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x, y, and z are integers and x<y<z. C is the set of all integers from x to y, inclusive, and D is the set of all integers from y to z, inclusive. The median of set C is \(\frac{5}{8}y\) and the median of set D is \(\frac{3}{4}z\). If E is the set of all integers from x to z, inclusive, then the median of set E is what fraction of z?

x, y, and z are integers and x<y<z. C is the set of all integers from x to y, inclusive, and D is the set of all integers from y to z, inclusive. The median of set C is \(\frac{5}{8}y\) and the median of set D is \(\frac{3}{4}z\). If E is the set of all integers from x to z, inclusive, then the median of set E is what fraction of z?

A. 1/2 B. 9/16 C. 5/8 D. 11/16 E. 3/4

C is the set of all integers from x to y

CONCEPT: For terms in Arithmatic progression (in which adjacent terms are equidistant) the MEAN = MEDIAN = Average of First and Last Term

i.e. Median of C = (x+y)/2 = (5/8)y (Given) i.e. 4x + 4y = 5y i.e. y = 4x

D is the set of all integers from y to z i.e. Median of D = (y+z)/2 = (3/4)z (Given) i.e. 2y + 2z = 3z i.e. z = 2y

Median of Set E = (x+z)/2 = [(y/4)+2y]/2 = 9y/8 = (9/8)*(z/2) = (9/16)z

ANswer: Option B
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Re: x, y, and z are integers and x<y<z. C is the set of all integers from [#permalink]

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21 May 2017, 21:44

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Lets take y=8, so take median of C = 5/8 * 8 = 5. So set C is a set of consecutive integers which goes upto 8 and its median is 5. This means C must be: C = {2, 3, 4, 5, 6, 7, 8}

Now set D should start from 8 and go upto z such that median = 3z/4 lets try for z=16, so median becomes 3*16/4 = 12 So set D goes from 8 to 16 and median is 12 (which fits in properly, no mistake) D = {8, 9, 10, 11, 12, 13, 14, 15, 16}

now Set E = {2, 3, 4, ....., 16} 15 terms, median will be the middle term, which is (15+1)/2 = 8th term 8th term of set E is 9.

We have to say 9 is what fraction of z. 9/16 Hence answer is B