% =============================================================
% Y11 EXTENDED — END OF YEAR ASSESSMENT (June 2026)
% PAPER 1: NON CALCULATOR (70 minutes, 60 marks)
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  \lhead{\textbf{MYP 11E} \;\textbar\; EOY Paper 2 --- Calculator}%
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\setlist[enumerate]{topsep=4pt, itemsep=4pt, parsep=0pt, leftmargin=1.6em}

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\begin{document}
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\noindent
\begin{tabular}{@{}p{8.2cm}p{8.2cm}@{}}
\textbf{Name:} \rule{5.5cm}{0.4pt} & \textbf{MYP Criteria:}\quad A\;\;\rule{1cm}{0.4pt}/8\quad C\;\;\rule{1cm}{0.4pt}/8 \\
\textbf{Date:} \rule{5.5cm}{0.4pt} & \textbf{Time:} 70 minutes \;\textbar\; \textbf{Total: 63 marks} \\
\bottomrule
\end{tabular}
\renewcommand{\arraystretch}{1}

\vspace{6pt}

\begin{infobox}
\textbf{\color{mypdark}Instructions for candidates}
\begin{itemize}[nosep, leftmargin=1.4em, topsep=3pt, itemsep=2pt]
  \item Do not open this examination paper until instructed to do so.
  \item A graphic display calculator \textbf{is required} for this paper. Ensure it is in the correct angle units (degrees / radians) for each question.
  \item All questions are to be answered in the space provided on this paper.
  \item Unless otherwise stated, all numerical answers should be given \textbf{exactly} or correct to 3 significant figures.
  \item The maximum mark for this examination paper is \textbf{[63 marks]}.
  \item Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations.
\end{itemize}
\end{infobox}

\ifsol\else
\vspace{4pt}
\noindent{\small\textit{\textcolor{markgray}{\textbf{A note on levels.}\; Each question is labelled \textbf{[Level X--Y]}. Questions marked \textbf{[Level 7--8]} are \emph{unfamiliar contexts}: they combine multiple Y11 topics with Y10 carryover skills (coordinate geometry, surds, indices, 3D thinking, perpendicular bisectors), require you to choose your own approach without method hints, and may ask you to define your own variables or recognise mathematical structure. Try the question even if the path isn't immediately obvious --- set up what you know and look for connections.}}}
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\ifsol
\vspace{6pt}
\begin{notebox}
\textbf{\color{mypamber}Mark scheme notes.}\; Award method marks (\textbf{M}) for correct strategy; accuracy marks (\textbf{A}) for correct answer; follow-through (\textbf{ft}) where an earlier error has been carried forward correctly. Equivalent forms accepted unless otherwise stated.
\end{notebox}
\fi

% =============================================================
\question{1}{3}{1--2}
Evaluate using your calculator. Give answers correct to 3 significant figures unless told otherwise.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item $\sin 124\degree$. \hfill\partmarks{1}
    \workspace{1.7cm}
  \item $(1.045)^{10}$. \hfill\partmarks{1}
    \workspace{1.7cm}
  \item An event $A$ has $P(A) = 0.35$. State the value of $P(A')$. \hfill\partmarks{1}
    \workspace{1.7cm}
\end{enumerate}
\begin{solution}
(a) $\sin 124\degree \approx \mathbf{0.829}$. \,\textit{[A1]}\\
(b) $(1.045)^{10} \approx \mathbf{1.55}$. \,\textit{[A1]}\\
(c) $P(A') = 1 - 0.35 = \mathbf{0.65}$. \,\textit{[A1]}
\end{solution}

% =============================================================
\question{2}{6}{3--4}
A factory has two machines. Machine $A$ produces $60\%$ of all items, of which $2\%$ are defective. Machine $B$ produces the remaining $40\%$, of which $5\%$ are defective. An item is selected at random from the day's output.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Draw a tree diagram showing the four outcomes. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find the probability that the item is defective. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Given the item is defective, find the probability it came from machine $B$. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) Tree: $A$ ($0.6$) $\to$ D ($0.02$), D$'$ ($0.98$); $B$ ($0.4$) $\to$ D ($0.05$), D$'$ ($0.95$). \,\textit{[A1, A1]}\\
(b) $P(D) = (0.6)(0.02) + (0.4)(0.05) = 0.012 + 0.020 = \mathbf{0.032}$. \,\textit{[M1, A1]}\\
(c) $P(B|D) = \dfrac{0.020}{0.032} = \mathbf{\dfrac{5}{8} = 0.625}$. \,\textit{[M1, A1]}
\end{solution}

% =============================================================
\question{3}{6}{3--4}
In a survey of $80$ people: $45$ like coffee, $38$ like tea, $28$ like juice. $22$ like coffee and tea, $15$ like tea and juice, $12$ like coffee and juice, and $8$ like all three.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Find the number of people who like at least one drink. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find the number who like exactly one drink. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item One person is selected at random. Find the probability they like \emph{none} of the three drinks. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) Inclusion--exclusion: $|C\cup T\cup J| = 45+38+28 - 22 - 15 - 12 + 8 = \mathbf{70}$. \,\textit{[M1, A1]}\\
(b) Pairs only (subtracting all-three): $14, 7, 4$. Singles: $19, 9, 9$. Total exactly one $= \mathbf{37}$. \,\textit{[M1, A1]}\\
(c) None: $80 - 70 = 10$, so $P = \mathbf{1/8 = 0.125}$. \,\textit{[M1, A1]}
\end{solution}

\newpage

% =============================================================
\question{4}{6}{5--6}
Triangle $ABC$ has $AB = 14$ m, $AC = 11$ m and $\angle BAC = 68\degree$.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Find the area of the triangle. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find the length $BC$. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find the angle $\angle ABC$. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) Area $= \tfrac{1}{2}(14)(11)\sin 68\degree \approx \mathbf{71.4}$ m\textsuperscript{2}. \,\textit{[M1, A1]}\\
(b) Cosine rule: $BC^2 = 196 + 121 - 2(14)(11)\cos 68\degree \approx 201.6$, $BC \approx \mathbf{14.2}$ m. \,\textit{[M1, A1]}\\
(c) Sine rule: $\dfrac{\sin B}{11} = \dfrac{\sin 68\degree}{14.2} \Rightarrow \sin B \approx 0.7177 \Rightarrow B \approx \mathbf{45.8\degree}$. \,\textit{[M1, A1]}
\end{solution}

% =============================================================
\question{5}{6}{5--6}
A circular pizza of radius $14$ cm is cut into a sector with central angle $\theta = \dfrac{5\pi}{12}$ radians.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Find the length of the curved edge of the sector. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find the area of the sector (the slice). \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find the perimeter of the slice. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) Arc $s = r\theta = 14 \cdot \dfrac{5\pi}{12} = \dfrac{35\pi}{6} \approx \mathbf{18.3}$ cm. \,\textit{[M1, A1]}\\
(b) Area $= \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(196)\dfrac{5\pi}{12} = \dfrac{245\pi}{6} \approx \mathbf{128}$ cm\textsuperscript{2}. \,\textit{[M1, A1]}\\
(c) Perimeter $= 2r + s = 28 + 18.3 \approx \mathbf{46.3}$ cm. \,\textit{[M1, A1]}
\end{solution}

% =============================================================
\question{6}{7}{5--6}
The graph of $y = f(x)$ has a minimum point at $(2, -3)$ and crosses the $y$-axis at $(0, 1)$.

State the coordinates of the corresponding point on each transformed graph:
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item $y = f(x) + 4$ (state the new minimum). \hfill\partmarks{1}
    \workspace{1.8cm}
  \item $y = f(x - 5)$ (state the new minimum). \hfill\partmarks{1}
    \workspace{1.8cm}
  \item $y = -f(x)$ (state the new $y$-intercept and whether the turning point becomes a max or min). \hfill\partmarks{2}
    \workspace{2.5cm}
  \item $y = f(2x)$ (state the new minimum, and the new $y$-intercept). \hfill\partmarks{3}
    \workspace{3.5cm}
\end{enumerate}
\begin{solution}
(a) Vertical translation $+4$: minimum at $\mathbf{(2, 1)}$. \,\textit{[A1]}\\
(b) Horizontal translation right by 5: minimum at $\mathbf{(7, -3)}$. \,\textit{[A1]}\\
(c) Reflection in $x$-axis: $y$-intercept at $\mathbf{(0, -1)}$;\; turning point becomes a \textbf{maximum} at $(2, 3)$. \,\textit{[A1, A1]}\\
(d) Horizontal stretch (factor $\tfrac{1}{2}$): minimum at $\mathbf{(1, -3)}$ \,\textit{[M1, A1]}; $y$-intercept at $f(0) = 1$, so still $\mathbf{(0, 1)}$. \,\textit{[A1]}
\end{solution}

\newpage

% =============================================================
\question{7}{7}{5--6}
A radioactive substance has initial mass $200$ grams. The mass decreases by $8\%$ each year.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Write a formula for the mass $M$ (in grams) after $t$ years. \hfill\partmarks{1}
    \workspace{1.8cm}
  \item Find the mass after $5$ years. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find the time, to the nearest year, when the mass first falls below $50$ g. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item By what percentage has the original mass decreased after $20$ years? \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) $\mathbf{M = 200(0.92)^t}$. \,\textit{[A1]}\\
(b) $M(5) = 200(0.92)^5 \approx \mathbf{132}$ g (3 sf). \,\textit{[M1, A1]}\\
(c) Solve $200(0.92)^t < 50$: $(0.92)^t < 0.25 \Rightarrow t > \dfrac{\ln 0.25}{\ln 0.92} \approx 16.6$. So $\mathbf{t = 17}$ years. \,\textit{[M1, A1]}\\
(d) $M(20) \approx 38.96$ g. Decrease $\approx 161$ g, i.e.\ $\mathbf{80.5\%}$ of original. \,\textit{[M1, A1]}
\end{solution}

% =============================================================
\question{8}{6}{5--7}
The function $f(x) = -2(x - 3)^2 + 5$ models the height (in metres) of a stunt rider above a ramp, where $x$ is the horizontal distance from the launch point.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item State the maximum height of the rider, and the horizontal distance at which it is reached. \hfill\partmarks{2}
    \workspace{2.5cm}
  \item Find the height of the rider when $x = 0$ (the launch point). \hfill\partmarks{1}
    \workspace{1.8cm}
  \item Find the values of $x$ at which the rider is at height $0$. Give your answers correct to 3 significant figures. \hfill\partmarks{3}
    \workspace{3.5cm}
\end{enumerate}
\begin{solution}
(a) Vertex form gives max height $\mathbf{5}$ m at $x = \mathbf{3}$ m. \,\textit{[A1, A1]}\\
(b) $f(0) = -2(9) + 5 = \mathbf{-13}$ m. \,\textit{[A1]}\\
(c) $-2(x-3)^2 + 5 = 0 \Rightarrow (x-3)^2 = 2.5 \Rightarrow \mathbf{x \approx 1.42}$ or $\mathbf{x \approx 4.58}$. \,\textit{[M1, A1, A1]}
\end{solution}

\newpage

% =============================================================
\question{9}{8}{7--8}
A vertical mast $PT$ stands on horizontal ground with foot $P$. From point $A$ on the ground, the mast is on a bearing of $060\degree$. From point $B$, which is $50$ m due east of $A$, the mast is on a bearing of $320\degree$. The angle of elevation of the top $T$ of the mast from $A$ is $35\degree$.
\begin{center}
\begin{tikzpicture}[scale=1.3]
  % Coordinates: P at geometrically correct position so bearings 060 from A and 320 from B both meet at P
  \coordinate (A) at (0,0);
  \coordinate (B) at (5,0);
  \coordinate (P) at (3.37,1.94);
  \coordinate (T) at (3.37,3.6);
  % North arrows (placed slightly outside vertices to avoid overlap)
  \draw[->, thin, gray] (A) ++(-0.2,0) -- ++(0,0.7) node[above, font=\scriptsize, gray] {N};
  \draw[->, thin, gray] (B) ++( 0.2,0) -- ++(0,0.7) node[above, font=\scriptsize, gray] {N};
  % Bearing arcs at A and B (small, interior to triangle)
  % At A: from N (math angle 90deg) clockwise to AP (math angle 30deg) = 60deg of arc, bearing 060
  \draw[thin] (A) ++(90:0.45) arc (90:30:0.45);
  \node[font=\tiny] at ($(A) + (60:0.62)$) {$060\degree$};
  % At B: from N (90deg) counterclockwise to BP (math angle 130deg) = 40deg of arc, bearing 320 (=360-40)
  \draw[thin] (B) ++(90:0.45) arc (90:130:0.45);
  \node[font=\tiny] at ($(B) + (110:0.62)$) {$320\degree$};
  % Ground triangle
  \draw[thick] (A) -- (B);
  \draw[thick] (A) -- (P);
  \draw[thick] (B) -- (P);
  % Vertical mast
  \draw[thick] (P) -- (T);
  % Vertex labels
  \node[below left=1pt] at (A) {$A$};
  \node[below right=1pt] at (B) {$B$};
  \node[right=3pt] at (P) {$P$};
  \node[above] at (T) {$T$};
  % Distance label below AB
  \node[below=3pt] at ($(A)!0.5!(B)$) {$50$ m};
\end{tikzpicture}
\end{center}
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Find the angles $\angle PAB$ and $\angle PBA$ in the ground triangle $PAB$. \hfill\partmarks{3}
    \workspace{3.5cm}
  \item Use the sine rule to find the distance $PA$. \hfill\partmarks{3}
    \workspace{3.5cm}
  \item Hence find the height of the mast. \hfill\partmarks{2}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) At $A$, $\angle PAB = 90\degree - 60\degree = \mathbf{30\degree}$. \,\textit{[M1, A1]}\\
At $B$, $\angle PBA = 320\degree - 270\degree = \mathbf{50\degree}$. \,\textit{[A1]}\\
(b) $\angle APB = 100\degree$. $PA = \dfrac{50\sin 50\degree}{\sin 100\degree} \approx \mathbf{38.9}$ m. \,\textit{[M1, M1, A1]}\\
(c) Height $= PA\tan 35\degree \approx \mathbf{27.2}$ m. \,\textit{[M1, A1]}
\end{solution}

% =============================================================
\question{10}{8}{7--8}
In a controlled experiment, the population $P$ (in thousands) of bacteria is observed at two times: at $t = 2$ hours, $P = 8$;\; at $t = 6$ hours, $P = 128$. The population is to be modelled by
\[ P(t) = a \cdot b^{\,t}, \]
where $a$ and $b$ are positive constants and $t$ is measured in hours.
\begin{enumerate}[label=(\alph*),leftmargin=1.6em, itemsep=2pt]
  \item Use the two observations to set up two equations in $a$ and $b$, and hence find the values of $a$ and $b$. \hfill\partmarks{4}
    \workspace{4.5cm}
  \item Hence state the initial population $P(0)$. \hfill\partmarks{1}
    \workspace{1.5cm}
  \item Sketch the graph of $P(t)$ for $0 \leq t \leq 6$, marking the $P$-intercept and the value of $P$ at $t = 6$. \hfill\partmarks{2}
    \workspace{6cm}
  \item Find, using the laws of logarithms, the time at which $P$ first reaches $1000$ (thousand). Give your answer correct to 3~s.f. \hfill\partmarks{1}
    \workspace{2.5cm}
\end{enumerate}
\begin{solution}
(a) Setting up: $a \cdot b^2 = 8$ \,(i),\quad $a \cdot b^6 = 128$ \,(ii). \,\textit{[M1, A1]}\\
Divide (ii) by (i): $b^4 = 16$, so $b = 2$. \,\textit{[M1, A1]}\\
From (i): $a \cdot 4 = 8$, so $a = 2$. \,\textit{[A1]}\\
\textit{(4 marks total: M1 setup, M1 division, A1 b, A1 a)}\\
(b) $P(0) = a = \mathbf{2}$ thousand. \,\textit{[A1]}\\
(c) Sketch through $(0, 2)$, increasing exponential; $P(6) = 2 \cdot 64 = \mathbf{128}$ thousand. \,\textit{[A1 shape with intercept, A1 endpoint]}\\
(d) $2 \cdot 2^t = 1000 \Rightarrow 2^t = 500 \Rightarrow t = \log_2 500 = \dfrac{\ln 500}{\ln 2} \approx \mathbf{8.97}$ hours. \,\textit{[A1]}
\end{solution}

\par\vspace*{\fill}
\begin{center}
\textit{End of Paper 2.}
\end{center}

\end{document}