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Practice Question #3Each employee of a certain company [#permalink]
29 May 2014, 08:18
Question Stats:
33% (00:55) correct
66% (00:26) wrong based on 3 sessions
Each employee of a certain company is in either Department X or Department Y, and there are more than twice as many employees in Department X as in Department Y. The average (arithmetic mean) salary is $25,000 for the employees in Department X and $35,000 for the employees in Department Y. Which of the following amounts could be the average salary for all of the employees of the company? Indicate all such amounts. (A) $26,000 (B) $28,000 (C) $29,000 (D) $30,000 (E) $31,000 (F) $32,000 (G) $34,000
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Re: Practice Question #3 [#permalink]
29 May 2014, 08:21
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Solution One strategy for answering this kind of question is to find the least and/or greatest possible value. Clearly the average salary is between $25,000 and $35,000, and all of the answer choices are in this interval. Since you are told that there are more employees with the lower average salary, the average salary of all employees must be less than the average of $25,000 and $35,000, which is $30,000. If there were exactly twice as many employees in Department X as in Department Y, then the average salary for all employees would be, to the nearest dollar, the following weighted mean, \(\frac{(2)(25,000)+(1)(35000)}{(2+1)}= 28,333\) where the weight for $25,000 is 2 and the weight for $35,000 is 1. Since there are more than twice as many employees in Department X as in Department Y, the actual average salary must be even closer to $25,000 because the weight for $25,000 is greater than 2. This means that $28,333 is the greatest possible average. Among the choices given, the possible values of the average are therefore $26,000 and $28,000. Thus, the correct answer consists of Choices A ($26,000) and B ($28,000).Intuitively, you might expect that any amount between $25,000 and $28,333 is a possible value of the average salary. To see that $26,000 is possible, in the weighted mean above, use the respective weights 9 and 1 instead of 2 and 1. To see that $28,000 is possible, use the respective weights 7 and 3.
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Re: Practice Question #3 [#permalink]
10 Oct 2018, 09:50
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more than twice as many
means that the number of X employees, Nx, is greater than 2 times the number of Y employees, Ny:
Nx > 2*Ny
The arithmatic means are:
(1) Mx = SUM(Sx)/Nx = 25k (2) My = SUM(Sy)/Ny = 35k Where S is the salary of employee i and the sums go over all employees in the respective departments
Entire Company
(3) Mtot = (SUM(Sx) + SUM(Sy))/(Nx + Ny) or solving for the sums in (1) and (2)
(3a) Mtot = (25k*Nx +35k*Ny)/(Nx + Ny) Plug in Nx = 2*Ny to get the limit on the maximum Mtot
Mtot = (50k + 35k)* Ny/(3*Ny) = 28.3k So the answers are a) 26k and b) 28k
The answer is a and b because the question says there are more than twice as many employees in company X than Y. If the maximum limit is 28.3k then plausible answers are anything below that as when you have more employees in company X it will drag down the average



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Re: Practice Question #3 [#permalink]
10 Oct 2018, 09:57
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There are x people in Department X and y people in Department Y.
Twice as many people in X as in Y would mean x = 2 y, so more than twice as many is simply x > 2y.
The total salary for everyone in Department X is $25,000 x and the total salary for everyone in Department Y is $35,000 y. The total salary for both departments is $25,000 x + $35,000 y.
The average salary for both departments is the total salary divided by the total number of people: *($25,000 x + $35,000 y) / (x + 2y).
Now, first some critical thinking before we go further. We could just try out a bunch of numbers, but we might miss something if we don't understand what's going on. Department Y people make more money than Department X people. So we can get a minimum average salary by having lots of X people and few Y people.
So let's try y=1 and x=1000. I get $25,009.99, which means that all of the answers are still fair game. Note that you could do this without actually trying numbers... Mathematically, you can solve for the average salary as the ratio of employees (x/y) goes to infinity (LOTS of X workers compared to Y workers). You would find the limit is $25,000. Or you could just think about it, and see that the more X workers you have, the closer the average will shift towards the Department X salary.
Now we need a maximum limit on average salary. This one's a bit trickier because of the "at least twice as many" thing. But the same basic principle applies. How do we get the maximum possible salary? By having as many Y workers and as few X workers as possible. We could take y=1 and therefore x=3.... But that's three times as many X workers. For higher numbers, we can lessen the difference between them. So let's try y=1000 and x=2001 (just barely more than twice as many). I get an average salary of $28,332.22. We could keep trying bigger numbers to fine tune things, but you'll quickly realize that's enough. An upper limit of about $28,333 means (a) and (b) are the answers.
That last one is harder to see by inspection... but you can still do it. Just need to realize that for every one $35,000 employe, you have at least two $25,000 employees. If you input y=1 and x=2 you get an average salary of $28,333.33. You just have to realize that this is a limit, meaning the highest average salary must be less than this number (because there will always be at least one more X worker in the mix to pull the average down). And of course you can do it mathematically as well, by solving for this limit. This time around you want the average salary as the ratio of employees (x/y) approaches 2. You'll get the same thing.




Re: Practice Question #3
[#permalink]
10 Oct 2018, 09:57





