1969-06-01 — Page 14

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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