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25 May, 16:40

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?

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  1. 25 May, 17:30
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    Step-by-step explanation:

    Given that X consists of 8 consecutive integers

    say X = {n, n+1, ... n+8}

    Set Y is formed by adding 4 to each and also subtracting 4 from each element of X

    Numbers got by addition = n+4, n+5, ... n+12

    Numbers got by subtraction = n-4, n-3, n-2, n, n+1, n+2, n+3, n+4

    We find that n+4 is repeated in both.

    Hence Y = {n-4, n-3, ... n+4, ... n+12}

    n (Y) = 15

    Y has 7 integers more than X
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