AQA AS Paper 1 2020 June — Question 10 12 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2020
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs ln(x) linear graph
DifficultyModerate -0.8 This is a standard logarithmic linearization question requiring routine application of log laws, plotting points, reading gradient/intercept from a graph, and substitution. All techniques are textbook exercises with no novel problem-solving required, making it easier than average for A-level.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines
Raj is investigating how the price, \(P\) pounds, of a brilliant-cut diamond ring is related to the weight, \(C\) carats, of the diamond. He believes that they are connected by a formula $$P = aC^n$$ where \(a\) and \(n\) are constants.
  1. Express \(\ln P\) in terms of \(\ln C\). [2 marks]
  2. Raj researches the price of three brilliant-cut diamond rings on a website with the following results.
    \(C\)0.601.151.50
    \(P\)49512001720
    1. Plot \(\ln P\) against \(\ln C\) for the three rings on the grid below. [2 marks] \includegraphics{figure_10b}
    2. Explain which feature of the plot suggests that Raj's belief may be correct. [1 mark]
    3. Using the graph on page 15, estimate the value of \(a\) and the value of \(n\). [4 marks]
  3. Explain the significance of \(a\) in this context. [1 mark]
  4. Raj wants to buy a ring with a brilliant-cut diamond of weight 2 carats. Estimate the price of such a ring. [2 marks]
Question 10:
AnswerMarks Guidance
10(a)Applies laws of logarithms to
obtain one correct term1.1a M1
ln P = ln a + n ln C
Obtains completely correct
AnswerMarks Guidance
expression1.1b A1
Subtotal2
AnswerMarks
10(b)(i)Calculates ln values. Condone
one slip. PI by any two correct
AnswerMarks Guidance
points1.1a M1
ln P = 6.20, 7.09, 7.45
(See graph below)
Correctly plots three points.
AnswerMarks Guidance
Line not required.1.1b A1
Subtotal2
AnswerMarks Guidance
10(b)(ii)Infers significance of straight
line2.2b E1
Subtotal1
AnswerMarks Guidance
10(b)(iii)Identifies ln a as the intercept. PI 3.4
ln a = 6.9
So a = 992
n is the gradient
= 1.37
Correctly calculates a
AnswerMarks Guidance
AWFW 960 to 10401.1b A1
Identifies n as the gradient. PI3.4 M1
Obtains correct n value. AWFW
AnswerMarks Guidance
1.35 to 1.411.1b A1
Subtotal4
AnswerMarks Guidance
10(c)Explains significance of a 2.4
Subtotal1
AnswerMarks
10(d)Substitutes their values into P
equation with C = 2
Or
Uses the graph to read off a
AnswerMarks Guidance
value for ln P3.4 M1
= £2560
Calculates correct value of P for
their values. AWFW 2440 to
2770
FT provided > 2000 and < 3000
AnswerMarks Guidance
must include units3.2a A1F
Subtotal2
Question Total12
QMarking Instructions AO 
Question 10:
--- 10(a) ---
10(a) | Applies laws of logarithms to
obtain one correct term | 1.1a | M1 | ln P = ln a + ln Cn
ln P = ln a + n ln C
Obtains completely correct
expression | 1.1b | A1
Subtotal | 2
--- 10(b)(i) ---
10(b)(i) | Calculates ln values. Condone
one slip. PI by any two correct
points | 1.1a | M1 | ln C = –0.51, 0.140, 0.405
ln P = 6.20, 7.09, 7.45
(See graph below)
Correctly plots three points.
Line not required. | 1.1b | A1
Subtotal | 2
--- 10(b)(ii) ---
10(b)(ii) | Infers significance of straight
line | 2.2b | E1 | The three points lie on a straight line
Subtotal | 1
--- 10(b)(iii) ---
10(b)(iii) | Identifies ln a as the intercept. PI | 3.4 | M1 | ln a is the intercept value
ln a = 6.9
So a = 992
n is the gradient
= 1.37
Correctly calculates a
AWFW 960 to 1040 | 1.1b | A1
Identifies n as the gradient. PI | 3.4 | M1
Obtains correct n value. AWFW
1.35 to 1.41 | 1.1b | A1
Subtotal | 4
--- 10(c) ---
10(c) | Explains significance of a | 2.4 | E1 | a is the price for a 1 carat diamond
Subtotal | 1
--- 10(d) ---
10(d) | Substitutes their values into P
equation with C = 2
Or
Uses the graph to read off a
value for ln P | 3.4 | M1 | 992 × 21.37
= £2560
Calculates correct value of P for
their values. AWFW 2440 to
2770
FT provided > 2000 and < 3000
must include units | 3.2a | A1F
Subtotal | 2
Question Total | 12
Q | Marking Instructions | AO | Marks | Typical Solution
Raj is investigating how the price, $P$ pounds, of a brilliant-cut diamond ring is related to the weight, $C$ carats, of the diamond.

He believes that they are connected by a formula
$$P = aC^n$$
where $a$ and $n$ are constants.

\begin{enumerate}[label=(\alph*)]
\item Express $\ln P$ in terms of $\ln C$. [2 marks]

\item Raj researches the price of three brilliant-cut diamond rings on a website with the following results.

\begin{tabular}{|c|c|c|c|}
\hline
$C$ & 0.60 & 1.15 & 1.50 \\
\hline
$P$ & 495 & 1200 & 1720 \\
\hline
\end{tabular}

\begin{enumerate}[label=(\roman*)]
\item Plot $\ln P$ against $\ln C$ for the three rings on the grid below. [2 marks]

\includegraphics{figure_10b}

\item Explain which feature of the plot suggests that Raj's belief may be correct. [1 mark]

\item Using the graph on page 15, estimate the value of $a$ and the value of $n$. [4 marks]
\end{enumerate}

\item Explain the significance of $a$ in this context. [1 mark]

\item Raj wants to buy a ring with a brilliant-cut diamond of weight 2 carats.

Estimate the price of such a ring. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2020 Q10 [12]}}
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