amcdrill | AMC8 AMC10 AMC12 Practice

Q1

Which of the following is the largest?

以下哪个最大？

(A) $\frac{1}{3}$ $\frac{1}{3}$ (B) $\frac{1}{4}$ $\frac{1}{4}$ (C) $\frac{3}{8}$ $\frac{3}{8}$ (D) $\frac{5}{12}$ $\frac{5}{12}$ (E) $\frac{7}{24}$ $\frac{7}{24}$

Correct Answer: D

Express all the choices as fractions using their least common denominator: $\frac{1}{3} = \frac{8}{24}$, $\frac{1}{4} = \frac{6}{24}$, $\frac{3}{8} = \frac{9}{24}$, $\frac{5}{12} = \frac{10}{24}$, $\frac{7}{24}$. Thus, $\frac{5}{12}$ is the largest.

将所有选项用最小公分母化为分数：$\frac{1}{3} = \frac{8}{24}$，$\frac{1}{4} = \frac{6}{24}$，$\frac{3}{8} = \frac{9}{24}$，$\frac{5}{12} = \frac{10}{24}$，$\frac{7}{24}$。因此，$\frac{5}{12}$ 是最大的。

Q2

$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10} =$

(A) $\frac{1}{4}$ $\frac{1}{4}$ (B) 6.4 6.4 (C) 9 9 (D) 10 10 (E) 11 11

Correct Answer: D

The sum of all the numerators is 100. Consequently, the sum of all the fractions is $100/10 = 10$.

所有分子之和为 100。因此，所有分数的和为 $100/10 = 10$。

Q3

Each day Maria must work 8 hours. This does not include the 45 minutes she takes for lunch. If she begins working at 7:25 AM and takes her lunch break at noon, then her working day will end at

Maria 每天必须工作 8 小时。这不包括她午餐的 45 分钟。如果她早上 7:25 开始工作，并在中午休息午餐，那么她的工作日将结束于

(A) 3:40 PM 下午3:40 (B) 3:55 PM 下午3:55 (C) 4:10 PM 下午4:10 (D) 4:25 PM 下午4:25 (E) 4:40 PM 下午4:40

Correct Answer: C

Maria’s working day ends 8 hours and 45 minutes later than 7:25 am. Eight hours later than 7:25 am is 3:25 pm. Forty-five minutes later than 3:25 pm is 4:10 pm.

Maria 的工作日结束时间是早上 7:25 后 8 小时 45 分钟。早上 7:25 后 8 小时是下午 3:25。再加 45 分钟是下午 4:10。

Q4

Which of the following represents the result when the figure shown at the right is rotated clockwise 120° about its center?

将右图顺时针绕其中心旋转 120° 后，哪个选项表示结果？

(A)

(B)

(C)

(D)

(E)

Correct Answer: B

Rotating the figure clockwise 120° about its center corresponds to choice (B).

图形绕中心顺时针旋转 120° 对应选项 (B)。

Q5

Given that 1 mile = 8 furlongs and 1 furlong = 40 rods, the number of rods in one mile is

已知 1 英里 = 8 弗隆，1 弗隆 = 40 杆，则 1 英里有多少杆？

(A) 5 5 (B) 320 320 (C) 660 660 (D) 1760 1760 (E) 5280 5280

Correct Answer: B

1 mile = 8 furlongs = 8 × 40 rods = 320 rods.

1 英里 = 8 弗隆 = 8 × 40 杆 = 320 杆。

Q6

The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is

任意六个连续正整数的乘积的个位数（个位）是

(A) 0 0 (B) 2 2 (C) 4 4 (D) 6 6 (E) 8 8

Correct Answer: A

Any selection of six consecutive positive whole numbers has at least one multiple of 5 and at least one multiple of 2. Thus, the product has at least one factor of 5 × 2 = 10 and must end in 0.

任意六个连续正整数中至少有一个5的倍数和至少有一个2的倍数。因此，该乘积至少有一个因子 $5 \times 2 = 10$，一定以0结尾。

Q7

If $\angle A = 60^\circ$, $\angle E = 40^\circ$ and $\angle C = 30^\circ$, then $\angle BDC =$

如果 $\angle A = 60^\circ$，$\angle E = 40^\circ$ 且 $\angle C = 30^\circ$，则 $\angle BDC =$

(A) 40° 40° (B) 50° 50° (C) 60° 60° (D) 70° 70° (E) 80° 80°

Correct Answer: B

Since the sum of the angles in any triangle is 180°, $\angle ABE = 180^\circ - (60^\circ + 40^\circ) = 80^\circ$. Since $\angle ABD$ and $\angle DBC$ together form a straight angle, their sum is 180°, so $\angle DBC = 180^\circ - 80^\circ = 100^\circ$. Thus $\angle BDC = 180^\circ - (100^\circ + 30^\circ) = 50^\circ$.

任意三角形的内角和为 $180^\circ$，因此 $\angle ABE = 180^\circ - (60^\circ + 40^\circ) = 80^\circ$。由于 $\angle ABD$ 和 $\angle DBC$ 组成一直角，它们的和为 $180^\circ$，所以 $\angle DBC = 180^\circ - 80^\circ = 100^\circ$。因此 $\angle BDC = 180^\circ - (100^\circ + 30^\circ) = 50^\circ$。

Q8

For how many three-digit whole numbers does the sum of the digits equal 25?

三位数中有多少个各位数字之和等于25？

(A) 2 2 (B) 4 4 (C) 6 6 (D) 8 8 (E) 10 10

Correct Answer: C

Since the sum of the three digits must equal 25 and 8+8+8 = 24, it follows that at least one of the digits must be a 9. If only one digit is 9, then the other two digits add up to 16 and are each 8. If two digits are 9, then the other digit is 7. There are six numbers: 988, 898, 889, 997, 979, 799.

三位数字之和必须等于25，且 $8+8+8 = 24$，因此至少有一个数字是9。如果只有一个9，则另外两个数字之和为16，每个是8。如果两个9，则另一个数字是7。有六个这样的数：988、898、889、997、979、799。

Q9

A shopper buys a $100 coat on sale for 20% off. An additional $5 is taken off the sale price by using a discount coupon. A sales tax of 8% is paid on the final selling price. The total amount the shopper pays for the coat is

一位顾客以20%折扣购买一件原价100美元的外套，使用折扣券额外减去5美元销售价。最终售价需缴纳8%的销售税。顾客为这件外套支付的总金额是

(A) $81.00 81.00美元 (B) $81.40 81.40美元 (C) $82.00 82.00美元 (D) $82.08 82.08美元 (E) $82.40 82.40美元

Correct Answer: A

The 20% discount lowers the price to $80. Then the $5 coupon reduces the price to $75. The sales tax is 8% of $75, or 0.08 × 75 = 6. Thus, the final cost is 75 + 6 = 81.00.

20%折扣后价格降至80美元。然后5美元优惠券减至75美元。销售税是75美元的8%，即 $0.08 \times 75 = 6$。因此，最终费用是 $75 + 6 = 81.00$ 美元。

Q10

For how many positive integer values of N (N > 0) is the expression $\frac{36}{N+2}$ an integer?

对于多少个正整数 N（N > 0），表达式 $\frac{36}{N+2}$ 是整数？

(A) 7 7 (B) 8 8 (C) 9 9 (D) 10 10 (E) 12 12

Correct Answer: A

N + 2 must be a positive integer divisor of 36 greater than 2. The positive divisors of 36 are 1,2,3,4,6,9,12,18,36. Excluding 1 and 2, there are 7 such divisors.

$N + 2$ 必须是大于2的36的正整数除数。36的正除数是1、2、3、4、6、9、12、18、36。排除1和2后，有7个这样的除数。

Q11

Last summer 100 students attended basketball camp. Of those attending, 52 were boys and 48 were girls. Also, 40 students were from Jones Middle School and 60 were from Clay Middle School. Twenty of the girls were from Jones Middle School. How many of the boys were from Clay Middle School?

去年夏天，有100名学生参加篮球夏令营。其中52名男孩，48名女孩。还有40名学生来自琼斯中学，60名来自克雷中学。20名女孩来自琼斯中学。克雷中学的男孩有多少名？

(A) 20 20 (B) 32 32 (C) 40 40 (D) 48 48 (E) 52 52

Correct Answer: B

Total girls 48, Jones girls 20, so Clay girls 28. Clay total 60, so Clay boys 60-28=32.

女孩总数48名，琼斯中学女孩20名，所以克雷中学女孩28名。克雷中学总数60名，所以克雷中学男孩60-28=32名。

Q12

Each of the three large squares shown below is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare?

下面的三个大正方形大小相同。穿过正方形边的线段都在边的中点相交。这些正方形的阴影面积如何比较？

(A) The shaded area in all three are equal. 三个阴影面积相等。 (B) Only the shaded areas of I and II are equal. 仅I和II阴影面积相等。 (C) Only the shaded areas of I and III are equal. 仅I和III阴影面积相等。 (D) Only the shaded areas of II and III are equal. 仅II和III阴影面积相等。 (E) The shaded areas of I, II and III are all different. I、II、III阴影面积均不同。

Correct Answer: A

The shaded area of square II is 1/4 of the total. In I, two triangles each 1/8 total, sum 1/4. In III, four triangles each 1/16, sum 1/4. All equal.

正方形II的阴影面积是总面积的1/4。I中两个三角形各占总面积的1/8，相加为1/4。III中四个三角形各占1/16，相加为1/4。全都相等。

Q13

The number halfway between $\frac{1}{6}$ and $\frac{1}{4}$ is

$\frac{1}{6}$ 和 $\frac{1}{4}$ 的中间数是

(A) $\frac{1}{10}$ $\frac{1}{10}$ (B) $\frac{1}{5}$ $\frac{1}{5}$ (C) $\frac{5}{24}$ $\frac{5}{24}$ (D) $\frac{7}{24}$ $\frac{7}{24}$ (E) $\frac{5}{12}$ $\frac{5}{12}$

Correct Answer: C

$\frac{1}{6} = \frac{4}{24}$, $\frac{1}{4} = \frac{6}{24}$，中间为 $\frac{5}{24}$。

$\frac{1}{6} = \frac{4}{24}$，$\frac{1}{4} = \frac{6}{24}$，中间为 $\frac{5}{24}$。

Q14

Two children at a time can play pairball. For 90 minutes, with only two children playing at one time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is

每次两个孩子玩配对球。90分钟内，每次只有两个孩子玩，五个孩子轮流玩，每人玩的时间相同。每个孩子玩的分钟数是

(A) 9 9 (B) 10 10 (C) 18 18 (D) 20 20 (E) 36 36

Correct Answer: E

Total playing time 2×90=180 minutes. Divided by 5 children: 180/5=36 minutes each.

总玩耍时间2×90=180分钟。5个孩子平分：180/5=36分钟每个。

Q15

If this path is to continue in the same pattern: then which sequence of arrows goes from point 425 to point 427?

如果这条路径继续相同的模式：那么从点425到点427的箭头序列是哪一个？

(A)

(B)

(C)

(D)

(E)

Correct Answer: A

The path repeats every four numbers. 425 mod 4 =1, 427 mod 4=3, same as from 1 to 3.

路径每四个数字重复一次。425 mod 4 =1，427 mod 4=3，与从1到3相同。

Q16

The perimeter of one square is 3 times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?

一个正方形的周长是另一个正方形周长的3倍。那么较大正方形的面积是较小正方形面积的几倍？

(A) 2 2 (B) 3 3 (C) 4 4 (D) 6 6 (E) 9 9

Correct Answer: E

Side ratio 3:1, area ratio 3²=9.

边长比3:1，面积比3²=9。

Q17

Pauline Bunyan can shovel snow at the rate of 20 cubic yards for the first hour, 19 cubic yards for the second, 18 for the third, etc., always shoveling one cubic yard less per hour than the previous hour. If her driveway is 4 yards wide, 10 yards long, and covered with snow 3 yards deep, then the number of hours it will take her to shovel it clean is closest to

Pauline Bunyan 第一小时铲雪20立方码，第二小时19立方码，第三小时18立方码，依此类推，每小时比前一小时少铲1立方码。如果她的车道宽4码，长10码，雪深3码，她铲干净需要的时间最接近

(A) 4 4 (B) 5 5 (C) 6 6 (D) 7 7 (E) 12 12

Correct Answer: D

Volume 4×10×3=120 cu yd. Sum first 7 hours: 20+19+...+14=119, closest to 120.

体积4×10×3=120立方码。前7小时总和：20+19+...+14=119，最接近120。

Q18

Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike's trip?

Mike离开家，慢慢向东穿过城市交通。到达高速公路后，他更快地向东开到购物中心，然后停下。他在商场购物1小时。Mike返回家时走同一条路，先快速向西沿高速公路行驶，然后慢慢穿过城市交通。每个图表纵轴表示离家的距离，横轴表示离开家后的经过时间。哪张图是Mike旅程的最佳表示？

(A)

(B)

(C)

(D)

(E)

Correct Answer: B

Graph (B) shows slow then fast outbound, stop, fast then slow return.

图(B)显示外出慢然后快，停顿，返回快然后慢。

Q19

Around the outsides of a 4 by 4 square, construct four semi-circles (as shown in the figure) with the four sides of the square as their diameters. Another square, ABCD, has its sides parallel to the corresponding sides of the original square, and each side of ABCD is tangent to one of the semicircle. The area of the square ABCD is

在一个4×4的正方形外侧，用正方形的四条边作为直径，构造四个半圆（如图所示）。另一个正方形ABCD，其边与原正方形对应边平行，ABCD的每条边与一个半圆相切。ABCD正方形的面积是

(A) 16 16 (B) 32 32 (C) 36 36 (D) 48 48 (E) 64 64

Correct Answer: E

Semicircle radius 2. ABCD side =4 +2×2=8, area 64.

半圆半径2。ABCD边长=4+2×2=8，面积64。

Q20

Let W, X, Y and Z be four different digits selected from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. If the sum $\frac{W}{X} + \frac{Y}{Z}$ is to be as small as possible, then $\frac{W}{X} + \frac{Y}{Z}$ must equal

让W、X、Y和Z是从集合{1, 2, 3, 4, 5, 6, 7, 8, 9}中选的四个不同数字。如果要使 $\frac{W}{X} + \frac{Y}{Z}$ 尽可能小，那么 $\frac{W}{X} + \frac{Y}{Z}$ 必须等于

(A) $\frac{2}{17}$ $\frac{2}{17}$ (B) $\frac{3}{17}$ $\frac{3}{17}$ (C) $\frac{17}{72}$ $\frac{17}{72}$ (D) $\frac{25}{72}$ $\frac{25}{72}$ (E) $\frac{13}{36}$ $\frac{13}{36}$

Correct Answer: D

$\frac{1}{8} + \frac{2}{9} = \frac{25}{72}$, smaller than $\frac{1}{9} + \frac{2}{8} = \frac{26}{72}$.

$\frac{1}{8} + \frac{2}{9} = \frac{25}{72}$，小于 $\frac{1}{9} + \frac{2}{8} = \frac{26}{72}$。

Q21

A gumball machine contains 9 red, 7 white, and 8 blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is

一个泡泡糖机里有9个红色的、7个白色的和8个蓝色的泡泡糖。一个人必须买至少多少个泡泡糖才能确保得到四个相同颜色的泡泡糖？

(A) 8 8 (B) 9 9 (C) 10 10 (D) 12 12 (E) 18 18

Correct Answer: C

Worst case: 3 each color=9, tenth forces 4 of one.

最坏情况：每种颜色各3个=9个，第十个将迫使某种颜色达到4个。

Q22

The two wheels shown at the right are spun and the two resulting numbers are added. The probability that the sum of the two numbers is even is

右侧所示的两个轮子被转动，得到两个数字相加。两个数字之和为偶数的概率是

(A) $\frac{1}{6}$ $\frac{1}{6}$ (B) $\frac{1}{4}$ $\frac{1}{4}$ (C) $\frac{1}{3}$ $\frac{1}{3}$ (D) $\frac{5}{12}$ $\frac{5}{12}$ (E) $\frac{4}{9}$ $\frac{4}{9}$

Correct Answer: D

Even sum: both even or both odd. P(odd1)*P(odd2) + P(even1)*P(even2) = (3/4)(1/3) + (1/4)(2/3) = 1/4 + 1/6 = 5/12.

偶数和：两个都是偶数或两个都是奇数。P(奇1)*P(奇2) + P(偶1)*P(偶2) = (3/4)(1/3) + (1/4)(2/3) = 1/4 + 1/6 = 5/12。

Q23

If X, Y and Z are different digits, then the largest possible 3-digit sum for XXX + YX + X has the form

如果X、Y和Z是不同的数字，则XXX + YX + X的最大可能3位和的形式是

(A) XXY XXY (B) XYZ XYZ (C) YYX YYX (D) YYZ YYZ (E) ZZY ZZY

Correct Answer: D

X max 8 (9 would carry over 4 digits). Y=9, Z=4: 888+98+8=994=YYZ.

X最大8（9会进位成4位）。Y=9, Z=4: 888+98+8=994=YYZ。

Q24

A 2 by 2 square is divided into four 1 by 1 squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.

一个2×2的正方形被分成四个1×1的小正方形。每个小正方形要么涂绿色要么涂红色。有多少种不同的涂色方式，使得没有绿色小方形与任何红色小方形共享其顶部或右侧？绿色小方形可以有0个到4个。

(A) 4 4 (B) 6 6 (C) 7 7 (D) 8 8 (E) 16 16

Correct Answer: B

Green squares force all above and right to be green. Possible configurations: all red; bottom-left green; bottom row green; left column green; bottom-left 2x2 green; all green. 6 ways.

绿色方形强制其上方和右侧全部为绿色。可能配置：全红；左下绿色；底行绿色；左列绿色；左下2×2绿色；全绿。6种方式。

Q25

Find the sum of the digits in the answer to $\frac{9999...99}{94 \text{ nines}} \times \frac{4444...44}{94 \text{ fours}}$ where a string of 94 nines is multiplied by a string of 94 fours.

求 $\frac{9999...99}{94 \text{ nines}} \times \frac{4444...44}{94 \text{ fours}}$ 的答案中各位数字之和，其中94个9的串乘以94个4的串。

(A) 846 846 (B) 855 855 (C) 945 945 (D) 954 954 (E) 1072 1072

Correct Answer: A

Product is 444...4 (93 fours) 3 555…5 (93 fives) 6. Digit sum 93×4 +3 +93×5 +6 =93×9 +9=846.

积是444...4（93个4）3 555…5（93个5）6。各位数字和93×4 +3 +93×5 +6 =93×9 +9=846。

AMC8 1994