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Kaplan GMAT Sample Problem: Advanced Rates

May 4, 2011 by

Try this advanced sample GMAT problem-solving question focusing on rates.  Remember, as complicated as the problems get, the key is using the distance = rate x time formula for all of the specific distances travelled.

Problem:

Shannon and Maxine work in the same building and leave work at the same time.  Shannon lives due north of work and Maxine lives due south.  The distance between Maxine’s house and Shannon’s house is 60 miles.  If they both drive home at the rate 2R miles per hour, Maxine arrives home 40 minutes after Shannon.  If Maxine rides her bike home at the rate of R miles per hour and Shannon still drives at a rate of 2R miles per hour, Shannon arrives home 2 hours before Maxine.  How far does Maxine live from work?

 

A) 15

B) 20

C) 30

D) 40

E) 45

 

Solution:

To solve for the distance between Maxine’s house and the building in which she works, we must first translate our question into a series of equations.  Remember that for each leg of a journey, we can use the distance = rate x time relationship.  First, we know that the distance between Maxine and Shannon’s houses is 60 miles.  If we call the distance between Maxine’s house & work DM, and the distance between Shannon’s house and work Ds, we can write the equation DM + DS = 60.  We can also write equations for the three types of trips.  For Shannon’s drive, we can say that DS = 2RT.  For Maxine’s drive, we can say that DM = 2R(T + 2/3) – Shannon’s time, which is T, plus 2/3 of an hour.  For Maxine’s bike ride, we can say DM = R(T + 2).  From these equations we can solve for DM.

DM = R(T + 2) and DM = 2R(T + 2/3), so R(T + 2) = 2R(T + 2/3). This can be simplified to RT + 2R = 2RT + 4R/3, which in turn becomes 3RT + 6R = 6RT + 4R.  We can then simplify even further to say 3RT = 2R, 3T = 2, T = 2/3.  Therefore, we know that it takes Shannon 2/3 of an hour to drive home.

Next, we can use the fact that Ds = 60 – DM.  This can be substituted into the equation Ds = 2RT, giving us 60 – Dm = 2RT.  Because we know that T = 2/3, we know that 60 = DM + 2(2/3)R.  We also know that DM = R(T + 2).  Again keeping in mind that T = 2/3, we can substitute and say that 60 = R(2/3 + 2) + 2(2/3)R.  We then solve for R:

60 = R(2/3 + 2) + 2(2/3)R

60 = 2R/3 + 2R + 4R/3

180 = 2R + 6R + 4R

180 = 12R

R = 15.

Once we know R is 15, we can plug back into the equation DM = R(T + 2), which becomes DM = 15(2/3 + 2) = 10 + 30 = 40.  Therefore, choice (D) is the correct answer.

 

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