In a set of five consecutive integers, which of the

[AAPL]

Princeton Review

In a set of five consecutive integers, which of the following must change the average of the set without changing its original median?

A. Multiplying each of the numbers in the set by 6.
B. Adding 10 to each of the numbers in the set.
C. Subtracting 3.5 from each of the numbers in the set.
D. Adding 8.2 to the 2 largest numbers and subtracting 8.2 from the 3 smallest numbers in the set.
E. Adding 5 to the 2 largest and to the 2 smallest numbers in the set.

OA E

[ceilidh.erickson]

OP: there is an error in this question. Are you sure it's transcribed correctly? As written there is no correct answer.

In any consecutive (or evenly spaced) set, the median is always equal to the average. If we want to change the average *without* changing the median, we must make it a non-evenly-spaced set.

You could test numbers to evaluate the answers choices, or you could just think conceptually:

A. Multiplying each of the numbers in the set by 6.
Conceptually: this will change the value of each term, so it will change both the median and the average. It will still be evenly spaced, though.

Testing the set [1, 2, 3, 4, 5]:
new set = [6, 12, 18, 24, 30]
median = 18, average = 18

B. Adding 10 to each of the numbers in the set.
Conceptually: this will change the value of each term, so it will change both the median and the average. It will still be evenly spaced, though.

Testing the set [1, 2, 3, 4, 5]:
new set: [11, 12, 13, 14, 15]
median = 13, average = 13

C. Subtracting 3.5 from each of the numbers in the set.
Conceptually: this will change the value of each term, so it will change both the median and the average. It will still be evenly spaced, though.

Testing the set [1, 2, 3, 4, 5]:
new set = [-2.5, -1.5, -0.5, 0.5, 1.5]
median = -0.5, average = -0.5

D. Adding 8.2 to the 2 largest numbers and subtracting 8.2 from the 3 smallest numbers in the set.
Conceptually: if we perform counter-balancing addition and subtraction, our overage will stay the same. So this will not change the average or the median.

Testing the set [1, 2, 3, 4, 5]:
new set = [-7.2, -6.2, 3, 12.2, 13.2]
median = 3, average = 3

E. Adding 5 to the 2 largest and to the 2 smallest numbers in the set.
Conceptually: The average must change, because we're adding 5 four times without any counter-balancing subtraction. However, the median will also change, because the original middle number will now be the lowest.

Testing the set [1, 2, 3, 4, 5]:
new set = [6, 7, 3, 9, 10] --> [3, 6, 7, 9, 10]
median = 7, average = 7

Thus, all 5 answer choices are incorrect.

OP, I'm assuming that the correct wording for answer choice E should have been something like:
"Adding 5 to the 2 largest and 1 to the 2 smallest numbers in the set."

If that were the case, the new set would be [2, 3, 3, 9, 10]. The median would stay the same but the average would change.

If the original wording is in fact what you transcribed, then the problem is flawed. Please advise.

[Scott@TargetTestPrep]

Princeton Review

In a set of five consecutive integers, which of the following must change the average of the set without changing its original median?

A. Multiplying each of the numbers in the set by 6.
B. Adding 10 to each of the numbers in the set.
C. Subtracting 3.5 from each of the numbers in the set.
D. Adding 8.2 to the 2 largest numbers and subtracting 8.2 from the 3 smallest numbers in the set.
E. Adding 5 to the 2 largest and to the 2 smallest numbers in the set.

OA E

Let's let the set be {1, 2, 3, 4, 5}. We see that the mean = 3 and the median = 3 also. Now let's analyze the answer choices.

We can skip choices A, B, and C since multiplying, adding and subtracting a constant to each number in the set will definitely change the average and median in the set (unless the median is 0). Let's look at choice D.

Under the conditions in choice D, the set becomes {-7.2, -6.2, -5.2, 12.2, 13.2}. We see that the new median is -5.2, which is not the same as the original median of 3. So D is not correct either.

Under the conditions in choice E, the set becomes {6, 7, 3, 9, 10}. We see the new median is 7, which is not the same as the original median of 3. So E is not correct either.

(Note: The original problem's choice E was the addition of 0.5 (instead of 5) to the two largest and the two smallest items in the set. This would result in the new set of {1.5, 2.5, 3, 4.5, 5.5}. The new average is 3.4, which is different from the original average of 3, but the median is unchanged; it is still 3.)

Answer: None