Question 1194484: A driver traveled from A to B. On the return trip, he averaged 10kph more and that trip took 30 minutes less. If he had averaged 10 kph less on the outgoing trip, that trip would have taken 40 minutes more. On the trip from A to B what was the average speed?
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39614)   (Show Source):
You can put this solution on YOUR website! ",... on the outgoing trip, that trip would have taken 40 minutes more. "
More than what?
SPEED TIME DISTANCE
A to B r d/r d
RETURN r+10 d/(r+10) d
DIFF. 1/2
If outgoing were slower
A to B r-10 d/(r-10) d
RETURN r+10 d/(r+10) d
DIFF. 2/3 (??)
If the incomplete description meant to say, "40 minutes more than was actually done", then
Now there can be the two equations in two unknown variables.
Other methods may be possible.
Answer by ikleyn(52756)  (Show Source):
You can put this solution on YOUR website! .
A driver traveled from A to B. On the return trip, he averaged 10kph more
and that trip took 30 minutes less.
If he had averaged 10 kph less on the outgoing trip, that trip would have taken 40 minutes more.
On the trip from A to B what was the average speed ?
~~~~~~~~~~~~~~~~~~
Let d be one way distance, r the average speed under the problem's question.
The time equation for the first scenario is
 -  = 1/2 of an hour (1)
The time equation for the second scenario is
 - = 2/3 of an hour (2)
To find "r", divide equation (1) by equation (2). The common factor "d" will be canceled,
and after simplification you will get this equation for r
 =  ,
 =  ,
4*(r-10) = 3*(r+10),
4r - 40 = 3r + 30
4r - 3r = 40 + 30
r = 70.
ANSWER. The average speed under the problem's question was 70 kilometers per hour.
Solved.
By the way, knowing this average speed, it is easy to find one-way distance from A to B.
It is 280 kilometers.
Answer by MathTherapy(10549)  (Show Source):
You can put this solution on YOUR website! A driver traveled from A to B. On the return trip, he averaged 10kph more and that trip took 30 minutes less. If he had averaged 10 kph less on the outgoing trip, that trip would have taken 40 minutes more. On the trip from A to B what was the average speed
Let outgoing speed and time be S and T, respectively
Then distance from A to B, or B to A = ST
Since the return trip took 30 minutes ( hour) less when traveling 10 km/h more, then the
return speed and return time are,  , respectively
We then get the following DISTANCE equation: 
Since the outgoing trip would've taken 40 minutes ( hour) more when traveling 10 km/h less, then the
return speed and return time would've been,  , respectively
We then get the following DISTANCE equation: 
 ------- Adding eqs (ii) & (i)
4S - 3S = 40 + 30 ----- Multiplying by LCD, 6
Outgoing speed, or 
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