南教聞港頁一第張七第 日三十月九年巳丁歷衰
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郭日僑
日五廿月十年七七九一公年六十六國民華中
二期星
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·緻试行壯年來經
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順風旅行社發展
本港新聞
1
僧,不各者
2. Let the
numbers be
育教僑華
a and ar
8+ ar 39
-)(a)(ar) = 729
學能推理練習專欄
From
智慧社主
(1)
(2)
འམ
文字推理練習十
關係的類比 ( 三 )
729
9
+9+9r
£2
Substitute into
39
3r 10r +3 = 0
五就通出下列各關最適當的答案,作為發龉: 序列的最後一項。
3 or r-
【例】天空龍島:除地、走獸;海洋(2
A船隻......黠波波
Ans
***
D
the three numbers are
3, 9 and 27
EM
鹽帽:天空有癮真;险地有走獸;海洋當然 是有游离了。是活動範圍典動物的照
此。答案感遇D。
【例二】渴、软料;队、食物:病
AALB
D休息
CRIN
題解:解渴常要飲料;光凱食要食物:治病 需要躲品。是所需物品的類比答案 AMA
選出下列各腦衰源當的簽墾,作為該語序列 的最後一項,並在它的下面畫一橫
辣汁:鲜橙、橙汁
C花生糖
D花生物
B.$
ES
天空、飛機:海洋,船隻;
DB 想星、太陽;行星、地球
A星球 DAX
B* EL
BAI
C銀河:
E城市 動物黼物:植物、楊柳;礦物
C
DZ
E
A酸
B苦
CR
D
E香
D 星期
A#
B#
D#
ER:
ARM
D夏天
洗衣機,洗衣:孃氣爐、煮食;冷氣機 BRI E
C工作
10.賞月、中秋節:登高、重陽節;拜年、
12.
A**
D包巧節
B端午節
E聖誕節
蝌蚪、青蝰;衄、蒼蠅、不
A
B***
E跳蚤
B悭
EM
叔
C元宵:
CAŁ
33,黎明、朝霞;黑夜、星星
B殘月:
E晚言:
DAY 14、九龍旺角:香港、北角;新界、
- Q R S C
我
冠中環
E赤柱
C油麻地
?).
A**
B
·D五行
EAT
DE
A
OD
OE
6 A
OB
9: E
·DA·
d) D..
02 B
.DE
14 A
1978
中學會考試題預習專欄
明德社主編
新數學(四)委榮家
MODERN MATHEMATICS (4)
Solutions to Test two
Section A
1
ronta
6x
is one of the
roots of the equation
1. (1) Since 3
2
13x + k = 0
6(3)2 + 13() + × × 0
k-- 5 (Ans.)
(ii) Therefore, the equa-
tion is
2
6x2 + 13x
As shown in the figure above
Area of sector AOB
-60 * ☛ x (70)2
360
2566.67 cm
Area of quaḍ. OATB
2x (Area of rt, AQAT)
: : 2 × ( + × 70 x AT)
70(AT)
But
AT
0Atan30 70tan30
。 Area of quad. OATB
70(70tan30)
2829.26 cm
Area of the shaded region.
(area of quad (ATB)
(area of sector OAB) -.2829.26 -2566.67
#2 262,59 cm (Aus.)
4. The even numbers between
2001 and 4001 which
divisible by 7 forme
arithmetic progression with
first term a = 2002
common difference d* 14 last term = 3990
Let the last term be the nth term of the A.P.
3990 - A+ (n-1)d.
- 2002 +(n-1)(14)
. 143
The sum of all the even numbers between 2001 and 4001 which are divisible by 7 is the sum of the 143
terms of the A, P.
$(143)=143(2(2002) + (143−1 )( 14 )
428428 (Ans.)
5. Since. T a mũ nữ
1+j - m(21
+n (1.
31)
-3)
► (2m+n)I ~ ( 3m+n) J3
2mn-1
*
n) 1 .....
1--2
n - 5 (Ans.)
6. (141)2+(1+1)+1
(1+1)
70
(1+i) + 1 +
(1+1)
+1 +
(1~1)
- + - +
INIS IN
(Ane.)
5)= 0
sine - cose - -
(sing + cose)
B) 2. 25.
sin28 + 2cosesino +
25
4. 2sin@coso m
2cosesine-m
-
5 - 0
(3x
-
1)(2x
·
言
5
or x.
Σ
the other root is
(fii) Let P
and P
be the
populations after a years and n+1 years.
The increased percentage
(aine sine
COB 0 -28inecoso.
1- 2sin@cose
1
(룩) 17
(Ans,)
As shown in the above
figure, AF represents the path and AE the line of greatest slope,
hAFsinë
h - AEsin30°
AEsin30
P
n+1
x 100%
P
n
k(1.15)"
n+1
k(1,15)" k(1,15)?
k(1,15)"
k(1.15) 100%
(1.15 - 1) x100%
0.15 x 100%
15% (Ang.)
14.1+3
+ 56
2(1)
to n terms
(n-1)(2)
(n+1)
1.9
#0(n+1)
(n+1);
n terrs
AE sine
AF
sin30
sine.
CO845
sin30
ain30°cos45°
0.3536
620942. (Ans)
The slope (or gradient)
the path
tane
tan 20742
0.3779
Section B
(Ans.)
9. Let the statement
proved hei
ar + ar
a (r^-1)
When n
L.S. & R.S.
2
s(x-1)
it is true for nl, Assuming that the statement ia true for nok, hence
aar + ar
a(rk - 1)
-1)
consider when n k+1
2
ล
ar
ar
شاه
*(-1) ark
ärk
ar
K+1
a (k+11)
it is true for n k+l if it is true for n➡ k.
By principle of mathematical
(a) if Yet then
a
.
(= -1)2.
(b) if (0,b) er
(Anh)
then
2(0)
+
2
(c)
* 2
2
2x + 2
2 (Ans.)
--x 44 is the
uired additional linear graph.
From the graph,
■ -1 for x - 2 (Ane.)
(d) 3 is the positive root
$
of the roots of the
qua
tion
- 2x +
3
From ther
-2x+5 is drawn
graph, the roots
3 are 1.73 and -1.73
- 1,73 (Ans.)
As shown in the diagram above Let 0Ch be the height of
induction, the statement is the tower. The angle of
true for all natural numbers
10.(i) The probablity that
both cards are odd
502 9C2
18 (Ans.) (ii) The probability that the sun is odd
The probability that one card is odd and one card is even
51.41 9C2
(Ang.)
(iii) The probability
that the sum is even
1 (probability that the sum is odd)
2
4
сов
9
(Ans.)
11.The graph of f is plotted
as below:
elevation of C from B is.
AB 4h
In rt, AUC,
¿ACO➡ 90° 25°
AO htan ACO
htan650-
In rt.ABOC,
ABCO =' (90°— .6).
BO - htàn(90°- 8)
In rt. AOB,
AB2 DA2 OD
Therefore
(4h)2=(htan65°)2+ htan(90°-0)2
16 - tan 65o+ tan2 (90°- 0)
*¢¢n{9% !
- 16 - tan 65°. -.11.399
tan(90°- 8) = 3.376
(90°-6) - 73°30
≈ 0.
16°30' (A)
13. (i): In 1970,
P: 4000000 n = 0 。°, 4000000 -
(ii) For 1980
k(1,15)
k 4000000 (Ans)
n = 10
P = k(115)10
4000000(1,15)10 4000000(4.046) 16184000 (Ans.)
1
n
-1.9(1+n)
n(0,ln - 1,9)
u)u
- 19) - 0
= 0 (rejected) or n=19
And in 19
(1) Since interest is credited yearly the total amount-returned after 10 years is given hy
10 10000(1+10%) -+10000 1+10%)? +10000(1 +10%)8
10000(1+10%) * 10000(1.1)1o+10000(1.
10000(1.1)
10000(1.1) (1.1)1o- ( 1.1 - 1)
10000(1.1) (1.594
(0.1)
175340 (Ans,)
Ans: $175340 will be returned
after 10 years;
(11) If interest is credited
half-yearly, then the total amount returned after ten years will be 10000 (1+5%) 20,
10000(1+5%)
16
10000(1+5%)18
+...+10000(1+5%)2
10000(1,05)2 +10000(1,05)18
+10000(1,05) 60
10000(1.05)2
10000(1,05)2 (1.05)20.
(1.05)2
10000(1,05) (1.653)
(0.1025)
177798 (Ans.)
Ans: 8177798 will be returned
after 10 years.
BR
-BA
BR
IBA
OR = OB BH
+XBA
- OB
(Ans.)
(b)(i)
According to the result of part (a), similarly, we have.
OR yoc (1 Y)OD (1K) 11 00 = fo and OD - 300 then OR - y(0) +(1−y)(308)
•₤0A + 3(1—y)08 But,
OR - ¤¤Â + (1 Therefore,
X =
x)01
I From which
3(1-y)
I
(Ans.)
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