QR- rate problem

Started by Ocean5
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Ocean5

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A ski-boat can travel 20 mph in calm water. On a river it can travel 45 miles with the current in the same time it can travel 27 miles against the currect. What is the current's rate in m/h?
 
Boat's speed is 20mph.
Current's speed is x mph.

So with the current the boat goes (20+x)mph.
Against the current the boat goes (20-x)mph.

R x T = D

Boat goes 45 miles with the current:

(20+x)*t = 45

Boat goes 27 miles against the current:

(20-x)*t = 27

So we have a system of equations:

20t + xt = 45
20t - xt = 27

Add the two:

40t + 0 = 72
t = 9/5 = 1.8

So plug that into one of the above equations:

(20+x)(9/5) = 45
20+x = 45(5/9) = 25
20+x = 25
x = 5

Current is 5 mph.

Plug in this answer to confirm.
 
Boat's speed is 20mph.
Current's speed is x mph.

So with the current the boat goes (20+x)mph.
Against the current the boat goes (20-x)mph.

R x T = D

Boat goes 45 miles with the current:

(20+x)*t = 45

Boat goes 27 miles against the current:

(20-x)*t = 27

So we have a system of equations:

20t + xt = 45
20t - xt = 27

Add the two:

40t + 0 = 72
t = 9/5 = 1.8

So plug that into one of the above equations:

(20+x)(9/5) = 45
20+x = 45(5/9) = 25
20+x = 25
x = 5

Current is 5 mph.

Plug in this answer to confirm.

👍 Thank You.
 
Ocean5,

These are good problems that you're posting. Where are you finding them?

REH.

Over time I have recoreded some problems that I had a harder time solving them, so I get them from my notebook. Sorry that I haven't recorded their original source. However, they are from all sources that we normally use to prepare for DAT.

I'm glad that you find them helpful.