Answer: D

Solution:

Jill gets to the pool in 1/10 of an hour, which is 6 minutes. Doing the same for Jack, we get that Jack arrives at the pool in 1/4 of an hour, which in turn is 15 minutes. Thus, Jill has to wait 15-6=9 minutes for Jack to arrive at the pool.

中文解析：

距离都是1英里， Jill的速度是10 mph, 因此需要的时间是1/10=0.1小时=6分钟。Jack的速度是4mph,因此需要的时间是1/4=0.25小时=15分钟。15-6=9分钟，答案是D 。

Problem 4

Answer: E

Solution:

There are 2 ways to order the boys on the end, and there are 6 ways to order the girls in the middle. We get the answer to be 2*6=12.

中文解析：

2个男孩站两边，有2种顺序， 3个女孩站中间，有3*2*1=6种顺序， 因此共有：2*6=12种顺序，答案是E。

Problem 5

Answer: A

Solution:

Because 40 is less than the score of every game they've played so far, the measures of center will never rise. Only measures of spread, such as the A(range) may increase.

中文解析:

40分比之前的比赛分数都低，因此增加40后，平均值会降低，中位数会降低，mode不会发生变化，Range会增加。 答案时A。

Problem 6

Answer: B

Solution:

Splitting the isosceles triangle in half, we get a right triangle with hypotenuse 29 and leg 21. Using the Pythagorean Theorem , we know the height is sqrt(29*29-21*21)=20. Now that we know the height, the area is (20*42)/2=420.

中文解析：

这个等腰三角形可以被分成两个面积相等的直角三角形，其中直角三角形的底是42/2=21， 斜边是29，根据勾股定理，高是20. 因此这个大三角形的面积是：1/2 * 42*20=420. 答案是B。

Problem 7

Answer: E

Solution:

We can instead calculate the probability that their product is odd, and subtract this from 1. In order to get an odd product, we have to draw an odd number from each box. We have a 2/3 probability of drawing an odd number from one box, so there is a 2/3 * 2/3=4/9 probability of having an odd product. Thus, there is a 1-4/9=5/9 probability of having an even product.

中文解析：

两个数的乘积是双数， 则其中一个数必须是双数。第一个数是双数的概率是1/3， 第一个数是单数，第二个数是双数的概率是：2/3*1/3 =2/9；因此乘积是双数的概率是：1/3+2/9=5/9. 答案是E。

Problem 8

Answer: D

Solution:

the third side of a triangle is less24. Thus, the perimeter is less than 48.

中文解析：

根据“两边之和大于第三边"规则, 三角形的第三条边一定小于24. 则这个三角形的边长小于48. 答案是D。

Problem 9

Answer: D

Solution:

The sum of 1,3,5,……39 is (1+39)*20/2= 400.

中文解析：

Janabel销售出去的商品数量分别是：1， 3， 5， 7， ....这是一个等差数列（Arithmetic Senquence）。第20项是：1+19*2=39. 前20项和是: (1+39)/2 * 20=400. 答案是D。

Problem 10

Answer: B

Solution:

The question can be rephrased to "How many four-digit positive integers have four distinct digits?", There are 9 choices for the first number, since it cannot be 0, there are only 9 choices left for the second number since it must differ from the first, 8 choices for the third number, since it must differ from the first two, and 7 choices for the fourth number, since it must differ from all three. This means there are 9*9*8*7=4536 integers.

中文解析：

千位数上的数字可以是：1，2，3，......9 中任选一个，有9中选择；百位数，可以从0，1，2，.....9中选，但不能和千位数重复，因此有9种选择， 十位数位数不能和千位数，百位数重复，因此有8种选择； 个位数有7种选择，因此： 9*9*8*7=4536. 答案是B。

Problem 11

Answer: B

Solution:

The probability of choosing A as the first letter is 1/5. The probability of choosing M next is 1/21. The probability of choosing C as the third letter is 1/20 (since there are 20 other consonants to choose from other than M). The probability of having 8 as the last number is 1/20. We multiply all these to obtain: 1/5 * 1/21 * 1/20 *1/10 =1/21000.

中文解析：

第一个从5个元音字母中选中”A“ 的概率是1/5. 第二个字母从21个辅音字母中选中”M“的概率是1/21. 第三个字母从剩下的20个辅音字母中选中”C"的概率是1/20. 第四个从0-9的数字中选中“8”的概率是1/10. 因此“AMC8”的概率是：1/5* 1/21 * 1/20*1/10=1/21000. 答案是B。

Problem 12

Answer: C

Solution:

We first count the number of pairs of parallel lines that are in the same direction as AB. There are 4 lines with same direction as AB. There are C(4,2)=6 pairs of parallel lines.

We can do the same for the lines in the same direction as AE and AD. This means there are 6*3=18 total pairs of parallel lines.

中文解析：

AB， EF， DC， HG这4条边都是平行的， 可以有C（4，2）对。同样的， AE， BF，CG，DH也是有C (4,2) 对平行边， AD， BC， FG， EH也是有C(4,2)对平行边。

共: 3*C(4,2)=3* 6=12对。答案是B。

Problem 13

Answer: D

Solution:

We can simply remove 5 subsets of 2 numbers, while leaving only 6 behind. The average of this one-number set is still 6, so the answer is D(5).

中文解析:

这组数的平均值是：（1+2+3+...+11）/11=6. 删除2个数后，剩下的数的平均值还要保持是6，因此删除的2个数平均值还是6就好。这个数据集中，平均值是6的2个数可以是：1和11， 2和10， 3和9， 4和8， 5和7， 共5组。因此答案是D。

Problem 14

Answer: D

Solution:

Let our 4 numbers be 2n-3,2n-1, 2n+1, 2n+3. Then our sum is 8n. The only answer choice that cannot be written as 8n is 100.

中文解析：

假设这4个连续的奇数分别是：2n-3, 2n-1, 2n+1, 2n+3, 则这4个数的和是8n. 也就是说，这个和是8的倍数， 这5个选项中只有100不是8的倍数，因此答案是D。

Problem 15

Answer: D

Solution:

There are 198 people. We know that 29 people voted against both the first issue and the second issue. That leaves us with 169 people that voted for at least one of them. 149+119-169=99.

中文解析：

198个学生中有29个学生两个议题都反对，因此有169个学生至少支持一个议题。149+119-169=99. 有99个学生两个议题都支持。答案是D。

Problem 16

Answer: B

Solution:

We see that the minimum number of ninth graders is 6, because if there are 3 then there is 1 ninth-grader with a buddy, which would mean 2/5 sixth graders with a buddy, and that's impossible. With 6 ninth-graders, 2 of them are in the buddy program, so there 5 sixth-graders total, two of whom have a buddy. Thus, the desired fraction is (2+2)/(5+6)=4/11.

中文解析：

1/3的9年级学生和2/5的6年级学生结成buddy. 我们也可以理解为有2/6 的9年级学生和2/5的6年级学生结成Buddy . 因此9年级的学生至少有6个， 6年级的学生至少有5个。 结成Buddy的9年级学生有2个，6年级学生有2个， 因此题目所求的占比是： （2+2）/（6+5）=4/11. 答案是B。

Problem 17

Answer: D

Solution:

Somehow we get d/v=1/3 ; d/(v+18)=1/5. Thus v=27. D=v*1/3=27*1/3=9.

中文解析：

假设正常情况下的速度是v, 则没有交通拥堵这天的速度是（v+18)， 距离是（v+18)* 12/60=v*20/60 可以算出v=27. 因此距离是：27*（20/60）=9 mile. 答案是D。

Problem 18

Answer: B

Solution:

The top row has a first term 1 and a fifth term 25, so we have the common difference is (25-1)/4=6. This means we can fill in the first row as(1,7,13, 19, 25). The fifth row has a first term of 17 and a fifth term of 81, so the common difference is 16. We can fill in the fifth row as (17,33,49,65,81): We must find the third term of the arithmetic sequence with a first term of 13 and a fifth term of 49. The common difference of this sequence is (49-13)/4=9, so the third term is 13+2*9=31.

中文解析：

每行，每列都是等差数列（Arithmetic equence) , 其中第一行应该是：1， 7， 13， 19， 25. 最后一行应该是：17， 33， 49， 65， 81. 第三列是: 13,22,31,40 , 49. 所以X是31. 答案是B。

Problem 19

Answer: A

Solution:

Using Pick's Theorem(B/2+I-1), the area of the triangle is 4/2 +4-1=5. Therefore, the triangle takes up 5/30=1/6 of the grid.

Problem 20

Answer: D

Solution:

The average cost of each pair of socks is 24/12=2.There are two ways to make packages of socks that average to $2. You can have:

Two $1 pairs and one $4 pair (package adds up to $6)

One $1 pair and one $3 pair (package adds up to $4)

So now we need to solve 6a+4b=24 where a is the number of $6 packages and b is the number of $4 packages. We see our only answer is a=b, b=3, which yields the answer of 2*2+3*1=7.

中文解析：

12双袜子24元钱，则平均每双2元。

每双袜子1元，3元或4元，为了平均一双袜子2元，则需要2双1元+1双4元 （共3双6元)。1双1元+1双3元(共2双4元）。Ralph每种价格的至少买一双，需要花费6元（2双1元+1双4元）+4元（1双1元+1双3元). 还剩下14元，需要买7双袜子。14=6+4+4， 因此需要再买2双1元+1双4元 +2 *(1双1元+1双3元)， 因此1元1双的袜子共买了：7双。答案是D。

Problem 21

Answer: C

Solution:

EF=sqrt(32) ; BC=EF=sqrt(32). JB=sqrt(18); Since JBK is equilateal, BK=sqrt(18).

Since ABCDEF is equiangular, <ABC=120. Thus, <KBC=360-120-90-60=90. So KBC is a right triangle with legs sqrt(32), sqrt(18). Now, its area is sqrt(32)*sqrt(18)/2=12.

中文解析：

在正方形ABJI中， 面积18， 因此边长JB是sqrt(18). 同理，在正方形中EFGH中， EF=sqrt(32).

六边形ABCDEF的每个角都相等，因此角ABC=（6-2）*180/6=120. 另外角JBA=90， 角JBK=60，因此角KBC=360-120-90-60=90. 三角形BCK是直角三角形。

BK =JB =sqrt(18）；BC=FE=sqrt(32). 因此根据直角三角形面积公式， BCK的面积是：1/2 * sqrt(18) *sqrt(32)=12. 答案是C。

Problem 22

Answer: C

Solution:

As we read through this text, we find that the given information means that the number of students in the group has 12 factors, since each arrangement is a factor. The smallest integer with 12 factors is 2^2 * 3*5=60.

中文解析：

根据题意， 这个数字有12个因子。12=3*2*2. 因此有12个因子的最小的数字是：2*2*3*5=60. 答案是C。

Problem 23

Answer: D

Solution:

The average of the 12 slips is 7, which means A, B, C, D, E will have values 5, 6, 7, 8, 9 respectively. Cup E has a sum of 9, since it already has a 2 slip, the others should be 4.5, 2.5. Cup B has a sum of 6, since it already has a 3 slip, the other slip should be 3. Cup A should be 2+3; Cup C should be 3+4; Thus, Cup D must be 3.5+2.5+2. The answer is D.

中文解析：

这12个数字的和是35， ABCDE是连续的5个整数，其和是35， 因此平均数是7， 故C=7， ABCDE这5个数分别是：　５，　６，　７，　８，　９.　B中已经有３了，　因此B中另外一个数也是３. E中已经有２，　还需要７.　７可以由４.５＋２.５组成. A是5, 可以用2+3 组成, C是7, 可以由4+3组成. D是8,可以由2.5+3.5+2组成. 因此3.5在D杯子里. 答案是D.

Problem 24

Answer: B

Solution:

On one team they play 3*n games in their division and 4*m games in the other. This gives 3n+4m=76.

Since m>4 we start by trying m=5.

We try m=7 this does work giving n=16, m=7 and thus 3*16=48 games in their division.

中文解析:

每个division有4个球队, 因此每个队在division内部的比赛场数应该是N*3=3N . 每个球队和division外部的比赛场数是: 4M. 因此: 3N+4M＝76. 其中N>2M>8, M>4. M和N是正整数. 因此我们从m=5,6,7,....逐个试,当M=7时,N=16符合题意. 因此内部比赛场数是3N =3*16=48. 答案是B.

Problem 25

Answer: C

Solution:

We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base 3 and height 1, so the combined area of the four triangles is 4*(1/2 * 3*1)=6. The area of the smaller square is 3*3=9. We add these to see that the area of the large square is 9+6=15.

中文解析:

如下图所示斜着的正方形面积最大. 其内部的正着放的小正方形的边长是: 5-1-1=3. 面积是: 3*3=9.

下面我们计算红色部分的小三角形的面积: 其底是3, 高是1, 因此面积是1/2 * 3*1=1.5.

因此题目所求的斜着的正方形的面积是: 9+ 4* 1.5=15. 答案是C.

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