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# Let [x]=(x^2+1)/2 and (y)=3y/2, for all integers x and y

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Intern
Joined: 22 Mar 2012
Posts: 4
Let [x]=(x^2+1)/2 and (y)=3y/2, for all integers x and y  [#permalink]

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Updated on: 23 Mar 2012, 14:25
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4
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Difficulty:

15% (low)

Question Stats:

81% (01:11) correct 19% (01:39) wrong based on 190 sessions

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Let $$[x]=\frac{x^2+1}{2}$$ and $$(y)=\frac{3y}{2}$$, for all integers $$x$$ and $$y$$. If $$m=(2)$$, [m] is equal to which of the following?

A. 13/8
B. 5/2
C. 15/4
D. 5
E. 37/2

Originally posted by Jakevmi80 on 23 Mar 2012, 14:00.
Last edited by Bunuel on 23 Mar 2012, 14:25, edited 2 times in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 60496

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23 Mar 2012, 14:22
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Jakevmi80 wrote:
symbol {x} = (x^2 +1)/2 and symbol (y) =3y/2, for all integers x and y. If m =2, the symbol for ({M}) is equal to?
(a) 13/8
(B) 3
(C) 15/4
(D) 5
(E) 37/2

Let $$[x]=\frac{x^2+1}{2}$$ and $$(y)=\frac{3y}{2}$$, for all integers $$x$$ and $$y$$. If $$m=(2)$$, [m] is equal to which of the following?
A. 13/8
B. 5/2
C. 15/4
D. 5
E. 37/2

Since $$(y)=\frac{3y}{2}$$ then $$m=(2)=\frac{3*2}{2}$$ --> $$m=(2)=3$$. So the question becomes: [m]=[3]=?

Now, since $$[x]=\frac{x^2+1}{2}$$ then $$[3]=\frac{3^2+1}{2}=5$$.

Hope it's clear.
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Re: Let [x]=(x^2+1)/2 and (y)=3y/2, for all integers x and y  [#permalink]

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14 Jan 2019, 02:25
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Re: Let [x]=(x^2+1)/2 and (y)=3y/2, for all integers x and y   [#permalink] 14 Jan 2019, 02:25
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# Let [x]=(x^2+1)/2 and (y)=3y/2, for all integers x and y

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