| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Linear modelling problems |
| Difficulty | Moderate -0.8 This is a straightforward linear modelling question requiring basic substitution into inverse proportion (part a), simple interpretation of model limitations (part b), solving two simultaneous linear equations (part c), and basic interpretation of gradient/intercept (parts d-e). All techniques are routine AS-level with no problem-solving insight needed. |
| Spec | 1.02r Proportional relationships: and their graphs1.02z Models in context: use functions in modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m = \frac{k}{t}\), so \(t=5\), \(m=2.1 \Rightarrow k=10.5\) | M1 | 2.1: Using algebraic expression to represent proportionality and one pair of values in attempt to find \(k\); oe with \(t=50, m=0.21\) |
| When \(t=50\), \(m = \frac{10.5}{50} = 0.21\), oe | M1 | 2.1: Uses the model to predict value of \(m\) for the other value of \(t\), or uses the other pair to check \(k\) |
| Either: the model fits because prediction agrees with given value; OR: model fits because same value of \(k\) obtained in each case | A1 [3] | 2.2a: Makes suitable statement about consistency of results; e.g. \(m = \frac{10.5}{t}\) |
| Alternative: When \(t\) is multiplied by 10, \(m\) is divided by 10, so mass is inversely proportional to time | M2, A1 [3] | Argument in words need not reference constant of proportionality; must make clear conclusion about inverse proportionality |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t\) is small, \(m = \frac{10.5}{t}\) is large, so the mass will not be modelled correctly | B1 [1] | 3.5a: Any suitable comment that identifies a problem at \(t=0\) or as \(t \to 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Melts completely when \(m=0\), but \(t = \frac{10.5}{0}\) is not defined so the model cannot be used | B1 [1] | 3.5a: Any suitable comment explaining that the model will not give a time for \(m=0\); allow "the model does not give a time for which \(m=0\)" oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(t = 5, m = 2.1 \Rightarrow 2.1 = 5a + b\) | M1 [3.3] | Uses either data point to form equation |
| Substitute \(t = 50, m = 0.21 \Rightarrow 0.21 = 50a + b\) | M1 [1.1a] | Uses the other data point to find 2nd equation and solve simultaneously |
| Solving simultaneously: \(a = -0.042\), \(b = 2.31\) | A1 [1.1] | Both values required. Solution may be BC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of line joining points \((5, 2.1)\) and \((50, 0.21)\): \(\frac{2.1 - 0.21}{5 - 50} = -0.042\) | M1 | Attempt to find gradient |
| So line is of the form \(y - 2.1 = -0.042(x - 5)\), giving \(y = -0.042x + 2.31\) | M1, A1 [3] | Uses either data point to complete the equation. Award for the correct equation without reference to \(a\) and \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a\) is the rate at which ice melts: \(0.042\) kg of ice is lost per hour | B1 [3.3] | Must refer to the value found; FT their negative \(a\). If no reference is made to the numerical values, allow SC1 for '\(a\) is the rate at which ice melts and \(b\) is the initial mass of the block' oe |
| \(b\) is the initial mass of the block: \(2.31\) kg | B1 [3.3] | FT their positive \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m = 0 \Rightarrow 0 = -0.042t + 2.31 \Rightarrow t = \dfrac{2.31}{0.042} = 55\) so time for block to melt is \(55\) hours | B1 [3.4] | FT their \(a\) and \(b\) only if \(t\) is positive |
## Question 11(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = \frac{k}{t}$, so $t=5$, $m=2.1 \Rightarrow k=10.5$ | M1 | 2.1: Using algebraic expression to represent proportionality and one pair of values in attempt to find $k$; oe with $t=50, m=0.21$ |
| When $t=50$, $m = \frac{10.5}{50} = 0.21$, oe | M1 | 2.1: Uses the model to predict value of $m$ for the other value of $t$, or uses the other pair to check $k$ |
| Either: the model fits because prediction agrees with given value; OR: model fits because same value of $k$ obtained in each case | A1 [3] | 2.2a: Makes suitable statement about consistency of results; e.g. $m = \frac{10.5}{t}$ |
| **Alternative:** When $t$ is multiplied by 10, $m$ is divided by 10, so mass is inversely proportional to time | M2, A1 [3] | Argument in words need not reference constant of proportionality; must make clear conclusion about inverse proportionality |
---
## Question 11(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t$ is small, $m = \frac{10.5}{t}$ is large, so the mass will not be modelled correctly | B1 [1] | 3.5a: Any suitable comment that identifies a problem at $t=0$ or as $t \to 0$ |
---
## Question 11(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Melts completely when $m=0$, but $t = \frac{10.5}{0}$ is not defined so the model cannot be used | B1 [1] | 3.5a: Any suitable comment explaining that the model will not give a time for $m=0$; allow "the model does not give a time for which $m=0$" oe |
## Question 11:
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $t = 5, m = 2.1 \Rightarrow 2.1 = 5a + b$ | **M1** [3.3] | Uses either data point to form equation |
| Substitute $t = 50, m = 0.21 \Rightarrow 0.21 = 50a + b$ | **M1** [1.1a] | Uses the other data point to find 2nd equation and solve simultaneously |
| Solving simultaneously: $a = -0.042$, $b = 2.31$ | **A1** [1.1] | Both values required. Solution may be **BC** |
**Alternative method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of line joining points $(5, 2.1)$ and $(50, 0.21)$: $\frac{2.1 - 0.21}{5 - 50} = -0.042$ | **M1** | Attempt to find gradient |
| So line is of the form $y - 2.1 = -0.042(x - 5)$, giving $y = -0.042x + 2.31$ | **M1**, **A1** [3] | Uses either data point to complete the equation. Award for the correct equation without reference to $a$ and $b$ |
---
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a$ is the rate at which ice melts: $0.042$ kg of ice is lost per hour | **B1** [3.3] | Must refer to the value found; FT their negative $a$. If no reference is made to the numerical values, allow **SC1** for '$a$ is the rate at which ice melts and $b$ is the initial mass of the block' oe |
| $b$ is the initial mass of the block: $2.31$ kg | **B1** [3.3] | FT their positive $b$ |
---
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = 0 \Rightarrow 0 = -0.042t + 2.31 \Rightarrow t = \dfrac{2.31}{0.042} = 55$ so time for block to melt is $55$ hours | **B1** [3.4] | FT their $a$ and $b$ only if $t$ is positive |11 David puts a block of ice into a cool-box. He wishes to model the mass $m \mathrm {~kg}$ of the remaining block of ice at time $t$ hours later. He finds that when $t = 5 , m = 2.1$, and when $t = 50 , m = 0.21$.
\begin{enumerate}[label=(\alph*)]
\item David at first guesses that the mass may be inversely proportional to time. Show that this model fits his measurements.
\item Explain why this model
\begin{enumerate}[label=(\roman*)]
\item is not suitable for small values of $t$,
\item cannot be used to find the time for the block to melt completely.
David instead proposes a linear model $m = a t + b$, where $a$ and $b$ are constants.
\end{enumerate}\item Find the values of the constants for which the model fits the mass of the block when $t = 5$ and $t = 50$.
\item Interpret these values of $a$ and $b$.
\item Find the time according to this model for the block of ice to melt completely.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q11 [11]}}