(a) Confetti for the party is packed in small boxes. Each box is in the shape of a triangular prism. The cross-section of each box is an isosceles triangle. The measurements are shown on the diagram below.
Diagram not drawn to scale
(i) Show that the perpendicular height of the cross-section of a confetti box is 40 mm. You must show all your working. [3]
(ii) This is the label on a confetti box..
The volume of this box is at least 20 000 mm3.
Calculate the volume of a confetti box to show that the statement on the label is correct. [3]
(b) Rhodri finds 2 venues that arrange party nights.
Friar Hall Party night special
£105 hall hire charge + £5 per person
Minfelin Lodge Party night special
£207 room hire charge + £3 per person
Rhodri calculates the total cost of organising the party at each venue. He finds that the total costs are the same. For how many people is Rhodri planning the 21st birthday party? You must show all your working. [3]
From the formula list: Volume of prism = area of cross-section × length
Mark scheme (a)
Scheme
Marks
Comments
(i) Correct statement of Pythagoras’ theorem
(Height2 =) 502 – (60 ÷ 2)2
(Height2 =) 502 – 302
502 = height2 + (60 ÷ 2)2
502 = height2 + 302
M1
Clear indication that all measurements have been converted to 3cm, 5cm, 4cm may be awarded all marks
Correct stage of evaluation
(Height2 =) 2500 – 900
(Height2 =) 1600
sight of \(\sqrt{1600}\)
(Height =) \(\sqrt{(50^{2} - 30^{2})}\)
A1
Working must be seen Allow M1 A1 for a slip in the initial notation then corrected at this evaluation stage
Mark final answer A0 for an incorrect statement, e.g. \(\sqrt{1600}\) = 402
1(a)(i) Alternative method 1
Scheme
Marks
Comments
Identifies the relationship ‘3, 4, 5’ and relates to the given (right-angled) triangle, e.g. sight of
3, 4, 5 and 30(mm), 40(mm), 50(mm)
3cm, 4cm, 5cm
3, 4, 5 and ‘× 10’
30, 40, 50 and ‘÷ 10’
AND a statement or conclusion, e.g.
Pythagorean triple
Right-angled triangle
3, 4, 5 triangle means it would be 30, 40, 50 triangle
B3
For B3 there must be an accompanying statement or conclusion
B2 for identifying the relationship ‘3, 4, 5’ and relates to the given(right-angled) triangle
without a conclusion or statement, or
with an incorrect conclusion or statement
B1 for sight of any one of the following:
‘3, 4, 5’
30 (mm) and 40 (mm) appropriately indicated on the diagram
A right-angled triangle drawn (with or without 90° indicated) appropriately labelled 30 (mm), 40 (mm) and 50 (mm)
1(a)(i) Alternative method 2
Scheme
Marks
Comments
Assuming height as 40mm with use of 50mm or 30mm within a correct statement of Pythagoras’ Theorem, e.g.
((½ base)2 =) 502 – 402
502 = 402 + x2
((hypotenuse)2 =) 402 + 302
M1
Clear indication that all measurements have been converted to 3cm, 5cm, 4cm may be awarded all marks
Correct stage of evaluation, e.g.
((½ base)2 = 502 – 402 =) 900
( ½ base =) \(\sqrt{900}\)
((hypotenuse)2 = 402 + 302 =) 2500
(hypotenuse =) \(\sqrt{2500}\)
A1
Working must be seen
Appropriate full evaluation, e.g.
( ½ base =) 30 (mm)
(hypotenuse =) 50 (mm)
A1
Mark final answer
Scheme
Marks
Comments
(ii) (Volume) ½ × 60 × 40 × 20 or equivalent
M2
M1 for sight of area of X-section possibly in stages, ½×60×40 or ½×30×40 + ½×30×40 (= 1200 mm2)
24 000 (mm3) (> 20 000 mm3)
A1
CAO
Corrected by Westie's Workshop
The board's mark scheme printed “(hypotenuse =) ( ½ base =) \(\sqrt{2500}\)”. Changed to “(hypotenuse =) \(\sqrt{2500}\)” because the square root of 2500 is the hypotenuse (50 mm), not half the base (30 mm), so the extra “( ½ base =)” was removed.
Original: see the board's mark scheme PDF (link above).
Mark scheme (b)
Scheme
Marks
Comments
Sight of or implication that: 5 × number of people + 105 = 207 + 3 × number of people or 5x + 105 = 207 + 3x
M1
Implication includes attempt to balance costing for the same number of people ≥ 3 at each venue, e.g.
(10 people) 5 × 10 + 105 with 207 + 3 × 10
(110, 115,) 120 with (210, 213,) 216
(5 – 3) × number of people = 207 – 105 or number of people = \(\frac{207-105}{5-3}\) or 5x – 3x = 207 – 105 or 2x = 102
m1
Includes correctly evaluated trial to attempt to balance costing for the same number of people at each venue provided ‘their trial for 30 ≤ the number of people ≤ 70’, e.g. correct costing for both venues for 40 people as (FH) (£)305 and (ML) (£)327
From M1, allow 1 slip in the rearrangement of ‘their equation’ provided ‘their equation’ is then simplified to ax = b, where a ≠ 0 and b ≠ 0
Sight of cost (£)360 for each venue implies M1 m1
51 (people)
A1
CAO
If no marks, award SC1 for finding the number of (whole) people for the same cost at each venue, provided the cost is > (£) 220, e.g.
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