The length of rectangle $B$ is $10$ percent less than the length of rectangle $A$, and the width of rectangle $B$ is $10$ percent greater than the width of rectangle $A$ .

Quantity A |
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | Quantity B |

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | ||

The area of rectangle $A$ | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | The area of rectangle $B$ |

- Quantity A is greater.
- Quantity B is greater.
- The two quantities are equal.
- The relationship cannot be determined from the information given.

So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 6 of the second Quantitative section of Practice Test 1. Those questions testing our knowledge of **Quadrilaterals** can be kind of tricky, but never fear, PrepScholar has got your back!

## Survey the Question

Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on our paper.

Since we’ll be comparing **areas of rectangles**, we should expect to use what we’ve learned about **Quadrilaterals** here. Let’s keep what we’ve learned about this skill at the tip of our minds as we approach this question.

## What Do We Know?

Let’s carefully read through the question and make a list of the things that we know.

- The length of rectangle $B$ is $10$ percent less than the length of rectangle $A$
- The width of rectangle $B$ is $10$ percent more than the width of rectangle $A$
- We want to compare the areas of rectangles $A$ and $B$

## Develop a Plan

We want to compare the areas of two rectangles, so let’s start by recalling the formula for the area of a rectangle

$$\Area \of \a \Rectangle = \Length ·\Width$$

Next, we know that the length of rectangle $B$ is $10$ **percent less** than the length of rectangle $A$ . We need to translate this into a math equation, but first let’s remind ourselves of the formula for calculating a new value from a percentage change.

$$\New \Value = (1+{\Percent \Change}/100)·\Original \Value$$

In this formula, percent change is positive for a percent increase, and negative for a percent decrease. Using this formula, we can translate the relationship given between the lengths of the rectangles into a math equation:

$\New \Value$ | $=$ | $(1+{\Percent \Change}/100)·\Original \Value$ |

$ $ | $ $ | |

$\Length \of B$ | $=$ | $(1+{-10\;}/100)·\Length \of \A$ |

$ $ | $ $ | |

$\Length \of B$ | $=$ | $(1-0.1)·\Length \of \A$ |

$ $ | $ $ | |

$\Length \of B$ | $=$ | $0.9·\Length \of \A$ |

We also know that the width of rectangle $B$ is $10$ percent more than the width of rectangle $A$. Let’s translate this into a math equation, just like how we did with the lengths of the rectangles.

$\New \Value$ | $=$ | $(1+{\Percent \Change}/100)·\Original \Value$ |

$ $ | $ $ | |

$\Width \of B$ | $=$ | $(1+{10\;}/100)·\Width \of \A$ |

$ $ | $ $ | |

$\Width \of B$ | $=$ | $(1+0.1)·\Width \of \A$ |

$ $ | $ $ | |

$\Width \of B$ | $=$ | $1.1·\Width \of \A$ |

Perfect! Looks like we have everything we need to calculate the areas of both rectangles in terms of the dimensions of rectangle $A$, so let’s do that right now.

## Solve the Question

In terms of the dimensions of rectangle A, the area of rectangle A is:

$$\Area_A = \Length_A·\Width_A$$

And the area of rectangle $B$, expressed in terms of the dimensions of rectangle $A$, is:

$\Area_B$ | $=$ | $\Length_B·\Width_B$ |

$ $ | $ $ | |

$\Area_B$ | $=$ | $0.9·\Length_A·1.1·\Width_A$ |

$ $ | $ $ | |

$\Area_B$ | $=$ | $0.9·1.1·\Length_A·\Width_A$ |

$ $ | $ $ | |

$\Area_B$ | $=$ | $0.99·\Length_A·\Width_A$ |

Well that’s interesting. Looks like the area of rectangle $B$ is the same as the area of rectangle $A$ EXCEPT that it is multiplied by $0.99$. And multiplying a value by $0.99$ decreases its magnitude, so the area of rectangle $B$ is slightly smaller than the area of rectangle $A$. So **the correct answer is A, Quantity A is greater**.

## What Did We Learn

While it would be nice and convenient if we could just add and subtract multiple percent changes, as we just saw, they don’t work that way. We can easily imagine someone believing that the two quantities would be the same, but it turns out that decreasing the length by $10$ percent and increasing the width by $10$ percent does NOT result in the area of the rectangle staying the same.

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