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Date November 2020 Marks available 1 Reference code 20N.2.SL.TZ0.T_5
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Write down Question number T_5 Adapted from N/A

Question

A large underground tank is constructed at Mills Airport to store fuel. The tank is in the shape of an isosceles trapezoidal prism, ABCDEFGH.

AB=70m , AF=200m, AD=40m, BC=40m and CD=110m. Angle ADC=60° and angle BCD=60°. The tank is illustrated below.

Once construction was complete, a fuel pump was used to pump fuel into the empty tank. The amount of fuel pumped into the tank by this pump each hour decreases as an arithmetic sequence with terms u1, u2, u3, , un.

Part of this sequence is shown in the table.

At the end of the 2nd hour, the total volume of fuel in the tank was 88200m3.

Find h, the height of the tank.

[2]
a.

Show that the volume of the tank is 624000m3, correct to three significant figures.

[3]
b.

Write down the common difference, d.

[1]
c.

Find the amount of fuel pumped into the tank in the 13th hour.

[2]
d.

Find the value of n such that un=0.

[2]
e.i.

Write down the number of hours that the pump was pumping fuel into the tank.

[1]
e.ii.

Find the total amount of fuel pumped into the tank in the first 8 hours.

[2]
f.

Show that the tank will never be completely filled using this pump.

[3]
g.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

sin60°=h40  OR  tan60°=h20           (M1)


Note:
 Award (M1) for correct substitutions in trig ratio.


OR

202+h2=402     402-202           (M1)


Note: Award (M1) for correct substitutions in Pythagoras’ theorem.


h= 34.6 m  1200, 203, 34.6410       (A1)(G2)


[2 marks]

a.

1270+11034.6410×200           (M1)(M1)


Note:
Award (M1) for their correctly substituted area of trapezium formula, provided all substitutions are positive. Award (M1) for multiplying by 200. Follow through from part (a).


OR


2×12×20×34.6410+70×34.6410×200           (M1)(M1)


Note: Award (M1) for the addition of correct areas for two triangles and one rectangle. Award (M1) for multiplying by 200. Follow through from part (a).


OR


70×34.6410×200+2×12×34.6410×20×200           (M1)(M1)


Note:
 Award (M1) for their correct substitution in volume of cuboid formula. Award (M1) for correctly substituted volume of triangular prism(s). Follow through from part (a).


623538         (A1)

624000m3            (AG)


Note:
Both an unrounded answer that rounds to the given answer and the rounded value must be seen for the (A1) to be awarded.


[3 marks]

b.

d=  -1800           (A1)


[1 mark]

c.

u13=   45000+13-1-1800           (M1)


Note:
Award (M1) for correct substitutions in arithmetic sequence formula.
OR
Award (M1) for a correct 4th term seen as part of list.


23400  m3        (A1)(ft)(G2)


Note:
Follow through from part (c) for their value of d.


[2 marks]

d.

0=45000+n-1-1800           (M1)


Note:
Award (M1) for their correct substitution into arithmetic sequence formula, equated to zero.


n=   26        (A1)(ft)(G2)


Note:
Follow through from part (c). Award at most (M1)(A0) if their n is not a positive integer.


[2 marks]

e.i.

25           (A1)(ft)


Note:
Follow through from part (e)(i), but only if their final answer in (e)(i) is positive. If their n in part (e)(i) is not an integer, award  (A1)(ft) for the nearest lower integer.


[1 mark]

e.ii.

S8= 822×45000+8-1×-1800           (M1)


Note:
Award (M1) for their correct substitutions in arithmetic series formula. If a list method is used, award (M1) for the addition of their 8 correct terms.


310000 m3 309600       (A1)(ft)(G2)


Note: Follow through from part (c). Award at most (M1)(A0) if their final answer is greater than 624000.


[2 marks]

f.

S25= 2522×45000+25-1×-1800  ,  S25= 25245000+1800           (M1)


Note:
Award (M1) for their correct substitutions into arithmetic series formula.


S25=585000 m3       (A1)(ft)(G1)


Note: Award (M1)(A1) for correctly finding S26=585000 m3, provided working is shown e.g. S26= 2622×45000+26-1×-1800 , S26= 26245000+0. Follow through from part (c) and either their (e)(i) or (e)(ii). If d<0 and their final answer is greater than 624000, award at most (M1)(A1)(ft)(R0). If d>0, there is no maximum, award at most (M1)(A0)(R0). Award no marks if their number of terms is not a positive integer.


585000 m3<624000 m3        (R1)

Hence it will never be filled        (AG)


Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.
For unsupported S25=585000 seen, award at most (G1)(R1)(AG). Working must be seen to follow through from parts (c) and (e)(i) or (e)(ii).


OR


Sn= n22×45000+n-1×-1800           (M1)


Note: Award (M1) for their correct substitution into arithmetic series formula, with n.


Maximum of this function 585225 m3       (A1)


Note: Follow through from part (c). Award at most (M1)(A1)(ft)(R0) if their final answer is greater than 624000. Award at most (M1)(A0)(R0) if their common difference is not 1800. Award at most (M1)(A0)(R0) if 585225 is not explicitly identified as the maximum of the function.


585225 m3<624000 m3        (R1)

Hence it will never be filled        (AG)


Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.


OR


sketch with concave down curve and labelled 624000 horizontal line           (M1)


Note: Accept a label of “tank volume” instead of a numerical value. Award (M0) if the line and the curve intersect.


curve explicitly labelled as Sn= n22×45000+n-1×-1800 or equivalent       (A1)

Note:
Award (A1) for a written explanation interpreting the sketch. Accept a comparison of values, e.g 585225 m3<624000 m3, where 585225 is the graphical maximum. Award at most (M1)(A0)(R0) if their common difference is not 1800.


the line and the curve do not intersect        (R1)

hence it will never be filled        (AG)


Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.


OR


624000=n22×45000+n-1×-1800           (M1)


Note: Award (M1) for their correctly substituted arithmetic series formula equated to 624000 (623538).


Demonstrates there is no solution       (A1)


Note: Award (A1) for a correct working that the discriminant is less than zero OR correct working indicating there is no real solution in the quadratic formula.


There is no (real) solution (to this equation)       (R1)

hence it will never be filled        (AG)


Note: At most (M1)(A0)(R0) for their correctly substituted arithmetic series formula =624000, 623538 or 622800 with a statement "no solution". Follow through from their part (b).


[3 marks]

g.

Examiners report

[N/A]
a.
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b.
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c.
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d.
[N/A]
e.i.
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e.ii.
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f.
[N/A]
g.

Syllabus sections

Topic 1—Number and algebra » SL 1.2—Arithmetic sequences and series
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Topic 1—Number and algebra
Prior learning

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