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Question Number 182461 by HeferH last updated on 09/Dec/22

$$Letxbeapositiveintegermultipleof17$$ $$thatsatisfiestheinequality:$$ $$0<\frac{5(x-120)}{x}<1$$ $$Findthevalueofx.$$

Answered by mr W last updated on 09/Dec/22

$$x=17k$$ $$0<\frac{5(17k-120)}{17k}<1$$ $$0<85k-600<17k$$ $$600<85k\Rightarrow k>\frac{600}{85}\Rightarrow k⩾8$$ $$68k<600\Rightarrow k<\frac{600}{68}\Rightarrow k⩽8$$ $$\Rightarrow k=8\Rightarrow x=17\times 8=136✓$$

Answered by MJS_new last updated on 09/Dec/22

$$x=17n$$ $$0<\frac{85n-600}{17n}<1$$ $$0<85n-600<17n$$ $$600<85n<17n+600$$ $$\frac{120}{17}<n<\frac{n}{5}+\frac{120}{17}$$ $$7.05...<n<\frac{n}{5}+7.05...$$ $$n_{\min }=8\mathrm{and}\mathrm{because}\mathrm{of}9>\frac{9}{5}+7.05...{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}$$ $$\mathrm{only}\mathrm{solution}$$ $$\Rightarrow x=8\times 17=136$$

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