SOLUTION: A student leaves the university at noon, bicycling south at a constant rate. At 12:30pm a second student leaves the same point and heads west, bicycling 7mph faster than the first

Question 16947: A student leaves the university at noon, bicycling south at a constant rate. At 12:30pm a second student leaves the same point and heads west, bicycling 7mph faster than the first student. At 2:00pm, they are 30 miles apart. How fast is each one going? Also, how would I properly set-up this equation?
Below is what I thought is the answer, please let me know where I went wrong?
r=d/t
30/2 = 15mph (first student)
30/1.5 = 20 + 7 = 27 mph (second student)
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
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

..we get x=9mph.

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