Data Sufficiency

Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
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IMO B is the answer here. What is the OA?

Yeah Your are very much correct!
We need to apply Weighted Average Concepts:

{(A1 * N1) +(A2 *N2)}/ (N1+N2) eqauls Total weighted average Aw.

St 1 gives only number of Men. Insufficient.

St2 : gives a relation between men & women . But not exact figures which we are looking for. Let us equate them as of now. N1=2*N2

Combining both of them, we can fing the Weighted average!

Pick C

gmatmachoman wrote:Yeah Your are very much correct!
We need to apply Weighted Average Concepts:

{(A1 * N1) +(A2 *N2)}/ (N1+N2) eqauls Total weighted average Aw.

St 1 gives only number of Men. Insufficient.

St2 : gives a relation between men & women . But not exact figures which we are looking for. Let us equate them as of now. N1=2*N2

Combining both of them, we can fing the Weighted average!

Pick C

Can you please solve this for me in more clarity, i have solved it in simple logical sense. but i want to know how you apply the weighted average concept. Thanks

gmatmachoman - As per the statement 2 N1= 2*N2 -- N2 actually cancels out. I don't think we need statement 1. agree?

boazkhan wrote:gmatmachoman - As per the statement 2 N1= 2*N2 -- N2 actually cancels out. I don't think we need statement 1. agree?

How ? i dint get u? please help me understand. I was expecting this answer since its mentioned that we dont require the statement 1 but im looking out for explanation.

sitsmegain wrote:

gmatmachoman wrote:Yeah Your are very much correct!
We need to apply Weighted Average Concepts:

{(A1 * N1) +(A2 *N2)}/ (N1+N2) eqauls Total weighted average Aw.

St 1 gives only number of Men. Insufficient.

St2 : gives a relation between men & women . But not exact figures which we are looking for. Let us equate them as of now. N1=2*N2

Combining both of them, we can fing the Weighted average!

Pick C

Can you please solve this for me in more clarity, i have solved it in simple logical sense. but i want to know how you apply the weighted average concept. Thanks

Now that from St 2 we know N1 = 2 *N2

N1 +N2 = 3*N2

Apply this one in the main equation

{(A1 * N1) +(A2 *N2)}/ (N1+N2) eqauls Total weighted average Aw.
{150 * 2N2 + 120 * N2}/3*N2

As Boazkhan said N2 cancels out: (Sorry for my mistake)

(150 * 2 +120) *N2/(3*N2)

=140

Lesson Learnt : Dont be so hasty when it comes to Weighted average concepts..Knowing them is one part but applying them is also as equal as "knowing them".

sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************

As you discovered, you don't need the actual number in each group to solve for a weighted average. Here's the rule:

To solve for a weighted average, you need to know the average of each subgroup and the weight of each subgroup.

Weighted Average = (avg group 1)(weight group 1) + (avg group 2)(weight group 2) + .... (avg group n)(weight group n)

When we combine the stem and statement (2), we get:

Weighted Average = (120)(1/3) + (150)(2/3)

which is certainly solvable.

We could also have solved using intuition: if there are twice as many men as women, then if we plot the weights on a number line, the average will be 2/3 of the way between women and men:

120 ---- 2x----avg--x--150

which is also solvable.

Nice Question. I got it wrong I thought we need both infomation.

sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************

When a DS question asks for a specific amount -- in this case, the average weight of the whole group -- a good way to check whether a statement is sufficient:

Plug in TWICE. If the value being sought stays the same, the statement is sufficient. If the value being sought changes, the statement is insufficient.

Looking at statement 2: twice as many men as women
Plug in M=2, W=1
Total weight of M = 2*150 = 300.
Total weight of W = 1*120 = 120.
Average of the whole group = 420/3 = 140.

Plug in M=4, W=2.
Total weight of M = 4*150 = 600.
Total weight of W = 2*120 = 240.
Average of the whole group = 840/6 = 140.

Since in each case the average weight of the whole group is 140, the statement is sufficient.

The correct answer is B.

DS questions become much clearer when we plug in actual values.

sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************

I don't know if this is the approved solution on how to solve it but it gave me the right answer.

F=Total of Females
M=Total of Males
TFW = Total Female Weight
TMW = Total Male Weight
Sum/n = Avg

Setup:
(TFW+TMW)/(F+M) = The Avg weight of all competitors (what we're trying to find)

Given: (TFW)/(F) = 120 or TFW = 120F
Given: (TMW)/(M) = 150 or TMW = 150M

They gave us a 2:1 ratio of men to women so I just said 2/1 = M/F, therefore M=2 and F=1

Plug all the numbers into the first equation...
(TFW+TMW)/(F+M) = The Avg weight of all competitors
(120F + 150M)/(F+M) = The Avg weight of all competitors
now plug in 2 & 1 for M & F...
(120*1+150*2)/(3)= The Avg weight of all competitors = 140; since I have an answer I know (B) is sufficient. Answer is B.

There is a neat little trick, which I learned in a MGMAT study hall from the great Lunarpower, for this problem. Basically the principle is, if we know the average weight of X and the average weight of Y, and the average weight of X+Y, we know the ratio of X to Y. The ratio of X to Y is the inverse of the distance X is from the average to the distance Y is from the average.

For example, let's say we have a room full of football and basketball players. The average weight of basketball players is 180 pounds and the average weight of football players is 210 pounds. The average of both basketball and football players is 200 pounds. Using the principle listed above, since the weight of basketball players is 20 away from 200, and the weight of basketball players is 10 away from 200, we know that the ratio of basketball players to football players is 10 to 20, or 1:2. We can always figure out the ratio of X to Y, as long as we know the average of X, average of Y, and the average of both.

We can apply that principle to this problem. The problem gives us the average weight of men, 150, and the average weight of women, 120, in the problem.

Statement 1 gives us the total # of men. Just using basic principles of weighted average we know that we need to know the # of women, thus insufficient.

Statement 2 gives us the ratio of men to women, 2 to 1. Since we know the average weight of men as 150, the average weight of women as 120, and we are given a ratio of 2 to 1 for men to women, we know we can calculate the average using the trick we just learned. Instead of figuring out a ratio, we are working "backwards" for an average. Sufficient.


Hopefully you guys can understand my rambling, if you have any questions feel free to ask.

Stuart Kovinsky wrote:

sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************

As you discovered, you don't need the actual number in each group to solve for a weighted average. Here's the rule:

To solve for a weighted average, you need to know the average of each subgroup and the weight of each subgroup.

Weighted Average = (avg group 1)(weight group 1) + (avg group 2)(weight group 2) + .... (avg group n)(weight group n)

When we combine the stem and statement (2), we get:

Weighted Average = (120)(1/3) + (150)(2/3)

which is certainly solvable.

We could also have solved using intuition: if there are twice as many men as women, then if we plot the weights on a number line, the average will be 2/3 of the way between women and men:

120 ---- 2x----avg--x--150

which is also solvable.

How did you get 1/3 and 2/3?

anirudhbhalotia wrote:

Stuart Kovinsky wrote:

sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************

As you discovered, you don't need the actual number in each group to solve for a weighted average. Here's the rule:

To solve for a weighted average, you need to know the average of each subgroup and the weight of each subgroup.

Weighted Average = (avg group 1)(weight group 1) + (avg group 2)(weight group 2) + .... (avg group n)(weight group n)

When we combine the stem and statement (2), we get:

Weighted Average = (120)(1/3) + (150)(2/3)

which is certainly solvable.

We could also have solved using intuition: if there are twice as many men as women, then if we plot the weights on a number line, the average will be 2/3 of the way between women and men:

120 ---- 2x----avg--x--150

which is also solvable.

How did you get 1/3 and 2/3?

1/3 is the ratio of women. 2/3 is for men. since there are twice as many men as women, that is why the ratio is as such.

Stuart Kovinsky wrote:

sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************

As you discovered, you don't need the actual number in each group to solve for a weighted average. Here's the rule:

To solve for a weighted average, you need to know the average of each subgroup and the weight of each subgroup.

Weighted Average = (avg group 1)(weight group 1) + (avg group 2)(weight group 2) + .... (avg group n)(weight group n)

When we combine the stem and statement (2), we get:

Weighted Average = (120)(1/3) + (150)(2/3)

which is certainly solvable.

We could also have solved using intuition: if there are twice as many men as women, then if we plot the weights on a number line, the average will be 2/3 of the way between women and men:

120 ---- 2x----avg--x--150

which is also solvable.

Stuart, when you say we need to know the "weight" of each group, you then use the fractions 1/3 and 2/3, which clearly refer to the w:m ratio of 1:2. To clarify, then, by "weight" you mean the ratio of one to the other?

Just want to make sure I'm clear, because it's a really elegant solution.

Thanks!