育融情帶其一第張八部日七廿月十年午戊睡
WAH, KIU: YAT PO
郭日橘
-
Section B
1979
中學會考試題預習專機
Modern Hathematics (8)
Suggested
test 3.
Section
'As shown in the figure,
0 is the centre of the
circle.
The area of the v-uliaped
polygon
={Area
of trapezium (ABFD).
*(Area of trapezium BCEF)
-3(2)(10×14)+}(2)(10614)
(Ans.)
2. Let the length of a side of
the cube be a
(1) Initial area of a face
of the cubewa
Final
of
24
06
men of the fuen
cube(142
*100 -(1.96)2 -(1.4a)
final length
of
side of the cube
2 CFD=180′′-(60°+55°)
4EOD-2BFD
=130°
But ZEOD+ZACB=180°
ZACN=180′′- LEOD
-180°-130"
(anu,)
30°
Let BD-xen, Aboliem
Inert. AABD
40°
In rt. BACD
h tan30. -(2)
Substitute (1) into (2)
10-htan40.
percentage change
1.40
1004
(a) Let T(n) and S(n) denote the nth tern and the sum
of first n terms of the
sequence repectively.
(1) _ _T(1)-8(1)
−(1)[2(1)+1)
(11) T(2)=$(2)-s(t):
-2{2(2)+1}-3
T(3)-8(3)-s(2)
-(2(3)-1)-2(2(2)+1)
-11
(414) The progression is on
arithmetic progression
becouse
T(n)=5(n)−8(n−1)
n+1)=(n-1) [2 (n-i) 2n2+n-(2n2-Jn+1)
▼{r}~T{r-1) -(hr-1)-[1(x-1)-1)
sconstant (meÐLIMON
difference)
(b) Assume a teras aust he
taken
(2n+1)> 171 +n÷17150 (2n+19)(n=9)> (
<} (Fbjected)
>9 (accepted)
6日七十月一十年八七九一公年七十六翼民豪中
-=2(15.5889)x(9.38)
Ans: The cost for 320
students is $2440,
(Ane.)
12.
14.
(a) From the given tal
huve
(a) Since the circle passes
through (1, 1).
sumber: probabilitý
1203 1000 25
100
1000
200
1000
100
1000:50
150 1000 20
250 1000 T
The required probability
-( { b ) ( t ) + ( } ) ( {} _ ) + ( } )2
The required probability
(중국) (조각)+(중국) (룩)+(주) (초급) +({}) (~ 5 ) + ( } ) ( } ) + (})(?)
+(_)(2)
2200 10000
(Aus)
The required probability
128
1~+12+2!(1)+2g(1)+3=0
21+2g+6+2=0~~~~~~(1) Also, the circle passes through (1, 3)
12+32+2? (1)+2g(3)+c=0
• 2f+6g+c+10=0---- (2)
The centre of the circle is (−f, -g) and is on 2x-3y+6=0.
2(−1)~3(−g)+6=0
21-32-6-0- (2)-(1) 4g+8=0
-(3)
g=-2 (ARS.)
substitute g=-2 into (3)
Lm0⋅ (Ans.)
substitute fTM0) and into (1),
2(0)+2(-2)+c+2=0
(Ans.)
(b) The centre of the circle
in (0, 2).
15.
radius- {{ 1-0)2+(1-2)2
-√2 (Ans.) (*1.4142)
(i) From the given table.
(ii) Final veline
cube #f1.4a)-
-2.744.3
The perCED!
change
sino.co
100%
-174.4%. An*,)
-0,7276(Aus
J
吉
m0.7276 (Ans.)
(1) Since T(n)sthe nth teru of
the A.P.
=5n-3.
~T(1)-5(1)~3
the first term 2.(ans) T(2)=5(2)~3
10tan30
ëtank0
10(0.5774)
*1+(0,8391) (0.5774)
#3.8805
AD=3.8895cm
(AMK,)
Since d and p are the roots
the equation
(1) Idiop
(ANS,)
(11)-
10 terms must be ken? (Ans.
(b) coxZIIT-
A?•¥Þ«(1)(−1)+(2)(1)
-cos EPF=0
[PT=94" (Ans.)
(a) Let C-cuat of the school
pienic
ninumber af students
3(3) 2(4)+3(+)
(Ans.)
vhe
kand
constanta.
When C-935, n=195
(0)
Since ya(ax-2)
97*»k, +105k^~~~~(1)
y=k(8x=2) where k in constant
When C-844, n#92.
-(2)
2=k(Ja=2)...
-(1)
(1)-(2).
When xeh, yuk
b=k(ba+2) Lək(2a-
(1) + (2) 1.
corimon, difference
-T(2)-T(1)
(AUB).
Substitute k„-7 into (2)
8b4=k,+7(92)
#7-2
(Ann.)
(ii)Substitute awl into (1)
2-k(3-2)
3-200
(11)5(10) the sum of first,
ten törm® --
-29(2(2)+(10−1)(5)
245 (Ans.)
(a) As shown in the figure,
is the centroid of AABC.
VO perpendicular to
plane ABC.
Since AABC is an
equilateral tria
BDLAC,
OL-D
In rt, AABD.
BDABOG 30°
Marks | Midɩvalue | Frequency
20-30
30-40
35
40-50
45
50-60
60-70-
GGGG
25.
10.
55
20-
28
70-80
00-100
mean score
~{4x25+6x33+10x45+20x55+28
(11)
x65+20x75+8x85+5x95)÷100
(Ans.)
(-5.196cm)
•*.
OD=3(3/3)
ii)From the graph,
(−1.7321cm)
the pass mark is 52.5
Inert. AADV AD=1zfcn=3c#
(Ans.
(n) (3x-4y-14:
-(1):
VD-JAVI-AD
- (2).
-91cm (*9.5394cm)
(2)x4 Rx-4y+16z=0-
-(3)
(3)-(1)
5x+30x-0
the formula is C=200+7B. (Ann.)
C=200+7(320)
#2440*
5
0.1816
ZODV=79°32' (Ans.)
(b) aren of & ABC
►ģ{AB} {a€) { gim£BAC) ➡k(6)(6)nin60o „9/Jcm3 (Ans.) (-15.5889cm VU-√VD -OD -√91–3
-9.35cm
Volume of the wolid
(1)x2
6x-8y-28z- (2)x3
6x-3y+12z=0.
-(5)
(5)-(4) 5y+40x=0
y;28:1 x:y:2=-6;−8:1 (Ans,);
(or 6:8:-1)
(b) Let x--6k, y=-8k, z-k. when kisa non-zero constant
3(~6k)3~(~8k)3+8(k)3 -16 -128k3-16 k3--
x=-6(-1)-3 y=−8(-4)-4
*--*
(Ans.)
(b) When n=320
When x3
[y=2(5=2}
(ang. )