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# At a certain company, a test was given to a group of men and

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At a certain company, a test was given to a group of men and [#permalink]  01 Jun 2012, 23:28
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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.
(2) The group consisted of more men than women.

Would someone please help me out with this question. I got no clue combining st.1 and 2 and the output of it.
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Jun 2012, 04:26, edited 1 time in total.
Edited the question
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Re: DS help needed!! [#permalink]  02 Jun 2012, 03:51
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rajman41 wrote:
At a certain company, a test was given to a group of men and women seeking promotions. If the average(artihmetic mean) score for the group was 80, was the average score for the women greater than 85?
1) The average score for the men was less than 75.
2) The group consisted of more men than women.

Thank you!!

Clearly both statement by itself are insufficient.

Let us say the group were equal.
Then if Men(x) average was 75, then women(x) average would be 85
As 75x + 85x = 80(2x)

IF Men were more (lets say x+1) and average of women's score was W
75(x+1) + Wx = 80(2x+1)
75x + 75 + Wx = 160x + 80
Wx - 85x = 5
x(W-85) = 5
As X cannot be zero or negetive
W-85>0
or W>85

BOth sufficient
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Re: At a certain company, a test was given to a group of men and [#permalink]  02 Jun 2012, 04:45
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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.

Hope it's clear.
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Re: At a certain company, a test was given to a group of men and [#permalink]  02 Jun 2012, 06:02
This is so far the clear and precise explanation, thanks Bunuel!!
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Re: At a certain company, a test was given to a group of men and [#permalink]  22 Dec 2012, 04:12
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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 [#permalink]  28 Dec 2012, 06:03
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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Re: At a certain company, a test was given to a group of men and [#permalink]  20 Feb 2013, 22:29
Bunuel wrote:
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.

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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Re: At a certain company, a test was given to a group of men and [#permalink]  01 Jul 2013, 09:50
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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 [#permalink]  06 Oct 2013, 10:31
Bunuel wrote:
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.

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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Re: At a certain company, a test was given to a group of men and [#permalink]  06 Oct 2013, 10:42
Expert's post
DivyanshuRohatgi wrote:
Bunuel wrote:
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.

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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Re: At a certain company, a test was given to a group of men and [#permalink]  06 Oct 2013, 10:48
Bunuel wrote:
DivyanshuRohatgi wrote:
Bunuel wrote:
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.

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?

Thanxs for the explanation
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Re: At a certain company, a test was given to a group of men and [#permalink]  02 Dec 2013, 21:41
rohanGmat wrote:
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

I like this approach. I used a similar approach using weighted averages.

The general formula is

[(average #1)(a) + (average #2)(b)] = weighted average, where:
a + b = 1, and

a and b represent the relative weightings of the two sub-groups

A similar approach is elaborated in this link
http://www.manhattangmat.com/blog/index ... -problems/
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Re: At a certain company, a test was given to a group of men and [#permalink]  02 Dec 2013, 22:18
rajman41 wrote:
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.
(2) The group consisted of more men than women.

Would someone please help me out with this question. I got no clue combining st.1 and 2 and the output of it.

Let A1 = Average score of men
n1= no. of men
A2= Average score of Women
n2= no.of women

Let A be the average of the group which is given as 80 in Q. stem
let n be the total no. of men and women. Clearly, n= n1+n2

Now we know that

n1*A1+n2*A2= n*A ---------------------- > 1

We need to find whether A2 > 85 or not

the above statement can be re-written as

A2= (n*A- n1*A1)/n2

or A2= (n1*A+n2*A- n1*A1)/n2

A2= (n1/n2)*80 + 80 - (n1/n2)*A1

Let n1/n2= x

So A2= 80+ (80-A1)*x -------------------------> 2

Now from st1 we have that A1<75.....

take any value of A1, lets take 74 then the equation 2 becomes

A2= 80 + 6*x --------> Now A2 will depend upon the ration of n1/n2 if n1/n2>/=1 then A2 >85 or else less than 85

St 1 is not sufficient as we don't know the value of x

From st 2 we know that n1/n2> 1 but we don't the value of A 1

hence not sufficient.

Combining both the equation we get that n1/n2>1 and A1 <75 and clearly for any value it will be sufficient.

Ans C.
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At a certain company, a test was given to a group of men and [#permalink]  20 Jul 2014, 20:32
1
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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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At a certain company, a test was given to a group of men and [#permalink]  15 Mar 2015, 09:44
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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Re: At a certain company, a test was given to a group of men and [#permalink]  23 Apr 2015, 11:07
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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At a certain company, a test was given to a group of men and [#permalink]  23 Apr 2015, 12:32
kirtivardhan wrote:
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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Re: At a certain company, a test was given to a group of men and [#permalink]  23 Apr 2015, 19:51
rajman41 wrote:
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.
(2) The group consisted of more men than women.

Would someone please help me out with this question. I got no clue combining st.1 and 2 and the output of it.

1) worthless without knowing the number of people, but lets us know that all men scored under 75 which can be applied to the next statement

2) If the group consisted of more men than women it is fair to assume that in order for LESS woman to pull the average to 80 from 75 they will need to have scored relatively high. We can test this just in case they didn't need to score above 85 though. Set the number of people at 10 and say there are 6 men and 4 women. We can make the following equation

(6*less than 75 + 4x)/10 = 80 ---->(choose the highest number below 75---> ( 74*6 + 4x)/10 = 80

444+4x = 800

4x = 800- 444

4x = 89 <---- This is the *lowest* the females needed to score in order to pull the average up to 80

Re: At a certain company, a test was given to a group of men and   [#permalink] 23 Apr 2015, 19:51
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