Let’s rethink this problem as one that involves the equation of a line. Station 5 is the origin (0,0), and all other stations take on the traditional values on the x and y axes. Thus, when the boat leaves station #2, the ordered pair is (0,3), and the line of the boat’s path passes through the point (station #7, station #4), or (2,1). The line that connects these two points has a slope of (3 – 1)/0 – 2) = 2/-2 = -1. Using y = -1x + b, we substitute the values from (0,3), obtaining 3 = (-1)(0) + b, and so b = 3. Thus, the equation of the line connecting the two points is y = -x + 3.

We now plug in the ordered pairs of the answer choices to determine which of their equivalent ordered pairs satisfies the equation y = -x + 3.

Choice A: station #7 is at (2,0). Does 0 = -2 + 3? No.

Choice B: station #8 is at (3,0). Does 0 = -3 + 3? Yes!

Answer: B
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Restructuring the problem using coordinate plane we get
Docking in #2 : (0,3)
east of docking station #4 and directly north of docking station #7: (2,1)
Equation of this st. line is y=-x+3 [slope = -1]
so, the docking in X-axis Y=0, we get
0=-x+3, which is x = 3 (dock #8) [option b]

East of 4 and North of 7 means it has traveled TWO ( 2 to 4) steps down and TWO(5 to 7) steps east
But it has to travel 3 steps from 2 to 5, so it will travel 3 to the east = 5+3=8

ofcourse it can be done by finding slope but easier to assimilate this way

B

Finding slope of line using two points
(5,2)- when the boat is docked on west harbour
(7,4) - given boat has passed through this point

then slope [(4-2)]/[(7-5)]=1

Then we can find the port where the boat is docked in the south harbour

1= [(4-5)]/[(7-s.port)]
7-s.port=-1
s.port=8

[quote="Bunuel"]
A certain harbor has docking stations along its west and south docks, as shown in the figure; any two adjacent docking stations are separated by a uniform distance d. A certain boat left the west dock from docking station #2 and moved in a straight line diagonally until it reached the south dock. If the boat was at one time directly east of docking station #4 and directly north of docking station #7, at which docking station on the south dock did the boat arrive?

A. #7
B. #8
C. #9
D. #10
E. #11

NEW question from GMAT® Quantitative Review 2019

(PS08375)


The slope of the line is given by tan theta
At the instance when boat is at intersection of 4&7 tan theta=2/2
Now as slope is constant and y is 3 stations down from dock 2 (tan theta=3/3)....the destination dock has to be 3 stations to right of dock 5
Hence #8

Hit kudos if you like the answer!

Coordinates of #2 \((0,3),\) #4&7 \((2,1)\)

The line is \(y=mx+b\)

\(m= \frac{(3-1)}{(0-2)} =-1\)
So,
\(y=-x+b\)

Now, when \(y=3\)

\(3=-1*0+b\)
\(b=3\)

So, the line intercept at \(3\) or dock \(8\) of the south dock.

Ans. B.
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This was the best explanation so far

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