Pcat Math Question

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217933

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A steamboat goes 24 miles upstream and then returns to its original position. The round trip takes six hours. The water flows at three miles per hour. What is the speed of the boat in still water?

Anyone know how to do this? The book I'm studying out of tells me how to do it but they don't explain why certain steps are taken and whatnot.
 
A steamboat goes 24 miles upstream and then returns to its original position. The round trip takes six hours. The water flows at three miles per hour. What is the speed of the boat in still water?

Anyone know how to do this? The book I'm studying out of tells me how to do it but they don't explain why certain steps are taken and whatnot.
Seems like this is a vector question. This is how I would solve it if it were on the PCAT:

The boat will have to put out enough power to counteract that of the stream. In still water, the speed of the boat should be equal to the vector sum of the speeds of boat and stream combined.
Boat speed = 48 miles/6 hours = 8mph
Stream speed = 3 mph
Speed of boat in still water = SQRT (8^2 + 3^2) = SQRT (73)
The speed of the boat should be greater than 8, but less than 9 (actually, it's 8.54)

Of course, I may be wrong...




Edit: this answer IS wrong. But I will leave it here, so people can see how NOT to solve this problem.
 
Last edited:
I think this is just a distance problem.

D = R * T

When you are going upstream (against the current) the rate is going to be R - 3 and when you are down steam (with the current) it will be R + 3

IT takes 6 hours to go a total distance of 48 miles (24 x 2)

So if you set up D = R * T

48 mi = { (R - 3) + (R + 3) } * 6

Solve for R and you get 8 miles per hour in still water. I may be wrong as well. lol
 
total time: 6 hours
speed of boat: x
speed of water: 3
t1(upstream):24/x-3
t2(downstream):24/x+3


(24/x-3) + (24/x+3)=6

so, the answer x is 9miles/hour.
 
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