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Bunuel
In a certain group, the average (arithmetic mean) age of the males is 28, and the average age of the females is 30. If there are 100 people in the group, how many of them are males?

(1) The average age of all 100 people is 28.9
We can find the number of men by 30-28.9/30-28 *100 = 1.1*50=55
and the women being 45
clearly sufficient

(2) There are 10 more males than there are females.
Since this is solving for 2 eqn in 2 variable we can conclusively arrive at an answer

Therefore IMO D
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In a certain group, the average (arithmetic mean) age of the males is 28, and the average age of the females is 30. If there are 100 people in the group, how many of them are males?

(1) The average age of all 100 people is 28.9

(2) There are 10 more males than there are females.

Okay in this situation we should start from statement 2.

(2) There are 10 more males than there are females. So females + (females+10)=100. females=X => X+(X+10)=100
2x=90=> X=45 is females. So males are 55.

Now let's calculate the average age.

(1) The average age of all 100 people is 28.9

Males- 55*28=1540
Females - 45*30=1350
Sum is 2890. Average is 28.9

The answer is C.
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Is there a better way to figure out for statement (1)?
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Shrutiv134
In a certain group, the average (arithmetic mean) age of the males is 28, and the average age of the females is 30. If there are 100 people in the group, how many of them are males?

(1) The average age of all 100 people is 28.9

(2) There are 10 more males than there are females.

Is there a better way to figure out for statement (1)?

The average age of males is 28, which is 0.9 below the overall average of 28.9.
The average age of females is 30, which is 1.1 above the overall average of 28.9.

This means the ratio of males to females must be in the reverse ratio of these differences:

males/females = 1.1/0.9 = 11/9.

So the group must have 55 males and 45 females to maintain the overall average. Basically, there must be more males to pull the overall average closer to their average than to that of the females.

General rule:

When two groups with different averages are combined, the sizes of the two groups are in the inverse ratio of the absolute differences between each group’s average and the overall average. The group whose average is closer to the overall average must be larger to pull the combined average in that direction.
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Bunuel
In a certain group, the average (arithmetic mean) age of the males is 28, and the average age of the females is 30. If there are 100 people in the group, how many of them are males?

(1) The average age of all 100 people is 28.9

(2) There are 10 more males than there are females.
Statement (1): The average age of all 100 people is 28.9
Let:
  • m = number of males
  • f = number of females
  • f=100−m
The total age of the group:
28×m + 30×(100−m) = 28.9×100

Solving:
28m+30(100−m)=2890
28m+3000−30m=2890
−2m=−110
m=55
Statement (1) alone is sufficient

Statement (2): There are 10 more males than females

Let:
  • m=f+10
  • We know that m+f=100. So, f=m-10.
Solving:
m+(m−10)=100
2m=110
m=55
Statement (2) alone is sufficient
Therefore, D.
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