SOLUTION: "Julio intends to drive his motorbike to a friend's house 200 kilometers away. If he increases his planned rate of speed by 10km/hr, he can decrease his travel time by 40 minutes.

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Question 1131120: "Julio intends to drive his motorbike to a friend's house 200 kilometers away. If he increases his planned rate of speed by 10km/hr, he can decrease his travel time by 40 minutes. What is his planned rate of speed?"
I made a RxT=D chart and I had an original rate and a new rate.
Original Rate | R = x - 10 | T = 200/(x-10) | D = 200
New Rate | R = x + 10 | T = 200/(x+10) | D = 200
I feel like I filled out the chart a bit off and thus resulting in solving my problem wrong. (I wasn't able to multiply & distribute all the way through so I didn't get an answer)
Found 4 solutions by josgarithmetic, ikleyn, stanbon, MathTherapy:
Answer by josgarithmetic(39630)   (Show Source):
You can put this solution on YOUR website!
r, the normal speed
r+10, the "if increase by 10 kilometer/hour" speed

SPEED TIME DISTANCE Normal r 200/r 200 If faster r+10 200/(r+10) 200 Difference 2/3

The equation to match the time difference for 40 minutes, of an hour,

Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website!
.
Julio intends to drive his motorbike to a friend's house 200 kilometers away. If he increases his planned rate
by 10km/hr, he can decrease his travel time by 40 minutes. What is his planned rate ?
~~~~~~~~~~~~~~~~~

            (!) NEVER say "rate of speed" (!)

            THERE IS NO such a term.   It DOES NOT EXIST and is NEVER USED (!)

            As soon as you use it, you demonstrate that you are not familiar with the correct usage of the terms in the subject !

Let x be the planned rate in kilometers per hour. The the hypothetical speed is (x+10) kilometer per hour The planned time is hours. The hypothetical time is hours. The difference is 40 minutes, which is of an hour. It gives you an equation - = . It is your basic equation, and when you got it, you completed the setup. To solve the equation, multiply its both sides by 3*x*(x+10). You will get 600*(x+10) - 600x = 2x*(x+10), 600x + 6000 - 600x = 2x^2 + 20x 2x^2 + 20x - 6000 = 0 x^2 + 10x - 3000 = 0 (x+60)*(x-50) = 0. Only positive root x= 50 is meaningful. The answer is: the planned speed is 50 kilometers per hour.

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To find many other similar solved problems, look into the lessons
    - Selected Travel and Distance problems from the archive
    - Had a car move faster it would arrive sooner
    - How far do you live from school?
in this site.

Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
"Julio intends to drive his motorbike to a friend's house 200 kilometers away. If he increases his planned rate of speed by 10km/hr, he can decrease his travel time by 40 minutes. What is his planned rate of speed?"
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Original Data::
rate = r kph ; time = t hrs ; distance = 200 km
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New DATA:
rate = r+10 kph ; time = t-(2/3) ; distance = 200 km
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Equations:
r*t = 200
(r+10)(t-2/3) = 200
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Substitute for "t" and solve for "r"
(r+10)((200/r)-(2/3)) = 200
(r+10)[(600-2r)/(3r)] = 200
(r+10)(600-2r) = 600r
600r -2r^2 + 1200 - 20r = 600r
r^2 +20r - 1200 = 0
r is approximately 26 kph
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Cheers,
Stan H.
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Answer by MathTherapy(10556)   (Show Source):
You can put this solution on YOUR website!

"Julio intends to drive his motorbike to a friend's house 200 kilometers away. If he increases his planned rate of speed by 10km/hr, he can decrease his travel time by 40 minutes. What is his planned rate of speed?"
I made a RxT=D chart and I had an original rate and a new rate.
Original Rate | R = x - 10 | T = 200/(x-10) | D = 200
New Rate | R = x + 10 | T = 200/(x+10) | D = 200
I feel like I filled out the chart a bit off and thus resulting in solving my problem wrong. (I wasn't able to multiply & distribute all the way through so I didn't get an answer)

Correct answer:
IGNORE STANBON's RIDICULOUS answer!!