I’m going to do a number of these posts. You can think of them as commentary on the 2016 *Official Guide for GMAT Review.* I’m going to use the abbreviations PS, DS, RC, CR, and SC for Problems Solving, Data Sufficiency, Reading Comprehension, Critical Reasoning, and Sentence Correction. Sometimes I’ll present alternative solutions, more detailed solutions, and, occasionally an example problem.

I can’t reproduce the complete problem here, so if you don’t have your official guide handy, now would be the time to break it out – page 167

*Working simultaneously at their respective constant rates, Machines A and B produce 800 nails…*

This is a classic combined rate problem with a VIC (variables in choices) twist. The algebraic solution isn’t crazy hard, so I’ll start with that as an appetizer, and of course I’ll show EVERY SINGLE STEP. For the main course I’m going to use back-solving to send the problem home crying to its mama “Mommy, that guy at Owl Test Prep hurt me.”

Let’s get started. First it’s important to know that the combined rate is the sum of the rates. In other words

*Rate of A and B together = Rate of A + Rate of B*

Based on the time – \(x\) hours – we can find the combined rate of A and B. By the same reasoning we can find the rate of machine A.. I’m a big fan of setting up rate problems with \( Work = Rate\times{Time}\) or \(Dsitance = Rate\times{Time}\) tables.

\(Work = \) | \(Rate\) | \(\times\) | \(Time\) |
---|---|---|---|

\(800 = \) | \(Rate_{ab}\) | \(\times\) | \(x\) |

\(800 = \) | \(Rate_{a}\) | \(\times\) | \(y\) |

\(800 = \) | \(Rate_{b}\) | \(\times\) | \(T\) |

A little bit of algebra gives us

\(Work = \) | \(Rate\) | \(\times\) | \(Time\) |
---|---|---|---|

\(800 = \) | \(\Large\frac{800}{x}\) | \(\times\) | \(x\) |

\(800 = \) | \(\Large\frac{800}{x}\) | \(\times\) | \(y\) |

\(800 = \) | \(Rate_{b}\) | \(\times\) | \(T\) |

Because combined rate is the sum of the rates, we know

\(\Large\frac{800}{x}\) \(=\) \(\Large\frac{800}{y}\)\( + \) \(Rate_{b}\)

\(\Large\frac{800}{x}\) \(–\) \(\Large\frac{800}{y}\) \(=\) \(Rate_{b}\)

The common denominator is \(xy\):

\(\Large\frac{800y}{xy}\) \(–\) \(\Large\frac{800x}{xy}\) \( = \) \( Rate_{b}\)

\(\Large\frac{800y – 800x}{xy} \) \(=\) \(Rate_{b}\)

Let’s factor our that 800:

\(\Large\frac{800(y – x)}{xy} \) \(=\) \(Rate_{b}\)

Now we can stick this back into the table to find the time it takes machine B to produce 800 nails:

\(Work = \) | \(Rate\) | \(\times\) | \(Time\) |
---|---|---|---|

\(800 = \) | \(\Large\frac{800(y – x)}{xy}\) | \(\times\) | \(T\) |

To solve for \(T\) we multiply by the reciprocal of \(R_{b}\)

\(\Large\frac{800xy}{800(y – x)}\) \( = \) \(T\)

Cancel the \(800\) and we get

\(T =\) \(\Large\frac{xy}{y – x}\)

The correct answer is E.

**BUT WE CAN DO BETTER!!!**

I think Pei Mai would agree with me when I tell you that the GMAT should fear you, not the other way around.

Because this is a VIC, it might be a good idea to try back-solving. Let’s pick some numbers and see what happens.

I’m going to go with \(x = 2\) and \(y = 4\) (be careful – \(y\) must be greater than \(x\) because \(x\) is the time they take together and will be less than the time one machine takes alone). Now the table looks like this:

\(Work = \) | \(Rate\) | \(\times\) | \(Time\) |
---|---|---|---|

\(800 = \) | \(400\) | \(\times\) | \(2\) |

\(800 = \) | \(200\) | \(\times\) | \(4\) |

\(800 = \) | \(Rate_{b}\) | \(\times\) | \(T\) |

We calculate \(R_{b}\) the same way we did before, but it’s a little bit easier this time:

\(R_{b} = 400 – 200 = 200\)Now let’s solve for \(T\)

\(800 = 200\times{T}\)Divide by \(200\) and we get \(T = 4\)

Now we plug our picks for \(x\) and \(y\) into the answer choices. The target number (the result we’re looking for) is 4. Remember that you have to check all of the answers because you might have picked some unfortunate numbers that produce two matches.

\((A) = \Large\frac{2}{2 + 4}\) Nope.

\((B) = \Large\frac{4}{2 + 4}\) Nope.

\((C) = \Large\frac{2\times{4}}{2 + 4}\) Nope.

\((D) = \Large\frac{2\times{4}}{2 – 4}\) Nope.

\((E) = \Large\frac{2\times{4}}{4 – 2}\) Yep!

Not bad, right?

Keep in mind that I’m laying it all out for you step by step. You could skip a lot of these steps and eyeball the answer choices – Fraction, Fraction, Fraction, Negative, No need to check – it’s got to be E.