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

Question

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

(PS00087)

ScottTargetTestPrep · Accepted Answer

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

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

EgmatQuantExpert · Answer

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

hero_with_1000_faces · Answer

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.

Raihanuddin · Answer

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.

septwibowo · Answer

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!

dollytaneja51 · Answer

Hi!!

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

Posted from my mobile device

KanishkM · Answer

How many more is the question

Set X : a b c d e f g h

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

Total 16

16-8 
C

himanics · Answer

this question is pretty straight. we do not even need to add 4 or subtract 4 to get set Y. set Y will have double the integer than X. so we can just subtract set X from Set Y.

BijayKru · Answer

Simple take x = (1,2,3,4,5,6,7,8,)

Y = X+4 ( 5,6,,8,9,10,11,12) AND 
 = X-4 (-3,-2,-1,0,1,2,3,4)

SO IN Y total 16 number and in x 8 number so , ans: 8 numbers more in Y compare to X.

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=WUZdcByyBn0

GMATinsight · Answer

Answer: Option C

Video solution by  GMATinsight

https://www.youtube.com/watch?v=hPa-v3mtUyY

ArnauG · Answer

Set X consists of eight consecutive integers, which means it includes 8 numbers. Let's assume the first number in Set X is N.

The eight consecutive integers in Set X can be represented as: N, N + 1, N + 2, N + 3, N + 4, N + 5, N + 6, N + 7.

For each number in Set X, we can add 4 and subtract 4 to create two new numbers in Set Y.

Adding 4 to each number in Set X: N + 4, (N + 1) + 4, (N + 2) + 4, (N + 3) + 4, (N + 4) + 4, (N + 5) + 4, (N + 6) + 4, (N + 7) + 4.

Subtracting 4 from each number in Set X: N - 4, (N + 1) - 4, (N + 2) - 4, (N + 3) - 4, (N + 4) - 4, (N + 5) - 4, (N + 6) - 4, (N + 7) - 4.

Thus, Set Y consists of 16 numbers (8 numbers obtained by adding 4 and 8 numbers obtained by subtracting 4).

To find the difference in the number of integers between Set Y and Set X, we subtract the size of Set X from the size of Set Y:

Number of integers in Set Y - Number of integers in Set X = 16 - 8 = 8.

Therefore, there are 8 more integers in Set Y than in Set X.

The correct answer is:

C. 8

Sa800 · Answer

Why is everyone doing the n+4 thing… since they are consecutive integers there wouldn’t be any overlaps right

TishaHBS · Answer

Hi, I’m confused about whether we’re just concerned with the overall size of the set (even if elements within it are repeated) or do we need the elements of the set Y to be different from set X or is it that we don’t want any element in set Y to be repeated because that then wouldn’t be a different integer?

I didn’t add or subtract 4 and create the list of elements, I simply understood it as set Y having 8 elements of +4 and 8 of -4, meaning total 16 elements and set X has 8, therefore difference in the number of elements is 16-8 = 8. I didn’t concern myself with the contents of each set as such.

Is this wrong? Would be really grateful if you could help me out.

Krunaal · Answer

Hi, let me take an example with smaller set to help put things in perspective, suppose Set X = {-1, 0, 1} and Set Y consists of all the integers that result from adding 1 to each of the integers in set X and all the integers that result from subtracting 1, therefore Set Y = {0, 1, 2, -2, -1} now we have 5 elements here i.e.   2 more than X  , if we had followed an approach where we didn't pay attention to possible contents Set X and Y can have, we erroneously could have got 6 elements for Y = {0, 1, 2, -2, -1,   0  } and got a wrong answer   3 more than X  . That's the reason why we are being cautious about the contents of the set, and making sure there are no overlaps, such as 0 we found in above example. Hope it helps.

TishaHBS · Answer

Yes, this clarified my doubt. Thank you so much!!  Krunaal

shaliny · Answer

hii Bunuel,
 i want to ask here that we have to find that how many more integers in Y than X or how many more different integers in Y than X?

Bunuel · Answer

The question is asking about  how many more total integers are in set Y than in set X , not about "different" ones separately.

Set X has 8 integers. 
 Set Y is formed by taking each integer in X, adding 4, and subtracting 4. So Y can have overlap, but we count the distinct integers. 
 After checking, Y ends up with 16 distinct integers.

So, compared to X, set Y has 16 - 8 =  8 more integers .

Answer: C.

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

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

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