廣東省廣州市華南師范大學(xué)(510631) 李湖南

1.What value ofxsatisfies

譯文方程的解x是多少?

解化簡(jiǎn)可得,故(E)正確.

2.The numbers 3,5,7,a,andbhave an average(arithmetic mean)of 15.What is the average ofaandb?

A.0 B.15 C.30 D.45 E.60

譯文數(shù)字3,5,7,a和b的平均值(算術(shù)平均值)是15.則a和b的平均值是多少?

解依題意有3+5+7+a+b=15×5,則a+b=60,平均值為30,故(C)正確.

3.Assuming34 and5,what is the value in simplest form of the following expression?

A.?1 B.1 C.D.E.

譯文設(shè)則表達(dá)式的最簡(jiǎn)形式是什么?

解約分即得故(A)正確.

4.A driver travels for 2 hours at 60 miles per hour,during which her car gets 30 miles per gallon of gasoline.She is paid$0.50 per mile,and her only expense is gasoline at$2.00 per gallon.What is her net rate of pay,in dollars per hour,after this expense?

A.20 B.22 C.24 D.25 E.26

譯文一位司機(jī)以60 英里/小時(shí)的速度駕車(chē)2 小時(shí),她的車(chē)每跑30 英里需要消耗1 加侖汽油.她能獲得0.50 美元/英里的報(bào)酬,唯一的花費(fèi)就是2 美元/加侖的汽油.問(wèn)她每小時(shí)除去消耗之后的凈收益是多少美元?

解1 個(gè)小時(shí)她能跑60 英里,獲得60×0.50=30 美元,汽油費(fèi)為60÷30×2 = 4 美元,故凈收益為26 美元,(E)正確.

5.What is the sum of all real numbersxfor which

A.12 B.15 C.18 D.21 E.25

譯文滿足方程的所有實(shí)數(shù)x之和是多少?

解分別解方程x2?12x+34=2 和x2?12x+34=?2,可得x1,2=4,8 和x3,4=6,故和為18,(C)正確.

6.How many 4-digit positive integers(that is,integers between 1000 and 9999,inclusive)having only even digits are divisible by 5?

A.80 B.100 C.125 D.200 E.500

譯文有多少個(gè)四位的正整數(shù)(也就是在1000 和9999之間的整數(shù))能被5 整除且所有數(shù)字均為偶數(shù)?

解依題意,符合條件的四位數(shù)的個(gè)位數(shù)只能是0,十位數(shù)和百位數(shù)可以是0,2,4,6,8,千位數(shù)只能是2,4,6,8,共有1×5×5×4=100 種選擇,故(B)正確.

7.The 25 integers from?10 to 14,inclusive,can be arranged to form a 5-by-5 square in which the sum of the numbers in each row,the sum of the numbers in each column,the sum of the numbers along each of the main diagonals are all the same.What is the value of this common sum?

A.2 B.5 C.10 D.25 E.50

譯文將25 個(gè)整數(shù)分別是從?10 到14,放入5×5 的格子中,使得格子里的每行、每列和兩條對(duì)角線的數(shù)字和均相等.問(wèn)這個(gè)數(shù)字和是多少?

解這是一個(gè)5 階幻方問(wèn)題,25 個(gè)數(shù)字之和是(?10)+(?9)+···+13+14 = 50,分別放入5 行,故每行的數(shù)字和是10,(C)正確.

8.What is the value of 1+2+3?4+5+6+7?8+···+197+198+199?200?

A.9800 B.9900 C.10000 D.10100 E.10200

譯文1+2+3?4+5+6+7?8+···+197+198+199?200的值是多少?

解原式= 1 +(2 +3?4)+5 +(6 +7?8)+···+197+(198+199?200)= 2×(1+5+···+197)=,故(B)正確.

9.A single bench section at a school event can hold either 7 adults or 11 children.WhenNbench sections are connected end to end,an equal number of adults and children seated together will occupy all the bench space.What is the least possible positive integer value ofN?

A.9 B.18 C.27 D.36 E.77

譯文在某學(xué)校的活動(dòng)中,一條長(zhǎng)凳可以坐7 個(gè)成人或者11 個(gè)兒童.當(dāng)N條長(zhǎng)凳首尾相接的時(shí)候,剛好坐滿了相同數(shù)量的成人和兒童.問(wèn)N的最小正整數(shù)值是多少?

解設(shè)有x條長(zhǎng)凳坐了兒童,則有N ?x條長(zhǎng)凳坐了成人,依題意有11x= 7(N ?x),因而故Nmin=18,(B)正確.

10.Seven cubes,whose volumes are 1,8,27,64,125,216,and 343 cubic units,are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top.Except for the bottom cube,the bottom face of each cube lies completely on top of the cube below it.What is the total surface area of the tower(including the bottom)in square units?

A.644 B.658 C.664 D.720 E.749

譯文七個(gè)立方體,體積分別是1,8,27,64,125,216,343個(gè)立方單位,依次按照體積大小由底到頂垂直地堆積成一座塔.除了最底部的立方體,每個(gè)立方體的底面都完全被下面的立方體的頂面覆蓋.問(wèn)這座塔的表面積(包括底面)是多少個(gè)平方單位?

解這七個(gè)數(shù)都是立方數(shù),則這七個(gè)立方體的棱長(zhǎng)分別是1,2,3,4,5,6,7,從而塔的側(cè)面積為4×(12+22+...+72)=560,而上、下底面積之和為2×72=98,共658,故(B)正確.

11.What is the median of the following list of 4040 numbers?

1,2,3,...,2020,12,22,32,...,20202

A.1974.5 B.1975.5 C.1976.5 D.1977.5 E.1978.5

譯文下列4040 個(gè)數(shù):1,2,3,...,2020,12,22,32,...,20202的中位數(shù)是多少?

解由于442=1936,452=2025,從而以上數(shù)列按遞增排列的話,就成為:

12.Triangle ?AMCis isosceles withAM=AC.Mediansandare perpendicular to each other,andMV=CU=12.What is the area of ?AMC?

A.48 B.72 C.96 D.144 E.192

譯文等腰?AMC中,AM=AC,中線和互相垂直,且MV=CU=12.則?AMC的面積是多少?

解如圖示,設(shè)MV交UC于點(diǎn)E,過(guò)A作AD ⊥ MC于D,交UV于F,則AD是MC的中垂線.依題意,U,V分別是AM,AC的中點(diǎn),則UV是?AMC的中位線,且UM=V C=,從而?UMC?V CM(SSS),可得∠UCM= ∠V MC= 45°,因 此EM=EC,點(diǎn)E在AD上.

圖1

于是?UV E～?CME,且均是等腰直角三角形,因而進(jìn)而?AMC,得故(C)正確.

13.A frog sitting at the point (1,2)begins a sequence of jumps,where each jump is parallel to one of the coordinate axes and has length 1,and the direction of each jump (up,down,left,right)is chosen independently at random.The sequence ends when the frog reaches a side of the square with vertices(0,0),(0,4),(4,0),and(4,4).What is the probability that the sequence of jumps ends on a vertical side of the square?

譯文一只青蛙坐在點(diǎn)(1,2)上,開(kāi)始一系列的跳躍,每次跳躍都平行于坐標(biāo)軸且長(zhǎng)度為1,方向(上、下、左、右)是隨機(jī)的且獨(dú)立,當(dāng)青蛙到達(dá)由點(diǎn)(0,0),(0,4),(4,0),(4,4)構(gòu)成的正方形的一條邊的時(shí)候,跳躍終止.問(wèn)跳躍終止于正方形豎直的兩條邊上的概率是多少?

解如圖示,青蛙在點(diǎn)F1處,它可以向四個(gè)方向跳躍,概率均為向左跳躍,立刻達(dá)成目標(biāo); 向上、向右、向下分別跳躍到點(diǎn)A1,C,A3處,再通過(guò)其它跳躍達(dá)成目標(biāo).根據(jù)對(duì)稱性,青蛙由點(diǎn)A1,A2,A3,A4出發(fā)達(dá)成目標(biāo)的概率是一樣的,設(shè)為a;青蛙由點(diǎn)B1,B2出發(fā)達(dá)成目標(biāo)的概率是一樣的,設(shè)為b;青蛙由點(diǎn)F1,F2出發(fā)達(dá)成目標(biāo)的概率是一樣的,設(shè)為x;青蛙由點(diǎn)C出發(fā)達(dá)成目標(biāo)的概率設(shè)為c.

圖2

因此,P(青蛙由F1出發(fā)達(dá)成目標(biāo))=P(青蛙向左)+P(青蛙向上)×P(青蛙由A1出發(fā)達(dá)成目標(biāo))+P(青蛙向右)×P(青蛙由C出發(fā)達(dá)成目標(biāo))+P(青蛙向下)×P(青蛙由A3出發(fā)達(dá)成目標(biāo)),即有同理,可得方程組成立,解得故(B)正確.

14.Real numbersxandysatisfiesx+y=4 andxy=?2.What is the value ofx++y?

A.360 B.400 C.420 D.440 E.480

譯文實(shí)數(shù)x,y滿足方程x+y= 4 和xy=?2.則的值是多少?

解依題意可得x2+y2= (x+y)2?2xy= 20,x3+y3=(x+y)(x2?xy+y2)=88,原式==440.故(D)正確.

15.A positive integer divisor of 12! is chosen at random.The probability that the divisor is a perfect square can be expressed as,wheremandnare relatively prime positive integers.What ism+n?

A.3 B.5 C.12 D.18 E.23

譯文隨機(jī)選取12!的一個(gè)正整數(shù)因子,該因子是一個(gè)完全平方數(shù)的概率可以表示為,其中m,n為互素的正整數(shù).則m+n是多少?

解由于12! = 210×35×52×7×11,則12! 有11×6×3×2×2=792 個(gè)正因子;

設(shè)k是12!的平方因子,則k=2a×3b×5c×7d×11i,其中a≤10,b≤5,c≤2,d≤1,i≤1,且a,b,c,d,i均為非負(fù)偶數(shù),即a=0,2,4,6,8,10,b=0,2,4,c=0,2,d=i=0,此時(shí)k有6×3×2=36 種選擇.從而所求概率為故m+n=23,(E)正確.

16.A point is chosen at random within the square in the coordinate plane whose vertices are(0,0),(2020,0),(2020,2020),and(0,2020).The probability that the point lies withindunits of a lattice point is.(A point(x,y)is a lattice point ifxandyare both integers.)What isdto the nearest tenth?

A.0.3 B.0.4 C.0.5 D.0.6 E.0.7

譯文坐標(biāo)平面上有一個(gè)以(0,0),(2020,0),(2020,2020)和(0,2020)為頂點(diǎn)的正方形.在正方形內(nèi)隨機(jī)選擇一個(gè)點(diǎn),該點(diǎn)位于格點(diǎn)的d個(gè)單位內(nèi)的概率是.(點(diǎn)(x,y)稱為格點(diǎn),若x和y均為整數(shù).)則d精確到十分位是多少?

圖3

解如圖示,以格點(diǎn)為圓心,d為半徑作一些圓,則正方形內(nèi)的圓內(nèi)部分就是符合條件的點(diǎn)集.因此,該點(diǎn)落在此區(qū)域的概率為求得故(B)正確.

17.DefineP(x)=(x ?12)(x ?22)···(x ?1002).How many integersnare there such thatP(n)≤0?

A.4900 B.4950 C.5000 D.5050 E.5100

譯文定義P(x)= (x ?12)(x ?22)···(x ?1002).則有多少個(gè)整數(shù)n使得P(n)≤0?

解解不等式P(n)=(n?12)(n?22)···(n?1002)≤0,得12≤n≤22,32≤n≤42,··· ,992≤n≤1002,因此符合條件的n有(22?12+1)+(42?32+1)+···+(1002?992+1)=2×(2+4+···+100)=5100 個(gè).故(E)正確.

18.Let(a,b,c,d)be an ordered quadruple of not necessarily distinct integers,each one of them in the set{0,1,2,3}.For how many such quadruples is it true thatad?bcis odd? (For example,(0,3,1,1)is one such quadruple,because 0·1?3·1=?3 is odd.)

A.48 B.64 C.96 D.128 E.192

譯文設(shè)(a,b,c,d)是一個(gè)四元數(shù),其中a,b,c,d ∈{0,1,2,3}且可以相同.則有多少個(gè)這樣的四元數(shù)使得ad ?bc是奇數(shù)?(例如,(0,3,1,1)就是一個(gè)符合條件的四元數(shù),因?yàn)?·1?3·1=?3 是奇數(shù).)

解要使得ad ?bc是奇數(shù),ad和bc必一奇一偶,分兩種情況:

(1)ad是奇數(shù),bc是偶數(shù):此時(shí)a,d ∈{1,3},b,c可以是一奇一偶、一偶一奇、兩個(gè)偶數(shù),共有2×2×(2×2×3)=48種選擇;(2)ad是偶數(shù),bc是奇數(shù):同理可得48 種選擇.故共有96 個(gè),(C)正確.

19.As shown in the figure below,a regular dodecahedron(the polyhedron consisting of 12 congruent regular pentagonal faces)floats in space with two horizontal faces.Note that there is a ring of five slanted faces adjacent to the top face,and a ring of five slanted faces adjacent to the bottom face.How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?

A.125 B.250 C.405 D.640 E.810

譯文如下圖左所示,一個(gè)正十二面體(由12 個(gè)完全相同的正五邊形面組成的多面體)放置在兩個(gè)水平面的空間中.注意到頂面附近有一個(gè)由五個(gè)斜面組成的圓環(huán),底面附近有一個(gè)由五個(gè)斜面組成的圓環(huán).問(wèn)有多少種方法可以通過(guò)一系列相鄰面從頂面移動(dòng)到底面,使得每個(gè)面最多經(jīng)過(guò)一次,并且不允許從底環(huán)移動(dòng)到頂環(huán)?

圖5

解將該圖簡(jiǎn)化成平面圖:頂面為點(diǎn)T,底面為點(diǎn)B,頂環(huán)依次為點(diǎn)T1,T2,T3,T4,T5,底環(huán)依次為點(diǎn)B1,B2,B3,B4,B5,相鄰的兩個(gè)面用線段連接,如上圖右所示.則原問(wèn)題相當(dāng)于:有多少條從點(diǎn)T出發(fā),經(jīng)過(guò)頂環(huán),再經(jīng)過(guò)底環(huán),最后到達(dá)點(diǎn)B的不重復(fù)路徑?

(1)點(diǎn)T到頂環(huán),有5 種選擇,即T1,T2,T3,T4,T5; (2)不妨設(shè)先到T1,又有9 種選擇到達(dá)底環(huán),分別是:T1→,T1→T2→,T1→T2→T3→,T1→T2→T3→T4→,T1→T2→T3→T4→T5→,T1→T5→,T1→T5→T4→,T1→T5→T4→T3→,T1→T5→T4→T3→T2→; (3)每一個(gè)頂環(huán)上的點(diǎn)有2 種方式到達(dá)底環(huán),如T1→B1或T1→B5; (4)以到達(dá)B1為例,有以下9 條路徑到達(dá)點(diǎn)B,分別是:B1→B,B1→B2→B,B1→B2→B3→B,B1→B2→B3→B4→B,B1→B2→B3→B4→B5→B,B1→B5→B,B1→B5→B4→B,B1→B5→B4→B3→B,B1→B5→B4→B3→B2→B.

綜上分析可得,共有5×9×2×9=810 條路徑,故(E)正確.

20.QuadrilateralABCDsatisfies ∠ABC= ∠ACD=90°,AC= 20 andCD= 30.DiagonalsACandBDintersects at pointEandAE= 5.What is the area of QuadrilateralABCD?

A.330 B.340 C.350 D.360 E.370

譯文四邊形ABCD滿足∠ABC= ∠ACD= 90°,AC= 20,CD= 30.對(duì)角線和交于點(diǎn)E,且AE=5.求四邊形ABCD的面積是多少?

圖6

解如圖示,以AC為直徑作一個(gè)圓,交BD與點(diǎn)F,依題意可得EC=15,設(shè)BE=x,依據(jù)相交弦定理AE ·EC=BE ·EF,則得再由切割線定理DC2=DF ·DB,得解得或x=而可得S?ABC=60,故SABCD=360,(D)正確.

21.There exists a unique strictly increasing sequence of nonnegative integersa1

A.117 B.136 C.137 D.273 E.306

譯文存在唯一嚴(yán)格遞增的非負(fù)整數(shù)列a1< a2<··· < ak使得則k是多少?

解令217=x,則

22.For how many positive integersn≤1000 isnot divisible by 3? (recall thatis the greatest integer less than or equal tox.)

A.22 B.23 C.24 D.25 E.26

譯文有多少個(gè)正整數(shù)n≤1000 使得不被3 整除?(注意表示小于等于x的最大整數(shù))

解當(dāng)n不是998,999 或1000 的因子時(shí),易得

(1)公因子:n=1 時(shí),N=2997 能被3 整除;n=2 時(shí),N=1498 不能被3 整除;

(3)999 的非公因子:N=不能被3 整除;

(4)1000 的非公因子:N=不能被3 整除.

綜上可得,符合條件的n有25?3=22 個(gè),故(A)正確.

23.LetTbe the triangle in the coordinate plane with vertices(0,0),(4,0),and(0,3).Consider the following five isometries(rigid transformations)of the plane:rotation of 90°,180°,and 270°counterclockwise around the origin,reflection across thex-axis,and reflection across they-axis.How many of the 125 sequences of three of these transformations(not necessarily distinct)will returnTto its original position? (For example,a 180°rotation,followed by a reflection across thex-axis,followed by a reflection across they-axis will returnTto its original position,but a 90°rotation,followed by a reflection across thex-axis,followed by another reflection across thex-axis will not returnTto its original position.)

A.12 B.15 C.17 D.20 E.25

圖7

譯文設(shè)T是坐標(biāo)平面上以(0,0),(4,0)和(0,3)為頂點(diǎn)的三角形.考慮以下五種平面上的等距變換(剛體變換):繞原點(diǎn)作90°,180°和270°的逆時(shí)針旋轉(zhuǎn),關(guān)于x軸或y軸的反射.任選三種變換(不必不同)可以組成125 種組合,有多少種組合將使得T變回起始位置?(例如,一個(gè)關(guān)于y軸的反射,接著一個(gè)關(guān)于x軸的反射,再接著一個(gè)180°的旋轉(zhuǎn),將會(huì)使得T變回起始位置;但一個(gè)關(guān)于x軸的反射,接著另一個(gè)關(guān)于x軸的反射,再接著一個(gè)90°的旋轉(zhuǎn),將不會(huì)使得T變回起始位置.)

解分兩種情況:(1)全部由旋轉(zhuǎn)組成:只要三次旋轉(zhuǎn)的角度和為360°或720°即可滿足要求,因此有90°+90°+180°,90°+180°+90°,180°+90°+90°,270°+270°+180°,270°+180°+270°,180°+270°+270°共6 種組合;(2)由旋轉(zhuǎn)和反射組合而成:有y軸+x軸+180°,y軸+180°+x軸,180°+x軸+y軸,180°+y軸+x軸,x軸+180°+y軸,x軸+y軸+180°,也是6 種組合.故(A)正確.

24.Letnbe the least positive integer greater than 1000 for which gcd(63,n+120)= 21 and gcd(n+63,120)= 60.What is the sum of digits ofn?

A.12 B.15 C.18 D.21 E.24

譯文設(shè)n是大于1000 的使gcd(63,n+120)= 21,gcd(n+63,120)=60 成立的最小正整數(shù).則n的數(shù)字和是多少?

解由gcd(63,n+120)= 21,可得n ≡6(mod 21);由gcd(n+63,120)= 60,可得n ≡57(mod 60).聯(lián)立解得n ≡237(mod 420),于是n= 1077,1497,1917,···.當(dāng)n= 1077 時(shí),gcd(63,n+120)= 63,不符;當(dāng)n= 1497 時(shí),gcd(n+63,120)= 120,也不符;當(dāng)n= 1917 時(shí),驗(yàn)證后符合條件,此時(shí)數(shù)字和為18,故(C)正確.

25.Jason rolls three fair standard six-sided dice.Then he looks at the rolls and chooses a subset of the dice(possibly empty,possibly all three dice)to reroll.After rerolling,he wins if and only if the sum of the numbers faces up on the three dice is exactly 7.Jason always plays to optimize his chances of winning.What is the probability that he chooses to reroll exactly two of the dice?

譯文詹森擲3 顆標(biāo)準(zhǔn)、均勻的骰子,他看了結(jié)果之后會(huì)選擇若干(可能是0,也可能是3)顆重?cái)S.當(dāng)3 顆骰子正面朝上的數(shù)字和為7 點(diǎn)的時(shí)候,他就贏了.詹森總是按照朝著他贏的最優(yōu)策略去擲.問(wèn)他剛好選擇2 顆骰子重?cái)S的概率是多少?

解擲1 顆骰子得1,2,3,4,5,6 點(diǎn)的概率均為;擲2顆骰子得3 點(diǎn)只有兩種情況:12 和21,概率為,···;擲3顆骰子得7 點(diǎn)有15 種情況:115,151,511,124,142,214,241,412,421,133,313,331,223,232,322,概率為,···.經(jīng)過(guò)計(jì)算,所有結(jié)果如下表所示:

分類/概率/結(jié)果1 2 3 4 5 6 7擲1 顆1 1 1 1 1 1 6 6 6 6 6 6擲2 顆1 2 3 4 5 36 36 36 36 36 1 3 3 10 15擲3 顆216 216 216 216 216

因此,詹森要選擇2 顆骰子重?cái)S,則上次擲的結(jié)果中,任意兩顆骰子的數(shù)字和不能小于7 點(diǎn),否則他將選擇重?cái)S1 顆骰子;且不能3 顆骰子都是4 點(diǎn)或者以上,要不然他將選擇重?cái)S3 顆骰子.根據(jù)以上分析,滿足條件的情況有:(1)擲出1 點(diǎn)、6 點(diǎn)、6 點(diǎn),3 種情況;(2)擲出2 點(diǎn)、5 點(diǎn)、5 點(diǎn),3 種情況;(3)擲出2 點(diǎn)、5 點(diǎn)、6 點(diǎn),6 種情況; (4)擲出2 點(diǎn)、6 點(diǎn)、6 點(diǎn),3 種情況;(5)擲出3 點(diǎn)、4 點(diǎn)、4 點(diǎn),3 種情況;(6)擲出3 點(diǎn)、4點(diǎn)、5 點(diǎn),6 種情況;(7)擲出3 點(diǎn)、4 點(diǎn)、6 點(diǎn),6 種情況;(8)擲出3 點(diǎn)、5 點(diǎn)、5 點(diǎn),3 種情況;(9)擲出3 點(diǎn)、5 點(diǎn)、6 點(diǎn),6 種情況;(10)擲出3 點(diǎn)、6 點(diǎn)、6 點(diǎn),3 種情況.