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Hi there!
[tex]\large\boxed{\approx 43.33 min}}[/tex]
Recall:
d = st, where:
d = distance
s = speed
t = time
We can set up an expression where the extra ten minutes is taken into account:
50x = 65(x - 10) <--- because J left 10 minutes after, we must subtract from the time variable, or "x".
Solve for x:
50x = 65x - 650
Subtract 65x from both sides:
-15x = -650
Divide both sides by -15:
x ≈ 130/3 or 43.33 min