1968-06-11 — Page 15

11 JUN 1963

WOW...

食三第張四节

CITY HA自五年申之腊罗

WAH. KUU YAT PO

日一十月六年八六九一層公年七十五廉民豪中

中中會冷普通數學試卷一解答

*#3%$#3 #E#E3%4 #5 #£5 UK 345%S #2%$#$%3 #3%S #SERAS #ENÉS / B53/363 ##

新橋

·

喬仲强。

`(1) *5.

*#*#*

*#$%

--用質因數之連惠式寫出7056元平方根..

217056/

-(3).

(2)×2, 3×1

213528

た

3#$%3#3%$#$%$$%$#$%#%3/

✡ 7056=2*x 37× 7*,

= $2

*(4)

21764

21882

3/441

(3)+(4)

1=57 •

= 2 * × 3 × 7

है.

金

2305 FO

英數洗後閒

31142

7149

"

71

二)求55,1320两數之最小公倍數

LESSON NINE

11-6-1968.)

5155 1320

2.c.m= 5 x 11 x 24 mm • 1320

264

24

*最小公倍數為1324..

代入

У

3-4

答:

nerefore, the required L.C.M

23

52

* 3 x 5 1.e. 600.

.II. LOWEST COMMON MULTIPLE (L.C.M.)

A number which is a multiple of two or more given numbers is called a COMMON MULTIPLE of these numbers.

A The least or lowest number which is a com- ' mon multiple of two or more given numbers is called their LOWEST COMMON MULTIPLE. (L.C.M.) Example.7

Find the L.C.M. of 12, SOLUTION:

12

22

15 and 18.

x 3 15- 3 x 5 Therefore,. the required L.Ç.M.

22 x 32 x 5 = 18

(三)港闻先貶值23%,後升值10%;問港圓实在貶值 百分之幾?

·( 1 − 14,3%)× (1+10%) = 1− 85.7% × 110%

== 1-94-27% 5.73%

答:实貶值5.73%

✡) # m+m⭑nk+n 34 m2+ m2x2+ n* £ ☎

'=== { a2+ 222b2+ b*)−a°f ̃*

18 *

2x32

@_a2+at+4”:

· (a2+ b2)2 (ab)2=

· (a+ab+b2) (a*−ab+b2)

#

n+m2ne+n=(m2+m*n*+nt}{m*—m*n*+nt

ANSWER: The L.C.M. of 12, 15 and 18 is 180..

Example 8

Find the L.C.M. of 15, 25, 40 and 75.

2 15 25 40 75

SOLUTION:

2 15

25 20 75

2 | 15 25 10 75

(m+m*nt+n) ÷ (m2+ m*n*+n+

— m2 m2x2+n2 (%)

(五)化簡下列兩式:

(a)

3

15 25

5

75

(b)

5 1

5 25

5 25

5

L

1 5 1

5

1

1

1 }

2

23

* 3 ; 25 x 32 33 x 52.

The required L.C.M. is 2

25 x 33

is

23 x 32

360.

3x+15

(2x-4)(x+54)′′

3x

>

(**

(3)

六將下到

(a) 1-273

3

3 [ (x+y) (x¬y)-2x+1]

¿ ~ 3 (3+1)

x

3 (x+ y−1) (x−y−1)

[答]

3+1

(答)

な

X+好整除,亦能被

答:

---(1).

NONOH

英值福同

* * - * 14 2 144) 1⁄2 viits, îi »ƒˆ œ ≥ 011 %

A平方公会

以A示!

# **** 42ð, #

477

=

-(U

A so 4π 2*. (2

~~~· v = &π(+)-*·· +-4A = 2

ATTA (%)

答

(+) Bko log, 2 = 0 301030

log101.221

Π

Kog,, 3=0.477121

3×2*

Pog 1.2 = log (1/2) = log "x2"

10

· Log, 3+2 log, 2- log![

0.477121 + 2 x 0.301030–1;

=0.07918 (%)

+-) E40 log, 2=0.301030, $log, 2 · log, 32 TI 由对數換底公式,

(log: 2) ( Logs 3) = 12 232 x 2

得

1号(对數以10為底)

118 Log 5 = log 10 == log 10-log 2

1-0.3-10 30 0.69 8970.

! (log, 2) (Logs 3);

0.30/030 0.698970

— 0.43068' 調和級数え第三項為13,其第四項為3; 共

対衣文等差級数的第三項ー店第四項 言:護等差級數之首項為口,公蕙為d

{a+ad=//=/

a+34=

#^(2) a+8=>

મા,

01) 4

̈£ $ 4• 16 $ * † - - - - - - / 答:第一項為一号(或-2) 未完待續

BRARIES

ANSWER: The L.C.M. of 15, 25, 40 and 75

Example 9

is 600.

Write down in index form the L.C.M. of

ANSWER:

Ixample 10

Find the least number such that the remainder

is 3 when it is divided either by 10, or by

18, or by 24. SOLUTION:

ANSWER:

Example 11

wwwww

The L.C.M. of 10, 18 and 24

x 5 i.e.

The required number is (360+ 3) had

i.e.-363.

Four bells toll at intervals of 3, 4, 7 and 9 seconds respectively.

They start together,

after what time will they next toll together? SOLUTION: The required time which they will next toll together (The L.C.M. of 3, 4, 7 and 9), 22 x

ANSWER

Example 12

32 * 7 seconds 252 seconds

4 minutes 12 seconds.

4

They will toll again after 4 minutes! 12 seconds.

Divide the L.C.M. of 15, 25 and 40 by their! H.C.F.

SOLUTION: The L.C.M. - 3 x 52 x 23

The H.C.F,

=

5

• 600

Therefore, the required quotient 600 = 5

• 120

ANSWER: The required quotient is 120.

EXERCISE 7

(1) Find the common factors of 42, 72 and 144.

(1)

3

(2) Find the H.C.F. of 48, 80 and 84.

(2)

(3) Find the N.C.F. of 464 and 1247 by the

Division Method.

(3)

(4) Find the largest number which is a factor

of each of the numbers, two hundred and forty-six, three hundred and eighty-four and five hundred and ninety-four.

(4)

(5) If 88, 144, and 452 are each divided by the same number, the remainder is 4. What is the greatest possible value of the divisor?_

(5)

(6) Find the L.C.M. of 15, 20, 36 and 42.

(6) (7) Write down in index form the H.C.F. of

27 x 35

35 x 73

x 17; 38 x 17.、

7)

"

8) Write down in index form the L.C.M. of

Question No. (7).

(8)

(9) Find the least number such that the re-

mainder is 5 when it is divided either by 8 or by 12 or by 18.

D) # (2-1) #14 (x2+x2+ax*-

#1(x) = x++ x*+a

x2 - (x + 1) {X+

X-1##14

Bp_f(0)=(0)*+ 4

B 1(1) m

(9)

x+6), ✡a è̟ lĹ.

(10) Express in prime factors the L.C.M. of,

36, 65 and 99.

(10)

LIB

Divide the L.C.M. of 24, 36 and 40 their H.C.F.

(11)

12) Find the sum of the LC

of 26, 39 and 65.

(12)

13) Find the smallest number which is a multiple of each of the numbers 35, 70 and 210,

(13)

'14) Four bells toll at intervals of 12 sec., 、 15 sec., 18 sec. and 28 sec. respectively

They start together; after how long will · they next toll together?

(14)

(15) What is the least number of

Man.

see

is an exact number of 5 gallons 2 pints; 2 gallons 5 pints; 1 gallon 4 pints and 4 gallons.

(15)

gali,

(21)

(ZZ) Write down the product of the prime

numbers between 50 and 60.

(22)

(23) Which of the following numbers is a

multiple of 11? 121, 133, 144

+4

298

(23)

362 488

·

•

077

24) Find the L.C.M. of the numbers 3, 4, 0.

9, 12, 16, 24 and 36.4

(24)

(3)最小公倍數

任何一轍,如其露两個或多個已知數之倍數,則爲此等日知概之公信𪖈• 两個或多個已知數之公倍數之最小數,卻當它們之最小公倍數•

谢715,LB之榖小公倍數。

CW: BEREMAAMM180.)

★15 : 25 › 40 † 752#/jZ® •

(:15, 25, 40 ⋅ 75ZIDANE600 • )

9 DEC 23×3125 × 32; ■2 2

X5之录小愿胗數

C: FORMARE 2* × 3′′ × 5• • }

10#18»?24%%% » #23ŻNE •

的浓次號3,4,7种,9嗨唱一次,現時開始,與藏何陶次再齊呢?

四於4分12才・)

10

(1

12 #15, 2572402H.C.F.

ZL.C.M. •

pta

(饗:廣東之窗鼯120)∗

練習七

(16) Find the smallest length of wire which can be divided into an exact number of pieces either of 6.in., 1 ft. 6 in.,

2 ft. 3 in.. 3 ft. 4 in, and 9 in.

(16)

(17) Find the sum and product of the L.C.M.

and H.C.F. of 34, 68 and 170.

(17)

(18) Find the least number which is a multipl of each of the numbers 56, 63 and 72

(18)

(19) Which of the following numbers is a

common factor of 55 and 1987 3,5 7. 11. 13, 17, 18 ,22

(19)

(20) Which of the following numbers is a

common multiple of 24 and 367

48 72 96.108

"

•

140 (20)

and 214

(21). Write down the sum of the prime number

between 70 and 80.

(1)

42,727 142AME.

(2)

48 0784ZH.C.F..

(3)試用輾轉相除法41247之H.C.F. (d) 一是大數,伴之爲246,384及594年一數之因。

(5)若884情及452因爲某一數所除·餘數皆爲4,間除數之最大值緒微 (6) 15, 20, 36742ŹL.C.M.

(7) KR 27 × 35; 35 3

x 7

* 173 38 (8) 試用指数方式出醫(7)各數之L.C.M.

”求一般小數,使不論其寫或12款所除,而餘數皆5#

(10) ZERNST36, 651299, ZL.C.M.

(11)

24.367540ZH.C.FAVREKŹL.C.M. (12):26 139765H.C.F.BL.C.M.Z« (13)求一雙小數,使其露35,70210三數之倍數

× 17 SIZE. €.17

(14)四大滿126,15及18秒及28的鳴一次,現同時期,間四歲何時才再喝? (15)只小牛奶,使之恰館以完整數目分為任份52品暸,2加3品脫一加3品說的

40.

(16),使之恰能以完整數目分為每段5吋或1呎吋22呎3吋3呎吋,滅?时,

GAZ •

(1734687170L.C.M.BH.C.F.ZMÉ#• (18)求一些小數,而該數儒56,5572之偏數

(19)下列數中,那一個是55198的公約數?315,7,11,13,17,18及22 (20)

247330992 ?48 72 96 108, 140214.

(21)您出70)80的質數之称。 (22)為出50與60間的質數之物。

(23) THA##-##114121193 144 290 362 488 677 (24) 23441 6 1.9 1 12, 16 247361L.C.M...

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日六十月五年电脑重

WAH KIU YAT PO