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Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4] [#permalink]
06 May 2012, 09:13
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Galiya wrote:
When X is even, [x]=a/2+1, when X is odd [x]=2a+1 then [7]*[4]=?
A. [22] B. [44] C. [45] D. [88] E. [90]
your thoughts? i went with C and was happy till didn't see the official answer
[7]*[4]=(2*7+1)(4/2+1)=45.
Now, notice that the answer choices are also in boxes, so we have that [x]=45, which means that for even x it should be 45=x/2+1 --> x=88=even and for odd x it should be 45=x*2+1 --> x=22=even, so not a valid solution.
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4] [#permalink]
26 Jun 2012, 02:42
Bunuel was reviewing my posts and got a question (im pretty sure, after the explanation ill think- OMG it was so obvious, but anyway): looking at the question above, why do we substitute [x], e.g. [7] or [4], instead of a and b? x and a (or b) are not the same numbers.
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4] [#permalink]
26 Jun 2012, 02:47
1
This post received KUDOS
Expert's post
Galiya wrote:
Bunuel was reviewing my posts and got a question (im pretty sure, after the explanation ill think- OMG it was so obvious, but anyway): looking at the question above, why do we substitute [x], e.g. [7] or [4], instead of a and b? x and a (or b) are not the same numbers.
It should be: "When X is even, [x]=x/2+1, when X is odd [x]=2x+1 then [7]*[4]=?" _________________
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4] [#permalink]
05 Oct 2013, 03:43
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Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4] [#permalink]
03 Nov 2014, 07:47
Bunuel wrote:
Galiya wrote:
When X is even, [x]=a/2+1, when X is odd [x]=2a+1 then [7]*[4]=?
A. [22] B. [44] C. [45] D. [88] E. [90]
your thoughts? i went with C and was happy till didn't see the official answer
[7]*[4]=(2*7+1)(4/2+1)=45.
Now, notice that the answer choices are also in boxes, so we have that [x]=45, which means that for even x it should be 45=x/2+1 --> x=88=even and for odd x it should be 45=x*2+1 --> x=22=even, so not a valid solution.
Why do we not consider 22 as an answer? Is it because 22 as an answer was even when equated to an odd equation, i.e 2a+1? Does the solution also have to be odd when equated to an odd equation (2a+1)? I hope my question makes sense ... not sure if I was able to explain myself clearly? =P Please let me know!
Re: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[4] [#permalink]
03 Nov 2014, 07:58
Expert's post
Rca wrote:
Bunuel wrote:
Galiya wrote:
When X is even, [x]=a/2+1, when X is odd [x]=2a+1 then [7]*[4]=?
A. [22] B. [44] C. [45] D. [88] E. [90]
your thoughts? i went with C and was happy till didn't see the official answer
[7]*[4]=(2*7+1)(4/2+1)=45.
Now, notice that the answer choices are also in boxes, so we have that [x]=45, which means that for even x it should be 45=x/2+1 --> x=88=even and for odd x it should be 45=x*2+1 --> x=22=even, so not a valid solution.
Why do we not consider 22 as an answer? Is it because 22 as an answer was even when equated to an odd equation, i.e 2a+1? Does the solution also have to be odd when equated to an odd equation (2a+1)? I hope my question makes sense ... not sure if I was able to explain myself clearly? =P Please let me know!
Thanks as always!!
We got that [7]*[4] = 45.
Now, [22], since 22 is even, is 22/2 + 1 = 12, not 45.
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