二張大赛
日三廿月正年巳乙
WAH KIU YAT PO
1
日四十月二年五六九一麽公年四十五中
a
華僑教育
te
(EE) RESOL 12, A£$ $$$*
(b+c=a)+(c+aub)
kt.l
एक्ट
QED
一九六五金
物理科
1.Solution
題預習専欄
(十二)
許藩芳·
1965
試題預防專欄
數學科( †二)
4a+ab+c= 44, 18 a=; sa+3b+c=4
(39) > * = 1, 2, 3 By = 5, 5, 4 sλ J=ax+bx+c 145
a+&+c=s
4(a) y=ax+bere &x=1, 2, 3 j je ** 18 §* 25.5.4. a, b, c ≥ 14
(a) The gelation between the density of a"liquid "at"t ̃C*and*that
at 0°C. in, terme of its cubical coefficient of expansion:
aw the kam as of the liquid
the volume of the liquid at o°c. *” the volume of the liquid at. t°C.
bet
d the density of liquid at 0°c.
--- b--- C-4 (1)
*
喬仲强。
310007mass
湯描绘上武士图线由x=3至6.并由图中求出==值 [@}} y==√x*+ 2x+4. 製賤下渡
d the density of liquid at t"c.
7 - kubical coefficient of expansion organquid,
volume z density,
and
Apta e mi
(+ =)
X-3 1-2 |-/ 4-5-1
2
3 4
$
612
2
u ←
2
2 -
--
模擬會攷試題(出組數學試卷二)
(a)作图(須依好設長麼作闔,保留作线,不須證明)
0闾半径3厘米,P為口园外已知奌OP=5厘米.求遍P作劃幾 PAB 而使弦 AB=4厘米,
(保)O.P两园相交於A,BGA,B分别作出直线CAD及EBF而以
国周為界,則CEIDF.
(a) 三角形頂角分角线典底边上之底线 弯成之角等於两底角之差
✰*
(4) ABC $42 = ¶ ̧ ã AB, BC ̧ ̧ CA ≈ $3] Æ P, Q, R 1$,
3(9) a, b, c † 4ABC 2 zié, ÌÈ SE
Æ ̧ = 4*+ £# +1. È
< C 90°
12
cmacoB+bcor Å
你描绘等地 Tan test 日之凰线(日由0°至90),並利用
AL BW' «X 28 †** 4 tano + 8 cot om 15.
以上四類金做,每題15分下列5-10韪送做四題,每題10分 新园内接四边形对角线互相垂直,则一边中澳对角线变身之联线
必垂直於其对论.
6. ABCD 為园之内接四边形,延長AB.DC使交持P AD. BC à # Q
PLQ角域相畠
7. 四边形ABCD中,E,F分别為AB.CD中奌,延長ADEFA
*G. X 804 EF & H. & AD=BC, B***
LEGALEHB
#+*
2À tom A + c®2à ct A+ 2 s cœÂ œœÂ+ cœÑA.
9.日在A之南1.在A気測得一飛機在其正柬没做角骂
$2048"的位置
高箔干吧?
10 解方程式
# B 3 £ løb É 42° 8' gut m
4 sin^x + cos x - £= 0 (A* » * 360°2 PRI
第十一次試題解答
86-72X3.057 × 5684
0.2>18 × STOX √.0.775
NO 0.8518
(A)
NO
LOG
36.72
7.9381
2057
0.4853
15664 [1.75 31 0.8766
3,3000
0.37726
2.7221
Ante log 2.7221-8ELL (A)
LOG
T9148
1.0000 0.tooo
$10.775, 7.8893 712/ 6
793) + (78893)
≈4× (5+2·8893)=7. 9631
(七)一木質园柱体長6吋,底径1吋直立浮於高6吋,底半佳品味和 內盛有水之园简中国柱上端距筒項1吋,水深5吋,今用力将园柱壁 下,使其下端達於筒底,闊水面昇高着干时?
(解)設如右图:园柱下端典筒底相距1吋
71
国体下時、排開之水え体
T产x)=(左方吋)
*******‡ (***-*, y=57 · £ Hath 127} $ ) 再用曲线连结之,得抛物线
11x-4 à ★ #⠀ * § P. ( 44, 3-4) BR (-24 -2.0) 放x=y之值有二分別為3.4或-2.4(答)
But vị~-~¢ ̧(lx+7t), and hence V1/****{11+Yt),
Therefore,
1++
(b)}Density;olimer curvlat 30°C., z51= 13.52°gm./e.c.
(3.4, 3.4)
(c)
ماتو
5.長21呎,潤13呎6吋的地面,鋪以完整之正方形階磚閉 最少用牛個?
(1) 21-26201,
13*6* == 162 *f. 2/162,252 正方形每边填為162.252之公约数;无階磚數目9181,126 最少時,則正方形每边颜善最大,故為基 H.-F.
$ 74.
HOD=2x) = 37) 3 18+f. BAIT LE (62+18)× (252=18)= 9×14=126 (*E) 6.某商人次8000元購入貨物器中,其中3件損壞不能售出其 餘增價:145%偕去,结果獲利80元,問原購物干?
(解)設購物体数笃心則原價每件50元,售價每件真
1000 × (1+25%) = 依題意得
元
*
(x-3)=1000+80 答:原購入貨物法件
解
某工程,甲獨作之,84日可成;已的二人合作,56日守箴;西
問乙一人獨做,幾日可成?
.........(0)
*** *30 (1+ Y × 30)
-13.52 (1+0.00018.x:30) -13.59 gm./c.c.
DensitVidimercuryTatt100°C.ais d100
From:(a)'and '(b)
£;{{1+¥*27100)= dz0(1: x 30)
0.00018 × 30)
$100- 12-53 (+0.00018.x100)
-13.35 g./c.c.
Ans:*Densities of mercury at 0°C. and 100°C, are 13.59 and 13.35 ./c...respectively,
(1) Coefficient or apparent expansion of mercury.
Mass expelled
mse remained r temp, rise,
(184.5 – 182,304) gm.
(182.304- 14.5) g. x (100 - 15) deg.C.
2.196
167.804 x 85
per deg.C.
0.000154 per deg.C.
(ii) Coefficient of apparent expansion;oz, mercury/perfidez.F.
-
- 3
x 0.000154
0.000086 pár deg.f.
(iii).Temperature rise
mass expelle
was remained z cvet.
(184.5 - 182.7) (182.7 -14.5) x 0,000154
- 69.51 deg.C.
Ans.Temperature eilis (15:4.69.511dug.C.
DO TANTOS
that is, 84.5.deg..
(a) In absolute ́scale of temperature, the zero-degree-28
equivalent to 273 deg.C. Thus 0 deg.6. is 273 degrets above the absolute zere, and 100 deg.C..is 373, degreen abovę Liktabsolute zero. That is.
1 × 275 • Ľ
where. tal the absolute'temperaturė
t is the Centigrade degree
(b) 'Let the absolute zero be'x'deg.C, below 0 deg..
Assume the change of volume Lakes place at constant pressure the volume is proportional to the absolute temperature. Hence
+25
1.035
+ S • 1.035(x + ̃15h
I - - 270.7
Ans. The absolute zero calculated"is-270;7°C.
Lay For a given mass of gan:
where"P. 76*en‚Tof ́Hg.
P2 70 em.of Hg.. V11 litre
T- 0°C. - 273°k.
T-273 100 -.373k.
而水面显高時,其截面面積為環狀纖鸟
*⠀⠀) 1) = T{R*X)=7(2°1')± 3r ( ¥à«)
故水面昇高燒味(答)
总(2)甲乙、丙三人合股營商,各出资本200,000元,30,000元
○, 咖元,因乙、丙三二人在店中雙職,年終獲利,先撥出一半,平 均分嶼人作為新金;其餘一半,則依照資本額比例分配。 計丙共線:8700元,問甲乙二人各得多少?
( ) = < # 2 tt = 200 000 : 30 000 : 20050=20:3:2
故丙佔溢利額+=湯
全部溢利為8700元÷106 = 36,000 À
$130,000×20 12,000 7; 219 30,000 2×(4+4*£)
2,3007 (4)
(30) Pa
乙一人每日做全工程之
2 + 2 6 = × 38
一就
2-KIỗ lk 3 1 5 = 72 (8)
=口之两根,又
8. £ 4.3x2-6x+k:
(角)由帳興係数閩係
x bf
(4)
ww
j k s l ÷ 1 h.
x+8=
-2 = -
کر کے ام
›
블
= 2
---3.
8=8=&
(1)式代入)式程
1..l-s.
❁༦)+ = --༨༤!>+༥€ = ནྡྷནཾ +ཨི, m ཨ;,
一般来維两鎮間,順流而下,速度每時15哩,逆流而上,速度 每時10哩,問此船往返之平均速度,每時幾哩?
Cha
(解)設兩鎮距離駡1時,則順流時間為本,逆流時間鸳
指数平均速度 全程距離÷全部時間・即
(+1)÷(7+7)= 2 =12 (2304) [*]
3. (a) = 9x-6x+a¢x®%£ox2+30x*+ ax+b 424fi ut
a b ¿ í 1 x À 275 TL 12
(解)
9x (3x=9x).
"-50x2+ 30x2+ ax + b
答: k&
R = $
(Ấ+TE) (A+B)+2NE
·4 x − 5 + 2/3x2-11x+6=(3x-2)+ (x-3)+2 [(3x-2) (x-3)
3x-2+ X-3) (*)
ac.bc.
1912 F (b-c) (c-a)
=(c)(c)
(20),
原式
cland
ac (a-l (t−c) (4~c) (at b
+
(b+c) (a+b)
be
-ca(a-b)
+
+
(b-c) (a+b)
(b-c) (p-a) (a+b)
<(a−b) (amb)+bc(c−q)=¢a(a-b)
(h-c)(c-a) (a+b)
= c(amb] [(a+b)_a}+bc(c-o).
*(b−c){c=a) (Q+6}
bcca-b+c-a)
(b+c)(c-a) (a+b)
t X
be (a−b) + beste-a)
"(b−c)/c-a) (a+b)
b e
bc(c-b) (b~e) (a¬a) (a+b)
23+2x-4 2x+24
M
7t
函
2.6+2m-the
(e-c) (a+b) (*)
試店
76 x 1 x 373
70 x 273
-1.48 litres
9ince 1.48 litre'óf gas wèighs 1.29 gm.
therefore, density in 1.29 gm./1. lith - 0.87120./litre Ans. Density of gas at 100°C.i1s 0,87% gm./litre.
(neatiopa to be answered
1. (a) Explain the terms specifio heat "thermal' capacity”” water
equivalent, and give the units in which they are measure!: (b) Describe an experiment to find the thermal capacity of a
lump of metal, giving full experimental details and point- ing out any precautions which should be taken. fel Account for the effect of high specific, heat of water on
climate,
(4) Assuming tuat there are no heat losses, what volume or
coml gas, having a calorific value of 480 9.Th.Ụ./cu.ft.; will be used to heat 30 gallons of water, contained in a copper boiler weighing 60 lb., from 52 deg. F. to boiling poiling point? Calculate the cost of heating this water by gas costing 10d.per therm, if there is a 50% lose in the heating apparatus. (1 gallon of water weighs 10.18.. specific heat of copper - 0.1 cal./m./ter.c.
2(a)derine the latent heat of fusion, and latent heat of
vaporisation of a substance.
[b] The following is a cooling curve for paraffin wax. (1)
What is the state of vax at A,BC and D? (11) What is tho, welting point of the wax? ('1') What.influences tha length of BC7
6*-2**+42 K
6-24-8 x
因此式孟完全平方式,
2--56,
(84) nic kau Xig **L***
2{2}+2x-4)+2{AK+RY−7)~(24+23−x)
·Z (2 x + 2y = 3 ) + 2 (23+2} = x)-(32+2*-YZ
2n+22
2984.
22 + 2m - př.
Q.ED.
Time (min.)
(6) Describe a method of measuring theuatent neat or steas
and explain any precautions you would take to improvë, the accuracy of the result.
(d) Steam ni 100 deg. C is admitted into an engiøsure
initially at 19 deg.0. and in which a cupper blook af maa 90 g. is supported by a thin wire. Water which condenser on the copper remains in contact with it, unti it rendbes 100 deg.C; Calculate the mass of water condensed.
(Latent heat or steam 740 cài./gm;;ï@pecificiḥcation. copper 0,1 cal,/gm./der.C.)