Sunny goes mountain biking every Saturday. The trail is $96$ km long, and is uphill all the way to the top.
His uphill speed is $16$ km per hour slower than his normal speed.
His downhill speed is $8$ km faster than his normal speed.
If a round trip on the trail takes $6$ hours, what is his normal speed (in km/h)?
+128475 Let the normal speed = S
And distance / rate = time....so
Time Up + Time Down = 6...so....
96 / ( S -16) + 96 / (S + 8) = 6 divide through by 6
16/ (S - 16) + 16/ ( S + 8) = 1
[16 ( S + 8) + 16 ( S -16) ] / [ (S -16) ( S + 8) ] = 1
[ 32S - 128 ] = (S -16) ( S + 8)
32S - 128 = S^2 - 8S - 128
32S = S^2 - 8S
40S = S^2
S^2 - 40S = 0
S ( S -40) = 0
S - 40 = 0
S = 40 km/h
