电子课本网 第73页

第73页

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解:设大容器的容积是$x$斛,小容器的容积是$y$斛,
根据题意,得$\begin{cases}5x + y = 3\\x + 5y = 2\end{cases}$
由$5x + y = 3$得$y = 3 - 5x,$
将$y = 3 - 5x$代入$x + 5y = 2$得:
$x + 5(3 - 5x) = 2$
$x + 15 - 25x = 2$
$-24x = 2 - 15$
$-24x = -13$
$x = \frac{13}{24}$
把$x = \frac{13}{24}$代入$y = 3 - 5x$得:
$y = 3 - 5×\frac{13}{24} = 3 - \frac{65}{24} = \frac{72 - 65}{24} = \frac{7}{24}$
答:大容器的容积是$\frac{13}{24}$斛,小容器的容积是$\frac{7}{24}$斛.
解:设甲种产品生产$x$个,乙种产品生产$y$个,
根据题意,得$\begin{cases}2x + 3y = \frac{40}{3}×60\\30x + 40y = 11×1000\end{cases}$
化简得$\begin{cases}2x + 3y = 800,①\\3x + 4y = 1100,②\end{cases}$
①$×3 -$②$×2$得:
$3(2x + 3y) - 2(3x + 4y) = 3×800 - 2×1100$
$6x + 9y - 6x - 8y = 2400 - 2200$
$y = 200$
把$y = 200$代入①得:
$2x + 3×200 = 800$
$2x + 600 = 800$
$2x = 800 - 600$
$2x = 200$
$x = 100$
答:甲种产品生产100个,乙种产品生产200个.
解:
(1)设足球的单价是$x$元,跳绳的单价是$y$元,
根据题意,得$\begin{cases}30x + 60y = 720\\10x + 50y = 360\end{cases}$
由$10x + 50y = 360$得$x = 36 - 5y,$
将$x = 36 - 5y$代入$30x + 60y = 720$得:
$30(36 - 5y) + 60y = 720$
$1080 - 150y + 60y = 720$
$-90y = 720 - 1080$
$-90y = -360$
$y = 4$
把$y = 4$代入$x = 36 - 5y$得:
$x = 36 - 5×4 = 36 - 20 = 16$
答:足球的单价是16元,跳绳的单价是4元.
(2)设该店的商品按原价的$m$折销售,
根据题意,得$16×\frac{m}{10}×100 + 4×\frac{m}{10}×100 = 1800$
$160m + 40m = 1800$
$200m = 1800$
$m = 9$
答:该店的商品按原价的9折销售.
解:
(1)设每本宣传册中$A$种彩页有$x$页,$B$种彩页有$y$页,
根据题意,得$\begin{cases}x + y = 10\\300x + 200y = 2400\end{cases}$
由$x + y = 10$得$x = 10 - y,$
将$x = 10 - y$代入$300x + 200y = 2400$得:
$300(10 - y) + 200y = 2400$
$3000 - 300y + 200y = 2400$
$-100y = 2400 - 3000$
$-100y = -600$
$y = 6$
把$y = 6$代入$x = 10 - y$得:
$x = 10 - 6 = 4$
答:每本宣传册中$A$种彩页有4页,$B$种彩页有6页.
(2)$2400+(2.5×4 + 1.5×6)×1500$
$=2400+(10 + 9)×1500$
$=2400 + 19×1500$
$=2400 + 28500$
$= 30900$(元)
答:印制这批宣传册制版费与印刷费的总和是30900元.