2013 AMC 8 真题与在线限时测验

计时

40:00

1.

丹妮卡想把她的模型车排成若干行，每行正好 $6$ 辆。她现在有 $23$ 辆模型车。为了能用这种方式排好所有车，她最少还需要买多少辆？

Danica wants to arrange her model cars in rows with exactly $6$ cars in each row. She now has $23$ model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?

$1$

$2$

$3$

$4$

$5$

答案：A

知识点：整除性估算

难度评级：370

小提示：

找出大于 $23$ 的下一个 $6$ 的倍数

Find the next multiple of $6$ after $23$ .

大提示：

只需要再买一辆车就能达到那个倍数

Only one more car is needed to reach that multiple.

视频讲解：

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文字解答：

为了让丹妮卡把模型车排成每行正好六辆，她拥有的模型车数量必须是 $6$ 的倍数。大于 $23$ 的最小这样的数是 $24$ ，所以她必须再买 $1$ 辆车。

所以正确答案是 A。

In order for Danica to arrange her model cars in rows of exactly six, the number of model cars she has must be a multiple of six. The smallest multiple of $6$ greater than $23$ is $24,$ so she must buy $1$ more car to attain this amount.

Thus, A is the correct answer.

2.

鱼市的招牌写着：“仅限今天， $50 \%$ 折扣：半磅装每包只要 $\$3$ 。”一整磅鱼的正常价格是多少美元？

A sign at the fish market says, “ $50 \%$ off, today only: half-pound packages for just $\$3$ per package.” What is the regular price for a full pound of fish, in dollars?

$6$

$9$

$10$

$12$

$15$

答案：D

知识点：百分数钱币

难度评级：450

小提示：

标出的价格是半磅打折后的价格

The quoted price is for a half pound after the discount.

大提示：

先撤销 $50\%$ 折扣，再加倍得到一磅价格

First undo the $50\%$ discount, then double to get one pound.

视频讲解：

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文字解答：

促销时半磅价格为 $\$3$ ，所以一磅促销价为 $2 \cdot \$3 = \$6$ 。

促销价是正常价格的 $50\%$ ，所以正常价格是促销价的两倍。因此一磅鱼的正常价格是 $2 \cdot \$6 = \$12$ 。

所以正确答案是 D。

During the sale, the price for half a pound is $\$3,$ and therefore, the price for a full pound is $2 \cdot \$3 = \$6.$

The sale price is $50\%$ off the regular price, and so the regular price is twice the sale price. Therefore, the regular price for a full pound of fish is $2 \cdot \$6 = \$12.$

Thus, D is the correct answer.

3.

下列表达式的值是多少？ $\begin{aligned} &4 \cdot (-1 + 2 - 3 + 4 - 5 + 6 \\ &\qquad {}- 7 + \cdots + 1000) \end{aligned}$

What is the value of $\begin{aligned} &4 \cdot (-1 + 2 - 3 + 4 - 5 + 6 \\ &\qquad {}- 7 + \cdots + 1000)? \end{aligned}$

$-10$

$0$

$1$

$500$

$2000$

答案：E

知识点：求和配对与分组

难度评级：770

小提示：

将括号内的项两两配对

Pair the terms inside the parentheses.

大提示：

每一对 $(-1+2),(-3+4),\ldots$ 的和相同

Each pair $(-1+2),(-3+4),\ldots$ has the same sum.

视频讲解：

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文字解答：

在括号内，将相邻项配对： $\begin{aligned} &(-1+2)+(-3+4) \\ &\quad {}+\cdots+(-999+1000) \end{aligned}\text{。}$

每一对的和都是 $1$ ，共有 $500$ 对，所以括号内的值是 $500$ 。再乘以 $4$ ，得到 $2000$ 。

所以正确答案是 E。

Within the parentheses, pair consecutive terms: $\begin{aligned} &(-1+2)+(-3+4) \\ &\quad {}+\cdots+(-999+1000). \end{aligned}$

Each pair has sum $1$ , and there are $500$ pairs, so the value inside the parentheses is $500$ . Multiplying by $4$ gives $2000$ .

Thus, E is the correct answer.

4.

八个朋友在餐馆吃饭，并同意平均分摊账单。因为朱迪忘带钱，她的七个朋友每人额外支付 $\$2.50$ 来替她支付她应付的部分。总账单是多少？

Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $\$2.50$ to cover her portion of the total bill. What was the total bill?

$\$120$

$\$128$

$\$140$

$\$144$

$\$160$

答案：C

知识点：钱币

难度评级：770

小提示：

七个朋友合起来支付了朱迪的那一份

The seven friends together paid Judi’s share.

大提示：

朱迪的那一份等于每个朋友额外支付金额的 $7$ 倍

Judi’s share is $7$ times the extra amount each friend paid.

视频讲解：

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文字解答：

朱迪应付的那一份由大家额外支付的钱完全覆盖，所以她应付的金额为 $7 \cdot \$2.50 = \$17.50\text{。}$

因为每个人原本应付相同金额，所以总账单为 $8 \cdot \$17.50 = \$140\text{。}$

所以正确答案是 C。

Since Judi’s portion was fully covered by everyone’s contributions, the portion of the bill she was responsible for was equal to $7 \cdot \$2.50 = \$17.50.$

As everyone paid the same amount, we can conclude that the total bill is $8 \cdot \$17.50 = \$140.$

Thus, C is the correct answer.

5.

汉米上 $6$ 年级，体重 $106$ 磅。他的四胞胎妹妹还是小婴儿，体重分别为 $5$ 、 $5$ 、 $6$ 和 $8$ 磅。这五个孩子的平均体重和中位数哪个更大，大多少磅？

Hammie is in the $6$ th grade and weighs $106$ pounds. His quadruplet sisters are tiny babies and weigh $5,$ $5,$ $6,$ and $8$ pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?

中位数，大 $60$ 磅

median, by $60$

中位数，大 $20$ 磅

median, by $20$

平均数，大 $5$ 磅

average, by $5$

平均数，大 $15$ 磅

average, by $15$

平均数，大 $20$ 磅

average, by $20$

答案：E

知识点：平均数中位数（数据）

难度评级：770

小提示：

先把五个体重排序，再找中位数

Order the five weights before finding the median.

大提示：

将中位数与总体重除以 $5$ 比较

Compare the median with the total weight divided by $5$ .

视频讲解：

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文字解答：

体重的中位数就是排序后正中间的那个值，可以这样找出： $\cancel{ 5 },\cancel{ 5 },\underline{6},\cancel{8},\cancel{106}$ 因此中位数是 $6$ 磅。

平均数可以计算如下： $\begin{align*} &\dfrac{106 + 5 + 5 + 6 + 8}{5} \\&= \dfrac{130}{5}\\ &= 26\text{。} \end{align*}$ 这个值比中位数大，具体来说，平均数比中位数大 $20$ 磅。

所以正确答案是 E。

The median weight is the middle value, which we can find as: $\cancel{ 5 },\cancel{ 5 },\underline{6},\cancel{8},\cancel{106}$ Therefore, the median is $6$ pounds.

The average (mean) can be calculated to be: $\begin{align*} &\dfrac{106 + 5 + 5 + 6 + 8}{5} \\&= \dfrac{130}{5}\\ &= 26. \end{align*}$ This is greater than the median, specifically, the mean is $20$ pounds greater than the median.

Thus, E is the correct answer.

6.

下方每个框中的数是它上方一行中与它相接的两个框中数字的乘积。例如， $30 = 6\times5$ 。

顶行缺失的数是多少？

The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5.$

What is the missing number in the top row?

$2$

$3$

$4$

$5$

$6$

答案：C

知识点：逆推法

难度评级：820

小提示：

用乘积 $600$ 求第二行缺失的数

Use the product $600$ to find the missing number in the second row.

大提示：

再利用 $5$ 乘顶行缺失的数得到那个第二行数

Then use $5$ times the top missing number to get that second-row value.

视频讲解：

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文字解答：

第二行两个数的乘积是 $600$ ，所以第二行右侧缺失的数是 $600 \div 30 = 20$ 。

现在知道 $5$ 与第一行缺失数的乘积为 $20$ 。因此缺失数是 $20 \div 5 = 4$ 。

所以正确答案是 C。

The product of the numbers in the second row is $600,$ so the missing number in the middle right box is $600 \div 30 = 20.$

Now we know that the product of $5$ and the missing number in the first row is $20.$ Therefore, the missing number is $20 \div 5 = 4.$

Thus, C is the correct answer.

7.

特雷和妈妈在铁路道口停下来让火车通过。火车刚开始通过时，特雷在前 $10$ 秒数到 $6$ 节车厢。火车以恒定速度完全通过道口用了 $2$ 分 $45$ 秒。下列哪一个最可能是这列火车的车厢数？

Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted $6$ cars in the first $10$ seconds. It took the train $2$ minutes and $45$ seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?

$\ 60$

$\ 80$

$\ 100$

$\ 120$

$\ 140$

答案：C

知识点：比与比例估算

难度评级：900

小提示：

将总通过时间转换为秒

Convert the total crossing time to seconds.

大提示：

利用 $10$ 秒通过 $6$ 节车厢列出比例式

Use a proportion from $6$ cars in $10$ seconds.

视频讲解：

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文字解答：

$2$ 分 $45$ 秒等于 $2 \cdot 60 + 45 = 165$ 秒。因为 $10$ 秒通过 $6$ 节车厢，所以设 $165$ 秒通过 $x$ 节车厢，满足 $\dfrac{6}{10}=\dfrac{x}{165}\text{。}$

解得 $x=\dfrac{165\cdot6}{10}=99\text{。}$ 最接近的选项是 $100$ 。

所以正确答案是 C。

Converting $2$ minutes and $45$ seconds into seconds, we get $2 \cdot 60 + 45 = 165$ seconds. Since $6$ cars passed in $10$ seconds, at the same rate the number of cars $x$ passing in $165$ seconds satisfies $\dfrac{6}{10}=\dfrac{x}{165}.$

Solving gives $x=\dfrac{165\cdot6}{10}=99.$ The closest answer choice is $100$ .

Thus, C is the correct answer.

8.

一枚公平硬币被抛 $3$ 次。至少出现两个连续正面的概率是多少？

A fair coin is tossed $3$ times. What is the probability of at least two consecutive heads?

$\dfrac{1}{8}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

答案：C

知识点：基本概率系统列举

难度评级：960

小提示：

列出三次抛硬币的 $8$ 种可能结果

List the $8$ possible outcomes of three coin tosses.

大提示：

只统计含有 $HH$ 的结果

Count only outcomes containing $HH$ .

视频讲解：

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文字解答：

至少有两个连续正面的结果是 $HHT$ 、 $THH$ 和 $HHH$ 。三次公平抛硬币共有 $2^3=8$ 个等可能结果，所以概率为 $\dfrac{3}{8}$ 。

所以正确答案是 C。

The outcomes with at least two consecutive heads are $HHT$ , $THH$ , and $HHH$ . There are $2^3=8$ equally likely outcomes from three fair coin tosses, so the probability is $\dfrac{3}{8}$ .

Thus, C is the correct answer.

9.

绿巨人浩克每次跳跃的距离都能比上一次翻倍。如果第一次跳 $1$ 米，第二次跳 $2$ 米，第三次跳 $4$ 米，依此类推，那么他第一次能跳超过 $1$ 千米是在第几跳？

The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is $1$ meter, the second jump is $2$ meters, the third jump is $4$ meters, and so on, then on which jump will he first be able to jump more than $1$ kilometer?

第 $9$ 跳

$9$ th

第 $10$ 跳

$10$ th

第 $11$ 跳

$11$ th

第 $12$ 跳

$12$ th

第 $13$ 跳

$13$ th

答案：C

知识点：2的幂等比数列

难度评级：960

小提示：

第 $n$ 跳的长度是 $2^{n-1}$

The $n$ th jump has length $2^{n-1}$ .

大提示：

将 $2$ 的幂与 $1000$ 比较

Compare powers of $2$ with $1000$ .

视频讲解：

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文字解答：

第 $n$ 跳时，浩克跳 $2^{n-1}$ 米。需要找最小的 $n$ ，使 $2^{n-1}>1000$ 。

因为 $2^9=512$ ，而 $2^{10}=1024$ ，所以第一次超过 $1000$ 米时 $n-1=10$ ，即 $n=11$ 。

所以正确答案是 C。

On the $n$ th jump, the Hulk jumps $2^{n-1}$ meters. We need the smallest $n$ for which $2^{n-1}>1000$ .

Since $2^9=512$ and $2^{10}=1024$ , the first jump longer than $1000$ meters has $n-1=10$ , so $n=11$ .

Thus, C is the correct answer.

10.

$180$ 和 $594$ 的最小公倍数与 $180$ 和 $594$ 的最大公因数之比是多少？

What is the ratio of the least common multiple of $180$ and $594$ to the greatest common factor of $180$ and $594?$

$110$

$165$

$330$

$625$

$660$

答案：C

知识点：质因数分解最大公约数最小公倍数

难度评级：1240

小提示：

将两个数都作质因数分解

Prime-factorize both numbers.

大提示：

最小公倍数使用较大的质因数幂，最大公因数使用较小的质因数幂

The LCM uses the larger prime powers, while the GCF uses the smaller ones.

视频讲解：

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文字解答：

将两个数质因数分解： $180=2^2\cdot3^2\cdot5\text{，}$ $594=2\cdot3^3\cdot11\text{。}$

最大公因数取较小的幂，所以为 $2\cdot3^2=18$ 。最小公倍数取较大的幂，所以为 $2^2\cdot3^3\cdot5\cdot11=5940\text{。}$ 所求比值为 $\dfrac{5940}{18}=330$ 。

所以正确答案是 C。

Prime-factorize the two numbers: $180=2^2\cdot3^2\cdot5,$ $594=2\cdot3^3\cdot11.$

The greatest common factor uses the smaller powers, so it is $2\cdot3^2=18$ . The least common multiple uses the larger powers, so it is $2^2\cdot3^3\cdot5\cdot11=5940.$ Therefore the desired ratio is $\dfrac{5940}{18}=330$ .

Thus, C is the correct answer.

11.

特德的祖父本周有 $3$ 天使用了跑步机。他每天走或跑 $2$ 英里。

星期一他以每小时 $5$ 英里的速度慢跑。星期三他以每小时 $3$ 英里的速度步行，星期五以每小时 $4$ 英里的速度步行。

如果祖父一直以每小时 $4$ 英里的速度步行，他会少花一些跑步机时间。少多少分钟？

Ted’s grandfather used his treadmill on $3$ days this week. He went $2$ miles each day.

On Monday he jogged at a speed of $5$ miles per hour. He walked at the rate of $3$ miles per hour on Wednesday and at $4$ miles per hour on Friday.

If Grandfather had always walked at $4$ miles per hour, he would have spent less time on the treadmill. How many minutes less?

$1$

$2$

$3$

$4$

$5$

答案：D

知识点：路程、速度与时间单位换算

难度评级：1280

小提示：

时间等于距离除以速度

Time equals distance divided by rate.

大提示：

比较实际总时间与以每小时 $4$ 英里的速度走 $6$ 英里所需的时间

Compare the actual total minutes with $6$ miles at $4$ mph.

视频讲解：

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文字解答：

祖父星期一慢跑用时 $\frac{2}{5}$ 小时，星期三步行用时 $\frac{2}{3}$ 小时，星期五用时 $\frac{2}{4}=\frac{1}{2}$ 小时。换算为分钟，星期一为 $\dfrac{2}{5} \cdot 60 = 24$ 分钟，星期三为 $\dfrac{2}{3} \cdot 60 = 40$ 分钟，星期五为 $\dfrac{1}{2} \cdot 60 = 30$ 分钟。因此实际总时间为 $94$ 分钟。

如果他一直以每小时 $4$ 英里的速度走，每天用时 $\frac{2}{4}=\frac{1}{2}$ 小时，即 $\frac{2}{4} \cdot 60 = 30$ 分钟， $3$ 天共 $3 \cdot 30 = 90$ 分钟。

因此少用的时间是 $94-90=4$ 分钟。

所以正确答案是 D。

Grandfather spent $\frac{2}{5}$ hours jogging on Monday. He spent $\frac{2}{3}$ hours walking on Wednesday and $\frac{2}{4}=\frac{1}{2}$ hours on Friday. Converting these times to minutes, we get $\dfrac{2}{5} \cdot 60 = 24$ minutes for Monday, $\dfrac{2}{3} \cdot 60 = 40$ minutes for Wednesday, and $\dfrac{1}{2} \cdot 60 = 30$ minutes for Friday. Therefore, Grandfather totaled $94$ minutes of exercise throughout the week.

If he walked at a pace of $4$ miles per hour each day, he would have spent $\frac{2}{4}=\frac{1}{2}$ hours each day walking. This equals $\frac{2}{4} \cdot 60 = 30$ minutes every day. If he did this for $3$ days, he would have totaled $3 \cdot 30 = 90$ minutes of exercise.

Therefore, Grandfather would have spent $94-90=4$ fewer minutes on the treadmill.

Thus, D is the correct answer.

12.

在 $2013$ 年威尼贝戈县集市上，一个摊贩正在对凉鞋推出“集市特价”。如果你以正常价格 $\$50$ 买一双凉鞋，第二双打 $40 \%$ 折扣，第三双半价。

哈维尔利用这个“集市特价”买了三双凉鞋。与 $\$150$ 的正常总价相比，他节省了百分之多少？

At the $2013$ Winnebago County Fair a vendor is offering a “fair special” on sandals. If you buy one pair of sandals at the regular price of $\$50,$ you get a second pair at a $40 \%$ discount, and a third pair at half the regular price.

Javier took advantage of the “fair special” to buy three pairs of sandals. What percentage of the $\$150$ regular price did he save?

$25$

$30$

$33$

$40$

$45$

答案：B

知识点：百分数

难度评级：1020

小提示：

计算每双凉鞋实际支付的价格

Compute the price paid for each pair of sandals.

大提示：

将总节省金额与正常总价 $\$150$ 比较

Compare the total savings with the regular total of $\$150$ .

视频讲解：

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文字解答：

哈维尔第二双节省 $0.40 \cdot 50 = 20$ 美元。第三双节省 $\frac{50}{2} = 25$ 美元。因此他一共节省 $45$ 美元。

因为 $\frac{45}{150} = \frac{3}{10} = 0.30\text{，}$ 所以与正常价格相比，他节省了 $30 \%$ 。

所以正确答案是 B。

Javier saved $0.40 \cdot 50 = 20$ dollars on the second pair. He also saved $\frac{50}{2} = 25$ dollars on the third pair. This shows that he saved a total of $45$ dollars.

As $\frac{45}{150} = \frac{3}{10} = 0.30,$ we can conclude that Javier saved $30 \%$ compared to the regular price.

Thus, B is the correct answer.

13.

克拉拉计算总分时，不小心把其中一个分数的个位数字和十位数字写反了。她的错误总和与正确总和可能相差下列哪一个数？

When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?

$45$

$46$

$47$

$48$

$49$

答案：A

知识点：位值整除性

难度评级：1020

小提示：

交换十位和个位会使一个数改变 $9$ 的倍数

Reversing tens and units changes a number by a multiple of $9$ .

大提示：

检查哪个选项能被 $9$ 整除

Check which answer choice is divisible by $9$ .

视频讲解：

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文字解答：

如果这个分数的后两位是 $10a+b$ ，交换后得到 $10b+a$ 。差为 $\begin{aligned} &(10a+b)-(10b+a) \\ &= 9(a-b)\text{，} \end{aligned}$ 因此错误总和与正确总和之差必须是 $9$ 的倍数。

选项中只有 $45$ 能被 $9$ 整除。

所以正确答案是 A。

If the last two digits of the score are $10a+b$ , then reversing those digits gives $10b+a$ . The difference is $\begin{aligned} &(10a+b)-(10b+a) \\ &= 9(a-b), \end{aligned}$ so the incorrect sum must differ from the correct sum by a multiple of $9$ .

Among the answer choices, only $45$ is divisible by $9$ .

Thus, A is the correct answer.

14.

艾布手里有 $1$ 颗绿色和 $1$ 颗红色软糖。比娅手里有 $1$ 颗绿色、 $1$ 颗黄色和 $2$ 颗红色软糖。两人各随机拿出一颗软糖给对方看。颜色相同的概率是多少？

Abe holds $1$ green and $1$ red jelly bean in his hand. Bea holds $1$ green, $1$ yellow, and $2$ red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\frac{1}{2}$

$\dfrac{2}{3}$

答案：C

知识点：基本概率分类讨论

难度评级：1140

小提示：

共有 $2\cdot4$ 种等可能的软糖配对

There are $2\cdot4$ equally likely jelly-bean pairings.

大提示：

统计绿色匹配和两个红色匹配

Count the green match and the two red matches.

视频讲解：

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文字解答：

艾布和比娅选择软糖共有 $2\cdot4=8$ 种等可能结果。颜色相同可能是两人都选绿色，有 $1$ 种；也可能两人都选红色，有 $1\cdot2=2$ 种。

因此 $8$ 个结果中有 $3$ 个颜色相同，概率为 $\dfrac{3}{8}$ 。

所以正确答案是 C。

There are $2\cdot4=8$ equally likely ways for Abe and Bea to choose their jelly beans. The colors match if they both choose green, which can happen in $1$ way, or if they both choose red, which can happen in $1\cdot2=2$ ways.

Thus $3$ of the $8$ outcomes match, for probability $\dfrac{3}{8}$ .

Thus, C is the correct answer.

15.

如果 $3^p + 3^4 = 90$ 、 $2^r + 44 = 76$ ，且 $5^3 + 6^s = 1421$ ，那么 $p$ 、 $r$ 、 $s$ 的乘积是多少？

If $3^p + 3^4 = 90$ , $2^r + 44 = 76$ , and $5^3 + 6^s = 1421$ , what is the product of $p$ , $r$ , and $s$ ?

$27$

$40$

$50$

$70$

$90$

答案：B

知识点：指数

难度评级：1070

小提示：

从每个方程中减去已知的数

Subtract the known number from each equation.

大提示：

将 $3^p$ 、 $2^r$ 和 $6^s$ 识别为熟悉的幂

Identify $3^p$ , $2^r$ , and $6^s$ as familiar powers.

视频讲解：

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文字解答：

从每个方程中减去已知项： $3^p=90-81=9\text{，}$ $2^r=76-44=32\text{，}$ $6^s=1421-125=1296\text{。}$

因此 $p=2$ 、 $r=5$ 、 $s=4$ 。它们的乘积是 $2\cdot5\cdot4=40$ 。

所以正确答案是 B。

Subtract the known number from each equation: $3^p=90-81=9,$ $2^r=76-44=32,$ $6^s=1421-125=1296.$

Therefore $p=2$ , $r=5$ , and $s=4$ . Their product is $2\cdot5\cdot4=40$ .

Thus, B is the correct answer.

16.

斐波那契中学的一些学生参加社区服务项目。 $8$ 年级学生与 $6$ 年级学生的比为 $5:3$ ， $8$ 年级学生与 $7$ 年级学生的比为 $8:5$ 。参加该项目的学生最少可能有多少人？

A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8$ th-graders to $6$ th-graders is $5:3,$ and the ratio of $8$ th-graders to $7$ th-graders is $8:5.$ What is the smallest number of students that could be participating in the project?

$16$

$40$

$55$

$79$

$89$

答案：E

知识点：比与比例最小公倍数

难度评级：1370

小提示：

八年级学生人数必须同时适合两个比例

The number of eighth-graders must fit both ratios.

大提示：

使用 $5$ 和 $8$ 的最小公倍数

Use the least common multiple of $5$ and $8$ .

视频讲解：

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文字解答：

$8$ 年级学生人数必须同时是 $5$ 和 $8$ 的倍数，所以最少为 $40$ 。这时 $6$ 年级人数为 $40 \cdot \dfrac{3}{5} = 24$ ， $7$ 年级人数为 $40 \cdot \dfrac{5}{8} = 25$ 。

因此学生总数为 $40 + 24 + 25 = 89$ 。

所以正确答案是 E。

The number of $8$ th-graders must be a multiple of $5$ and $8,$ which means that it is at least $40.$ Using this number, we get that the number of $6$ th-graders is $40 \cdot \dfrac{3}{5} = 24.$ Similarly, the number of $7$ th-graders is $40 \cdot \dfrac{5}{8} = 25.$

The total number of students is therefore $40 + 24 + 25 = 89.$

Thus, E is the correct answer.

17.

六个连续正整数的和是 $2013$ 。这六个整数中最大的是多少？

The sum of six consecutive positive integers is $2013.$ What is the largest of these six integers?

$335$

$338$

$340$

$345$

$350$

答案：B

知识点：等差数列一次方程

难度评级：1070

小提示：

六个连续整数的平均数正好在中间两个数的正中间

The average of six consecutive integers is halfway between the middle two.

大提示：

最大数比平均数大 $2.5$

The largest is $2.5$ more than the average.

视频讲解：

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文字解答：

设这六个连续整数为 $x$ 、 $x+1$ 、 $x+2$ 、 $x+3$ 、 $x+4$ 、 $x+5$ 。它们的和为 $6x+15=2013\text{。}$

因此 $6x=1998$ ，所以 $x=333$ 。最大整数为 $x+5=338$ 。

所以正确答案是 B。

Let the six consecutive integers be $x,$ $x+1,$ $x+2,$ $x+3,$ $x+4,$ $x+5$ . Their sum is $6x+15=2013.$

Thus $6x=1998$ , so $x=333$ . The largest integer is $x+5=338$ .

Thus, B is the correct answer.

18.

伊莎贝拉用一英尺见方的立方块搭建一个长方体堡垒，长 $12$ 英尺、宽 $10$ 英尺、高 $5$ 英尺。地板和四面墙都厚一英尺。这个堡垒包含多少块立方块？

Isabella uses one-foot cubical blocks to build a rectangular fort that is $12$ feet long, $10$ feet wide, and $5$ feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?

$204$

$280$

$320$

$340$

$600$

答案：B

知识点：体积长方体

难度评级：1480

小提示：

从外部长方体体积中减去内部空心体积

Subtract the empty inside volume from the outside rectangular prism.

大提示：

内部尺寸在长和宽上各减少二英尺，在高上减少一英尺

The inside dimensions are reduced by two feet in length/width and one foot in height.

视频讲解：

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文字解答：

外表面确定的长方体体积为 $12\cdot10\cdot5=600$ 立方英尺。内部空心部分在长上少 $2$ 英尺、宽上少 $2$ 英尺、高上少 $1$ 英尺，所以尺寸为 $10\times8\times4$ 。

空心体积为 $10\cdot8\cdot4=320$ 立方英尺，所以堡垒含有 $600-320=280$ 个一英尺见方的立方块。

所以正确答案是 B。

The rectangular prism determined by the outside faces has volume $12\cdot10\cdot5=600$ cubic feet. The empty inside is smaller by $2$ feet in length, $2$ feet in width, and $1$ foot in height, so its dimensions are $10\times8\times4$ .

The empty volume is $10\cdot8\cdot4=320$ cubic feet, so the fort contains $600-320=280$ one-foot cubical blocks.

Thus, B is the correct answer.

19.

布里奇特、卡西和汉娜正在讨论上次数学测试的结果。汉娜给布里奇特和卡西看了她的试卷，但布里奇特和卡西没有把自己的试卷给任何人看。卡西说：“我不是班里的最低分。”布里奇特补充说：“我不是最高分。”三位女生从高到低的排名是什么？

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don’t show theirs to anyone. Cassie says, “I didn’t get the lowest score in our class,” and Bridget adds, “I didn’t get the highest score.” What is the ranking of the three girls from highest to lowest?

汉娜、卡西、布里奇特

Hannah, Cassie, Bridget

汉娜、布里奇特、卡西

Hannah, Bridget, Cassie

卡西、布里奇特、汉娜

Cassie, Bridget, Hannah

卡西、汉娜、布里奇特

Cassie, Hannah, Bridget

布里奇特、卡西、汉娜

Bridget, Cassie, Hannah

答案：D

知识点：逻辑推理

难度评级：1030

小提示：

卡西能和汉娜比较，因为她看到了汉娜的试卷

Cassie can compare herself with Hannah because she sees Hannah’s test.

大提示：

卡西高于汉娜，而布里奇特低于汉娜

Cassie is above Hannah, while Bridget is below Hannah.

视频讲解：

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文字解答：

因为卡西能推断自己不是最低分，她一定比汉娜分数高。同样，布里奇特的话说明她一定比汉娜分数低。

因此从高到低的排名是卡西、汉娜、布里奇特。

所以正确答案是 D。

Since Cassie deduced she didn’t get the lowest score, she must have got a higher score than Hannah. Similarly, Bridget’s statement reveals that she must have got a lower score than Hannah.

Therefore, the ranking from highest to lowest is Cassie, Hannah, and then Bridget.

Thus, D is the correct answer.

20.

一个 $1 \times 2$ 长方形内接于一个半圆，较长的一边在直径上。这个半圆的面积是多少？

A $1 \times 2$ rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?

$\dfrac{\pi}{2}$

$\dfrac{2\pi}{3}$

$\pi$

$\dfrac{4\pi}{3}$

$\dfrac{5\pi}{3}$

答案：C

知识点：勾股定理圆面积

难度评级：1370

小提示：

半径从直径中点连到长方形的上方顶点

The radius goes from the midpoint of the diameter to a top corner of the rectangle.

大提示：

这条半径是一个 $1$ 乘 $1$ 直角三角形的斜边

That radius is the hypotenuse of a $1$ by $1$ right triangle.

视频讲解：

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文字解答：

由勾股定理，半圆半径为 $\sqrt{2}$ 。因此半圆面积为 $\dfrac{1}{2} \cdot \pi \cdot \sqrt{2}^2 = \pi\text{。}$ 所以正确答案是 C。

Using the Pythagorean theorem, we get that the radius of the semicircle is $\sqrt{2}.$ This means that the area of the semicircle is $\dfrac{1}{2} \cdot \pi \cdot \sqrt{2}^2 = \pi.$ Thus, C is the correct answer.

21.

萨曼莎住在城市公园西南角以西 $2$ 个街区、以南 $1$ 个街区的地方。

她的学校在城市公园东北角以东 $2$ 个街区、以北 $2$ 个街区的地方。上学日她先沿街骑车到城市公园的西南角，再走公园里的对角线小路到东北角，然后沿街骑车到学校。

如果她的路线尽可能短，她有多少种不同路线可走？

Samantha lives $2$ blocks west and $1$ block south of the southwest corner of City Park.

Her school is $2$ blocks east and $2$ blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school.

If her route is as short as possible, how many different routes can she take?

$3$

$6$

$9$

$12$

$18$

答案：E

知识点：格路乘法原理

难度评级：1540

小提示：

分别统计进公园前和出公园后的最短街道路线路数

Count shortest street routes before and after the park separately.

大提示：

进公园前有 $3$ 种，出公园后有 $6$ 种

There are $3$ ways before the park and $6$ ways after it.

视频讲解：

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文字解答：

为了最快到达公园西南角，萨曼莎必须向东走 $2$ 个街区、向北走 $1$ 个街区，这些步骤可排列成 $3$ 种方式。经过公园对角线小路后，她必须再向东走 $2$ 个街区、向北走 $2$ 个街区到学校，这些步骤可排列成 $\binom{4}{2}=6$ 种方式。

公园前后的选择相互独立，所以最短路线总数为 $3\cdot6=18$ 。

所以正确答案是 E。

To reach the southwest corner of the park as quickly as possible, Samantha must go $2$ blocks east and $1$ block north, which can be arranged in $3$ ways. After the diagonal path through the park, she must go $2$ blocks east and $2$ blocks north to reach school, which can be arranged in $\binom{4}{2}=6$ ways.

The choices before and after the park are independent, so the total number of shortest routes is $3\cdot6=18$ .

Thus, E is the correct answer.

22.

用牙签搭成一个长 $60$ 根牙签、高 $32$ 根牙签的网格。一共用了多少根牙签？

Toothpicks are used to make a grid that is $60$ toothpicks long and $32$ toothpicks high. How many toothpicks are used altogether?

$1920$

$1952$

$1980$

$2013$

$3932$

答案：E

知识点：基本计数

难度评级：1280

小提示：

分别统计竖直牙签和水平牙签

Count vertical and horizontal toothpicks separately.

大提示：

一个 $60$ 乘 $32$ 的网格有 $61$ 条竖直线和 $33$ 条水平线

A $60$ by $32$ grid has $61$ vertical lines and $33$ horizontal lines.

视频讲解：

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文字解答：

一个 $60$ 乘 $32$ 网格有 $61$ 条竖直网格线，每条由 $32$ 根牙签组成；还有 $33$ 条水平网格线，每条由 $60$ 根牙签组成。

因此牙签总数为 $\begin{aligned} 61\cdot32+33\cdot60 &= 1952+1980 \\ &= 3932 \end{aligned}\text{。}$

所以正确答案是 E。

A $60$ -by- $32$ grid has $61$ vertical grid lines, each made of $32$ toothpicks, and $33$ horizontal grid lines, each made of $60$ toothpicks.

Therefore the total number of toothpicks is $\begin{aligned} 61\cdot32+33\cdot60 &= 1952+1980 \\ &= 3932. \end{aligned}$

Thus, E is the correct answer.

23.

在 $\triangle ABC$ 中， $\angle ABC$ 是直角。如图， $\triangle ABC$ 的三边分别作为半圆的直径。以 $\overline{AB}$ 为直径的半圆面积为 $8\pi$ ，以 $\overline{AC}$ 为直径的半圆的弧长为 $8.5\pi$ 。以 $\overline{BC}$ 为直径的半圆半径是多少？

In $\triangle ABC$ , $\angle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi,$ and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi.$ What is the radius of the semicircle on $\overline{BC}?$

$7$

$7.5$

$8$

$8.5$

$9$

答案：B

知识点：勾股定理圆面积弧

难度评级：1720

小提示：

用 $AB$ 上的半圆求出 $AB$

Use the semicircle on $AB$ to find $AB$ .

大提示：

$AC$ 上的弧长给出半径，从而给出直径 $AC$

The arc length on $AC$ gives the radius, hence diameter $AC$ .

视频讲解：

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文字解答：

$\overline{AB}$ 上的半圆面积为 $8\pi$ ，所以对应整圆面积为 $16\pi$ 。其半径为 $4$ ，因此 $AB=8$ 。

半径为 $r$ 的半圆弧长为 $\pi r$ 。因为 $\overline{AC}$ 上的弧长为 $8.5\pi$ ，其半径为 $8.5$ ，所以 $AC=17$ 。

在直角三角形 $ABC$ 中，由勾股定理， $BC=\sqrt{17^2-8^2}=\sqrt{225}=15\text{。}$ 因此 $\overline{BC}$ 上半圆的半径为 $\frac{15}{2}=7.5$ 。

所以正确答案是 B。

The semicircle on $\overline{AB}$ has area $8\pi$ , so its full circle would have area $16\pi$ . Its radius is therefore $4$ , and $AB=8$ .

A semicircle of radius $r$ has arc length $\pi r$ . Since the arc on $\overline{AC}$ has length $8.5\pi$ , its radius is $8.5$ , so $AC=17$ .

Using the Pythagorean theorem in right triangle $ABC$ , $BC=\sqrt{17^2-8^2}=\sqrt{225}=15.$ The radius of the semicircle on $\overline{BC}$ is $\frac{15}{2}=7.5$ .

Thus, B is the correct answer.

24.

正方形 $ABCD$ 、 $EFGH$ 和 $GHIJ$ 面积相等。点 $C$ 和 $D$ 分别是边 $IH$ 和 $HE$ 的中点。阴影五边形 $AJICB$ 的面积与三个正方形面积之和的比是多少？

Squares $ABCD,$ $EFGH,$ and $GHIJ$ are equal in area. Points $C$ and $D$ are the midpoints of sides $IH$ and $HE,$ respectively. What is the ratio of the area of the shaded pentagon $AJICB$ to the sum of the areas of the three squares?

$\dfrac{1}{4}$

$\dfrac{7}{24}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{5}{12}$

答案：C

知识点：面积分割三角形面积

难度评级：1790

小提示：

设每个正方形边长为 $1$

Let each square have side length $1$ .

大提示：

从总面积 $3$ 中减去非阴影的长方形和三角形

Subtract the unshaded rectangle and triangle from the total area $3$ .

视频讲解：

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文字解答：

设每个正方形边长为 $1$ ，则三个正方形总面积为 $3$ 。从总面积中减去对角线 $AJ$ 左侧的非阴影区域会更容易。

根据解答图，长方形 $EDKF$ 面积为 $1\cdot\dfrac12=\dfrac12$ ，三角形 $AKJ$ 面积为 $\dfrac12\cdot\dfrac32\cdot2=\dfrac32\text{。}$ 因此非阴影面积为 $\dfrac12+\dfrac32=2$ ，阴影五边形面积为 $3-2=1$ 。

所求比值为 $\dfrac{1}{3}$ 。

所以正确答案是 C。

Let each square have side length $1$ , so the total area of the three squares is $3$ . It is easier to subtract the unshaded region to the left of diagonal $AJ$ from this total.

Using the solution diagram, rectangle $EDKF$ has area $1\cdot\dfrac12=\dfrac12$ , and triangle $AKJ$ has area $\dfrac12\cdot\dfrac32\cdot2=\dfrac32.$ Thus the unshaded area is $\dfrac12+\dfrac32=2$ , so the shaded pentagon has area $3-2=1$ .

The requested ratio is $\dfrac{1}{3}$ .

Thus, C is the correct answer.

25.

一个直径为 $4$ 英寸的球从点 $A$ 出发沿图示轨道滚动。轨道由 $3$ 段半圆弧组成，半径分别为 $R_1 = 100$ 英寸、 $R_2 = 60$ 英寸、 $R_3 = 80$ 英寸。球始终与轨道接触且不打滑。从 $A$ 到 $B$ 的过程中，球心行进的距离是多少英寸？

A ball with diameter $4$ inches starts at point $A$ to roll along the track shown. The track is comprised of $3$ semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance in inches the center of the ball travels over the course from $A$ to $B$ ?

$238 \pi$

$240 \pi$

$260 \pi$

$280 \pi$

$500 \pi$

答案：A

知识点：弧圆周长

难度评级：1930

小提示：

球心沿着偏移后的半圆弧运动

The ball center follows offset semicircular arcs.

大提示：

对外侧圆弧减去球半径，对内侧圆弧加上球半径

Subtract the ball radius from outside arcs and add it to the inside arc.

视频讲解：

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文字解答：

球的半径为 $2$ 。滚动时，球心沿着与轨道平行的半圆弧运动。第一段和第三段圆弧中球心路径在轨道圆弧内侧，所以半径分别为 $100-2=98$ 和 $80-2=78$ 。中间圆弧中球心路径在轨道圆弧外侧，所以半径为 $60+2=62$ 。

球心行进的总距离是三段半圆弧长度之和： $(98+62+78)\pi=238\pi\text{。}$

所以正确答案是 A。

The ball has radius $2$ . As the ball rolls, its center follows semicircular arcs parallel to the track. On the first and third arcs the center path is inside the track arcs, so those radii are $100-2=98$ and $80-2=78$ . On the middle arc the center path is outside the track arc, so that radius is $60+2=62$ .

The total distance traveled by the center is the sum of the three semicircle lengths: $(98+62+78)\pi=238\pi.$

Thus, A is the correct answer.