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Thanks for the response, so following that thought...
Journey 1
v = 174/t
Journey 2
v+ 0.5 = 174/t-0.4 thus
(t - 0.4)(v + 0.5) = 174
replace v with 174/t
(t - 0.4)(174/t + 0.5) = 174
174 + 0.5t - 348/5t - 0.2 = 174
t/2 -348/5t = 0.2 + 174 - 174
5t^2 - 696
-------------- = 0.2
10t
5t^2 - 696 = 0.2(10t)
5t^2 - 2t -696 = 0
t = 12 or t = -11.6
t cannot be negative so t = 12
v = 174/12
v = 14.5
if v = 15 then t = 174/15
t = 11.6
thus 0.4 of an hour less than Journey 1 or 24 minutes less than Journey 1
Thank you for your response!
Hi all,
I had this in an exam at Uni today and am baffled. If anyone could answer this I would be most grateful...
A cyclist travels a distance of 174 km on the outward leg of a road trip. The following day, the cyclist travels the return leg of 174 km at an average speed of 0.5 km/hr faster than the outward leg. Overall the journey time on the return is 24 minutes less than the outward leg.
Express the above information as a single mathematical equation or as a pair of simultaneous equations. Hence find the cyclist's speed on the outward leg.
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