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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:

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  1. anonymous
    • 5 years ago
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    vt=d (v+3)(t-2/3)=d (v-2)(t+2/3)=d 3 equations and three unknowns so you can find d.

  2. anonymous
    • 5 years ago
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    any other easy way guys? its too time consuming . This question came for 1 mark and need to solve in 1 minute max.

  3. anonymous
    • 5 years ago
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    d=40 km

  4. anonymous
    • 5 years ago
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    ya its 40 km but how? plz solve

  5. anonymous
    • 5 years ago
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    \[d=(r+3)\left(t-\frac{40}{60}\right),d=(r-2)\left(t+\frac{40}{60}\right) \]Expand\[d=-2-\frac{2 r}{3}+3 t+r t,d=-\frac{4}{3}+\frac{2 r}{3}-2 t+r t \]

  6. anonymous
    • 5 years ago
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    Solve 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\[\left\{r\to 12,t\to \frac{10}{3}\right\} \]\[r t = 12 \frac{10}{3}=40\text{km}/\text{hr} \]

  7. anonymous
    • 5 years ago
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    Note: 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 \[0=-2-\frac{2 r}{3}+3 t \]

  8. anonymous
    • 5 years ago
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    The same simplification holds for the second equation.

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