Higher June 2023 Paper 4 Q14
14 A college offers 41 different subjects including 9 different languages.
Students are asked to choose one subject from Option A, one subject from Option B and one subject from Option C.
Each of the 41 different subjects appears only once, either in Option A, or in Option B or in Option C.
Option A : 14 subjects including 2 languages
Option B : 12 subjects including 3 languages
Option C : 15 subjects including 4 languages
Work out the proportion of all the possible subject combinations that include at least one language.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{1332}{2520}\) oe or 0.53 or 0.529 or 0.5286 or 0.52857… or 53% or 52.9%, 52.86% or 52.857% with correct working | 5 | M4 for \(1 - \frac{12}{14} \times \frac{9}{12} \times \frac{11}{15}\) oe OR M3 for \(\frac{12}{14} \times \frac{9}{12} \times \frac{11}{15}\) OR M1 for \(\frac{2}{14} \times \frac{3}{12} \times \frac{4}{15}\) OR M1 for [total choices=] 14 × 12 × 15 implied by 2520 and M3 for (2 × 12 × 15) + (12 × 3 × 15) + (12 × 9 × 4) or 360 + 540 + 432 implied by 1332 or M2 for (2 × 12 × 15) + (14 × 3 × 15) + (14 × 12 × 4) or 360 + 630 + 672 implied by 1662 or M1 for one bracketed term correct from the M3 expression e.g 2 × 12 × 15 OR M1 for [total choices=] 14 × 12 × 15 implied by 2520 and M2 for [choices no languages=] (14 – 2) × (12 – 3) × (15 – 4) implied by 1188 or M1 for this expression with one error and M1 for their 2520 − their 1188 or 1332 If 0 or M1 scored SC2 for correct answer with no or insufficient working Note : For MR see appendix | “Correct working requires evidence of at least M3 or M1M1M1 Equivs. Include \(\frac{666}{1260}, \frac{333}{630}, \frac{111}{210}, \frac{37}{70}\) condone \(\frac{1}{14} \times \frac{1}{12} \times \frac{1}{15} \left[= \frac{1}{2520}\right]\) for M1 [total choices] Alternative equivalent methods M4 for fully correct method leading to their 2520 and their 1332 or M1 for [total choices=] 14 × 12 × 15 implied by 2520 and M3 for (2×3×4) + (12×3×4+2×9×4+2×3×11) + (2×9×11+12×3×11+12×9×4) or (24) + (144 + 72 + 66) + (198 + 396 + 432) implied by 1332 or M2 for this expression with at least two of the bracketed terms correct or M1 for at least one of the bracketed terms correct or any three of the individual terms correct e.g 2 × 3 × 4 Use of tree diagrams See appendix |
Appendix: Question 14
Using tree diagrams :
M4 for method A: \(1 - \frac{12}{14} \times \frac{9}{12} \times \frac{11}{15}\) oe
e.g. method B : \(\frac{2}{14} \times \frac{3}{12} \times \frac{4}{15} + \frac{2}{14} \times \frac{3}{12} \times \frac{11}{15} + \frac{2}{14} \times \frac{9}{12} \times \frac{4}{15} + \frac{12}{14} \times \frac{3}{12} \times \frac{4}{15} + \frac{2}{14} \times \frac{9}{12} \times \frac{11}{15} + \frac{12}{14} \times \frac{3}{12} \times \frac{11}{15} + \frac{12}{14} \times \frac{9}{12} \times \frac{4}{15}\)
M3 for \(\frac{12}{14} \times \frac{9}{12} \times \frac{11}{15}\) or method B with at least five relevant branches correct
or
M2 for method B with at least four relevant branches correct
or
M1 for method B with at least one relevant branch correct
Misreads
Some candidates read this as 16, 15 and 19 subjects. Treat this as a misread (MR) so the mark scheme for them will be (max. of 4 marks):
M4 for \(1 - \frac{14}{16} \times \frac{12}{15} \times \frac{15}{19}\) oe implied by answer \(\frac{2040}{4560} = \frac{102}{228} = \frac{51}{114} = \frac{17}{38}\) or 0.447368…..etc rot to at least 3 s.f.
OR
M3 for \(\frac{14}{16} \times \frac{12}{15} \times \frac{15}{19}\) oe implied by \(\frac{2520}{4560} = \frac{126}{228} = \frac{63}{114} = \frac{21}{38}\) or 0.55263…etc rot to at least 3 s.f.
OR
M1 for [total choices=] 16 × 15 × 19 implied by 4560
AND
M3 for 2 × 15 × 19 + 14 × 3 × 19 + 14 × 12 × 4 or 570 + 798 + 672 implied by 2040
or M2 for 2 × 15 × 19 + 16 × 3 × 19 + 16 × 15 × 4 or 570 + 912 + 960 implied by 2442
or M1 for this expression with one term correct or at least 5 numbers correct
OR
M1 for [total choices=] 16 × 15 × 19 implied by 4560
M2 for [choices no langs =] (16 – 2) × (15 – 3) × (19 – 4) implied by 2520 or M1 for this expression with one error
and
M1 for their 4560 − their 2520
Alternative equivalent methods
M4 for fully correct method leading to their 4560 and their 2040
or
M1 for [total choices=] 16 × 15 × 19 implied by 4560
and
M3 for (2×3×4) + (14×3×4 + 2×12×4 + 2×3×15) + (2×12×15 + 14×3×15 + 14×12×4) or (24) + (168 + 96 + 90) + (360 + 630 + 672) or 2040
or
M2 for two of the bracketed terms
or
M1 for one of the bracketed terms or any three of the individual terms correct