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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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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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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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Each of these 20 integers can be 5 so there average can also be 5.
Option E is the answer.

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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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Bunuel

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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Bunuel

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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chondro48
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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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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Bunuel

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
Even if we take 5 for each value, it's still maximum
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