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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by AJWILL » Sun Jul 22, 2012 3:13 am
In compny x the median annual salary is $50000. what percent of the employees earned less than $80000 and more than $50,000?
1)the employees who earned less than $80,000 and more than $50,000 is equal to the number of the employees who earned no less than $80,000
2)30% of the total salary was earned by the employees who earned less than $80000 and more than $50000
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by theCEO » Sun Jul 22, 2012 6:22 am
AJWILL wrote:In compny x the median annual salary is $50000. what percent of the employees earned less than $80000 and more than $50,000?
1)the employees who earned less than $80,000 and more than $50,000 is equal to the number of the employees who earned no less than $80,000
2)30% of the total salary was earned by the employees who earned less than $80000 and more than $50000
We are asked to find % of employees... which means no. of employees/ total number of employees

Glancing through the choices, I do not see any number of employees given, so the question cannot be answered. Also no infomation is given about employees with salaries under 50,000
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by Jim@StratusPrep » Mon Jul 23, 2012 6:44 am
First thing is that wording here is pretty poor.

Second

Statement 1 says that the number of people between 50k and 80k is equal to the number above 80k
Statement 2 says that the total dollar figure earned by people earning between 50k and 80k is 30% of the total dollar amount.

The answer is E because we have no idea the income distribution.
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by piti » Sun Jul 29, 2012 5:17 am
I think the answer is A. We need to solve for (50K < Num < 80K)/N, where N is the total number of employees, and Num is the number of employees that earn between 50K and 80K.

Since the median is 50K, half earn less/eq than 50K, half earn more/eq than 50K, so there are N/2 employees that earn above 50K. Statement #1 tells us that the number above 80K equals that between 50K and 80K, so we know that between 50K and 80K we have N/4. Therefore, the answer is 1/4, and #1 is sufficient. #2 isn't sufficient - it gives no info on the number. Thus, the answer is A.
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by theCEO » Sun Jul 29, 2012 8:40 am
piti wrote:I think the answer is A. We need to solve for (50K < Num < 80K)/N, where N is the total number of employees, and Num is the number of employees that earn between 50K and 80K.

Since the median is 50K, half earn less/eq than 50K, half earn more/eq than 50K, so there are N/2 employees that earn above 50K. Statement #1 tells us that the number above 80K equals that between 50K and 80K, so we know that between 50K and 80K we have N/4. Therefore, the answer is 1/4, and #1 is sufficient. #2 isn't sufficient - it gives no info on the number. Thus, the answer is A.
We need to find the percentage of employees who earned less than $80000 and more than $50,000?
If we don't know the total number of employees we can't solve this problem!
We don't even know the highest salary or the lowest salary!

In compny x the median annual salary is $50000. what percent of the employees earned less than $80000 and more than $50,000?
1)the employees who earned less than $80,000 and more than $50,000 is equal to the number of the employees who earned no less than $80,000
What if there were employees who earned over 80,000?


2)30% of the total salary was earned by the employees who earned less than $80000 and more than $50000
What if there were employees who earned over 80,000?

Hope I am right :)
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by willrc » Thu Aug 02, 2012 12:58 pm
The wording is indeed rather slack. But it's just possible to infer C as the answer, as follows:

1) tells us that employees earning more than the median can be split equally into those earning >80k and those earning <=80k. Since 50% earn more than the median, it appears that 25% earn between 50k and 80k. Looking good... BUT! If everyone earned exactly 50k, all these conditions are satisfied with no-one earning >50k-80k. So because of the edge case, this is INSUFFICIENT.

2) on its own, this is INSUFFICIENT, because knowing the total salary fraction of this 50-80k group does not tell us how many there are.

However, statement 2 does rule out the edge case in statement 1 -- i.e. taking the statements together, we know some employees earn >50k. Hence the answer is C.[/spoiler]
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