Ray is $2$ inches taller than Lin, and Ray is $3$ inches taller than Sam.

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

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

The average (arithmetic mean) height of Ray, Lin, and Sam | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | The median height of Ray, Lin, and Sam |

- 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 3 of the second Quantitative section of Practice Test 1. Those questions testing our knowledge of **Numerical Methods for Describing Data** 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.

This question asks us to compare the **average** and the **median** heights of three individuals, both quantities we learned how to calculate from the **Numerical Methods for Describing Data** GRE math skill. 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.

- Ray is $2$ inches taller than Lin
- Ray is $3$ inches taller than Sam
- We want to compare the average to the median of the heights for these three people

## Develop a Plan

While we could use a variable to represent the height of one of these people and then calculate the heights of the others,

let’s go ahead and just randomly choose a number that we can plug in to work with. This should make it easier to think about this problem. So let’s arbitrarily choose Ray to be… $63$ inches tall. Works as good as any other number!

We can next use the algebraic sentences in this question to find the heights of Lin and Sam, and then afterwards we’ll calculate the average and medians of their heights.

## Solve the Question

The question tells us that “Ray is 2 inches taller than Lin.” Since we chose Ray to be **$63$ inches tall**, then Lin must be $63-2$ inches tall, or **$61$ inches tall**. Similarly, if “Ray is $3$ inches taller than Sam,” then Sam must be $63-3$ inches tall, or **$60$ inches tall**.

To find the average of a list of numbers, we take their sum and divide by the number of values. Let’s calculate the average of these heights right now.

$\Average$ | $=$ | ${\Sum}/{\Number \of \Values}$ |

$ $ | $ $ | |

$\Average$ | $=$ | ${60+61+63}/3$ |

$ $ | $ $ | |

$\Average$ | $=$ | $184/3$ |

$ $ | $ $ | |

$\Average$ | $=$ | $61.333$ |

So **the height is $61.333$ inches**. To find the median of a list of numbers, we must put them in order from least to greatest, then choose the number in the middle. Putting the heights in order from least to greatest, we get:

$$60, 61, 63$$

The number in the middle is $61$, so we know that **the median is $61$ inches**.

Since the average is greater than the median, **the correct answer is A, Quantity A is greater**.

## What Did We Learn

We shouldn’t hesitate to put in a trial number for a specific variable in a question if: 1) it’ll help us solve the question easier and 2) it won’t affect the final answer. From what we know about **Numerical Methods for Describing Data** we can shift all of the values in a list of numbers up or down by the same amount without affecting whether the average or the median of that list is greater. Choosing a height for Ray definitely made the problem easier to work with, as we didn’t have to think about averages and medians that contained a variable.

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