13、（10分）8个人乘速度相同的两辆小汽车同时赶往火车站，每辆车乘4人（不包括司机），其中一辆小汽车在距火车站15km的地方出现故障，此时，距停止检票的时间还有42分钟，这时，唯一可以利用的交通工具是另一辆小汽车，已知包括司机在内这辆车限乘5人，且这辆车的平均速度是60km/时，人步行的平均速度是5km/时，试设计两种方案，通过计算说明这8个人能够在停止检票前赶到火车站．
（全国初中数学竞赛题）
七年级数学拔高测试 第 6 页 共 6页
7 页 6页
七年级数学拔高测试 第 共
2014年河南省鹤壁市淇滨中学7年级3、4班数学拔高测试
期末测评试卷参考答案及评分标准
第1卷 基础夯实（满分20分）
题号 答案 1 5 2 2 3 3；2；9 4 3 5 d?c 2第2卷 能力扩展（满分80分）
6、解：由题意得 3a+2b=5-c 2a+b=1+3c············································3分 解得 a=7c-3 b=7-11c··············································5分 ∴m=3c-2···············································6分 a≥0 ∵ b≥0
七年级数学拔高测试答案 第 1 页 共 5 页
c≥0··············································7分 7c-3≥0 ∴ 7-11c≥0
c≥0··············································8分
7············································9分 1151∴-≤m≤-··········································10分
711∴≤c≤
37
7、∵S△PBC+S△PAD=S四边形ABCD=S△ABD····························2分 ∴S△PAC=S△ABD-S△PAB-S△PAB···································4分 ∴S△PAC=S△PBC-S△PAB·········································6分 ∵S△PBC=5，S△PAB=2········································8分 ∴S△PAC=5-2=3···········································10分
8、连FG、DG、FB、DB，···································2分 ∵S△EGF=S△EGD,S△HFG=S△HFB····································3分 ∴S四边形DGBF=2S四边形GHFE ①···································4分 ∵DE=EF=FC,AG=GH=HB····································5分 ∴S△DBC=3S△FBC,S△DBA=3S△DGA···································6分 ∴S四边形ABCD=3(S△FBC+S△DGA)··································7分 得S△FBC+S△DGA=S四边形ABCD ②································8分 ①+②，得S四边形ABCD=2S四边形GHFE+S四边形ABCD·····················9分 故S四边形GHFE=S四边形ABCD.···································10分
13131312 七年级数学拔高测试答案 第 2 页 共 5 页
9、由题意可得
49+S1+35+13+S2=S3+S阴+S4 ①
49+S3+13+35+S4=S1+S阴+S2 ②·····························3分 由①、②得
49+35+13=S阴+S3+S4-S1-S2 ③·····························6分 49+35+13=S阴+S1+S2-S3-S4 ④·····························7分 ③+④得
2S阴=2×（49+35+13）·····································8分 S阴=49+35+13············································9分 S阴=97·················································10分
10、∵△ABC为直角三角形 又∵BC=AB=4
又∵平移的距离为3······································2分 ∠B=∠A′B′C=90°································3分 ∴ ∠A=∠ACB=∠A′=∠C′=45°························4分 BB′=CC′=3cm·····································5分 ∴∠BOC=∠A′B′B-∠ACB=90°-45°=45°·················6分 ∴B′O=B′C············································7分 ∴B′C=BC-BB′=4-3=1(cm)·······························8分 ∴B′O=1cm·············································9分
七年级数学拔高测试答案 第 3 页 共 5 页