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In a music contest, a panel of 100 audience members ranked each of 4 performers with integer ranks–from 1 to 4. For each performer, the table indicates the number of audience members who assigned each rank to the given performer. For example, exactly 49 audience members assigned rank 4 to Performer A. Two methods–the Mode Method and the Median Method–are used to assign final ranks to the performers based on these audience rankings.
Mode Method: Assign to each performer a final rank equal to the audience rank the performer received most often.
Median Method: Compute the median audience ranking for each performer. The ordering of the final rankings will be the same as the ordering of these medians.
| Rank | ||||
|---|---|---|---|---|
| Performer | 1 | 2 | 3 | 4 |
| A | 51 | 0 | 0 | 49 |
| B | 23 | 5 | 72 | 0 |
| C | 5 | 95 | 0 | 0 |
| D | 21 | 0 | 28 | 51 |
For each of the following final ranks, select Same if the two final ranking methods assigned the same performer the given rank. Otherwise, select Not same.
Rank 1
Rank 2
Rank 3
Let's start by understanding the dataset with the intention of "owning" it completely. We have a table showing how audience members ranked four performers (A, B, C, D) on a scale of \(1\text{-}4\), with percentages indicating what portion of the audience gave each performer a particular rank.
When we examine this ranking data, we immediately notice a striking pattern:
Key insight: Each performer has one rank that received more than \(50\%\) of the audience votes. This observation will be crucial for our analysis because it creates a mathematical certainty about both the mode (most common value) and median (middle value).
Statement 1 Translation:
Original: "For Performer A, the mode of audience rankings is [greater than/less than/the same as] the median."
What we're looking for:
In other words: Is the most common rank for Performer A different from the middle rank?
Now that we understand what we need to find, let's use our key insight about the dataset. For Performer A, \(51\%\) of the audience gave rank \(1\). This means:
Why this works: When more than half of the values in a dataset are the same, that value must be the median. To visualize this, imagine lining up all \(100\) audience members by the rank they gave. If \(51\) people gave rank \(1\), then when lined up in order, the \(50\mathrm{th}\) position (the median) must be rank \(1\).
Therefore, for Performer A, the \(\mathrm{mode} \,(1) = \mathrm{median}\, (1)\).
Answer for Statement 1: Same
Statement 2 Translation:
Original: "For Performer B, the mode of audience rankings is [greater than/less than/the same as] the median."
What we're looking for:
In other words: Is the most common rank for Performer B different from the middle rank?
For Performer B, we see that \(72\%\) of the audience gave rank \(3\). This is an even stronger majority than we saw for Performer A.
Teaching callout: Notice how we didn't need to calculate anything or arrange values in order. The mathematical property that "when a value appears more than \(50\%\) of the time, it must be the median" allows us to make this determination instantly.
Therefore, for Performer B, the \(\mathrm{mode}\, (3) = \mathrm{median}\, (3)\).
Answer for Statement 2: Same
Statement 3 Translation:
Original: "For Performer C, the mode of audience rankings is [greater than/less than/the same as] the median."
What we're looking for:
In other words: Is the most common rank for Performer C different from the middle rank?
For Performer C, we observe the most dramatic majority: \(95\%\) of the audience gave rank \(2\).
Strategic insight: The pattern holds for all performers, but Performer C shows it most clearly. With such an overwhelming majority (\(95\%\)), it's almost impossible for the median to be anything other than \(2\).
Therefore, for Performer C, the \(\mathrm{mode}\, (2) = \mathrm{median}\, (2)\).
Answer for Statement 3: Same
Let's compile our findings for all three statements:
Answer: All three statements are Same
Remember: On GMAT Table Analysis questions, before diving into calculations, always look for dominant patterns that might create mathematical shortcuts! The ability to recognize when more than half of a dataset shares the same value instantly tells you that both the mode and median must equal that value.
Rank 1
Rank 2
Rank 3