A dataset S consists of 9 distinct integers. The median of S is 20, and the mean of S is 2… — GRE General Test

A dataset $S$ consists of $9$ distinct integers. The median of $S$ is $20$ , and the mean of $S$ is $24$ . The mean of the $4$ smallest integers in $S$ is $10$ . If the largest integer in $S$ is removed to form a new dataset $R$ consisting of $8$ integers, what is the maximum possible median of dataset $R$ ?

A
$18.5$
B
$19.0$
$19.5$ Cevap
D
$20.0$
E
$20.5$

Cevap

The maximum possible median of dataset

R

19.5

In the sorted set of

9

distinct integers, the median is the

5

th element, which is

20

. Removing the largest value leaves

8

integers, whose median is the average of the

4

th and

5

th elements. Because all integers are distinct, the

4

th element can be at most

19

. The maximum median is therefore

\frac{19 + 20}{2} = 19.5

Adım Adım Çözüm

Identify the position of the median in the original dataset

S

In a dataset of

9

sorted distinct integers

x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9

, the median is the

5

th element, so

x_5 = 20

For an odd number of ordered elements

n=9

, the median is at index

\frac{9+1}{2} = 5

Determine the formula for the median of the modified dataset

R

Removing the largest element

x_9

leaves

8

ordered elements

x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8

. The median of

R

\frac{x_4 + x_5}{2} = \frac{x_4 + 20}{2}

For an even number of ordered elements

n=8

, the median is the average of the two middle terms at indices

4

and

5

Maximize the value of

x_4

Since all integers are distinct and

x_5 = 20

x_4

must be an integer strictly less than

20

. Thus, the maximum integer value

x_4

can take is

19

Maximizing

x_4

maximizes the average

\frac{x_4 + 20}{2}

Verify that

x_4 = 19

is achievable under the given mean constraint.

The sum of the smallest

4

integers is

4 \times 10 = 40

. With

x_4 = 19

, we require

x_1 + x_2 + x_3 = 21

. Choosing distinct integers such as

x_1 = 1

x_2 = 2

, and

x_3 = 18

satisfies

1 + 2 + 18 = 21

and

x_1 < x_2 < x_3 < x_4

Confirming feasibility ensures the upper bound

19

is valid.

Calculate the maximum median of dataset

R

\frac{19 + 20}{2} = 19.5

Substitute the maximum value of

x_4 = 19

into the median expression.

Anahtar Kavram

Properties of median for odd vs. even datasets and optimization under distinctness constraints

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