Often, the hardest part of a GMAT quantitative problem is taking all the information and organizing it in a meaningful way so that you can actually **start** the math part of the problem. (How many of you have faced this on the Grizzly Peak problem in the Arithmetic lecture?)

Let’s look at a particular type of problem that’s common on the test: the multiple rate problem.

We have three basic pieces of any rate problem: Rate, Time, and Distance (or work).

I’ll show you this technique using an example.. (It’s VERY much like the Work/Rate challenge problem from page 123 in the Word Problems text)

Albert takes 8 hours to complete a certain job and starts that job at noon.

After 2 hours working alone, Albert is joined by Bob and together they complete the job at 4pm. What is Bob’s rate working alone?

Here’s our chart:

Rate | Time | Work (or Distance) | |

Albert | |||

Bob |

Now, Albert’s rate is 1/8, since we always take our time to complete the job and put it under 1. You learned this in the lecture.

Bob’s rate is 1/B because of the same logic.

The time Albert works is 4 hours, and Bob, since he starts 2 hours later, has a time of 4-2, or 2 hours. Let’s put those into our chart:

Rate | Time | Work (or Distance) | |

Albert | 1/8 | 4 | |

Bob | 1/B | 2 |

Now, remembering our formulas, we know that Rate * Time = Work (or distance), so Albert’s rate times Albert’s time should give us Albert’s work: 4 * 1/8 = 4/8 = 1/2.

Bob’s rate times Bob’s time should give us Bob’s work: 1/B * 2 = 2/B

So let’s put those into the chart as well.

Rate | Time | Work (or Distance) | |

Albert | 1/8 | 4 | 1/2 |

Bob | 1/B | 2 | 2/B |

Total: | 1 |

You’ll notice here I’ve added another row at the bottom for our total. We know that the total amount of work done is 1 job, so I’ve put a 1 in the bottom right square.

This gives us our formula:

1/2 + 2/B = 1

Solving for B gives us B=4 here, and since our rate is 1/B, our answer is ¼.

Let’s try it on one more example:

Trains A and B are traveling toward each other from two cities 200 miles apart. Train A has a speed of 40 miles per hour and Train B has a speed of 60 miles per hour.

If Train A leaves 3 hours before train B, how many miles will B have traveled when they meet?

Here’s our chart setup:

Rate | Time | Distance | |

Train A | |||

Train B | |||

Now, plug in what we know:

Rate | Time | Distance | |

Train A | 40 | T+3 | 40T + 120 |

Train B | 60 | T | 60T |

Total: | 200 |

I’ve assigned T+3 to the time for Train A, and T for the time for Train B here. It’s easiest if we stick with the smallest number of variables possible. (Also note that I’m assigning T to Train B here, since we’ll be solving for Train B’s distance later. We could also put T in for Train A, and then T-3 for Train B, of course, but this adds complication in the final step.)

Now, the distance traveled by Train A is R*T, or 40(T+3), which is 40T + 120.

The distance traveled by Train B is also R*T, or 60T here.

You can fill these in on our chart in the distance column.

We also know that the collective distance covered by our trains is 200 miles. Here’s our basic algebra formula from our information:

40T + 120 + 60T = 200

100T + 120 = 200

100T = 80

T = 80/100 = 4/5 hour

Train B will have traveled 60 * 4/5 miles or 48 miles when the two meet.

Keep in mind that in a basic distance or work problem, there’s no need for a chart, but when you have multiple times and multiple travelers or workers, it’s often helpful to keep our information in a simpler format, and it will result in fewer formula errors in the end. I’ve found that my chart layout solves about 90% of my rate problems when they’re more complicated than your basic R*T=D setup.

Good luck!

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*Valerie Browning has been teaching GMAT for Veritas Prep for 10 years. After graduating from the McCombs School of Business at UT Austin, Valerie is now based in Houston. Since graduating, she has been interviewing applicants to McCombs as an alumni volunteer.*