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The Table Tactic for Rates

by Economist GMAT on May 6th, 2012
11 comments
Click here to learn more about The Economist GMAT Tutor and its products & services.
Posted in
  • Data Sufficiency
  • GMAT Math
  • Problem Solving
  • Speed/Distance
  • Word Problems
  • Work/Rate

Rate problems are not intrinsically difficult. Fundamentally, the only formula you need to understand is the simple rate-time-work equation:

Rate × Time = Work

Yet in spite of its seeming simplicity, GMAT rate questions are rarely so easy to solve. Instead, GMAT test-makers go to extreme lengths to obfuscate the data provided for each problem, usually by presenting the information in the form of a long and grueling word problem. The onus is on the student to sort out the resulting jumble:

Martin and Wood are hired to perform a copywriting job. Working alone, Martin could finish 2/3 of the the job in 15 days, and if Wood would work alone, he could finish half the job in 9 days. If Martin and Wood work together, how many days would it take them to finish the entire job?

Chances are good that without a systematic approach, you will quickly get bogged down by three common errors:

  1. Forgetting a vital piece of information: The problem then becomes unsolvable.
  2. Re-reading the question stem multiple times: This wastes valuable time
  3. Struggling to retrace your steps: Double-checking your work becomes impossible

Master GMAT’s solution for dealing with chaotic word problems is to organize the information into a table. This technique allows us to approach each problem systematically to reduce careless mistakes and to solve the problem quickly.

Let’s set up one such table which we’ll use for this problem. Label the headers: Rate, Time, and Work:

A Generic Rates Table

Before we fill in the rows of the table, let’s use a simple trick to avoid dealing with fractions.

When dealing with rates problems, look to see what the units of work are. Often, if the work is a job or a project, students choose to use the value “1″ for work. The problem with this approach is it that you will end up with lots of messy fractions when calculating the rate and the time. So instead, let’s plug-in a number for the work that will be multiple by all the numbers mentioned in this problem.

This plug-in will contain all the numbers you see in the question stem as factors. We have two fractions in the question stem (2/3 and 1/2) and two integers (15 and 9). A good number, then, be a multiple of all four integers, such as 2·9·15 = 2·135 = 270. So for this problem, we’ll pretend that the job is actually 270 pages of copywriting.

Our next step is to translate each clause in the question stem into a row in our table:

“Working alone, Martin could finish 2/3 of the the job in 15 days“: This corresponds to 2/3 × 270 = 180 for the work, and 15 for the time

“if Wood would work alone, he could finish half the job in 9 days“: Fill in 1/2 × 270 = 135 for the work, and 9 for the time

Rates Table with First Two Rows

To calculate the rate for each employee, we simply use the rate-time-work equation. For example, for Martin working alone,

Rate x 15 = 180, so
Rate = 180/15 = 12

Using the same equation, we get a rate of 15 for Wood alone. Use these values to fill in the left-most cell of each row.

“If Martin and Wood work together, how many days would it take them to finish the entire job?” — The combined rate is the sum of the two individual rates, or 27, and the work is 270.

Rates Table Filled In

We then look for the amount of time it takes the duo to work together:

27 x Time = 270
Time = 270/27 = 10


The table tactic is all you need to solve any speed or rate problem on the GMAT. To test your understanding, here is a challenging “gap” problem, similar to one you may see on the GMAT. Try it using the table tactic:

Sari and Ken climb up a mountain. At night, they camp together. On the day they are supposed to reach the summit, Sari wakes up at 06:00 and starts climbing at a constant pace. Ken starts climbing only at 08:00, when Sari is already 700 meters ahead of him. Nevertheless, Ken climbs at a constant pace of 500 meters per hour, and reaches the summit before Sari. If Sari is 50 meters behind Ken when he reaches the summit, at what time did Ken reach the summit?

Hints:

  1. The equation Speed x Time = Distance is analogous to Rate x Time = Work
  2. Treat the “gap” that is being opened/closed as an object deserving its own row

If you liked this article, let Economist GMAT know by clicking Like.

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11 comments

  • Robin on May 8th, 2012 at 5:16 am

    "Sari and Ken climb up a mountain. At night, they camp together. On the day they are supposed to reach the summit, Sari wakes up at 06:00 and starts climbing at a constant pace. Ken starts climbing only at 08:00, when Sari is already 700 meters ahead of him. Nevertheless, Ken climbs at a constant pace of 500 meters per hour, and reaches the summit before Sari. If Sari is 50 meters behind Ken when he reaches the summit, at what time did Ken reach the summit?"

    Could you solve the problem with the table tactics? I tried but failed.

    Reply to this comment
    • Asmita on May 8th, 2012 at 9:28 pm

      Do you have the answer to this? I want to verify my answer. 

  • Darren on May 8th, 2012 at 7:50 am

    I solved this question in two ways. I was unable to immediately "see" the author's suggested approach until after solving via a more complex path.

    Initial approach:
    Determined rate for Sari as 700m/2hr=350m/h
    Ken's time = T
    Sari's time = (T+2)

    Equation to solve:
    (550T) - 50 = 350 * (T+2)
    550T - 50 = 350T + 700
    150T = 750
    T=5
    Answer = Ken's start plus 5 = 8am+5hrs = 1pm

    Shorter Approach (Gap):

    Rate = Ken's "catch-up" rate = 150m/hr
    Distance = initial distance plus his lead = 700+50 = 750

    150T = 750
    T = 5

    Reply to this comment
    • Robin on May 12th, 2012 at 7:00 am

      Hello there,
      if you could explain this part:
      "Equation to solve:
      (550T) - 50 = 350 * (T+2)"

      Thanks for your assistance.

  • Vineet on May 8th, 2012 at 5:18 pm

    Another easy approach to the problem which I was able to solve in 30 seconds. 

    "Ken starts climbing only at 08:00, when Sari is already 700 meters ahead of him" - You should instantaneously calculate Sari's rate of climb which is 350 m/hr

    Create a table with Sari's and Ken's altitude at each hour. The time when Ken reaches summit will be when Ken is 50 meters ahead of Sari. 

    Reply to this comment
    • Robin on May 12th, 2012 at 7:07 am

      Hello there,
      Could you show the solution in details?

  • Vineet on May 9th, 2012 at 2:13 am

    I still find the method of taking 1 as work, reliable; maybe because I am too used to it now. Table is good though.

    Reply to this comment
  • Aaron on May 11th, 2012 at 4:38 pm

    This was definitely a challenging problem, wasn't it? Let's walk through it together now. First, let's break the problem into two stages:

    Stage I: Sari climbs 700 meters alone from 06:00 to 08:00 while Ken sleeps. 700 meters in two hours places Sari's rate at 350 m/hr. This helps "open" up a gap — the distance between Sari and Ken — which they will later help close.

    Stage II: Although Sari climbs at 350m/hr, Ken starts climbing at 500m/hr to "close" the gap between him and Sari. Eventually, he "closes" the 700m gap, but then opens a new 50m gap when he reaches the summit. All being said, the total size of the gaps is 750m.

    Treat the gap as an object of its own; it deserves its own row in the table. The "distance" of the gap is 700m+50m = 750m, and its speed (the rate at which it is being closed) is 500m/hr − 350m/hr = 150m/hr.

    Draw up the table with 3 rows: Sari, Ken, and the "Gap" row.

    To calculate how long it took for Ken to reach the summit, we simply apply the equation to the last row to find the time:

    Speed × Time = Distance
    150m/hr × Time = 750m
    Time = 750m ÷ 150m/hr
    Time = 5hr

    Since Ken started climbing at 08:00, he will reach the summit 5 hours later, at 13:00 (1PM).

    Reply to this comment
  • Robin on May 12th, 2012 at 7:05 am

    Thanks for the details. It was really helpful.
    Could you solve the math as exactly as the table tactic shown in the article?

    Reply to this comment
  • Ihsan Saleheen on June 3rd, 2012 at 7:54 am

    Thanx to everyone sharing their own tactics

    Reply to this comment
  • mangal bajracharya on July 3rd, 2012 at 1:57 am

      for the first question given above, we can solve this way also,
        martin can do 2/3 work in 15 days.
         he can do 2/3*1/15 work in 1 day .
       so, he can do 2/45 work in 1 day .
     then,
       wood can do 1/2 work in 9 days.
     wood can do 1/2*1/9 work in 1 day.
     so, wood can do 1/18 work in 1 day.
     combining 1 day's work of each 
     we can say ,
       martin and wood can do 2/45 +1/18 work in 1 day.
     therefore, they can do 1/10 work in 1 day together .
     so they can complete 1 work in 10 days working together. 

    Reply to this comment

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