三道小学六年级奥数题,要求有解答过程,快快快 六年级上册奥数题(越难越好,要有解答过程的)

\u95ee3\u9053\u516d\u5e74\u7ea7\u5965\u6570\u9898

1.
\u72d7\u53ea\u6570 = (390 - 160*2)/2 = 35
\u730e\u624b\u6570 = (160*4 - 390)/2 = 125
2.
\u724c\u70b9\u6570\u7684\u603b\u6570\u662f13, \u6bcf\u79cd\u90fd\u53ea\u6709\u4e24\u5f20\u65f6\u662f2*13
\u6700\u5c11\u53d6 2*13 + 1 = 27 \u5f20\u724c\u5c31\u53ef\u4ee5\u4fdd\u8bc1\u5176\u4e2d3\u5f20\u724c\u7684\u70b9\u6570\u76f8\u540c
3.
\u2460 ... ... ... ... ....\u2461 ... ... ... ... ....\u2462


... ... ....\u2463 ... ...\u2464 ... ...\u2465


\u2466 ... ... ... ... ....\u2467 ... ... ... ... ....\u2468
\u7b2c\u4e00\u6392\uff1a\u2460\u2014\u2014\u2461\u2014\u2014\u2462
\u7b2c\u4e8c\u6392\uff1a\u2463\u2014\u2014\u2464\u2014\u2014\u2465
\u7b2c\u4e09\u6392\uff1a\u2466\u2014\u2014\u2467\u2014\u2014\u2468
\u7b2c\u56db\u6392\uff1a\u2461\u2014\u2014\u2464\u2014\u2014\u2467
\u7b2c\u4e94\u6392\uff1a\u2460\u2014\u2014\u2463\u2014\u2014\u2467
\u7b2c\u516d\u6392\uff1a\u2462\u2014\u2014\u2465\u2014\u2014\u2467
\u7b2c\u4e03\u6392\uff1a\u2466\u2014\u2014\u2463\u2014\u2014\u2461
\u7b2c\u516b\u6392\uff1a\u2468\u2014\u2014\u2465\u2014\u2014\u2461
\u7b2c\u4e5d\u6392\uff1a\u2460\u2014\u2014\u2464\u2014\u2014\u2468
\u7b2c\u5341\u6392\uff1a\u2462\u2014\u2014\u2464\u2014\u2014\u2466

\u7b2c\u5341\u4e8c\u8bb2 \u5bb9\u65a5\u539f\u57cb


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\u5982\u679c\u96c6\u5408A\u4e0eB\u6709\u516c\u5171\u5143\u7d20\uff0c\u4f8b\u5982

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第一题
第一小时走了全程的 30%,第二小时多走了 3千米,这样就走了全程的 60% 加 3千米,然后再走 15千米才去到目的地,就是 60% 加 18千米就等于全程了。那么 18千米就是全程的 40%,计算全程走了多远,就是 18/ 0.4 = 180/4 = 45千米。

第二题
把当前堰塞湖,水位到达坝顶的水量,也就是堰塞湖可增加的水量,看作整体1。
不开辟泄洪渠道,水位 20天后到达坝顶,每天的流入水量,就是 1/20。
泄洪渠道开通后,水位到达坝顶推迟 30天,就是 50天后。这样每天增加的水量,就是 1/50。
毕竟每天流入的水量不变,计算每天泄洪流出的水量,就是 1/20 - 1/50 = 5% - 2% = 3%。
每天流出的这个 3%,相对于每天流入的 1/20 又是多少呢?3% / (1/20) = 20X3% = 60%,每天泄出的水量,就是流入水量的 60%。

第三题
想象一下画个图,这一题应该相当简单。原来排成 n X n 的正方形,多余了 3颗棋子,可是正方形如果要纵横加一行,加一列,还缺少 8颗。我们就知道,原来一个“口”,要加一横加一竖,变成“可”,就需要 3+8= 11颗棋子。11= 5X2+1= 5+6,我们就知道,原来 nXn 的正方形是由 5X5= 25颗棋子组成,棋子还有 3颗多余,棋子就是 25+3= 28颗。检查一下儿,正方形纵横加一行加一列,就要 6X6= 36颗棋子,缺少 8颗棋子,棋子就是 36-8= 28颗。

1.解:第二小时比第一小时多走了3千米
说明走了60%后剩40% 就是3+15=18千米
18÷40%=45千米

2.[(30+20)*1-20]/(30+20)
=30/50
=60%

3.3+8=11
(11-1)÷2=5

说明以前每排5个
则5x5+3=28

手打啊 有几个抄袭我的
谢谢采纳啊
不会追问

1、解:设共走x千米,则有
30%x+(30%x+3)+15=x
解得x=45(千米)

2、解:设每天泻出水量是流入湖中水量的x,
则(20+30)(1-x)=20
解得x=60%

3、解:设有x个棋子

x=n^2+3
x=(n+1)^2-8
解得n=5
x=28
答第二小时比第一小时多走了28千米

1、解:设共走x千米,则有
30%x+(30%x+3)+15=x
解得x=45(千米)
2、解:设每天泻出水量是流入湖中水量的x,
则(20+30)(1-x)=20
解得x=60%
3、解:设有x个棋子

x=n^2+3
x=(n+1)^2-8
解得n=5
x=28

(15+3)/(1-30%-30%) 15千米是40%少3千米。
=18/40%
=45千米

设每天注入量1,则湖的现容水量为20。
[(30+20)*1-20]/(30+20)
=30/50
=60%

(3+8+1)/2=6 求出现在每边几个。
6*6-8=28

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