An automated manufacturing unit employs N experts such that the range

Question

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.

(A) 10
(B) 12
(C) 14
(D) 15
(E) 20

Bunuel · Accepted Answer

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.

Zhenek · Answer

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

Yash12345 · Answer

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.

sananoor · Answer

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)..?

EMPOWERgmatRichC · Answer

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

Vinayprajapati · Answer

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 the lowest/first no is L.
Max will be 10000+L, Median will be L+5000.
L................L+5000...............L+10000 =====> Mean = (7000+L)

......(N-1)/2 nos....1(i.e. L+5000).........(N-1)/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 +{(N-3)/2}*(L+5000) +(L+5000) +{(N-1)/2}*(L+10000) = N*(L+7000)
Solving we get N = 15. Rest all cancels out.

Hence D

kivalo · Answer

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?

KarishmaB · Answer

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)

kivalo · Answer

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!

AliciaSierra · Answer

IanStewart ,  MathRevolution

Hello Ian, MathRevolution,

Is there any easier way to do this question. I couldn't understand above mentioned approaches. Could you please help me here.

Thanks,

IanStewart · Answer

The information we're given is purely based on distances, so our actual numbers don't matter. So we can assume the smallest salary is $0, and then the largest must be $10,000, because of the range. If that were our set, we'd get a median of $5000, but our mean wouldn't be large enough (our mean needs to be $7000). So we need to add elements that will raise the mean without changing the median. We want to add as few elements as possible, and since the mean is based on the sum of our values, we want to add values that are as large as possible.

Roughly half the values in a list are less than or equal to the median, so we'll want all of those values to be $5000, so they can be as large as possible. Roughly half the values are greater than or equal to the median, so we'll want to make all of those $10,000, so they can be as large as possible. So our list will look like this, if we're trying to maximize our mean with the fewest possible elements:

0, 5000, 5000, 5000, ...., 5000, 10000, 10000, 10000, ... 10000

Say we have n values equal to 5000. Then if the list has an odd number of values, for the median to be 5000 we can have at most n values equal to 10,000, and one value equal to zero. So using as many 10,000s as we can, the sum of the list is 5000n + 10,000n, and the list will have 2n+1 values in total. Since the mean is equal to 7000, we have

(5000n + 10,000n)/(2n + 1) = 7000
5000n + 10,000n = 14,000n + 7000
1000n = 7000
n = 7

and since we have 2n + 1 values in the list, 2(7) + 1 = 15 is the smallest number of values we can have.

I assumed we had an odd number of values, but it's harder to make the mean large with an even number of values, because you need to include an extra 5000 in the middle of the set, which drags the mean down. Of course, if one wanted to confirm that, one could try making a set with 14 values (one zero, seven 5000s, and six 10,000s) to confirm you don't get a mean as large as 7000.

It's not a question I'd worry about too much though.

bhupendersingh27 · Answer

take lowest = 0
median= 5000
average= 7000
highest possible = 10000

0---------------5000------7000-------10000

for any set if average is greater than median, the observation greater than median should be high enough or equal to greatest observation, to make average greater than median, and to restrict the greatest possible value (as in this case it is restricted by range) observation below median should be equal to median value.

now from answer choice
A, n-10
2x median = 5th+6th = 10000
lowest = 0, 2nd=3rd=4th = 5000
total of 6 values = 25000
balance = 10x7000 - 25000 = 45000,
greatest possible value = 45000/4= 11250... not possible as it is >10000

lets try option C, N=14
2xmedian = 7th + 8th=10000
lowest=0, 2nd to 6th all values=5000
total upto 8 values= 35000
balance = 14x7000-35000=63000
greatest possible = 63000/6= 10500... >10000 not possible

Try with option D, N=15
greatest possible values is {15x7000-(0+7x5000)}/7 = 10000.. possible

answer option D

ThatDudeKnows · Answer

Let's start with a really basic case to start seeing how things work.

If there were only three employees, we could make them 0, 5000, and 10000. That settles our range and median, but our mean is too low. Okay, we need to add at least one more employee over 7000, which means we will need to add one more employee at or below the median. Let's go with 0, 5000, 5000, 10000, 10000. That's a total of 30000 with 5 employees, so the mean is 6000. Still too low. We can add two more employees...yadda yadda yadda...every time we add two employees, the number of employees will still be odd. Only one answer choice is odd.

Answer choice D.

arraj · Answer

the range of the set is 10000.

So 0..........10000

median is given as 5000 greater than lowest value

So 0.........5000..........10000

mean is given as 7000. If we calculate the mean of the existing 3 values(0+5000+10000/3 = 5000)
so each of three values should increase by 2000(2000*3=6000) in order for the mean to become 7000.

Since the median is 5000 we should add value to both the sides of 5000.
We cannot just add on one side. If so median will not be 5000. In order to maximize the value added(this is because we need to make the mean as 7000 for the first three numbers), we have to choose largest value in btw 0 to 5000 and in btw 5000 to 10000, which is 5000,10000.

0.....5000,5000.......10000,10000

each time we add pair of 10000,5000. we are adding 15000 which is 7500*2. So totally they are having 500*2=1000 extra than the required mean(7000). This 1000 can be added to the first three values(0,5000,10000) in order to make their mean 7000. Earlier we had found that those three values should increase by 6000 inorder for to get 7000 as mean. So total of 6 pairs or 6*2=12 numbers should be added for mean to become 7000. Now the total number is 12+3=15

Freddy12 · Answer

Doesn't this approach be valid only if N is odd?

Posted from my mobile device

Adarsh_24 · Answer

I think may be we can use algebra as below too.

N odd = 2n+1
N even = 2n
We will ignore 1000 from all values and take 5,7,10 instead.

Lowest say x
Largest 10+x

We want to increase every unknown as large as possible. So everything below median will be same as median and above median same as largest

Odd:
x ————5+x———-10+x

Each side of 5+x has n items.
So we have 
x———-(n-1) times (5+x) ——-(5+x)———(n-1) times (10+x)———-(10+x)

x+(5+x)n+n(10+x)=(7+x)(2n+1)
Which will give n=7
N= 2*7+1=15

Even:
2 (5+x) in the middle(we pick both equal so lower numbers can be as max as possible to reduce N)
x——-(5+x)(5+x)————(10+x)
2n-4 Items remaining => n-2 both sides
x+n(5+x)+(n-1)(10+x) = 2n(7+x)
n=10 or N=20

Odd is lower

I checked others solutions again. X won’t have any effect since it’s all differences. We can consider x as 0 initially itself like others did. It will avoid all the mess.

Kinshook · Answer

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 the lowest salary be $L per month

Average monthly salary = L+7000
Median monthly salary = L + 5000
Highest monthly salary = L + 10000

Since median is below average, more number of salaries are below average

Let the distribution = {L, L+5000, L+5000, L+5000,..... N-x-1 times, L+10000, L+10000, .... x times}
Average monthly salary = {NL + 10000x + 5000(N-x-1)}/N = L + 10000x/N + 5000(N-x-1)/N = L + 7000
10000x + 5000(N-x-1) = 7000N
10x + 5N - 5x - 5 = 7N
5x - 5 = 2N
x = (2N + 5)/5; 
Since x & N are both positive integers and x < N/2
(x, N) = {7,15}

The distribution = {L, L+5000, L+5000, L+5000, L+5000, L+5000, L+5000, L+5000, L+5000, L+5000, L+10000, L+10000, L+10000, L+10000, L+10000, L+10000, L+10000}

Minimum value of N = 15

IMO D

shonik · Answer

This only applies if the members of the set are all same or when they are in AP.
It doesn't apply as purely in all other sets of numbers

mehulnarang010 · Answer

I've solved it like this. Since the median is 5000 so at least half the datapoints will be <=5000. now one of them we know is at 0 salary. The rest of the ones to the "left" of 5000, we will keep at 5000 only (to be closes to the mean of 7000 to get the minimum N.

Let's assume there are  x  employees at 5000. so that makes a total of x + 1 that have <=5000 salary. To have the median as 5000, we will need equal number of employees to the right of 5000 as well. so

Employees to the right will also be  x

Now we want these x employees to be as far away from the mean as possible to make up for the deficit on the left. So we will place all these at 10.

So we have,
1 employee with salary = 0
x employee with salary = 5000
 x employee with salary = 10000

Hence, the total amount will be 1*0 +x*5000 + x*10000 = x*15000

Since the average is 7000, the total amount = 7000*(2x+1)

Equating both we get x*15000 = x*14000 + 7000

x= 7
total = 2x+1 = 15

Do note how the number has to be odd to ensure the minimum value of N.

An automated manufacturing unit employs N experts such that the range

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