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Question Number 119386 by ZiYangLee last updated on 24/Oct/20

$$\mathrm{The}\mathrm{length}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{is}\mathrm{decreased}\mathrm{by}20\mathrm{\%},$$$$\mathrm{and}\mathrm{the}\mathrm{width}\mathrm{is}\mathrm{increased}\mathrm{by}x\mathrm{\%},$$$$\mathrm{but}\mathrm{the}\mathrm{area}\mathrm{remains}\mathrm{the}\mathrm{same}.$$$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}x.$$

Answered by mr W last updated on 24/Oct/20

$$(1-\frac{20}{100})L\times (1+\frac{x}{100})W=L\times W$$$$(1-\frac{20}{100})(1+\frac{x}{100})=1$$$$1+\frac{x}{100}=\frac{100}{80}$$$$x=\frac{100}{4}=25$$

Commented by ZiYangLee last updated on 24/Oct/20

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Answered by som(math1967) last updated on 24/Oct/20

$$letlength=lunit$$$$width=bunit$$$$\therefore \frac{80l}{100}\times \frac{(100+x)b}{100}=l\times b$$$$\Rightarrow (100+x)=\frac{100\times 100}{80}$$$$\therefore x=125-100=25ans$$

Commented by ZiYangLee last updated on 24/Oct/20

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