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PQR region

Expert replies
by gmatblood » Fri Nov 04, 2011 9:22 am
P, Q and R are located in a flat region of a certain state. Q is x miles due east of P and y miles due north of R. Each pair of points is connected by a straight road. What is the number of hours needed to drive from Q to R and then from R to P at a constant rate of z miles per hour, in terms of x, y and z? (Assume x, y, and z are positive)

a. (√x² + y²) / z
b. (x + √ x² + y²) / z
c. (y + √x² + y²) / z
d. z / (x + √x² + y²)
e. z / (y + √x² + y²)
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Source: — Problem Solving |

by Tani » Fri Nov 04, 2011 9:34 am
The problem gives you a right triangle. You need to use the Pythagorean theorem
PQ = x, PR = y and RQ = sqrt (x^2 + y^2)

To go from q to P by way of R, you must travel y + sqrt (x^2 + y^2) miles.

Dividing your miles by your miles per hour gives you the hours needed.
the answer then is
(y + sqrt (x^2 + y^2))/z
C is the closest although it is unclear the way it is typed that the square root symbol applies to the entire quantity (x^2 + y^2)
Tani Wolff
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by Scott@TargetTestPrep » Wed Jan 03, 2018 4:07 pm
gmatblood wrote:P, Q and R are located in a flat region of a certain state. Q is x miles due east of P and y miles due north of R. Each pair of points is connected by a straight road. What is the number of hours needed to drive from Q to R and then from R to P at a constant rate of z miles per hour, in terms of x, y and z? (Assume x, y, and z are positive)

a. (√x² + y²) / z
b. (x + √ x² + y²) / z
c. (y + √x² + y²) / z
d. z / (x + √x² + y²)
e. z / (y + √x² + y²)
We see that P, Q, and R form the vertices of a right triangle, with Q as the vertex of the right angle. Furthermore, PQ = x and QR = y are the legs of the right triangle, and RP is the hypotenuse of the right triangle.

Thus, if we let RP = n, then by the Pythagorean theorem we have:

n^2 = x^2 + y^2

n = √( x^2 + y^2)

Since time = distance/rate, it takes y/z hours to drive from Q to R and √( x^2 + y^2)/z hours to drive from R to P.

So, the total driving time is [y + √( x^2 + y^2)]/z.

Answer: C

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