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
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An automated manufacturing unit employs N experts such
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We may as well assume our lowest salary is 0, our median is 5, and our largest is 10. So our set looks something like this:
0, a, b, c, ..., 5, d, e, f, ... 10
Now if we want to make the mean of this set as large as possible, we want to make every element as large as possible, so that the sum is as big as possible. So we want the set to look like:
0, 5, 5, 5,.... 5, 10, 10, 10, ..., 10
You'd do this with an odd number of elements, or with an even number - if the set has an even number of elements, making the two middle values both 5's allows you to make all of the smallest values as large as possible.
Now, if you can see that you want an odd number of elements in the set, you can answer the question immediately (since only answer C is odd). What we're really trying to do here is to make the mean as large as possible using as few elements as possible. Say we look at a set with the right range and mean, where we maximize all our values, and where we have an even number of elements. Let's use six elements:
0, 5, 5, 5, 10, 10
Notice if we just remove one of the 5's, we'd get a five-element set with the right median and range:
0, 5, 5, 10, 10
but the mean of this set must be larger than before, because we removed a '5' from the first set, and 5 is below the average of that set. If you remove something that's below average, your average goes up.
So if we can get a mean of 7 in this question using k elements, where k is even, we can easily get a mean of 7 using k-1 elements. So the minimum number of elements must be odd, and only 15 could be correct.
Or you can construct the set. Say we have this set, and we assume we have an odd number of elements:
0, 5, 5, ..., 5, 10, 10, 10, ...., 10
If we have n values equal to 5, and n values equal to 10, then we have 2n+1 values in total. If the mean is 7, then the sum is 7(2n+1). But because we have n values equal to 5 and n values equal to 10 the sum is also 5n + 10n. So
5n + 10n = 7(2n + 1)
15n = 14n + 7
n = 7
so we want seven 5s, seven 10s, and one 0, for 15 elements in total.
If you can see that the answer should be odd, you don't need to check for an even number of elements, but if you do, I think you'll find you need 20 elements to get the same mean.
0, a, b, c, ..., 5, d, e, f, ... 10
Now if we want to make the mean of this set as large as possible, we want to make every element as large as possible, so that the sum is as big as possible. So we want the set to look like:
0, 5, 5, 5,.... 5, 10, 10, 10, ..., 10
You'd do this with an odd number of elements, or with an even number - if the set has an even number of elements, making the two middle values both 5's allows you to make all of the smallest values as large as possible.
Now, if you can see that you want an odd number of elements in the set, you can answer the question immediately (since only answer C is odd). What we're really trying to do here is to make the mean as large as possible using as few elements as possible. Say we look at a set with the right range and mean, where we maximize all our values, and where we have an even number of elements. Let's use six elements:
0, 5, 5, 5, 10, 10
Notice if we just remove one of the 5's, we'd get a five-element set with the right median and range:
0, 5, 5, 10, 10
but the mean of this set must be larger than before, because we removed a '5' from the first set, and 5 is below the average of that set. If you remove something that's below average, your average goes up.
So if we can get a mean of 7 in this question using k elements, where k is even, we can easily get a mean of 7 using k-1 elements. So the minimum number of elements must be odd, and only 15 could be correct.
Or you can construct the set. Say we have this set, and we assume we have an odd number of elements:
0, 5, 5, ..., 5, 10, 10, 10, ...., 10
If we have n values equal to 5, and n values equal to 10, then we have 2n+1 values in total. If the mean is 7, then the sum is 7(2n+1). But because we have n values equal to 5 and n values equal to 10 the sum is also 5n + 10n. So
5n + 10n = 7(2n + 1)
15n = 14n + 7
n = 7
so we want seven 5s, seven 10s, and one 0, for 15 elements in total.
If you can see that the answer should be odd, you don't need to check for an even number of elements, but if you do, I think you'll find you need 20 elements to get the same mean.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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To make the math easier, use smaller numbers, as follows:gmat_guy666 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
Range of the salaries = 10.
Lowest salary = 0.
Average salary = 7.
Median salary = 5.
Highest salary = 10.
We can PLUG IN THE ANSWERS, which represent the smallest possible number of employees.
Since we want the smallest possible value, start with the smallest answer choice.
Answer choice A: 10
Here, the sum of the 10 salaries = (number of employees)(average salary) = 10*7 = 70.
Since the median salary = 5, the sum of the 2 salaries in the middle = 10.
In ascending order, the 10 salaries could look as follows:
0, a, b, c, 5, 5, d, e, f, 10.
If a=b=c=5 and d=e=f=10, the greatest possible sum for the 10 salaries = 0 + (5*5) + (4*10) = 65.
Since the sum of the 7 salaries must by $70, eliminate A.
Answer choice B: 12
Here, the sum of the 12 salaries = (number of employees)(average salary) = 12*7 = 84.
Since the median salary = 5, the sum of the 2 salaries in the middle = 10.
In ascending order, the 12 salaries could look as follows:
0, a, b, c, d, 5, 5, e, f, g, h, 10.
If a=b=c=d=5 and e=f=g=h=10, the greatest possible sum for the 10 salaries = 0 + (6*5) + (5*10) = 80.
Since the sum of the 7 salaries must by $84, eliminate B.
Answer choice C: 14
Here, the sum of the 14 salaries = (number of employees)(average salary) = 14*7 = 98.
Since the median salary = 5, the sum of the 2 salaries in the middle = 10.
In ascending order, the 14 salaries could look as follows:
0, a, b, c, d, e, 5, 5, f, g, h, i, j, 10.
If a=b=c=d=e=5 and f=g=h=i=j=10, the greatest possible sum for the 14 salaries = 0 + (7*5) + (6*10) = 95.
Since the sum of the 7 salaries must by $98, eliminate C.
Answer choice D: 15
Here, the sum of the 15 salaries = (number of employees)(average salary) = 15*7 = 105.
In ascending order, the 15 salaries would look as follows:
0, a, b, c, d, e, f, 5, g, h, i, j, k, l, 10.
If a=b=c=d=e=f=5 and g=h=i=j=k=l=10, the greatest possible sum for the 15 salaries = 0 + (7*5) + (7*10) = 105.
Success!
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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