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Set X consists of eight consecutive integers. Set Y consists of all th

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Set X consists of eight consecutive integers. Set Y consists of all th  [#permalink]

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New post 25 Jun 2018, 04:35
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Set X consists of eight consecutive integers. Set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X. How many more integers are there in set Y than in set X ?


A. 0

B. 4

C. 8

D. 12

E. 16


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Re: Set X consists of eight consecutive integers. Set Y consists of all th  [#permalink]

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New post 25 Jun 2018, 04:45

Solution


Given:
    • Set X consists of eight consecutive integers
    • Set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X

To find:
    • How many more integers are there in set Y than in set X

Approach and Working:
    • Let us assume the elements in Set X = {n, n+1, n+2, n+3, n+4, n+5, n+6, n+7}
    • Therefore, the elements in Set Y = {n+4, n+5, n+6, n+7, n+8, n+9, n+10, n+11, n-4, n-3, n-2, n-1, n, n+1, n+2, n+3}
    = {n-4, n-3, n-2, n-1, n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8, n+9, n+10, n+11}

The number of more elements in set Y than in set X = 16 – 8 = 8

Hence, the correct answer is option C.

Answer: C

Note: The question can also be solved assuming any value in place of n


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Re: Set X consists of eight consecutive integers. Set Y consists of all th  [#permalink]

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New post 28 Jun 2018, 17:06
Bunuel wrote:
Set X consists of eight consecutive integers. Set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X. How many more integers are there in set Y than in set X ?


A. 0

B. 4

C. 8

D. 12

E. 16


Since set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X, it must have 8 integers each is 4 more than the ones in X and other 8 integers each is 4 less than the ones in X, unless there are overlaps in the elements obtained in each step. To verify that there are no overlaps, let the smallest element in X be x. After adding 4 to each element in the set, the smallest element obtained in this way is x + 4. If x is the smallest element in X, then x + 7 is the largest element in X (since there are 8 consecutive integers in X). Thus, the largest element obtained after subtracting 4 from each element is x + 3. In other words, there are no overlaps and Y has 8 + 8 = 16 integers, so it has 16 - 8 = 8 integers more than X.

Alternate Solution:

Let’s assume that set X contains 1, 2, 3, 4, 5, 6, 7, 8. Adding 4 to each term yields 5, 6, 7, 8, 9, 10, 11, 12.

Now, we do the same thing, but instead we subtract 4 from each term of the original set; thus, we have -3, -2, -1, 0, 1, 2, 3, 4.

We see that we have a total of 8 + 4 + 4 = 16 elements in set Y. This is 16 - 8 = 8 more elements than the number of elements in set X.

Answer: C
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Re: Set X consists of eight consecutive integers. Set Y consists of all th  [#permalink]

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New post 03 Jul 2018, 07:52
Hi ScottTargetTestPrep

The question says Set X consists of eight consecutive integers and Set Y contains all i.e "8 integers + 4" to each and again all i.e "8 integers + 4" to each. So its pretty clear set Y has 8 more integers in set X ?

Isn't it very clear that the answer is 8. Also, I was unable to understand what you meant by overlap. :)
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Re: Set X consists of eight consecutive integers. Set Y consists of all th  [#permalink]

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New post 30 Sep 2018, 19:25
X = 11, 12, 13, 14, 15, 16, 17, 18,

Subtract 4 from each= 7, 8, 9, 10, 11, 12, 13, 14

Add 4 with each = 15, 16, 17, 18, 19, 20, 21, 22


So, Y= 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22


The bold numbers are in also x. So they are repeated. The new more numbers are the remaining 8.
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Re: Set X consists of eight consecutive integers. Set Y consists of all th  [#permalink]

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New post 04 Nov 2018, 14:46
Bunuel wrote:
Set X consists of eight consecutive integers. Set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X. How many more integers are there in set Y than in set X ?


A. 0

B. 4

C. 8

D. 12

E. 16


NEW question from GMAT® Official Guide 2019


(PS00087)


Bunuel,

For a question like this, I believe that we don't care about the repetition. A set {1,1,1,1,1} contains 5 integers.
So we can easily answer C in this case.

Is my understanding correct?

Thanks!
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Re: Set X consists of eight consecutive integers. Set Y consists of all th  [#permalink]

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New post 05 Nov 2018, 06:25
septwibowo wrote:
Bunuel wrote:
Set X consists of eight consecutive integers. Set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X. How many more integers are there in set Y than in set X ?


A. 0

B. 4

C. 8

D. 12

E. 16


NEW question from GMAT® Official Guide 2019


(PS00087)


Bunuel,

For a question like this, I believe that we don't care about the repetition. A set {1,1,1,1,1} contains 5 integers.
So we can easily answer C in this case.

Is my understanding correct?

Thanks!


Hi!!

The question says set X has consecutive integers. Therefore all the integers in the set will be different integers.

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Re: Set X consists of eight consecutive integers. Set Y consists of all th &nbs [#permalink] 05 Nov 2018, 06:25
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