SAT One-Variable Data Question 48 | Free SAT Prep | Aniko

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Question 48·Hard·One-Variable Data Distributions; Measures of Center and Spread

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A set of 9 distinct integers has a mean of $12$ and a median of $10$ . When the greatest integer in the set is removed, the mean of the remaining 8 integers decreases to $10$ and the range decreases by $9$ .

What is the greatest integer in the original set?

For SAT questions involving mean and removing or adding numbers, immediately convert each mean into a total sum by multiplying mean × number of values. Then compare the totals: the difference between the original total and the new total is exactly the value that was removed (or added). Only after you have that value do you briefly check extra information like median or range to ensure it’s consistent, but don’t let that extra data distract you from the quick, direct calculation using sums.

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Turn means into totals

Use $\text{mean} = \dfrac{\text{sum}}{\text{number}}$ . What is the total sum of the 9 integers if their mean is $12$ ? What is the total sum of the remaining 8 integers if their mean is $10$ ?

Connect the two sums

Let $G$ be the greatest integer that gets removed. How are the sum of all 9 integers, the sum of the remaining 8 integers, and $G$ related by a simple subtraction?

Think about what was removed

Once you write an equation involving the two sums and $G$ , solve for $G$ . You do not actually need to use the median or range to compute $G$ ; they just provide consistency checks.

Desmos Guide

Compute the two sums from the means

In one expression line, type 9*12 to see the total of the 9 integers. In another line, type 8*10 to see the total of the remaining 8 integers.

Find the removed (greatest) integer

In a new line, type 9*12 - 8*10. The value Desmos displays for this expression is the integer that was removed, which is the greatest integer in the original set.

Step-by-step Explanation

Use the mean of 9 integers to find their total

"Mean" (average) is defined as

\text{mean} = \frac{\text{sum of values}}{\text{number of values}}.

You are told that a set of 9 distinct integers has a mean of $12$ . So the sum of all 9 integers is

9 \times 12 = 108.

You are also told the median is $10$ , which means that if the integers are ordered from least to greatest, the 5th number is $10$ . (We don’t actually need the median to compute the greatest integer, but it is consistent information.)

Use the new mean after removing the greatest integer

When the greatest integer is removed, there are now 8 integers left, and their mean is $10$ .

Using the same mean formula,

\text{sum of remaining 8 integers} = 8 \times 10 = 80.

Relate the two sums to the removed integer

Originally, the sum of all 9 integers was $108$ . After removing the greatest integer (call it $G$ ), the sum of the remaining 8 integers is $80$ .

So the removed integer must account for the difference between these two sums:

108 - 80 = G.

Now we just need to evaluate this difference to find $G$ . (Notice we still haven’t used the range information; we’ll see that it checks out after we find $G$ .)

Compute the greatest integer and check the range condition

Compute the difference:

G = 108 - 80 = 28.

So the greatest integer in the original set is $28$ .

As a quick check using the range information: let the original maximum be $28$ and the original minimum be $m$ . The original range is $28 - m$ . After removing $28$ , the new maximum is the second-largest number. Because the range decreases by $9$ , the new maximum must be $28 - 9 = 19$ , and the new range is $19 - m$ , which is indeed $9$ less than the original range.

Therefore, the greatest integer in the original set is $28$ , which corresponds to answer choice C.

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation