2014美国数学竞赛AMC10A、10B试题及答案
2014 AMC 10A Problems
2014 AMC 10A ProblemsProblem 1 What is 1)1015121(10-++∙ ? （A ）3 （B ）8 （C ）225 （D ）3170 （E ）170 Problem 2Roy's cat eats 1/3 of a can of cat food every morning and 1/4 of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing 6 cans of cat food. On what day of the week did the cat finish eating all the cat food in the box?（A ）Tuesday （B ）Wednesday （C ）Thursday （D ）Friday （E ）Saturday Problem 3Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $2.50 each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $0.75 for her to make. In dollars, what is her profit for the day?（A ）24 （B ）36 （C ）44 （D ）48 （E ）52 Problem 4Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?（A ）2 （B ）3 （C ）4 （D ）5 （E ）6 Problem 5On an algebra quiz, 10% of the students scored 70 points, 35% scored 80 points, 30% scored 90 points, and the rest scored 100 points. What is the difference between the mean and median score of the students' scores on this quiz?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5 Problem 6Suppose that cows give gallons of milk in days. At this rate, how many gallons ofmilk will cows give in days?（A ）ac bde （B ）bde ac （C ）c abde （D ）abcde （E ）de abc Problem 7Nonzero real numbers ,,,a y x and b satisfy a x < and b y <. How many of the following inequalities must be true?（I ）b a y x +<+ （II ）b a y x -<- （II ）ab xy < （IV ）b a y x //<（A ）0 （B ）1 （C ）2 （D ）3 （E ）4Problem 8Which of the following numbers is a perfect square?（A ）2!15!14 （B ）2!16!15 （C ）2!17!16 （D ）2!18!17 （E ）2!19!18 Problem 9 The two legs of a right triangle, which are altitudes, have lengths 32 and 6. How long is the third altitude of the triangle?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5 Problem 10Five positive consecutive integers starting with have average . What is the averageof 5 consecutive integers that start with ?（A ）3+a （B ）4+a （C ）5+a （D ）6+a （E ）7+a Problem 11A customer who intends to purchase an appliance has three coupons, only one of which may be used:Coupon 1: 10% off the listed price if the listed price is at least $50Coupon 2: $20 off the listed price if the listed price is at least $100Coupon 3: 18% off the amount by which the listed price exceeds $100For which of the following listed prices will coupon 1 offer a greater price reduction than either coupon 2 or coupon 3?（A ）$179.95 （B ）$199.95 （C ）$219.95 （D ）$239.95 （E ）$259.95 Problem 12A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown What is the area of the shaded region?（A ）π9327-（B ）π6327-（C ）π18354-（D ）π12354-（E ）π93108- Problem 13Equilateral △ABC has side length 1, and squares ABDE, BCHI, CAFG lie outside the triangle. What is the area of hexagon DEFGHI ?（A ）43312+ （B ）29 （C ）33+ （D ）2336+ （E ）6 Problem 14The y -intercepts, P and Q , of two perpendicular lines intersecting at thepoint A (6,8) have a sum of zero. What is the area of △APQ ?（A ）45 （B ）48 （C ）54 （D ）60 （E ）72Problem 15David drives from his home to the airport to catch a flight. He drives 35 miles in the first hour, but realizes that he will be 1 hour late if he continues at this speed. He increases his speed by 15 miles per hour for the rest of the way to the airport and arrives 30 minutes early. How many miles is the airport from his home?（A ）140 （B ）175 （C ）210 （D ）245 （E ）280 Problem 16In rectangle ABCD, AB =1, BC =2, and points E, F , and G are midpoints of CD BC ,, and AD , respectively. Point H is the midpoint of GE . What is the area of the shaded region?（A ）121 （B ）183 （C ）122 （D ）123 （E ）61 Problem 17Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?（A ）61（B ）7213 （C ）367 （D ）245 （E ）92 Problem 18A square in the coordinate plane has vertices whose -coordinates are 0, 1, 4, and 5.What is the area of the square?（A ）16 （B ）17 （C ）25 （D ）26 （E ）27Problem 19Four cubes with edge lengths 1, 2, 3, and 4 are stacked as shown. What is the length of the portion of XY contained in the cube with edge length 3?（A ）5333 （B ）32 （C ）3332 （D ）4 （E ）23Problem 20The product (8)(888…8), where the second factor has digits, is an integer whose digitshave a sum of 1000. What is ?（A ）901 （B ）911 （C ）919 （D ）991 （E ）999Problem 21Positive integers a and b are such that the graphs of 5+=ax y and b x y +=3 intersect the -axis at the same point. What is the sum of all possible-coordinates of these points of intersection?（A ）-20 （B ）-18 （C ）-15 （D ）-12 （E ）-8Problem 22In rectangle ABCD, AB =20 and BC =10. Let E be a point on CD such that ︒=∠15CBE . What is AE ?（A ）3320 （B ）310 （C ）18 （D ）311 （E ）20 Problem 23 A rectangular piece of paper whose length is 3 times the width has area A . The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area B . What is the ratio B:A ?（A ）1:2 （B ）3:5 （C ）2:3 （D ）3:4 （E ）4:5Problem 24A sequence of natural numbers is constructed by listing the first 4, then skipping one, listing the next 5, skipping 2, listing 6, skipping 3, and, on the th iteration, listing3+n and skipping . The sequence begins 1,2,3,4,6,7,8,9,10,13.What is the 500,000thnumber in the sequence?（A ）996,506 （B ）996507 （C ）996508 （D ）996509 （E ）996510 Problem 25The number 8675 is between 20132 and 20142. How many pairs of integers ),(n m are there such that 20121≤≤m and 125225++<<<n m m n ?（A ）278 （B ）279 （C ）280 （D ）281 （E ）2822014 AMC 10A SolutionsProblem 1 We have 1)1015121(10-++∙Making the denominators equal gives2254510)108(10)101102105(1011⇒∙⇒∙⇒++∙⇒-- Problem 2 Each day, the cat eats 1274131=+ of a can of cat food. Therefore, the cat food will last for 7726127= days, which is greater than 10 days but less than 11 days. Because the number of days is greater than 10 and less than 11, the cat will finish eatingin on the 11th day, which is equal to 10 Problem 3She first sells one-half of her 48 loaves, or 48/2=24 loaves. Each loaf sells for $2.50, so her total earnings in the morning is equal to 24·$2.50=$60.This leaves 24 loaves left, and Bridget will sell 162432=⨯ of them for a price of 25.1$250.2$=. Thus, her total earnings for the afternoon is 16·$1.25=$20. Finally, Bridget will sell the remaining 24-16=8 loaves for a dollar each. This is a total of $1·8=$8.The total amount of money she makes is equal to 60+20+8=$88.However, since Bridget spends $0.75 making each loaf of bread, the total cost to make the bread is equal to $0.75·48=$36.Her total profit is the amount of money she spent subtracted from the amount of money she made, which is ?????Problem 4The following problem is from both the 2014 AMC 12A #3 and 2014 AMC 10A #4Solution 1Attack this problem with very simple casework. The only possible locations for the yellow house (Y ) is the 3rd house and the last house.Case 1: Y is the 3rd house.The only possible arrangement is B-O-Y-RCase 2: Y is the last house.There are two possible ways: B-O-R-Y and O-B-R-YSolution 2There are 24 possible arrangements of the houses. The number of ways with the blue house next to the yellow house is 3!2!=12, as we can consider the arrangements of O, (RB), and Y . Thus there are 24-12 arrangements with the blue and yellow houses non-adjacent.Exactly half of these have the orange house before the red house by symmetry, and exactly half of those have the blue house before the yellow house (also by symmetry), so our answer is 3212112=∙∙. Problem 5The following problem is from both the 2014 AMC 12A #5 and 2014 AMC 10A #5Without loss of generality, let there be 20 students(the least whole number possible) who took the test. We have 2 students score 70 points, 7 students score 80 points, 6 students score 90 points and 5 students score 100 points.The median can be obtained by eliminating members from each group. The median is 90 points.The mean is equal to the total number of points divided by the number of people, which gives 87Thus, the difference between the median and the mean is equal to 90-87=3. Problem 6The following problem is from both the 2014 AMC 12A #4 and 2014 AMC 10A #6Solution 1We need to multiply by a d / for the new cows and c e / for the new time, so theanswer is acbde c e a d b =∙∙, or . Solution 2We plug in ,5,4,3,2====d c b a and 6=e . Hence the question becomes "2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?"If 2 cows give 3 gallons of milk in 4 days, then 2 cows give 3/4 gallons of milk in 1 day, so 1 cow gives 3/(4*2) gallons in 1 day. This means that 5 cows give (5*3)/(4*2) gallons of milk in 1 day. Finally, we see that 5 cows give (5*3*6)/(4*2) gallons of milk in 6 days.Substituting our values for the variables, this becomes ac dbe / , which is Solution 3We see that the the amount of cows is inversely proportional to the amount of days and directly proportional to the gallons of milk. So our constant is b ac /.Let be the answer to the question. We haveacbde g bde gac b ac g de =⇒=⇒=Problem 7SolutionFirst, we note that (I) must be true by adding our two original inequalities. b a y x b y a x +<+⇒<<,Though one may be inclined to think that (II) must also be true, it is not, for we cannot subtract inequalities.In order to prove that the other inequalities are false, we only need to provide one counterexample. Let's try substituting 1,1,2,3==-=-=b a y x(II) states that 0111)2(3<⇒-<---⇒-<-b a y x Since this is false, (II) must also be false.(III) states that 1611)2)(3(<⇒∙<--⇒<ab xy . This is also false, thus (III) is false. (IV) states that 15.12123<⇒<--⇒<b a y x . This is false, so (IV) is false.One of our four inequalities is true, hence, our answer is 1. Solution 2Also, with some intuition, we could have plugged ,3,1,0Y A X =-== and B =-2 and then plugged these values into the equations to see which ones held.Problem 8Note that for all positive , we have21)!(2)1()!(2)!1(!22+∙⇒+⇒+n n n n n n We must find a value of such that21)!(2+∙n n is a perfect square. Since 2)!(n is a perfect square, we must also have 21+n be a perfect square. In order for21+n to be a perfect square, 1+n must be twice a perfect square. From theanswer choices, 181=+n works, thus, 17=n and our desired answer is 2!18!17 Problem 9Solution 1 We find that the area of the triangle is 3636=⨯. By the Pythagorean Theorem , we have that the length of the hypotenuse is 346)32(22=+ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.Let h be the third height of the triangle. We have 336234=⇒⨯=h h Problem 10The following problem is from both the 2014 AMC 12A #9 and 2014 AMC 10A #10Solution 1Let 1=a . Our list is {1,2,3,4,5} with an average of 15/5=3. Our next set starting with 3 is {3,4,5,6,7}. Our average is 25/5=5.Therefore, we notice that 5=1+4 which means that the answer is 4+a . Solution 2We are given that 254321+=⇒++++++++=a b a a a a a b We are asked to find the average of the 5 consecutive integers starting from in termsof . By substitution, this is4565432+=+++++++++a a a a a aThus, the answer is 4+a . Problem 11The following problem is from both the 2014 AMC 12A #8 and 2014 AMC 10A #11Solution 1Let the listed price be . Since all the answer choices are above $100, we can assume100>x . Thus the discounts after the coupons are used will be as follows:Coupon 1: x x 1.0%10=∙Coupon 2: 20Coupon 3: 1818.0)100(%18-=-∙x xFor coupon 1 to give a greater price reduction than the other coupons, we must have 200201.0>⇒>x x and 2251818.01.0<⇒->x x x .From the first inequality, the listed price must be greater than $200, so answer choices(A) and (B) are eliminated.From the second inequality, the listed price must be less than $225, so answer choices(D) and (E) are eliminated.The only answer choice that remains is $219.95. Problem 12The area of the hexagon is equal to 35423)6(32= by the formula for the area of a hexagon. We note that each interior angle of the regular hexagon is 120º which means that each sector is 1/3 of the circle it belongs to. The area of each sector is ππ33/9=. The area of all six is ππ1836=⨯.sectors, which is equal to π18354- Problem 13Solution 1The area of the equilateral triangle is 4/3. The area of the three squares is 3*1=3. Since 360=∠C , 120609090360=---=∠GCH .Dropping an altitude from C to GH allows to create a 30-60-90 triangle since GCH ∆ is isosceles. This means that the height of GCH ∆ is 1/2 and half the length of GH is 2/3. Therefore, the area of each isosceles triangle is 432321=⨯. Multiplying by 3 yields 4/33 for all three isosceles triangles.Therefore, the total area is 33433433+=++ Solution 2As seen in the previous solution, segment GH is 3 . Think of the picture as one large equilateral triangle, JKL ∆ with the sides of 132+, by extending EF ,GH, and DI to points J, K, and L , respectively. This makes the area of JKL ∆ 431312)132(432+=+Triangles △DIJ , △EFK , and △GHL have sides of3, so their total area is439))3(43(32=.Now, you subtract their total area from the area of △JKL : 3439431312+=-+Problem 14Solution 1Note that if the -intercepts have a sum of 0, the distance from the origin to each of theintercepts must be the same. Call this distance . Since the ∠PAQ=90º, the length of themedian to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is 108622=+, this means 10=a , and the length of the hypotenuse is 202=a . Since the of A is the same as the altitude to thehypotenuse, []602/620=∙=APQ Solution 2We can let the two lines be b x my b mx y --=+=1,. This is because the lines are perpendicular, hence theandm1- , and the sum of the y-intercepts is equal to 0, hence the b b -,. Since both lines contain the point (6,8), we can plug this into the two equations to obtain b m +=68and b m--=168. Adding the two equations gives m m 6616-+=. Multiplying by m gives06166661622=--⇒-=m m m m . Factoring gives 0)3)(13(=-+m m .We can just let 3=m , since the two values of do not affect our solution - one is theslope of one line and the other is the slope of the other line.Plugging 3=m into one of our original equations, we obtain 10)3(68-=⇒+=b bSince △APQ has hypotenuse 202=b and the altitude to the hypotenuse is equal tothe x-coordinate of point A , or 6, the area of △APQ is equal to 20·6/2=60. Problem 15The following problem is from both the 2014 AMC 12A #11 and 2014 AMC 10A #15 Solution 1 (Algebra)Note that he drives at 50 miles per hour after the first hour and continues doing so until he arrives.Let be the distance still needed to travel after the first 1 hour. We have that 355.150d d =+, where the 1.5 comes from1 hour late decreased to 0.5 hours early.Simplifying gives ,105257d d =+ or 175=d .Now, we must add an extra 35 miles traveled in the first hour, giving a total of (C) 210 miles.Solution 2 (Answer Choices)Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. Quickly checking, we know that neither choice(A) or choice(B) work, but(C) does. We can verify as follows. After 1 hour at 35mph , David has 175 miles left. This then takes him 3.5 hours at 50mph . But 210/35=6 hours. Since 1+3.5=4.5 hours is 1.5 hours less than 6, our answer is 210.Problem 16Solution 1Denote D=(0,0). Then A=(0,2), F=(1/2,0), H=(1/2,1). Let the intersection of AF and DH be X , and the intersection of BF and CH be Y . Then we want to find the coordinates of X so we can find XY . From our points, the slope of AF is 4)2(21-=-, and its -intercept isjust 2. Thus the equation for AF is 24+-=x y . We can also quickly find that the equation of DH is x y 2=. Setting the equations equal, we have 3/1242=⇒+-=x x x . Becauseof symmetry, we can see that the distance from Y to BC is also 1/3, so 313121=∙-=XY . Now the area of the kite is simply the product of the two diagonals over 2. Since the lengthHF =1, our answer is 612131=∙ .Solution 2Let the area of the shaded region be . Let the other two vertices of the kitebe I and J with I closer to AD than J . Note that [][][][][]BCJ ADI x DCH ABF ABCD ++-+= . The area of ABF is 1 and the area of DCH is 1/2. We will solve for the areas of ADI and BCJ in terms of x by noting that the area of each triangle is the length of the perpendicularfrom I to AD and J to BC respectively. Because the area of IJ x *21=based on the areaof a kite formula, 2/ab for diagonals of length and , x IJ 2=. So each perpendicularis length 221x -. So taking our numbers and plugging them into [][][][][]BCJ ADI x DCH ABF ABCD ++-+= gives us x 3252-=. Solving thisequation for gives us 6/1=x .Solution 3From the diagram in Solution 1, let be the height of XHY and f be the height of XFY . Itis clear that their sum is1 as they are parallel to GD . Let be the ratio of the sides of thesimilar triangles XFY and AFB , which are similar because XY is parallel to AB and the triangles share angle F . Then 2/f k = , as 2 is the height of AFB . Since XHY and DHC are similar for the same reasons as XFY and AFB , the height of XHY will be equal to the base, like in DHC , making e XY =. However, XY is also the base of XFY , so AB e k /= where AB =1 so e k =. Subbing into 2/f k = gives a system of linear equations, 1=+f e and 2/f e =. Solving yields 3/1==XY e and 3/2=f , and since the area of the kite is simply the product of the two diagonals over 2 and HF =1, our answerIs 612131=∙. Solution 4Let the unmarked vertices of the shaded area be labeled I and J , with I being closer to GD than J . Noting that kite HJFI can be split into triangles HJI and JIF . Because HJI and JIF are similar to HDC and ABF , we know that the distance from line segment JI to H is half the distance from JI to F . Because kite HJFI is orthodiagonal, we multiply6/12/))3/1(*1(= Problem 17Solution 1 (Clean Counting)First, we note that there are 1,2,3,4 and 5 ways to get sums of 2,3,4,5,6 respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is 725)3654321(61=++++. Since there are ⎪⎪⎭⎫ ⎝⎛13 ways to choose which die will be theone with the sum of the other two, our answer is 2457253=∙ . Solution 2 (Casework)Since there are 6 possible values for the number on each dice, there are 21663= total possible rolls.The possible results of the 3 dice such that the sum of the values of two of the die is equal to the value of the third die are, without considering the order of the die, (1,1,2), (1,2,3), (1,3,4), (1,4,5), (1,5,6), (2,2,4), (2,3,5), (2,4,6), (3,3,6). There are 3!/2=3 ways to order the first, sixth, and ninth results, and there are 3!=6 ways to order the other results.Therefore, there are a total of 3*3+6*6=45 ways to roll the dice such that 2 of the dice sumto the other die, so our answer is 45/216=5/24. Problem 18Let the points be )5,(),1,(),0,(321x C x B x A ===, and )4,(4x D =Note that the difference in value of B and C is 4. By rotational symmetry of the square,the difference in value of A and B is also 4. Note that the difference in valueof A and B is 1. We now know that AB , the side length of the square, is equal to174122=+, so the area is 17. .Problem 19By Pythagorean Theorem in three dimensions, the distance XY is 3321044222=++ . Let the length of the segment XY that is inside the cube with side length 3 be . By similartriangles, 10/3323/=x , giving 5/333=x . Problem 20The following problem is from both the 2014 AMC 12A #16 and 2014 AMC 10A #20Note that for 2≥k , , which has a digit sum of k k +=++-+94027. Since we are given that said number has a digit sum of 1000,we have 99110009=⇒=+k k Problem 21Note that when 0=y , the values of the equations should be equal by the problemstatement. We have that a x ax /550-=⇒+=,3/30b x b x -=⇒+=.Which means that 153//5=⇒-=-ab b a .The only possible pairs ),(b a then are )1,15(),3,5(),5,3(),15,1(),(=b a . These pairs give respective x -values of 3/1,1,3/5,5----Problem 22Note that 3102010/15tan -=⇒=︒EC EC . (If you do not know the tangent half-angle formula, it is aa sin cos 1-. Therefore, we have 310=DE . Since ADE is a30-60-90 triangle, 201022=∙=∙=AD AE . Problem 23Solution 1I've no clue how to draw pictures on here, so I'll give instructions. Find the midpoint of the dotted line. Draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. These two lines plus the dotted line form a triangle which is the double-layered portion of the folded paper. WLOG, assume the width of the paper is 1 and the length is 3 . The triangle we want to find has side lengths3321)33(,3322=+ , and 3321)33(2=+. It is an equilateral triangle with height 1333=∙, and area 3321332=∙. The area of the paper is 331=∙, and the folded paper has area 32333=-. The ratio of the area of the folded paper to that of theoriginal paper is thus 2:3. Solution 2 Our original paper can be divided like this:After the fold across the dotted line, our paper becomes:Since our original sheet of paper has six congruent 30-60-90 triangles and and our newone has four, the ratio of the area B:A is equal to 4:6=2:3 Problem 24Solution 1If we list the rows by iterations, then we get1,2,3,46,7,8,9,1013,14,15,16,17,18 etc.so that the 500,000th number is the 506th number on the 997th row.(4+5+6+7+…+999=499,494). The last number of the 996th row (when including the numbers skipped) is 499,494+(1+2+3+4+…+996)=996,000 , (we add the 1—996 becauseof the numbers we skip) so our answer is 996,000+506=996,506. Solution 2Let's start with natural numbers, with no skips in between. 1,2,3,4,5,…, 500000All we need to do is count how many numbers are skipped, , and "push" (add on to)500000 however many numbers are skipped.Clearly, 000,5002)1000(999≤ . This means that the number of skipped number "blocks" in the sequence is 999-3=996 because we started counting from 4.Therefore 506,4962)997(996==n Problem 25The following problem is from both the 2014 AMC 12A #22 and 2014 AMC 10A #25Between any two consecutive powers of 5 there are either 2 or 3 powers of 2 (because 312252<<). Consider the intervals ).5,5),...(5,5(),5,5(8678662110 We want the number of intervals with 3 powers of 2.From the given that 20148672013252<<, we know that these 867 intervals together have 2013 powers of 2. Let of them have 2 powers of 2 and of them have 3 powers of 2.Thus we have the system 201332,867=+=+y x y x from which we get 279=y , sothe answer is2014 AMC 10A Answer Key 1. C 2. C3.E4. B5. C6.A7.B8.D 8. C 10.B 11.C 12. C13.C 14.D 15. C 16.E 17. D 18.B 19.A 20.D 21.E 22.E23.C 24.A 25.B。
2014年美国“数学大联盟杯赛”(中国赛区)初赛七年级(初一)详解
(七年级) 一、 选择题 1. B. No even number can be written as the product of two odd integers. Since 11 is the product of 1 and 11, Skip may have run 11 kilometers. A) 10 B) 11 C) 12 D) 14
27. D. As shown, only choice D is not a product of a divisor of 24 and a divisor of 35.
A) 1 × 1 B) 6 × 7 C) 8 × 7 D) 6 × 11 28. B. The sum of the dimensions is 40 ÷2 = 20. Its dimensions are 16 × 4, and its area is 64. A) 100 B) 64 C) 40 D) 24 29. A. Her average score for 6 tests was 82. So her total was 6 × 82 = 492. Adding 2 × 98, her total for 8 tests was 688. Her average score was 86. A) 86 B) 88 C) 90 D) 94 30. D. From 1 to 99 there are 9; in every 100 #s after there are 19. Include 1000. A) 162 B) 171 C) 180 D) 181 二、填空题 31. 8. 32. 2257. 33. 21. 34. 37. 35. 120. 36. 7410. 37. 2. 38. 81. 39. 32451. 40. 671.
AMC10 2014最新版 (完整版)
6 Suppose that a cows give b gallons of milk in c days. At this rate, how many gallons of milk
will d cows give in e days?
(A) 24 (B) 36 (C) 44 (D) 48 (E) 52
4 Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?
(A) Tuesday (B) Wednesday (C) Thursday (D) Friday (E) Saturday
3 Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $2.50 each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $0.75 for her to make. In dollars, what is her profit for the day?
2014AMC10A试题及简答(繁体中文)
(D) 12
(E) 8
22. 在長方形 ABCD 中， AB 20 、 BC 10 。若點 E 在 CD 上使得 CBE 15 ，則 AE 長度
是多少？
(A)
20 3 3
(B) 10 3
(C) 18
(D) 11 3
(E) 20
23. 一張長方形紙片的長是寬的 3 倍，其面積為 A。將兩個長邊三等分，如圖所示；圖中的
19. 四個邊長分別為 1、2、3、4 的正立方體堆疊如圖所示。試問 XY 包含在那個邊長為 3 的
正立方體的截線段之長度為何？
(A)
3 33 5
(B) 2 3
(C)
2 33 3 X 1
(D) 4
(E) 3 2
2 3 4
Y
20. 考慮 8 8888 兩項的乘積，其中第二項是 k 位數，若乘積為一整數，其各位數字的和
16. 在長方形 ABCD 中， AB 1 、 BC 2 ，點 E、F、G 分別為 BC 、 CD 、 AD 的中點，點 H
為 GE 的中點。試問圖中陰影區域的面積為多少？
(A)
1 12
(B)
3 18
(C)
2 12
(D)
3 12
(E)
1 6
A
B
1
G
H
E
1
D 1 F 1 C 2 2
17. 投擲三顆公正的骰子，試問出現在其中兩顆骰子上數字的和等於剩下那顆骰子出現的數 字之機率為多少？ 1 13 7 5 2 (B) (C) (D) (E) 6 72 36 24 9 18. 若坐標平面上的某個正方形，它各頂點的 y 坐標分別為 0、1、4、5，則此正方形的面積 為多少？ (A) (A) 16 (B) 17 (C) 25 (D) 26 (E) 27
-年amc 10a和b竞赛真题及答案(英文版)
2010-2015年AMC 10A和B竞赛真题及答案（英文版）
你好，目前，只分享了2010-2013年AMC 10A和B竞赛真题及答案（英文版），2014-2015年的暂时还没有上传，等采纳后再私信我吧！需要的话，我可以将Word文档中的原图文件一同发给你，求采纳，毕竟我花了差不多一个下午才整理完，谢谢！(@_@)
2010年AMC 10Aห้องสมุดไป่ตู้赛真题及答案（英文版）
2010年AMC 10B竞赛真题及答案（英文版）
2011年AMC 10A竞赛真题及答案（英文版）
2011年AMC 10B竞赛真题及答案（英文版）
Problem 4
2012年AMC 10A竞赛真题及答案（英文版）
go去wentgone
2012年AMC 10B竞赛真题及答案（英文版）
get得到gotgot
sink下沉sank / sunksunk / sunken
give给gavegiven
hit打hithit
不规则动词表
2013年AMC 10A竞赛真题及答案（英文版）
mistake误认mistookmistaken
grow成长grewgrown
think思考thoughtthought
2014 年美国大学生数学建模竞赛 MCM
2014 年美国大学生数学建模竞赛MCM、ICM 试题2014 MCM A: The Keep-Right-Except-To-Pass RuleIn countries where driving automobiles on the right is the rule (that is, USA, China and most other countries except for Great Britain, Australia, and some former British colonies), multi-lane freeways often employ a rule that requires drivers to drive in the right-most lane unless they are passing another vehicle, in which case they move one lane to the left, pass, and return to their former travel lane.Build and analyze a mathematical model to analyze the performance of this rule in light and heavy traffic. You may wish to examine tradeoffs between traffic flow and safety, the role of under- or over-posted speed limits (that is, speed limits that are too low or too high), and/or other factors that may not be explicitly called out in this problem statement. Is this rule effective in promoting better traffic flow? If not, suggest and analyze alternatives (to include possibly no rule of this kind at all) that might promote greater traffic flow, safety, and/or other factors that you deem important.In countries where driving automobiles on the left is the norm, argue whether or not your solution can be carried over with a simple change of orientation, or would additional requirements be needed.Lastly, the rule as stated above relies upon human judgment for compliance. If vehicle transportation on the same roadway was fully under the control of an intelligent system –either part of the road network or imbedded in the design of all vehicles using the roadway – to what extent would this change the results of your earlier analysis?2014 MCM B: College Coaching LegendsSports Illustrated, a magazine for sports enthusiasts, is looking for the “best all time college coach” male or fem ale for the previous century. Build a mathematical model to choose the best college coach or coaches (past or present) from among either male or female coaches in such sports as college hockey or field hockey, football, baseball or softball, basketball, or soccer. Does it make a difference which time line horizon that you use in your analysis, i.e., does coaching in 1913 differ from coaching in 2013? Clearly articulate your metrics for assessment. Discuss how your model can be applied in general across both genders and all possible sports. Present your model’s top 5 coaches in each of 3 different sports.In addition to the MCM format and requirements, prepare a 1-2 page article for Sports Illustrated that explains your results and includes a non-technical explanation of your mathematical model that sports fans will understand.。
2014-2015年度美国”数学大联盟杯赛“(中国赛区)初赛-(十、十一、十二年级).doc
2014-2015年度美国”数学大联盟杯赛“（中国赛区）初赛（十、十一、十二年级）一、选择题（每小题10分，答对加10分，答错不扣分，共100分，请将正确答案A 、B 、C 或者D 写在每题后面的圆括号内。
）正确答案填写示例如下：=-⨯⨯20522 ？ （A ）A)5 B)15 C)25 D)301. Meg loves her megaphone! The large circular end has a circumference that is the reciprocal of its diameter. What is the area of the circle? ( )A)π14 B) π12 C) 14 D) 122. How many solutions does the equation x x +=233 have? ( )A)0 B)1 C)2 D)43. If y x =-1, which of the following is always true for any value of x ? ( )A) ()()x y -=-2211B) ()()x x y y -=-222211 C) ()()x x y y --=-222211 D) ()()()()x x y y -+=-+22221111 4. Lee the crow ate a grams of feed that was 1% seed, b grams of feed that was 2% seed, and c grams of feed that was 3% seed. If combined, all the feed he ate was 1.5% seed. What is a in terms of b and c ?( )A)b c +3B)b c +3 C)b c +23 D)b c +32 5. If <x 0 and <.x 2001, then x -1 must be ( )A)less than -10B)between-0.1 and 0 C)between 0and 0.1 D) greater than 106. At 9:00 A.M., the ratio of red to black cars in a parking lot was 1 to 5. An hour later the number of red cars had increased by 2, the number of black cars had decreased by 5, and the ratio of red to black cars was 1 to 4. How many black cars were in the lot at 10:00 A.M.? ( )A)13 B)15 C)60 D)657. If x ≠1and x ≠-1, then ()()()x x x x x --++-32241111=( ) A)x -21 B) x +21 C) x -241 D) x -341 8. The Camps are driving at a constant rate. At noon they had driven 300 km.At 3:30 P.M. they had driven 50% further than they had driven by 1:30 P.M.What is their constant rate in km/hr? ( )A)150 B)120 C)100 D)909. The letters in DIGITS can be arranged in how many orders without adjacent I ’s? ( )A)240 B)355 C)600 D)71510. Al, Bea, and Cal each paint at constant rates, and together they are painting a house. Al and Bea togethercould do the job in 12 hours; Al and Cal could to it in 15, and Bea and Cal could do it in 20. How many hours will it take all three working together to paint the house?( )A)8.5 B)9 C)10 D)10.5二、填空题（每小题10分，答对加10分，答错不扣分，共200分）11. What is the sum of the degree-measures of the angles at the outer points ,,,A B C D and E of a five-pointed star, as shown? Answer: . 12. What is the ordered pair of positive integers (,k b ), with the least value of k , which satisfiesk b ⋅⋅=34234?Answer: .13. A face-down stack of 8 playing cards consisted of 4 Aces (A ’s) and 4 Kings (K ’s).After I revealed and then removed the top card, I moved the new top card to thebottom of the stack without revealing the card. I repeated this procedure until thestack without revealing the card. I repeated this procedure until the stack was leftwith only 1 card, which I then revealed. The cards revealed were AKAKAKAK ,in that order. If my original stack of 8 cards had simply been revealed one card at atime, from top to bottom (without ever moving cards to the bottom of the stack),in what order would they have been revealed?Answer: .14. For what value of a is one root of ()x a x a -+++=222120 twice the other root?Answer: .15. Each time I withdrew $32 from my magical bank account, the account ’sremaining balance doubled. No other account activity was permitted. My fifth$32 withdrawal caused my account ’s balance to become $0. With how manydollars did I open that account?Answer: .16. In how many ways can I select six of the first 20 positive integers, disregarding the order in which these sixintegers are selected, so that no two of the selected integers are consecutive integers?Answer: .17. If, for all real ,()()xx f x f x =-21, what is the numerical value of f (3)?Answer: .18. How many pairs of positive integers (without regard to order) have a least common multiple of 540?Answer: .19. If the square of the smaller of consecutive positive integers is x , what is the square of the larger of thesetwo integers, in terms of x ?Answer: .20. A pair of salt and pepper shakers comes in two types: identical and fraternal.Identical pairs are always the same color. Fraternal pairs are the same colorhalf the time. The probability that a pair of shakers is fraternal is p andthat a pair is identical is .q p =-1 If a pair of shakers is of the same color, AE DCBword 格式-可编辑-感谢下载支持 determine, in terms of the variable q alone, the probability that the pair is identical. Answer: .21. As shown, one angle of a triangle is divided into four smaller congruentangles. If the lengths of the sides of this triangle are 84, 98, and 112, as shown,how long is the segment marked x ?Answer: .22. How long is the longer diagonal of a rhombus whose perimeter is 60, if threeof its vertices lie on a circle whose diameter is 25, as shown?Answer: .23. The 14 cabins of the Titanic Mail Boat are numbered consecutively from1 through 14, as are the 14 room keys. In how many different ways canthe 14 room keys be placed in the 14 rooms, 1 per room, so that, for everyroom, the sum of that room ’s number and the number of the key placed inthat room is a multiple of 3?Answer: .24. For some constant b , if the minimum value of ()x x b f x x x b -+=++2222is 12, what is the maximum value of ()f x ? Answer: .25. If the lengths of two sides of a triangle are 60cos A and 25sin A , what is the greatest possible integer-length of the third side?Answer: .26. {}n a is a geometric sequence in which each term is a positive number. If a a =5627, what is the value oflog log log ?a a a +++3132310Answer: . 27. What is the greatest possible value of ()=sin cos ?f x x x ++3412Answer: .28. Let C be a cube. Triangle T is formed by connecting the midpoints of three edges of cube C . What is the greatest possible measure of an angle of triangle T ?Answer: .29. Let a and b be two real numbers. ()sin f x a x b x =++34 and (lg log )f =3105. What is the value of (lg lg )f 3?Answer: .30. Mike likes to gamble. He always bets all his chips whenever the number of chips he has is <=5. He always bets n （10-）chips whenever the number of chips he has is greater than 5 and less than 10. He continues betting until either he has no chips or he has more than 9 chips. For every round, if he bets n chips. The probability that he wins or loses in each round is 50%. If Mike begins with 4 chips, what is the probability that he loses all his chips?Answer: .1129884xword格式-可编辑-感谢下载支持。
(参考资料)2014年美国“数学大联盟杯赛”(中国赛区)初赛五、六年级试卷
A) 100
B) 160
C) 200
D) 250
二、填空题(每小题 5 分，答对加 5 分，答错不扣分，共 50 分，答案请填涂在答题卡上)
31. The sum of the digits of 2014 is 2 + 0 + 1 + 4 = 7. Let n be a natural number.
m = n + 2014. The sum of the digits of m is half the sum of the digits of n.
What is the minimum value of n?
Answer: ______.
32. The sum of 5 different prime numbers is 200. Each of the 5 prime
1
2
26. If 5 of the 200 stripes on Frank’s giant shell are blue, 5 of the remaining
stripes are brown, and the rest are white, there are ? more white stripes
D) 110
22. The average of 2014 sixes is equal to the average of 4028 ? .
A) threes
B) sixes
C) nines
D) twelves
23. What is 0.625% of 8% of 500?
A) 0.25
B) 2.5
数字为三个连续的偶数，个位数字为三个连续的奇数。如果将四位数的
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2014 AMC 10B Problems Problem 1 Leah his 13 coins all Cif Vzhtch are penhiss 占rd r 亡kels . It ske had 亡fim rTicre ritkel than che na5 r aw d ^hefl she would navi the $arrii@ number or pemi&& and Enhow much are Leah 1^ coins 询crthT (A)瞬 (B) .^5(G) 37 (T>) 3f> (E) 41 Problem 22-J +2-^?Problem 3R^ndy dra^s tha first third or his trip an a gravel void, the nsxt 20 miFaf on pdvamsnt, and tha retridinirig one-fifth an a dirt raadL In how long Pandy 1! trip?Problem 4Susi'A pays for 4 mufhin? emd 3 ban^rei5i Calvin twee a? much 口ayring Kor 2 nuuffris ^nd 16 h^nsflas,A muffin 乜 h 口w many t mos as Q^ponsi^e 3^ 口 banana? (A) | (n) | (c) J (D) 2 (E) y(A) 16(B) 24 {C) 32 (r>) 48 (E) 54400 (B)罟 (C >5(A) - 2014 （E ）M4 (C)磊(D) 1 (E) 2014Problem 5daug construets J squara 碘kido 站 u&irtq & ^qujl-fl2B panas of glass^ as shown. Ttia ratio of the height XQ width tor each pane i? ： 2, -and the borcers aro 」nd sne between the pane? are 2 inches wide In inches, what Is tie side length of the squim 「9 win dow?(A) 2fi (B) 28 (C) M (D) 32 (E) 34Problem 6Orvin writ to the stoTB with just snDiigh muhBy to buy 30 ball 口口口弓，When he airivad 〉he disc^overed th/ the h^d A spacial on bd'looni; buy 1 b 訓loan At the ragular pries and get t Escond it 壬 off th« regular p 科亡o. /Vhat is the gr&at*t? F-Hjnrber of bal □亡勺r ：s Orvin cniuld bu^?(A) 33 (B) 34 (C) 36 (D)典 (E)曲 Problem 7Scppo^s A > D > 0 dnd A is u:% grt=dl«i Lhan B What 旨 x7J _ RJ _L R d -L X? a _ ft J W 伽土泸j (B) 1叫爭)(C) lOG(i-) (D) 100(^) M 100(却Sclut anProblem 8 代 truck travels — feet every t seccnds. There are ^eet h a yard How nanv yaiTls does the truck travel Oin 3 minutes?Z.clut anProblem 9F 亡「real rumbe-'S UJ andi±| = 2014it £(A )w^of (H)竽 （C ）7 （D ）7 冋丫TVnat isProblem 10[n ihc adch：ian shawn belcrw .1^ B.C, ond D zine istinct ciqrts. Haw man/ dFttGrent value-z 3r e pas^ibls fcrZ??ABECB+ BCADADBDDD(A) 2 (B) 4 (C)7 (D) 8 (E) 9Problem 1111. For the consurriw- A diicouni uF HA is more idv^r td.w-iii thmn AH/ of tha following dif:□ jn：£：< 1 ) twd 5LCIZE5£」TE 15"足(115COUnti(2) t n ee sui utfMw I〔瞋disCLunli<3> m 2合冥discocrit fallowed by a S% d scuuntWhat i£the smallest pa^^ble pasitice integer valLe t4 ri?(A) 27 (B) 28 (C) 29 (D) 31 (E) 33Problem 12IB. The hrgast div也or of Z014.00Ct<XX) ; itself u its fifth largest divisor?(A) 125.番氐000 (B) 201+ 900* DOO (C) 35l t750.000 (D) W, S00, DOG (E) 5OJ, 500” 000Problem 13Six rugula kti^dgoi ib sur tcui id a _egu ch of siUt t?iig th 1 TZ shown. Whdi 0、C is area of A.4J?C A(A) ?\/3 (B) (C) 1 + Sx/2 (D)2+2jA (E) 3 + 2亦Problem 14droi/e her new c^r on 右trip for a \wno:s ni_mber of hourc, ^v^r^girg 55 milet per hour. At the^Agirning nf ths trip, mlz WA^ rrisplaypd nn thp rdnnprRr, A/hPr=可hr is a3-rigir numtipr wit+i /i > 1 and « + ?J + n < 7 AI the end of itie trip, tre JchmEtFt shc^ea 伽tniiss WMt iE J + M + r" ?(A) 26 (B) 27(C) 36 (D> 笄(£} 41Problem 15Tn rur ijlw DC = 2CB and points E and F lie on AB n th^t ED and FJ? trisect ZADCas shown. What is the ratio nf the area nF Z^DEF tn the area of rectangle ABCD T(A)晋(B)罟(C)攀(D) 1 (E)晋Problem 16Nv fan-s\> ; ded dice are relied. Wnal II die probab) itv that a: (east three of the four dies show t issame⑷籟g ◎箱囲Problem 17^hat is the g-eatest 口亡吨r at 2 that ts a factor of >- l^1?(A) 2'1102(B) 2KIU?l(C) 2LmM(D) 2"J,)S(E) 2山阳Problem 18A Ii3t □' 11 positive integers has a mean of 10』a mEdiari of 瞬and a unicue mode of 8.列h日t is the langes t possible value of an inteqe^ in the Ktl(A) 24 (B)加(C) 31 (D) 33 (E) 35Problem 丄9T IAO con匚wrt|-,c c neies r^dil and 2. Two pcints cn the outer circle ai'e chosen inde^enderit y ard uniformly at randoin. What iu the probability lhat ths chord joiriing the two points ntere&ctc the irirer cin cle?⑴！9)字(D>3(E)|Problem 20Fdf how nsn/ int =g?r5 -r is the number r1—+ SO 师第Tiw?(A) y (B) 10 (C) 12 (D) 14 (E) IdProbfem 211 rap^zold AECD has parallel sices AB of lengTh 33 and CD of length 21 I ne 匚ther rwo sid=s are of lengths 1U and ll. The mn glmw at A mnd B mre 己二“土日.Whmt『w ths length of the shorter di^igonsl of ABCDl(A) 1fl"5 (B) 25 (C) fiTin (D) 1 用迈(E) 2fiSolutionProblem 22Ciqhr 弓=厂,匕 rim:；li-i? the irsice nf a 三、“* 讨i「h 5 re =- gti- 2 mm shz./■.■■-i. i= t-|p ra dii.^ 十the : re I? tan^nr to mil of the?e semictrclss?(A)气匹(B) ^2 (C)^±l (D)军(E)^Problem 23A ipliuid 5 ri^Lilibud In a t uriLdLyd right cilLUldi uui it ihuwhi・ ill匕vuluum uT the truiiLditsdl cune I、twice that of "lhe sphere. What 石the ratio of the radios nF th= bottom hasp of "the truncatprl 匚匚no to the radius of the top oase D F the truncated cone?Problem 2斗The nunber? L 2d 3』4, 5 are ;□ be arrangeG in a circle. An aiTangemer^t is 匕ed f it b riot true that for every n from 1 to 15 one can find a subset of the numbers that appear consscutiMel/ on the crirde that wum to n.Arran gemerts that diffe r arl/ by a ratation or a refection are cjnsidEned the same. How many difF&rort □rr-angemont; aro th&ro?(A) 1 (B) 2 (C) 3 (D) 4 (E) 5Problem 25In a smell pond t*iene are eleLer lily pads ri a 口许libelee 0 through 10L b frog is sitting on pad 1 //hen the frog is on pad N)U < N < 1© it wilf jump tc pad JV -1 with probability 百and to pad /V +1 with probability 1 - Each jump is independent of the previa us jumps, If the frog reaches pad 0 it 嘶II be eaten by a patientl/ waiting snake. If tne freg reaches pad =0 it w II ezit the pend,, ne\jer to Tturn. jvhat is the probability that the frog will escape being oaten by ttie snake?吨但熄(唸(D圧砒答案：1. C2. E(c) V3 (DJ 2 (E)汇兰633. E4. B5. A6. C7. A8. E9. A10. C11. C12. C13. B14. D15. A16. B17. D18. E19. D20. C21. B22. B23. E24. B25. CProblem 12C 17(1(A) 3 (B) 8 (C) —(D) —(E) 170Problem 21亠IRoi/'s c at eats —of a cai of c^t feed evs-r^ nnorniriq and 了口f t匚耳n 口f cat fccri ever^ pypnifiq.日戶fuirs ife!i?diFiij 1右&L on Munday rnarniB iy^ Ruy npi=*H^d 划；bux itsntdiHiiriy 6 &di is 酎u^st luud^・Di i A h dI dsy uf tliw wttfk did thy c di finish 邑^Hng dll thid edit Food m Ihw btM?(A) Tuesday (B) V^Wric&day (C) ThuTsday (D) Fridas" (R) SaturdayProblem 3Bndq=t bafkes 斗吕a = \jes ot bread for her bakery- SH B sell= half of t hem in t-~ie merninq tor each. I rnthe afternoon she s：三||生tv/o tkirids Df v/hat eke nas left^ and because they are not fresh^ 三hm charges only ha if price in rhp ate aftfimoan she selk rhe r&imiining loav 已弓at a dcxlsr each. Each loaf costs SO. 7F> for h ar to iriAkm・in dollars H wh<3t ig hor praHt for the >dlay?(A) 21 (B)眺(C) 44 (D> 犒(K) 52Problem 4lAijIking down Jane stnget, Ra^ph ft ur dailies In 2 row* aach pointed m dirfisrent color. Heihu riDij_.i=»bufufL ihti r^o rtou^ti, and F IE pmaPiJ thu Lluu hou^u teforb th® yJlu^1 hDu!=u. Thu blu^jhu Jsy *3=rtet tu 七’im yellow hu 卜uvf hiai iy urd&rir»^b o f th«L ul 口Itfd liacl士弋0 Trx posslb w *(A) 2 (B) 3(C) 4 (D) 5 {E) 6Problem 5On I jlc^bTd qui2, LC/JJi af studtritb 70 pOiritL. )5坏 沁」3 80 pan ：£, 30'兀 tLcrud 90 puinti. jnd the ro?t scored 100 points, what lg the dlfi^rsnc& batman the maar and median sc ora or ths Student^' 2匚ure5 u” this quiz^(A) 1 (B) 2 (C) 3 (D) 4 (E) 5Pnobl^m ES jppose that a cow5 give b gallons of milk in 栏 dav s ' At this rate, how man/ qaJIans of rmlk will d cows qive n r si.ai/E?Pnoblem 7Man zero raal numb erg# ci … andl ii 亍曰 ti^Fy _r V 口 and y V b How m-any of the following inequalities EU 五 tbe t 『ue? (I) jr + y < ^ + 6(IT) yr — ir <(/ — />(III)刊 < “(A) (] (H> 1 (C) 2 (n) {E )4Problem 8whicn of ths following numtiars is a perfect squara? (C)空空(D)辿 rl (E)(A)呼an 1M161 HiM7! P (D ) (E) 1 KT19JProblem 9The two cf a right irjangle, which are atiitud^ hau? lengths 2\/3 and G. How long is the third昌hitudw of the tr dangle?(A) 1 (R> 2 (C) 3 (D) 4 (E) 5Problem 10Five positive consecutive irn卸日巧startrig with(i have average b、What is the average of 5 corsecutive integers that stare with(A) u + 3 (B) u+4 (C)«+5 (D) u+& (E)a + 7Problem 11A custumsr whe intends to purchase an 日匚has three： cgupons. only one of 宵hHh may used;Coupon 1: 10% off Hhe listed price if the listed price is at least 830CHijp in 2： S2fl \\~f iS'H i price f rli& pfir? S < If 礼祁00Coiipnn m：nff tfie ^niciJITt fcy wh rh thp I F T PC! prirp PK「PP*卜匚「Aihizh of the foliating I tted pnces will caupan 1 offer a greater priem radLiction than coupon 2 ar ccupon 3?(A) $179』5 (B) $190.95 (C) S219.95 (D) S23D D5 (E) S2M.^jProblem 12h rsgiiFar hsitagon hs= z r?ngth □. Congrucnl arcs w th radi 3 z徑drawn wsth tha center ac ^ach of the verti匚匚resting circular sectors as shown, rhe reg on iiisidE the hexagon but outside the sectn^s is shsdAd as shown v>Fi*『；s「F IB吕『吕n o>r the shaded re-gion?"(A) 271/3 - S TJ(B) 27?3-6TU(C) &4S/5-18JT(D) 54/3 —12工(E)丄08爲-陥Problem 13Equilateral △ABU has fide length 1』snd squares AB[>E. t CAF<^is cuUida Vie triarnjie,& th-= am4 oF 卜口n r^EFCi HF?(A) 严 (D)专(C)3+v^ (D)今超(E) 0Problem 14rhe ij-intarcapts. J P arid Q f of two perpondiculBr knes intarsectiinq at ths point A( G, 8) ti.avs a ^um of i&^o Wh竜t is th© 目「吕目of △A J PQ T(A) 45 (B) 48 (C) 54 (D) 60 (E) 72Prablem 15D.3Vin nl-ix/RS from his l^cnE tc thm sirpnrr ―口ratch a fliqhr. He dr\i^^ :i5 in the fir=；r hour,, but nerizRT Ihd L 1 it: will tiu 1 t uur IdLe ir ht: LunLinuE^ 口L l和in J. hit: hin spuuLl Ly 1G rrilt?i pu，i uur fur(ha rest of the to lhe airport mrid arrivs^ 30 minuw^ early, How many miss is the airport 耐onn hit hom»?(A) 140 (B) 175 (C) 210 (D) 245 (E) 260Problem 16In r»ctangl»AHC^rt, AI3 —1』F3C7 ■ 2}and points E, F,and G are nnidponts of HC7, CD,, and AD r rEipeetivtl^. F^clnt II s th* imdpaint of GE.What s cht area of the shaded region fD L L C2窗⑷召咗⑹春⑼容®lProblem 17Th res *air s M-sided dice are ■ol ed ・内h 曰t s the prot ability that the values shown on two nf t 卜旧dice SUT to the value shown on the remaining dis?i n 7 q 空S讣㈣箱(C)36 (D暢(瞻Problem 18A sqja^e in the coord nate plane has vertices whose ir-coorciiriares are 0, L 4」and 5 what Is ihe ansa of the square?(A) 16 (B) 17 (C) 2& (D) 26 (E) 27Problem 19Four cubss with edge lengths L 2, 3f and 4 are stacked as shown. Thwt is the length of the poftcn of .Y V eontoH&d H tbid cubo with adgo length 3?(A)響(B) 2v/3 (C)警(D) 4 (E) 3迈Problem 20rre aroduct ；&){ES8 一 -. inhere Uia second rector ha? k uigt® IM an integer 罰口證zlgitz hjvm 扌surr- ct 1000. Wiiat is it?(A) iW)l (D) <>11 (C) 919 (D) 901 (E) 999Problem 21PositivB integers a arid b are such that ths graphs d y = ax +5 jnd j/ = Jr + A intersec: ths z~axis mt the sdiTfi point. Wl idt is the S^ITII□ f dll fjnsziibl匕X"Luurdin.diteb of Ihe^e poihts J T intersHcticn?(A) - 20 (B) -18 (C) - 15 (D) -12 (E) - &Problem 22in rectangle A BCD, AB= 20 mnd HC= 10, Let £? he pciint on C7D such thmt ZCJ3/17 = 15°. w|i3t is AE1 (A)卑岁(B) 30^ (C) 18 (D) 11^/3 (E) 20Problem 23A rsctang」ln「pne^s D F P a P B" whose length is times the width has =raa A,Th© paper is divided inta three equal sections alonq the 口pposite enqths-, and then a dotted line is drawn from the first divider to the secorid divider on the? o-Dpcsiie 宜ick 35 shsv^in. The paper is then fcldsd flat slong this dotted fine to create d new slape wtth arsaB Wha t is the ratio D: .4?Problem 24A ■■querce of naturBi numbers is ranitrucked hy listing the frit 4. lii 曰i skipping one, listing ll it rit^：d 3 sluppinci 2. Iistifi^ 6, shipping 3,吕nd 」cn the retn iteratorij listing M +3 and skipping a. "■ he seq 」三"匕巳 begHs l n L?n 3,丄 G, 7. K. 9,10n 13. What is the 500* 000th nutrrber in the “quencm 亍(A) 996 f>ft6 (B) ^^7 (C)怖5俯 (D) 996509 (E) 99651C1 Problem 25T TB n limber 5*"丁 © botwoon 2?111 and 2'Q14. How manv p^lrs of integers (rn. JA ' are th are ^uch tnat 1 兰 ru 兰 jMJl 空 ai id5* < 2fa < 2*1*2 < 5rt+1 ?答案：1.C 2.D 3.D 4.B 5.E 6.B 7.B 8.B 9.D 10.C 11.C 12.C 13.C 14.C 15.B 16.B 17.D 18.E 19.D 20.B 21.C 22.A25.A(D) 3 "(A) 27B (B) 2TO (C) 2S0 (D) 281 (E) 2&2。