OCR MEI Further Pure Core 2024 June — Question 17 20 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2024
SessionJune
Marks20
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeTank/reservoir mixing problems
DifficultyStandard +0.8 This is a standard Further Maths integrating factor question with a mixing problem context. While it requires multiple steps (setting up the DE, applying integrating factor, finding maximum), each step follows a well-established procedure taught in Further Pure. The integration and logarithmic manipulation are routine for this level, making it moderately challenging but not exceptional.
Spec4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations

17 In an industrial process, a container initially contains 1000 litres of liquid. Liquid is drawn from the bottom of the container at a rate of 5 litres per minute. At the same time, salt is added to the top of the container at a constant rate of 10 grams per minute. After \(t\) minutes the mass of salt in the container is \(x\) grams, and you are given that \(x = 0\) when \(t = 0\). In modelling the situation, it is assumed that the salt dissolves instantly and uniformly in the liquid, and that adding the salt does not change the volume of the liquid.
    1. Show that the concentration of salt in the liquid after \(t\) minutes is \(\frac { \mathrm { X } } { 1000 - 5 \mathrm { t } }\) grams per litre.
    2. Hence show that the mass of salt in the container is given by the differential equation $$\frac { d x } { d t } + \frac { x } { 200 - t } = 10$$
  1. Show by integration that \(\mathrm { x } = 10 ( 200 - \mathrm { t } ) \ln \left( \frac { 200 } { 200 - \mathrm { t } } \right)\).
    1. Hence determine the mass of salt in the container when half the liquid is drawn off.
    2. Determine also the time at which the mass of salt in the container is greatest.
  2. When the process is run, it is found that the concentration of salt over time is higher than predicted by the model. Suggest a reason for this.

Question 17:
AnswerMarks Guidance
17(a) (i)
x
so concentration =
AnswerMarks Guidance
1 0 0 0 − 5 tB1
[1]3.1b AG Volume of liquid (allow “volume”, “amount of
liquid”, “capacity” or “liquid”) in the container must be
given in first step. Do not allow “volume leaving”.
AnswerMarks Guidance
`17(a) (ii)
= 10…
d𝑡
𝑥
…[−]5( )
1000−5𝑡
d x x
 + = 1 0
AnswerMarks
d t 2 0 0 − tB1*
B1*
B1dep
AnswerMarks
[3]3.3
3.3
AnswerMarks
3.3Not oe
AG successful completion
AnswerMarks Guidance
17(b)  1 d t
IF is e 2 0 0 − t
1
= e − ln ( 2 0 0 − t ) =
2 0 0 − t
1 d x x 1 0
 + =
2 0 0 − t d t ( 2 0 0 − t ) 2 2 0 0 − t
d  x  1 0 𝑥 10
 = or = ∫ d𝑡
d t 2 0 0 − t 2 0 0 − t 200−𝑡 200−𝑡
𝑥
= −10ln(200−𝑡)[+𝑐]
200−𝑡
t =0,x=0c=10ln200
x  2 0 0 
= − 1 0 ln ( 2 0 0 − t ) + 1 0 ln 2 0 0 = 1 0 ln
2 0 0 − t 2 0 0 − t
 200 
x=10(200−t)ln
AnswerMarks
200−tM1
A1
M1
M1*
A1
B1FT
M1dep
A1
AnswerMarks
[8]3.1a
2.1
2.1
2.1
2.1
3.1b
2.1
AnswerMarks
2.1Finding integrating factor
1
Allow 𝐴 where 𝐴 is clearly an unevaluated constant
200−𝑡
of integration and not from incorrect integration
Multiplying equation by their IF, soi by next line
Either, ft their IF (i.e. their IF × 𝑥)
First A1 must have been scored. Condone missing +𝑐 and
bracketing errors.
Correct 𝑐 for their integral
Combining their log terms correctly, do not condone
bracketing errors without clear recovery
AG www
AnswerMarks Guidance
17(c) (i)
⇒ 𝑥 = 10×100×ln2 = 693 [g]M1
A1
AnswerMarks
[2]3.4
3.4May be embedded but must be seen
Accept 690 or better, or 1000ln2
AnswerMarks Guidance
17(c) (ii)
[x is maximised when] =0
dt
200
10(200−𝑡)ln( ) = 10(200−𝑡)
200−𝑡
 2 0 0 
 l n = 1
2 0 0 − t
2 0 0
 = e
2 0 0 − t
AnswerMarks
⇒ 𝑡 = 126.42…M1
M1*
A1
M1dep
A1
AnswerMarks
[5]3.1a
1.1
1.1
1.1
AnswerMarks
3.4soi. Implied by 𝑥 = 10(200−𝑡).
d𝑥 200
Or finding = 10(1−ln( ) ) correctly
d𝑡 200−𝑡
Eliminating logarithms correctly
130 (2sf) or better. Do not accept exact value. Ignore
units.
AnswerMarks Guidance
17(d) In reality the salt would not
dissolve/mix/spread/diffuse/disperse/distribute instantlyB1
[1]3.5b Or “the salt does not
dissolve/mix/spread/diffuse/disperse/distribute
uniformly”. Must be explicit that modelling assumption is
unrealistic.
Ignore subsequent comments
Do not accept reference to volume or exogenous factors,
e.g. evaporation/temperature
AnswerMarks
ResponseMark
1
is undefined when 𝑥 = 2
AnswerMarks
3 x − 2B1
There is an asymptote at 𝑥 = 2B1
The function is not continuous at 𝑥 = 2B1
1
When 𝑥 = 2 the denominator will be 0 and is undefined/impossible/invalid/improper/indeterminate/an error/cannot be computed
AnswerMarks
0B1
When 𝑥 = 2 we get 1 divided by 0 which is undefined/impossible/invalid/improper/indeterminate/an error/cannot be computedB1
1
When 𝑥 = 2 we get which makes the integral undefined/impossible/invalid/improper/indeterminate/an error/incomputable
AnswerMarks
0B1
When 𝑥 = 2 the denominator of the function is 0 and you cannot divide by 0B1
1
When 𝑥 = 2 it becomes = undefined
AnswerMarks
0B1
1
When 𝑥 = 2 we get which tends to infinity
AnswerMarks
0B0
1
When 𝑥 = 2 we get which is divergent
AnswerMarks
0B0
1
When 𝑥 = 2 we get which has no solutions
AnswerMarks Guidance
0B0
When 𝑥 = 2 we get 1 divided by 0 which is an error functionB0
When 𝑥 < 2 the value is complex/not a real numberB0
It is undefined when 𝑥 → 2B0
It is undefined when 𝑥 = 2B1
ResponseMark
[The series] converges/works only when −1 < 𝑥 ≤ 1A1
The series requires/works/converges when −1 < 𝑥 ≤ 1A1
The series converges only when −1 < 𝑥 < 1A0
The series converges when −1 ≤ 𝑥 ≤ 1A0
The series does not converge when 𝑥 > 1A1
The series does not converge when 𝑥 is greater than 1A1
The series does not converge when 𝑥 ≥ 1A0
The series converges only when𝑥 < 1
The series does not converge when𝑥 > 1
The series does not converge when𝑥 ≥ 1
The series does not converge when𝑥 ≥ 1 unless 𝑥 = 1
𝑥 = 2 is greater than 1 so outside the range for convergenceA1
2 ≥ 1A0
For series to converge 𝑥 < 1A0
2 > 1 so the series does not convergeA1
The series converges when −1 < 𝑥 < 1
The series converges when 𝑥 is between −1 and 1
For the series to converge 𝑥 must be between −1 and 1
AnswerMarks Guidance
The series converges when𝑥 < 1
2 > 1
The series does not converge when 𝑥 = 2
PMT
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Question 17:
17 | (a) | (i) | After t minutes volume of liquid = 1000 − 5t [litres]
x
so concentration =
1 0 0 0 − 5 t | B1
[1] | 3.1b | AG Volume of liquid (allow “volume”, “amount of
liquid”, “capacity” or “liquid”) in the container must be
given in first step. Do not allow “volume leaving”.
`17 | (a) | (ii) | d𝑥
= 10…
d𝑡
𝑥
…[−]5( )
1000−5𝑡
d x x
 + = 1 0
d t 2 0 0 − t | B1*
B1*
B1dep
[3] | 3.3
3.3
3.3 | Not oe
AG successful completion
17 | (b) |  1 d t
IF is e 2 0 0 − t
1
= e − ln ( 2 0 0 − t ) =
2 0 0 − t
1 d x x 1 0
 + =
2 0 0 − t d t ( 2 0 0 − t ) 2 2 0 0 − t
d  x  1 0 𝑥 10
 = or = ∫ d𝑡
d t 2 0 0 − t 2 0 0 − t 200−𝑡 200−𝑡
𝑥
= −10ln(200−𝑡)[+𝑐]
200−𝑡
t =0,x=0c=10ln200
x  2 0 0 
= − 1 0 ln ( 2 0 0 − t ) + 1 0 ln 2 0 0 = 1 0 ln
2 0 0 − t 2 0 0 − t
 200 
x=10(200−t)ln

200−t | M1
A1
M1
M1*
A1
B1FT
M1dep
A1
[8] | 3.1a
2.1
2.1
2.1
2.1
3.1b
2.1
2.1 | Finding integrating factor
1
Allow 𝐴 where 𝐴 is clearly an unevaluated constant
200−𝑡
of integration and not from incorrect integration
Multiplying equation by their IF, soi by next line
Either, ft their IF (i.e. their IF × 𝑥)
First A1 must have been scored. Condone missing +𝑐 and
bracketing errors.
Correct 𝑐 for their integral
Combining their log terms correctly, do not condone
bracketing errors without clear recovery
AG www
17 | (c) | (i) | Half drawn off when t = 1 0 0
⇒ 𝑥 = 10×100×ln2 = 693 [g] | M1
A1
[2] | 3.4
3.4 | May be embedded but must be seen
Accept 690 or better, or 1000ln2
17 | (c) | (ii) | dx
[x is maximised when] =0
dt
200
10(200−𝑡)ln( ) = 10(200−𝑡)
200−𝑡
 2 0 0 
 l n = 1
2 0 0 − t
2 0 0
 = e
2 0 0 − t
⇒ 𝑡 = 126.42… | M1
M1*
A1
M1dep
A1
[5] | 3.1a
1.1
1.1
1.1
3.4 | soi. Implied by 𝑥 = 10(200−𝑡).
d𝑥 200
Or finding = 10(1−ln( ) ) correctly
d𝑡 200−𝑡
Eliminating logarithms correctly
130 (2sf) or better. Do not accept exact value. Ignore
units.
17 | (d) | In reality the salt would not
dissolve/mix/spread/diffuse/disperse/distribute instantly | B1
[1] | 3.5b | Or “the salt does not
dissolve/mix/spread/diffuse/disperse/distribute
uniformly”. Must be explicit that modelling assumption is
unrealistic.
Ignore subsequent comments
Do not accept reference to volume or exogenous factors,
e.g. evaporation/temperature
Response | Mark
1
is undefined when 𝑥 = 2
3 x − 2 | B1
There is an asymptote at 𝑥 = 2 | B1
The function is not continuous at 𝑥 = 2 | B1
1
When 𝑥 = 2 the denominator will be 0 and is undefined/impossible/invalid/improper/indeterminate/an error/cannot be computed
0 | B1
When 𝑥 = 2 we get 1 divided by 0 which is undefined/impossible/invalid/improper/indeterminate/an error/cannot be computed | B1
1
When 𝑥 = 2 we get which makes the integral undefined/impossible/invalid/improper/indeterminate/an error/incomputable
0 | B1
When 𝑥 = 2 the denominator of the function is 0 and you cannot divide by 0 | B1
1
When 𝑥 = 2 it becomes = undefined
0 | B1
1
When 𝑥 = 2 we get which tends to infinity
0 | B0
1
When 𝑥 = 2 we get which is divergent
0 | B0
1
When 𝑥 = 2 we get which has no solutions
0 | B0
When 𝑥 = 2 we get 1 divided by 0 which is an error function | B0
When 𝑥 < 2 the value is complex/not a real number | B0
It is undefined when 𝑥 → 2 | B0
It is undefined when 𝑥 = 2 | B1
Response | Mark
[The series] converges/works only when −1 < 𝑥 ≤ 1 | A1
The series requires/works/converges when −1 < 𝑥 ≤ 1 | A1
The series converges only when −1 < 𝑥 < 1 | A0
The series converges when −1 ≤ 𝑥 ≤ 1 | A0
The series does not converge when 𝑥 > 1 | A1
The series does not converge when 𝑥 is greater than 1 | A1
The series does not converge when 𝑥 ≥ 1 | A0
The series converges only when |𝑥| < 1 | A0
The series does not converge when |𝑥| > 1 | A1
The series does not converge when |𝑥| ≥ 1 | A0
The series does not converge when |𝑥| ≥ 1 unless 𝑥 = 1 | A1
𝑥 = 2 is greater than 1 so outside the range for convergence | A1
2 ≥ 1 | A0
For series to converge 𝑥 < 1 | A0
2 > 1 so the series does not converge | A1
The series converges when −1 < 𝑥 < 1
The series converges when 𝑥 is between −1 and 1
For the series to converge 𝑥 must be between −1 and 1
The series converges when |𝑥| < 1
2 > 1
The series does not converge when 𝑥 = 2
PMT
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touch with our customer support centre.
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/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2024 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
17 In an industrial process, a container initially contains 1000 litres of liquid. Liquid is drawn from the bottom of the container at a rate of 5 litres per minute. At the same time, salt is added to the top of the container at a constant rate of 10 grams per minute. After $t$ minutes the mass of salt in the container is $x$ grams, and you are given that $x = 0$ when $t = 0$.

In modelling the situation, it is assumed that the salt dissolves instantly and uniformly in the liquid, and that adding the salt does not change the volume of the liquid.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the concentration of salt in the liquid after $t$ minutes is $\frac { \mathrm { X } } { 1000 - 5 \mathrm { t } }$ grams per litre.
\item Hence show that the mass of salt in the container is given by the differential equation

$$\frac { d x } { d t } + \frac { x } { 200 - t } = 10$$
\end{enumerate}\item Show by integration that $\mathrm { x } = 10 ( 200 - \mathrm { t } ) \ln \left( \frac { 200 } { 200 - \mathrm { t } } \right)$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence determine the mass of salt in the container when half the liquid is drawn off.
\item Determine also the time at which the mass of salt in the container is greatest.
\end{enumerate}\item When the process is run, it is found that the concentration of salt over time is higher than predicted by the model.

Suggest a reason for this.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2024 Q17 [20]}}