1、 P a s t E x a m i n a t i o n P a p e r s 2 0 0 8 / 2 0 0 9 , 2 0 0 9 / 2 0 1 0 84 2009/2010學年 入學考試試題 ADMISSION EXAMINATION PAPER 第一部份 Part I Write your answers in the space provided for each question. Answers written elsewhere will not be marked. 將答題步驟及答案寫在每題所提供的空間內，寫在該空間以外的答案不會被評分。 Part I 1. (a)
2、(3 marks) Solve the following equation for x: (3 分) 對g1488g1344方程求解x：g691g691g691 g691 xx =+ 2122 g691g691g691g699g757g700g691(3 marks) Let 23)( 2 = xxf . For what values of x is )(xf a real number? (3 分) g3707g691 23)( 2 = xxf 。求g2220有g2060 )(xf g2657實數g691的 x 值。 2. (4 marks) Express )1)(1( 25 22+x
3、xxx as sum of partial fractions. (4 分) g3378 )1)(1( 25 22+xxxx 化g2657部分分式和。g691g6913. (6 marks) The following data show the selling prices of 9 models of mobile phones: (6 分) g1488g1344數g5525g2220g1602g26579g3955不g1653g2496g4550g1451g3900g4669g4585的g3335g5085: $2,130 $3,250 $4,760 $1,800 $2,770 $6,0
4、50 $2,340 $1,980 $1,560 Find the (a) mean, (b) median, and (c) standard deviation of the set of data listed above. 求g4576組數g5525的(a)平g1857數，(b)中g1775數，(c)g5201g4407差。 4. The marked price of a television is $7,500. A customer may choose one of the following three methods of payment: 一部g4669g4142g5554
5、的g5201g5085g2594$7,500，有g1343種付g3955方法可供顧客選擇 S y l l a b u s 2 0 1 0 / 2 0 11 85 Method 1 : Pay by one single payment with a 5% discount. 方法一： g1488現金一次付清g2220有g3955項，可享g5201g5085的九五折優惠。 Method 2 : First, pay 20% of the marked price as initial payment. Then pay the remainder together with a 7% simpl
6、e interest by 12 monthly equal installments. 方法二： 先g1488現金支付g5201g5085的 20%，餘額要另g1507 7%單利息，可分 12g2849月每月支付g2700g1653的g3955項清還。 Method 3 : At first, no initial payment is needed. Afterwards, pay the marked price together with an 8% simple interest by 12 monthly equal installments. 方法g1343： g4210始時不用
7、付g3955，g2220有欠g3955要另g1507 8%單利息，可分 12 g2849月每月支付g2700g1653的g3955項清還。 (a) (4 marks) What is the percentage increase of the total amount paid by using Method 2 instead of Method 1? (4 分) g1488方法二付g3955，比g1488方法一付g3955，g2220需g3955項增g1507的百份比g2594多少？ (b) (4 marks) What is the difference between the amo
8、unt of a monthly installment when using Method 2 and that when using Method 3? (4 分) 方法二和方法g1343每月支付的分期g3955項g2700差多少？ 5. A squad of soldiers are marching from a barracks to a mountain area. They carry a load of 4,800 kg and it is shared equally by all soldiers. In the middle of the journey, 22 sold
9、iers left and brought away 320 kg of load. The rest of the soldiers continue their journey and the load carried by each remaining soldier now is 3 kg heavier than before. 一隊軍人奉召由軍營出發g2459往一g2849山區。每人背負g2700g1653重g4198的g2338資，g1636隊攜帶了4,800 kg。在途中，22人g6182g4210大隊並帶走共重320 kg 的g2338資，其餘軍人繼g6491g2459往g4
10、576山區，每人須背負的g2338資比出發時重了3 kg。 (a) (4 marks) Let the number of soldiers setting out from the barracks be x. In the context of this question, establish a quadratic equation of x. (4 分) g3707g4576隊軍人由軍營出發時有x人，根g5525題意，g4578g3707立一條有關 x的二次方程式。 (b) (4 marks) After 22 soldiers have left, how many soldiers
11、 remain in the squad? What is the load that each remaining soldier must carry? (4 分) 當22人在途中g6182隊後，還剩g1344軍人多少名？餘g1344各人g2220攜g2338資有多重？ 6. Box A contains 1 black ball and 3 white balls. Box B contains 3 black balls and 2 white balls. A ball is drawn out at random from A and put into B. Then a ball
12、 is drawn out at random from B. Find the probabilities of the following P a s t E x a m i n a t i o n P a p e r s 2 0 0 8 / 2 0 0 9 , 2 0 0 9 / 2 0 1 0 86 events: A 盒載有一g2849黑球和g1343g2849白球，B 盒載有g1343g2849黑球和g2079g2849白球。若從 A盒中g5727g5554抽取一球放入B盒g1411，然後又從B盒g1411g5727g5554抽取一球，求g1488g1344g2051g1625的概
13、g3575: (a) (4 marks) both balls drawn out are white in color,g691 (4 分) 抽出的g2079g2849球皆g2657白色， (b) (4 marks) the second ball drawn out is white in color. (4 分) 第二g2849抽出的球g2657白色。 第二部份 Part II This part contains 20 multiple choice questions. Please select the desired answers on the pink answer shee
14、t appeared on the last page of Part I. Please follow the instruction given on that page. 本部份有20條選擇題，請在本試卷第一部份尾頁的粉紅色答案紙上選擇答案。請依照該頁內的指示填寫答案。 1. If xxxf += 11)( , then =+afaf1)( 若 xxxf += 11)( ，g2462 =+afaf1)( A. 0 B. 1 C. a1 2 D. aa12 E. aa14 2. If the equation 052 2 =+ kxx has no real roots, find the
15、 range of the values of k. 若方程 052 2 =+ kxx 沒有實根，求k值的範圍。 A. 825=k S y l l a b u s 2 0 1 0 / 2 0 11 87 B. 825k D. 825k E. 825k 3. If )(1( axx and bx + 2)2( are identical, find a and b. 若 )(1( axx 與 bx + 2)2( g2700g1653，求a 及b。 A. a = 2 and b = 0 a = 2 及b = 0 B. a = 3 and b = 1 a = 3 及 b = 1 C. a = 3 a
16、nd b = 0 a = 3 及 b = 0 D. a = 3 and b = 1 a = 3 及 b = 1 E. a = 3 and b = 1 a = 3 及 b = 1 4. The figure below shows the graph of cbxaxy += 2 . The coordinates of the vertex are (3, 5). With reference to the figure below, which of the following must be correct? g1344圖g2220g1602g2657 cbxaxy += 2 的圖像，其頂
17、g6042的坐g5201g2594g691g6993,g6915g700。參看g1344圖，g1488g1344何者必g2657正確？g691I. The y-intercept of the graph 0. 圖像的y軸g4786g4175 0。 II. 042 acb . III. The equation of the axis of symmetry of the graph is x = 3. o x y (3, 5) cbxaxy += 2P a s t E x a m i n a t i o n P a p e r s 2 0 0 8 / 2 0 0 9 , 2 0 0 9 /
18、2 0 1 0 88 圖像的對g4896軸的方程g2594x = 3。 A. II only 只有II B. I and II only 只有I及II C. I and III only 只有I及III D. II and III only 只有II及III E. I, II and III I、II及III 5. What is the sum of all the different five-digit integers formed by using the digits 1, 2, 3, 5, 9 without repetition? 用1, 2, 3, 5, 9 組成數g1775
19、不g1653的g2220有五g1775數總和g2657多少？g691A. 5,333,280 B. 4,666,640 C. 3,999,960 D. 3,555,520 E. 2,666,640 6. Mr. Chan has deposited a sum in a bank. The interest rate is 6% per annum, compounded monthly. If the interest is $5,720 after 2 years, find the original sum. 陳先生把一筆g3955項存入銀g1769，年利g3575g2657 6%，每月
20、g2784算複利息一次。若g2079年後他可得利息$5,720，求原來那筆存g3955。 A. $44,983 B. $45,220 C. $45,574 D. $46,278 E. $48,200 7. If a is greater than b by 20% and a is greater than c by 50%, find a : b : c. 若a比b大20%，a比c大50%，求a : b : c。 A. 18 : 15 : 10 B. 15 : 12 : 10 C. 10 : 8 : 5 D. 8 : 6 : 3 E. 6 : 5 : 4 8. Solve .21071353
21、251243+xxxxS y l l a b u s 2 0 1 0 / 2 0 11 89 解 +21071353251243xxxx。 A. 2x B. 2x C. 22 k . Find the quotient. 若多項式 754 2 + xx 除g1488 kx + 的餘數g2594 15 k ，其中 0k ，求g3320式。 A. 4x 7 B. 4x 5 C. 4x + 5 D. 4x 3 E. 4x + 3 19. An arithmetic series is such that its first term is a and its third term is b. Let
22、 the sum of its first n terms be denoted by nS . Given that 754 , , SSS are consecutive terms of a geometric series, then 有一等差數g1640其g2824項g2657a及第g1343項g2657b。g3707 nS g2657其g2824n項g1392和。若754 , , SSS g2657一等比數g1640的g2700g5411項，g2462 A. 22 2528 ba = B. 22 716 ba = C. 22 1011 ba = D. 22 137 ba = E.
23、ba 2528 = 20. In the figure ABC is an equilateral triangle inscribed in a circle as shown. The chord MN bisects two sides of ABC at D, E respectively. If 2 | =AB , the length of EN is: 圖中 ABC g2657g4285g1411g3447g1636等g1343g2017g1902。g2192MN把 ABC g2079g6320等分g2253g6042g691D、E。若 2 | =AB , ENg691的長g25
24、39g2657g691S y l l a b u s 2 0 1 0 / 2 0 11 93 A. 5 15 + B. 2 15 C. 510 D. 4 12 + E. 46 P a s t E x a m i n a t i o n P a p e r s 2 0 0 8 / 2 0 0 9 , 2 0 0 9 / 2 0 1 0 94 2009/2010學年 參考答案 MODEL ANSWER Part I 1. (a) 2)2(122 2122 +=+ +=+ xx xx g691g691 g691 0)4)(2( 082 2=+=+xxxx g6914or ,2 = xx g699g7
25、73g760g765g760g758g775g760g759g700g705g691g691 g691Thus, there is only one real root, namely, 2. (b) )(xf is a real number if 022 x and 023 2 x . 022 x 0)2)(2( + xx ) 2or 2( xx . (*) 23 2 x 29 2 x 112 x 1111 x . . (*) Combining (*) and (*), we see that 112or , 211 xx 2. Let 11)1)(1( 25 222+=+xCBxxAx
26、xxx . Then )1)()1(25 22 +=+ xCBxxAxx for all 1x . In particular, CAx += 20 BCACBAx +=+= 42)(281 . 852542 =+= CACBAx Solving the above equations gives ,2,1 = BA and 3=C . Thus, 13211)1)(1( 25 222+=+xxxxxxx . S y l l a b u s 2 0 1 0 / 2 0 11 95 3. (a) $2,960 (b) $2,340 (c) 91.1423$918247600$ = 4. (a)
27、Total amount paid by using Method 1 = 125,7$)95.0(7500$ = Total amount paid by using Method 2 920,7$)07.01()2.01(75002.07500 $ =+= Percentage increase of the total amount paid %16.11%100712571257920 = . (b) The month installment when using Method 2 535$12 )07.01()2.01(7500$ =+= The month installment
28、 when using Method 3 675$12 )08.01(7500$ =+= Thus, monthly installment of the latter method is $140 higher than the former. 5. (a) 320480034800)22( = +xx , or 010560025432 =+ xx . (b) The two roots of the quadratic equation formulated in (a) are 150, and 32234 (rejected). So the number of soldiers r
29、emaining in the squad 12822150 = . Each remaining soldier must carry 3531504800 =+ kg. 6. (a) 836343 = , or 0.375. (b) 241162416343 =+ , or 0.458. P a s t E x a m i n a t i o n P a p e r s 2 0 0 8 / 2 0 0 9 , 2 0 0 9 / 2 0 1 0 96 Part II 1. A 2. D 3. E 4. D 5. A 6. A 7. E 8. E 9. D 10. C 11. D 12. B
30、 13. E 14. E 15. B 16. E 17. A 18. A 19. D 20. B S y l l a b u s 2 0 1 0 / 2 0 11 97 2008/2009學年 入學考試試題 ADMISSION EXAMINATION PAPER 第一部份 Part I Write your answers in the space provided for each question. Answers written elsewhere will not be marked. 將答題步驟及答案寫在每題所提供的空間內，寫在該空間以外的答案不會被評分。 1. Solve the
31、equations for x: 對以下方程求解 x： (i) (2 marks) 12212 654 + = xxx (2 分) 12212 654 + = xxx (ii) (3 marks) 11000 log52 = xx (3 分) 11000 log52 = xx 2. (5 marks) Find the values of k if the sum of square of the roots of the following equation is equal to 25. (5 分) 若下面方程式的根的平方和為 25，求 k 的值。 0)1(2)2(2 =+ kxkx 3.
32、 Let AxAxxxf += 4)( 23 . When f(x) is divided by 2x 1, the remainder is 5. 設 AxAxxxf += 4)( 23 。當 f(x) 除以 2x 1 時，其餘數為 5。 a. (3 marks) Find the value of A. (3 分) 求 A 的值。 b. (2 marks) Find the remainder when f(2x 1) is divided by 2x 1. (2 分) 求當 f(2x 1) 除以 2x 1 時的餘數。 P a s t E x a m i n a t i o n P a p
33、 e r s 2 0 0 8 / 2 0 0 9 , 2 0 0 9 / 2 0 1 0 98 4. By selling a shirt at $84, a man makes a profit of 40%. Now, the cost of the shirt increases by 25%, 若某商人以 $84 出售一件恤衫， 盈利百分率為 40%。現恤衫的成本增加了 25%， a. (3 marks) if the man sells the shirt at the same price as before, find the new profit percentage. (3
34、分) 若恤衫的售價不變，求新的盈利百分率。 b. (3 marks) what should the new selling price be so that he still makes a profit of 40%? (3 分) 若要保持盈利百分率為 40%， 問每件恤衫的新售價是多少？ 5. The manufacturing cost of a video game console is $C. It is known that C is the sum of two parts: one part is fixed and the other part varies inverse
35、ly as the number N of game consoles produced. When 3,000 game consoles are produced, the cost of each game console is $328. When 6,000 game consoles are produced, the cost of each game console is $224.5. 生産一部遊戲機的成本是$C，已知 C 是兩部分的和，而其中一部分固定不變，另一部分則隨遊戲機的生産量 N 而反變。當生産 3000 部時，每部的成本是$328，當生産 6000 部時，每部的成
36、本是$224.5。 a. (3 marks) If the cost of each game console is $293.5, how many game consoles are produced? (3 分) 若每部的成本是$293.5，共生產了多少部遊戲機？ b. (3 marks) If the least number of game consoles to be produced is 2,300, what will be the highest cost of each game console? (3 分) 如果最少須生産 2300 部，每部的成本最高是多少? 6. (
37、6 marks) One hour after a train had departed Station A and heading for Station B, a fault occured which required the train to reduce its speed to 3/4 of its normal time-table speed. As a result the train reached Station B three hours late. If the fault had occurred after traveling for another 60 kil
38、ometers, the train would have been late for two hours and forty minutes. Find the distance between Stations A and B. (6 分) 有一列火車從甲站開往乙站一個小時後在途中發生故障，需要以正常速度的 3/4 行駛， 故比平時晚了三個小時到達乙站。若列車多行 60 公g2042g1385發生故障，則列車g3378g4366比平時晚了兩小時g1540g1340分g6442g1385到達乙站。 求甲g917乙兩站g1392g4212的g4175g6182。 S y l l a b u s
39、 2 0 1 0 / 2 0 11 99 7. In throwing two dice, the outcome is called a double when the numbers on the two dice are equal. 在g1931g6058兩g6029g1857g4896的g5071g1367時，若兩g6029g5071g1367的g6042數g2700g1653，g1914g2842g2433g4896g1392g16125g984g3822g5071g985。 a. (2 marks) Find the probability of getting a double
40、 when throwing two dice. (2 分) 求g1931g6058兩g6029g5071g1367一g1723而g2103g3411g984g3822g5071g985的g4375率。 b. (5 marks) In a game, two dice are thrown three times in succession. Find the probability of getting only one double in the three throws. (5 分) 求g3729g6491g1931g6058兩g6029g5071g1367三g1723，g1533g34
41、11一g1723g984g3822g5071g985的g4375率。 第二部份 Part II Please select the answers on the answer sheet on the last page of Part I. 請在本試卷第一部份尾頁的答案紙上選擇答案。 1. Which of the following is an identity / are identities? 以下g1782g2371是g2561g4059式？ I. 21 =+ xx II. baabba +=+ 11 III. mnnm :1:1 = A I, II and III Ig917II
42、g1431 III B. II and III only g1533有 II g1431 III C. I only g1533有 I D. II only g1533有 II E. III only g1533有 III P a s t E x a m i n a t i o n P a p e r s 2 0 0 8 / 2 0 0 9 , 2 0 0 9 / 2 0 1 0 100 2. Simplify 3235logloglogxxx g1423g61133235logloglogxxx A. 43 B. 67 C. xlog43 D. xlog67 E. xlog67 3. A m
43、edal is made of 12g of pure gold and a large amount of pure silver. If the ratio of the weights of gold and silver is 3 : 37, find the weight of the medal. 一個g5243g4008g1591 12 g g3146g2414和g1365量g3146g5036g6583g3735而成。如果g2414g4950g5036g2809量的比是 3：37， 求g4576g5243g4008的g2809量。 A. 148g B. 154g C. 160g
44、 D. 166g E. 170g 4. For all real numbers a and b, the expression 2 | abba + is equal to 對g2220有g4756數 ag917b， 2 | abba + g4059g2253 A. the average of a and b a 和 b 的平g1857值 B. the average of a and b a 和 b 的平g1857值 C. the maximum of a and b a 和 b 中的g4373g1365值 D. the minimum of a and b a 和 b 中的g4373小
45、值 E. | ba + S y l l a b u s 2 0 1 0 / 2 0 11 101 5. In a summer camp, there are x male and y female campers. On each day, every male camper and every female camper have to run 3 kilometers and 2 kilometers respectively so that the campers have run a total of 255 kilometers. It is known that the numb
46、er of male campers is less than that of female campers by 25%. According to the above information, we have 在某g2905g1495g5852中，有 x g1660g1994g5852g1430和 y g1660g1366g5852g1430。每g1660g1994g5852g1430和每g1660g1366g5852g1430每g1434須分g1809g4178g1953 3 公g2042和 2 公g2042， g1636g6639g5852g1430共g4178g1953 255 公g
47、2042。已知g1994g5852g1430的人數比g1366g5852g1430少 25%。g2057g6194g4320， g1521g3411 A. =+%)251(25532yxyx B. =+%)251(25532xyyx C. =+%2525523xyyx D. =+%)251(25523xyyx E. =+%)251(25523yxyx 6. Let R be the region in the xy-plane satisfying the inequalities ,0x 3x , 0 yx , and 2+ yx . The minimum value of yx 56 +
48、 on this region R is 設 R 為在 xy-平面g1346的g3316g3345，g1482g4840g2025不g4059式 0x ， 3x ， 0 yx ，g1431 2+ yx 。在g3316g3345 R g1346 yx 56 + 的g4373小值是 A. 9 B. 10 C. 11 D. 12 E. 33 7. When 6662 is divided by 13, the remainder is 6662 除以 13 的餘數為 i. 1 ii. 3 iii. 4 iv. 7 v. 12 P a s t E x a m i n a t i o n P a p e r s 2 0 0 8 / 2 0 0 9 , 2 0 0 9 / 2 0 1 0 102 8. Find the area of the shaded region in the figure below. 求以下g4729中g3757g5145g3316g3345的面g5608。 A. 32.5
