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The arithmetic mean of a set of 15 integers is 78. The largest value o 28 Oct 2023, 04:47
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The arithmetic mean of a set of 15 integers is 78. The largest value of the set is five times the smallest value of the set. If the mean of the set is equal to the median, then what is the maximum possible value for the range of the set?

A. 52
B. 104
C. 112
D. 208
E. 260
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D
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The arithmetic mean of a set of 15 integers is 78. The largest value o 28 Oct 2023, 06:07
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Given: The arithmetic mean of a set of 15 integers is 78. The largest value of the set is five times the smallest value of the set.
Asked: If the mean of the set is equal to the median, then what is the maximum possible value for the range of the set?

Let the largest value and smallest value be l & s respectively
l = 5s

Range = l - s = 5s - s = 4s

To maximize range, we have to maximise l=5s and minimise s.

Let 7 integers be equal to s, 7 integers equal to 78 and 1 integer equal to l=5s

7s + 7*78 + 5s = 15*78
12s = 8*78
s = 52
Range = l -s = 5s - s = 4s = 4*52 = 208

IMO D
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The arithmetic mean of a set of 15 integers is 78. The largest value o 27 Sep 2025, 01:11
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But when we are solving like this, we are maximising the range, however since mean = median then in that case d should be same but in the below scenario d doesnt remain same

x x x x x x x 78*7 5x
So here we will have 7x +78*7+5x = 15*78

But the above scenario is not equally spaced out. Can you please help here?

GMATNinja karishma
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The arithmetic mean of a set of 15 integers is 78. The largest value o 27 Sep 2025, 06:46
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nikitathegreat
But when we are solving like this, we are maximising the range, however since mean = median then in that case d should be same but in the below scenario d doesnt remain same

x x x x x x x 78*7 5x
So here we will have 7x +78*7+5x = 15*78

But the above scenario is not equally spaced out. Can you please help here?

GMATNinja karishma
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Your question isn't clear.

You have 7 numbers below the mean/median. 7X

Then the mean/median of 78.Then 6 more values of 78 for a total of 7*78 as you indicate.

And finally 5X as the largest of the 15 values.

The mean and median haven't changed - 78 is both the middle value of the ordered list and the average, as long as you continue to use the sum of the 15 values as equal to 15*78 when solving for X, as described in the above posts.
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The arithmetic mean of a set of 15 integers is 78. The largest value o 27 Sep 2025, 09:19
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nikitathegreat
But when we are solving like this, we are maximising the range, however since mean = median then in that case d should be same but in the below scenario d doesnt remain same

x x x x x x x 78*7 5x
So here we will have 7x +78*7+5x = 15*78

But the above scenario is not equally spaced out. Can you please help here?

GMATNinja karishma
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The problem does NOT require the set to be an Arithmetic Progression (equally spaced).

To find the maximum range, we only need to satisfy the non-decreasing condition (T1<=T2<=⋯<=T15) while keeping the sum (=1170) constant.
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The arithmetic mean of a set of 15 integers is 78. The largest value o 28 Sep 2025, 22:41
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nikitathegreat
But when we are solving like this, we are maximising the range, however since mean = median then in that case d should be same but in the below scenario d doesnt remain same

x x x x x x x 78*7 5x
So here we will have 7x +78*7+5x = 15*78

But the above scenario is not equally spaced out. Can you please help here?

GMATNinja karishma
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nikitathegreat You're absolutely right that the arrangement you proposed (\(x, x, x, x, x, x, x, 78\cdot7, 5x\)) doesn't create equal spacing - and that's perfectly fine! The problem never requires the integers to form an arithmetic sequence.

What "Mean = Median" Actually Means:
- Mean = \(78\) → Sum of all values divided by \(15\) equals \(78\)
- Median = \(78\) → The \(8\)th value (middle value) equals \(78\)

This does not mean the values must be equally spaced! It only means the middle value happens to equal the average.

The Correct Approach to Maximize Range:

To maximize range (\(5x - x = 4x\)), we need to maximize \(x\). Here's the optimal arrangement:

  1. Place \(7\) values at the minimum: \(x\)
  2. Place the \(8\)th value (median) at: \(78\)
  3. Place \(7\) values at the maximum: \(5x\)

Sum equation: \(7x + 78 + 7(5x) = 15 \cdot 78\)
\(7x + 78 + 35x = 1170\)
\(42x = 1092\)
\(x = 26\)

Therefore, maximum range = \(4x = 4(26) = 104\)

Process Diagnosis:
You were conflating two different concepts - thinking that for mean to equal median, the data must be symmetrically distributed with equal gaps. In GMAT problems, mean = median simply constrains one relationship; it doesn't dictate the entire structure of the data set.

Strategic Pattern Recognition:
For intermediate-level range maximization problems:
- If median is fixed → Pack values at the extremes
- If mean = median → This is just one constraint, not a distribution requirement
- Decision tree: Check if median divides the dataset symmetrically → If yes, use extreme packing strategy

You can practice similar questions here (you'll find a lot of OG questions) - select Statistics under Number Properties.
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