag of sweets by its weight in grams and then multiply this by $100$.

\begin{align}
\displaystyle\frac{\var{cost}}{\var{weight}} \times 100 &= \var{(100*cost/weight)}\\ &= \var{dpformat(100*cost/weight,2)} \; (\text{rounded to $2$ decimal places}).
\end{align}

The sweets cost £1.76 per 100g.

b)

To convert the cost from pounds per $100$ grams to pounds per kilogram we need to use the fact that $1\text{g} = \displaystyle\frac{1}{1000}\text{kg}$.

This means that $100\text{g} = \displaystyle\frac{1}{10}\text{kg}$.

We know from a) that sweets cost £1.76 per 100g, which is the same as £1.76 per $\frac{1}{10}$kg.

We want the price per one kilogram of sweets, so we multiply by $10$.

Note that we use the actual value of $\displaystyle\frac{\var{cost}}{\var{weight}} \times 100 = \var{100*cost/weight}$ here to ensure that our final answer is accurate.

\begin{align}
\var{100*cost/{weight}} \times 10 &= \var{dpformat(1000*{cost}/{weight},2)} \; (2 \; \text{d.p})
\end{align}

So, the sweets cost £17.63 per kg.

c)

We worked out in part a) that sweets cost £1.76 per 100g when bought in the bag, so at £1.64 per 100g the Pick'n'Mix is cheaper than buying the bag.

a)

We are told that the baby's weight is $\var{weight_kg}$kg and we would like to convert this to pounds and ounces.

To do this, we first obtain the weight in ounces using the conversion given:

\begin{align}
\var{weight_kg} \text{ kg}  &= \var{weight_kg} \times \var{oz_to_kg} \text{ oz} \\
&= \var{weight_oz}\text{ oz.}
\end{align}

We can now divide by $16$ to find out how many whole pounds we have:

\[ \var{weight_oz}\text{ oz} \div 16 =  \var{dpformat(weight_oz/16,1)}\text{ lb.} \]

There are $\var{weight_lb}$ whole pounds, which leaves a remainder

\[ \var{weight_oz} - (16 \times \var{weight_lb}) =  \var{weight_oz_rem} \text{ oz.} \]

The baby weighs $\var{weight_lb}$ pounds and $\var{weight_oz_rem}$ ounces.

b)

We are given the weight of the baby in stones, pounds and ounces. To convert to kg we first convert the baby's weight to ounces.

$1$ stone is $14$ pounds, so the baby weighs $\var{news['lb']+14}$ pounds and $\var{news['oz']}$ ounces, or

\[  (\var{news['lb']+14} \times 16) + \var{news['oz']} = \var{(news['lb']+14)*16+news['oz']} \text{ oz.}\]

We can now convert to kg:

\begin{align}
\var{(news['lb']+14)*16+news['oz']} \text{ oz} &= \var{(news['lb']+14)*16+news['oz']} \div \var{oz_to_kg}  \\
&= \var{dpformat(newskg,1)}\text{ kg.}
\end{align}

a)

The advertised engine size is $\var{cc}$ cubic centimetres. To convert cubic centimetres to litres, we divide by $1000$. 

\[\var{cc}\div 1000= \var{litres}\text{ litres.}\]

b)

In order to convert to cubic metres, we first note that 

\[ 1 \text{cm} = 0.01 \text{m.} \]

An example of a volume of $1\text{cm}^3$ is a cube with $1$cm sides. Converting each side into metres,

\begin{align}
1\text{cm}^3 &= 1\text{cm}\times1\text{cm}\times1\text{cm} \\
&= 0.01\text{m}\times0.01\text{m}\times0.01\text{m} \\
&= 0.000001\text{m}^3 \text{.}
\end{align}

Therefore $\var{cc}\text{cm}^3$ is 

\[ \var{cc} \times 0.000001 = \var{metres_cubed}\text{m}^3\text{.} \]

a)

The price per banana at Fine Fare supermarket is

\[ \var{b} \div 5 = \var{b_single}\text{p.}\]

b)

As the price at Hintons supermarket is cheaper, he should shop there.

c)

The new price per banana at Fine Fare Supermarket is

\begin{align}
\var{b} \div 6 &= \var{if(isint(b/6),b/6,dpformat(b/6,1))}\text{p.} \\
\end{align}

Since the price at Hintons Supermarket is still cheaper, Ayaan should not change his decision.