Question 16947
let the speed of the 1st.student=x mph.
time of start of 1st.student = 12=00 noon....time at end is 2=00 pm.
duration of travel =2 hrs.
distance travelled by 1st.student from start in south direction ...........   
 =speed * duration = 2*x miles 
speed of 2nd.student is  7 mph more.hence it is x+7 mph.
time of start of 2nd.student is 12=30 noon....time at end is 2=00 pm.
duration of travel =1.5 hrs.
distance travelled by 2nd.student from start in west direction ................= speed * duration = 1.5*(x+7) miles 
if we call their 2 end positions as A and B and start point as S,we can see that ASB is a right angled triangle at S with SA=2x and SB=1.5(x+7).and AB=30 as given in the problem.hence using pythogarus theorem in right angled triangle ASB , we have SA^2+SB^2=AB^2
  (2x)^2+[1.5(x+7)]^2=30^2
4x^2+2.25(x^2+14x+49)=900
6.25x^2+31.5x+110.25=900
6.25x^2+31.5x-789.75=0  using the standard formula for quadratic equation 
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
..we get x=9mph.