Using my very basic VOR Distance/Time formulas, I managed to get as close as possible to the "Correct Answer" you provided.
First we can conclude that the Aircraft has flown from NW to NE and as it has flown across 90 degrees of Radials (310-040), we can say the aircraft was almost perfectly flying East (090). Its a perfect Isosceles triangle.
First Example: We understand from IFR ground school that a scale of 1 dot is = 200 feet when 1 nm from the station. So 1 degree at 1 nm = 100 feet.
At 40nm 1 degree would be 4000 feet (40nmX100feet).
The airplane flew through 90 Degrees of Radials so 90X4000 = 360,000 feet.
1nm = 6076 feet. Therefore, the Distance flown from the 310 radial to the 040 is 59.24nm(360,000/6076).
Now we can say that in 15 mins (0020-0035) the airplane covered a distance of 59.24nm. The aircraft must be traveling at 236 knots.
Second Example:
60 x minutes flown between
bearing change
Time to station = ----------------------------------
degrees of bearing change
X = 60 x 15
--------
90
X is 10 minutes: Therefore the aircraft will cover 40nm to the station in 10 minutes. This gives a speed of 240 knots
TAS x minutes flown
Distance to station = ------------------------------
degrees of bearing change
Now 240knots x 15
-----------------
90
= 40nm from station.
I guess I can rest my case to say
problems solved. At least to the nearest "Correct Answer".