A triangular field $\text{ABC}$ is such that $\text{AB}=56 \text{m}$ and $\text{BC}=82 \text{m}$, each measured correct to the nearest metre, and the angle at $\text{B}$ is equal to $105°$, measured correct to the nearest $5°$.

Calculate the maximum possible area of the field.

attempt to find any relevant maximum value         (M1)

largest sides are $56.5$ and $82.5$         (A1)

smallest possible angle is $102.5$         (A1)

attempt to substitute into area of a triangle formula         (M1)

$\frac{1}{2}\times 56.5\times 82.5\times \sin \left(102.5°\right)$

$=2280 \left({\text{m}}^2\right) \left(2275.37\dots \right)$             A1

[5 marks]

Write down the value of the iron in the form $a\times {10}^k$ where $1\le a<10 , k\in \mathrm{\mathbb{Z}}$.

Calculate James’s estimate of its volume, in ${\text{km}}^3$.

The actual volume of the asteroid is found to be $6.074\times {10}^6 {\text{km}}^3$.

Find the percentage error in James’s estimate of the volume.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$8.97\times {10}^{18} \left(\text{EUR}\right) \left(8.973\times {10}^{18}\right)$        (A1)(A1)  (C2)

Note: Award (A1) for $8.97 (8.973)$, (A1) for $\times {10}^{18}$. Award (A1)(A0) for $8.97\text{E}18$.
Award (A0)(A0) for answers of the type $8973\times {10}^{15}$.

[2 marks]

$\frac{4\times \mathrm{\pi }\times {113}^3}{3}$       (M1)

Note: Award (M1) for correct substitution in volume of sphere formula.

$6 040 000 \left({\text{km}}^3\right) \left(6.04\times {10}^6, \frac{5771588\mathrm{\pi }}{3}, 6 043 992.82\right)$       (A1)  (C2) 

[2 marks]

$\left|\frac{6 043 992.82-6.074\times {10}^6}{6.074\times {10}^6}\right|\times 100$       (M1)

Note: Award (M1) for their correct substitution into the percentage error formula (accept a consistent absence of “$\times {10}^6$” from all terms).

$0.494 \left(\%\right) \left(0.494026\dots \left(\%\right)\right)$       (A1)(ft)  (C2) 

Note: Follow through from their answer to part (b). If the final answer is negative, award at most (M1)(A0).

[2 marks]

Find the equation of the line that the road follows.

Town $\text{C}$ is due north of town $\text{A}$ and the road passes through town $\text{C}$.

Find the $y$-coordinate of town $\text{C}$.

midpoint $(1, 2.5)$           A1

$m_{AB}=\frac{6-\left(-1\right)}{8-\left(-6\right)}=\frac{1}{2}$           (M1)A1

Note: Accept equivalent gradient statements including using midpoint.

$m_{\perp }=-2$           M1

Note: Award M1 for finding the negative reciprocal of their gradient.

$y-2.5=-2\left(x-1\right)$  OR  $y=-2x+\frac{9}{2}$  OR  $4x+2y-9=0$           A1

[5 marks]

substituting $x=-6$ into their equation from part (a)           (M1)

$y=-2\left(-6\right)+\frac{9}{2}$

$y=16.5$           A1

Note: Award M1A0 for $\left(-6, 16.5\right)$ as their final answer.

[2 marks]

A large proportion of candidates seemed to be well drilled into finding the gradient of a line and the subsequent gradient of the normal. But without finding the coordinates of the midpoint of AB, no more marks were gained.

Many candidates worked out the value of $y$ correctly (or “correct” following the value they found in part (a)) but then incorrectly gave their answer as a coordinate pair.

Write down the size of the angle of depression from $\text{H}$ to $\text{C}$.

Find the distance from $\text{A}$ to $\text{C}$.

Find the distance from $\text{B}$ to $\text{C}$.

Find Minta’s speed, in metres per hour.

$25°$            A1

[1 mark]

$\text{AC}=\frac{380}{\tan 25°}$  OR  $\text{AC}=\sqrt{{\left(\frac{380}{\sin 25°}\right)}^2-{380}^2}$  OR  $\frac{380}{\sin 25°}=\frac{\text{AC}}{\sin 65°}$         (M1)

$\text{AC}=815 \text{m} \left(814.912\dots \right)$           A1

[2 marks]

METHOD 1

attempt to find $\text{AB}$         (M1)

$\text{AB}=\frac{380}{\tan 40°}$

$=453 \text{m} \left(452.866\dots \right)$           (A1)

$\text{BC}=814.912\dots -452.866\dots$

$=362 \text{m} \left(362.046\dots \right)$          A1

METHOD 2

attempt to find $\text{HB}$         (M1)

$\text{HB}=\frac{380}{\sin 40°}$

$591 \text{m} \left(=591.175\dots \right)$           (A1)

$\text{BC}=\frac{591.175\dots \times \sin 15°}{\sin 25°}$

$=362 \text{m} \left(362.046\dots \right)$          A1

[3 marks]

$362.046\dots \times 4$

$=1450 {\text{m\hspace{0.05em}h}}^{-1} \left(1448.18\dots \right)$          A1 

[1 mark]

Explain why Tim will receive the strongest signal from tower $\text{T4}$.

Write down the coordinates of tower $\text{T4}$.

Tower $\text{T1}$ has coordinates $(-13, 3)$.

Find the gradient of the edge of the Voronoi diagram between towers $\text{T1}$ and $\text{T2}$.

every point in the shaded region is closer to tower $\text{T4}$                     R1

Note: Specific reference must be made to the closeness of tower $\text{T4}$.

 
[1 mark]

$\left(-9, 1\right)$                    A1A1

Note: Award A1 for each correct coordinate. Award at most A0A1 if parentheses are missing.

 
[2 marks]

correct use of gradient formula                   (M1)

e.g. $\left(m=\right)\frac{5-3}{-9--13} \left(=\frac{1}{2}\right)$

taking negative reciprocal of their $m$ (at any point)                    (M1)

edge gradient $=-2$                    A1

 
[3 marks]

Express $\overrightarrow{\text{AB}}$ in terms of $m$.

Find the value of $m$.

Consider a unit vector $\mathit{u}$, such that $\mathit{u}=p\mathit{i}-\frac{2}{3}\mathit{j}+\frac{1}{3}\mathit{k}$, where $p>0$.

Point $\text{C}$ is such that $\overrightarrow{\text{BC}}=9\mathit{u}$.

Find the coordinates of $\text{C}$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach to find $\overrightarrow{\text{AB}}$        (M1)

eg     $\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}} , \text{A}-\text{B}$

$\overrightarrow{\text{AB}}=\left(\begin{array}{c}8\\ m-1\\ -8\end{array}\right)$       A1     N2

[2 marks]

valid approach        (M1)

eg     $L=\left(\begin{array}{c}9\\ m\\ -6\end{array}\right) , \left(\begin{array}{c}9\\ m\\ -6\end{array}\right)=\left(\begin{array}{c}-3\\ -19\\ 24\end{array}\right)+s\left(\begin{array}{c}2\\ 4\\ -5\end{array}\right)$

one correct equation        (A1)

eg       $-3+2s=9, -6=24-5s$

correct value for $s$            A1

eg       $s=6$

substituting their $s$ value into their expression/equation to find $m$       (M1)

eg       $-19+6\times 4$

$m=5$       A1     N3

[5 marks]

valid approach        (M1)

eg     $\overrightarrow{\text{BC}}=\left(\begin{array}{c}9p\\ -6\\ 3\end{array}\right), C=9\mathit{u}+B , \overrightarrow{\text{BC}}=\left(\begin{array}{c}x-9\\ y-5\\ z+6\end{array}\right)$

correct working to find $\text{C}$        (A1)

eg     $\overrightarrow{\text{OC}}=\left(\begin{array}{c}9p+9\\ -1\\ -3\end{array}\right), \text{C}=9\left(\begin{array}{c}p\\ -\frac{2}{3}\\ \frac{1}{3}\end{array}\right)+\left(\begin{array}{c}9\\ 5\\ -6\end{array}\right), y=-1$ and $z=-3$

correct approach to find $\left|\mathit{u}\right|$ (seen anywhere)            A1

eg     $p^2+{\left(-\frac{2}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2 , \sqrt{p^2+\frac{4}{9}+\frac{1}{9}}$

recognizing unit vector has magnitude of $1$        (M1)

eg     $\left|\mathit{u}\right|=1 , \sqrt{p^2+{\left(-\frac{2}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2}=1 , p^2+\frac{5}{9}=1$

correct working        (A1)

eg     $p^2=\frac{4}{9} , p=\pm \frac{2}{3}$

$p=\frac{2}{3}$            A1

substituting their value of $p$        (M1)

eg     $\left(\begin{array}{c}x-9\\ y-5\\ z+6\end{array}\right)=\left(\begin{array}{c}6\\ -6\\ 3\end{array}\right), \text{C}=\left(\begin{array}{c}6\\ -6\\ 3\end{array}\right)+\left(\begin{array}{c}9\\ 5\\ -6\end{array}\right), \text{C}=9\left(\begin{array}{c}\frac{2}{3}\\ -\frac{2}{3}\\ \frac{1}{3}\end{array}\right)+\left(\begin{array}{c}9\\ 5\\ -6\end{array}\right), x-9=6$

$\text{C}\left(15, -1, -3\right)$  (accept $\left(\begin{array}{c}15\\ -1\\ -3\end{array}\right)$)     A1     N4

Note: The marks for finding $p$ are independent of the first two marks.
For example, it is possible to award marks such as (M0)(A0)A1(M1)(A1)A1 (M0)A0 or (M0)(A0)A1(M1)(A0)A0 (M1)A0.

[8 marks]

Calculate the total surface area of one piece of candy.

The total surface of the candy is coated in chocolate. It is known that $1$ gram of the chocolate covers an area of $240 {\text{mm}}^2$.

Calculate the weight of chocolate required to coat one piece of candy.

$\frac{1}{2}\times 4\times \mathrm{\pi }\times 6^2+\mathrm{\pi }\times 6^2$  OR  $3\times \mathrm{\pi }\times 6^2$                       (M1)(A1)(M1)

Note: Award M1 for use of surface area of a sphere formula (or curved surface area of a hemisphere), A1 for substituting correct values into hemisphere formula, M1 for adding the area of the circle.

$=339 {\text{mm}}^2 \left(108\mathrm{\pi }, 339.292\dots \right)$               A1

[4 marks]

$\frac{339.292\dots }{240}$                      (M1)

$=1.41 \left(\text{g}\right) \left(\frac{9\mathrm{\pi }}{20}, 0.45\mathrm{\pi }, 1.41371\dots \right)$              A1

[2 marks]

Find an expression for the velocity of $P_1$ at time $t$.

Particle $P_2$ also moves in a straight line. The position of $P_2$ is given by $\mathit{r}=\left(\begin{array}{c}-1\\ 6\end{array}\right)+t\left(\begin{array}{c}4\\ -3\end{array}\right)$.

The speed of $P_1$ is greater than the speed of $P_2$ when $t>q$.

Find the value of $q$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognizing velocity is derivative of displacement     (M1)

eg    $v=\frac{\text{d}s}{\text{d}t} , \frac{\text{d}}{\text{d}t}\left(10-\frac{7}{4}{t}^2\right)$

velocity$=-\frac{14}{4}t \left(=-\frac{7}{2}t\right)$        A1 N2

[2 marks]

valid approach to find speed of $P_2$     (M1)

eg    $\left|\left(\begin{array}{c}4\\ -3\end{array}\right)\right| , \sqrt{4^2+{\left(-3\right)}^2}$ , velocity$=\sqrt{4^2+{\left(-3\right)}^2}$

correct speed     (A1)

eg   $5 {\text{m\hspace{0.05em}s}}^{-1}$

recognizing relationship between speed and velocity (may be seen in inequality/equation)        R1

eg   $\left|-\frac{7}{2}t\right|$ , speed = | velocity | , graph of $P_1$ speed ,  $P_1$ speed $=\frac{7}{2}t , P_2$ velocity $=-5$

correct inequality or equation that compares speed or velocity (accept any variable for $q$)      A1

eg   $\left|-\frac{7}{2}t\right|>5 , -\frac{7}{2}q<-5 , \frac{7}{2}q>5 , \frac{7}{2}q=5$

$q=\frac{10}{7}$ (seconds) (accept $t>\frac{10}{7}$ , do not accept $t=\frac{10}{7}$)       A1   N2

Note: Do not award the last two A1 marks without the R1.

[5 marks]

Find the area of the field.

The farmer would like to divide the field into two equal parts by constructing a straight fence from $\text{A}$ to a point $\text{D}$ on $\text{[BC]}$.

Find $\text{BD}$. Fully justify any assumptions you make.

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

Area$=\frac{1}{2}\times 110\times 85\times \sin 55°$        (M1)(A1)

        $=3830 \left(3829.53\dots \right) {\text{m}}^2$        A1

Note: units must be given for the final A1 to be awarded.

[3 marks]

${\text{BC}}^2={110}^2+{85}^2-2\times 110\times 85\times \cos 55°$        (M1)A1

$\text{BC}=92.7 (92.7314\dots ) \left(\text{m}\right)$        A1

METHOD 1

Because the height and area of each triangle are equal they must have the same length base         R1

$\text{D}$ must be placed half-way along $\text{BC}$        A1

$\text{BD}=\frac{92.731\dots }{2}\approx 46.4 \left(\text{m}\right)$        A1

Note: the final two marks are dependent on the R1 being awarded.

METHOD 2

Let $\text{C}\hat{\text{B}}\text{A}=\theta °$

$\frac{\sin \theta }{110}=\frac{{\displaystyle \sin 55°}}{{\displaystyle 92.731\dots }}$        M1

$\Rightarrow \theta =76.3° \left(76.3354\dots \right)$

Use of area formula

$\frac{1}{2}\times 85\times \text{BD}\times \sin \left(76.33\dots °\right)=\frac{3829.53\dots }{2}$        A1

$\text{BD}=46.4 (46.365\dots ) \left(\text{m}\right)$        A1 

[6 marks]

Given that $\text{A}\hat{\text{B}}\text{C}$ is acute, find $\sin \theta$.

Find $\cos \left(2\times \text{C}\hat{\text{A}}\text{B}\right)$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 – (sine rule)

evidence of choosing sine rule       (M1)

eg   $\frac{\sin \hat{A}}{a}=\frac{\sin \hat{B}}{b}$

correct substitution       (A1)

eg   $\frac{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{10}=\frac{\sin {\displaystyle }{\displaystyle \theta }}{15} , \frac{\sqrt{3}}{30}=\frac{{\displaystyle \sin \theta }}{15} , \frac{{\displaystyle \sqrt{3}}}{{\displaystyle 30}}=\frac{{\displaystyle \sin \text{B}}}{15}$

$\sin \theta =\frac{\sqrt{3}}{2}$        A1  N2

METHOD 2 – (perpendicular from vertex $\mathbf{\text{C}}$)

valid approach to find perpendicular length (may be seen on diagram)       (M1)

eg    , $\frac{h}{15}=\frac{\sqrt{3}}{3}$

correct perpendicular length       (A1)

eg    $\frac{15\sqrt{3}}{3} , 5\sqrt{3}$

$\sin \theta =\frac{\sqrt{3}}{2}$        A1  N2

Note: Do not award the final A mark if candidate goes on to state $\sin \theta =\frac{\mathrm{\pi }}{3}$, as this demonstrates a lack of understanding.

[3 marks]

attempt to substitute into double-angle formula for cosine       (M1)

$1-2{\left(\frac{\sqrt{3}}{3}\right)}^2, 2{\left(\frac{\sqrt{6}}{3}\right)}^2-1, {\left(\frac{\sqrt{6}}{3}\right)}^2-{\left(\frac{\sqrt{3}}{3}\right)}^2, \cos \left(2\theta \right)=1-2{\left(\frac{\sqrt{3}}{2}\right)}^2, 1-2 {\sin }^2\left(\frac{\sqrt{3}}{3}\right)$

correct working       (A1)

eg  $1-2\times \frac{3}{9}, 2\times \frac{6}{9}-1, \frac{6}{9}-\frac{3}{9}$

$\cos \left(2\times \text{C}\hat{\text{A}}\text{B}\right)=\frac{3}{9} \left(=\frac{1}{3}\right)$          A1  N2

[3 marks]

A storage container consists of a box of length $90 \text{cm}$, width $42 \text{cm}$ and height $34 \text{cm}$, and a lid in the shape of a half-cylinder, as shown in the diagram. The lid fits the top of the box exactly. The total exterior surface of the storage container is to be painted.

Find the area to be painted.

$2\times 90\times 34 \left(=6120\right)$  AND  $2\times 42\times 34 \left(=2856\right)$             (A1)

$90\times 42 \left(=3780\right)$             (A1)

$r=21$             (A1)

$\mathrm{\pi }\times {21}^2 \left(=441\mathrm{\pi }, 1385.44\dots \right)$               (M1)

use of curved surface area formula               (M1)

$21\mathrm{\pi }\times 90 \left(=1890\mathrm{\pi }, 5937.61\dots \right)$             (A1)

$20100 {\text{cm}}^2 (20079.0\dots )$                A1

[7 marks]

Write down the perimeter of the base of the hat in terms of $\mathrm{\pi }$.

Find the value of $\theta$.

Find the surface area of the outside of the hat.

$13\mathrm{\pi } \mathrm{cm}$           A1

Note: Answer must be in terms of $\mathrm{\pi }$.

[1 mark]

METHOD 1

$\frac{\theta }{360}\times 2\mathrm{\pi }\left(18\right)=13\mathrm{\pi }$  OR  $\frac{\theta }{360}\times 2\mathrm{\pi }\left(18\right)=40.8407\dots$               (M1)

Note: Award (M1) for correct substitution into length of an arc formula.

$\left(\theta =\right) 130°$           A1

METHOD 2

$\frac{\theta }{360}\times \mathrm{\pi }\times {18}^2=\mathrm{\pi }\times 6.5\times 18$               (M1)

$\left(\theta =\right) 130°$           A1

[2 marks]

EITHER

$\frac{130}{360}\times \mathrm{\pi }{\left(18\right)}^2$               (M1)

Note: Award (M1) for correct substitution into area of a sector formula.

OR

$\mathrm{\pi }\left(6.5\right)\left(18\right)$               (M1)

Note: Award (M1) for correct substitution into curved area of a cone formula.

THEN

(Area$=$) $368 {\text{cm}}^2 \left(367.566\dots , 117\mathrm{\pi }\right)$             A1

Note: Allow FT from their part (a)(ii) even if their angle is not obtuse.

[2 marks]

Although most candidates understood what to do in part (a), many of them wrote a decimal approximation instead and did not give their answer in terms of $\mathrm{\pi }$ as required in this part. Many candidates were able to use the length of arc formula in (a)(ii). Some candidates went wrong with this question by confusing between the two values: $6.5 \mathrm{cm}$ and $18 \mathrm{cm}$ for the radius. In part (b), although candidates were able to find the surface area of the outside of the hat, several added the surface area of the base to their calculation. A few candidates used the cosine rule to find chord $\mathrm{AB}$ which was then used as the circumference of the base of the cone.

Although most candidates understood what to do in part (a), many of them wrote a decimal approximation instead and did not give their answer in terms of $\mathrm{\pi }$ as required in this part. Many candidates were able to use the length of arc formula in (a)(ii). Some candidates went wrong with this question by confusing between the two values: $6.5 \mathrm{cm}$ and $18 \mathrm{cm}$ for the radius. In part (b), although candidates were able to find the surface area of the outside of the hat, several added the surface area of the base to their calculation. A few candidates used the cosine rule to find chord $\mathrm{AB}$ which was then used as the circumference of the base of the cone.

Although most candidates understood what to do in part (a), many of them wrote a decimal approximation instead and did not give their answer in terms of $\mathrm{\pi }$ as required in this part. Many candidates were able to use the length of arc formula in (a)(ii). Some candidates went wrong with this question by confusing between the two values: $6.5 \mathrm{cm}$ and $18 \mathrm{cm}$ for the radius. In part (b), although candidates were able to find the surface area of the outside of the hat, several added the surface area of the base to their calculation. A few candidates used the cosine rule to find chord $\mathrm{AB}$ which was then used as the circumference of the base of the cone.

Show that the new station should be built at $\text{P}$.

Write down the equation of the perpendicular bisector of $\text{[PC]}$.

Hence draw the missing boundaries of the region around $\text{P}$ on the following diagram.

(the best placement is either point $\text{P}$ or point $\text{Q}$)
attempt at using the distance formula              (M1)

$\text{AP}=\sqrt{{\left(10-6\right)}^2+{\left(6-2\right)}^2}$  OR

$\text{BP}=\sqrt{{\left(10-14\right)}^2+{\left(6-2\right)}^2}$  OR

$\text{DP}=\sqrt{{\left(10-10.8\right)}^2+{\left(6-11.6\right)}^2}$  OR

$\text{BQ}=\sqrt{{\left(13-14\right)}^2+{\left(7-2\right)}^2}$  OR

$\text{CQ}=\sqrt{{\left(13-18\right)}^2+{\left(7-6\right)}^2}$  OR

$\text{DQ}=\sqrt{{\left(13-10.8\right)}^2+{\left(7-11.6\right)}^2}$  

($\text{AP}$ or $\text{BP}$ or $\text{DP}$$=$)  $\sqrt{32}=5.66 \left(5.65685\dots \right)$  AND

($\text{BQ}$ or $\text{CQ}$ or $\text{DQ}$$=$)  $\sqrt{26}=5.10 \left(5.09901\dots \right)$            A1

$\sqrt{32}>\sqrt{26}$  OR  $\text{AP}$ (or $\text{BP}$ or $\text{DP}$) is greater than $\text{BQ}$ (or $\text{CQ}$ or $\text{DQ}$)            A1

point $\text{P}$ is the furthest away            AG

Note: Follow through from their values provided their $\text{AP}$ (or $\text{BP}$ or $\text{DP}$) is greater than their $\text{BQ}$ (or $\text{CQ}$ or $\text{DQ}$).

[3 marks]

$x=14$           A1

[1 mark]

           A1A1

Note: Award A1 for each correct straight line. Do not FT from their part (b)(i).

[1 mark]

In part (a) many candidates realized that distances were required. Many candidates seemed to have an idea about Voronoi diagrams. However, several candidates did not realize that they had to consider point $\mathrm{Q}$ as well in their comparison. Hence, several candidates only calculated distances from $\mathrm{P}$. The numerical comparison of the distance from $\mathrm{P} \left(\mathrm{AP}/\mathrm{BP}/\mathrm{DP}\right)$ and from $\mathrm{Q} \left(\mathrm{BQ}/\mathrm{CQ}/\mathrm{DQ}\right)$ need to be clearly shown. It was a pity to see that some candidates lost marks due to incorrect rounding of the values to three significant figures. The most common error being $5.09$. In part (b)(i) not many candidates seemed to understand what was required. A significant number of candidates wrote down the equation of the line through $\mathrm{PC}$, $y=6$, rather than the required line. In part (b)(ii), it seemed that much time was lost as many candidates attempted to find the equation of the perpendicular bisector of $\mathrm{DP}$ to draw the boundaries.

In part (a) many candidates realized that distances were required. Many candidates seemed to have an idea about Voronoi diagrams. However, several candidates did not realize that they had to consider point $\mathrm{Q}$ as well in their comparison. Hence, several candidates only calculated distances from $\mathrm{P}$. The numerical comparison of the distance from $\mathrm{P} \left(\mathrm{AP}/\mathrm{BP}/\mathrm{DP}\right)$ and from $\mathrm{Q} \left(\mathrm{BQ}/\mathrm{CQ}/\mathrm{DQ}\right)$ need to be clearly shown. It was a pity to see that some candidates lost marks due to incorrect rounding of the values to three significant figures. The most common error being $5.09$. In part (b)(i) not many candidates seemed to understand what was required. A significant number of candidates wrote down the equation of the line through $\mathrm{PC}$, $y=6$, rather than the required line. In part (b)(ii), it seemed that much time was lost as many candidates attempted to find the equation of the perpendicular bisector of $\mathrm{DP}$ to draw the boundaries.

In part (a) many candidates realized that distances were required. Many candidates seemed to have an idea about Voronoi diagrams. However, several candidates did not realize that they had to consider point $\mathrm{Q}$ as well in their comparison. Hence, several candidates only calculated distances from $\mathrm{P}$. The numerical comparison of the distance from $\mathrm{P} \left(\mathrm{AP}/\mathrm{BP}/\mathrm{DP}\right)$ and from $\mathrm{Q} \left(\mathrm{BQ}/\mathrm{CQ}/\mathrm{DQ}\right)$ need to be clearly shown. It was a pity to see that some candidates lost marks due to incorrect rounding of the values to three significant figures. The most common error being $5.09$. In part (b)(i) not many candidates seemed to understand what was required. A significant number of candidates wrote down the equation of the line through $\mathrm{PC}$, $y=6$, rather than the required line. In part (b)(ii), it seemed that much time was lost as many candidates attempted to find the equation of the perpendicular bisector of $\mathrm{DP}$ to draw the boundaries.

The front view of a doghouse is made up of a square with an isosceles triangle on top.

The doghouse is $1.35 \text{m}$ high and $0.9 \text{m}$ wide, and sits on a square base.

The top of the rectangular surfaces of the roof of the doghouse are to be painted.

Find the area to be painted.

height of triangle at roof $=1.35-0.9=0.45$          (A1)

Note: Award A1 for $0.45$ (height of triangle) seen on the diagram.

$\text{slant height}=\sqrt{0.{45}^2+0.{45}^2}$  OR  $\sin \left(45°\right)=\frac{0.45}{\text{slant height}}$           (M1)

$=\sqrt{0.405} \left(0.636396\dots , 0.45\sqrt{2}\right)$          A1

Note: If using $\sin \left(45°\right)=\frac{0.45}{\text{slant height}}$ then (A1) for angle of $45°$, (M1) for a correct trig statement.

area of one rectangle on roof $=\sqrt{0.405}\times 0.9 \left(=0.572756\dots \right)$           M1

area painted $=\left(2\times \sqrt{0.405}\times 0.9 =2\times 0.572756\dots \right)$

$1.15 {\text{m}}^2 \left(1.14551\dots {\text{m}}^2, 0.81\sqrt{2} {\text{m}}^2\right)$          A1

[5 marks]

Although the first question on the paper, with appropriate low-level mathematics, the interpretation required seems to have been quite high, and many candidates found this challenging. Many candidates scored only the one mark for the height of the triangle. The most common wrong method seen was calculation of the area of the triangle and adding their result to the calculation 2 × 0.9 × 0.9. Stronger candidates lost the final mark either through premature rounding or incorrect units; both are aspects that can and do occur throughout candidate responses and hence clearly require focus in the classroom.

Determine the angle of depression from Camera $1$ to the centre of the cash register. Give your answer in degrees.

Calculate the distance from Camera $2$ to the centre of the cash register.

Without further calculation, determine which camera has the largest angle of depression to the centre of the cash register. Justify your response.

$\sin \theta =\frac{2.1}{2.8}$   OR   $\tan \theta =\frac{2.1}{1.85202\dots }$           (M1)

$\left(\theta =\right) 48.6° \left(48.5903\dots °\right)$           A1

[2 marks]

METHOD 1

$\sqrt{2.8^2-2.1^2}$   OR   $2.8 \cos \left(48.5903\dots \right)$   OR   $\frac{2.1}{\tan \left(48.5903\dots \right)}$           (M1)

Note: Award M1 for attempt to use Pythagorean Theorem with $2.1$ seen or for attempt to use cosine or tangent ratio.

$1.85 \left(\text{m}\right) \left(1.85202\dots \right)$           (A1)

Note: Award the M1A1 if $1.85$ is seen in part (a).

$\left(6.4-1.85202\dots \right)$

$4.55 \mathrm{m} \left(4.54797\dots \right)$           (A1)

Note: Award A1 for $4.55$ or equivalent seen, either as a separate calculation or in Pythagorean Theorem.

$\sqrt{{\left(4.54797\dots \right)}^2+2.1^2}$

$5.01 \text{m} \left(5.00939\dots \text{m}\right)$           A1

METHOD 2

attempt to use cosine rule           (M1)

$\left(c^2=\right) 2.8^2+6.4^2-2\left(2.8\right)\left(6.4\right) \cos \left(48.5903\dots \right)$           (A1)(A1)

Note: Award A1 for $48.5903...°$ substituted into cosine rule formula, A1 for correct substitution.

$\left(c=\right) 5.01 \text{m} \left(5.00939\dots \text{m}\right)$           A1

[4 marks]

camera $1$ is closer to the cash register (than camera $2$ and both cameras are at the same height on the wall)           R1

the larger angle of depression is from camera $1$           A1

Note: Do not award R0A1. Award R0A0 if additional calculations are completed and used in their justification, as per the question. Accept “$1.85<4.55$” or “$2.8<5.01$” as evidence for the R1.

[2 marks]

Many candidates calculated the angle from vertical rather than the angle of depression.

Candidates could successfully use their vertical angle from (a) or other correct trigonometry, such as Pythagorean theorem or cosine rule, to find the distance from camera 2 to the cash register. This question is a good example of how premature rounding can affect a final answer, and some had an inaccurate final answer because they had rounded intermediate values.

Many provided reasonable justification for their response, even though they often followed correct reasoning with an incorrect conclusion about the larger angle of depression.

Find the distance between stations $\text{A}$ and $\text{B}$.

Find the coordinates of station $\text{M}$.

Write down the height of station $\text{M}$, in metres, above the ground.

attempt at substitution into 3D distance formula              (M1)

$\text{AB}=\sqrt{{\left(140-20\right)}^2+{\left(15-5\right)}^2+{250}^2} \left(=\sqrt{77 000}\right)$

$=277 \text{m} \left(10\sqrt{770}, 277.488\dots \right)$                   A1

[2 marks]

attempt at substitution in the midpoint formula            (M1)

$\left(\frac{140+20}{2}, \frac{15+5}{2}, \frac{0+250}{2}\right)$

$\left(80, 10, 125\right)$                   A1

[2 marks]

$125 \text{m}$                   A1

[1 mark]

Calculate the gradient of the line segment AE.

Find the equation of the line which would complete the Voronoi cell containing site E.

Give your answer in the form $ax+by+d=0$ where $a$, $b$, $d\in \mathbb{Z}$.

In the context of the question, explain the significance of the Voronoi cell containing site E.

$\frac{3-1}{7-3}$       (M1)

= 0.5        A1

[2 marks]

$y-2=-2\left(x-5\right)$          (A1) (M1)

Note: Award (A1) for their −2 seen, award (M1) for the correct substitution of (5, 2) and their normal gradient in equation of a line.

$2x+y-12=0$      A1

[3 marks]

every point in the cell is closer to E than any other snow shelter     A1

[1 mark]

A footpath is to be laid around the curved side of the lawn. Find the length of the footpath.

Find the area of the lawn.

$135°\times \frac{12\mathrm{\pi }}{360°}$                 (M1)(A1)

$14.1 \left(\text{m}\right) \left(14.1371\dots \right)$                 A1

[3 marks]

evidence of splitting region into two areas                 (M1)

$135°\times \frac{\mathrm{\pi }{6}^2}{360°}-\frac{6\times 6\times \sin 135°}{2}$                 (M1)(M1)

Note: Award M1 for correctly substituting into area of sector formula, M1 for evidence of substituting into area of triangle formula.

$42.4115\dots -12.7279\dots$

$29.7 {\text{m}}^2 \left(29.6835\dots \right)$                A1

[4 marks]

Find CÂB.

Point B on the ground is 5 m from point E at the entrance to Ollie’s house. He is 1.8 m tall and is standing at point D, below the sensor. He walks towards point B.

Find the distance Ollie is from the entrance to his house when he first activates the sensor.

$\frac{\text{sin}\text{C}\stackrel{\wedge }{\text{A}}\text{B}}{6}=\frac{\text{sin}{15}^{\circ }}{4.5}$        (M1)(A1)

CÂB = 20.2º (20.187415…)    A1

Note: Award (M1) for substituted sine rule formula and award (A1) for correct substitutions.

[3 marks]

$\text{C}\stackrel{\wedge }{\text{B}}\text{D}=20.2+15={35.2}^{\circ }$       A1

(let X be the point on BD where Ollie activates the sensor)

$\text{tan}35.18741{\dots }^{\circ }=\frac{1.8}{\text{BX}}$       (M1)

Note: Award A1 for their correct angle $\text{C}\stackrel{\wedge }{\text{B}}\text{D}$. Award M1 for correctly substituted trigonometric formula.

$\text{BX}=2.55285\dots$       A1

$5-2.55285\dots$       (M1)

= 2.45 (m) (2.44714…)       A1     

[5 marks]

The diagram below is part of a Voronoi diagram.

Diagram not to scale

A and B are sites with B having the co-ordinates of (4, 6). L is an edge; the equation of this perpendicular bisector of the line segment from A to B is $y=-2x+9$

Find the co-ordinates of the point A.

Line from A to B will have the form $y=\frac{1}{2}x+c$      M1A1

Through $\left(4\text{,}6\right)\Rightarrow c=4$   so line is $y=\frac{1}{2}x+4$      M1A1

Intersection of $y=-2x+9$ and $y=\frac{1}{2}x+4$ is (2, 5)      M1A1

Let $A=\left(p\text{,}q\right)$ then $\left(2\text{,}5\right)=\left(\frac{p+4}{2},\frac{q+6}{2}\right)\Rightarrow p=0\text{,}q=4$      M1A1A1

$A=\left(0,4\right)$ 

[9 marks]

Calculate the value of $\theta$, the measure of angle $\text{MŜB}$.

Find the area of the Bermuda Triangle.

attempt at substituting the cosine rule formula           (M1)

$\cos \theta =\frac{{1660}^2+{1550}^2-{1670}^2}{2\left(1660\right)\left(1550\right)}$           (A1)

$\left(\theta =\right) 62.6° \left(62.5873\dots \right)$  (accept $1.09$ rad $\left(1.09235\dots \right)$)                       A1

[3 marks]

correctly substituted area of triangle formula         (M1)

$A=\frac{1}{2}\left(1660\right)\left(1550\right)\sin \left(62.5873\dots \right)$

$\left(A=\right) 1140 000 \left(1.14\times {10}^6, 1142 043.327\dots \right) {\text{km}}^2$                      A1

Note: Accept $1150 000 \left(1.15\times {10}^6, 1146 279.893\dots \right) {\text{km}}^2$ from use of $63°$. Other angles and their corresponding sides may be used.

[2 marks]

Most candidates were successful at selecting the cosine rule formula in part (a). In most cases, the cosine rule formula was correctly substituted. Some candidates found it hard to choose the correct side to obtain the required angle. Most of the candidates scored one or two marks out of three in this part. In part (b) some candidates assumed the triangle was right angled and used $\frac{1}{2}bh$ instead of $\frac{1}{2}ab \sin C$. In part (b) many candidates who answered part (a) incorrectly were able to recover. Many candidates managed to score full marks in this part despite an incorrect answer in part (a). Some found angle $\mathrm{BMS}$ or $\mathrm{SBM}$ in part (a) but used the correct sides to obtain the correct area. Final answers given in calculator notations (such as $1.14\mathrm{E}10$) scored at most one mark out of two. Calculator notation should generally be avoided; it is considered too informal to earn A marks, and although it can imply a method and earn M marks, we advise that candidates still provide the necessary commentary to support any GDC notation.

Most candidates were successful at selecting the cosine rule formula in part (a). In most cases, the cosine rule formula was correctly substituted. Some candidates found it hard to choose the correct side to obtain the required angle. Most of the candidates scored one or two marks out of three in this part. In part (b) some candidates assumed the triangle was right angled and used $\frac{1}{2}bh$ instead of $\frac{1}{2}ab \sin C$. In part (b) many candidates who answered part (a) incorrectly were able to recover. Many candidates managed to score full marks in this part despite an incorrect answer in part (a). Some found angle $\mathrm{BMS}$ or $\mathrm{SBM}$ in part (a) but used the correct sides to obtain the correct area. Final answers given in calculator notations (such as $1.14\mathrm{E}10$) scored at most one mark out of two. Calculator notation should generally be avoided; it is considered too informal to earn A marks, and although it can imply a method and earn M marks, we advise that candidates still provide the necessary commentary to support any GDC notation.

Find a vector equation for L₁.

The vector $\left(\begin{array}{c}2\\ p\\ 0\end{array}\right)$ is perpendicular to $\overrightarrow{\text{AB}}$. Find the value of p.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

any correct equation in the form r = a + tb (accept any parameter for t)

where a is $\left(\begin{array}{c}2\\ 1\\ 3\end{array}\right)$, and b is a scalar multiple of $\left(\begin{array}{c}1\\ 3\\ 1\end{array}\right)$     A2 N2

eg r = $\left(\begin{array}{c}2\\ 1\\ 3\end{array}\right)=t\left(\begin{array}{c}1\\ 3\\ 1\end{array}\right)$, r = 2i + j + 3k + s(i + 3j + k)

Note: Award A1 for the form a + tb, A1 for the form L = a + tb, A0 for the form r = b + ta.

[2 marks]

METHOD 1

correct scalar product     (A1)

eg  (1 × 2) + (3 × p) + (1 × 0), 2 + 3p

evidence of equating their scalar product to zero     (M1)

eg  a•b = 0, 2 + 3p = 0, 3p = −2

$p=-\frac{2}{3}$       A1 N3

METHOD 2

valid attempt to find angle between vectors      (M1)

correct substitution into numerator and/or angle       (A1)

eg  $\text{cos}\theta =\frac{\left(1\times 2\right)+\left(3\times p\right)+\left(1\times 0\right)}{\left|a\right|\left|b\right|},\text{cos}\theta =0$

$p=-\frac{2}{3}$       A1 N3

[3 marks]

Find a vector equation of the line that passes through P and Q.

The line through P and Q is perpendicular to the vector 2i $+$ nk. Find the value of $n$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid attempt to find direction vector     (M1)

eg$$$\overrightarrow{\text{PQ}},\overrightarrow{\text{QP}}$

correct direction vector (or multiple of)     (A1)

eg$$6i $+$ j $-$ 3k

any correct equation in the form r $=$ a $+$ tb (any parameter for $t$)     A2     N3

where a is i $+$ 2j $-$ k or 7i $+$ 3j $-$ 4k , and b is a scalar multiple of 6i $+$ j $-$ 3k

eg$$r $=$ 7i $+$ 3j $-$ 4k $+$ t(6i $+$ j $-$ 3k), r $=\left(\begin{array}{c}1+6s\\ 2+1s\\ -1-3s\end{array}\right),r=\left(\begin{array}{c}1\\ 2\\ -1\end{array}\right)+t\left(\begin{array}{c}-6\\ -1\\ 3\end{array}\right)$

Notes: Award A1 for the form a $+$ tb, A1 for the form L $=$ a $+$ tb, A0 for the form r $=$ b $+$ ta.

[4 marks]

correct expression for scalar product     (A1)

eg$$$6\times 2+1\times 0+(-3)\times n,-3n+12$

setting scalar product equal to zero (seen anywhere)     (M1)

eg$$u $\bullet$ v $=0,-3n+12=0$

$n=4$    A1     N2

[3 marks]

Find the volume of the money box.

A second money box is in the shape of a sphere and has the same volume as the cylindrical money box.

Find the diameter of the second money box.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(V =)  $\pi {\left(4.43\right)}^2\times 12.2$     (M1)(A1)

Note: Award (M1) for substitution into volume of a cylinder formula, (A1) for correct substitution.

752 cm³  (752.171…cm³)     (A1)(C3)

[3 marks]

$752.171\dots =\frac{4}{3}\pi {\left(r\right)}^3$       (M1)

Note: Award (M1) for equating their volume to the volume of a sphere formula.

$\left(r=\right)$ 5.64169…cm      (A1)(ft)

Note: Follow through from part (a).

$\left(d=\right)$ 11.3 cm  (11.2833…cm)      (A1)(ft)   (C3)

[3 marks]

Calculate the radius of the base of the cone which has been removed.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\sqrt{{15}^2-{12}^2}$     (M1)

Note: Award (M1) for correct substitution into Pythagoras theorem.

OR

$\frac{\text{radius}}{21}=\frac{15}{35}$     (M1)

Note: Award (M1) for a correct equation.

= 9 (cm)     (A1) (C2)

[2 marks]

Find the length of EB.

Write down the angle of elevation of B from E.

Find the vertical height of B above the ground.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

Units are required in parts (a) and (c).

$\frac{\text{EB}}{\sin 53{}^{\circ }}=\frac{1.2}{\sin 7{}^{\circ }}$     (M1)(A1)

Note:     Award (M1) for substitution into sine formula, (A1) for correct substitution.

$(\text{EB}=)7.86\text{ m}$$$OR$$$786\text{ cm }(7.86385\dots \text{ m}$$$OR$$$786.385\dots \text{ cm})$     (A1)     (C3)

[3 marks]

34°     (A1)     (C1)

[1 mark]

Units are required in parts (a) and (c).

$\sin {34}^{\circ }=\frac{\text{height}}{7.86385\dots }$     (M1)

Note:     Award (M1) for correct substitution into a trigonometric ratio.

$(\text{height}=)4.40\text{ m}$$$OR$$$440\text{ cm }(4.39741\dots \text{ m}$$$OR$$$439.741\dots \text{ cm})$     (A1)(ft)     (C2)

Note:     Accept “BT” used for height. Follow through from parts (a) and (b). Use of 7.86 gives an answer of 4.39525….

[2 marks]

Find the length of the rope connecting $\text{A}$ to $\text{F}$.

Find $\text{FÂO}$, the angle the rope makes with the ground.

$\sqrt{3.2^2+4.5^2+5.8^2}$           (M1)

$=8.01 \left(8.00812\dots \right) \text{m}$          A1

[2 marks]

$\text{FÂO}={\sin }^{-1}\left(\frac{5.8}{8.00812\dots }\right)$  OR  ${\cos }^{-1}\left(\frac{5.52177\dots }{8.00812\dots }\right)$  OR  ${\tan }^{-1}\left(\frac{5.8}{5.52177\dots }\right)$           (M1)

$46.4° \left(46.4077\dots °\right)$          A1

[2 marks]

This question was the first of its type to be tested and the problem was reduced (erroneously) to 2D trigonometry. Whilst a minority of candidates did tackle this part correctly, many simply arrived at an incorrect length with $\sqrt{{(5.8)}^2+{(4.5)}^2}=7.34$ metres.

The incorrect interpretation of the diagram as being 2D, meant that there was an assumption that OA was of length 3.2 metres and so ${\tan }^{-1}\left(\frac{5.8}{3.2}\right)=61.1$° was seen on many scripts. If, using their incorrect answer for part (a), the candidate had used ${\sin }^{-1}\left(\frac{5.8}{\text{their part (a)}}\right)$ it was possible to award “follow through” marks.

Calculate the length of the arc made by $\text{B}$, the end of the rubber blade.

Determine the area of the windscreen that is cleared by the rubber blade.

attempt to substitute into length of arc formula         (M1)

$\frac{140°}{360°}\times 2\mathrm{\pi }\times 56$

$137 \text{cm} \left(136.833\dots , \frac{392\mathrm{\pi }}{9} \text{cm}\right)$         A1

[2 marks]

subtracting two substituted area of sectors formulae        (M1)

$\left(\frac{140°}{360°}\times \mathrm{\pi }\times {56}^2\right)-\left(\frac{140°}{360°}\times \mathrm{\pi }\times {10}^2\right)$   OR   $\frac{140°}{360°}\times \mathrm{\pi }\times \left({56}^2-{10}^2\right)$        (A1)

$3710 {\text{cm}}^2 \left(3709.17\dots {\text{cm}}^2\right)$         A1

[3 marks]

There was some difficulty determining the correct radius to substitute, with several candidates substituting a radius of 46.

It was common to see candidates subtracting the radii before substituting into the area formula, rather than subtracting the sector areas after calculating each. Using the π key on the calculator rather than an approximated value was prevalent and pleasing to see.

Write down an equation for the area, $A$, of the curved surface in terms of $r$.

Find $\frac{\text{d}A}{\text{d}r}$.

Find the value of $r$ when the area of the curved surface is maximized.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$A=2\pi r\left(12-r\right)$  OR  $A=24\pi r-2\pi r^2$        (A1)(M1)  (C2)

Note: Award (A1) for  $r+h=12$  or  $h=12-r$  seen. Award (M1) for correctly substituting into curved surface area of a cylinder. Accept $A=2\pi r\left(12-r\right)$  OR  $A=24\pi r-2\pi r^2$.

[2 marks]

$24\pi -4\pi r$       (A1)(ft)(A1)(ft)  (C2)

Note: Award (A1)(ft) for $24\pi$ and  (A1)(ft) for $-4\pi r$ . Follow through from part (a). Award at most (A1)(ft)(A0) if additional terms are seen.

[2 marks]

$24\pi -4\pi r=0$       (M1)

Note: Award (M1) for setting their part (b) equal to zero.

6 (cm)       (A1)(ft)  (C2)

Note: Follow through from part (b).

[2 marks]

Calculate the volume of the balloon.

Calculate the radius of the balloon following this increase.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

Units are required in parts (a) and (b).

$\frac{4}{3}\pi \times 6^3$    (M1)

Note:     Award (M1) for correct substitution into volume of sphere formula.

$=905\text{ c}{\text{m}}^3(288\pi \text{ c}{\text{m}}^3,904.778\dots \text{ c}{\text{m}}^3)$    (A1)     (C2)

Note:     Answers derived from the use of approximations of $\pi$ (3.14; 22/7) are awarded (A0).

[2 marks]

Units are required in parts (a) and (b).

$\frac{140}{100}\times 904.778\dots =\frac{4}{3}\pi r^3$ OR $\frac{140}{100}\times 288\pi =\frac{4}{3}\pi r^3$ OR $1266.69\dots =\frac{4}{3}\pi r^3$     (M1)(M1)

Note:     Award (M1) for multiplying their part (a) by 1.4 or equivalent, (M1) for equating to the volume of a sphere formula.

$r^3=\frac{3\times 1266.69\dots }{4\pi }$ OR $r=\sqrt[3]{\frac{3\times 1266.69\dots }{4\pi }}$ OR $r=\sqrt[3]{(1.4)\times 6^3}$ OR $r^3=302.4$     (M1)

Note:     Award (M1) for isolating $r$.

$(r=)6.71\text{ cm }(6.71213\dots )$     (A1)(ft)     (C4)

Note:     Follow through from part (a).

[4 marks]

Find the value of $\text{tan}\theta$.

Line $L$ passes through the origin and has a gradient of $\text{tan}\theta$. Find the equation of $L$.

Find the derivative of $f$.

The following diagram shows the graph of $f$  for 0 ≤ $x$ ≤ 3. Line $M$ is a tangent to the graph of $f$ at point P.

Given that $M$ is parallel to $L$, find the $x$-coordinate of P.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of valid approach       (M1)

eg   sketch of triangle with sides 3 and 5, $\text{co}{\text{s}}^2\theta =1-\text{si}{\text{n}}^2\theta$

correct working       (A1)

eg  missing side is 4 (may be seen in sketch), $\text{cos}\theta =\frac{4}{5}$,  $\text{cos}\theta =-\frac{4}{5}$

$\text{tan}\theta =-\frac{3}{4}$       A2 N4

[4 marks]

correct substitution of either gradient or origin into equation of line        (A1)

(do not accept $y=mx+b$)

eg   $y=x\text{tan}\theta$,   $y-0=m\left(x-0\right)$,   $y=mx$

$y=-\frac{3}{4}x$     A2 N4

Note: Award A1A0 for $L=-\frac{3}{4}x$.

[2 marks]

$\frac{\text{d}}{\text{d}x}\left(\frac{-3x}{4}\right)=-\frac{3}{4}$  (seen anywhere, including answer)       A1

choosing product rule       (M1)

eg   $u{v}^{\prime }+v{u}^{\prime }$