1968-06-06 — Page 23

– 6 JUN 1968 ·

瓦三第强六第

日一十月五年申展工

WAH KIU YAT PO

CITY HALL

英中會考數學(二)答案

(. (3 marke)

·歐陽鉻文·

日六月六年八六九一曆公年七十五國民 華帶

Solve the simultaneous equations.

LESSON EIGHT'

CRAFTER SEVEN

6-1968.

I, FRIME NUMBERS AND COMPOSITE NUMBERS

[(A) FACTORS

Since 21 - 3 x 7, 3 and 7 are each called FACTOR of 21; and 21 is called the MULTIPLE cf 3 and also the MULTIPLE of 7.

(B) NUMBERS

(1) Numbers, of which 2 is a factor,"are"

called EVEN NUMBERS e.g. 2, 4, 6, 8.....~*

etc.

(11) Any number which is not even is called an

ODD NUMBER e.g. 1, 3, 5, 7, .

. etc.

(C) PRIMÉ NUMBERS

+

A number 18 said to be prime if it is divisi- ble only by itself and ɑne..

'D) COMPOSITE NUMBERS

A number which is not prime is called s Composite Number, It can be written as the product of two or more Prime Factors.

for example

NOTE 1 e.gi

1.e. etc.

1 that is

et cetera

(E) INDEX NOTATION

(i) 4 x 4 is written as 42, called "

squared" or "the square of 4”. (11) 4 x 4 x 4 is written as 13, called~)

"A cubed" or "the cube of i'

(ija) 4 x 4 x 4 x 4 x 4 x á is written as 4 called "4 to the power 6" or "the 6th power of 4".

In the symbol, 4“, 6 is called the) index.

(F) PRIME MUMBERS within 200 (To be memorized)

2

3

5

7

11

13

17

}

19

23

29

31

37

41

43

47

53

59

61.

67

71

73

79

83

89

97

101

103

107

109

113

127

131

137

139

149

151

157

163

167 173

179

181

191

193

· 197

199

Example 1

JAYASE DOK Answer.

R.K. Certificate of Education Examination 1968

Mathematica

L

Paper II Algebra

Section A (40 marka)

Credit will only be given for the correct answer which should be given in ita simplest terms. Put each answer in the space provided on the question paper. Any working required may be done on the last few pages of the answer book but will not be marked.)

1. (2 marks)

-

Make g the subject of the formula t - 21

2. (2 marke)

+7 (4)

28/2

Ans. J=

Ng

72 is the product of two consecutive integers, Find the two pairs of such integers.

Let x, x+1 be the reg'd pair.

Then *(2+1)=72′′

2*+ 1 -72 = 0

Ans. 8.1 cv = 9, -8

if x = 8

X=-9

then

.. 28 ov - 9; x+1=9

XH =-8.

3. (3 marka

Find the 4(x

factore

3

x - Y

I

x + y

+ 33 - 3 kusukume-(2) âne.”.

From Eploy

فه

**

ax=-44 x=-24

( Subat €13) into Eg (a):

2 (−29)+34 = 3)

-= 3

From $9 (3)) !x=-2y= 6

{1=4

y=-31

(3)

check: In Epus y=4;-&-1⁄2

In Eq (a) ; 22 + 24 = 2(6) + 3(-3)= 3

[9. (3 marka)-

If 2x+

- 4 A

2 x

find 'X.)

(Ans.

4=

X = 4

2

Go 2 x 3

your

*(x*- y3)=` 2*× (z+9}{2=y)

(x-3

L. C. M. = 2′′x3(X+H){ 2-9]{X+39)

- 1

* - - - • (x + 2 + 3 ).

*ABB.

10. (3 marka 7

There is a misprint in the coefficient of y in› one of the three simultaneoue equations,

* + 3y+ 22 5, 2x - 4y - 3 - 2,

3x-y-4s - 5. The correct solutions are

x = 4, y = -1,

2. Write down the value of) the corrected coefficient. bet

*+32 + 2 = 5° - - - - - (1) Ane. 7X-4

- * * * ---- (*)

3 x - - ** - 5 - - mm (3)

Pat x=4, 2 = 2. EgtD) becomes 4*33+2(5)=$ £y=-1)

#

Ep (a) Eg (3)

3(4) − 5 ~ 4())=$ «*==» By the known value y=-1, we find that then is a misprint in the coeff, of y in Eg (x). The Con, 1844] 11. (3 marks)

Given 10810

* 0.5uru30 and 106103 find to 5 places of decimals 108101.2

1.2= log. 40 (

-0.477121,)

Ans." 0.079 13

3 T3

= Log (2'x37 - Log 10

ok

log 2+ Lag

≈ 2x0.3010 3:0 + 0.477728*9.

0,07.9.18 1

4. (3 marko]

Simplify

Exp.

Express 240 in prise factors.

SOLUTION 240 - 2 x 120

- 2 x 2 x 60

2 x 2 x 2 x 30

* 2 x 2 x 2 x 2 * 15

.2 * 2 x 2 4 2 × 3 × 5 ..24 x 3 x 5

Therefore, 240 - 2o × 3 × 5

Example 2

Find the sum of the prime numbers between, 50 and 70-

SOLUTION The required sum

53 59 61 - 67 - 240

ANSWER: The required sum of the prime numbers,

between 50 and 70 is 240.

II. HIGHEST COMMON_FACTOR (H.C.F.)

A number which is a factor of two or more gaven, numbers is called a COMMON FACTOR of thesa numbers.

The greatest or nighest number which is a common factor of two or more given numbers is called their HIGHEST COMMON FACTOR or H.C.F. Example 3

Find the common factors of 36 and £8.

SOLUTION:

Since 36 - 2 x 18 - 2 x 2 x.9

- 2 x 2 x 3 x 3

22

32

48 -

x

7

5. (3 marka)

HONG KO

The cost of y catties of vegetables la o cents. If the price is increased by 5 cents par catty, find the number of catties that can be bought for a centa

the cast of a cathies is ek. Alba

The increased price is

sinh năm bại

5 cents

*

+

3 cents per catty

Sathes

67 and

in the formula

G-FUBL

ވ

„Salve =

2

+243m 0.

www.

*** dax+û** a2=b.

(x*a) ̃= a*- b

1× = -a± √a-b

OR, by applying the formula. x=£

CIRKARI

13..() marke}

ZA

If the 5th term of an arithmetinal progression is 16 and its 13th term ia 64, find the first term.

Lex

a = the 1st term of the A.Ş. gās.

the common diff.

"

&

Then, the 5th term = a+4d= 16 midi tip

the 13th terms at 12 d = 64

4. (*)-GU) - 8 d=48

.. 4 = 6

+

ay

63V++9x 63=44V

V} = 7 y

RV ==44)

14. (3 marks)

À geometrical progression nas seven terma, its 3rd term is 324 and its 5th term la 144. Find its last term

ANSWER:

Example 4

The required common factors are 2. 3. 4, 6 and 12.

Find the 8.C.F. of 24, 36 and 96,””

SOLUTION

.

·24 - 23 x 3 36 24 x

96-25 x 3

32

The highest power of 2 which is a factor of -each number is 2; the highest power of 3 vhies is a factor of each number is 1 and there is no other common factor.

Therefore, the Ħ.C.F. is 22 x 3. 1.0. 12. ANSWER: The required H.C.F. of 24, 35 and 96,

Example 5

is 121

Find the H.C.P. of 437 and 667 by the fivis" ion Method.

SOLUTION..

7 | 6.67

1. (3 marks)

Factorize 27

27 -

6

I

Ane.

•^ 3* − {x*)3 · =(3 × ×){ 3* + (3M*$$ (29)*]

NOTE :, A3- B3 = (A + B ) ( A2‍ + AB + B↑)

Let

as the 1st term of the GP. Ans,

the common ratio of the GP

64

Then the third termm are 324 vice-IS

the 54 term = Q1*= {**

Hence

14407

324

"the" 7" ferm= ar

SAT (HR)

(B)合偶數(雙數)

o.G. A L... W etc @ke•

日4×4可挑款 4 爲4平方

792 - 2

96 2

2

198

(C)質數或素豉

(D)組合數

2 x

3 x 11

(E)指數送

12016-

23% 211

x 11

F64

r=1

2.3.0 43.7

2.0.7 230

2.07 207

2.3

ANSWER:

The required H.C.F.

23.

Example 6

Find the largest number which the numbers 504, 792 and: 2016.

SOLUTION 504

factor

2.x

21

2 x 120

2x64

ANSWER 1

The required largest number 1.6. 8,

Answers to Exercise 6

(1) $22 (2): 99; 151200,(3) 15 min. (4) 32 m.p.h. (5) 120 sq. in. (6) $15.40 (7): 9 hr. 30 min. (8) 1 ton 364 lb. 12 oz.

(9) $1,80 (10) 480 students; 49 empty seats"

11) 44 egge left; $25 (12) 4 hr. 37 min. 30 sec..

13) 486 miles (14) 40 marbles (15) 3gts.

(16) 0 (17) 2 (18) 80 (19) 337 (20) $0.80

第七章(1)賞由廠

CADER

C

例對

:

@4x4x4x4x4×4

200以內的質數。

以贊因敬之述乘養表示240

210-21 × 3 × 5)

MM 2507702A972 ROSÉ •.

1、用篇4之大米”或“之第六次方”

C啦:50與20之間的質數之和露240)

(2)最大公約數

某敬假如其爲兩個或多個已知觀之因數,則該數種牒此等之已知數的公因數成公約數 #

安個,多個已知數之易約數之最大數,稱爲最大公約數

33648794

OFFERZAN2 - 3 - 4 - 68112.)

1942436.

保送

(答:24:熱約數為12

437笈667之微大公約數: 試用無棒相法

(答:所之最大公购*器23)

例磁石宋一夜大的數,而被數爲504 7922016簽數之因數

答:所求之歐露