1976-01-28 — Page 28

7. OX-

=24cm

1976中學會考試題預習專欄

·Since OP = 25cm

.*. DX — (25, = 24)cm

=1cm

女長波。

(11)

pectively

x+y=11

fron (2) x-y=1

(1)+ (3) 2x= 12

x=6:

8. Let the tenis digit and the unit

digit of the number be x and y res-

(10x+y)-(10y + x) = 9

新數學 (十)

MODERN MATHEMATICS

Solution to exercise 4

Section A

Section

日八廿月一年六七九一屣公年五十六國民華中 育教僑華 (ii) Let y=号== k

AB along the great circle.

博四第張七第 日八廿月二十年卯乙艇

WAH KIU YAT PO

報日僑華

三期星 c)//A0B= 2204*=0.3852

is -

Equation of the re-

=3960x0.3852 mi

x-2

quired straight line ig-

ive.5x+4y = 22

√25272 cm

數學科

(十一)

葉柏芳。

M

TRADITIONAL MATHEMATICS

(11)

1525 mi (2 mask")

-(1)

(2)

(3)

(1) OA = a

OB= a sec30°

OC = OR Bec 300

OE = OD sec300

日

x = 3k, y = 5k, 2= 4k,

and ky+ yz + zx = 47

(3k)(5k) + (5k)(4k)'+ (4k)(3k)

=47-

1.8. 13x2 +20x2 + 12k2 - 47

i.e. 47k2 = 47

i.e.k=+1

y=5, z = 4 or x=-3:y= -5, 24

11. Let the cost of the handbags be

$P perdozen.

the manufacturer sold them at

$1.3P per dozen to the retailery

The amount sold to the retailer =$1.39x1,000 $1,300P

The retailer returned 400 dozen at

half price, i.e. at $0.65P perdozen

=

the amount returned

$0.65P x 400

= $260P

The remainding amount

=

$(1,300-260)P = $1.040P

The amount was paid at 55 cents in

the dollar, i.e. at 55%

OD = OC

sec

300 = 2.

8

*33*

(ii) OEt OA = = 16 a: a = (111) AOAB=J0A. OB=

16:9

ive.

1

za a tan 300

300 tan

▲0BC = 70B• BC = 1⁄2• a •—a tan 30

tan 30° (4)

AOCD = doc

.CD =

tan 300

a2tan 30° (4)2

AODE = 20D+DE

8

cash paid to the manufacturer. $1040P. x 55% = $572P

the manufacturer sold 600.

dozen hand bags and received $572P

But the cost of 600 dozen. handbage iS $600P

he lost $(600-572)P

Now his loss was $11,200

28P 11,200 or P = 400

$28P

i.e. the cost of the hand bags was

$400 per dozen. Cost of 600 dozen handbags

1. Let =号=k

2.a)

a=.3k and t=5k

2 -262 (3k)2 - 2 (5k) 2

2ab

2(3k) (5k)

(x + 1 ) 22 - 2

-2 14

b) x2 + 1 = (x2 + − 1 ) 2 - 2

2 = 194.

3. The arithematic mean of pe and

2

The geometric mean of p and

2.2

is p qor pq

Since (p-q)=

2pq + qx?m

2

+q2pq

2.

土

2

ice. P+ ≥ pq

The arithmatic mean is not less

than the geometric mean.

4. Area of sector ACB

-162x sq.cm.

-3 sq. cm.

Area of AACD

3x6x4xsin 300 sq•cm*

-6 sq.cm.

Area of shaded part

(3 -:6)sq.cm.

5. Since 5000:7500:2500

=

2:3:1.

2

A receives $300 x.

pr $100.

6. Slope of line

4x-5y= 18. is

slope of PQ

(1) - (3) 2y = 10

y=5

The number is 65

Section B

9. Capacity of the test-tube

2.

Ta b +

2

3

-Ta2 (b+za) (2muka)

when the test-tube is half-full, the height

of the mark above the lowest

point of the test tube is

(a + b). – 2 722 (b + ₤a)

顥+如

ra

2

48+30: (2/2iks)

when the test-tube is one-quarter. full, the height of the mark above the lowest point of the test-tube is

(a+b)

za2(b+a)

=28+0 (5 marks)

when the test-tube is three-quarter full, the height of the mark above: the lowest point of the test tube is

(a+b)

5日

1 4

ma? (b + {e)

3D 4

10a +

12

(3 maks)

10. AC = CB=radius of the parallel

of latitude =3960 cos 67030 mi

1516 mi (A mock)

/ACB=70°-100=60°=1,04720

AB = 1516x1.0472 mi

1588 mi (3 maiks)

AD 1516 sin 30°

+

758 mi

AD 758

̇sin AOD = K0 - 3960

ZAOD=11921

ZAOB=22°4′ (3mok)

導務作鉏精鋼務在

(3 mask.)

3√3 35

12.

tan 300

-za tan 30o (4)3

the areas of AOAB,AOBC, NocD, and AODE are in G.P.

10.(i) 9x + y - 8ż = 0

4x8y+7z = 0

Applying the method of cross

multiplication

器

(1) (2)-(-8) (-8) = (-8)(4)(7) (99)

Z

= (9) (8) (4) (1)

ivex q#q= =32%63 = =72-4

1.e. -57-9576

x: y: z 35: 4.

$400 x 600 = $240,000

his loss percent =

11 240,

x100%

12.(i) (0.782) 0.1

..n log 0.7821og 0. i.e. n(T.893) 7.000 i.e. -0.107b < -1.

-0-107

i.e. n9-36

. least integral value satis-

fying the inequality is n-10) (ii) (10829+106,3) (108,4 + 108,2)

5

108.2+

log 2 log 4: log 9

·log 2

210g 2.

+

4108 3+ 10g 3) (1062 + 1082)

21og 2

log

log

、國團嚴及,會務

校一

譯由中國 茲部

際內謹實領培出租 及服,在導長會馓 交,分。能更更 業區部務友過宗爲 指股工團空服旨年

服六。的,導故發

四名照二人晴暴暴非法

上半 從的工計

一八三助工及在研陶之里一開

界並筵此

校學慈長佩管

雅七車選工数本程本箱

校長受持

用"

完曌對列綫起

愛的教育小組

檢討工作成績

(RE) TE

一。爲使選小組能更切震地去體驗,發缎問題及 | 向及目標,它越發明 我們要切實 一的教育一小粒日後之蚴】關心他們,與他們合作 寫對穬饔務。從中「愛一位敎師,接納他們, 筆辦過兩次研討會,五特別是校方,應葉重每 「青」已意半載,並先後其重要性是不容忽視,

【青」小組推行「愛的教」作,眞是任重而道送,

提出五點心得贤抛

救

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|教師本赛悬推行之動力」小組熱切希望各界不

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▲全面發證他們的花雞|意識,放藥小學教科醫 ◎認潸兒童飛別之獏具|推行一些有意娄之運動 (一)敦師敦學時及合作將神。 臺通童愛心以兒童食中「支持及協助 ,例如,培養學生寮深

給別人,這算是小

|從工作或經檢分享中香

我們若能將這個意

吝指教,俾能有效地實

基督敎學生合唱團

,它已擒起了醉世紀人

女士任指挥

一,向社會人士見證基督 ..厨合基督徒事由之力,一對養小姐為總統。在林 勵身努力規習,使合唱一外,更著風顧展之品游 彭博士在致阿中幼,一峯風音質,樂理之游赚 出战合唱團成立目的在一鄉群推導,黃進昌、並 y_韦羞洪雅會舉行成立典痛經,讀爲興

·乡孝顺博士致詞,指一何莫燕宜女士任指挥, 錄。由萬股神校長主持學證基協會主辦,敦 黃陽東伴奏,歐斯任

| 大學聯合書院扶輪青年」,設有幼稚爲及小一至

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