Questions - A-Level Maths

OCR FP1 AS 2021 June Q1

1

Find a vector which is perpendicular to both $\left( \begin{array} { r } 1
3
- 2 \end{array} \right)$ and $\left( \begin{array} { r } - 3
- 6
4 \end{array} \right)$.
The cartesian equation of a line is $\frac { x } { 2 } = y - 3 = 2 z + 4$. Express the equation of this line in vector form.

OCR FP1 AS 2021 June Q2

2 In this question you must show detailed reasoning.
The complex numbers $z _ { 1 }$ and $z _ { 2 }$ are given by $z _ { 1 } = 2 - 3 \mathrm { i }$ and $z _ { 2 } = a + 4 \mathrm { i }$ where $a$ is a real number.

Express $z _ { 1 }$ in modulus-argument form, giving the modulus in exact form and the argument correct to 3 significant figures.
Find $z _ { 1 } z _ { 2 }$ in terms of $a$, writing your answer in the form $c + \mathrm { i } d$.
The real and imaginary parts of a complex number on an Argand diagram are $x$ and $y$ respectively. Given that the point representing $z _ { 1 } z _ { 2 }$ lies on the line $y = x$, find the value of $a$.
Given instead that $z _ { 1 } z _ { 2 } = \left( z _ { 1 } z _ { 2 } \right) ^ { * }$ find the value of $a$. In this question you must show detailed reasoning.
Express $( 2 + 3 \mathrm { i } ) ^ { 3 }$ in the form $a + \mathrm { i } b$.
Hence verify that $2 + 3 \mathrm { i }$ is a root of the equation $3 z ^ { 3 } - 8 z ^ { 2 } + 23 z + 52 = 0$.
Express $3 z ^ { 3 } - 8 z ^ { 2 } + 23 z + 52$ as the product of a linear factor and a quadratic factor with real coefficients.

OCR FP1 AS 2021 June Q4

27 marks

4
Prove by induction that $2 ^ { n + 1 } + 5 \times 9 ^ { n }$ is divisible by 7 for all integers $n \geqslant 1$. \section*{Total Marks for Question Set 3: 30} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

OCR FP1 AS 2021 June Q1

1 In this question you must show detailed reasoning.
The cubic equation $2 x ^ { 3 } + 3 x ^ { 2 } - 5 x + 4 = 0$ has roots $\alpha , \beta$ and $\gamma$. By making an appropriate substitution, or otherwise, find a cubic equation with integer coefficients whose roots are $\frac { 1 } { \alpha } , \frac { 1 } { \beta }$ and $\frac { 1 } { \gamma }$.

OCR FP1 AS 2021 June Q2

2 The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { r r c } 2 & 1 & 2
1 & - 1 & 1
2 & 2 & a \end{array} \right)$.

Show that $\operatorname { det } \mathbf { A } = 6 - 3 a$.
State the value of $a$ for which $\mathbf { A }$ is singular.
Given that $\mathbf { A }$ is non-singular find $\mathbf { A } ^ { - 1 }$ in terms of $a$.

OCR FP1 AS 2021 June Q3

30 marks

3 The matrices $\mathbf { A }$ and $\mathbf { B }$ are given by $\mathbf { A } = \left( \begin{array} { r r } t & 6
t & - 2 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { r r } 2 t & 4
t & - 2 \end{array} \right)$ where $t$ is a constant.

Show that $| \mathrm { A } | = | \mathrm { B } |$.
Verify that $| \mathrm { AB } | = | \mathrm { A } | | \mathrm { B } |$.
Given that $| \mathbf { A B } | = - 1$ explain what this means about the constant $t$. The $2 \times 2$ matrix $A$ represents a transformation $T$ which has the following properties.
- The image of the point $( 0,1 )$ is the point $( 3,4 )$.
- An object shape whose area is 7 is transformed to an image shape whose area is 35 .
- T has a line of invariant points.
- Find a possible matrix for $\mathbf { A }$.
The transformation $S$ is represented by the matrix $B$ where $B = \left( \begin{array} { l l } 3 & 1
2 & 2 \end{array} \right)$.
Find the equation of the line of invariant points of S .

Show that any line of the form $y = x + c$ is an invariant line of S . \section*{Total Marks for Question Set 4: 30} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

OCR FP1 AS 2021 June Q1

1 In this question you must show detailed reasoning.
Use an algebraic method to find the square roots of $- 77 - 36 \mathrm { i }$.

OCR FP1 AS 2021 June Q2

2 In this question you must show detailed reasoning.
The complex number $7 - 4 \mathrm { i }$ is denoted by $z$.

Giving your answers in the form $a + b \mathrm { i }$, where $a$ and $b$ are rational numbers, find the following.
1. $3 z - 4 z ^ { * }$
2. $( z + 1 - 3 \mathrm { i } ) ^ { 2 }$
3. $\frac { z + 1 } { z - 1 }$
Express $z$ in modulus-argument form giving the modulus exactly and the argument correct to 3 significant figures.
The complex number $\omega$ is such that $z \omega = \sqrt { 585 } ( \cos ( 0.5 ) + \mathrm { i } \sin ( 0.5 ) )$. Find the following.
- $| \omega |$
- $\arg ( \omega )$, giving your answer correct to 3 significant figures

OCR FP1 AS 2021 June Q3

3 In this question you must show detailed reasoning.
The cubic equation $5 x ^ { 3 } + 3 x ^ { 2 } - 4 x + 7 = 0$ has roots $\alpha , \beta$ and $\gamma$.
Find a cubic equation with integer coefficients whose roots are $\alpha + \beta , \beta + \gamma$ and $\gamma + \alpha$.

OCR FP1 AS 2021 June Q4

24 marks

4 Prove that $n ! > 2 ^ { 2 n }$ for all integers $n \geqslant 9$. \section*{Total Marks for Question Set 5: 30} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

Question

Answer

Marks

AO

Guidance

\multirow[t]{6}{*}{4}

If $n = 9$, LHS $= 9$ ! $= 362880$

RHS $= 2 ^ { 18 } = 262144 <$ LHS

[So true for $n = 9$ ]

B1

2.1

Basis case. Comparison must be explicit and correct

Bod sight of "true for $\mathrm { n } = 1$ "

Assume that $k ! > 2 ^ { 2 k }$ for some $k \geq 9$.

M1

2.1

Inductive hypothesis set up

Might see $2 ^ { 2 k } = 4 ^ { k }$ throughout

$( k + 1 ) ! = ( k + 1 ) k ! > ( k + 1 ) \times 2 ^ { 2 k } \ldots$

M1

1.1

Considering for $k + 1$ and using inductive hypothesis correctly

Allow M1 for use of inductive $n = k$ step when showing $P _ { \mathrm { k } + 1 } \rightarrow P _ { \mathrm { k } }$

\(\begin{aligned}

\ldots > 9 \times 2 ^ { 2 k } > 4 \times 2 ^ { 2 k } = 2 ^ { 2 } \times 2 ^ { 2 k } = 2 ^ { 2 + 2 k } = 2 ^ { 2 ( k + 1 ) }

\text { ie } ( k + 1 ) ! > 2 ^ { 2 ( k + 1 ) } \end{aligned}\)

A1

2.2a

Showing enough working to establish statement for $k + 1$

Might see $\geq 10$ or $\geq 9$ for $> 9$. A0 here if $k > 9$ stated earlier Do not allow if implication shown in the direction $P _ { \mathrm { k } + 1 } \rightarrow P _ { \mathrm { k } }$

So true for $n = k \Rightarrow$ true for $n = k + 1$. But true for $n = 9$. So true for all integers $n \geq 9$

A1

2.4

Clear and complete conclusion

[5]

OCR FP1 AS 2021 June Q2

2 You are given the system of equations $$\begin{array} { r } a ^ { 2 } x - 2 y = 1
x + b ^ { 2 } y = 3 \end{array}$$ where $a$ and $b$ are real numbers.

Use a matrix method to find $x$ and $y$ in terms of $a$ and $b$.
Explain why the method used in part (a) works for all values of $a$ and $b$.

OCR FP1 AS 2021 June Q3

3 The equations of two intersecting lines are
$\mathbf { r } = \left( \begin{array} { c } - 12
a
- 1 \end{array} \right) + \lambda \left( \begin{array} { l } 2
2
1 \end{array} \right) \quad \mathbf { r } = \left( \begin{array} { l } 2
0
5 \end{array} \right) + \mu \left( \begin{array} { c } - 3
1
- 1 \end{array} \right)$
where $a$ is a constant.

Find a vector, $\mathbf { b }$, which is perpendicular to both lines.
Show that b. $\left( \begin{array} { c } - 12
a
- 1 \end{array} \right) =$ b. $\left( \begin{array} { l } 2
0
5 \end{array} \right)$.
Hence, or otherwise, find the value of $a$.

OCR FP1 AS 2021 June Q4

24 marks

4 Two loci, $C _ { 1 }$ and $C _ { 2 }$, are defined by $$\begin{aligned} & C _ { 1 } = \left\{ z : | z | = \left| z - 4 d ^ { 2 } - 36 \right| \right\}
& C _ { 2 } = \left\{ z : \arg ( z - 12 d - 3 \mathrm { i } ) = \frac { 1 } { 4 } \pi \right\} \end{aligned}$$ where $d$ is a real number.

Find, in terms of $d$, the complex number which is represented on an Argand diagram by the point of intersection of $C _ { 1 }$ and $C _ { 2 }$.
[0pt] [You may assume that $C _ { 1 } \cap C _ { 2 } \neq \varnothing$.]

Explain why the solution found in part (a) is not valid when $d = 3$. \section*{Total Marks for Question Set 6: 30} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

OCR FS1 AS 2021 June Q1

1 The probability distribution for the discrete random variable $W$ is given in the table.

$w$	1	2	3	4
$\mathrm { P } ( W = w )$	0.25	0.36	$x$	$x ^ { 2 }$

Show that $\operatorname { Var } ( W ) = 0.8571$.
Find $\operatorname { Var } ( 3 W + 6 )$.

OCR FS1 AS 2021 June Q2

23 marks

2 In the manufacture of fibre optical cable (FOC), flaws occur randomly. Whether any point on a cable is flawed is independent of whether any other point is flawed. The number of flaws in 100 m of FOC of standard diameter is denoted by $X$.

State a further assumption needed for $X$ to be well modelled by a Poisson distribution. Assume now that $X$ can be well modelled by the distribution $\operatorname { Po } ( 0.7 )$.
Find the probability that in 300 m of FOC of standard diameter there are exactly 3 flaws. The number of flaws in 100 m of FOC of a larger diameter has the distribution $\mathrm { Po } ( 1.6 )$.
Find the probability that in 200 m of FOC of standard diameter and 100 m of FOC of the larger diameter the total number of flaws is at least 4 . Judith believes that mathematical ability and chess-playing ability are related. She asks 20 randomly chosen chess players, with known British Chess Federation (BCF) ratings $X$, to take a mathematics aptitude test, with scores $Y$. The results are summarised as follows. $$n = 20 , \Sigma x = 3600 , \Sigma x ^ { 2 } = 660500 , \Sigma y = 1440 , \Sigma y ^ { 2 } = 105280 , \Sigma x y = 260990$$
Calculate the value of Pearson's product-moment correlation coefficient $r$.
State an assumption needed to be able to carry out a significance test on the value of $r$.
Assume now that the assumption in part (ii) is valid. Test at the $5 \%$ significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by $$\text { ELO rating } = 8 \times \text { BCF rating } + 650$$ Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion.
Calculate the value of Pearson's product-moment correlation coefficient $r$.
State an assumption needed to be able to carry out a significance test on the value of $r$.
Assume now that the assumption in part (b) is valid. Test at the $5 \%$ significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.

There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by $$\mathrm { ELO } \text { rating } = 8 \# \mathrm { BCF } \text { rating } + 650 .$$ Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion. An environmentalist measures the mean concentration, $c$ milligrams per litre, of a particular chemical in a group of rivers, and the mean mass, $m$ pounds, of fish of a certain species found in those rivers. The results are given in the table. \end{table}

Question

Answer

Marks

AO

Guidance

1

(a)

\(\begin{aligned}

0.25 + 0.36 + x + x ^ { 2 } = 1

x ^ { 2 } + x - 0.39 = 0

x = 0.3 \text { (or } - 1.3 \text { ) }

x \text { cannot be negative }

\mathrm { E } ( W ) = 2.23

\mathrm { E } \left( W ^ { 2 } \right) = \Sigma w ^ { 2 } \mathrm { p } ( w ) \quad [ = 5.83 ]

\text { Subtract } [ \mathrm { E } ( W ) ] ^ { 2 } \text { to get } \mathbf { 0 . 8 5 7 1 } \end{aligned}\)

\(\begin{gathered} \text { M1 }

\text { A1 }

\text { B1ft }

\text { B1 }

\text { M1 }

\text { A1 }

{ [ 7 ] } \end{gathered}\)

3.1a

1.1b

2.3

1.1b

1.1

2.1

Equation using $\Sigma p = 1$

Correct simplified quadratic Correctly obtain $x = 0.3$

Explicitly reject other solution

2.23 or exact equivalent only Use $\Sigma w ^ { 2 } \mathrm { p } ( w )$

Correctly obtain given answer, www

Can be implied

Method needed ft on their quadratic Allow for $\mathrm { E } ( W ) ^ { 2 } = 4.9729$

Need 2.23 or 4.9729 and 5.83 or full numerical $\Sigma w ^ { 2 } \mathrm { p } ( w )$

1

(b)

$9 \times 0.8571 = 7.7139$

B1

[1]

1.1b

Allow 7.71 or 7.714

2

(a)

Flaws must occur at constant average rate (uniform rate)

B1

[1]

1.2

Context (e.g. "flaws") needed

Extra answers, e.g. "singly": B0

Not "constant rate" or "average constant rate".

2

(b)

$\operatorname { Po(2.1)~or~ } e ^ { - \lambda } \frac { \lambda ^ { 3 } } { 3 ! }$

M1

A1

[2]

1.1

1.1b

Po(2.1) stated or implied, or formula with $\lambda = 2.1$ stated Awrt 0.189

2

(c)

Po(3)

$1 - \mathrm { P } ( \leq 3 )$

M1

A1

[3]

1.1

1.1b

$\operatorname { Po } ( 2 \times 0.7 + 1.6 )$ stated or implied

Allow $1 - \mathrm { P } ( \leq 4 ) = 0.1847$, or from wrong $\lambda$

Awrt 0.353

Or all combinations $\leq 3$

$1 -$ above, not just $= 3$

Question

Answer

Marks

AO

Guidance

3

(a)

0.4(00)

B2

[2]

1.1

1.1b

SC: if B0, give SC B1 for two of $S _ { x x } = 12500 , S _ { y y } = 1600 , S _ { x y } = 1790$ and $S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)$

Also allow SC B1 for equivalent methods using Covariance \

SDs

3

(b)

Data needs to have a bivariate normal distribution

B1

[1]

1.2

Needs "bivariate normal" or clear equivalent. Not just "both normally distributed"

Allow "scatter diagram forms ellipse"

3

(c)

$\mathrm { H } _ { 0 }$ : higher maths scores are not associated with higher BCF grading; $\mathrm { H } _ { 1 }$ : positively associated

CV 0.3783

$0.400 > 0.3783$ so reject $\mathrm { H } _ { 0 }$

Significant evidence that higher maths scores are associated with higher BCF grading

B1

M1ft

A1ft

[4]

2.5

1.1b

2.2b

3.5a

Needs context and clearly onetailed $O R \rho$ used and defined Not "evidence that ..."

Allow 0.378

Reject/do not reject $\mathrm { H } _ { 0 }$

Contextualised, not too definite Needn't say "positive" if $\mathrm { H } _ { 1 } \mathrm { OK }$

SC 2-tail: B0; 0.4438, or 0.3783 B1; then M1A0

$\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho > 0$ where $\rho$ is population pmcc (not $r$ )

FT on their $r$, but not CV

Not "scores are associated

...". FT on their $r$ only

3

(d)

It makes no difference as this is a linear transformation

B1

[1]

2.2a

Need both "unchanged" oe and reason, need "linear" or exact equivalent

"oe" includes "their 0.4"

4

(a)

Neither

B1

[1]

2.5

OE

Not "neither is independent of the other"

4

(b)

$c = 2.848 - 0.1567 m \quad \mathbf { B C }$

B1

[3]

1.1

Correct $a$, awrt 2.85

Correct $b$, awrt 0.157

Letters correct from correct method

(If both wrongly rounded, e.g. $c = 2.84 - 0.156 m$, give B2)

$\mathrm { SC } : m$ on $c$ :

$m = 15.65 - 4.832 c$ : B2

$y = 15.65 - 4.832 x$ : B1

$c = 15.65 - 4.832 m : \mathrm { B } 1$

If B0B0, give B1 for correct letters from valid working

Question

Answer

Marks

AO

Guidance

4

(c)

$a$ unchanged, $b$ multiplied by 2.2 (allow " $a$ unchanged, $b$ increases", etc)

B1 [1]

2.2a

oe, e.g. $c = 2.848 - 0.345 m$; $m = 7.114 - 2.196 c$

SC: $m$ on $c$ in (b): Both divided by 2.2 B1

4

(d)

Draw approximate line of best fit

Draw at least one vertical from line to point

Say that "Best fit" line minimises the sum of squares of these distances

M1

A1

[3]

1.1

2.4

Needs M2 and "minimises" and "sums of squares" oe

SC: Horizontal(s):

full marks (indept of (b))

OCR FS1 AS 2021 June Q1

1 On any day, the number of orders received in one randomly chosen hour by an online supplier can be modelled by the distribution $\mathrm { Po } ( 120 )$.

Find the probability that at least 28 orders are received in a randomly chosen 10 -minute period.
Find the probability that in a randomly chosen 10-minute period on one day and a randomly chosen 10-minute period on the next day a total of at least 56 orders are received.
State a necessary assumption for the validity of your calculation in part (b).

OCR FS1 AS 2021 June Q2

2 The members of a team stand in a random order in a straight line for a photograph. There are four men and six women.

Find the probability that all the men are next to each other.
Find the probability that no two men are next to one another.

OCR FS1 AS 2021 June Q3

28 marks

3 Sixteen candidates took an examination paper in mechanics and an examination paper in statistics.

For all sixteen candidates, the value of the product moment correlation coefficient $r$ for the marks on the two papers was 0.701 correct to 3 significant figures. Test whether there is evidence, at the $5 \%$ significance level, of association between the marks on the two papers.
A teacher decided to omit the marks of the candidates who were in the top three places in mechanics and the candidates who were in the bottom three places in mechanics. The marks for the remaining 10 candidates can be summarised by
$n = 10 , \Sigma x = 750 , \Sigma y = 690 , \Sigma x ^ { 2 } = 57690 , \Sigma y ^ { 2 } = 49676 , \Sigma x y = 50829$.
1. Calculate the value of $r$ for these 10 candidates.
2. What do the two values of $r$, in parts (a) and (b)(i), tell you about the scores of the sixteen candidates? A bag contains a mixture of blue and green beads, in unknown proportions. The proportion of green beads in the bag is denoted by $p$.
Sasha selects 10 beads at random, with replacement. Write down an expression, in terms of $p$, for the variance of the number of green beads Sasha selects. Freda selects one bead at random from the bag, notes its colour, and replaces it in the bag. She continues to select beads in this way until a green bead is selected. The first green bead is the $X$ th bead that Freda selects.
Assume that $p = 0.3$. Find
1. $\mathrm { P } ( X \geqslant 5 )$,
2. $\operatorname { Var } ( X )$.

In fact, on the basis of a large number of observations of $X$, it is found that $\mathrm { P } ( X = 3 ) = \frac { 4 } { 25 } \times \mathrm { P } ( X = 1 )$. Estimate the value of $p$. \section*{Total Marks for Question Set 4: 29} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'. We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s . f . }$ unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
oe	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

Question

Answer

Marks

AO

Guidance

1

(a)

$1 - \mathrm { P } ( \leq 27 )$

$= 0.0525 \quad \mathbf { B C }$

M1

A1

[2]

3.4

1.1

Allow M1 for $1 - 0.9657 = 0.0343$

In range [0.0524, 0.0525] BC

1

(b)

Po(40)

$1 - \mathrm { P } ( \leq 55 )$

$= 0.00968$

M1*

depM1

A1

[3]

3.1b

1.1a

1.1

$\operatorname { Po( } 2 \times$ their 20) stated or implied

Allow M1 for $1 - \mathrm { P } ( \leq 56 ) = 0.00658$ or $1 - 0.990 = 0.01$ or 0.0097

Awrt 0.00968

1

(c)

Orders on one day are independent of orders on the other

B1

[1]

3.2b

Use "orders independent", clearly referred to the two different days, needs context [not "events"], and nothing else

Not anything affecting given separate Poissons, such as "orders must be independent" or "constant average rate".

\multirow[t]{3}{*}{2}

\multirow[t]{3}{*}{(a)}

\multirow{3}{*}{}

$7 ! \times 4$ !

$\div 10$ ! $= \frac { 120960 } { 3628800 } = \frac { 1 } { 30 }$

M1

A1

2.1

1.1a

1.1

Allow for $6 ! \times 4$ ! or $6 ! \times 4 ! \times 2$

Divide by 10!, needs at least one factorial in numerator

Answer, exact or awrt 0.0333

3/3 for $\frac { 1 } { 30 }$ www

Alternative: $7 \times \frac { 6 } { 10 } \times \frac { 4 } { 9 } \times \frac { 5 } { 8 } \times \frac { 3 } { 7 } \times \frac { 4 } { 6 } \times \frac { 2 } { 5 } \times \frac { 3 } { 4 } \times \frac { 1 } { 3 } \times \frac { 2 } { 2 } \times \frac { 1 } { 1 }$

M1

A1

no 7, one other error

only one error

correct answer

[3]

\multirow[t]{3}{*}{2}

\multirow[t]{3}{*}{(b)}

\multirow{3}{*}{}

Women placed in 6 ! ways, men in 4 ! $[ = 720 \times 24 ]$

4 slots $m$ in $m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m = { } ^ { 7 } C _ { 4 }$ ${ } ^ { 7 } C _ { 4 } \times \frac { 6 ! \times 4 ! } { 10 ! }$

$= \frac { 1 } { 6 }$

B1

М1

A1

2.1

3.1b

1.1a

1.1

$6 ! \times 4 !$ anywhere, or $6 ! \times$ attempt at ${ } ^ { 7 } P _ { 4 }$

Or ${ } ^ { 7 } P _ { 4 }$. Allow for $m$ and $W$ reversed

Needs attempt at both terms

Or 0.167 or 0.1667 etc

Or $6 ! \times 7 \times 6 \times 5 \times 4 \quad$ B2

${ } ^ { 7 } P _ { 4 } \times 6$ !: B1M1

$4 \times ( 6 ! \times 4 ! ) / 10 ! = 2 / 105 :$ В 1 М 1

Alternative: PIE

$( 10 ! - 12 \times 9 ! + ( 3 \times 4 \times 8 ! + 12 \times 2 \times 8 ! ) - 24 \times 7 ! ) / 10 !$

$[ = ( 3628800 - 4354560 + 1451520 - 120960 ) / 10 ! ]$

M2

A1

Signs alternating, at least one term $\sqrt { }$ Allow one term omitted or wrong Correct answer

[3]

Three together: $7 \times 6 \times \frac { 6 ! 4 ! } { 10 ! } = \frac { 1 } { 5 }$

Two pairs: $\frac { 7 \times 6 } { 2 } \times \frac { 6 ! 4 ! } { 10 ! }$

$\frac { 1 } { 10 }$

One pair: $7 \times \frac { 6 \times 5 } { 2 } \times \frac { 6 ! 4 ! } { 10 ! } = \frac { 1 } { 2 }$

Question

Answer

Marks

AO

Guidance

3

(a)

$\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho \neq 0$, where $\rho$ is population pmcc

0.701 > 0.4973

Reject $\mathrm { H } _ { 0 }$. There is significant evidence of association between the marks on the two papers

B1

M1ft

A1

[4]

2.5

1.1a

1.1

2.2b

Must use symbols. Allow no definition of letter if $\rho$ used

Correct CV stated, allow 0.497

FT on wrong CV

Not FT. Needs context, and not too definite.

Not " $\mathrm { H } _ { 0 }$ : there is no assoc' n , $\mathrm { H } _ { 1 }$ : there is association"

Not There is association ..."

3

(b)

(i)

-0.534

B2

[2]

1.1a

1.1

SC: if B0, give B1 for two of 1440, 2066, -921 and $S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)$

-0.53: B1

3

(b)

(ii)

6 candidates did very well or very badly on both papers; middle 10 tended to do badly on one paper and well on the other

B1

[1]

2.4

Correct inference about scores oe, not "correlation/association/value of $r$ ". Not "outliers" or "anomalies".

Allow inference for one group only, provided it is clearly for only one group \

any ref to other group is not wrong

4

(a)

$10 p ( 1 - p )$

B1

[1]

1.2

Allow $10 p q$ oe, e.g. $10 p - 10 p ^ { 2 }$

Not just $n p ( 1 - p )$

4

(b)

(i)

$0.7 ^ { 4 }$

= 0.240(1)

M1

A1

[2]

1.1a

1.1

$0.7 ^ { 5 } = 0.168$ or $0.7 ^ { 6 } = 0.118$ : M1

Allow 0.24

Or $1 - 0.3 \left( 1 + 0.7 + 0.7 ^ { 2 } + 0.7 ^ { 3 } \right)$ Allow M1 if also $0.3 \times 0.7 ^ { 4 }$ [0.15 is from binomial]

4

(b)

(ii)

$q / p ^ { 2 } = \frac { 70 } { 9 }$ or $7.777 \ldots$

B1

[1]

1.1

Allow 7.78, 7.778, etc

Allow 8 only if evidence, e.g. ( $1 - 0.3$ )/ $0.3 ^ { 2 }$

4

(c)

\(\begin{aligned}

( 1 - p ) ^ { 2 } p = \frac { 4 } { 25 } p

p = 0 \text { or } ( 1 - p ) ^ { 2 } = \frac { 4 } { 25 } \quad ( p \neq 0 )

( 1 - p ) = \pm \frac { 2 } { 5 }

p \neq \frac { 7 } { 5 }

p = \frac { 3 } { 5 } \end{aligned}\)

B1

M1

B1ft

A1

[5]

1.1

1.1a

1.1

2.3

2.1

Correct equation

Reduce to quadratic/cubic and solve

Obtain two non-zero solutions

Explicitly discard one solution, either here or in line 2 (not enough to give 2 answers and then only 1 )

Exact final answer exact (0.6) no others left, allow from ± omitted

e.g. $p \left( p ^ { 2 } - 2 p + \frac { 21 } { 25 } \right) = 0$ ± omitted: M0B0A1
Allow " $p = 0 , \frac { 3 } { 5 } , \frac { 7 } { 5 }$ but $p \leq 1$ "
SC binomial: B0 then \(75 p ^ { 2 } = ( 1 - p ) ^ { 2 } \	\) solve M1 $0.104 [ 0.1035 ] \quad$ A1 Explicitly reject 0 or - 0.13 B1 SC Poisson: 0

OCR FS1 AS 2021 June Q1

1 Five observations of bivariate data $( x , y )$ are given in the table.

$x$	7	8	12	6	4
$y$	20	16	7	17	23

Find the value of Pearson's product-moment correlation coefficient.
State what your answer to part (a) tells you about a scatter diagram representing the data.
A new variable $a$ is defined by $a = 3 x + 4$. Dee says "The value of Pearson's product-moment correlation coefficient between $a$ and $y$ will not be the same as the answer to part (a)." State with a reason whether you agree with Dee. An investor obtains data about the profits of 8 randomly chosen investment accounts over two one-year periods. The profit in the first year for each account is $p \%$ and the profit in the second year for each account is $q \%$. The results are shown in the table and in the scatter diagram.
Account A B C D E F G H
$p$ 1.6 2.1 2.4 2.7 2.8 3.3 5.2 8.4
$q$ 1.6 2.3 2.2 2.2 3.1 2.9 7.6 4.8
$n = 8 \quad \Sigma p = 28.5 \quad \Sigma q = 26.7 \quad \Sigma p ^ { 2 } = 136.35 \quad \Sigma q ^ { 2 } = 116.35 \quad \Sigma p q = 116.70$
\includegraphics[max width=\textwidth, alt={}, center]{4c7546b9-03ee-47a1-915f-41e2b4ca19c0-03_762_1248_906_260}
State which, if either, of the variables $p$ and $q$ is independent.
Calculate the equation of the regression line of $q$ on $p$.
1. Use the regression line to estimate the value of $q$ for an investment account for which $p = 2.5$.
2. Give two reasons why this estimate could be considered reliable.
Comment on the reliability of using the regression line to predict the value of $q$ when $p = 7.0$.

OCR FS1 AS 2021 June Q3

38 marks

3 At a cinema there are three film sessions each Saturday, "early", "middle" and "late". The numbers of the audience, in different age groups, at the three showings on a randomly chosen Saturday are given in Table 1. \begin{table}[h] \end{table}

Question

Solution

Marks

AOs

Guidance

1

(a)

-0.954 BC

B2 [2]

1.1 1.1

SC: If B0, give B1 if two of 7.04, 29.0[4], -13.6[4] (or 35.2, 145[.2], -68.2) seen

1

(b)

Points lie close to a straight line Line has negative gradient

B1 B1 [2]

2.2b 1.1

Must refer to line, not just "negative correlation"

1

(c)

No, it will be the same as $x \rightarrow a$ is a linear transformation

B1 [1]

2.2a

OE. Either "same" with correct reason, or "disagree" with correct reason. Allow any clear valid technical term

2

(a)

Neither

B1 [1]

1.2

2

(b)

$q = 1.13 + 0.620 p$

B1B1 B1 [3]

1.1,1.1 1.1

0.62(0) correct; both numbers correct Fully correct answer including letters

2

(c)

(i)

2.68

B1ft [1]

1.1

awrt 2.68, ft on their (b) if letters correct

2

(c)

(ii)

2.5 is within data range, and points (here) are close to line/well correlated

B1 B1 [2]

2.2b 2.2b

At least one reason, allow "no because points not close to line" Full argument, two reasons needed

2

(d)

Not much data here/points scattered/ possible outliers

So not very reliable

M1 A1 [2]

2.3 1.1

Reason for not very reliable (not "extrapolation") Full argument and conclusion, not too assertive (not wholly unreliable!)

3

(a)

Expected frequency for Middle/25 to 60 is 4.4 which is < 5 so must combine cells

B1*ft depB1 [2]

2.4 3.5b

Correctly obtain this $F _ { E }$, ft on addition errors " < 5" explicit and correct deduction

3

(b)

Early	Middle	Late
29.4	23.1	31.5
26.6	20.9	28.5

Early	Middle	Late
0.9918	0.4160	2.2937
1.0962	0.4598	2.5351

B1

1.1

Both, allow 28.4 for 28.5

awrt 2.29, but allow 2.3 In range [2.53, 2.54]

Question

Solution

Marks

AOs

Guidance

3

(c)

$\mathrm { H } _ { 0 }$ : no association between session and age group. $\mathrm { H } _ { 1 }$ : some association

$\Sigma X ^ { 2 } = 7.793$

$v = 2 , \chi ^ { 2 } ( 2 ) _ { \text {crit } } = 5.991$

Reject $\mathrm { H } _ { 0 }$.

Significant evidence of association between session attended and age group.

B1

M1ft

A1ft [5]

1.1

2.2b

Both. Allow "independent" etc

Correct value of $X ^ { 2 }$, awrt 7.79 (allow even if wrong in (b))

Correct CV and comparison

Correct first conclusion, FT on their TS only

Contextualised, not too assertive

3

(d)

The two biggest contributions to $\chi ^ { 2 }$ are both for the late session ... ... when the proportion of younger people is higher, and of older people is lower, than the null hypothesis would suggest.

M1ft

A1ft

[2]

1.1

2.4

Refer to biggest contribution(s), FT on their answers to (b), needs "reject $\mathrm { H } _ { 0 }$ "

Full answer, referring to at least one cell (ignore comments on next highest cells)

\multirow[t]{2}{*}{4}

\multirow{2}{*}{}

\multirow{2}{*}{OR:}

$\frac { { } ^ { 2 m } C _ { 2 } \times m } { { } ^ { 3 m } C _ { 3 } }$

$= \frac { 2 m ( 2 m - 1 ) } { 2 } \times m \div \frac { 3 m ( 3 m - 1 ) ( 3 m - 2 ) } { 6 }$

$= \frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) }$ $\frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) } = \frac { 28 } { 55 }$

$\Rightarrow 16 m ^ { 2 } - 71 m + 28 = 0$

$m = 4$ BC

Reject $m = \frac { 7 } { 16 }$ as $m$ is an integer

M1

A1

M1

A1

M1

A1

[7]

3.1b

2.1

3.1a

2.1

1.1

3.2a

Use ${ } ^ { 2 m } C _ { 2 }$ and $m$
Divide by ${ } ^ { 3 m } C _ { 3 }$
Correct expression in terms of $m$ (allow with $m$ not cancelled yet)
Equate to $\frac { 28 } { 55 }$ \	simplify to three-term quadratic
Correct simplified quadratic, or (quadratic) $\times m , = 0$, aef Solve to get both 4 and $\frac { 7 } { 16 }$
Explicitly reject $m = \frac { 7 } { 16 }$

$\frac { 2 m ( 2 m - 1 ) \times m \times 3 ! } { 3 m ( 3 m - 1 ) ( 3 m - 2 ) \times 2 }$ then as above

Multiplication method can get full marks, but if no 3 or 3 !, max

M1M0A0 M1A0M0A0

OCR FS1 AS 2021 June Q1

1 Every time a spinner is spun, the probability that it shows the number 4 is 0.2 , independently of all other spins.

A pupil spins the spinner repeatedly until it shows the number 4 . Find the mean of the number of spins required.
Calculate the probability that the number of spins required is between 3 and 10 inclusive.
Each pupil in a class of 30 spins the spinner until it shows the number 4 . Out of the 30 pupils, the number of pupils who require at least 10 spins is denoted by $X$. Determine the variance of $X$.

OCR FS1 AS 2021 June Q2

27 marks

2 After a holiday organised for a group, the company organising the holiday obtained scores out of 10 for six different aspects of the holiday. The company obtained responses from 100 couples and 100 single travellers. The total scores for each of the aspects are given in the following table. \end{table}

Question

Answer

Mark

AO

Guidance

1

(a)

$\frac { 1 } { 0.2 } = 5$

M1 A1 [2]

3.3 1.1

Geometric distribution soi 5 (or $5.00 \ldots$ ) only

1

(b)

$0.8 ^ { 2 } - 0.8 ^ { 10 }$ $= \mathbf { 0 . 5 3 3 } \quad ( 0.5326258 \ldots )$

М1 A1 [2]

1.1 3.4

Allow for powers 2, 3, 4 and 9, 10, 11 .

Awrt 0.533, www. [5201424/976562]

Or $0.2 \left( 0.8 ^ { 2 } + \ldots . + 0.8 ^ { 9 } \right) , \pm 1$ term at either end [0.506, 0.378, 0.275, 0.405, 0.302, 0.554, 0.426, 0.324]

1

(c)

$\mathrm { P } ( \geq 10 ) = 0.8 ^ { 9 }$

$= 0.1342 \ldots$

B(30, 0.1342...)

Variance $= n p q$ = 3.486…

M1

A1

М1

A1ft [4]

3.1b

1.1

3.1b

1.1

Or $0.8 ^ { 10 }$. Can be implied by correct $p$

[0.10737… is M1A0 here]

Stated or implied, their $0.8 ^ { 9 }$ or $0.8 ^ { 10 }$

In range [3.48, 3.49]

SC: 0.134(2) oe not properly shown: B2 for correct final answer.

SC: 2.875 from $0.8 ^ { 10 }$ : M1A0M1A1ft

Question

Answer

Mark

AO

Guidance

2

(a)

Test is for rankings/rankings arbitrary/not bivariate normal etc

B1 [1]

2.4

OE

2

(b)

$\mathrm { H } _ { 0 } : \rho _ { s } = 0 , \mathrm { H } _ { 1 } : \rho _ { s } > 0$, where $\rho _ { s }$ is the population rank correlation coefficient

Ranks 543612

512643

$\Sigma d ^ { 2 } = 20$

$r _ { s } = 1 - \frac { 6 \times 20 } { 6 \times 35 }$

$= 3 / 7$ or $0.42857 \ldots$

<0.9429

B1

M1

A1

B1

1.1

Allow $\rho _ { s }$ not defined; allow $\rho$.

Allow: $\mathrm { H } _ { 0 }$ : no association between rankings.

$\mathrm { H } _ { 1 }$ : positive association (but not $\mathrm { H } _ { 1 }$ : association)

Do not reject $\mathrm { H } _ { 0 }$

Insufficient evidence of association between ranking given by the two categories

M1ft

A1ft

[7]

1.1

2.2b

FT on their $\Sigma d ^ { 2 }$ only

2

(c)

Not dependent on any distributional assumptions

B1

[1]

1.2

Oe (cf. Specification, 5.08f)

Question

Answer

Mark

AO

Guidance

3

(a)

Failures occur to no fixed pattern/are not predictable

B1 [1]

1.1

OE. NOT "independent"

3

(b)

Failures occur independently of one another and at constant average rate

B1

[2]

1.1

Not recoverable from (a) if independence not restated here; must be contextualised

Ignore "singly". Allow "uniform" rate, not "constant rate" or "constant probability"; must be contextualised

3

(c)

Variance (1.6384) $\approx$ mean

So suggests that it is likely to be well modelled

M1

A1

[2]

1.1

3.5a

Compare variance (or SD). Allow square/square-root confusion

Correct comparison and conclusion, 1.64 or better seen

3

(d)

$\mathrm { e } ^ { - 1.61 }$

B1

[1]

3.4

Exact needed, allow even if $0 !$ or $1.61 ^ { 0 }$ or both left in

3

(e)

1	$\geq 2$
0.322	0.478

B1

[2]

3.4

1.1

One correct e.g. 0.3218

Other correct e.g. 0.4783

3

(f)

$\mathrm { P } ( F = 1 )$ will be smaller as single failures are less likely

B1*

depB1

[2]

3.5c

3.3

OE. Partial answer: B1

OCR Further Mechanics 2021 June Q2

2 A solenoid is a device formed by winding a wire tightly around a hollow cylinder so that the wire forms (approximately) circular loops along the cylinder (see diagram).
\includegraphics[max width=\textwidth, alt={}, center]{91098ecb-fb4a-44aa-9e59-6c6fe3704966-02_164_697_1484_233} When the wire carries an electrical current a magnetic field is created inside the solenoid which can cause a particle which is moving inside the solenoid to accelerate. A student is carrying out experiments on particles moving inside solenoids. His professor suggests that, for a particle of mass $m$ moving with speed $v$ inside a solenoid of length $h$, the acceleration $a$ of the particle can be modelled by a relationship of the form $a = k m ^ { \alpha } v ^ { \beta } h ^ { \gamma }$, where $k$ is a constant. The professor tells the student that $[ k ] = \mathrm { MLT } ^ { - 1 }$.

Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
The mass of an electron is $9.11 \times 10 ^ { - 31 } \mathrm {~kg}$ and the mass of a proton is $1.67 \times 10 ^ { - 27 } \mathrm {~kg}$. For an electron and a proton moving inside the same solenoid with the same speed, use the model to find the ratio of the acceleration of the electron to the acceleration of the proton.
The professor tells the student that $a$ also depends on the number of turns or loops of wire, $N$, that the solenoid has. Explain why dimensional analysis cannot be used to determine the dependence of $a$ on $N$.

OCR Further Mechanics 2021 June Q3

3 A right circular cone $C$ of height 4 m and base radius 3 m has its base fixed to a horizontal plane. One end of a light elastic string of natural length 2 m and modulus of elasticity 32 N is fixed to the vertex of $C$. The other end of the string is attached to a particle $P$ of mass 2.5 kg .
$P$ moves in a horizontal circle with constant speed and in contact with the smooth curved surface of $C$. The extension of the string is 1.5 m .

Find the tension in the string.
Find the speed of $P$.

OCR Further Mechanics 2021 June Q4

33 marks

4 Two particles $A$ and $B$, of masses $m \mathrm {~kg}$ and 1 kg respectively, are connected by a light inextensible string of length $d \mathrm {~m}$ and placed at rest on a smooth horizontal plane a distance of $\frac { 1 } { 2 } d \mathrm {~m}$ apart. $B$ is then projected horizontally with speed $v \mathrm {~ms} ^ { - 1 }$ in a direction perpendicular to $A B$.

Show that, at the instant that the string becomes taut, the magnitude of the instantaneous impulse in the string, $I \mathrm { Ns }$, is given by $I = \frac { \sqrt { 3 } m v } { 2 ( 1 + m ) }$.
Find, in terms of $m$ and $v$, the kinetic energy of $B$ at the instant after the string becomes taut. Give your answer as a single algebraic fraction.

In the case where $m$ is very large, describe, with justification, the approximate motion of $B$ after the string becomes taut. \section*{Total Marks for Question Set 1: 36} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

Question

Answer

Marks

AOs

Guidance

4

(a)

$\cos \theta = \frac { 1 } { 2 }$ or $\sin \theta = \frac { \sqrt { 3 } } { 2 }$

M1

3.1b

Must be clear where it comes from ie. Shown in diagram or clear use of distances. May be other way round

$\theta$ is the angle between the string and the direction of $A B$ when the string initially becomes taut

A1

3.1b

Working must be seen

\(\begin{aligned}

v \sin \theta = ( 1 + m ) V

V = \frac { \sqrt { 3 } v } { 2 ( 1 + m ) }

I = m V \text { or } I = \pm ( V - v \sin \theta )

V = \frac { v \times \frac { 1 } { 2 } \sqrt { 3 } } { 1 + m } \Rightarrow I = \frac { \sqrt { 3 } m v } { 2 ( 1 + m ) } \end{aligned}\)

A1

2.2a

AG Justification of using Impulse = change in momentum

[4]

4

(b)

\(\begin{aligned}

\mathrm { KE } = \frac { 1 } { 2 } \left( V ^ { 2 } + ( v \cos \theta ) ^ { 2 } \right) \text { soi }

\mathrm { KE } = \frac { 1 } { 2 } \left( \left( \frac { \sqrt { 3 } v } { 2 ( 1 + m ) } \right) ^ { 2 } + \left( v \times \frac { 1 } { 2 } \right) ^ { 2 } \right)

\mathrm { KE } = \frac { v ^ { 2 } \left( 4 + 2 m + m ^ { 2 } \right) } { 8 ( 1 + m ) ^ { 2 } } \end{aligned}\)

M1

3.1b

Transverse component unchanged and using their longitudinal component

Condone consideration of speed squared or inclusion of m in KE

М1

1.1

Substituting in for $V$ and $\cos \theta$ (could just be in speed equation)

A1

1.1

Or any equivalent single algebraic fraction

[3]

4

(c)

$V = \frac { \sqrt { 3 } v } { 2 ( 1 + m ) } \rightarrow 0 \text { as } m \rightarrow \infty$

$B$ moves (approximately) in a circle around $A$

B1

3.2a

If m is very large then $A$ is approximately stationary or $B$ has only its transverse velocity of $\frac { 1 } { 2 } v$ after the string becomes taut

B1

2.2b

[2]

\(w\)	1	2	3	4
\(\mathrm { P } ( W = w )\)	0.25	0.36	\(x\)	\(x ^ { 2 }\)

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Account	A	B	C	D	E	F	G	H
\(p\)	1.6	2.1	2.4	2.7	2.8	3.3	5.2	8.4
\(q\)	1.6	2.3	2.2	2.2	3.1	2.9	7.6	4.8