買二第張六第
日二十月三年申庚麽至
WAH KIU YAT PO
郭日僑華
六期星
日六廿月四年〇八九一腊公年九十六國民餘中
1980
(x5)(5x16) K0
MC
Or
<3 or
*> 16
6x+2y
sine
(b)
AC
中學會考試題預習專欄
SECTION B
新數學 C卌D
Combine the results of the two inequality, we obtain
9(a) Let a = the first term
of the progres-
2.052
10
0.2052
sion
11 50
明德出版社魯榮家提供資料
20 < x <- 2
(Ans.)
the common
•.[CAC
20
MODERN MATHEMATICS (30)
W.K. LO
(MILL & DALE PRESS LTD.)
2y
(1)
ratio of the progression
(b) AM
AC,coso
SUGGESTED SOLUTIONS TO
243
REVISION EXERCISE, PAPER I.
10co8l1°50 9.787 m
SECTION A
47
27
8.
1. (a) A ADE CO A BEC
From (2)
a = 27(1-r)
~27(1+r) (r)
2
9r2-9r + 2 = 0
COS / DAN = 9.787
* 0.8174
ZDAM = 35°10'
the angle between AD and the plane ACC' is 35°10. (Ans.)
(c) If 60% 6 of students
passed in the test, then 20 students with scores less than the pase mark. From the graph, the pass mark is 60.
4(a) The equation of L is:
0
(Ans.)
3
347
Area of BEC
Area of ADE
(4)2-4
·(2)
From (2):
Area of BBC
2. (31 −2)(3r-1)
2
4 x (Area of A ADE)
4 x 4 cm2
16 cm
Let the area of AAEF
be X cm2 and the area
of A CFE be Y cm
3(4y-5)
or
(Ans.)
(Ans.) (Ans.)
As r is less than therefore r
3.
Substitute
15-2
245
(b) Assume
5
(3-1) (2+3).
(3+1)
13-1 (3+1)(3−1)
↑ (Ans.)
be taken
a(1–
into (1)
(Ane.)
në must
(1-1) > 26.9 18[1-(3) *] -
> 26.
271-26.9
27-27 26.9
0.1727(3
(3)<270
nlog(3) < 198270
2.4314
-0.4771n
n>
(where k is
constant)
n>5.096
terma
taken (Ans.)
10(a)
If y is increased by 10% then
(1.1y)
1-21 (2)
0.8264
it is decreased by (1-0.8264)x100%
S
X
Let E = the set of
students passed
in English
= the get of
students passed
in Chinese
the set of students passed in Mathematica
the number of
students passed all subjects Since there are altoge 60 students passed in English, therefore (10−x)+(x)+(25-x)+(30)
x = 5
(a)(1) The required
probability
60
(b) Radius of the circle
(16–3)2 + (7-3)2
The equation of the circle
18
(x-6)2+(y-7)2= 25
(Ana.)
2x + y = 14 = 0 ...(1) (x-6)2+(y~7)2 = 25....(2)
From (1):
Y = 14
Substitute (3) into
(3)
(x−6) + (14-21-7)———
(x-6)2+(7-2x)?
25
−12x+36+49~28x+4x2=25
2-40x+60=0
-8x+12 = 0 (x−6)(x-2)
X2 or
when x 2,
J = 10
when x
y 2
P and Q are (2,10) and
(6,2)
15(a)(1). CF:FD = 1:2
and BA - 153
(3)- (3) 2 = 4
(1)
,4X-57+16 = 0.
XY+16 (A) 2 (+)2=
8X-Y-16
(2) x 5:
40X-51-80.
(3) - (1)
36X-96 -
(cm2)
(Ans.)
2(a) Let the cost of A be
SP
..C paid:
20
x (1+206)
1.08P
1.08P-P = 3600
0.082 3600
P45000,
(1-100)
paid $45000 (Ans.)
B paid $45000 x
120 100
854000 (Ana.)
C paid $45000 x (1-100)
$48600 (Ans.)
(b) Profit of A
$54000 $45000
$9000
(Ans.)
B losses $54000-$48600
Multiply
$5400 (Ane.))
both sides by (x-3)2 2(x-3)2 < (x-3) 2(x-3)2-(x-3)< 0 (x-3)[2 (x-3)-1]<0 (x-3)(2x-7) <0
3<^<} ===3<5 Multiply
both sides by (x-3)2 (x-3)<5(x-3)? 5(x-3)2-(x-3) < 0 (x-3)[ 5(x-3)-1] < 0
2
17.36%
17.49%
LAOB
(Ans.)
2 L ACD = 30°
Area of sector AOB
300 x (71102)
2 cm
Area of ▲ BOC
7.(08)(00).sin LB0C 1.(10)(10).sin 1500
§. (10)(10).sin30°
25 cm2
Area of the shaded region
*(3π + 25) c¤
[(32) (3) + 25] on
50 em
(Ans.)
8. The centre of the circle:
(24 2-4)
Radius of the circle.
= (3-2)2+(-1-2)2
√10
the circle is.
(x-3)2+(y+1) 2 (x-3)2+(y+1)2 10
(10)?
(Ans.)
(b) From the graph, they
at 1127 (p.p.) 3 mest miles from A.
11(a)
In the figure, M is the mid-point of co'.
and
... DM_LCO
AC
CDM -
CDC!
CA in 200
2052
AB +BO
20
(Ans.)
(11) The required.
probality.
20+
(Ans.)
(111) The required
probability
150-[(60)+20+(15-x)+40]
1505130
(Ans.)
(b) The required probab-
1lity
- 30 - € (Ans.)
13(a) The mean acore
x25 + 3x35 + 5x45 +
10x55 + 14x65 + 10x75
4x85 + 2x95
62.5
(Ane,)
53 -103
101 + 5) (Ans.)
DE
AF » AD
101 -10) (Ans.) Since EM:MF = 3:2 apply the formula P
AM
2A3 + A
2(−15,5)+3(103−101)
123 (Ane.)
(11)
AM JE
12
13.42 (Ans.)
BF -√102 + 52
(111) AM.BR
11.18 (Ans.)
(61-12)) (101+53)
☐ 60-60
AML BE
(b) If BE: BC then
900
• K(101-15))+(-15.)
- ( 20 ) - 157
*
AE L BE A.
0
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