A boy rides a bike 8 miles. The trip would have taken 20 min less if he had ridden 2 miles per hour faster?

Question

His actual speed 6 miles an hour but I have no idea how to get to it. Can someone explain it to me? Thanks.

Answer 1

let v be the speed of the boy, the time he covers 8 miles is t.

8 = vt

if he increase his speed by 2 mi/hr faster the the one before, the he can save 20 minutes or1/3hr
speed is v + 2
time is t - 1/3

8 = (v+2) (t - 1/3)


you have two equations
8 = vt
8 = (v + 2) (t - 1/3)

solve fot t
t = 8/v

8 = (v + 2) (8/v - 1/3)
8 = 8 - v/3 + 16/v - 2/3

multiply 3v for both sides
24v = 24v - v^2 + 48 - 2v
0 = -v^2 + 48 - 2v
v^2 + 2v - 48 = 0
(v + 8) (v - 6) = 0
v = 6 or -8

take the positive number. V = 6mi/hr

Answer 2

Let t = the time to ride the 8 miles
Then 8/t was his rate of speed
8/t + 2 = rate if he rides 2 mph faster
So 8/(8/t+2) = t-1/3
8t/(8+2t) = t-1/3
8t = 8t-8/3 +2t^2 -2t/3
2t^2 -2t/3 -8/3 = 0
6t^2 -2t -8 = 0
3t^2 -t -4 = 0
(3t-4)(t+1)=0
So 3t-4 = 0 --> t= 4/3 hour --> original speed = 8/4/3 = 6mph

If he rides 2 mph faster his rate is 6+2 = 8 mph
so time = 8/8= 1 hour which is 1/3hour = 20 min leess

Answer 3

At 6 miles per hour, that means he travelled:

20 mins: 2 miles
40 mins: 4 miles
60 mins: 6 miles
1 hour and 20 minutes : 8 miles

If he had travelled 8 miles per hour, he would have been there in 1 hour.

Answer 4

distance (d) = velocity (v) x time (t)
d = vt = 8, for the real trip
t = 8/v

(v+2)(t-1/3) = 8, for the imaginary trip
vt+2t-(1/3)v-2/3 = 8

Plug in vt = 8 and t = 8/v,
2t-(1/3)v-2/3 = 0
6t-v-2 = 0
6(8/v)-v-2 = 0
v^2+2v-48 = 0
(v-6)(v+8) = 0
v = 6 mph, the only answer since the speed must be positive.