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第100页

信息发布者：

$\frac {1}{12}$

$\frac {1}{8}$

$\frac {1}{4}$

$1-\frac {63}{64}$

$\frac {25}{33}$

$= (1-\frac {1}{2})+(1- \frac {1}{4} )+(1- \frac {1}{8} )+···+ (1-\frac {1}{256})$

$=8-( \frac {1}{2} +\frac {1}{4}+\frac {1}{8}+···+\frac {1}{256})$

$=8-(1-\frac {1}{256} )$

$=7 \frac {1}{256}$

=0.5+1+1.5+2+···+4.5

=(0.5+4.5)×9÷2

=22.5

$10$分钟$=\frac {1}{6}$小时

$\frac {5}{7}-\frac {1}{6}+\frac {5}{7}=\frac {53}{42}($小时$)$

答：这名游客上山、下山共用了$\frac {53}{42}$小时。

$\frac {11}{12}-(\frac {1}{4}+\frac {1}{4}+\frac {1}{6})=\frac {1}{4}($吨$)$

答：原来乙仓库存粮$\frac {1}{4}$吨。

$1−(\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\frac {1}{16})=1−(1−\frac {1}{16})=\frac {1}{16} $

答：还剩总数的$\frac {1}{16}。$

$ \frac {1}{2} +\frac 13+ \frac {1}{6} =1$

$[19，$$5]=95$

$\frac {x}{19}＜\frac {4}{5}＜\frac {y}{19}$可以改写为$\frac {5x}{95}＜\frac {76}{95}＜\frac {5y}{95}$

$5x＜76＜5y$

所以$x=15，$$y=16$