1980-06-18 — Page 26

頁二第張七第 日六初月五年申庚圈夏

WAH KIU YAT PO

EXAMPLES:

Revision Exercise for Junior Forms

1. Find to the nearest cent the interest on $3600 at 24% per annum for 150 days.

3.

(H,初中課程綜合練習

雄風社主線

中二數學

BANK INTERESTS

NOTES-

1. Principal (P) Th sum of money deposited in a bank.

2. Interest (1) - The money

- paid by the bank for a

principal.

3. Rate (R%) - The percent

principal that is paid as interest.

4. Time (T) - The period for

Lac principal put in a bank.

5. Amount (A)- The sum of-prin-

cipal and interest

6 p.a. per annum -

1980

中學會考試題預習專欄

新數學建議參考資料 明德出版社提供

Suggested Solutions sto

Modern Mathematics

SECTION A

1 (2x+40)

(Ext. Z of A)

40-2x

x=20

a(3b-c) (3b-c)

=(a+1)(3b-c)

(b) x 1

150

I ='$3600 x 24 x 33 25 $36.98

報日僑

三期星

2. If the interest for $7200

at 41% 18 $59, find the number of days the sum of money has been deposited. Find the sum of the prin- cipal if the interest on. it for one year is increased

from 4% to 42%.

日八十月六年〇八九一层公年九十六國民華中育教僑華

2. Subtract the first from the second

EXAMPLES:

1. Subtract 5x3-8x2-5x from

4x+2x

(5x3-8x2-5x)

4.

4x3-

The value of rice increased by 3% every year at a comp ounded rate. If the present 2, value of rice is $1.50 per catty, find ite value 10 years later.

1546

Add

to 4x2+2x2+5 (4x3+ +5)

2x2-5x+5

2-3p+0

~~4x; x4-4) -4x+6; 10-8x+13x2 5. Simplify the following:

4x2-5x+9x-8x2-3x+7

b) (x*-8x+7)=(5x2+3x)+2x2

2. IT the interest for putting. $1600 at 6% per annum is $120, find the time the principal has been deposited.

= P 1 R% IT

120 160026%

The time is one year and three months.

QUESTIONS (1): –

If the interest for $7200 for 24 years is $81, find the interest on $5000 for 5 months at the same rate

Substitute y-26 into (1);

we obtain

x=36-y

10/

a(1+13g)=b(1-100)

(Ans.)

bx

··1:00

bx

100 100 -

a+b)=(b−8)

100

100(L-a)

a+b

(Ans,)

The daily wage for a seni- skilled worker

$120 x

-890

(Ans

The daily wage for an unskilled worker

$120 x

-860

(Ans.) —

Mean daily wage 10x120420x90+30x60

10+20+30

-(x2+1)(x^−1)

*(x +1)(x+1)(x-1) (Ans.)

$80

SECTION D

(a) (1) In rt. 4 PCA,

ANSWERS QUESTIONS(1));* 1.89.375

2. 3. 84000

66days 4. $2.016"

SIMPLE POLYNOMIALS

CUESTIONS

1. Add together

Addition & Subtraction: Like other algebraic express- ions, polynomials may be added & subtracted.

=2(3)-2(2)

CD

cos/RCD=CR

/RCD=60°

:60

(дns.)

The common ratïc

100k ∙10k

-6x+2

3y +47 −5y+1; 25-47+6 |c) 2+5x+6x2+8x3; 5x=2x2+3x3

d) 7x2-3x+4; 5x2-6x+7;2x2-4x+5

From the graph

13

y=8

number of first seats 8

number of economy

class seats - 48(Ans.)

(9) (51.43) (x1+y])

=3x+4y

(11) [04|

(Ans.)

(Ans.)

(ADS.)

OP

(iii) cos LAOP

10 Ans

(Ans.

-10

Sum of first n terus

(b) cos/BUP

OB OP

k(10o-1)

10-1

(Ans.)

10

-1)k

(Ans.)

ANSWERS

b)55+

+10x+2

-13x+16

2. a)zz2,

+4x+1

d}4x2-4x+4

a) −4x2+X+

b) 3x2-5x3-9x2-112+9

A is (2,0) and H is. (8

(Ans

+8y+16=0

8

Slop

(0,-4) (Ans.)

of TB

00

AC//TB

slope of AC=

Apply point-slope

form, the equation AC 18 ven by

(ANS,

(H)

+160.

From (1),

_gf_roots==> Ans.)

the other root

product of roots

tan X

10k-1

10k

tanx

(Ans.)

(ii) In rt.4 PCB

81010

on

(Ans.)

10610

100k-log 10%

Lang

(An's-)

100k

10 10k

(b) In rt▲ ACB, ZCAB=rt

log.

· *. 5+(-1)=-

(Ans.)

gine=cos120

=-00560 0.5

0-180o

'+30°-210° (Ans.) 0-360°-30°=330"

the length:

AC=\cos30° BCI-AC

-cos30%

es30°

50 k{1cpg 0

50ME

BC AC +AB.

tant

+4002

44005

tan 60

(Ans.)

+4:002

b2x4002

-x4002

correct:to

length of the rod is 373mm

Suppose x mothers lost only

one of their children.

y mothere lost two children.

(x+2y=62

y=26

(2)-(1)

(Ans.)

and.

400

-200

-200(1.2247)

#24h94

-245- conect -sig.

AO-AP

LAOP=LOPA-O

[PAX=LAOP+_ OPA

-20

Similarly.

(b) The

QBX-20

RCX-20

(âns.)

Ana

OCR are similar to each other ... Therefore

secrRNA AP MOBO and·

Area of sector OAP:area of Bector OBQ:area of sector TOCR

A0B0÷GO

9:16M

CR-CO4: _CD=0D=0C:

-280-20A

(Ansz)

Toto

10810k, log,10k and 10810100

100k are in A P.

(ii) S ̧ [2a+(n−1)d]

=10810k, d=10g1010–1

Sum of first n terms

-[210g1k+(n−1)(1))

[2108+(-1) (Ann.)

When n=10

80-210g+(10–1)

=101

0810

k+45 (Ans.)

Let x be the number of

economy-class seats installed and

be the number of first-class seats installed.

x+1.5y≤ 60

10x+30y ≤ 720 DIEN

YEN

Let the profit of an economy-class ticket be

profit=(px+2py)=p(x+ -p(x+2y)

(c) If OP➡xi+yi represents

the bisector of LAOB then

ZAOP -ZBOP

or COBLAOP=cos¿BOP

3x+4y

-7y=0

(Ans.)

(a) The required probability

(Ans.)

(b) The required probability

− ( 8 ) ( 1 —— ) ( - ) + (1 - 18 ) ( ) (78)

+30

∙13

(Ang.)

(c) Probability that à does

not obtain the

qualification by sitting each paper once -probability that B does

not obtain the qualification by. each paper once

itting

tobability that BOTH A and B do not obtain the

lification by sitting ch paper once

The required

ability

(2y+2)+y=10(2y+2)

y(5y-4)=0

y=0 is rejected for corresponds to pos When

-2()+2

is

(2o, §) (A¤s.)

-5-0

#2x+5

the required straight

line is y=2x+5

1-3.4

(Ans.)

Fron

graph

·1.4"

(b) x2-2x

XG-1.4

or x>3.4 (Ans,)

(c)

(Ans.)

=2x-5=0

x2-x--0

0x+8y+16ml

15

0x+16-0

(x-2)(x~8)~0

the required straigh: line is y=x+r

From the graph *--1.2 or

x=3,2°