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**slaineskelton****Member**- Registered: 2018-01-16
- Posts: 2

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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Suppose that the cyclist's average speed when he traverses the outward leg is km/hr and he does this in time , where is measured in hours.The cyclist then travels the return leg at a speed of km/hr and he does this in time . (24 minutes is 0.4 hours.)Given that the legs are both the same distance, how can you now use this information to find equations relating and which you can solve?

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**slaineskelton****Member**- Registered: 2018-01-16
- Posts: 2

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!

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