4

I have a question based on the GMAT exam regarding work rate problems. In particular, problems that involve multiple entities having different work rates, but those work rates are defined relative to each other. My goal is to find a general algebraic solution to solve problems like this.

Here's an example question:

Machine A is twice as fast as Machine B in completing a task. Together they can complete the task in 10 hours. How long will it take each of the machines to complete the task separately?

To solve general work rate problems, most GMAT books use the formula below like so:

$\frac{1}{A} + \frac{1}{B} = \frac{1}{t}$

where A and B are the work rates of two separate machines required to complete a task. This is simple enough when we have a question like

$\text{A takes 2 hours to complete a task. B takes 3 hours to complete a task. How many hours does it take for A and B completed to task together?}$

The solution is simply:

$\frac{1}{2} + \frac{1}{3} = \frac{1}{t}$

Solving for t gives $\frac{6}{5}$ hours.

However, I'm probably missing something when it comes to related rates like the first question above.

My approach is as below:

  1. Define A in terms of B.
  2. Then solve using the formula provided above.

The problem I face is translating (1) to (2).

If Machine A is twice as fast as Machine B, then it should be:

$\text{A = 2B}$.

Substituting this into the work rate formula yields:

$\frac{1}{2B} + \frac{1}{B} = \frac{1}{10}$.

Solving for B gives 15 hours. Taking this result into $\text{A} = 2B$ gives 30 hours, which is the inverse result I'm looking for.

Instead, the solution should be $\text{A = 15 hours, and B = 30 hours.}$ Pretty sure it's something subtle, but would appreciate some greater clarity in approaching problems like these.

N. F. Taussig
  • 76,571
John
  • 43
  • 3
    I think that in the course of memorising formulas for an exam, you have lost sight of the logic of the problem itself. Remind yourself: what are A and B? Are they the times taken by A and B, or the work speed of A and B? And if they represent time, does being twice as fast not mean that the time taken by A is HALF, and not double, the time taken by B, so that A=B/2? – insipidintegrator Dec 10 '22 at 12:17
  • Thanks a bunch for the clarification! You're right, I was trying to rely too much on a mechanical solution instead of problem solving. Will need more practice! – John Dec 10 '22 at 13:13
  • Regarding the comment by @insipidintegrator about losing sight of the underlying mathematical principles, my answer to Combined work problem without joined time spent in total might be of interest. However, given that it's gotten 0 upvotes in the past 3 years, it apparently didn't go over all that well in MSE . . . – Dave L. Renfro Dec 10 '22 at 13:41

1 Answers1

3

Since machine A is twice as fast as machine B, machine B will take twice as long as machine A to perform the same task. For instance, if machine A can perform a task in one hour, then machine B will take two hours; if machine A can perform a task in two hours, then machine B will take four hours.

Thus, if $A$ is the amount of time machine $A$ takes to perform a task and $B$ is the amount of time machine $B$ takes to perform the same task, then $B = 2A$. Hence, you should have $$\frac{1}{A} + \frac{1}{2A} = \frac{1}{10~\text{h}}$$ If you solve that equation for $A$, you should obtain $A = 15$ hours, so $B = 2A = 30$ hours.

N. F. Taussig
  • 76,571