OCR Further Pure Core 2 Specimen — Question 11 17 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
SessionSpecimen
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeModeling context with interpretation
DifficultyChallenging +1.2 This is a structured Further Maths question on coupled differential equations with clear scaffolding. While it involves multiple techniques (exponential models, first-order linear ODEs, optimization), each part guides students through the solution methodically. The conceptual demand is moderate—recognizing conservation laws and solving a standard integrating factor problem—but doesn't require novel insight beyond applying taught methods to a multi-variable system.
Spec4.10a General/particular solutions: of differential equations4.10h Coupled systems: simultaneous first order DEs
During an industrial process substance \(X\) is converted into substance \(Z\). Some of the substance \(X\) goes through an intermediate phase, and is converted to substance \(Y\), before being converted to substance \(Z\). The situation is modelled by $$\frac{dy}{dt} = 0.3x + 0.2y \text{ and } \frac{dz}{dt} = 0.2y + 0.1x$$ where \(x\), \(y\) and \(z\) are the amounts in kg of \(X\), \(Y\) and \(Z\) at time \(t\) hours after the process starts. Initially there is 10 kg of substance \(X\) and nothing of substances \(Y\) and \(Z\). The amount of substance \(X\) decreases exponentially. The initial rate of decrease is 4 kg per hour.
  1. Show that \(x = Ae^{-0.4t}\), stating the value of \(A\). [3]
    1. Show that \(\frac{dx}{dt} + \frac{dy}{dt} + \frac{dz}{dt} = 0\). [2]
    2. Comment on this result in the context of the industrial process. [2]
  3. Express \(y\) in terms of \(t\). [5]
  4. Determine the maximum amount of substance \(Y\) present during the process. [3]
  5. How long does it take to produce 9 kg of substance \(Z\)? [2]
Question 11:
AnswerMarks Guidance
11(i) A(cid:32)10
dx
kt kt
x(cid:32) Ae (cid:159) (cid:32) Ake giving Ak (cid:32)(cid:16)4
dt
AnswerMarks
10k = –4 so k = –0.4B1
M1
E1
AnswerMarks
[3]3.3
3.3
AnswerMarks
1.1dx
Differentiate and use (cid:32)(cid:114)4
dt
when t = 0
AG
AnswerMarks Guidance
11(ii) (a)
(cid:14) (cid:14) (cid:32)(cid:16)4e(cid:16)0.4t
dt dt dt
(cid:14) (0.3x(cid:16)0.2y)(cid:14)(0.2y(cid:14)0.1x)
AnswerMarks
(cid:32)(cid:16)0.4x(cid:14)0.3x(cid:14)0.1x(cid:32)0M1
E1
AnswerMarks
[2]3.4
1.1n
e
AG
AnswerMarks Guidance
11(ii) (b)
(x(cid:14) y(cid:14)z)(cid:32)0(cid:159)x(cid:14) y(cid:14)z(cid:32)constant
dt
Throughout the industrial process the total amount
AnswerMarks
of the three substances is constantB1
c
E1
AnswerMarks
[2]m
2.2a
i
AnswerMarks
3.4i.e. no loss or gain of amounts
Must be in context
AnswerMarks Guidance
11(iii) dy
(cid:14)0.2y(cid:32)3e(cid:16)0.4t p
dt
(cid:117)e0.2t
S
ye0.2t (cid:32)3(cid:179)e(cid:16)0.2tdt
y(cid:32)(cid:16)15e(cid:16)0.4t (cid:14)ce(cid:16)0.2t
AnswerMarks
c(cid:32)15e
M1
M1
A1
M1
A1
AnswerMarks
[5]3.4
3.1a
1.1
1.1
AnswerMarks
1.1Substitute x and rearrange
Alternative CF and PI
Allow arithmetic slips in
coefficients of exponential terms
and/or sign errors
AnswerMarks Guidance
11(iv) (cid:16)3e(cid:16)0.2t (cid:14)6e(cid:16)0.4t (cid:32)0
t(cid:32)3.466
y (cid:32)3.75kg
AnswerMarks
maxM1
A1
A1
AnswerMarks
[3]3.4
1.1
AnswerMarks
3.2ady
FT their soi
dt
BCn
All three marks may be implied
AnswerMarks
by 3.75kg BCOR
M1 0.3x(cid:32)0.2y so
0.3(cid:117)10e(cid:16)0.4t
(cid:32)0.2(cid:117) (cid:11) (cid:16)15e(cid:16)0.4t (cid:14)15e(cid:16)0.2t(cid:12)
A1t(cid:32)3.466 BC
A1y (cid:32)3.75kg BC
max
AnswerMarks Guidance
11(v) 9 kg of substance Z impliesx(cid:14) y(cid:32)1
t(cid:32)13.4(cid:11)25(cid:12) hours to 3 s.f. or betterM1
A1
AnswerMarks
[2]m3.4
1.1e
Or by z(cid:32)10(cid:14)5e(cid:16)0.4t (cid:16)15e(cid:16)0.2t (cid:32)9
BC
PPMMTT
Y541 Mark Scheme June 20XX
Assessment Objectives (AO) Grid
n
e
m
i
c
e
PS = Problem Solving
p
M = Modelling
S
13
PPMMTT
Y541 Mark Scheme June 20XX
BLANK PAGE
n
e
m
i
c
e
p
S
14
PPMMTT
Y541 Mark Scheme June 20XX
BLANK PAGE
n
e
m
i
c
e
p
S
15
PPMMTT
Y541 Mark Scheme June 20XX
BLANK PAGE
n
e
m
i
c
e
p
S
16
Question 11:
11 | (i) | A(cid:32)10
dx
kt kt
x(cid:32) Ae (cid:159) (cid:32) Ake giving Ak (cid:32)(cid:16)4
dt
10k = –4 so k = –0.4 | B1
M1
E1
[3] | 3.3
3.3
1.1 | dx
Differentiate and use (cid:32)(cid:114)4
dt
when t = 0
AG
11 | (ii) | (a) | dx dy dz
(cid:14) (cid:14) (cid:32)(cid:16)4e(cid:16)0.4t
dt dt dt
(cid:14) (0.3x(cid:16)0.2y)(cid:14)(0.2y(cid:14)0.1x)
(cid:32)(cid:16)0.4x(cid:14)0.3x(cid:14)0.1x(cid:32)0 | M1
E1
[2] | 3.4
1.1 | n
e
AG
11 | (ii) | (b) | d
(x(cid:14) y(cid:14)z)(cid:32)0(cid:159)x(cid:14) y(cid:14)z(cid:32)constant
dt
Throughout the industrial process the total amount
of the three substances is constant | B1
c
E1
[2] | m
2.2a
i
3.4 | i.e. no loss or gain of amounts
Must be in context
11 | (iii) | dy
(cid:14)0.2y(cid:32)3e(cid:16)0.4t p
dt
(cid:117)e0.2t
S
ye0.2t (cid:32)3(cid:179)e(cid:16)0.2tdt
y(cid:32)(cid:16)15e(cid:16)0.4t (cid:14)ce(cid:16)0.2t
c(cid:32)15 | e
M1
M1
A1
M1
A1
[5] | 3.4
3.1a
1.1
1.1
1.1 | Substitute x and rearrange
Alternative CF and PI
Allow arithmetic slips in
coefficients of exponential terms
and/or sign errors
11 | (iv) | (cid:16)3e(cid:16)0.2t (cid:14)6e(cid:16)0.4t (cid:32)0
t(cid:32)3.466
y (cid:32)3.75kg
max | M1
A1
A1
[3] | 3.4
1.1
3.2a | dy
FT their soi
dt
BCn
All three marks may be implied
by 3.75kg BC | OR
M1 0.3x(cid:32)0.2y so
0.3(cid:117)10e(cid:16)0.4t
(cid:32)0.2(cid:117) (cid:11) (cid:16)15e(cid:16)0.4t (cid:14)15e(cid:16)0.2t(cid:12)
A1t(cid:32)3.466 BC
A1y (cid:32)3.75kg BC
max
11 | (v) | 9 kg of substance Z impliesx(cid:14) y(cid:32)1
t(cid:32)13.4(cid:11)25(cid:12) hours to 3 s.f. or better | M1
A1
[2] | m3.4
1.1 | e
Or by z(cid:32)10(cid:14)5e(cid:16)0.4t (cid:16)15e(cid:16)0.2t (cid:32)9
BC
PPMMTT
Y541 Mark Scheme June 20XX
Assessment Objectives (AO) Grid
n
e
m
i
c
e
PS = Problem Solving
p
M = Modelling
S
13
PPMMTT
Y541 Mark Scheme June 20XX
BLANK PAGE
n
e
m
i
c
e
p
S
14
PPMMTT
Y541 Mark Scheme June 20XX
BLANK PAGE
n
e
m
i
c
e
p
S
15
PPMMTT
Y541 Mark Scheme June 20XX
BLANK PAGE
n
e
m
i
c
e
p
S
16
During an industrial process substance $X$ is converted into substance $Z$. Some of the substance $X$ goes through an intermediate phase, and is converted to substance $Y$, before being converted to substance $Z$. The situation is modelled by

$$\frac{dy}{dt} = 0.3x + 0.2y \text{ and } \frac{dz}{dt} = 0.2y + 0.1x$$

where $x$, $y$ and $z$ are the amounts in kg of $X$, $Y$ and $Z$ at time $t$ hours after the process starts.

Initially there is 10 kg of substance $X$ and nothing of substances $Y$ and $Z$. The amount of substance $X$ decreases exponentially. The initial rate of decrease is 4 kg per hour.

\begin{enumerate}[label=(\roman*)]
\item Show that $x = Ae^{-0.4t}$, stating the value of $A$. [3]

\item \begin{enumerate}[label=(\alph*)]
\item Show that $\frac{dx}{dt} + \frac{dy}{dt} + \frac{dz}{dt} = 0$. [2]

\item Comment on this result in the context of the industrial process. [2]
\end{enumerate}

\item Express $y$ in terms of $t$. [5]

\item Determine the maximum amount of substance $Y$ present during the process. [3]

\item How long does it take to produce 9 kg of substance $Z$? [2]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2  Q11 [17]}}
This paper (11 questions)
View full paper
Q1 4 Q2 4 Q3 4 Q4 5 Q5 4 Q6 8 Q7 7 Q8 8 Q9 6 Q10 8 Q11 17