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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
Quote:
The average (arithmetic mean) of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average (arithmetic mean) of these 20 integers?

A. 3
B. 3.5
C. 4
D. 4.5
E. 5


n=30, avg=5, sum=5*30=150
n=20: 20 integers are <= to 5
to max the avg of these n=20
we must min the avg of the other n=10
if exactly 20 don't exceed, then the other 10 must exceed
so the min 10 integer values > 5 is 6
10*6=60, 150-60=90, 90/20=9/2=4.5

Ans (D)
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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
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for the avg of 20 integers which didnt exceed 5 to be maximum......the avg of the rest 10 integers must be minimum.....for that to be min.......all of them must be 6....which is the next least integer.....

for the sum to be max they all must be equal and should be max integer
20*x+10*6=5*30
20*x=90
x=4.5

OA:D
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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
Expert Reply
Quote:
The average (arithmetic mean) of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average (arithmetic mean) of these 20 integers?

A. 3
B. 3.5
C. 4
D. 4.5
E. 5


Answer: Option D

The detailed explanation can be seen in the video

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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
sum: n*avg = 30*5=150
less than 5 = 4, highest possible.(integer)
4*20 = 80 => 150-80=70
70/10=7..so 10 numbers with value 7.
Ans C
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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
Each of these 20 integers can be 5 so there average can also be 5.
Option E is the answer.

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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
for these 20 to be maximum other 10 should be min.
the min number exceeding 5 is 6
so the max mean is
\(\frac{30*5 - 10*6 }{20}\)
= 4.5
Thus D
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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
Bunuel wrote:

Competition Mode Question



The average (arithmetic mean) of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average (arithmetic mean) of these 20 integers?

A. 3
B. 3.5
C. 4
D. 4.5
E. 5




Are You Up For the Challenge: 700 Level Questions


Given: The average (arithmetic mean) of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5.

Asked: What is the highest possible value of the average (arithmetic mean) of these 20 integers?

Let the average of 20 integers be x and average of remaining 10 integers be A

30*5 = 150 = 20x + 10A
A = 15 - 2x > 5
2x < 10
x < 5

If we take x = 5 ; 15 - 2x = 5 = 5; Not feasible
Next lower value of x = 4.5; 15-2x = 15- 9 =6 >5: Feasible

IMO D
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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
Expert Reply
Bunuel wrote:

Competition Mode Question



The average (arithmetic mean) of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average (arithmetic mean) of these 20 integers?

A. 3
B. 3.5
C. 4
D. 4.5
E. 5


Since the average of the 30 integers is 5, the sum of these integers is 30 x 5 = 150. To maximize the average of the integers that do not exceed 5, let’s minimize the integers that do exceed 5; i.e. let’s assume that all the integers that exceed 5 are actually equal to 6. Then, the sum of the integers that exceed 5 is 6 x 10 = 60 and thus, the sum of the integers that do not exceed 5 is 150 - 60 = 90. Thus, the highest possible value of the average of the integers that do not exceed 5 is 90/20 = 4.5

Answer: D
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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
chondro48 wrote:
Sum of the 30 integers =5*30 = 150

Minimum possible sum of the 10 integers, each of which exceeds 5 = 10*6 = 60

So, maximum possible sum of the remaining 20 integers, each of which doesn't exceed 5 = 150 – 60 = 90. Thus, the corresponding average = 90/20 = 4.5.

FINAL ANSWER IS (D)

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if none of the integers of the list exceeds 5 then the max possible value of all the 20 integers is 4 ..so how the average is 4.5 ??
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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
For the average to be 4.5 what will be the distribution of the numbers. Because each of the 20 numbers can max take the value of 4.
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Re: The average (arithmetic mean) of 30 integers is 5. Among these 30 inte [#permalink]
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