A man travels a distance of $300$ km. On his return journey his average speed increased by $20$ km/h and his journey time decreased by $1\frac{1}{4}$ hours. If $v$ is the average speed of his outward journey how can we show that: $$\frac{300}{v} - \frac{300}{v+20} = 1.25$$ I'm very stuck.
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By the information you have, using $\text{speed}\cdot \text{time} = \text{distance}$ you can create two equations: $$\text{I:}\quad v\cdot t = 300\\ \text{II:}\quad (v+20)\cdot (t-1.25) = 300$$
Edit: Express $t$ from each equation, you get $$\text{I:}\quad t = \frac{300}{v}\\ \text{II:}\quad t = \frac{300}{v+20} + 1.25$$
Now compare substitute $t$ from the first equation into the second one. $$\frac{300}{v} = \frac{300}{v+20} + 1.25$$ Now you only have to subtract $\frac{300}{v+20}$.
Jakube
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how to show that the equation in the question is true from that equation? – imulsion Mar 09 '13 at 12:16
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THANK YOU! I have literally torn two pages of scribbled out equations out of my maths book – imulsion Mar 09 '13 at 12:23
So, $$\frac{300}v-\frac{300}{v+20}=\frac54$$
– lab bhattacharjee Mar 09 '13 at 11:54