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The arithmetic mean of a set of 12 numbers is 65, and the median of Updated on: 12 Jun 2024, 02:01
Originally posted by EgmatQuantExpert on 25 Oct 2018, 01:26.
Last edited by Bunuel on 12 Jun 2024, 02:01, edited 2 times in total.
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C
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55% (hard)

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68% (02:36) correct 32% (02:47) wrong based on 680 sessions

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The arithmetic mean of a set of 12 numbers is 65, and the median of the set is 40. What is the minimum possible value of the largest number of this set?

A. 65
B. 80
C. 100
D. 120
E. 130


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C
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The arithmetic mean of a set of 12 numbers is 65, and the median of 29 Oct 2018, 01:45
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Solution


Given:
    • The arithmetic mean of a set of 12 numbers = 65
    • Median of the set = 40

To find:
    • The minimum possible value of the largest number of the set

Approach and Working:
    • Given, the arithmetic mean of 12 numbers = 65
      o Implies, the sum of 12 numbers = 12 * 65 = 780
         \(t_1 + t_2 + … + t_6 + t_7 + …. + t_{11} + t_{12} = 780\)

    • We are also given that the median = 40,
      o That implies, \(t_6 + t_7 = 2 * 40\)

    • So, for the largest number to be minimum, all other numbers must take the maximum possible value
      o The maximum value of all the five numbers before the median term = the median itself (since they cannot be greater than the median value)
      o The maximum value of all the five numbers after the median term = the value of the largest number

    • Thus, 5 * 40 + 2 * 40 + 5 * x =780, where x is the largest number
      o Implies, \(x = \frac{500}{5} =100\)

    • Therefore, the minimum possible value of the largest number = 100

Hence, the correct answer is option C.

Answer: C

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The arithmetic mean of a set of 12 numbers is 65, and the median of 25 Oct 2018, 03:38
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I considered the set from number x1to x12

As the median of the set is 40 I have considered that all numbers from x1 to x7 are 40. Thus 40x7=280

On the other hand, the sum of all the set is 65*12=780

The subtract is
780-280=500

As we have 5 numbers between x8 to x12, I divided 500/5= 100

C. 100

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The arithmetic mean of a set of 12 numbers is 65, and the median of 16 Mar 2026, 01:24
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The arithmetic mean of a set of 12 numbers is 65, and the median of the set is 40. What is the minimum possible value of the largest number of this set?

Average = Sum/Number

Number × Average = Sum

12 × 65 = 780

So, the 12 numbers must add to 780.

To minimize the largest number, maximize the values of the others.

The median is 40. So, each of the lowest 6 numbers must be at or below 40, and the middle two numbers, the 6th and 7th, must have an average of 40.

Thus, the maximum total value of the first 7 numbers must be 7 × 40 = 280.

We have 5 numbers and 780 - 280 = 500 in total value remaining.

To minimize the value of the largest number, make the other 4 remaining numbers as large as possible. The way to do so is to make all 5 equally large.

500/5 = 100

So, the largest will be minimized when all 5 remaining numbers are 100.

Note: Even though 5 numbers are 100, we can still say that the largest is 100 since 100 is the largest value.

A. 65
B. 80
C. 100
D. 120
E. 130

Correct answer: C
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The arithmetic mean of a set of 12 numbers is 65, and the median of 16 Mar 2026, 01:40
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To minimize the largest number we maximize the sum of the first 7 numbers here.
Why? Because that max sum would make sure the compensation in the sum is minimal over the right side of median.
Now median of 12 elements is 6th+7th/2
So lets consider first 7 elements to be 40
Now the trailing sum is 25*7=175 which is to be compensated other side of median which now has 5 elements
So to minimize this value simply divide:
175/5=35
So 35+Mean would give the minimum value = 35+65 = 100

Answer: Option C
EgmatQuantExpert
The arithmetic mean of a set of 12 numbers is 65, and the median of the set is 40. What is the minimum possible value of the largest number of this set?

A. 65
B. 80
C. 100
D. 120
E. 130


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To read all our articles:Must read articles to reach Q51

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