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The average weight of the women in a room is 120 lbs, and [#permalink]

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10 Jan 2011, 17:55

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The average weight of the women in a room is 120 lbs, and the average weight of the men in the room is 150 lbs. What is the average weight of the people in the room?

(1) There are twice as many men as women in the room.

(2) There are a total of 120 people in the room.

Bunuel, please can you clarify when to take a problem as average weight and when to take it as a normal average problem. I have to divulge details so i apologize as I just wanted to clarify a concept. So i understand that if you have to average the total of 2 quantities you need the total of A and total of B. Hence i am confused here. so here would you not need the total women and total men then get the respective weights and then average. Would appreciate your help.

The average weight of the women in a room is 120 lbs, and the average weight of the men in the room is 150 lbs. What is the average weight of the people in the room?

(1) There are twice as many men as women in the room.

(2) There are a total of 120 people in the room.

Bunuel, please can you clarify when to take a problem as average weight and when to take it as a normal average problem. I have to divulge details so i apologize as I just wanted to clarify a concept. So i understand that if you have to average the total of 2 quantities you need the total of A and total of B. Hence i am confused here. so here would you not need the total women and total men then get the respective weights and then average. Would appreciate your help.

Any average equals to \(average=\frac{total \ weight}{sum \ of \ values}\) (I think this is what you call "normal average"). When you have two values (boys and girls for example) then the formula can be written as \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) note that it's the same \(\frac{total \ weight}{# \ of \ values}\).

Back to the question:

The average weight of the women in a room is 120 lbs, and the average weight of the men in the room is 150 lbs. What is the average weight of the people in the room?

The average weight of the women in the room is 120 lbs --> \(average_{women}=120=\frac{total \ weight}{# \ of \ women}\), let # of women be \(w\) --> \(total \ weight=120w\); The average weight of the men in the room is 1500 lbs --> \(average_{women}=150=\frac{total \ weight}{# \ of \ women}\), let # of men be \(m\) --> \(total \ weight=150m\);

Question: \(average_{all}=\frac{total \ weight}{# \ of \ values}=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}=\frac{120w+150m}{w+m}\)?

(1) There are twice as many men as women in the room --> knowing individual averages (120 and 150) and the ratio of the values (w/m=1/2) is always sufficient to get the weighted average: \(m=2w\) --> \(average_{all}=\frac{120w+150*2w}{w+2w}=\frac{120+150*2}{3}=140\). Sufficient.

(2) There are a total of 120 people in the room --> \(w+m=120\), not sufficient.

Re: The average weight of the women in a room [#permalink]

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12 Jan 2011, 17:49

thanks! Bunuel. So here i have a doubt. If we have to calculate the total average value of 2 quantities and we are given the average of both the quantities then don't we find the the total of first quant and the total of second quant and then the grand sum total/ total(a) + total(b). This is where my confusion is (ie one cannot take the average of both quantities and then average them out). Please advise
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thanks! Bunuel. So here i have a doubt. If we have to calculate the total average value of 2 quantities and we are given the average of both the quantities then don't we find the the total of first quant and the total of second quant and then the grand sum total/ total(a) + total(b). This is where my confusion is (ie one cannot take the average of both quantities and then average them out). Please advise

Not sure understood your question.

Please read this again: any average equals to \(average=\frac{total \ weight}{# \ of \ values}\) (for example: average speed in mile/hour=total distance covered in miles/# of hours spent or average salary=total salary/# of employees ... I think this is what you call "normal average"). When you have two values (boys and girls for example) then the formula can be written as \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) note that it's the same \(\frac{total \ weight}{# \ of \ values}\) (nominator gives total weight and denominator gives total # values).
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Re: The average weight of the women in a room is 120 lbs, [#permalink]

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26 Nov 2011, 01:26

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Clearly (A).

Using statement (1), we can easily calculate the weighted average of the people in the room. Sufficient. Using statement (2), we don't know the split between the number of men/women so we cannot calculate the overall average. Insufficient.
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Re: The average weight of the women in a room is 120 lbs, [#permalink]

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26 Nov 2011, 03:56

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Average of the Group = (weight of all women + weight of all men)/size of group

From 1. we can get: weight of all women = 120 x (size of group x 1/3) = 40 x size of group weight of all women = 150 x (size of group x 2/3) = 100 x size of group

therefore Average of the Group = (40 x size of group + 100 x size of group)/size of group = 140

SUFFICIENT

2. the total number of people in the group does not give us enough information to figure out the average weight of the group. For example: If 1 woman and 119 men average will be very close to 150. If 119 women and 1 man, average will be very close to 120. INSUFFICIENT

The average weight of the women in a room is 120 lbs, and [#permalink]

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22 Aug 2015, 12:13

[quote="Bunuel"][quote="ajit257"]The average weight of the women in a room is 120 lbs, and the average weight of the men in the room is 150 lbs. What is the average weight of the people in the room?

(1) There are twice as many men as women in the room.

(2) There are a total of 120 people in the room.

Weighted average: \(\frac{120W+150M}{W+M}\)

A) Sufficient: It is subtle but a good point to note here is that the comparison of women to men is in absolute value "twice". Had the statement mentioned that the ratio of women to men then it would've been insufficient.

We're given two pieces of information to work with: 1) The average weight of the women in a room is 120 pounds 2) The average weight of the men that same room is 150 pounds

We're asked for the average weight of ALL the people in that room.

Fact 1: There are twice as many men as women in the room.

With this ratio, we can prove that the average weight is always the same. Here's how...

IF we had.... 2 men and 1 woman the average would be [2(150)+1(120)]/3 = 420/3 = 140 pounds

4 men and 2 women the average would be [4(150)+2(120)]/3 = 840/6 = 140 pounds

6 men and 3 women the average would be [6(150)+3(120)]/9 = 1260/9 = 140 pounds

The average is ALWAYS 140 pounds. Fact 1 is SUFFICIENT

Fact 2: There are a total of 120 people in the room.

The average will vary depending on how many men and women are in the room...

IF we had.... 60 men and 60 women the average would be exactly 'in the middle' = 135 pounds

61 men and 59 women the average would be something other than 135 pounds Fact 2 is INSUFFICIENT

Where I get confused is with Fact 1 (There are twice as many men as women in the room.)

Edit:

To me that translates to a ratio of W:2M. However, as posted with answers to this question, the ratio is supposed to be 2W:M. I don't understand how.

Hi,

There are twice as many men as women in the room... this means there are twice men, that is the number of men is MORE.. How much twice so M = 2W
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Re: The average weight of the women in a room is 120 lbs, and [#permalink]

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