HKDSE-MATH - 2018

2018-I-01

Ans: $b = \dfrac{2a+5}{3}$
$\begin{array}{rcl}
\dfrac{a

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2018

2018-I-02

Ans: $\dfrac{x^9}{y^5}$
$\begin{array}{cl}
& \dfr

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2018

2018-I-03

Ans: (a) $266$ (b) $265.4$ (c) $270$

$265.473 = 266 \text{ (r

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2018

2018-I-04

Ans: $7$
$\begin{array}{rcl}
P(\text{drawing a red ball}

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2018

2018-I-05

Ans: (a) $9r^2(r-2s)$ (b) $(r-2s)(3r+s)(3r-s)$

$\begin{

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2018

2018-I-06

Ans: (a) $-8 \le x < \dfrac{-11}{5}$ (b) $-3$

Consider $\dfrac{3-x}{2} 2x + 7$.
$\begin

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2018

2018-I-07

Ans: $\$520$
Let $\$x$ be the cost of the vase.

Then the marke

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2018

2018-I-08

Ans: $x = 180^\circ – \theta$, $y = 2\theta – 18

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2018

2018-I-09

Ans: $1.75$ hours
Let $x \text{hours}$ be the time needed fr

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2018

2018-I-10

Ans: (a) $a = 48$, $b = 62$ (b) No

Since the range of the distrib

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2018

2018-I-11

Ans: (a) (i) $1$ (ii) $8$ (b) (i) $3$ (ii) $19$ (c) $9$

1. Since $k

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2018

2018-I-12

Ans: (a) $a = -19$, $b = -135$ (b) No

Since $x – 3$ is a fac

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2018

2018-I-13

Ans: (b) (i) $48\text{ cm}$ (ii) $750\text{ cm}^2$ (iii) No

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2018

2018-I-14

Ans: (a) $4096\pi\text{ cm}^3$ (b) $48\text{ cm}$ (c) No

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2018

2018-I-15

Ans: (a) $40320$ (b) $8640$

The required number
$\begin{ar

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2018

2018-I-16

Ans: (a) $500$ (b) $75$

Let $a$ and $r$ be the 1st term and the c

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2018

2018-I-17

Ans: (a) $31.9\text{ cm}$ (b) (i) $38.0^\circ$ (ii) $71.9^

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2018

2018-I-18

Ans: (a) $f(x) = 3x^2 + 24x$ (b) (i) $(-4, -48)$ (ii) $(-4, 75)

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2018

2018-I-19

Ans: (a) $C: (x – 8)^2 + (y – 2)^2 = r^2$, $r^2 = \d

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2018

2018-II-01

Ans: B
$\begin{array}{cl}
& \dfrac{8^{2n+1}}{4^{3

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2018