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Mathematics · Oct / Nov 2021 · Paper 31 · Question 3

YMA01

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The question as text

3. G O k t Figure 2 The total mass of gold, G tonnes, extracted from a mine is modelled by the equation G = 40 – 30e1–0.05t t  k G  0 where t is the number of years after 1st January 1800. Figure 2 shows a sketch of G against t. Use the equation of the model to answer parts (a), (b) and (c).

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  • 3(a)[3]

    40

    1 0.05

    −

    t

    0

    =

    40

    −

    30e

    ⇒−

    1 0.05

    t

    =

    ln 30

    (

    ) awrt 14.2

    ⇒

    k

    =

    (ii)

    March 1814

    1 0.05 70

    −

    ×

    Attempts

    G

    =

    40

    −

    30e

    =

    37.5

    tonnes

    40 (tonnes)

    Mark (a)(i) and (ii) together

    (a) (i)

    G =

    0

    M1

    Sets

    and proceeds to a linear equation in t or k using a correct method by

    taking logs of both sides, but condone slips in the rest of their rearrangement.

    4

    (

    ) awrt 14.2

    A1

    k =

    or 14.25 but allow the exact value

    20 1 ln 3

    −

    or other

    4

    ln

    1

    −

    4

    3

    3

    equivalent exact expressions such as

    or

    20 1 ln 3

    −

    or

    20ln

    e

    . Isw

    −

    0.05

    4

    after a correct answer.

    (a)(ii)

    A1

    March 1814 but allow April 1814 following ''correct'' k.

    Condone “the third month of 1814” or “the fourth month of 1814” following

    ''correct'' k

    Withhold this mark if they have two answers of which one is incorrect.

    (b)

    M1

    Attempts to find G with t =70 (condone t =69) which may be implied by awrt 37.5

    1 0.05 70

    −

    ×

    The numbers embedded in the equation is sufficient eg

    G

    =

    40

    −

    30e

    awrt 37.5tonnes (requires units)

    A1

    (c)

    B1

    40 (tonnes). Condone

    G <

    40

    or

    G „

    40

    . Do not allow

    G

    >

    40

    (

    )

  • 3(b)[2]

    1 0.05 70

    −

    ×

    Attempts

    G

    =

    40

    −

    30e

    =

    37.5

    tonnes

  • 3(c)[1]

    40 (tonnes)

    Mark (a)(i) and (ii) together

    (a) (i)

    G =

    0

    M1

    Sets

    and proceeds to a linear equation in t or k using a correct method by

    taking logs of both sides, but condone slips in the rest of their rearrangement.

    4

    (

    ) awrt 14.2

    A1

    k =

    or 14.25 but allow the exact value

    20 1 ln 3

    −

    or other

    4

    ln

    1

    −

    4

    3

    3

    equivalent exact expressions such as

    or

    20 1 ln 3

    −

    or

    20ln

    e

    . Isw

    −

    0.05

    4

    after a correct answer.

    (a)(ii)

    A1

    March 1814 but allow April 1814 following ''correct'' k.

    Condone “the third month of 1814” or “the fourth month of 1814” following

    ''correct'' k

    Withhold this mark if they have two answers of which one is incorrect.

    (b)

    M1

    Attempts to find G with t =70 (condone t =69) which may be implied by awrt 37.5

    1 0.05 70

    −

    ×

    The numbers embedded in the equation is sufficient eg

    G

    =

    40

    −

    30e

    awrt 37.5tonnes (requires units)

    A1

    (c)

    B1

    40 (tonnes). Condone

    G <

    40

    or

    G „

    40

    . Do not allow

    G

    >

    40

    (

    )

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