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Question Number 114615 by I want to learn more last updated on 19/Sep/20

Answered by Olaf last updated on 19/Sep/20

$$t\mathrm{is}\mathrm{the}\mathrm{scheduled}\mathrm{time}$$$$x=v_1{t}_1=45(t+\frac{1}{20})\left(1\right)$$$$x=v_2{t}_2=60(t-\frac{3}{20})\left(2\right)$$$$(\frac{1}{20}h=3min\mathrm{and}\frac{3}{20}h=9min)$$$$\left(1\right)-\frac{3}{4}\times \left(2\right):$$$$x-\frac{3}{4}x=45(t+\frac{1}{20})-\frac{3}{4}\times 60(t-\frac{3}{20})$$$$\frac{1}{4}x=\frac{45}{20}+\frac{45\times 3}{20}=9km$$$$\Rightarrow x=4\times 9=36km$$$$\mathrm{with}\left(1\right):t+\frac{1}{20}=\frac{x}{45}=\frac{36}{45}=\frac{4}{5}$$$$t=\frac{4}{5}-\frac{1}{20}=\frac{3}{4}h=45min$$$$$$

Commented by I want to learn more last updated on 20/Sep/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}$$

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