SOLUTION: I keep getting a negative answer so I'm not doing something right. On the first part of a 350-kilometer trip, a salesperson travels 2 hours and 15 minutes at an average speed of

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Question 199675This question is from textbook Algebra and Trigonometry
: I keep getting a negative answer so I'm not doing something right.
On the first part of a 350-kilometer trip, a salesperson travels 2 hours and 15 minutes at an average speed of 100 kilometers per hour. The salesperson needs to arrive at the destination in another hour and 20 minutes. Find the average speed required for the remainder of the trip. This question is from textbook Algebra and Trigonometry

Found 2 solutions by nerdybill, checkley77:
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
On the first part of a 350-kilometer trip, a salesperson travels 2 hours and 15 minutes at an average speed of 100 kilometers per hour. The salesperson needs to arrive at the destination in another hour and 20 minutes. Find the average speed required for the remainder of the trip.
.
Applying the distance formula: d = rt
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distance traveled during the first part:
2.25(100) = 225 km
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remainder of trip:
350 - 225 = 125 km left to go
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average speed he needs to travel:
4/3 is equivalent to 1 hr 20 minutes
125/(4/3) = (3/4)(125) = 93.75 kilometers per hour

Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
350-100*2.25=350-225=125 MILES LEFT AFTER 2 HOURS & 15 MINUTES.
125/1.333=93.77 KMH IS THE REQUIRED SPEED FOR THE REMAINDER OF THE TRIP TO ARRIVE ON TIME.