三、读一读，填一填。
1. What kind of sweet is always LATE for class? It's .
2. There are two brothers. One is on one side, and the other is on the other side. When people say something, they can hear. But they can't see each other. They're .
3. Mike is standing in Group One. There are seven students in front of him and five students behind him. How many students are there in Group One? There are students.
4. Mary is twelve years old. Her father is four times(四倍) as old as her. Her grandfather is twice(两倍) as old as her father. How old is her grandfather? He is .
5. Ann and Bill have five sweets. Bill and Clare have four sweets. Clare and Ann have three sweets. Ann has sweet(s). Bill has sweet(s). Clare has sweet(s). Ann, Bill and Clare have sweet(s).
6. The Chinese sentence “忽如一夜春风来，千树万树梨花开。” describes scenery in .
(write down the season in English)

1. chocolate 2. ears 3. thirteen/13
4. ninety-six/96 5. two/2; three/3; one/1; six/6
6. winter

【分析】
本题为综合类基础题，需分别结合英语谜语、数学计算、古诗理解的思路解题：1. 英语谜语需结合甜食类词汇的逻辑/谐音特征推导；2. 身体部位谜语需匹配“两侧、能听不能见”的描述；3. 排队计数需包含自身，总人数=前人数+后人数+1；4. 年龄倍数计算需依次推导父亲、祖父的年龄；5. 三元一次方程应用题通过两两之和求各量；6. 古诗理解需明确诗句描写的季节。
【解析】
1. 谜语逻辑：chocolate（巧克力）是甜食，符合谜语要求，故填chocolate；
2. 耳朵位于头部两侧，能听声音但互相看不到，故填ears；
3. 计算：7（前）+5（后）+1（自身）=13，故填thirteen/13；
4. 计算：父亲年龄=12×4=48，祖父年龄=48×2=96，故填ninety-six/96；
5. 设Ann有A个，Bill有B个，Clare有C个，列方程：A+B=5，B+C=4，C+A=3；三式相加得2(A+B+C)=12→总和6，再解得A=2，B=3，C=1，故依次填two/2、three/3、one/1、six/6；
6. 诗句出自《白雪歌送武判官归京》，描写冬季雪景，故填winter。
【答案】
1. chocolate 2. ears 3. thirteen/13 4. ninety-six/96 5. two/2; three/3; one/1; six/6 6. winter
【知识点】
英语谜语、数学应用题、古诗理解
【点评】
本题综合考察多学科基础知识点，题型多样，难度适中，适合基础巩固练习。
【难度系数】
0.6

四、逻辑推理。阅读，将人物按照游戏中的身份分类。
Five children—Colleen, Jake, Hendrik, Vito and Tandeka—play a game of cops(警察) and robbers(强盗). The robbers' statements are always false while the cops' statements are always true.
a)Colleen says that Jake is a cop. b)Hendrik says that Vito is a robber.
c)Tandeka says that Colleen is not a robber. d)Jake says that Hendrik is not a cop.
e)Vito says that Tandeka and Colleen play on different sides.
Cop(s):
Robber(s):

Cop(s): Hendrik
Robber(s): Colleen, Jake, Vito, Tandeka

【分析】
首先明确核心规则：游戏中警察的陈述永远为真，强盗的陈述永远为假。我们通过假设法结合各人物的陈述推导身份：先假设Hendrik的身份，若假设出现矛盾则排除，再根据确定的身份推导其余人物的身份，确保所有陈述符合真假规则。
【解析】
根据规则：警察陈述为真，强盗陈述为假。
1. 假设Hendrik是强盗：由d)，Jake称“Hendrik不是警察”为真→Jake是警察；由a)，Colleen称“Jake是警察”为真→Colleen是警察；由c)，Tandeka称“Colleen不是强盗”为真→Tandeka是警察；由e)，Vito称“Tandeka与Colleen不同边”→因T、C均为警察，故Vito的话假→Vito是强盗；由b)，Hendrik称“Vito是强盗”→Hendrik是警察，与假设矛盾，因此Hendrik是警察。
2. 由Hendrik是警察，b)为真→Vito是强盗；
3. 由d)，Jake称“Hendrik不是警察”为假→Jake是强盗；
4. 由a)，Colleen称“Jake是警察”为假→Colleen是强盗；
5. 由e)，Vito是强盗，其陈述为假→Tandeka与Colleen同边→Tandeka是强盗。
综上，身份分类符合所有规则。
【答案】
Cop(s): Hendrik
Robber(s): Colleen, Jake, Vito, Tandeka
【知识点】
逻辑推理、命题真假判断
【点评】
本题需结合警察与强盗的陈述规则，通过假设法推导身份，关键在于利用矛盾点排除错误假设，逐步确定每个人的身份，能有效锻炼逻辑分析能力。
【难度系数】
0.3