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Re: The average (arithmetic mean) of 4 positive integers is 50. [#permalink]
Sum(4)=200
Sum(2)=90
Hence um(2 leftovers)=200-90=110

Since there is no restriction of the median
Let one of them be 1 => other one will be 109

Hence D
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Re: The average (arithmetic mean) of 4 positive integers is 50. [#permalink]
Bunuel wrote:
iyersu wrote:
The average (arithmetic mean) of 4 positive integers is 50. If the average of 2 of these integers is 45, what is the greatest possible value that one of the other 2 integers can have?

A. 55
B. 65
C. 100
D. 109
E. 115


The sum of 4 integers is 4*50 = 200.
The sum of 2 integers is 2*45 = 90.

Thus, the sum of the other 2 integers is 200 - 90 = 110. Since we know that the integers are positive then the greatest one from these two can be 109, the other one being 1.

Answer: D.

Hope it's clear.


Sorry I do not understand why sum of two remaining integers = 110 becomes 109?
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Re: The average (arithmetic mean) of 4 positive integers is 50. [#permalink]
Expert Reply
kanusha wrote:
Bunuel wrote:
iyersu wrote:
The average (arithmetic mean) of 4 positive integers is 50. If the average of 2 of these integers is 45, what is the greatest possible value that one of the other 2 integers can have?

A. 55
B. 65
C. 100
D. 109
E. 115


The sum of 4 integers is 4*50 = 200.
The sum of 2 integers is 2*45 = 90.

Thus, the sum of the other 2 integers is 200 - 90 = 110. Since we know that the integers are positive then the greatest one from these two can be 109, the other one being 1.

Answer: D.

Hope it's clear.


Sorry I do not understand why sum of two remaining integers = 110 becomes 109?


x + y = 110, where both x and y are positive integers. The max value of say x there is 109, when y = 1.
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Re: The average (arithmetic mean) of 4 positive integers is 50. [#permalink]
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Re: The average (arithmetic mean) of 4 positive integers is 50. [#permalink]
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