買二第張四第二日七十月四年西己壓 WAH, KIU YAT PO.
夏季橋
僑
ま
英中會考數學科答案 (穎),歐陽錯交。
.E.E.(English) 1969:
Arithmetic & Trigonometry
ested Answers
1. (a) This de a problem on compound interest. In which,
the investment is equivalent to the principal. (denoted by $P)
the sum $3C000 is equivalent to the amount
11 P.k. 1n 3 years time By the compound interest formule (1)
郭日僑翠
日期星1日一月六年九六九一番公年八十五國民華中育教儒藜
When a certain amount of zino is added so that the percentage of zano 18 to be 665
665
Then, the % of tin is 1 - 655% - 33%
33% of the weight of the new alloy
260 10.
The total weight the new alloy
260 lb. 33
260 lb. 3
780 lb.
aight of zino in the new alloy
780 1b,- 260 lb. Weight of zinc added
NOTE
520 lb.
5201b.-
- 380. lb.
The problem can also be solved by Algebra.NE
There are 1401 b. zino in 4001b, alloy. If x lb. of zine must be added to change the final percentage of zinc to be 66-%
5.{a} As shown in the figure, Suppose that the longer
string is x on long."
By sine formula,
50
sin: 75
sin360
508in
No.: க.
1.6990
Ste 75*. 7. 48.49:
6719
19147
Nath
-82.17
82 (cm)(to the nearest cm.
In the figurë
Q: Queen Mary Hospital
Si Sek Kong
C: Cheung Chauss
Such that Sq = 10.8 mi)
∙CQ.83mx
30000 (17+
- 30000
1,083.
*23820
3000
Rust
The present investment
$23800 (comm, to the osarest $100)
(b) In the business contribute
$100,000
Y contributes $60000
contributes $50,000
The proportion of their contributions
$100000 $60000 $50000.
· 10 6
dence, 2 shares.
the business
2 acte as manager, then he receivOR
the profite first.
2 totally receives
profita
receives $56000
21 of 1 ) or the
(¦ • 24 ) of the profite = $56000
the total profits of the busines0
$56000 ± 1 -
$56000 1
$168000
radius of the cylindrical
6 inches
-sectional area of the jaz
367 #q.in.
Radius of each iron sphere. Volume of each iron spher
-T(1))- cu.in.
22curin
140+ I
6636 -
400. +
30140
2(400
380
(b) The 1st train 250 ft. is moving at 60 m.)
The 2nd train 200. ft. is moving at 45 m.pad (1) The let train moves 60 mi. in 1 nr.
The Ist train movem
60 160
60 x 1760 x 3
The 1st train moves 250 ft. in
60 x 60 x 250 Back 2.0 880. 60 x 1760 x 3-
The 1st train will take 2.8 sec. to pass
a milestone,
1 hr., the let train catches 15-mil-sa i... To overtake 15 miles, it takes 1 hra
To overtake 1 ft., 11 takes
15x
·x 3
60 x 60 1517
вес
1601ort it takes
15x1760x3.
20.5 sec.
Thesist train will take 20. pass the 2nd train.
50x60
iii) The lst train travels lrt, in 60x1760x3
The 1st train travels 7001
60x60x700 a
6011760x3
08.0 sec.
the time takanato Apasal conplate platform 150 vd. 3one da 8.0 sec,
NOTE: The answers are given to the nearest tenth
of a second.
(4) As shown in the figura
P be the window in the house
so that PQ- 20 ft.:
TB regresents the tower
Aasc
(8 24
658
By coa ine
SQ
2.50.CQ.com 116 58
10,8 x 8.1 x cos116°59'
116,6+68,89+2x10.8x8. ixcos63°2
116.6+ 68.89 81.32 266.81
SC- $266.81 10.33.
The extra distance flow
(SCCQ)-SQUA
(16.33 1.48; 3 mi) 10.5m1
0.3010
9.3
Product-19102
13.8 mi to the nearest tenth of a mile.)
700
sin?o - ninoܟ
38569 + 1)) (26 ing
15- 0 or 2 sir
eing
If sino
aine 0.50 -0.3133
0.5 then 30% ör
-073333 then
030 or 150 o 199
360° 0=0.
340 321
or 340
When the 8 equal spheron are completely immersed, the rise water-level der *
• (~2 = 3677 ) inones -
BP
W
BR CBC25 32*3
20 #025 32 (rt
L
- 90° - 58°47' -
Tom Pt.ATPR
Apply the sine formula ATPB
the base-diameter of the right cone.
3.62 com
bace-area of the cons
TB sinTPB
PB
Bin T
TB
PB ein TPB sinT
20.os c25°32
sin84
89.06
40
Tanza
"Cat 25°31'} 0.3846
Si No TA ZE. Num. 1:6643 Son 3}°13′- | T. 7146) Exp 7.9447
height of the right cone
Volume of the solid cone
x 1,812 x 5.14 0.0..
2
1.81 x 8.00.
5.14 cm
the length of the solid rectangular block
4.58 cm.
Area of its cross-section.
1.815
5.14 4.58) aq.on.
The length of a side of the square (base
81
·3 x 4.58
1.962 cm.
·44971 03154 -0.7/10
∙COTT.
Num.
1.7435
3
FACT
0.477) 10.6609· Y-370 10.5855
10 +927
3.(a) In 400 lb. of the original alloy,
the weight of zing is 35% of 400 lb. 140 lb.
the weight of tin is 400 lb. -
140 6 260.1b.
89 ft. (to the nearest foot)
(b) Lat C, P be the positions of the church, school resp
PYA
·300
In \ PXI |
PXX
90
AFXT is
equilateral
100, IP – X – 400 và.
From Ft. A CXY 1
CX = XY tan(315° -270°),
400tan45 (rd)
400(ya)
AXCP is isos. The length CF can be found
by cosine formula or by the following
methods
Drap XKLCP at K
Then CP 20K
2.CX.sin15
•NO | fog
800 2.953 Sen 12o 741-3p
£
123161
CP. 2x4008in15
207
corr. to the nearest yd.)
2. Binx
+ 3cosx.; 90x0
(1) För y 2 sinx +300ax
we find, 33.7
11) If sinx + 1.500BX-
Then y 28inx
Зсовк
FTOP the graph, we find
10 sinx + 15 cos
3008x
48inx
from the graph,
6.1 of 60
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