A) 20
B) 25
C) 30
D) None of these
Correct Answer: B
Solution :
[b] Distance \[=(\text{Time}\times \text{Speed})=t\times x.\] Let the "required increases in speed be p%. Then, \[(80%\,\,of\,\,t)\times \frac{(100+p)}{100}\times x=t\times x\] \[\Rightarrow \] \[\frac{80}{100}\times t\times \frac{(100+p)}{100}\times x=t\times x\] \[\Rightarrow \] \[\frac{4(100+p)}{500}=1\]\[\Rightarrow \]\[4p=100\] \[\Rightarrow \] \[p=25\] \[\therefore \] Required increase in speed \[=25%\] |
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