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Is n= 2m+ 3?

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Is n= 2m+ 3?  [#permalink]

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New post 28 Sep 2018, 10:31
2
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

29% (01:33) correct 71% (01:56) wrong based on 93 sessions

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Is \(n= 2m+ 3\) ?

1)\((n-3)^2\) = \(4m^2\)
2)\(n^2\) =\(4m^2 +12m+9\)

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Re: Is n= 2m+ 3?  [#permalink]

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New post 28 Sep 2018, 12:37
From statement 1:

\((n−3)^2\) = 4\(m^2\)

\(n^2\)+9-6n = 4\(m^2\)

Rearranging gives \(n^2\) = 4\(m^2\)+6n-9.
A is insufficient.

From statement 2:

\(n^2\) = 4\(m^2\)+12m+9

or \(n^2\) = \((2m+3)^2\).

n can be +(2m+3) or -(2m+3).
Hence B is insufficient.

Combining both gives
4\(m^2\)+12m+9 = 4\(m^2\)+6n-9

or 12m-6n+18 = 0
2m-n+3 = 0
or n = 2m+3.

C is the answer.
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Re: Is n= 2m+ 3?  [#permalink]

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New post 01 Oct 2018, 08:17
1
Please explain why C? I chose D.
(n-3)^2 = 4m^2
Moving RHS to LHS and using a^2 -b^2 = (a+b)(a-b)
We get
N-3-2m =0 and n- 3+2m =0
Therefore
N=2m+3 and 3-2m
Therefore the question is statistfied.

Similarly for statement B.

Therefore answer should be D
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Re: Is n= 2m+ 3?  [#permalink]

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New post 01 Oct 2018, 08:36
You yourself have mentioned in the solution that there are 2 values one is positive and other is negative. If i a statement is giving two possible answers, it's clearly not sufficient.
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Re: Is n= 2m+ 3?  [#permalink]

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New post 01 Oct 2018, 09:06
Abhi077 wrote:
You yourself have mentioned in the solution that there are 2 values one is positive and other is negative. If i a statement is giving two possible answers, it's clearly not sufficient.


The question asks if n = 2m +3. The solution states that n can hold this value. The question never states that n should only have this value.
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Re: Is n= 2m+ 3?  [#permalink]

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New post 01 Oct 2018, 09:29
Abhi077 wrote:
Is \(n= 2m+ 3\) ?

1)\((n-3)^2\) = \(4m^2\)
2)\(n^2\) =\(4m^2 +12m+9\)


Statement I;

Take m = 0, n = 3.

Take m = -1, n = 1,5.... Insufficient.

Statement II:

Take m = 0, n = 3,-3... Insufficient.

Combining the two Statements,

n = 2m + 3.
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Re: Is n= 2m+ 3?  [#permalink]

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New post 01 Oct 2018, 09:29
Top Contributor
Bhumsa wrote:
Please explain why C? I chose D.
(n-3)^2 = 4m^2
Moving RHS to LHS and using a^2 -b^2 = (a+b)(a-b)
We get
N-3-2m =0 and n- 3+2m =0
Therefore
N=2m+3 and N = 3-2m
Therefore the question is statistfied.

Similarly for statement B.

Therefore answer should be D


You're correct to say that statement 1 yields 2 possible solutions (in red above)
However, if N = 2m + 3, then the answer to the target question is "YES, n does equal 2m + 3)
And if N = 3 - 2m , then the answer to the target question is "NO, n does not equal 2m + 3)

Since we have 2 DIFFERENT answers to the target question, the statement is NOT sufficient.
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Re: Is n= 2m+ 3?  [#permalink]

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New post 09 Oct 2018, 06:50
n = 2m + 3 <=> n - 3 = 2m

1) --> |n-3| = |4m| --> insufficient
2) --> |n| = |2m + 3| --> insufficient

1-2) --> (n-3)^2 - n^2 = (4m)^2 - (2m+3) ^2
<=> n = 2m + 3

--> sufficient --> C
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Re: Is n= 2m+ 3? &nbs [#permalink] 09 Oct 2018, 06:50
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