Answer: Option B
:
B
Using $\frac{1}{x}\&\frac{1}{x+1}$
We know that the speed is increasing by $\frac{1}{2}$ hence the time has to decrease by $\frac{1}{3}$ (using the $\frac{1}{x}\&\frac{1}{x+1})$
shortcut (refer demo tutorial) but the time is actually decreasing by 2 hrs.
Hence the actual time of travel is 6 hrs hence he leaves at 7:00 AM.
The total distance travelled is $10\times 6=60kms$
To reach at noon he will have to cover 60 km in 5 hours. i.e., he has to travel at 12 km/hr.
Approach 2: Conventional Approach:
Let the person take t hrs to cover the distance at 10 km/hr.
If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m, which means he takes (t-2)hr.
Let the distance covered be D.
So we have $15\times (t-2)=10\times \left(t\right)$
T=6 , or he started 6 hours before 1, i.e., at 7.
Hence the total distance now is $10\times 6=60$ km.
To reach at noon he will have to cover 60 km in 5 hours. He was to travel at 12 km/hr.