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anonymous
 5 years ago
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
anonymous
 5 years ago
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0vt=d (v+3)(t2/3)=d (v2)(t+2/3)=d 3 equations and three unknowns so you can find d.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0any other easy way guys? its too time consuming . This question came for 1 mark and need to solve in 1 minute max.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya its 40 km but how? plz solve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[d=(r+3)\left(t\frac{40}{60}\right),d=(r2)\left(t+\frac{40}{60}\right) \]Expand\[d=2\frac{2 r}{3}+3 t+r t,d=\frac{4}{3}+\frac{2 r}{3}2 t+r t \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solve the following simultaneous equations for r and t.\[0=2\frac{2 r}{3}+3 t,0=\frac{4}{3}+\frac{2 r}{3}2 t \] I got\[\left\{r\to 12,t\to \frac{10}{3}\right\} \]\[r t = 12 \frac{10}{3}=40\text{km}/\text{hr} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Note: r t = d\[d=2\frac{2 r}{3}+3 t+r t \]simplifies to\[d=2\frac{2 r}{3}+3 t+d\]subtracting d from each side \[0=2\frac{2 r}{3}+3 t \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The same simplification holds for the second equation.
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