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Re: An automated manufacturing unit employs N experts such that the range
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13 Apr 2015, 06:03
Bunuel wrote: An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?
(A) 10 (B) 12 (C) 14 (D) 15 (E) 20
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Let’s first assimilate the information we have. We need to find the minimum number of experts that must be there. Why should there be a minimum number of people satisfying these statistics? Let’s try to understand that with some numbers. Say, N cannot be 1 i.e. there cannot be a single expert in the unit because then you cannot have the range of $10,000. You need at least two people to have a range – the difference of their salaries would be the range in that case. So there are at least 2 people – say one with salary 0 and the other with 10,000. No salary will lie outside this range. Median is $5000 – i.e. when all salaries are listed in increasing order, the middle salary (or average of middle two) is $5000. With 2 people, one at 0 and the other at 10,000, the median will be the average of the two i.e. (0 + 10,000)/2 = $5000. Since there are at least 10 people, there is probably someone earning $5000. Let’s put in 5000 there for reference. 0 … 5000 … 10,000 Arithmetic mean of all the salaries is $7000. Now, mean of 0, 5000 and 10,000 is $5000, not $7000 so this means that we need to add some more people. We need to add them more toward 10,000 than toward 0 to get a higher mean. So we will try to get a mean of $7000. Let’s use deviations from the mean method to find where we need to add more people. 0 is 7000 less than 7000 and 5000 is 2000 less than 7000 which means we have a total of $9000 less than 7000. On the other hand, 10,000 is 3000 more than 7000. The deviations on the two sides of mean do not balance out. To balance, we need to add two more people at a salary of $10,000 so that the total deviation on the right of 7000 is also $9000. Note that since we need the minimum number of experts, we should add new people at 10,000 so that they quickly make up the deficit in the deviation. If we add them at 8000 or 9000 etc, we will need to add more people to make up the deficit at the right. Now we have 0 … 5000 … 10000, 10000, 10000 Now the mean is 7000 but note that the median has gone awry. It is 10,000 now instead of the 5000 that is required. So we will need to add more people at 5000 to bring the median back to 5000. But that will disturb our mean again! So when we add some people at 5000, we will need to add some at 10,000 too to keep the mean at 7000. 5000 is 2000 less than 7000 and 10,000 is 3000 more than 7000. We don’t want to disturb the total deviation from 7000. So every time we add 3 people at 5000 (which will be a total deviation of 6000 less than 7000), we will need to add 2 people at 10,000 (which will be a total deviation of 6000 more than 7000), to keep the mean at 7000 – this is the most important step. Ensure that you have understood this before moving ahead. When we add 3 people at 5000 and 2 at 10,000, we are in effect adding an extra person at 5000 and hence it moves our median a bit to the left. Let’s try one such set of addition: 0 … 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000 The median is not $5000 yet. Let’s try one more set of addition. 0 … 5000, 5000, 5000, 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000, 10000, 10000 The median now is $5000 and we have maintained the mean at $7000. This gives us a total of 15 people. Answer (D) Granted, the question is tough but note that it uses very basic concepts and that is the hallmark of a good GMAT question! Try to come up with some other methods of solving this.
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An automated manufacturing unit employs N experts such that the range
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Updated on: 13 Apr 2015, 06:19
Bunuel wrote: An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?
(A) 10 (B) 12 (C) 14 (D) 15 (E) 20
Kudos for a correct solution. let s take lowest salary to be 1000. therefore highest is 11000 ( because range = highest  lowest). median is 6000.(lowest +5000) now we need to minimize N and we know that avg is(1000+7000) 8000, which is above the median (6000). to get the mean to 8000 we ll have to maxmize the salaries of the ppl above the mean. but max we can take is 11000. at the same time maximize salaries of people below the median therefore , max we can take is 6000. now we have values of 3(1000+6000+11000) people which totals 18000. thus avg is 6000 as of now . now we ll start adding 2 ppl at a time , one with sal above the mean and one with below. therefore 6000+11000=17000 so now for 5 people the avg is (17000+18000)/5=7000. so this way we see the avg is moving up towards 8000. similarly we keep adding 2 ppl and by the time we cross 15 ppl the avg crosses 8000. hence IMO the ans is D.
Originally posted by Yash12345 on 07 Apr 2015, 10:15.
Last edited by Yash12345 on 13 Apr 2015, 06:19, edited 1 time in total.



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08 Apr 2015, 03:19
Am  minimum salary, Am + 10000 = max salary, Am + 5000  median, Am + 7000  average What the guy above said is indeed correct regarding the way to approach our goal. Median must be constant, so we need to add people accordingly, as in one person "to the left" from median and 1 person "to the right" from the average, to offset the first person's salary. With just 2 people we got average at Am + 5,5 so we need to get that 1,5 somewhere to match Am+7000. The quickest way to increase our average is to pick the biggest possible salaries for the new people, thus using information from our question I came up with this equation. \(A_m+7000=\frac{(A_m+a*(A_m+5000)+(b+1)*(A_m+10000))}{(a+b+2)}\) a  number of people with salary equal to median b  number of people with salary equal to max salary in order to preserve our median's value, \(a >= b + 1\) (coz if we have to many people with max salary, our median will shift upwards) solving the equation we end up with \(2a + 4 = 3b\) taking into account the \(a >=b + 1\) with a simple plug method we determine that a = 7, b = 6 are the first possible values we can use and thus \(a + b + 2 = 15\)  minimum amount of people, the answer to our question. 15 = answer D



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Re: An automated manufacturing unit employs N experts such that the range
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08 Feb 2016, 21:33
Okay i need an advise... the median value is $5000, doesnt this shows that the number of experts must be some Odd number (answer choice D)..?
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12 Feb 2016, 17:49
Hi sananoor, When a set has an EVEN number of values, the median is equal to the AVERAGE of the two 'middle terms.' As such it can still be an integer. Here are some examples: {5, 5, 5, 5} > the median is 5 {4, 6, 8, 9} > the median is 7 GMAT assassins aren't born, they're made, Rich
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An automated manufacturing unit employs N experts such that the range
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25 Feb 2016, 22:32
Quote: Question: An automated manufacturing unit employs N experts. Their average monthly salary is $7000 while the median monthly salary is only $5000. If the range of their monthly salaries is $10,000, what is the minimum value of N?
(A)10 (B)12 (C)14 (D)15 (E)20
Hey Bunuel, I am quoting a vairation of this question from statisticsmadeeasyallinonetopic203966.html#p1563225 and I see why there was a tweak in the verbiage introduced as in this thread (mostly to avoid below). Please validate my solution and confirm if I am thinking in the right direction: We are given the salary sequence as: ... 5000(median),...,7000(mean),.... I choose $15000 as the higher limit to the salary, so that gives me: ..., 5000,7000,15000 To balance out the mean and satisfy the range condition, I add 5000s on left side of the mean: 5000,5000,5000,5000,7000,15000. (Mean: 7000, Range:10000, Median: 5000). So the least possible value of N satisfying all conditions is 6. (which is lower than any answer choices mentioned) If above is correct, may be it would be a good idea to edit the question here statisticsmadeeasyallinonetopic203966.html#p1563225 to make it more full proof.
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15 Mar 2016, 21:06
5000,5000,5000,5000,5000,5000,6000,9000,10000,15000 ( TOTAL 10 NUMBERS ) RANGE= 150005000=10000 MEDIAN=(5000+5000)/2 = 5000 AVERAGE= SUM/TOTAL=70000/10=7000 OPTION : A minimum



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Re: An automated manufacturing unit employs N experts such that the range
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19 Mar 2016, 09:59
MuraliPappala wrote: 5000,5000,5000,5000,5000,5000,6000,9000,10000,15000 ( TOTAL 10 NUMBERS ) RANGE= 150005000=10000 MEDIAN=(5000+5000)/2 = 5000 AVERAGE= SUM/TOTAL=70000/10=7000 OPTION : A minimum Read carefully the question says ABOVE the LOWEST VALUE i.e in your example the mean has to be 12000 and Median has to be 10000
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Re: An automated manufacturing unit employs N experts such that the range
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20 Mar 2016, 02:28
An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?
Let the lowest/first no is L. Max will be 10000+L, Median will be L+5000. L................L+5000...............L+10000 =====> Mean = (7000+L)
......(N1)/2 nos....1(i.e. L+5000).........(N1)/2 Nos..... In total N nos
The nos between L and (L+5000) are assumed to be same as (L+5000) while the numbers between (L+5000) and (L+10000) are assumed to be (10000+L). I have taken (L+5000) and (L+10000) as the mean is 7000+L which may be achieved only if such values are taken.
Sum = N*Mean L +{(N3)/2}*(L+5000) +(L+5000) +{(N1)/2}*(L+10000) = N*(L+7000) Solving we get N = 15. Rest all cancels out.
Hence D



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Re: An automated manufacturing unit employs N experts such that the range
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03 May 2016, 13:04
Hello everyone. I ended up getting a totally different result: a minimum of 3 experts. Suppose you have three experts with the salaries of $3000, $5000 and $13000. Is the median $5000? CHECK Is the average $7000? $3000 + $5000 + $13000 = $21000 / 3 = $7000 CHECK Is the range $10000? $13000  $3000 = $10000 CHECK I really think the author's answer explanation for this one takes a lot of things for granted. For example, where does the assumption of "at least 10 experts" come from early on? Besides, you do realize that you don't necessarily have to assume the minimum and maximum salaries to be medianrange/2 and median+range/2, right? As long as you have one expert earning 5k, an odd number of experts, and the expert's salaries evenly spread between the 5k mark, the median will remain 5k. So how did I come up with these values? Let's analyse each variable: A median of 5k means you'll have at least one expert. A range of 10k means at least two experts. You could have only two experts, one earning 10k and another earning zero, and you would still have a 5k median. An average of 7k means the sum of the salaries will be a multiple of 7. This also means you will need more than two experts to meet all the criteria, because if you had only 2, you would need to meet the median criteria which means the sum of the two salaries divided by two, AND the average criteria which is, under this context, calculated the same way, but has a different value. Combining all the three values, you know that you must have at least one expert earning 5k to meet the 5k criteria, but since the range is 10k and you must express that with at least 2 more experts, the sum of the salaries is at least 15k. Noting that the sum of the salaries must also be a multiple of 7, the first number that meets all the conditions is 21k, where the median salary is 5k, the lowest salary could be anything between 0 and 5k, exclusive, and the maximum salary is 21k  5k  lowest salary.
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Re: An automated manufacturing unit employs N experts such that the range
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03 May 2016, 13:45
samirabrahao1 wrote: Hello everyone.
I ended up getting a totally different result: a minimum of 3 experts.
Suppose you have three experts with the salaries of $3000, $5000 and $13000.
Is the median $5000? CHECK Is the average $7000? $3000 + $5000 + $13000 = $21000 / 3 = $7000 CHECK Is the range $10000? $13000  $3000 = $10000 CHECK
I really think the author's answer explanation for this one takes a lot of things for granted. For example, where does the assumption of "at least 10 experts" come from early on? Besides, you do realize that you don't necessarily have to assume the minimum and maximum salaries to be medianrange/2 and median+range/2, right? As long as you have one expert earning 5k, an odd number of experts, and the expert's salaries evenly spread between the 5k mark, the median will remain 5k.
So how did I come up with these values? Let's analyse each variable:
A median of 5k means you'll have at least one expert.
A range of 10k means at least two experts. You could have only two experts, one earning 10k and another earning zero, and you would still have a 5k median.
An average of 7k means the sum of the salaries will be a multiple of 7. This also means you will need more than two experts to meet all the criteria, because if you had only 2, you would need to meet the median criteria which means the sum of the two salaries divided by two, AND the average criteria which is, under this context, calculated the same way, but has a different value.
Combining all the three values, you know that you must have at least one expert earning 5k to meet the 5k criteria, but since the range is 10k and you must express that with at least 2 more experts, the sum of the salaries is at least 15k. Noting that the sum of the salaries must also be a multiple of 7, the first number that meets all the conditions is 21k, where the median salary is 5k, the lowest salary could be anything between 0 and 5k, exclusive, and the maximum salary is 21k  5k  lowest salary. Reread the question : the average is not 7000, but 7000 above the least value



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Re: An automated manufacturing unit employs N experts such that the range
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03 May 2016, 14:35
Alex75PAris wrote: Reread the question : the average is not 7000, but 7000 above the least value It appears someone has modified the original question, which was: statisticsmadeeasyallinonetopic203966.html#p1563225"Question: An automated manufacturing unit employs N experts. Their average monthly salary is $7000 while the median monthly salary is only $5000. If the range of their monthly salaries is $10,000, what is the minimum value of N? (A)10 (B)12 (C)14 (D)15 (E)20" The discussion link contained in that thread leads to this one, so I assumed this was related to the same question.
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An automated manufacturing unit employs N experts such that the range
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10 Aug 2016, 09:15
Bunuel wrote: An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?
(A) 10 (B) 12 (C) 14 (D) 15 (E) 20
Kudos for a correct solution. 0 5000 5000 5000 5000 5000 9800 9800 9800 9800 9800 10000 Count=12, Average 7000. median=5000 , Hence answer can be 12 as well. Many options like this can be worked out as this is a complete hit & trial question. I would rate this question as extremely poor w.r.t GMAT standards



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An automated manufacturing unit employs N experts such that the range
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10 May 2017, 13:19
Hi VeritasPrepKarishma . Could you check would my reasoning be correct, If the questiion was the original one. I want to check this concept on a similar problem. So here is the old version: An automated manufacturing unit employs N experts. Their average monthly salary is $7000 while the median monthly salary is only $5000. If the range of their monthly salaries is $10,000, what is the minimum value of N? (so there is no condition saying that mean is 7000 above lowest salary) Here is how I would do it: Start with 3000, 5000, 13000, try to get N>10 and keep mean at 7000 and mode at 5000(If I am right starting with mean 7000 minimizes number of additions needed to be done  I want to contrast this with starting with 0, 5000, 10000). We want to add the least number of elements possible, so in each step we add 13000 (+6000) and offset it with adding 5000 three times (2000,2000,2000) to keep the median. So each step adds minimum 4 members. Therefore, we would have 3, 7, 11, 15... members after each addition, and 11 is lowest number greater than 10 that satisfies initial conditions. Is my thinking right here?



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Re: An automated manufacturing unit employs N experts such that the range
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11 May 2017, 07:52
kivalo wrote: Hi VeritasPrepKarishma . Could you check would my reasoning be correct, If the questiion was the original one. I want to check this concept on a similar problem. So here is the old version: An automated manufacturing unit employs N experts. Their average monthly salary is $7000 while the median monthly salary is only $5000. If the range of their monthly salaries is $10,000, what is the minimum value of N? (so there is no condition saying that mean is 7000 above lowest salary) Here is how I would do it: Start with 3000, 5000, 13000, try to get N>10 and keep mean at 7000 and mode at 5000(If I am right starting with mean 7000 minimizes number of additions needed to be done  I want to contrast this with starting with 0, 5000, 10000). We want to add the least number of elements possible, so in each step we add 13000 (+6000) and offset it with adding 5000 three times (2000,2000,2000) to keep the median. So each step adds minimum 4 members. Therefore, we would have 3, 7, 11, 15... members after each addition, and 11 is lowest number greater than 10 that satisfies initial conditions. Is my thinking right here? In case your mean and median are not with reference to the lowest value, then 3000, 5000, 13000 is a case that satisfies all conditions. Why would you want N to be greater than 10? N certainly cannot be 1 and 2 (with 2 elements, mean = median)
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Re: An automated manufacturing unit employs N experts such that the range
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11 May 2017, 08:27
VeritasPrepKarishma wrote: kivalo wrote: Hi VeritasPrepKarishma . Could you check would my reasoning be correct, If the questiion was the original one. I want to check this concept on a similar problem. So here is the old version: An automated manufacturing unit employs N experts. Their average monthly salary is $7000 while the median monthly salary is only $5000. If the range of their monthly salaries is $10,000, what is the minimum value of N? (so there is no condition saying that mean is 7000 above lowest salary) Here is how I would do it: Start with 3000, 5000, 13000, try to get N>10 and keep mean at 7000 and mode at 5000(If I am right starting with mean 7000 minimizes number of additions needed to be done  I want to contrast this with starting with 0, 5000, 10000). We want to add the least number of elements possible, so in each step we add 13000 (+6000) and offset it with adding 5000 three times (2000,2000,2000) to keep the median. So each step adds minimum 4 members. Therefore, we would have 3, 7, 11, 15... members after each addition, and 11 is lowest number greater than 10 that satisfies initial conditions. Is my thinking right here? In case your mean and median are not with reference to the lowest value, then 3000, 5000, 13000 is a case that satisfies all conditions. Why would you want N to be greater than 10? N certainly cannot be 1 and 2 (with 2 elements, mean = median) No particular reasons, the older version of the question from this topic that I found had all the answer choices greater than 10, even though N is minimized when n = 3 (probably that's why wording is now changed). I just wanted to check whether I would be able to make n>10 anyway, to practice this concept (I wasn't able to ind similar questions). Thank you for replying!



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06 Aug 2018, 10:23
Bunuel wrote: Bunuel wrote: An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?
(A) 10 (B) 12 (C) 14 (D) 15 (E) 20
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Let’s first assimilate the information we have. We need to find the minimum number of experts that must be there. Why should there be a minimum number of people satisfying these statistics? Let’s try to understand that with some numbers. Say, N cannot be 1 i.e. there cannot be a single expert in the unit because then you cannot have the range of $10,000. You need at least two people to have a range – the difference of their salaries would be the range in that case. So there are at least 2 people – say one with salary 0 and the other with 10,000. No salary will lie outside this range. Median is $5000 – i.e. when all salaries are listed in increasing order, the middle salary (or average of middle two) is $5000. With 2 people, one at 0 and the other at 10,000, the median will be the average of the two i.e. (0 + 10,000)/2 = $5000. Since there are at least 10 people, there is probably someone earning $5000. Let’s put in 5000 there for reference. 0 … 5000 … 10,000 Arithmetic mean of all the salaries is $7000. Now, mean of 0, 5000 and 10,000 is $5000, not $7000 so this means that we need to add some more people. We need to add them more toward 10,000 than toward 0 to get a higher mean. So we will try to get a mean of $7000. Let’s use deviations from the mean method to find where we need to add more people. 0 is 7000 less than 7000 and 5000 is 2000 less than 7000 which means we have a total of $9000 less than 7000. On the other hand, 10,000 is 3000 more than 7000. The deviations on the two sides of mean do not balance out. To balance, we need to add two more people at a salary of $10,000 so that the total deviation on the right of 7000 is also $9000. Note that since we need the minimum number of experts, we should add new people at 10,000 so that they quickly make up the deficit in the deviation. If we add them at 8000 or 9000 etc, we will need to add more people to make up the deficit at the right. Now we have 0 … 5000 … 10000, 10000, 10000 Now the mean is 7000 but note that the median has gone awry. It is 10,000 now instead of the 5000 that is required. So we will need to add more people at 5000 to bring the median back to 5000. But that will disturb our mean again! So when we add some people at 5000, we will need to add some at 10,000 too to keep the mean at 7000. 5000 is 2000 less than 7000 and 10,000 is 3000 more than 7000. We don’t want to disturb the total deviation from 7000. So every time we add 3 people at 5000 (which will be a total deviation of 6000 less than 7000), we will need to add 2 people at 10,000 (which will be a total deviation of 6000 more than 7000), to keep the mean at 7000 – this is the most important step. Ensure that you have understood this before moving ahead. When we add 3 people at 5000 and 2 at 10,000, we are in effect adding an extra person at 5000 and hence it moves our median a bit to the left. Let’s try one such set of addition: 0 … 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000 The median is not $5000 yet. Let’s try one more set of addition. 0 … 5000, 5000, 5000, 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000, 10000, 10000 The median now is $5000 and we have maintained the mean at $7000. This gives us a total of 15 people. Answer (D) Granted, the question is tough but note that it uses very basic concepts and that is the hallmark of a good GMAT question! Try to come up with some other methods of solving this. As bunnel mentions "Now the mean is 7000 but note that the median has gone awry. It is 10,000 now instead of the 5000 that is required" but how has the median went to 10,000 shouldn't it have went 2 points up which is 7000 ?




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