At 0636, when the faster aircraft passes Point E, the slower aircraft is already 25 nm beyond Point E. The difference in speed between the two aircraft means that the faster aircraft is closing with the slower one at a rate of 50 kts, or 5nm per 6 minutes. This means that the gap will be closed in (25/5*6) or 30 minutes. In 30 minutes after Point E (i.e. at 0706hrs), the faster aircraft will have travelled 150 nm.
You can then double check by calculating where the slower aircraft would also be at 0706 hrs by a simple time and distance calculation (i.e. 250 kts travelled in 30 minutes equals 125 nm plus the 25 miles already travelled beyond Point E between 0630 and 0636 brings the answer to 150 nm).
Last edited by Curtis E Carr; 14th February 2010 at 19:24.
Reason: Beat me to it!