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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

5

This post received KUDOS

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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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