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# Set J consists of 18 consecutive even numbers. If the smallest term in

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Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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30 Oct 2015, 08:44
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75% (hard)

Question Stats:

47% (01:14) correct 53% (01:15) wrong based on 499 sessions

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Set J consists of 18 consecutive even numbers. If the smallest term in the set is -10, what is the range of the positive integers in set J?

A. 22
B. 24
C. 34
D. 36
E. 38

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Re: Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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30 Oct 2015, 08:55
1
2
shasadou wrote:
Set J consists of 18 consecutive even numbers. If the smallest term in the set is -10, what is the range of the positive integers in set J?

A. 22
B. 24
C. 34
D. 36
E. 38

Straightforward question that can be solved either by listing all 18 numbers and then finding the range or

18th term of the sequence with first term = -10 and consisting only of even integer = -10+(18-1)*2 = -10+34=24.

Thus the range asked = 24-2 = 22 .

A is the correct answer.

Make sure to NOT include 0 as 0 is neither positive nor negative.

Hope this helps.
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Re: Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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30 Oct 2015, 09:05
3
1
shasadou wrote:
Set J consists of 18 consecutive even numbers. If the smallest term in the set is -10, what is the range of the positive integers in set J?

A. 22
B. 24
C. 34
D. 36
E. 38

Since there are only 18 integers, another approach is the just list all 18.
We get: -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
Range of POSITIVE INTEGERS = 24 - 2 = 22

Answer: A

Cheers,
Brent
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Re: Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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18 Sep 2016, 00:09
GMATPrepNow wrote:
shasadou wrote:
Set J consists of 18 consecutive even numbers. If the smallest term in the set is -10, what is the range of the positive integers in set J?

A. 22
B. 24
C. 34
D. 36
E. 38

Since there are only 18 integers, another approach is the just list all 18.
We get: -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
Range of POSITIVE INTEGERS = 24 - 2 = 22

Answer: A

Cheers,
Brent

howcome 0 is even integer ? 0 is not integer and it is not an even number as i know.
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Re: Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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18 Sep 2016, 00:13
shasadou wrote:
Set J consists of 18 consecutive even numbers. If the smallest term in the set is -10, what is the range of the positive integers in set J?

A. 22
B. 24
C. 34
D. 36
E. 38

-10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18 20 22 24 26

26-2=24 my answer is B 24
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Posts: 7844
Re: Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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18 Sep 2016, 00:45
azamaka wrote:
shasadou wrote:
Set J consists of 18 consecutive even numbers. If the smallest term in the set is -10, what is the range of the positive integers in set J?

A. 22
B. 24
C. 34
D. 36
E. 38

-10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18 20 22 24 26

26-2=24 my answer is B 24

Hi,

0 is an even integer, however it is neither positive nor negative
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Re: Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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18 Sep 2016, 06:42
Top Contributor
1
azamaka wrote:
howcome 0 is even integer ? 0 is not integer and it is not an even number as i know.

An EVEN integer is an integer that can be written as 2k, where k is an integer.
So, 6 is EVEN, because 6 = 2(3) and 3 is an integer
Also, -8 is EVEN, because -8 = 2(-4) and -4 is an integer
And 40 is EVEN, because 40 = 2(20) and 20 is an integer

Likewise, 0 is EVEN, because 0 = 2(0) and 0 is an integer

Cheers,
Brent
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Re: Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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03 Aug 2018, 20:31
1
shasadou wrote:
Set J consists of 18 consecutive even numbers. If the smallest term in the set is -10, what is the range of the positive integers in set J?

A. 22
B. 24
C. 34
D. 36
E. 38

The formula for n-th member of arithmetic progression is:

a(n) = a(1) + d*(n-1)
Here d = 2.
So, a(18) = -10 + 2*(18-1) = 24.
The first positive number in the set is 2.

Range = 24-2 = 22, option A.
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Set J consists of 18 consecutive even numbers. If the smallest term in  [#permalink]

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26 Sep 2018, 09:38
1
This problem is small enough that simple counting may be sufficient for a rapid answer. There is no reason to spend time coming up with the equation for a general relationship when listing out the numbers is equally effective.
The problem tells us that set J consists of 18 consecutive even numbers, starting with -10.
Quote:
Therefore, set J looks like this: {−10,−8,−6,−4,−2,0,2,4,6,8,10,12,14,16,18,20,22,24}

If you don't want to list them out realize that, because of the inclusive set rule, the range of 18 consecutive even integers will not be 18 x 2 but rather 17 x 2 or 34. Starting at -10 and adding 34 puts you at 24 for the largest number.

At this point, you must read carefully and realize that there must be some other type of difficulty in the problem. If the question were asking for the range of the whole set, you would be done and the answer would be 34.
But it is not...it is asking for the range of the POSITIVE numbers.
Since 0 is neither positive nor negative, the first positive integer in the set is 2. Therefore, the range is 24 – 2 = 22 and the correct answer is A,

Quote:
Take Away: The major mistake that people make here in this question after the one already mentioned is regarding 0, 0 is neither positive nor negative.

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Set J consists of 18 consecutive even numbers. If the smallest term in   [#permalink] 26 Sep 2018, 09:38
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# Set J consists of 18 consecutive even numbers. If the smallest term in

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