# Question 10:

## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a$ represents the proportion of juvenile chimpanzees that survive and **remain** juvenile chimpanzees the next year | B1 | Must be clear it is the proportion of juveniles that remain as juveniles; do not accept "surviving juveniles"; accept "do not become adults" |

## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Determinant $= 0.82a - 0.08 \times 0.15$ | M1 | Attempts determinant in terms of $a$; allow $0.82a - 0.12$ as a slip |
| $\begin{pmatrix} a & 0.15 \\ 0.08 & 0.82 \end{pmatrix}^{-1} = \cdots\begin{pmatrix} 0.82 & -0.15 \\ -0.08 & a \end{pmatrix}$ | M1 | Attempts form of inverse, swapped leading diagonals and sign changed on off-diagonals; signs must be correct |
| $\begin{pmatrix} a & 0.15 \\ 0.08 & 0.82 \end{pmatrix}^{-1} = \dfrac{1}{0.82a-0.012}\begin{pmatrix} 0.82 & -0.15 \\ -0.08 & a \end{pmatrix}$ | A1 | Correct inverse matrix |

## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} a & 0.15 \\ 0.08 & 0.82 \end{pmatrix}^{-1}\begin{pmatrix} 15360 \\ 43008 \end{pmatrix} = \dfrac{1}{0.82a-0.012}\begin{pmatrix} 0.82\times15360 - 0.15\times43008 \\ (-0.08)\times15360 + 43008a \end{pmatrix}$ | M1 | Uses inverse matrix to find initial juvenile and adult populations; may have determinant 1; alternatively sets up simultaneous equations $15360 = aJ_0 + 0.15\times A_0$ and $43008 = 0.08\times J_0 + 0.82\times A_0$ |
| $\dfrac{1}{0.82a-0.012}\left[6144 + (43008a - 1228.8)\right] = 64000$ leading to $a = \ldots$ | M1 | Uses sum of initial populations equals 64000 to find $a$; if using alternative, uses $A_0 = 64000 - J_0$ |
| $a = \dfrac{5683.2}{9472} = 0.60$ | A1 | Correct value; accept $0.60$ or $\dfrac{3}{5}$ |

## Part (b)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Initial juvenile population $= \dfrac{\text{"6144"}}{\text{"0.48"}} = 12800$ | M1 | Uses their $a$ to find $J_0$; mark may be gained from work in (ii) if alternative used |
| Change of 2560 juvenile chimpanzees | A1 | Correct difference found; no contradictory statement; "decrease of 2560" is A0 |

## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| As the number of juveniles has increased, the model is not initially predicting a decline, so is not suitable in the short term | B1ft | Follow through on answer to (b); must have attempted to find at least a value for $J_0$; if decrease found, allow comment that model is suitable |

## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Introduces a third category for chimpanzees aged 40 and above (Mature $M_n$) and sets up a matrix multiplication of increased dimension | M1 | Dimension of matrix should be 3 in at least either row or column; $3\times1$ column vector required; accept $3\times3$, $3\times2$, or $2\times3$ matrices with all three categories |
| $\begin{pmatrix} J_{n+1} \\ A_{n+1} \\ M_{n+1} \end{pmatrix} = \begin{pmatrix} a & b & \underline{0} \\ 0.08 & c & 0 \\ 0 & d & e \end{pmatrix}\begin{pmatrix} J_n \\ A_n \\ M_n \end{pmatrix}$ | A1 | Underlined zero must be correct (no new juveniles from mature); overlook other values; ideally other zeros shown to indicate mature chimpanzees do not regress and juveniles cannot proceed directly to mature |

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# Appendix: Alternatives to Question 8(b):

## Alt 1 — Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| As per main scheme | B1, M1 | |
| $d^2 = (-400+600\lambda)^2 + (200-100\lambda)^2 + (-250+100\lambda)^2$ $= 380000\lambda^2 - 570000\lambda + 262500$ $= 380000\!\left(\lambda - \dfrac{3}{4}\right)^2 + 48750 \Rightarrow \lambda = \ldots$ | dM1 | Attempts distance or distance squared of $\overrightarrow{MX}$, expands and completes the square to find value of $\lambda$ for minimum distance; look for $A(B\lambda-C)^2 - D + \text{"262500"}$ where $A,B,C,D \neq 0$ |
| As per main scheme | M1, A1 | |

## Alt 1 — Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Length of tunnel is $\sqrt{\text{"48750"}} = \ldots$ | M1 | |
| Awrt 221m from correct working; completion of square must have been correct | A1 | Must include units |

## Alt 2 — Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| As per main scheme | B1, M1 | |
| $d^2 = (-400+600\lambda)^2+(200-100\lambda)^2+(-250+100\lambda)^2$ $= 380000\lambda^2 - 570000\lambda + 262500$ $\dfrac{d}{dx}(d^2)=0 \Rightarrow 760000\lambda - 570000 = 0 \Rightarrow \lambda = \ldots$ | dM1 | Differentiates and sets to zero to find $\lambda$ |
| As per main scheme | M1, A1 | |

## Alt 2 — Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Length of tunnel is $\sqrt{(150-100)^2+(325-200)^2+(-75-100)^2} = \ldots$ | M1 | |
| Awrt 221m from correct working; differentiation must have been correct | A1 | Must include units |

## Alt 3 — Part (b)(i) ($k=200$):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k=200$ | B1 | |
| $\overrightarrow{MP} = \begin{pmatrix}-400\\200\\-250\end{pmatrix} \Rightarrow \cos\theta = \dfrac{\begin{pmatrix}-400\\200\\-250\end{pmatrix}\cdot\begin{pmatrix}600\\-100\\100\end{pmatrix}}{\sqrt{(-400)^2+200^2+(-250)^2}\,\sqrt{600^2+(-100)^2+100^2}}$ | M1 | Finds $\overrightarrow{MP}$ (or $\overrightarrow{MQ}$) and attempts scalar product formula with direction of line to find angle |
| $\Rightarrow \|\overrightarrow{PX}\| = \|\overrightarrow{MP}\|\cos\theta = \ldots$ | dM1 | Uses angle with cosine to find length of $\overrightarrow{PX}$; accept equivalent trigonometric methods |
| $\overrightarrow{OX} = \begin{pmatrix}-300\\400\\-150\end{pmatrix} + \dfrac{\|\overrightarrow{PX}\|}{\left\|\begin{pmatrix}600\\-100\\100\end{pmatrix}\right\|}\begin{pmatrix}600\\-100\\100\end{pmatrix} = \ldots$ | M1 | Uses length of $\overrightarrow{PX}$ to find coordinates of point on line at shortest distance from $M$ |
| Coordinates of $X$ are $(150, 325, -75)$ | A1 | |

## Alt 3 — Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Length of tunnel is $\|\overrightarrow{MP}\|\sin\theta = \ldots$ | M1 | Correct method; may use sine ratio with angle between line and $\overrightarrow{MP}$ |
| Awrt 221m from correct working | A1 | Must include units; accept awrt 221m or $25\sqrt{78}$ m |

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