NYGH 2018-S3EOY-IM1 Ans Key

Sec 3 EOY IM1 2018 Answer Key 1(i) 𝑃 = 38 𝑉 (ii) 𝑉 ≈ 0.00204 m3 (iii) 2(a) (i)& (ii) (b) Any equation of the form (i) y = ax3 + 2, where a < 0 (ii) y = 2akx, where a > 0, k < 0 3 Perimeter of shaded region ≈ 30.4 cm 4(i) 𝑎 = 0.75, 𝑏 = 1.5 (ii) 𝑥 = 0.915, −2.91 (iii) The x-intercept is ̶ 2 5(i) 𝑦 = − 8 5 𝑥 + 56 5 𝑜𝑜 5𝑦 + 8𝑥 = 56 (ii) Since Gradient of 𝐷𝐷 × Gradient of 𝐴𝐴 = − 8 7 ≠ −1, DF is not perpendicular to AB. (iii) B is (−4, −4) (iv) C is (12, −8) (v) Area = 90 sq units 6(i) 𝐏𝐏 = �120 140 160 120� �0.25 0.3 0.45 0.35 0.25 0.4 � = �79 71 110 82 78 120� The elements in the first row of PQ represent the number of boys who are healthy, ill and carriers respectively. The elements in the second row of PQ represent the number of girls who are healthy, ill and carriers respectively. OR There are 79 boys and 82 girls who are healthy, 71 boys and 78 girls who are sick and 110 boys and 120 girls who are carriers of the disease. P V O O y x O y x

(ii) 𝐑 = � 10 25 15 � The total medical cost for the boys and girls are $4215 and $4570 respectively. (iii) The total medical costs incurred is $7414. 7(i) ∠𝐴𝐴𝐷 ≈ 71.4° 𝑜𝑜 108.6° (ii) Distance student walked ≈ 327m (Accept 328m) 8(a) Gradient ≈ −7.85 (Accept -10 ≤ 𝑚 ≤ −6.36) (b) 𝑥 ≈ 2.9 (Accept 2.8 ≤ 𝑥 ≤ 3.1) (c)(i) 0.35 ≤ 𝑥 ≤ 2.45 (Accept 0.35-0.4 for lower bound and 2.4-2.45 for upper bound) (c)(ii) 𝑥 ≈ 0.55 or 3 (Accept 0.55 or 0.6 / 2.95 or 3 or 3.05) 9(a)(i) angle BJC = 40° (ii) angle EBA = 65° (iii) angle JCE = 25° (iv) angle BFC = 65° (b) Angles in the same segment (c) EB = 3cm since equal arcs in the same circle subtend equal angles at the circumference 10(ii) Eqn of circle is (𝑥 + 3)2 + (𝑦 − 2)2 = 25 (iii) Centre of circle C2 is B (−3, 4) Centre of circle C3 is C (3, 4) (iv) A circle can be drawn as the three points are not collinear. Since AB is perpendicular to BC, ABC forms a semi-circle as angle in semi-circle = 90°. The centre of circle C4 is the midpoint of AC. Centre = (0, 3) 11(i) ∠𝐴𝐴𝐴 ≈ 54.9° (ii) 𝑃𝑃 ≈ 4.05km 12 ∠𝑃𝑃𝑄 = ∠𝑆𝑃𝑄 (alternate segment thm) ∠𝑃𝑄𝑃 = ∠𝑃𝑄𝑆 (common angle) ∴ ∆𝑃𝑃𝑄 is similar to ∆𝑆𝑃𝑄 ⇒ 𝑃𝑄 𝑆𝑄 = 𝑃𝑄 𝑃𝑄 𝑃𝑄2 = 𝑃𝑄 × 𝑆𝑄 𝑃𝑄2 − 𝑃𝑃2 = 𝑃𝑄 × 𝑆𝑄 (𝑃𝑃 ⊥ 𝑃𝑄 since 𝑃𝑄 is a tangent to the circle) ∴ 𝑃𝑄2 = 𝑃𝑃2 + 𝑃𝑄 × 𝑆𝑄 OR 𝑃𝑄2 = 𝑃𝑄 × 𝑆𝑄 (tangent-secant theorem) Since 𝑃𝑄 is a tangent to the circle, 𝑃𝑃 ⊥ 𝑃𝑄. Using Pythagoras’ Theorem, 𝑃𝑄2 = 𝑃𝑃2 + 𝑃𝑄2 ∴ 𝑃𝑄2 = 𝑃𝑃2 + 𝑃𝑄 × 𝑆𝑄