頁二第張八第 日五初月四年未己愿 WAH KIU YAT PO
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日十三月四年九七九一层公年八十六國民德中 VT ANY TIE ME
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MODERN MATHEMATICS (30)
SUGGESTED SOLUTIONS TO
REVISION EXERCISE, PAPER
SECTION
1(a)AÅDE∞ A BEC
BC
AD1
Area of ABEC-
rea of Aah (2)
Area of ABEC
x (Aren
X-4 cm
16 cm
▲ADE)
(Ans,)
let the area of AAEF be-
Xem and the area.
ACFE be Yem
X
4X
5Y
16
(麵)”
(2)
(2) x
40X
- (3)
·(3)
(1):
36X
96 = 0
(cm2)
2(a) Let the cos
× (1+20) x (1-180)
1.08p
08P P 3600
cost of A be $P
C paid:
20
From
27(1
9,2
9r
(3r
2)(r
Lot E
·(2)
From (2)
Aa r ís less than
∙the
Substi
into (1)
18 (Ans.)
1
(Ans.)
(Ans.)
(b) Ansume ʼn terms must
be taken.
a(1.
(5-1)(2
18 (1 - (3)")
30
25-*
。
the set of students
assed in English the set of students passed in Chinese
the set of students
passed in Mathematics
the sumber of students
all three
Since
are altogether
stüdeuts passed in English, therefore (1−x)+(x)+(25−x)+(30)
60
when
when x
15(a)(i
and Q are and (6, 2)
CF. FD
and BA 151
10)
DF --103
53
(Ans.)
AF
AD DE
107 (Ans.
Since BM : ME - 3: 2
apply the formula p
27 [1 − (4)")> 26.9
13- (/3 + 1)(/3
27 - 27(3)" > 26.9
0.1 > 27(3)”
5-1
B-1
- -- + 1/5
(4)" <
ning (4) < 108276
log
kahb
(a)(i) The required
probability
.. AMA + SAF
150
2(~15])+3(10f-103)
30
The
(Áns.)
required
0.477in < −2.4314
2,4314.
probability
20 30
150:
(Ans.)
5.096
(Ans,)
6 terms are taken.(Ans.
(iii) The required
is
(where k is constant)
ncreased by 10%,
then
Ty
1.21
(
0.8264
it is ecreased by
0.8264) 100%
17.36 %
17.4% (Ans.)
2/ACB 30
· LA OB
of sector 40B
(ans.)
(b) From the gra they meet
at 1927
13(a) The mean score
10k
miles
from
11(a)A
6
240°
3x35 + 5x4.5
50
10x55 + 14x65
50
4x85 +2x95
62.5
157
L BF
10x75
BF
(Ans.)
-)1-15j]·(101453)
750
probability 150–【(60)+20+(15-x)+40]
150 130 150
(Ans.)
The required probability
10
12
13.52 (ADS.)
102
* 11.18 (Ans.)
(61-123) (101+57) 60 60
AM BF
(b) If BE BC - k
then
ARKAC - AB
ZAMP
90%
k(101- 153) + (−153)
(111) AM • BE
08P
3600
45000
paid $45000
paid $45000 x
paï
(Ans.)
120 100
$54000 (Ans.)
$45000 × (1–100)
#48600 (Ans.)
Profit of A
= £54000 - $45000
Blosses $54000 $48600
(Ang.)
Multiply
both sides
3)
2(x
312
2(x
3) [2(x − 3) - 1]
(x − 3)(2x - 7)<
360 × (x102)
x
CM
area of A BOCƐ
≥ (OB)(OC)-sin/BOC
(10)(10) sin150°
g* (10)(10) •sin30°
25 cm.
Area of the shaded region
(惡化+25) m? (()(3)+25) cm2
50cm: (Ans.)
The centre of the circle
-(2,、,3,4)
(3-1)
Hadius of the circle
(3-2)
In the figure, If is the
mid-point of CG'
and [CUM = }/CDC'
MC - CDsin20"
2.052
+BC
10. m
sine
MC AC
2.052
10
0.2052
(c) If 60% of students passed
in the test, then 20
students with scores less
than the pass mark. From
the graph, the pass mark.
is 60.
14(a) The equation of L
(Ans,
100k
16(a) cost m
100k - 75(k+1).
(Ans.) 0x + 60y)
* 15y > 180 140x200y » 1400
xyEN
(c) The constraints in (b)
can
be
simplified as
12x + 5y > 60 7x + 10y 3 70
YEN
/DAM 35 10
the angle between AD
and the plane ACC!
35" "10".
(Ans.)
Ay - 14. we.
(b) Radius of the circle
✓(16~3)2
The equation of the circle is
(x-6)2
(c) [2x +
{(x-6)2 From (1)
25 (Ans
14
7)2
25.(2)
2x
142
(3) Substitute (3) into (2)
(x-6)2 + (14-2x-7)2 25 (x-6)2
(7-2x)
36+49–28x+4)
460
As shown in the graph; tha wolution set if the constrainta is: the set of lattice points..
(4) From the graph,
xw5 and
Therefore, the most economical way is ta hire 3 bús A and Ga
The coat.
80(3) + 60(6)
• 18500:
(ano.). ANSWERS TO REVISION EXERCISE
PAPER IT
D. 12, B
· 13. C.
23. D 34. C 45. 8
24 A 35.7
A
14. D
25. A 36.
g 15.
26. ▲, 37, G
5. C J. D. 27, E 38, E°
17% C 28. D 39.0 50.
29,040. C
G. E
7. E 1R, G.
30. A
8. E. 19. B
9. E 20. E 317 * 42, c 10. 8 21. D 11. D 22. D
31. D
33.
32. D 43. B 33.A
54
NA, D.
END
3 < x < //
Multiply
19
11 50
the
rele
[CAC
29
both sides by
(x − 3) < 5(x − 3)2
(/10)2
23o 40 (b) AM = AC•cos?
5(x-3)2 - (x
10
9:787
· (x − 3) [5(x - 3 } – 1
(x - 5)(5x - 16)
COS/DAM
9.787
SECT FON D.
9(a) Let
the first term of
Combine the results of the
two inequality, we obtain
(Ang.)
the
progressi ion,
common ratio
the progression
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