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°