A) 60
B) 70
C) 80
D) 120
Correct Answer: A
Solution :
Equating two parts of useful relation for\[{{V}_{1}}\]and\[{{V}_{2}}\] \[\frac{{{V}_{1}}{{V}_{2}}}{d}=\frac{{{V}_{1}}-{{V}_{2}}}{{{t}_{1}}-{{t}_{2}}}\] where\[\Delta V=\]difference in two speeds \[=10\text{ }km/h={{V}_{1}}-{{V}_{2}}\] \[\Delta t=\]difference in time\[={{t}_{1}}-{{t}_{2}}=1\] hours \[{{V}_{1}}=\]normal speed = V \[{{V}_{2}}=\]changed speed \[=(V+10)\] \[420=\frac{V(V+10)}{10}\times 1\] \[{{V}^{2}}+10V-4200=0\] \[V=60\,km/h\]
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