AQA C4 2011 January — Question 5 7 marks

Exam BoardAQA
ModuleC4 (Core Mathematics 4)
Year2011
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeHalf-life and doubling time
DifficultyModerate -0.3 This is a straightforward application of an exponential decay model with clear structure. Part (a) is direct substitution, part (b) requires simple equation solving using index laws (recognizing 1/16 = 2^{-4}), and part (c) involves logarithms or trial with inequalities. All parts follow standard textbook patterns for exponential decay with no novel problem-solving required, making it slightly easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context

5 A model for the radioactive decay of a form of iodine is given by $$m = m _ { 0 } 2 ^ { - \frac { 1 } { 8 } t }$$ The mass of the iodine after \(t\) days is \(m\) grams. Its initial mass is \(m _ { 0 }\) grams.
  1. Use the given model to find the mass that remains after 10 grams of this form of iodine have decayed for 14 days, giving your answer to the nearest gram.
  2. A mass of \(m _ { 0 }\) grams of this form of iodine decays to \(\frac { m _ { 0 } } { 16 }\) grams in \(d\) days. Find the value of \(d\).
  3. After \(n\) days, a mass of this form of iodine has decayed to less than \(1 \%\) of its initial mass. Find the minimum integer value of \(n\).

Question 5:
Part (a)
AnswerMarks
\(m = 10 \times 2^{-\frac{14}{8}} = 10 \times 2^{-1.75}\)M1
\(m \approx 3\) gramsA1
Part (b)
AnswerMarks
\(\frac{m_0}{16} = m_0 \cdot 2^{-\frac{d}{8}}\)M1
\(2^{-4} = 2^{-\frac{d}{8}} \Rightarrow d = 32\)A1
Part (c)
AnswerMarks
\(m_0 \cdot 2^{-\frac{n}{8}} < \frac{m_0}{100}\)M1
\(2^{-\frac{n}{8}} < 0.01 \Rightarrow -\frac{n}{8}\ln 2 < \ln(0.01)\)M1
\(n > \frac{8\ln 100}{\ln 2} \approx 53.15\)
Minimum integer \(n = 54\)A1
# Question 5:

## Part (a)
| $m = 10 \times 2^{-\frac{14}{8}} = 10 \times 2^{-1.75}$ | M1 | |
|---|---|---|
| $m \approx 3$ grams | A1 | |

## Part (b)
| $\frac{m_0}{16} = m_0 \cdot 2^{-\frac{d}{8}}$ | M1 | |
|---|---|---|
| $2^{-4} = 2^{-\frac{d}{8}} \Rightarrow d = 32$ | A1 | |

## Part (c)
| $m_0 \cdot 2^{-\frac{n}{8}} < \frac{m_0}{100}$ | M1 | |
|---|---|---|
| $2^{-\frac{n}{8}} < 0.01 \Rightarrow -\frac{n}{8}\ln 2 < \ln(0.01)$ | M1 | |
| $n > \frac{8\ln 100}{\ln 2} \approx 53.15$ | | |
| Minimum integer $n = 54$ | A1 | |
5 A model for the radioactive decay of a form of iodine is given by

$$m = m _ { 0 } 2 ^ { - \frac { 1 } { 8 } t }$$

The mass of the iodine after $t$ days is $m$ grams. Its initial mass is $m _ { 0 }$ grams.
\begin{enumerate}[label=(\alph*)]
\item Use the given model to find the mass that remains after 10 grams of this form of iodine have decayed for 14 days, giving your answer to the nearest gram.
\item A mass of $m _ { 0 }$ grams of this form of iodine decays to $\frac { m _ { 0 } } { 16 }$ grams in $d$ days.

Find the value of $d$.
\item After $n$ days, a mass of this form of iodine has decayed to less than $1 \%$ of its initial mass.

Find the minimum integer value of $n$.
\end{enumerate}

\hfill \mbox{\textit{AQA C4 2011 Q5 [7]}}