SOLUTION: a man covers a certain distance on scooter.had he moved 3 kmph faster, he would have taken 40 min.less. if he had moved 2 kmph slower he would have taken 40 min.more .the distance

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Question 900249: a man covers a certain distance on scooter.had he moved 3 kmph faster, he would have taken 40 min.less. if he had moved 2 kmph slower he would have taken 40 min.more .the distance is?
Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39800) About Me  (Show Source):
You can put this solution on YOUR website!
Be aware: I likely made a mistake in this, and have not successfully finished solving....

RT=D uniform rates for travel

_____________speed_______time______distance
IF FASTER____r+3_________t-2/3_____d=(r+3)(t-2/3)
IF SLOWER____r-2_________t+2/3_____d=(r-2)(t+2/3)
Rate Used____r____________t________d=rt

A possible place in the analysis that someone could become stuck is to not recognize the very simple r*t=d, while each of these three variables is still unknown.

(Removed the previous shown solution because of continuing mistakes while making corrections. See the following.)


***CONTINUING WITH GOOD SOLUTION PROCESS---------------------------------------------------------

FAST
rt%2B3t-%282%2F3%29r-2=d
rt%2B3t-%282%2F3%29r-2=rt
3t-%282%2F3%29r-2=0
9t-2r-6=0
highlight_green%289t-2r=6%29

SLOW
rt-2t%2B%282%2F3%29r-4%2F3=d
-2t%2B%282%2F3%29r-4%2F3=0
2t-%282%2F3%29r%2B4%2F3=0
6t-2r%2B4=0
3t-r%2B2=0
highlight_green%283t-r=-2%29

Start with Elimination Method to eliminate r.
Multiply SLOW equation by 2 and use subtraction, FAST-SLOW.
9t-2r-%286t-2r%29=6-%28-4%29
9t-2r-6t%2B2r=10
3t=10
highlight%28t=10%2F3=3%261%2F3%29, hours

Use t in either equation, here being SLOW as the chosen equation, to find r.
-r%2B3%2810%2F3%29=-2
-r%2B10=-2
-r=-12
highlight%28r=12%29, miles per hour

The distance, according to r*t?
40 miles.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
r*t=d,
(r+3)*(t-40/60)=d,
(r-2)*(t+40/60)=d'
r*t=d,
(r+3)*(t-2/3)=d,
(r-2)*(t+2/3)=d
d = r* t,
d = 1/3 (r+3) (3*t-2),
d = 1/3 (r-2) (3*t+2)
d = r*t,
3d=3rt
3d = (r+3)*(3*t-2),
3d = (r-2)*(3*t+2)
3rt+9t-6-2r=3d
3rt-6t-4+2r=3d
9t-6-2r=0
-6t-4+2r=0
3t-10=0
3t=10
t=10/3
9*10/3-6-2r=0
30-6=2r
24=2r
r=12
d=12*10/3=40
distance is 40
d = 40, r = 12, t = 10/3
check
r*t=d,
12*10/3=40
(r+3)*(t-40/60)=d,
15*8/3=40
5*8=40
(r-2)*(t+40/60)=d
10*12/3=40
10*4=40
ok