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Set X consists of eight consecutive integers. Set Y consists

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by BTGmoderatorDC » Fri Dec 14, 2018 5:32 pm

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Answers

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C

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E

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Difficulty—

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

OA C

Source: Official Guide
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by deloitte247 » Sun Dec 16, 2018 11:22 am
$$Let\ set\ x=\ 11,12,13,14,15,16,17,18$$
subtract 4 from the elements of set x
$$=7,8,9,10,11,12,13,14$$
Add 4 to elements of set x
$$=15,16,17,18,19,20,21,22$$
so set Y is
$$=7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22$$
New element of set Y that are not in set x
$$=7,8,9,10,19,20,21,22$$ $$=8\ elements$$
$$answer\ is\ Option\ C$$
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by Scott@TargetTestPrep » Thu Mar 14, 2019 3:56 pm
BTGmoderatorDC 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

OA C

Source: Official Guide
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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