Set S is a 7 positive integer set with an average of 25 and a median

Question

Set S is a 7 positive integer set with an average of 25 and a median of 30. If n is in S, what is the least possible value of n?

A) 1
B) 2
C) 3
D) 4
E) 5

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NextstopISB · Answer

Given:
- Set S has 7 positive integers.
- Average = 25 ⟹ Sum of elements = 25 × 7 = 175.
- Median = 30 ⟹ The 4th smallest number in the ordered set is 30.

To minimize n, we maximize the other numbers while keeping the sum 175.

Let the ordered set be:  .

Maximizing x, y, z:
- Choose the largest reasonable values: x = 30, y = 30, z = 30.
- Sum constraint: n + a + b + 30 + 30 + 30 + 30 = 175.
- Simplifies to: n + a + b = 55.

Minimizing n:
- To keep numbers positive and increasing, set a = 29, b = 29.
- Then: n + 29 + 29 = 55 ⟹ n = 55 - 58 = 1.

Thus, the least possible value of n is:

Answer: (A) 1

AVMachine · Answer

The Nos. Could be like that:

1, 1, 1, 30, x, y, z;

25*7 = 175; x+y+z = 175 - 33 = 142

142 / 3 > 30;

So Minimum Value Could be 1.

GMATinsight · Answer

AVerage of 7 integers = 25
i.e. Sum of 7 integers = 25*7 =175 
Median = 30

n_{min} = ?

Let's place the numbers at 7 places 
n _ _ 30 _ _ _

our simple objective is to try to make n 1 (if possible) so we distribute other values conveniently as mentioned below

n _ _ 30 30 30 30

175 - 4*30 = 55
55 can be distributed as {1, 27, 27} in the first three places hence the minimum value of  n may be 1

Answer: Option A

P.S. Somehow I feel that the question was meant to ask something else but instead what we have got was much simpler.

A harder and related question:

https://www.youtube.com/watch?v=Ic-4Icc8pv0

Capablanca · Answer

S = {A1, A2, A3, 30, A4, A5, A6}

To find the min value of A1, we start by maximizing A2 and A3, which can take the highest value of median 30, so the new set is:

S = {A1, 30, 30, 30, A4, A5, A6}

Now we can set up the equation:

A1 + 90 + A4 + A5 + A6 = 175
A1 + A4 + A5 + A6 = 85, not possible since A4 + A5 + A6 can not be less than 90.

Since we want to find the least possible value, we can try the equation by changing A2 and A3 for 27 each since we need A1 + A4 + A5 + A6 to be equal to at least 91.

S = {A1, 27, 27, 30, A4, A5, A6}

Now the equation is

A1 + 84 + A4 + A5 + A6 = 175
A1 + A4 + A5 + A6 = 91

And since A4, A5, and A6 can all take the value of 30, we can confidently say that the least value of the set could be 1.

Answer (A)

ManifestDreamMBA · Answer

With no other constraints like distinct positive integers, etc. The number could be 1 but let's see

Let the set be a,b,c,30,d,e,f - with each of these numbers a positive integer, i.e. they can be min 1
To minimise the first number, we need to max the other 5 numbers
Sum of the 7 terms = 7*25 = 175
Given 1 term is 30, the rest can only have max sum of 145
d = e = f = 30 minimum, we are left with 145 - 90 = 55 max for a,b,c we can divide this in multiple ways with each value <=30
For instance, 1,27,27 or 1,26,28, etc..
Therefore, a = 1

Gmat750aspirant · Answer

Set S has 7 integers with: 
- Average = 25 → Total sum = 7 * 25 = 175 
- Median = 30 → 4th number is 30

Let the integers in S be: 
a1 ≤ a2 ≤ a3 ≤ a4 ≤ a5 ≤ a6 ≤ a7 
a4 = 30, and a5 = a6 = a7 = 30 (to maximize larger numbers). 
Sum of a4, a5, a6, a7 = 120.

Sum of a1, a2, a3 = 175 - 120 = 55. 
Maximizing a2 and a3: a2 = a3 = 27 → a1 = 55 - 54 = 1.

Thus, the least possible value of n is 1.

Answer: (A) 1

Set S is a 7 positive integer set with an average of 25 and a median

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Set S is a 7 positive integer set with an average of 25 and a median

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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