HKDSE-MATH - 2019

2019-I-01

Ans: $h=4-27k$
$\begin{array}{rcl}
9(h + 6k) & =

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2019

2019-I-02

Ans: $\dfrac{x}{(7x-6)(5x-4)}$
$\begin{array}{cl}
&am

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2019

2019-I-03

Ans: $-3$
By using the Pythagoras Theorem, we have

$\begin{

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2019

2019-I-04

Ans: (a) $(2m+3)(2m-3)$ (b) $n(2m-3)(m+5)$ (c) $(2m-3)(2

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2019-I-05

Ans: (a) $\$920$ (b) $\$600$

Let $\$x$ be the marked price of

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2019-I-06

Ans: (a) $x \ge 2$ (b) $8$

$\begin{array}{rcl}
\dfrac{7x + 2

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2019-I-07

Ans: $39$
Let $x$ and $y$ be the original numbers of adults an

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2019-I-08

Ans: (a) $2$ (b) $4$ (c) $\dfrac{9}{20}$

Since the sector of

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2019-I-09

Ans: (a) $288\pi\text{ cm}^3$ (b) $180\pi\text{ cm}^2$

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2019-I-10

Ans: (a) $24x-48$ (b) $4$

Let $h = k_1 + k_2 x$, where $k_1$ and

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2019-I-11

Ans: (a) $5x – 3$ (b) $1$

Since $p(x)$ is divisible by $

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2019-I-12

Ans: (a) $4$ (b) (i) $\left\{ \begin{array}{l} a = 0 \\ b = 7 \e

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2019-I-13

Ans: (a) $25^\circ$ (b) No

Consider quadrilateral $OAED$,

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2019-I-14

Ans: (b) (i) $(\sqrt{2}-1)\ell$ (ii) Yes

1. In $\Delta BCG$ an

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2019-I-15

Ans: $200\ 914$
The required number

$\begin{array}{cl}
= &

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2019-I-16

Ans: (a) $\alpha = 9$, $\beta = 27$ (b) $60$

$\left\{ \begin{

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2019-I-17

Ans: (b) (i) Angle bisector (ii) $8x-y-60=0$

Sketch $\Delt

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2019-I-18

Ans: (a) (i) $61.4^\circ$ (ii) $11.4\text{ cm}$ (b) No

1. By ap

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2019-I-19

Ans: (b) (i) $(3k-1, 28-9k)$ (ii) $3$ (iii) No

$\begin{arra

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2019

2019-II-01

Ans: C
$\begin{array}{cl}
& (a-b)(a^2 +ab – b^

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2019