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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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Question Stats:
59% (02:10) correct
41% (01:21) wrong based on 813 sessions
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.
Re: DS help needed!! [#permalink]
02 Jun 2012, 03:51
4
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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
Re: At a certain company, a test was given to a group of men and [#permalink]
02 Jun 2012, 04:45
23
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Expert's post
15
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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.
Re: At a certain company, a test was given to a group of men and [#permalink]
22 Dec 2012, 04:12
1
This post received KUDOS
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
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 \(Aw>85, Nw> Nm.\)
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.
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. _________________
___________________________________________ Consider +1 Kudos if my post helped
Re: At a certain company, a test was given to a group of men and [#permalink]
01 Jul 2013, 09:50
2
This post received KUDOS
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 _________________
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.
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
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.
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? _________________
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.
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?
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
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. _________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
At a certain company, a test was given to a group of men and [#permalink]
15 Mar 2015, 09:44
1
This post was BOOKMARKED
To me taking example is easy. Let total men and women = 6 808080808080
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
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
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]
26 Jun 2015, 16:51
I didn't like the long and complicated answer in the book, I like to use logic on this one. I think Bunel explained it really well and it's similar to my thought process as well when looking at the question.
1) This really doesn't give you much information. You know men have an average score of less than 75, but that doesn't really give you any context. Were there more compared to the women? Less compared to the women? Were they equal? Because of that I do not have enough information to accurately say this is sufficient. Therefore (1) is INSUFFICIENT
2) This doesn't give you any information either. So what if there are more men than women? I have no numbers aside from the fact that I know the average is 80. This statement on its own is INSUFFICIENT
Together: You know that the men make up the majority of the test takers. You also know that they have a total average of less than 75. This MUST mean that the smaller portion of women need to pull the average with the man in it higher, to bring the group average up to 80. The only way this can happen is if the women's average is greater than 80- and in case greater than 85.
Re: At a certain company, a test was given to a group of men and [#permalink]
30 Sep 2015, 22:40
A1n1+A2n2=80(n1+n2) question is asking us is A2>85? so simply the original question stem for A2 A2=(80-A1)*(n1/n2)+ 80 so.. if A2 has to be greater than 85.. 1st term has to be greater than 5 if A1<75 , n1/n2 is unknown, if N1/n2 is know still we dont know value of 1st term comibing the two we know that 1st term has be greater than 6 (suppose we take extreme case for lowest value of 80-74=6) so A2 will be some value greater than 6 + 80 which is >85 so c is the answer.
gmatclubot
Re: At a certain company, a test was given to a group of men and
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30 Sep 2015, 22:40
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