Question Below!

Question

On a particular day, the wind added 4 miles per hour to Alfonso's rate when he was cycling with the wind and subtracted 4 miles per hour from his rate on his return trip. Alfonso found that in the same amount of time he could cycle 57 miles with the wind, he could go only 33 miles against the wind.What is his normal bicycling speed with no wind?

Answer 1

Let S_0 be Alphonso's normal speed. Then

(1) S_0 +4    is his speed in mi/ hr with the wind in the direction of travel

(2) S_0 - 4    is his speed in mi/ hr with the wind in the opposite direction  of travel

 

we will use the distance and rate formula in the following format:

(3) Rate = Distance / Time

 

we are give that he traveled 57 mi ( call it D1)  with the wind

we are also given that he traveled 33 mi ( call it D2 ) against the wind 

Note: we are not given the time but are told it took the SAME amount of time 

So we can write the equations:

 

(4) S_0 + 4 = D1 / Time = 57  units are mi/hr

(5) S_0  - 4 = D2/  Time = 33  units are mi/hr

 

now we perform some algebra!

 

multiply thru by Time

 

(6) (S_0 + 4)*Time = D1= 57 mi
(7) (S_0 - 4 )*Time = D2= 33 mi

 

subtract to solve for Time in hrs

 

(8) 8*Time =24  or Time = 24/8 = 3 hrs

 

now go back to (6) and (7) and add to solve for S_0 because you now know Time

 

(9) 2*S_0*3 hrs = (57+33)mi =90  or S_0 15 =  mi/ hr is Alphonso's usual speed without wind

 

CHECK Your Answer

with wind (15+4)mi/ hr*3 hr = 19 mi / hr

against wind (15-4)mi/hr*3 hr = 11 mi /hr

 

 

 

 

Answer 2

This question looks harder than it really is.  The windless rate would be 45 miles per time period (since you add 4 or subtract 4, you can just look for the middle number).  Since 57-45=12, and 12/4=3, the time period is 3 hours.  45=3=15, so without wind he cycles 15 miles per hour.  With the wind he cycles 19/hr, against the wind 11 miles/hour.