OCR H240/01 2022 June — Question 8 9 marks

Exam BoardOCR
ModuleH240/01 (Pure Mathematics)
Year2022
SessionJune
Marks9
PaperDownload PDF ↗
TopicExponential Functions
TypeExponential growth/decay model setup
DifficultyModerate -0.3 This is a straightforward exponential decay question requiring standard techniques: finding the decay model from given data points, using logarithms to solve for time at half-life, differentiation for rate of decay, and basic comparison. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context1.07b Gradient as rate of change: dy/dx notation

8
  1. Substance \(A\) is decaying exponentially such that its mass is \(m\) grams at time \(t\) minutes. Find the missing values of \(m\) and \(t\) in the following table.
    \(t\)01050
    \(m\)1250750450
  2. Substance \(B\) is also decaying exponentially, according to the model \(m = 160 \mathrm { e } ^ { - 0.055 t }\), where \(m\) grams is its mass after \(t\) minutes.
    1. Determine the value of \(t\) for which the mass of substance \(B\) is half of its original mass.
    2. Determine the rate of decay of substance \(B\) when \(t = 15\).
  3. State whether substance \(A\) or substance \(B\) is decaying at a faster rate, giving a reason for your answer.

Question 8(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(20\) (minutes)B1 3.3
\(97.2\) (grams)B1 3.4
[2]
Question 8(b)(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(160e^{-0.055t} = 80\)B1 3.4
\(e^{-0.055t} = 0.5\)M1 3.4
\(-0.055t = \ln 0.5\)
\(t = 12.6\) (minutes)A1 1.1
[3]
Question 8(b)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dm}{dt} = -8.8e^{-0.055t}\)B1 3.4
\(-8.8e^{-0.055 \times 15}\)M1 3.4
\(= -3.86\), hence rate of decay is \(3.86\) grams/minuteA1 1.1
[3]
Question 8(c):
AnswerMarks Guidance
AnswerMarks Guidance
For \(A\), \(\frac{dm}{dt} = -63.9e^{-0.0511t}\). Rate of decrease at \(t = 15\) is \(29.7\) g/min, hence \(A\) decaying at a faster rateB1 3.4
# Question 8(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $20$ (minutes) | B1 | 3.3 | Obtain $t = 20$. Allow $[19.9, 20.1]$ from setting up and using exponential model |
| $97.2$ (grams) | B1 | 3.4 | Obtain $m = 97.2$. Allow $[97.1, 97.3]$ from setting up and using exponential model |
| | **[2]** | |

---

# Question 8(b)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $160e^{-0.055t} = 80$ | B1 | 3.4 | Equate given model to 80. soi, so could be $e^{-0.055t} = 0.5$ |
| $e^{-0.055t} = 0.5$ | M1 | 3.4 | Attempt correct process to find value of $t$, as far as dealing with exponential term. Rearrange to $e^{-0.055t} = k$, and hence obtain $-0.055t = \ln k$. If introducing logs straight away then need to get as far as $\ln 160 - 0.055t = \ln(\text{their } 80)$ |
| $-0.055t = \ln 0.5$ | | |
| $t = 12.6$ (minutes) | A1 | 1.1 | Obtain $t = 12.6$, or better. If more sig fig given, allow answers which round to $12.60$ (more accurate answer is $12.602676...$) |
| | **[3]** | |

---

# Question 8(b)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dm}{dt} = -8.8e^{-0.055t}$ | B1 | 3.4 | Correct derivative soi. Allow unsimplified. No need to see $\frac{dm}{dt} =$ |
| $-8.8e^{-0.055 \times 15}$ | M1 | 3.4 | Substitute $t = 15$ into their derivative. Must be of the form $ke^{-0.055t}$, with $k \neq 160$. Possibly still with $k$ unsimplified. Substitution sufficient, no need to evaluate for M1 |
| $= -3.86$, hence rate of decay is $3.86$ grams/minute | A1 | 1.1 | Units required and positive answer. Must follow correct derivative i.e. negative coefficient. No need to see $-3.86$ first, but A0 if clear error. Accept $3.9$ grams/minute. Accept g/m for grams/minute |
| | **[3]** | |

---

# Question 8(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For $A$, $\frac{dm}{dt} = -63.9e^{-0.0511t}$. Rate of decrease at $t = 15$ is $29.7$ g/min, hence $A$ decaying at a faster rate | B1 | 3.4 | State $A$, with clear comparison. Insufficient to just say $A$ has greater initial mass – needs to consider decay factor as well. Allow solutions identifying $B$ is decaying faster with supporting evidence. e.g. after 10 min, $B$'s mass is $92.3$g which is $58\%$ of initial mass whereas $A$ is $60\%$ of initial mass so $B$ decaying faster. e.g. $A$'s half-life is $13.6$ so $B$ is decaying faster. e.g. compare coefficients of $t$ (for $A$, coeff is $-0.0511$); $B$'s is of greater magnitude hence decaying faster. For either solution, conclusion and supporting evidence must be consistent. Numerical supporting evidence must be correct, allowing for slight inaccuracies from using different numbers of sig fig (see appendix) |
8
\begin{enumerate}[label=(\alph*)]
\item Substance $A$ is decaying exponentially such that its mass is $m$ grams at time $t$ minutes. Find the missing values of $m$ and $t$ in the following table.

\begin{center}
\begin{tabular}{ | l | l | l | l | l | }
\hline
$t$ & 0 & 10 &  & 50 \\
\hline
$m$ & 1250 & 750 & 450 &  \\
\hline
\end{tabular}
\end{center}
\item Substance $B$ is also decaying exponentially, according to the model $m = 160 \mathrm { e } ^ { - 0.055 t }$, where $m$ grams is its mass after $t$ minutes.
\begin{enumerate}[label=(\roman*)]
\item Determine the value of $t$ for which the mass of substance $B$ is half of its original mass.
\item Determine the rate of decay of substance $B$ when $t = 15$.
\end{enumerate}\item State whether substance $A$ or substance $B$ is decaying at a faster rate, giving a reason for your answer.
\end{enumerate}

\hfill \mbox{\textit{OCR H240/01 2022 Q8 [9]}}