真二第張六第
日一廿月二年酉辛曆夏
WAH KIU YAT PO
郭日橋華
學分
生小的
慨與
,以一八〇的優耎成楂榮獲項目十九,初殽祖
在唱一十三歲以以下)冠军的故素女小學。
四期星日六廿月三年一八九一曆公年十七國民華中
..攢月二四六中級組小提鄴二交保勇
兩組六級鋼琴獨奏
校浩斌 朱以八十八分奪得冠军
以八十六分的成精名列第三。八馬、梁),
昨日在聖 下月二日六組會師大會堂角逐 安得烈大教堂行之項
曾浩輝李洵華分奪冠軍
一三甲優勝者,而香港 -位參賽者中,共有七名 |複賽事,在尝尝十一
仁中三學生智浩則..
「佔鰲頭,以八十七分的,
成绡勇奪冠軍,並將做
·女優影對於四月二日會第一中學陳偉基及撰文塔,季軍 屠英文中學
與其他六組六級婚|佛教大光中學黃精華,雙保良局理聯誼會
,十三做以下單簧管插在任務者合影左忠冠軍
| 師大會 音樂+角,寶珠
| 亳無椒,只示有
| 的六級高手, 得獎後脚
「十三歲,却是一個熱
沙梨牌行盃:
愛的活,今年只有
第一次參加比
和音樂感,「做得好! 一個困難,有良好的節奏 ,可是他却可以夠服造 會在愅癸時感到緊
一,在這小小的年紀多少 一团,在演奏中未有出
|然是經過仔細小心的考
·碳示?曾浩的演奏
· 釺判李察格夫斯
少臘撕
八十六分縢次于李軍四立學校(一百七十分)八參加只得一 小孽五至六年級中文歌曲合唱:大埔
晾
,手耳赤柱产士提及中學恩志照後爲郝 翌士提反中學黃鴿昌,亞軍馬利亞書院產藍系
·低音大揭开優勝音與判合影左起冠軍赤
百六十四分。(注:參只用一葉) 中學程中文歌曲合唱·面:黃鋸枝中華(
·成、亞革SVZY, WHANG 及拳軍立、
小提猱協委曲怪勝者,由左至右,軍鐡
欣
玲別
露
天:
工藏院,單有蔬去
莉霞九以名名時
及深珑十分列水 救,利七別后营:
恩額分由:
院撤院成港·市佳 健的摩
永鵝可屈利保蹊 的天音集
室利欣居諾 羅堡母佑藥中
育教刊特蓢樂音
奏獨琴鋼級五
名一十軍季亞冠 上以準水多大者賽參
善改法設須塲怯犯有惜
鮮雞
第三屆
冠本一較高下,送這利琴行杯。 二日在大會獄音樂奧與其他四罷五級強獨 八 十八分的超卓能按點雄,有機會於四月 由德生,而位居榜首的聖保羅院林天佑过 烈大教堂舉行之項目一一七五版獨奏祖 一共有十一名冠亞季軍的運踪者在昨日奥安簽·
香港學校音樂節特輯
.1在法医的橋形底下,平白損失了分慨,故在 技術療營之外,公開弹奏的勇氣培強亦不可忽 在水準之上,可惜在臨場演奏愛精神不能集中 杯判李寨里格夫斯戏示,大部分發賽者都
是李察格里夫斯的作
林天佑
· 冠軍林天佑在弹奏時有
殼其特別的风格,待神集中,群
| 粗女子合唱團-〔本報 灣區中學組合唱團比賽組離位置的漸明中學初級 ",以一百七十六分的速成;與安柱中學暫列基
; 左是黎靜怡、旁爲陸乏芝,獲八級雙鋼琴
本(本報記者編:
1981
中學會考試題預習專欄
數學
(四)
明德出版社交長波提供資料
12.
But
13.
Mathematics 24
C. P. Man
MILL & DALE PRESS LTD.
Solution to exerc Section H.
11.
Area of AAOR
3.P.tan 54°
0.344p2
Area of AAOB
siu72
0.4755mm
2
Area of regular pentage
5×0.344p0.
$79 5×0元
Increased percentage
the first
49-25
25×100%
- 96%
Increased percentage of
the second
42-30 X, 100%
30
(0.3)(1-2) - (1-0-3)(1-2)
(L-0.3)(L-1-2) = (0.3)(1-2)
(rejected)
(iii) Let the temperature of
the copper vessel and its content ríves e c real energy aborted by the copper vessel and its content
*0.75A
the length of the wire
is 1.5m
(Answ
2:25A
(Ans")
(b) Substitute the value
Linto
(b)(3)
fil) The current throu
the
1.20.0
•40cm X
Given: A, B are centres
of the cireles APLXY,BPLE
TOTOVE(a) XP.PY+HP.PK.
· {b}· XY=HK⠀⠀
Proof (a) CP, PD XI. PY
CP,PD=ur.PK (Intersecting
chords)
XP. PYP PK (Subst.) (b) APLXY(Given)
XP=PY (line from
centre 1. to
the chord)
Similarly HP = PK
(EXY)2 = (38K)
(Given)
A:P
Solution: Let be the
centre of the circle.
radius r.
AG, BO, P9 are tangents
of the circle
LCAO = ZCB0. * 90°
·ZAOB - 90°
AORC is a rectangle.
PRI HORO
(tangents from
ext. pt.).
·APRO
r-2+r-1.
2x-3
Let Pq be one of the
sides of the regular
decagon
¿POO=
360
36.
10 180
ZOPO
OR - PRtan72" - tan72°
ea of Apoq
Id.stan72°
0.7694d
0.2959r
Area of the regular. decagon
1080:7694d
10x0.2939
7.694d
0
2.939r
(e)
Solution: AB = AC = • BC.
•
Ares of AABC]
16×6×gin60·8q.cm 15.59 sq.cm
Area of sector AEF
Area of sector: FBD) Area of sector CDE
(3)78q.cm
4.4.71 aq.om
Area of shaded part 15.59-3x4.71 9q.cm
- 1,46-sq.cms.
Volume of the space
between the cylinders,
1,46X160 cù.em
233.6 cu.cm
As shown in the figure,
resistance of AB
=16
x 20
Kesistance of
20% +60
BOBOL
電物
理
(四)
明德出版社機榮家提供資料
(2)
7.694
0.4V
.3827-
(AnÒe)
2V
PLY (24)
至马。 Solution:
2x-by-720 5x+4y-8z
LA DARK VAKE IT.)
Suggested soluti
Exercise 12
39+10-8-30
Let the length of the wir be 1. (a)
0.3m
JAV
As shown in the figure,
me.c.es of copper
3.3 x 1077 kgo
Energy elven out by the heating coil in the period
ht
1.5) (10) (5600) 81000J.
(0-5 × 4200 ×
+ (0.5 x 420 x
23109 J
(Arisa)
heat energy given out by the heater
Assuming no heat losses; or
gains from the surroundings.
23100 81000.
(Ans.)
0 = 35.
the rise in temperature of the copper vessel and its cantent" in 35c.
(iv) the balence.length is
'rom
hesistance of
華(20)
sistance
20
the
(Aps.)
balance length is
Apply the principle
wheatstone bridge
R
- XQ
=120d; and
the resistance of bul measured
the circuit.
-R
100 - 40
= 1800
(11) The current, through B
· 120. + 180
0.01A
(Ans.)
(iii) The power dissipated
by 6
(0.01) (180) -0.018 W
(Ans:)
(c) the current through Bin (a) is much greater than the current in :(b);
therefore, trie temperature: of 1 in (a) is much higher
Consequent-
thần In (b)..
ly, the resistance of
in (a) woul be greater
for resistance increases
as temperature increases..
4(1) Let the current through
the battery be 1, the
currents through the
voltameter and the heate
2.618 en.
Solution: (a) Let AB be one
of the sides of the
regular pentage
LAOR 2.
LOAB =
3600
72°
180,-72°
X:y:24:-1:2
x=4k; yamk; z2k
k) ^ - 16 (− k ) " + ( 24 ) - 48K-10K2+442
3uk2
-22; y=71; z=21.
--༔
15. Solution: Weights of, three
clumps of metal before
cliange
5:h:7:
23:30:35
Weights of three lumps.
of metal after change
Apply the principle
Wheatstone bridge
們
7:05
49:42:35
3 x 10x3600
0.75A
The current I taken from
(2) x (1)
the battery is riven by.
3(*) Apply
200 - 200.
be
I and 1, respectively Apply the formula
נת
zIt
= the existance of the
bulb
200
(11) Apply
(ans)
= mass of copper
deposited
8.91 10-4
time taken
(Ans.)
= 3600