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

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In a certain group of people, the average (arithmetic mean) weight of the males is 180 pounds and of the females, 120 pounds. What is the average weight of the people in the group?

(1) The group contains twice as many females as males.
(2) The group contains 10 more females than males.

Answer; A
Source: Official guide
Source: — Data Sufficiency |

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BTGModeratorVI wrote:
Fri Jun 05, 2020 11:41 am
In a certain group of people, the average (arithmetic mean) weight of the males is 180 pounds and of the females, 120 pounds. What is the average weight of the people in the group?

(1) The group contains twice as many females as males.
(2) The group contains 10 more females than males.

Answer; A
Source: Official guide
We can solve this question using weighted averages
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Given: In a certain group of people, the average (arithmetic mean) weight of the males is 180 pounds and of the females, 120 pounds.

Target question: What is the average weight of the people in the group?

Statement 1: The group contains twice as many females as males.
This statement is SUFFICIENT because it provides us with information about the proportion of people in each group.
The above information tells us that, out of every three people, two are females and one is male.
In other words 2/3 of the group are female and 1/3 are male.
Adding this to our weighted average formula we get: average weight of the people in the group = (2/3)(120) + (1/3)(180)
Since we COULD evaluate this expression, we COULD answer the answer the target question with certainty
Statement 1 is SUFFICIENT

Statement 2: The group contains 10 more females than males.
This information does not tell us the proportion of people in each group (males and females)
For example, it could be the case that there are 14 people all together, with a breakdown of 12 females and 2 males. In this case, 12/14 of the people are females and 2/14 are males.
Or it could be the case that there are 16 people all together, with a breakdown of 13 females and 3 males. In this case, 13/16 of the people are females and 3/16 are males.
As you might imagine, the two scenarios, when plugged into are weighted averages formula, would yield different average weights for the group.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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Given that:
The average weight of males = 180
The average weight of females = 120
Let the number of males = m, and the number of females = f

Target question => What is the average weight of the people in the group?
$$Average\ weight=\frac{sum\ of\ all\ weight}{number\ of\ people\ in\ the\ group}$$ $$=\frac{180\left(cm\right)+120\left(f\right)}{m+f}$$

Statement 1 => The group contains twice as many females as males
i.e f = 2m $$substituting\ \ this\ into\ the\ \exp ression\ from\ the\ question\ stem
$$\frac{180\left(cm\right)+120\left(f\right)}{m+f}$$
$$\frac{180\left(m\right)+120\left(2m\right)}{m+2m}=>\frac{180m+240m}{3m}=>\frac{420m}{3m}$$
$$average=140$$
Statement 1 is SUFFICIENT

Statement 2 => The group contains 10 more females than males
i.e f = m + 10
Substituting this into the expression from the question stem
$$\frac{180\left(m\right)+120\left(f\right)}{m+f}$$
$$\ \frac{180\left(m\right)+120\left(m+10\right)}{m+m+10}=>\frac{180m+120m+1200}{2m+10}$$
$$=>\frac{300m+1200}{2m+10}$$
$$if\ m=2\ then\ \frac{300\left(2\right)+1200}{2\left(2\right)+10}=\frac{600+1200}{4+10}$$
$$=\frac{1800}{14}\ \ \ \ average=128.6$$
$$but\ if\ m=6\ then\frac{300\left(6\right)+1200}{2\left(6\right)+10}=\frac{1800+1200}{12+10}$$
$$=\frac{3000}{22\ \ }\ \ \ \ \ \ average=136.4$$
The statement provided is not enough to arrive at a definite answer. Statement 2 is NOT SUFFICIENT

Since only statement 1 is sufficient,
Answer = A