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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
z and m are integers such that z > -7 and m > 3
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23 May 2019, 12:00
23 May 2019, 12:00
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\(z\) and \(m\) are integers such that \(z > -7\), and \(m > 3\).
If \(k = \frac{6m+2z}{2m+z}\), which of the following COULD be the value of k?
i) 5
ii) 6
iii) 7
A) ii only
B) i and ii
C) i and iii
D) ii and iii
E) i, ii and iii
If \(k = \frac{6m+2z}{2m+z}\), which of the following COULD be the value of k?
i) 5
ii) 6
iii) 7
A) ii only
B) i and ii
C) i and iii
D) ii and iii
E) i, ii and iii
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: z and m are integers such that z > -7 and m > 3
[#permalink]
24 May 2019, 07:42
24 May 2019, 07:42
1
GreenlightTestPrep wrote:
\(z\) and \(m\) are integers such that \(z > -7\), and \(m > 3\).
If \(k = \frac{6m+2z}{2m+z}\), which of the following COULD be the value of k?
i) 5
ii) 6
iii) 7
A) ii only
B) i and ii
C) i and iii
D) ii and iii
E) i, ii and iii
If \(k = \frac{6m+2z}{2m+z}\), which of the following COULD be the value of k?
i) 5
ii) 6
iii) 7
A) ii only
B) i and ii
C) i and iii
D) ii and iii
E) i, ii and iii
Let's examine each statement, starting with....
i) 5
If k = 5, then we get the equation: \(5 = \frac{6m+2z}{2m+z}\)
We can write: \(5(2m+z) = 6m + 2z\)
Expand: \(10m+5z = 6m + 2z\)
Subtract 6m from both sides: \(4m+5z = 2z\)
Subtract 5z from both sides: \(4m = -3z\)
Divide both sides by z to get: \(\frac{4m}{z} = -3\)
Divide both sides by 4 to get: \(\frac{m}{z} = -\frac{3}{4}\)
Since \(m > 3\) and \(z > -7\), it's IMPOSSIBLE for \(\frac{m}{z} = -\frac{3}{4}\)
So, k CANNOT equal 5
Check the answer choices . . . ELIMINATE B, C and E, since they that k COULD equal 5
IMPORTANT: We now know that it MUST be possible for k to equal 6.
We know this because the two remaining answer choices (A and D) both have statement ii is true
This means we need not test statement ii, and we can head straight to...
iii) 7
If k = 7, then we get the equation: \(7 = \frac{6m+2z}{2m+z}\)
We can write: \(7(2m+z) = 6m + 2z\)
Expand: \(14m+7z = 6m + 2z\)
Subtract 6m from both sides: \(8m+7z = 2z\)
Subtract 7z from both sides: \(8m = -5z\)
Divide both sides by z to get: \(\frac{8m}{z} = -5\)
Divide both sides by 8 to get: \(\frac{m}{z} = -\frac{5}{8}\)
Since \(m > 3\) and \(z > -7\), it's IMPOSSIBLE for \(\frac{m}{z} = -\frac{5}{8}\)
So, k CANNOT equal 7
Answer: A
Cheers,
Brent
gmatclubot
Re: z and m are integers such that z > -7 and m > 3 [#permalink]
24 May 2019, 07:42
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