1966-02-23 — Page 11

TEX.

日四初月二年牛芮膺墓

華僑教育

中考試題預習專欄

數學科

(八)

喬仲強。

模擬會放試題解合

1.(a)利用對數表,求下式之值:

(-330x

+

0:230

0.672 3707

(解)在未解本題之前,我們先要認識清楚加減的運算是不能 利用对数的所以本題利用对數的部分只限于乘除乘方開

No

LOG

330 12:5185

49.

1.6902

108283

22

1.34.24

184510

0:678718312

0.3859

Z4mm Ne

POI B

Q2 + b2 + c2+3abe

(al+bm+cn)(a+b+c^-ab-be-ca)xill+m+ x} {a}+b2+C-3a

== (buonon) (a+b+c){a+b+c^~ab-be-ca)

Babcing th a'rvic"-ab-be-ca =0, KX = 1# *

(a+b+c)(bemon) ·

QE.D. (4)布若干匹着慣10え則本4%,如要利5%毎 4/12a1??

(解)布每碼成本為10元÷(1-4%)故管價液韻為

107 ÷ (1−4%] × (1+5%) = 10x100x-02 7 = 1015 2 (*) (七)甲乙两三人合服營商,分别出資本 50,000元 30,000元及

25,000元,因两任司理;訂明先以可獲利益先撥素给典術品 其 金額資本額分配、現年終結等丙若井線 2800元問

中乙二人各得多少?

(31) = ^ † † 11 <= = 50,000! 30000: 25, 30: =10:6.5

+27 16 34 A 25-2

a★/12411 ÷ }+(-4) = = = 共賺

84007

28007 ==!

故得 84002731 35007

z 11 8,4007 × 7×

=21007

}(答

6.(0) A 18, 27, 36. 28 @ $4.59] - -整數東之,而所得之積均相 同、問這回數最小是要少?

9.01 0.9547.

234 7 369 2 (-

5.11

1.4167(2

·0.7084

X BA=1781+51

€ AB) 8 13 244 % 18, 27, 36, 28 46 £ c m,

3×9×7—756

-756-18-42,

756-36=21,

(5) £40 * 104055, Tekash § 6937 ££££

茜数的和二大谷×浦屋質數的和 104055

15 −1+14 = 2+ 13 = 3+1 2 ( 7 ♬ 3 4 $)=4+// =5410(不見る質数)=6(不是種数)=3+8

故本题答案共有四维

(a) {

6937*1=6937.

6937714-97118

64:37x4=27748.

6937 × 2=13874

6937×13=90181

(d){

6937×7=48559

6933 x8 = 55491

756-27=28

756-28=27.

2)18 27.36 28

19, 27,18:34

*: 13433) 42 28, 21 21 1 2 85***ƒ.

坂上小路間之鶴馬環境一畑直送 40名的図形水池如な小路上

鋪以了时厚的沙,間用沙多少?(=3.4x) NG

(39) = si «| Mil = TR-TA =T(R-A°)

34 63 80 43 = π ( 22′- 20')=841 #768

LOG

756

2.8785 04972+

33757

0.60.216

27736

月二年六六九一公年五十五國民黨中

***** HA † Y SU

英中会考

化學科 (八)

王錦釗。

̧CHEMISTRY (8)

(4) A golution o: potassium nitrate is obtained by neutralising

a given. volume of a normal solution of nitric acid with the same volume of a pomal solution of potassium hydroxide,

→ H2 O

KOH

KHO

potassium potassiu hydroxide nitrate:

nitric monacid The solution is then evaporated to dryness, it being necessary

to carry out the final stages over a water bath. Pure dry crystals of potassium nitrate will be left in the evaporating dish,

(b) Calcium sulphate is obtained by double decomposition between

soluble calcium salt, such as calcium chloride, and a soluble sulphate, such as sodium sulphate. Calcium sulphate, which is only sparingly soluble in water, is precipitated out of the solution.

2NaCl Caso

CaChy Calcium

chloride

Plág Sopp sodium sulphate

qalelum

sulphate

The calcium sulphate is separated out of the solution by filtra- tion. It is then thoroughly rinsed and the residue is removed to an evaporating dish and dried over a gentle, bunsen flama, Pure dry calcium sulphate will be obtained in the dish,

(c) Anhydrous ferricchlorida is obtained by direct synthesis of iron

"and-chlorine, When a stream of dry chlorine is passed over soMS.

hot iron wire in a combustion tube, the iron glows and black crystals of anhydrous. ferric éhloride are formed in the cooler parts of the tube.

2Fe... • 3 C1 = 2FeCl fron Chlorine ferric

chloride

(a) When dry ammonia is passed over heated cupric oxide, tha ammonia is oxidised by the cupric oxide to nitrogen. Equation 2H

armonia

capric nitrogen copper mater

oxide

(b) Nitrogen is obtained when a solution of an ammonium salt, such

es ammoniwr chloride is heated gently with a solution of sadium nitrite:

Equationt-

Na NO2

Naci

armonium

chloride

sodium nitrite

3. (a) when ammonium chloride is heated in an evaporating dish, dense

white fumes of ammonium chloride, will be given off. When a filter funnel is placed over the dish, a sublimate of ammonius“ chloride will be obtained on the funnel,

69 37 ×/1=76307

2 (a) 74 6 x 1 341947 ± 24021 2£6 348a, A-14 €0¶ 獨做則可旗工費3鋏15光合由で獨做可護工資3銭7先合6便 問語工程由こん会襭共得工資負

(解)(錘≥30先令:

= 2便士 1先令

̇故甲獨做光日數=3鎊15先令-3先令9便士=75元今:

- 20

210 6 2 = a § −338 716 648± – 246 3}R+= 673 4/7 = 2

30.

=^^4 £=== $k = 1-(2+32)=12

**^on #1| 2 † (3£§ 912+ +2ƒ§ 3 12 1) × 12

你有直径了时的錯球,現将之熔解罵三個鉛球,其中的球的直播 分别总

吋,来第三球的直接

(**) ** 4*** = 3 r l = { π d2 ( d $ = ‡ 214)

{ π × (4) - {π x 23

* (27-7---8)

故第三球工体精 =

第三球的直

8年破ズピーx+2整除試求

a b 18 # # # NA

= 593.7 ±±* (*)

7.10 % 741x x2+ px + q = ? p2= q(3p-1)+g2 =

角)報道根則中根敏之間係得

又田篤源方程式を放

(4-9)*

£50, £+£3, 4'-3 p`q + 34 g - g3 = p2g = 3p′′q +3pg= g

(1-8)

q ( g − 1 ) − g < g + 1 ) ( 8 − 1) = 4-$38+8 +8)=0

QE

xax

(解)因被除式為五次式,除式為二次式,故感光勇三次越

$ x2 + 8 x2 + C x + D ? *) = 0 # § à (Inspection)☎** A=1.

2

4 1⁄2 Ã A 1⁄2 x2+ Bx2+ Cx

比較两边獠數,得,B-1=

* a=3 b =

4-C+18=b

-4+2

=6,其两根均為1

2x − 32x + 13x+8x-4=0 [x°218 ££ $ 13 74]

(解)利用用書定理及嘗試法触将原方程式 32+ 分解為

(X-2) (24−1) (2x + 1) (3x-2) = 1

成量(答)

x + 2 ) ( x 3 + B x2 + Cx+0)

8

28

3x

(解光製成でき

Y

8

+3x+4

Q

- (x+p), (x+g) **

$(4) 1} C=3 wx 21 x (2) (3)}} a=3 ̈f=5

(92) * x−x +2 2 14 ✯ SA

91 * 12 54) 9.44 ★ 278, dp 1k } (p) = 0, ƒ(-q)=0,#1 a,b

西元

狼聨立方程式以解

12/6-4

- 2x-4 65 || 14 (x K {[• −21 •2) 54}@< V!

ANMANAM

DEAR WHITE FUMES DI AMMONIA COMRADE

·MEATTM

CHLORIDE

When ammonium chloride is heated in a hard glass tube as show in the diagram, a piece of red litmus paper placed obove the asbestos wool will cum blue, while a piece of blue litmus paper placed below the ammonium salt will Aurn red.)

NH2CT NH3 HCI.

The lighter ammonia diffuses through the asbestos wool faster than the hearter hydrogen chloride which flows down the tubs

કાર પ્રમો

RED LITHUA

The experiment illustrates the fact that ammonium chloride will dissociate into ammonia and hydrogen chloride,

Questions for next week.

20.1.Describe how you would snow by experiment the presence

(a) oxygen in-nitric acio

(b) carbon in calcium carbonate:

(c) chlorine in potassium chlorate.

(d) copper in copper nitrate:

(e) nitrogen in armonia.

19.2. 1.00 gram of an alloy containing silver was dissolved in nitric acid, Excess sodium chloride solution was then added and the resulting precipitate was found to weigh 0.717 gram. What was the percentage of silver in the alloy?

A white powder,x, colours a bunsen flame yellow and on heating

evolves a gas which turns lime water milky.

(b) On adding a dilute acid to the residuey, more of the same gas

15 produced. A

Reduce the nature of x and y and write equations for the

reactions that take place in (a) and (b).

(14) * * _ (x+4=2}}(x−y+2}} | (y+23~x) ( Y−23+x)

+

(5z+x+y) (234x=y} {x+y+2}}(x+y-23).

(y+23+x) (Y+23−X).

(x+Y+23)+(8+23.

再像次将谷美

x+4=23

成一曲得

4. (a) [ log (x+4)+ Logz (x−4)=

log x + Log = I + logia"

(AB) (1) ☆ & 16 3 logz (x + y) (x-4) == 2

(x+4) (x− y) = 3′

Ep

(a) The B long, X. 8 = log, 10+ log. 2 = log,, (10×2)

由(4)式得

xy=20

-(4)

g== x < 4x(3) * £#

dum 13x-98 - 400—0

但是负数及虚数均無对数*可言故

THXW*48 4= 4

a+b+c)(l+m+n)

(解)体和比定理 2

-5 4-4

£ a b c z # f

C + b m + c n

+(b_ca)+(c2=ab).

al+fm+on

$ (Parabola) 又於同一圈中再

14y=-3x-1

Q

因此弟一直线

故祇描两点

於團中得的交奌

(1-4) (-1.5,3.5)

yu 3.5

(附註),本題亦可利用該抛物线以解一元二次方程,24+x+30 其形加作で直方程式 佐下法中之 A=2x2x4(原地物制 0= 2x2+x−3 (31) 4-18# y=-3x-1 kk Bp 為用加工直线方程式。

2.因抛物线太狭長,故橫軸以一吋(即10日格)表1單位,是由以

工事部(金)

3.(2)在等三角形の一条正任頂

(陽)設△ABC三*長各為abc並設:$zard

之手切园在三边之切奌分三边开成各线段z長

210) 本作一園切一定貞並切一定重律於一を知ら

(b) A ABC 2 PALE AB AC=AY #ES 12 20HÍ ABOE, ACFG

BG ICE

3. (a) Af ****

* 8 = 3 sine cose + 2 cm D ({ # * 3654

cos

4 # A B C D 1 22 #1 + A8 = 7/6 BC=74 0D=/2.8 DAZULLE

AC=1 # BD x k

I 2

(1242 2.20)

5:10: 7/1 AB 18 1

*

CAB2+ BC" = 3%¢

(b) to AE= BF ACIEGH

1 EG + FH-AC

6 (0) ABC a=

C=184 求各角

(b)縦42呎高之塔や測襍池一七て仰角煮

底角測之則御雀為23 求

71a) { x = r sin cos &

x2+ y2 + z2 = 7°

thia Ance a betžeik

*

art to ca = BRA

13 香織

8 (a) A ABC 2 1⁄2 16 BC 2 × 121 - £ +#m, SIE RTI PARA IT

切道其外公切线上任一头至此二海心作一劃鑄劍 ¤ § * £ - BR -