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In a certain group, the average (arithmetic mean) age of the

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In a certain group, the average (arithmetic mean) age of the

by BTGmoderatorLU » Tue Jul 30, 2019 3:05 pm

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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 male?

1) The average age of all 100 people is 28.9.
2) There are 10 more males than there are females.

The OA is D
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Source: — Data Sufficiency |
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by deloitte247 » Sun Aug 04, 2019 10:11 am

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For statement 1, the average age of all 100 people is 28.9.
$$\left[\left(30-28.9\right):\left(30-28\right)\right]\cdot100= males$$
(1.1 : 2) * 100 = males
$$\frac{1.1}{2}\cdot100=males$$
$$\frac{110}{2}=55=males$$
Statement 1 is sufficient.

For statement 2, there are 10 more males than females.
$$So,\ m=f+10\ ---\ \left(i\right)$$
$$m+f=100\ ---\ \left(ii\right)$$
From equation (i),
$$m-f=10\ ---\ \left(iii\right)$$
Add eqn 2 & 3 together; we have
2m = 110
$$m=\frac{110}{2}=55$$
Statement 2 is also sufficient.
Each statement alone is sufficient. Thus, option D is correct.
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by swerve » Wed Aug 07, 2019 3:15 pm

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Statement 1:

Since total 100 people, so total age \(=100 \cdot 28.9 =2890\)
Men's mean \(=28\) women's mean +30, looking at the total mean it is clear that there are more men than women.

Using basic math, if there are 50 men \(- 50\cdot 28=1400\) and 50 women \(-50\cdot 30 = 1500\)

total\(=1500+1400 =2900\) which is greater than 2890 totaled above. So increasing one man decreases the average by 2, so total men \(=55\), women\(=45\)
Hence Sufficient \(\color{green}{\checkmark}\)

Statement 2:

Suppose there are \(w\) women. So men\(=10+w\)
Also \(m+w=100\), so using this equation it is clear to find the answer which is not required.
Hence Sufficient \(\color{green}{\checkmark}\)

Answer __D__ \(\color{green}{\checkmark}\)
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