1978-04-25 — Page 26

頁二第張七第 日九十月三年午戊夏

WAH KIU YAT PO

郭日僑華

二期星

21

育敎僑華

200 (Anas)

13. (i)

Range of

Number of candi-

dates falling in

-(2k

B

marks

1978

this range

「中學會考試題預習專欄

0-4

6

UA1 Admi

5-9

9

·Hit & Den D`rans)

10- 14

20

15- 19

10

新數學 三十

MODERN MATHEMATICS (30):

20 - 24

the product of root

of the equation

8)

1-2k8

k--9/B (Ann.)

(11)The equation is then

(−2+2)× - 2(-{})+8

+ 5x

日五廿月四年八七九一层公年七十六國民華中

51 D -52 C 53 B 54 C

*******THE END **********

數學

(三十)女長波

Mathematics 30

Solution to exercise 14.

育教儒華

.The height of the tower 18

471m.

14.Solution:

(a) Let x and d be the 1st term.

and the common difference

of the A.P. respectively.

<n=x+(x−1 ) d=

---(1)

m=x+ (n-1) {~-----(2) (1)~(2)--(x-1)d-(n-1)d n-(-n)d

.The common difference.

of the A.P. is 1.

(b)x-n-(x-1)(−1)

金榮家

(11)

clues

Range

mid-

frequ-

Ex

п 1

of marks

ency f value x

Answers to Revision Exer- cise, Paper I (cont.)

11.

Let the required

0-4

2

6

12

59

7

9

63.

equation be 2

10 - 14:

12

20

240

15 19

17

10

170

120 - 24

22

5

110

X-50

2fx

51

-5/2

sum of roots

Proof:

595

(a) DE EC 1

DE

ADDE

BF: FC

63 (AD

BF - - BC - 23

AB + BF

50

23

401

*

AB • BC

(b) AC

Let

63

=(43+2])+n(1463)

=(4m+n)] + (2m+6n)J

40+ n = 4

2m 4 6 6.

From (1) and (2)

G'D P

9/11

8/11

AC

(c)(i) EF - AF – AË

30

(1)

10

(AUS)

(ii) If BK = k then. BK

mean score

595

11.9

(iii).

(Ans:

the students failed

guidation means

27 students having marka

above the pass score

Eran

the graph the estimated pass mark is

12 (Ans.)

-

mn

product of roots

the required equation

0

Given: AQ-EP, AY-CX, RQ-BY

RO//CX, AB- PQ-RP BC.

To prove: (a) ABPQ and BACXY are

//gram

(b) Area of ABPQ

Area of ACXY

RB PQ-NPBC (Given) RB EP

BC PQ

BP//CQ (Converse, proportional division) AQ=BP (Given)

ABPQ ia a gram opp sides equal and //}. AB//HQ (properties of

// grap) CX//RQ (Given) .CX//BY (both // to RQ)

CX-AY (Given)

ACXY is a // gram (opp. sides equal and 7/). (b)ZPRB-¿ABC; (corr Zs RQ//BY

PBR-CACB (corr 48 PB//QC) ZRPB=<BAC (3rd 2 of *)

-RPB

are similar BAC

(Equiangular de) (corr. sides similar As)

RP_PB

(An.

16. (a) Since the points (0,5), (1,0) are on the graph

(0+a)2+ b

5.

(1)

·(X+4)

0..(2)

(1):

,,:

+2a +1

Substitute into (1)

AC

PB-AQ

*AB-FQ (opp. siden // gram)

RQ=BY (Given)

•RP-AY

AY AQ

AB AC

or AY• AC-AB-‍AQ.

Area of ABPQ-AB*AQ sin BAQ. Area of ACXY=AC®AY·ɛin-CAY,

ZCAY=¿BAQ (vert, opp.

área of ABPQ=Area of

ACXY.

-+-1

The 1st term of the A.P. ia n+m-1.

(c)The value of (men)th term

-n+m-! (m+n−1)(-1)

15,Solution:

(a)The ratio of their diameter

1+21%

1.1:1

The increase percentage of ita diameter..

1.1-1×100%-10%

(b)The ratio of their volumes

=(1+10%) 3

-1.331:1

The increase percentage ite volume

1.331-1

-33.1%

16.Solution:

x100%

The work can be finished by 48x5x8 men-hour or 1920 men-hour.

At the end of 15 days, the work remains

(1920-5x8x12-4x8x3)men-hour

-1744men-hour

The rate of work aust be increase by 134k-3x9x33.

3x9x33

x100%

51% correct to the

nearest percent)

學能推理練習專欄

智慧社主藕

數字推理練習(十八)

- A + BK

✦ k} (An■.)

(iii) Since AK LEF

(41 + kJ) (31 – 43)

BK. KC

3 6-3 - 1: 1 (An■.) 12(a)(i)The probability of getting exactly one gold- coin

10

TO (Ams.) (ii) The probability of

getting a gold coin

and a silver coin

100

the probability of

getting at least one

gold coin.

5.

18 10*100

20. (Ans.)

A.

b) The probability of getting 2 gold coins and

1 silver coin.

·

(100) (180) + (-5) (10)

(Ans.)

(ii) The probability of

getting no coin in 2

<dravo

-1-(10+20 +2)

100

As shown in the figure

let the height of the

balloon be h metera. htan60°.

BObtan45.

In rt.A A0B

2 40

AB - 20km m 20000m

2 x 10-

108-A02

·204

10000 m

The speed of the balloon

10000m

Smin.

2000 meter per min. ̈2000.

60 meter per second 33.3-ma

-1

33.3 (Ans.)

15. (a)(i) Since 2 is one

of the roots of the quad- ratic equation.

(k+2)x

2

2

- 2k +

-4 (Ans.)

the curve is

= (x-3)=.

When y

(x-3)2

-(x−1)(x-5)

1 or x 5.

5 (Ans.)

(b) The value of y is

lasat when x = -3

Given:AD//EP//BC.

To prove:(a)EO-OF.

8

Proof:

}If AD=8øm,

BC-12cm, find

the length of EP.

(a)Since AD//EF//BC.

·LABO=LABC;

ZAOB=ZACBEO//BC (corr. La

LEAO-BAC (3rd 4 of

are similar ~ (equiangular As)

BC-AC (corr. sides, similar ås)

【列】

數字推理練習一八

̇圖形的類比推理之

下列兩類左方的每組圖形都缺去

一個,在右方列舉的圖形中,選出 適當的補上。

AA

A

·B

塑解:上列第一、二丙組圖形都是在第2

中的小圖形上加塗黑色成第3陶;所 以第三虹的第3圖應是A

【例二】

A

願解:上列第一、二兩組圖形都是第1.

兩圓對稱的;所以第三組的第3題也 該是與蕭姐弟1圈對稱的。答案應斑

E

* −4 (Ann.)

AEO

As

ABC

the least value of

y = (3-3)2

EO AO

EO-BC

AC

Similarly OF=BC»

DO HD

AO DO AC BD

(proportionel

division)

(c) Since the graph is

y

(x-3)2

6x45

0

or y = x

(1)

4x

-6x+5= -2x+10

the required linear

graph is you −2x+10(Ans.)

2

(11) 2x - x

3. 0

x2-6x+5=11x+7

-E0-0F (Substitution) (b)20AD=20CB

(Altźs AD//BC)

ZOBC-ZODA 4A0DB0C(3rd 2 of 4)

下列各窿左方的每和圖形都缺去一個,就在 右方列舉的圖形中,選出適當的輔上,填在所溫, 答案的下面裹一橫隸。

AOD

Are similar

COB

(equiangular As)

A A

40-OD-4D AD

- AO-

AD

·AO+CO AD4BC

AO

8

i.e.

the required Lincer

graph is

EF=2x43cm

4471 m.

20

The probability of

getting exactly one gold coin and one .silver

coin is

(180) (20) + (206) (10) +(-3) (23) + (13) (157)

(ii) The equation is

-(−2+2)= −2(-2)+8

(x−2)(x+2)

x - 2 or

2

the other root is -2 (b) If m ̧n are the roots

of the equation

2_{k+2)x (1) ́ m =

- 2k + 8

- - 11x + 7 (Ans.)

*******THE END ***********

Answers to Revision Exer- cise Paper II.

1.D 2.8 3.B 4.A 5,C 6.T 7. 8. 9. 109

11 D 128 130 140 15B 168 17R 180 190 20T 21D 22D 23D 24A 25A 26P 27T 28S 298 30P. 31E 32D 33A 34C 35A 36P 37B 38T 398 40R 41 D 420 430 44D 45B 460 478 489 49P 50R

Solution:

Let b. be the height of the tower CD.

AC-h-cot45-ha,

BC=hcot30° x-/3bn.

In AABC.

471-h2+(√3h)2-2h(J3h) · cos30°

471-b

b2-4712

h-471

D

Π

D

B

C

3.

A

B

D

E

?

A

D

A

A S

B

C

E

MA

D

E

A

?

答案:E @D

QE

OB 6C