几道小学升初中的数学题（要过程和答案)
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\u94fe\u63a5\uff1ahttps://pan.baidu.com/s/1znmI8mJTas01m1m03zCRfQ
?pwd=1234
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1\u3001\u7565\uff1b2\u3001\u8131\u5f0f\u8ba1\u7b97\u24605\uff1b\u246128.21\uff1b\u24627 \uff1b\u24633 \uff1b\u246498.1\uff1b\u2465900
3\u3001\u89e3\u65b9\u7a0b\uff1a4\uff0c5\uff1b
4\u3001\u5217\u5f0f\u8ba1\u7b97\uff1a3\uff0c33.3%\uff1b
\u4e94\u3001\u9634\u5f71\u9762\u79ef\uff1a48\u5e73\u65b9\u5206\u7c73\uff1b
\u516d\u3001\u5e94\u7528\u9898
1\u300154\uff08\u6b21\uff09\uff1b2\u300124\uff08\u5929\uff093\u300111.1%\uff1b4\u30019023.2\uff08\u5e73\u65b9\u5206\u7c73\uff09\uff1b5\u30011540\uff08\u672c\uff09
6\u30015\uff08\u5929\uff09\uff1b7\u3001576\uff08\u5343\u7c73\uff09\uff1b
2、设第一堆有12x吨煤，即第二堆存煤量为7x吨。于是有：
[12x×（1-1/4）]-7x=2.4
得：x=1.2,所以第一堆原有煤1.2吨×12=14.4吨。
3、依题意 （快车的速度+慢车的速度）×8小时=总路程，即每小时（快车的行程+慢车的行程）=总路程/8，所以从两车相遇到两小时后两车共走了 总路程/4，而“这时快车离乙地还有250千米，慢车离甲地还有350千米”，
所以有 总路程-总路程/4=250+350，得：总路程=800千米，即甲乙两地相距800千米.
第2题:
解:因为第一堆煤和第二堆煤存煤量之比是12:7,我们可设其中一份为X,则第一堆煤有12X吨,第二堆煤有7X吨.
则可列式:
12X-12X*1/4-7X=2.4
解得:X=1.2
则第一堆煤有12*1.2=14.4(吨).
答:第一堆煤原有煤14.4吨.
第3题:
解:设快车的速度为X.则:
250+2X=8X*4-350*4
解得:X=55
则甲乙两地全程为:
55*(8+2)+250=800(千米)
答:甲乙两地相距800千米.
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