A cyclist moves the first-half of the distance with 10 km`h^(-1)` speed and the second-half of the distance with speed V km `h^(-1)`. If the average speed of the cycle is 15 km `h^(-1)`, then the value of V is _____ km `h^(-1)`.

Answer

30

Explanation

`v_(1)=10km h^(-1)`
Here `v_(1)=(s/(2))/t_(1)rArrt_(1)=(s/(2))/(10)=s/(20)`
`v_(2)=(s/(2))/(t_(2))rArrt_(2)=(s/(2))/(2v_(2))`
Average speed `V=("Total distance")/("Total time taken")`
`rArr15=(s/(2)+s/(2))/(t_(1)+t_(2))=s/(s//20+s//2v_(2))`
`15=s/(s((2v_(2)+20)/(40v_(2))))=(40v_(2))/(s(v_(2)+10))`
`40v_(2)=30v_(2)+300`
`rArrv_(2)=30 km h^(-1)=v` .
Hence, the correct answer is 30.