辽宁省2003年中等学校招生考试
数学试题参考答案及评分标准
一、（选两个或两个以上答案不给分） 1．B 2．C 3．B 4．C 5．C 二、
11．x≥1且x≠2 16．2 19．（1?12．3
6．C
7．B
8．D 9．B 10．A
13．8，7.5 14．42 18．3.6
15．七
17．y2-8y-20=0（或写成y2-20=8y）
3）米 20．15°或75° 2 （注：13题错1个扣1分，顺序错不给分；17题写成分式方程不给分； 19题不写单位扣1分；20题只写对一解扣1分） 三、 21．
xx?y?yx?y?x(x?x?yy)?y(x?x?yy) ·································· 2分
?x?y ···························································································· 4分 x?y2?3························································ 6分 ??5 ·
2?3（注：不化简，直接代数求值，按相应步骤给分） 22．如图
M （O点找对） ·················································· 3分 （切线画对） ·················································· 6分
A B （注：不用尺规作图，不给分，没有保留作图痕迹不给分） C O 当x=2，y=3时，原式?23． （1）（频数）12，（频率）0.24 ······························································ 2分 （2）补全频率分布直方图 ··································································· 4分 （3）50···························································································· 6分 （4）80.5～90.5 ················································································· 8分 （5）216人 ····················································································· 10分 （注：（1）中每空1分，（2）中直方图1个1分，（3）中样本容量写单位的扣1分）
四、说明：本题给分点由两部分组成，一部分是图形设计（满分5分），按设计合理性和测量数据多少给分（5分、3分、1分、0分）；另一部分是依据图形计算（满分5分）。对不同设计方案（如1、2、3），同一图形可能字母标记不同，但只要计算正确即可得5分。根据不合理方案（如图d、e），计算正确给3分。 24．解： H 方案1：（1）如图a（测三个数据） ··················· 5分 （2）解：设HG=x
在Rt△CHG中 CG=x?cotβ A D α M 在Rt△DHM中 DM=（x－n）?cotα n β ∴x?cotβ=（x－n）?cotα ························ 8分 C G B n?cot?方案1图a
∴x= ···································· 10分
cot??cot?H 方案2：（1）如图b（测四个数据） ··················· 3分
（2）解：设HG=x
在Rt△AHM中 AM=(x－n) ?cotγ 在Rt△DHM中 DM=（x－n）?cotα ∴(x－n) ?cotγ=（x－n）?cotα+m ·············· 6分
m?n?cot??n?cot? ∴x= ························ 8分
cot??cot?方案3：（1）如图c（测五个数据） ··················· 1分 （2）参照方案1（2）或方案2（2）给分 ······ 6分
注：①如果在设计和计算中，考虑了测倾器高度，参照以上标准给分.
②以下两种方案（图d、e）或其他与其相似的图形不给分，但如果计算正确给3分．
H H A γ D α m n B C 方案2图b
M G H
A γ D α m n β C B 方案3图c
M G A B D αβ n C 图d
M G A B m D γ n C β M G 图e
五、 25．解：
（1）设s与t的函数关系式为s=at2+bt+c
?a?b?c??1.5?a?b?c??1.5??由题意得?4a?2b?c??2 （或?4a?2b?c??2）
?25a?5b?c?2.5?c?0??
1?a??2?解得?
b??2???c?0
1∴s=t2?2t ····················································································· 4分
21（2）把s=30代入s=t2?2t
21得30=t2?2t
2解得t1=10，t2=-6（舍）
答：截止到10月末公司累积利润可达到30万元 ······································· 7分 （3）把t=7代入，得 121s=?72?2?7??10.5 22把t=8代入，得
1s=?82?2?8?16 216-10.5=5.5
答：第8个月公司获利润5.5万元． ······················································ 10分
六、
26．解：设每周参观人数与票价之间的一次函数关系式为y=kx+b
?10k?b?7000由题意得? ····································································· 2分
15k?b?4500?
?k??500解得?
b?12000?∴y=-500x+12000 ··············································································· 4分 根据题意，得 xy=40000 即 x(-500x+12000)=40000 ··································································· 6分 x2-24x+80=0
解得 x1=20 x2=4 ············································································· 8分 把x1=20，x2=4分别代入y=-500x+12000中 得 y1=2000，y2=10000 ······································································ 10分 因为控制参观人数，所以取x=20，y=2000 ·············································· 11分 答：每周应限定参观人数是2000人，门票价格应是20元． ······················· 12分 （注：其他方法按相应步骤给分）
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