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Let number of men and women be m and w.

[m*Avg(m) + w*Avg(w)]/[m+w] = 80 ---from the problem statement
or, m*Avg(m) + w*Avg(w) = 80m + 80w

From 1st option, Avg(m) < 75

substitute this in the equation of problem statement:
75m + w*Avg(w) > 80m + 80w
or w*Avg(w) > 5m + 80w
or Avg(w) > 5m/w + 80 ----- (1)
this isn't sufficient.
From 2nd statement:
m>w
this means, m/w > 1
substitute this in eq (1):
Avg(w) > 5 + 80
or Avg(w) > 85

So, both statement together can answer the problem. Hence, C
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Re: At a certain company, a test was given to a group of men and women see [#permalink]
This is so far the clear and precise explanation, thanks Bunuel!!
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Stem :
\((Am + Aw) / (Nm + Nw) = 80\)

Where is average and n is number
This is a weighted average problem, where we are asked to find whether Aw>85


Statement one

I shall use the see-saw method

\(75----------------------80---------------------85\\
(5) (5)\)
\(Nw:Nm = 5:5 = 1:1\)
Hence we can see that for\(Aw=85,\) number of men and women should be equal.

We can conclude that since we want \(Aw>85, Nw> Nm.\)

Since we are not given this info. Insuff.

(2)
Clearly Insuff


(1) + (2) together

Bingo! we have exactly what we need

Hence Suff
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Bunuel
You can solve this question without any algebra.

At a certain company, a test was given to a group of men and women seeking for promotions. If the average (artihmetic mean) score for the group was 80, was the average score for the women more than 85?

(1) The average score for the men was less than 75. Since the average score for the group was 80 then the average score for women must be more than 80, but we don't know whether it's more than 85. Not sufficient.

(2) The group consisted of more men than women. Clearly insufficient.

(1)+(2) From (1) we have that the distance between the average score for the men and the average score for the whole group is more than 5 and from (2) we have that there are more men than women. Now, in order to compensate that difference and to make the average for the whole group 80 the average of the smaller group (women) must be further from 80 than the average for the larger group (men), so the average score for the women must be more than 85. Sufficient.

Answer: C.

Hope it's clear.

Once again a wonderful explanation by the best. Bunuel makes me wonder how did u became so good at all this problems.
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How about using wt avg (differences) --
Let
men = m
women = w

St 1 --->
Suppose avg score for men is 74.99
Difference for women = +X
(-5.01)*m + (+X)*w = 0 (differences should cancel out)
we cannot deduce X

St 2 --->
m>w : clearly insufficient

-------------------------------------------------------------
Combining --
(-5.01)*m + (+X)*w = 0
X = (5.01 * m)/w; since m > w => m/w > 1 ; therefore X will be greater than 5
Thus women's average = 80 + (>5) i.e greater than 85
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Re: At a certain company, a test was given to a group of men and women see [#permalink]
Bunuel
You can solve this question without any algebra.

At a certain company, a test was given to a group of men and women seeking for promotions. If the average (artihmetic mean) score for the group was 80, was the average score for the women more than 85?

(1) The average score for the men was less than 75. Since the average score for the group was 80 then the average score for women must be more than 80, but we don't know whether it's more than 85. Not sufficient.

(2) The group consisted of more men than women. Clearly insufficient.

(1)+(2) From (1) we have that the distance between the average score for the men and the average score for the whole group is more than 5 and from (2) we have that there are more men than women. Now, in order to compensate that difference and to make the average for the whole group 80 the average of the smaller group (women) must be further from 80 than the average for the larger group (men), so the average score for the women must be more than 85. Sufficient.





Answer: C.

Hope it's clear.


Hi Bunuel

Can you please explain why C is correct . I am finding it difficult to understand how the number of men and women helps to answer the question

Regards
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Bunuel
You can solve this question without any algebra.

At a certain company, a test was given to a group of men and women seeking for promotions. If the average (arithmetic mean) score for the group was 80, was the average score for the women more than 85?

(1) The average score for the men was less than 75. Since the average score for the group was 80 then the average score for women must be more than 80, but we don't know whether it's more than 85. Not sufficient.

(2) The group consisted of more men than women. Clearly insufficient.

(1)+(2) From (1) we have that the distance between the average score for the men and the average score for the whole group is more than 5 and from (2) we have that there are more men than women. Now, in order to compensate that difference and to make the average for the whole group 80 the average of the smaller group (women) must be further from 80 than the average for the larger group (men), so the average score for the women must be more than 85. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel

Can you please explain why C is correct . I am finding it difficult to understand how the number of men and women helps to answer the question

Regards

The average score for the men is 75.
The average score for the women is x.
The average score for the group is 80.

There are more men than women.

Now, ask yourself if x is less than 85, how can the average be 80?
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m = average score for the men
w = average score for the women

Statement 1: m<75

if the number of men = the number of women
80-m=w-80
since m<75
80-m=w-80>5
w>85

if the number of men < the number of women
say 1 men and 2 women
80-m=2(w-80)
since m<75
80-m=2(w-80)>5
w>82.5

if the number of men > the number of women
say 2 men and 1 women
2(80-m)=w-80
since m<75
2(80-m)=w-80>10
w>90
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To me taking example is easy.
Let total men and women = 6
80 80 80 80 80 80

Statement-1. Let 2 men each got 74. So 12 less from the total. To cover that 12, 4 women have to get 12 more in total. So each is 3 more than 80. It means avg of women is 83. But if no of men is equal to or more than no. of women then avg will be more than 85. So not sure- NS

Statement -2. NS

By combining 2nd example works. Sufficient. Ans C
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Hi Bunnel,

When you say the average of men is 75 , the average of women is X and the average of the group as 80 can X be less that 85.

Are we trying to combine the averages of the two groups i.e (75+X)/2=80 ?

If so doesn't this take the form of weighted average?
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kirtivardhan
Hi Bunnel,

When you say the average of men is 75 , the average of women is X and the average of the group as 80 can X be less that 85.

Are we trying to combine the averages of the two groups i.e (75+X)/2=80 ?

If so doesn't this take the form of weighted average?


Hello kirtivardhan

Yes, this is weighted average task. And in case than we have equal quantity of men and women (for example 2) when we have formula that you wrote:

\(\frac{(75m+Xw)}{2}=80\) and X equal to 85.
And we have more men than women (for example 2 men and 1 woman) and from this information we know that X should be more than 85, because if X equal to 85 we will have

\(\frac{(75*2+85*1)}{3}=78\) but this is wrong we should have average 80

so X will be equal to 90:

\(\frac{(75*2+90*1)}{3}=80\)
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kirtivardhan
Hi Bunnel,

When you say the average of men is 75 , the average of women is X and the average of the group as 80 can X be less that 85.

Are we trying to combine the averages of the two groups i.e (75+X)/2=80 ?

If so doesn't this take the form of weighted average?

Hi, its indeed a wted average problem. We know how weights affect the wted average. For instance, if avg for men and women were 75 and 85 respectively and if there were more men than women, the weighted avg would be closer to 75 and hence would be less than 80. But since its given that the weighted average is 80, to make 75 relatively closer to 80 we will have to move the avg for women farther away i.e. more than 85. Does that make sense?
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This question can easily be solved using the alligation diagram.

Before I jump in to solve this question, let me provide you a brief explanation of how to approach weighted average questions on the GMAT. Weighted average questions can be easily solved by making use of the alligation/mixture diagram given below.

Attachment:
Mixtures 1.png
Mixtures 1.png [ 6.07 KiB | Viewed 19303 times ]

Putting in values in the alligation/mixture diagram and subtracting along the diagonals gives us a ratio in which two quantities are mixed. This ratio can now be used to find out what specific amounts of two quantities need to be mixed to obtain a particular mixture.

The only thing that you need to keep in mind here is that the values you need to use, that is the higher value, lower value and mean value have to be values which are associated with the word 'per' (percents, average, per km, per kg etc.).

The alligation/mixture diagram proves useful not only when mixing solutions or combining solids but also to explain the weighted average concept (the word average is also associated with the word per i.e. if the average marks of the class is 80, then it can be understood as 80 marks per student). Say if we have a class A where the average marks is 80 and another class B where the average marks is 70 and the combined average of both class A and B is 74, then we can definitely comment upon which class has the greater number of students. If we represent the average values in the mixture diagram, the ratio of students of Class A and Class B will be 2 : 3. This clearly indicates that class B has the greater number of students.

Now the question here tells us that the overall average (mean value) of the group is 80 and we are asked to answer if the average score of the women is greater than 85.

Statement 1 : The average score for the men was less than 75

Attachment:
Mixtures 2.PNG
Mixtures 2.PNG [ 17.56 KiB | Viewed 19287 times ]

Here we do not have any information about the ratio of men to women. So we ca consider any value for the average score of women. Insufficient.

Statement 2 : The group consisted of more men than women

Attachment:
Capture 3.PNG
Capture 3.PNG [ 14.98 KiB | Viewed 19261 times ]

So if the ration of W : M is 1 : 2, then the average score of the women will be 82 and the average score of the men will be 81 which gives us a NO.

If the ratio of W : M is 1 : 7, then the average score of the women will be 87 and the average score of the men will be 81 which gives us a YES. Insufficient.

Combining 1 and 2 : We know that the average score for the men was less than 75 and the number of men is greater than the number of women.

Attachment:
Combined.PNG
Combined.PNG [ 34.91 KiB | Viewed 19261 times ]

So the average score for the women will always be greater than 85. Sufficient.
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Re: At a certain company, a test was given to a group of men and women see [#permalink]
CrackverbalGMAT Was confused how to use weighted average technique in this. This helped a lot to understand the basics.Thanks!
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TBT
CrackverbalGMAT Was confused how to use weighted average technique in this. This helped a lot to understand the basics.Thanks!
Hi TBT
We are glad we could help you :)

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Video solution here, using the Teeter Totter Timesaver Method (3:11):



­
Teeter Totter Basic Examples Playlist: https://www.youtube.com/playlist?list=PL2exXfCUscn8Hvafet5-IPH1eNNLSjQBP
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At a certain company, a test was given to a group of men and women see [#permalink]
­Let there be 'x' men. (not X-Men) lol
and y women.
Let M = avg Marks of Men
Let W = avg Marks of Women.

so (xM+yW)/(x+y)=80
From S1) we know M<75, but what about W?
Solving the above two eqn yields the following results:
W> 80 + 5(x/y)

So, if x>=y We can safely say W>85
but if x<y, then 5(x/y) will be less than or equal to 4.5, hence W>84.5.

Clearly insufficient

For S2) we see x>y

Clearly insufficient

Combinng, 5(x/y) is greater than 5, and W is greater than 85!

C is sufficient.­
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