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CEO  V
Joined: 12 Sep 2015
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z and m are integers such that z > -7 and m > 3  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 34% (03:07) correct 66% (02:45) wrong based on 27 sessions

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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

_________________

Originally posted by GMATPrepNow on 23 May 2019, 07:59.
Last edited by GMATPrepNow on 24 May 2019, 07:22, edited 1 time in total.
Senior Manager  P
Joined: 09 Jun 2014
Posts: 269
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
Re: z and m are integers such that z > -7 and m > 3  [#permalink]

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GMATPrepNow 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

I liked this questions.First I tried an approach by plugging in numbers and I got 6 as one of the option and then it was taking time to figure out others.

Here is a better approach which I figured out after my first attempt.

Lets plugin in values of from options and make sure the values the z and m stay as integers and follow the condition stated in question stem
which are as below

z > -7, and
m > 3

Options 1:

Substituting K=5

k=(6m+2z)/(2m+z)

5*(2m+z) = 6m+2z

10m+ 5z = 6m + 2z

4m= - 3z

m= - 3/4 * z

for m to be an integer z has to be a multiple of 4

Now z=-4 (makes m=3 which is not possible since m > 3)

((Z=-8 .No need to try since condition is Z>-7))

Now z= 4 (makes m=-3 which is not possible since m > 3)

So we conclude 5 is not possible.

Now Option 2:

Substituting K=6

k=(6m+2z)/(2m+z)

6*(2m+z)= 6m+2z

12m + 6z = 6m +2z

6m = -4z

3m = -2z

m= -2/3*z

Now z=3 (makes m=-2 which is not possible since m > 3)
Now z= -3 (makes m=2 which is not possible since m > 3)
Now z= -6 (makes m=4 which is possible since m > 3)

So we conclude 6 is possible.

Option 3:

Substituting K=7

7=(6m+2z)/(2m+z)

7*(2m+z)= 6m+2z

14m + 7z = 6m +2z

8m = -5z

m= -5/8*z

Now z=8 (makes m=-5 which is not possible since m > 3)

Now z= -8 (makes m=5 which is possible since m > 3)

This is not possible since Z>-7 as stated before.

So we conclude 7 is also NOT possible.

So I think option A is correct now and K can be only 6 but not 7 ,5.

Press Kudos if it helps!!

Originally posted by prabsahi on 23 May 2019, 10:22.
Last edited by prabsahi on 23 May 2019, 19:13, edited 1 time in total.
Senior Manager  P
Joined: 19 Oct 2018
Posts: 465
Location: India
Re: z and m are integers such that z > -7 and m > 3  [#permalink]

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z>-7, hence z can't be equal to -8

Now z=8 (makes m=-5 which is not possible since m > 3)
Now z= -8 (makes m=5 which is possible since m > 3)

So we conclude 7 is also possible.
SVP  V
Joined: 26 Mar 2013
Posts: 2229
Re: z and m are integers such that z > -7 and m > 3  [#permalink]

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prabsahi wrote:
GMATPrepNow 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

I liked this questions.First I tried an approach by plugging in numbers and I got 6 as one of the option and then it was taking time to figure out others.

Here is a better approach which I figured out after my first attempt.

Lets plugin in values of from options and make sure the values the z and m stay as integers and follow the condition stated in question stem
which are as below

z > -7, and
m > 3

Options 1:

Substituting K=5

k=(6m+2z)/(2m+z)

5*(2m+z) = 6m+2z

10m+ 5z = 6m + 2z

4m= - 3z

m= - 3/4 * z

for m to be an integer z has to be a multiple of 4

Now z=-4 (makes m=3 which is not possible since m > 3)

((Z=-8 .No need to try since condition is Z>-7))

Now z= 4 (makes m=-3 which is not possible since m > 3)

So we conclude 5 is not possible.

Now Option 2:

Substituting K=6

k=(6m+2z)/(2m+z)

6*(2m+z)= 6m+2z

12m + 6z = 6m +2z

6m = -4z

3m = -2z

m= -2/3*z

Now z=3 (makes m=-2 which is not possible since m > 3)
Now z= -3 (makes m=2 which is not possible since m > 3)
Now z= -6 (makes m=4 which is possible since m > 3)

So we conclude 6 is possible.

Option 3:

Substituting K=7

7=(6m+2z)/(2m+z)

7*(2m+z)= 6m+2z

14m + 7z = 6m +2z

8m = -5z

m= -5/8*z

Now z=8 (makes m=-5 which is not possible since m > 3)

Now z= -8 (makes m=5 which is possible since m > 3)

So we conclude 7 is also possible.

So I think option D is correct and K can be 6,7 but not 5.

Press Kudos if it helps!!

Hi,
How z could be -8 in the highlighted part while z > -7??? I think it is not possible.
Senior Manager  P
Joined: 09 Jun 2014
Posts: 269
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
Re: z and m are integers such that z > -7 and m > 3  [#permalink]

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nick1816 wrote:
z>-7, hence z can't be equal to -8

Now z=8 (makes m=-5 which is not possible since m > 3)
Now z= -8 (makes m=5 which is possible since m > 3)

So we conclude 7 is also possible.

Thanks for pointing out Nick.I have made the correction.
Senior Manager  P
Joined: 09 Jun 2014
Posts: 269
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
Re: z and m are integers such that z > -7 and m > 3  [#permalink]

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Mo2men wrote:
prabsahi wrote:
GMATPrepNow 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

I liked this questions.First I tried an approach by plugging in numbers and I got 6 as one of the option and then it was taking time to figure out others.

Here is a better approach which I figured out after my first attempt.

Lets plugin in values of from options and make sure the values the z and m stay as integers and follow the condition stated in question stem
which are as below

z > -7, and
m > 3

Options 1:

Substituting K=5

k=(6m+2z)/(2m+z)

5*(2m+z) = 6m+2z

10m+ 5z = 6m + 2z

4m= - 3z

m= - 3/4 * z

for m to be an integer z has to be a multiple of 4

Now z=-4 (makes m=3 which is not possible since m > 3)

((Z=-8 .No need to try since condition is Z>-7))

Now z= 4 (makes m=-3 which is not possible since m > 3)

So we conclude 5 is not possible.

Now Option 2:

Substituting K=6

k=(6m+2z)/(2m+z)

6*(2m+z)= 6m+2z

12m + 6z = 6m +2z

6m = -4z

3m = -2z

m= -2/3*z

Now z=3 (makes m=-2 which is not possible since m > 3)
Now z= -3 (makes m=2 which is not possible since m > 3)
Now z= -6 (makes m=4 which is possible since m > 3)

So we conclude 6 is possible.

Option 3:

Substituting K=7

7=(6m+2z)/(2m+z)

7*(2m+z)= 6m+2z

14m + 7z = 6m +2z

8m = -5z

m= -5/8*z

Now z=8 (makes m=-5 which is not possible since m > 3)

Now z= -8 (makes m=5 which is possible since m > 3)

So we conclude 7 is also possible.

So I think option D is correct and K can be 6,7 but not 5.

Press Kudos if it helps!!

Hi,
How z could be -8 in the highlighted part while z > -7??? I think it is not possible.

You are right.I have made the necessary corrections.Thanks for pointing out. CEO  V
Joined: 12 Sep 2015
Posts: 3777
Re: z and m are integers such that z > -7 and m > 3  [#permalink]

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1
1
Top Contributor
1
GMATPrepNow 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

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

Cheers,
Brent
_________________ Re: z and m are integers such that z > -7 and m > 3   [#permalink] 24 May 2019, 08:42
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