2024 NCHS Prelim Math P2 Suggested Solutions

NCHS 2024 EM Paper 2 Solutions 1 (a) 9𝑎2−4𝑏2 3𝑎𝑝+6𝑎𝑞−4𝑏𝑞−2𝑏𝑝 = (3𝑎+2𝑏)(3𝑎−2𝑏) (3𝑎𝑝+6𝑎𝑞)−(2𝑏𝑝+4𝑏𝑞) = (3𝑎+2𝑏)(3𝑎−2𝑏) 3𝑎(𝑝+2𝑞)−2𝑏(𝑝+2𝑞) = (3𝑎+2𝑏)(3𝑎−2𝑏) (𝑝+2𝑞)(3𝑎−2𝑏) = 3𝑎+2𝑏 𝑝+2𝑞 (b) 𝑥 6−7𝑥−5𝑥2 + 2 5𝑥−3 = 𝑥 −(5𝑥−3)(𝑥+2) + 2 5𝑥−3 = −𝑥+2(2+𝑥) (5𝑥−3)(𝑥+2) = −𝑥+4+2𝑥 (5𝑥−3)(𝑥+2) = 𝑥+4 (5𝑥−3)(𝑥+2) 1(c) − 1 2 < 2𝑦 5 − 1+𝑦 3 ≤ 5 6 − 1 2 < 3(2𝑦)−5(1+𝑦) 15 𝑎𝑛𝑑 3(2𝑦)−5(1+𝑦) 15 ≤ 5 6 − 1 2 < 6𝑦−5−5𝑦 15 6𝑦−5−5𝑦 15 ≤ 5 6 − 1 2 < 𝑦−5 15 𝑦−5 15 ≤ 5 6 −15 < 2(𝑦 − 5) 𝑦 − 5 ≤ 25 2 −15 < 2𝑦 − 10 𝑦 ≤ 17 1 2 −5 < 2𝑦 −2 1 2 < 𝑦 ≤ 17 1 2 2(a) 𝑟 19 = 25 40 𝑟 = 25 40 × 19 = 11.875 𝑉 = 1 3 𝜋(11.875)2(25) = 3691.78 𝑐𝑚3 = 3.69 𝑙𝑖𝑡𝑟𝑒𝑠 (b) Method 1 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟 = 1 3 𝜋(19)2(40) ( ℎ1 ℎ2 ) 3 = 𝑉1 𝑉2 ( ℎ1 40) 3 = 1 3𝜋(19)2(40)−1 3𝜋(11.875)2(25) 1 3𝜋(19)2(40) ℎ1 40 = √ (19)2(40)−(11.875)2(25) (19)2(40) 3 ℎ1 = 40 × √ (19)2(40)−(11.875)2(25) (19)2(40) 3 = 36.437 𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 40 − 36.437 = 3.56 𝑐𝑚 (3𝑠𝑓) Method 2 𝑟1 19 = ℎ1 40 𝑟1 = 19 40 ℎ1 1 3 𝜋(19)2(40) − 1 3 𝜋(𝑟1)2(ℎ1) = 3691.78 1 3 𝜋(𝑟1)2(ℎ1) = 11429.75 r 19 25 40 h1 40 r1 19 h1 h2

( 19 40 ℎ1) 2 (ℎ1) = 10914.61 ℎ1 3 = 48375.00277 ℎ1 = 36.437 𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 40 − 36.437 = 3.56 𝑐𝑚 (3𝑠𝑓) 3(a) 108888−100176.96 108888 × 100% = 8% (b) Balance owed = 85 100 × 108888 = $92554.80 Interest = 92554.80×2.98×5 100 = $13790.6652 Monthly instalment = 92554.80+13790.6652 5×12 = $1772.42 (2𝑑𝑝) (c) 𝑃 = 108888 − 18888 = $90000 𝐴 = 960 × 10 × 12 = $115200 𝐴 = 𝑃 (1 + 𝑅 100) 𝑛 115200 = 90000 (1 + 𝑅 100) 10 (1 + 𝑅 100) 10 = 115200 90000 1 + 𝑅 100 = √ 115200 90000 10 𝑅 = 100 ( √ 115200 90000 10 − 1) = 2.499 = 2.50 (3sf) 4(a) 𝑄 = ( 1.20 1.50 0.95 ) (b) 𝑅 = ( 60 68 55 49 56 71 53 70 80 ) ( 1.20 1.50 0.95 ) = ( 226.25 210.25 244.60 ) (c) The elements in R represent the total cost of baking caramel, strawberry and mint cupcakes in outlet A, B and C respectively (d) Method 1 Selling Price = ( 3 0 0 0 2.5 0 0 0 4 ) ( 1.20 1.50 0.95 ) = ( 3.60 3.75 3.80 ) Method 2 Profit for each type of cupcake = ( 2 0 0 0 1.5 0 0 0 3 ) ( 1.20 1.50 0.95 ) = ( 2.40 2.25 2.85 )

Profit earned in a week = ( 60 68 55 49 56 71 53 70 80 ) ( 3.60 3.75 3.80 ) − ( 226.25 210.25 244.60 ) = ( 680 656.20 757.30 ) − ( 226.25 210.25 244.60 ) = ( 453.75 445.95 512.70 ) Total Profit earned in a month = 4 [(1 1 1) ( 453.75 445.95 512.70 )] = 4(1412.40) = (5649.60) Total Profit = $5649.60 Profit earned in a week = ( 60 68 55 49 56 71 53 70 80 ) ( 2.40 2.25 2.85 ) = ( 453.75 445.95 512.70 ) Total Profit earned in a month = 4 [(1 1 1) ( 453.75 445.95 512.70 )] = 4(1412.40) = (5649.60) Total Profit = $5649.60 5(a) 𝑊𝑌 sin 102.6° = 12.5 sin 34° 𝑊𝑌 = 12.5 sin 34° × sin 102.6° = 21.815 𝑋𝑌2 = 10.32 + 21.8152 − 2(10.3)(21.815)(cos 45.3°) 𝑋𝑌 = 16.306 𝑜𝑟 − 16.306(𝑁𝐴) = 16.3 (3𝑠𝑓) (b) Method 1 ∠𝑁1𝑊𝑍 = 102.6° + 34° (𝑒𝑥𝑡 ∠ 𝑜𝑓 ∆) = 136.6° ∠𝑊𝑍𝑁2 = 180° − 136.6° (𝑖𝑛𝑡 ∠, 𝑁1𝑊//𝑁2𝑍) = 43.4° 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑜𝑓 𝑊 𝑓𝑟𝑜𝑚 𝑍 = 360° − 43.4° (∠ 𝑎𝑡 𝑎 𝑝𝑡) = 316.6° Method 2 ∠𝑌𝑊𝑍 = 180° − (34° + 102.6°) (∠ 𝑠𝑢𝑚 𝑜𝑓 ∆) = 43.4° ∠𝑊𝑍𝑁2 = 43.4° (𝑎𝑙𝑡 ∠, 𝑁1𝑊//𝑁2𝑍) 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑜𝑓 𝑊 𝑓𝑟𝑜𝑚 𝑍 = 360° − 43.4° (∠ 𝑎𝑡 𝑎 𝑝𝑡) = 316.6° (c) ∠𝑌𝑊𝑍 = 43.4° (𝑎𝑙𝑡 ∠, 𝑁1𝑌//𝑁2𝑍) 𝑙 = 12.5 sin 43.4° OR 1 2 × 21.815 × 𝑙 = 1 2 × 12.5 × 21.815 × sin 43.4° = 8.5886 𝑙 = 8.5886 Let the greatest angle of elevation be θ tan 𝜃 = 8−1.9 12.5 sin 43.4° 𝜃 = 35.38° = 35.4° W 12.5 m N1 X Y Z 102.6° 34° 45.3° 10.3 m N2 8.5886 1.9 8 θ W 12.5 Y Z 102.6° 34° 21.815 l

6(a) m = 4.2 (b) (c) Draw tangent at (1, 0.2) Gradient = −0.5 to −0.7 (d) 𝑥2 + 5 𝑥 − 4𝑥 − 4 = 0 𝑥2 5 + 1 𝑥 − 4 5 𝑥 − 4 5 = 0 𝑥2 5 + 1 𝑥 − 1 = 4 5 𝑥 − 1 5 𝑦 = 4 5 𝑥 − 1 5 Plot graph of 𝑦 = 4 5 𝑥 − 1 5 𝑥 = −1.35 ± 0.05 𝑜𝑟 0.75 ± 0.05 𝑜𝑟 4.6 ± 0.05 7(a) Method 1 ∠𝐵𝑅𝐶 = ∠𝑃𝑅𝑄 (𝑐𝑜𝑚𝑚𝑜𝑛 ∠) 𝐶𝑅 𝑄𝑅 = 1 2 (𝐶 𝑖𝑠 𝑚𝑖𝑑𝑝𝑡 𝑜𝑓 𝑄𝑅) 𝐵𝑅 𝑃𝑅 = 1 2 (𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑠 𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 𝑏𝑖𝑠𝑒𝑐𝑡 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟) ∆𝑅𝐵𝐶 𝑎𝑛𝑑 ∆𝑅𝑃𝑄 𝑎𝑟𝑒 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 (𝑆𝐴𝑆 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦) Method 2 Since diagonals of a parallelogram bisect each other, B is the midpoint of PR. In addition C is given as the midpoint of QR, using midpoint theorem, BC is parallel to PQ. ∠𝐵𝑅𝐶 = ∠𝑃𝑅𝑄 (𝑐𝑜𝑚𝑚𝑜𝑛 ∠) ∠𝐵𝐶𝑅 = ∠𝑃𝑄𝑅 (𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 ∠, 𝐵𝐶//𝑃𝑄) ∆𝑅𝐵𝐶 𝑎𝑛𝑑 ∆𝑅𝑃𝑄 𝑎𝑟𝑒 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 (𝐴𝐴 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦)

(b) Method 1 𝑃𝐶⃗⃗⃗⃗⃗ = 𝑃𝑄⃗⃗⃗⃗⃗ + 𝑄𝐶⃗⃗⃗⃗⃗ = (𝑃𝐵⃗⃗⃗⃗⃗ + 𝐵𝑄⃗⃗⃗⃗⃗ ) + 1 2 𝑄𝑅⃗⃗⃗⃗⃗ = −5𝒂 − 3𝒃 + 1 2 (−5𝒂 + 3𝒃) = − 15 2 𝒂 − 3 2 𝒃 = − 3 2 (5𝒂 + 𝒃) (𝑆ℎ𝑜𝑤𝑛) Method 2 𝑃𝐶⃗⃗⃗⃗⃗ = 𝑃𝐵⃗⃗⃗⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗ = 𝑃𝐵⃗⃗⃗⃗⃗ + 1 2 𝑃𝑄⃗⃗⃗⃗⃗ = −5𝒂 + 1 2 (−5𝒂 − 3𝒃) = − 15 2 𝒂 − 3 2 𝒃 = − 3 2 (5𝒂 + 𝒃) (𝑆ℎ𝑜𝑤𝑛) (c) Method 1 𝐴𝑄⃗⃗⃗⃗⃗ = 2 3 (−3𝒃) = −2𝒃 𝐴𝐶⃗⃗⃗⃗⃗ = 𝐴𝑄⃗⃗⃗⃗⃗ + 𝑄𝐶⃗⃗⃗⃗⃗ = −2𝒃 + 1 2 (−5𝒂 + 3𝒃) = − 5 2 𝒂 − 1 2 𝒃 𝒐𝒓 − 1 2 (5𝒂 + 𝒃) Method 2 𝐴𝐶⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗ = 𝒃 + 1 2 (−5𝒂 − 3𝒃) = − 5 2 𝒂 − 1 2 𝒃 𝒐𝒓 − 1 2 (5𝒂 + 𝒃) (d) 𝑃𝐶⃗⃗⃗⃗⃗ = − 3 2 (5𝒂 + 𝒃) & 𝐴𝐶⃗⃗⃗⃗⃗ = − 1 2 (5𝒂 + 𝒃) 𝑃𝐶⃗⃗⃗⃗⃗ = 3𝐴𝐶⃗⃗⃗⃗⃗ ⸫ C is a common point, P, A and C are collinear. (e) Method 1 𝐵𝐶//𝑃𝑄 (∆𝑅𝐵𝐶 𝑎𝑛𝑑 ∆𝑅𝑃𝑄 𝑎𝑟𝑒 𝑠𝑖𝑚𝑖𝑙𝑎𝑟) 𝐴𝑟𝑒𝑎 ∆𝐵𝑃𝑄 = 1 2 × 𝑃𝑄 × ℎ or 𝑨𝒓𝒆𝒂 ∆𝑷𝑩𝑪 = 𝟏 𝟐 × 𝑩𝑪 × 𝒉 = 𝐴𝑟𝑒𝑎 ∆𝐶𝑃𝑄 = 𝑨𝒓𝒆𝒂 ∆𝑸𝑩𝑪 𝑆𝑖𝑛𝑐𝑒 ∆𝐴𝑃𝑄 𝑖𝑠 𝑐𝑜𝑚𝑚𝑜𝑛 𝑎𝑟𝑒𝑎 or ∆𝑨𝑩𝑪 𝒊𝒔 𝒄𝒐𝒎𝒎𝒐𝒏 𝒂𝒓𝒆𝒂 ∴ 𝐴𝑟𝑒𝑎 ∆𝑃𝐴𝐵 𝑎𝑛𝑑 𝐴𝑟𝑒𝑎 ∆𝑄𝐴𝐶 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 Method 2 ∠𝑃𝐴𝐵 = ∠𝑄𝐴𝐶 (𝑣𝑒𝑟𝑡 𝑜𝑝𝑝 ∠) 𝐴𝐵 𝑄𝐴 = 1 2 (𝑓𝑟𝑜𝑚 (𝑐)) & 𝑃𝐴 𝐴𝐶 = 2 1 (𝑓𝑟𝑜𝑚 (𝑑)) 𝐴𝑟𝑒𝑎 ∆𝑃𝐴𝐵 𝐴𝑟𝑒𝑎 ∆𝑄𝐴𝐶 = 1 2×𝑃𝐴×𝐴𝐵×sin ∠𝑃𝐴𝐵 1 2×𝑄𝐴×𝐴𝐶×sin ∠𝑄𝐴𝐶 = 𝑃𝐴 𝐴𝐶 × 𝐴𝐵 𝑄𝐴 P Q R S B C 5a 3b P Q R S B A C 5a 3b P Q R S B A C 5a 3b

= 2 1 × 1 2 = 1 1 ∴ 𝐴𝑟𝑒𝑎 ∆𝑃𝐴𝐵 𝑎𝑛𝑑 𝐴𝑟𝑒𝑎 ∆𝑄𝐴𝐶 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 Method 3 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑃𝐴𝐵 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑄𝐴𝐶 = 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑃𝐴𝐵 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 × 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑄𝐴𝐶 = 1 2×𝑃𝐴×ℎ1 1 2×𝐴𝐶×ℎ1 × 1 2×𝐴𝐵×ℎ2 1 2×𝑄𝐴×ℎ2 = 𝑃𝐴 𝐴𝐶 × 𝐴𝐵 𝑄𝐴 = 2 1 × 1 2 = 1 1 ∴ 𝐴𝑟𝑒𝑎 ∆𝑃𝐴𝐵 𝑎𝑛𝑑 𝐴𝑟𝑒𝑎 ∆𝑄𝐴𝐶 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 (f) (i) 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑄𝐵𝐶 = 1 3 (ii) 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑅𝑃𝑄 = 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑄𝐵𝐶 × 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑄𝐵𝐶 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑅𝐵𝐶 × 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑅𝐵𝐶 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑅𝑃𝑄 = 1 3 × 1 1 × ( 1 2) 2 = 1 12 8(a) (i) Middle position = 400th Median Length = 10.6 cm (ii) Q1 position = 200th Q3 position = 600th Q1 = 9.95 Q3 = 11.25 or 11.2 Interquartile Range = 11.25 – 9.95 or 11.2 – 9.95 = 1.3 cm or 1.25 cm (b) 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 = 15 + 25 800 × 100% 𝑜𝑟 20 + 25 800 × 100% = 5% 𝑜𝑟 5.625% (c) Lobsters caught in June are shorter in length as the median length of 9.3 cm is lower as compared to 10.6 cm in May. The lengths of the lobsters caught in June has a wider spread as its ITR of 3.7 cm is higher as compared to 1.3 cm in May. (d) I disagree with Jill as 25% of lobsters needed to be released in June because they are shorter than 8.3 cm which is higher as compared to the total percentage released of 5.625% in May. Or I disagree with Jill as 75% of the lobsters caught in June are within the legal length which is lower as compared to 94.375% of the lobster caught in May are within the legal length.

9(a) (i) ∠𝑀𝐿𝑂 = 47° (𝑏𝑎𝑠𝑒 ∠ 𝑖𝑛 𝑖𝑠𝑜𝑠 ∆) ∠𝑀𝑁𝐻 = 180° − 47° (∠ 𝑖𝑛 𝑜𝑝𝑝