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Math Expert V
Joined: 02 Sep 2009
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Set M consists of 50 consecutive integers. If the range of the negativ  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 68% (01:59) correct 32% (02:03) wrong based on 112 sessions

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Set M consists of 50 consecutive integers. If the range of the negative integers in Set M is 36, what is the sum of all positive integers in Set M?

A. 78
B. 91
C. 105
D. 112
E. 120

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Re: Set M consists of 50 consecutive integers. If the range of the negativ  [#permalink]

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Bunuel wrote:
Set M consists of 50 consecutive integers. If the range of the negative integers in Set M is 36, what is the sum of all positive integers in Set M?

A. 78
B. 91
C. 105
D. 112
E. 120

If set M has no positive integers, then the range of the negative integers is 49 (if set M has zero) or 50 (if all elements in set M is negative).

Hence, set M must have negative integers and positive numbers.

The range of the negative integers is 36, so that range is {-37, -36, ..., -1}
Next elements of set M are 0, 1, 2,...

Set M consists of 50 consecutive integers, so the last number is 12.
Set M is {-37, -36, -35, ..., 0, ..., 11, 12} (50 numbers)

The sum of all positive numbers in set M is: 12 * (1+12) / 2 = 78

The answer is A
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Re: Set M consists of 50 consecutive integers. If the range of the negativ  [#permalink]

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Bunuel wrote:
Set M consists of 50 consecutive integers. If the range of the negative integers in Set M is 36, what is the sum of all positive integers in Set M?

A. 78
B. 91
C. 105
D. 112
E. 120

The greatest negative number is -1
Let k = least negative number
So, if the range of the negative integers is 36, then -1 - k = 36
This tells us that k = -37
Since the negative values go from -1 all the way to -37, we can see that there are 37 negative values in total.

So, set M = {-37, -36, -35, . . . -1, 0, 1, 2, ...}
What is the GREATEST value in set M?

Well, we know that the 50 integers consist of 37 negative integers and 1 zero. So, the remaining 12 numbers must be positive.
So, set M = {-37, -36, -35, . . . -1, 0, 1, 2, 3, ..., 11, 12}

What is the sum of all positive integers in Set M?
There's a nice formula that says: 1 + 2 + 3 + . . . + n = (n)(n+1)/2
So, 1 + 2 + 3 + + 11 + 12 = (12)(13)/2 = 78

Cheers,
Brent
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Re: Set M consists of 50 consecutive integers. If the range of the negativ  [#permalink]

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For consecutive negative integers with range 36 the negative integer set should be from -1 to -37 as range=max value - min value. The +ive integers will be from 1 to 12. The total count of integers will be 50(inc 0) the sum of positive integers will be n*(n+1)/2 (n=12) hence correct answer is A 78

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Re: Set M consists of 50 consecutive integers. If the range of the negativ  [#permalink]

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1
Bunuel wrote:
Set M consists of 50 consecutive integers. If the range of the negative integers in Set M is 36, what is the sum of all positive integers in Set M?

A. 78
B. 91
C. 105
D. 112
E. 120

Since the range of the negative integers in set M is 36, the negative integers are from -1 to -37 inclusive. Thus, there are 37 negative integers in set M, and after accounting for zero, there are 12 positive integers left, i.e., the integers from 1 to 12 inclusive.

Thus, the sum of the consecutive integers from 1 to 12 is:

Sum = average x quantity

Sum = [(1 + 12)/2] x 12

Sum = 13/2 x 12

Sum = 13 x 6 = 78

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Re: Set M consists of 50 consecutive integers. If the range of the negativ  [#permalink]

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Quote:
This problem tests a handful of concepts related to statistics. First, of course, is the concept of range. The range of a set of numbers is the greatest value minus the least value. So when you're told that the range of the negative integers is 36, and you know that the largest possible negative integer is -1, you know that that calculation is then:
Quote:
-1 - x = 36
So -x = 37, meaning that x = -37
.

Quote:
That then tells you that you have 37 negative numbers, and you should make sure that you include zero in your set of consecutive integers, remembering that zero is neither positive nor negative. With zero, you have 38 numbers that are not positive, out of 50 total; that leaves 12 positive values. That means that your positive set is the integers 1 through 12.

Here you can use the quick method of summing evenly-spaced sets, derived from the formula for the Mean.
Since Mean = Sum of Values / Number of Values, and you know that in an evenly-spaced set the median equals the mean, you can multiply 12 (the number of values) by 6.5 (the middle value, which you can calculate by just averaging the first and last terms, 1 and 12) to get 78.

The correct answer is A.
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Please award kudos, If this post helped you in someway.  Re: Set M consists of 50 consecutive integers. If the range of the negativ   [#permalink] 26 Sep 2018, 09:55
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