z and m are integers such that z -7 and m 3

Question

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

prabsahi · Accepted Answer

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!!

nick1816 · Answer

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.

Mo2men · Answer

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

prabsahi · Answer

Thanks for pointing out Nick.I have made the correction.

prabsahi · Answer

You are right.I have made the necessary corrections.Thanks for pointing out.

BrentGMATPrepNow · Answer

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

prabsahi · Answer

Let's examine each statement, starting with.... i) 5 Option A: Lets say k=5 Therefore expression becomes \frac{6m+2z}{2m+z} 10m+5z = 6m + 2z 4m = -3 z m = -3/4 z Now in the question its given m>3 and z<-7 so lets substitute -3/4 z > 3 -z > 4 z < -4 Now this can holds true. What am I missing ? I guess while writing this query I got the solution intuitively .It seems when I plug bac z=-5 it doesnt satisfy the condition. But still how to avoid such approach on the exam day. We generally dont pluggin the numbers back once we have the equation of this kind. Can you please help. Thanks a ton!!

vishalkazone · Answer

Option A
Can we solve this question by logic?

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

As m and z are integers,
So value of numerator has to br EVEN.
And
Value of denominator can be Either ODD or EVEN

Now EVEN / EVEN has to be EVEN
And EVEN/ ODD has to be EVEN

So value of k=6m+2z/2m+z cant be odd.

And from answer choice we need to select one option.

So Answer has to be A.( Even)

Posted from my mobile device

prabsahi · Answer

This approach looks real good!!

It can definitely help save some time..

BrentGMATPrepNow · Answer

Once you conclude that   z < -4 , we can combine that with the given information:  x > -7 
Since z is an INTEGER, this means that either z = -5 or z = -6

When plug both of these possible z-values into your equation m = (-3/4)z, we see that, in both cases, the value of m is a NON-INTEGER, which breaks the condition that m is an INTEGER.

Does that help?

Cheers,
Brent

prabsahi · Answer

Once you conclude that   z < -4 , we can combine that with the given information:  x > -7 
Since z is an INTEGER, this means that either z = -5 or z = -6

When plug both of these possible z-values into your equation m = (-3/4)z, we see that, in both cases, the value of m is a NON-INTEGER, which breaks the condition that m is an INTEGER.

Does that help?

Cheers,
Brent

Thanks for replying back Brent.

I was able to conclude that ..but one thing that is bothering me..During that day say 2 mns of time .This could have been an issue.

Generally the tendency is not to put the back values and check.

I really liked this question.It can be solved in multiple ways !!

BrentGMATPrepNow · Answer

Your approach of testing each value takes a lot more work. So, timing would be an issue on test day. 
That pretty much sums up the GMAT. Almost all GMAT math questions can be solved using at least 2 different approaches. Typically, one approach is much faster than the other(s). So, the key to speed improvement is recognizing the fastest approach for each question.

Cheers,
Brent

prabsahi · Answer

Your approach of testing each value takes a lot more work. So, timing would be an issue on test day. 
That pretty much sums up the GMAT. Almost all GMAT math questions can be solved using at least 2 different approaches. Typically, one approach is much faster than the other(s). So, the key to speed improvement is recognizing the fastest approach for each question.

Cheers,
Brent

This is so true and this is the whole point I am looking into question with different approaches ..I guess you mentioned in your solution to skip checking point 2 makes a lot of sense..some 45 seconds fo sure and I need to recognize these things more ....Thanks a lot !! Your posts and particularly your questions are very helpful .

Posted from my mobile device

nick1816 · Answer

This approach is not correct.
Even/Even can be odd
6/2=3

bebs · Answer

Given: z and m are integers where z > -7 and m >3

I: k = 5 ==> 10m + 5z = 0 ==> 2m + z = 0 
m = 1 and z = -2 will make this equation true, but we are told m > 3, therefore k cannot be 5
II: k = 6 ==> 6m + 4z = 0 ==> 3m + 2z = 0
m = 4 and z = -6 will make this equation true, therefore k could be 6
III: k = 7 ==> 8m + 5z = 0
m = 5 and z = -8 will make this equation true, but we are told z > -7, therefore k cannot be 7

Answer is   A

GMATGuruNY · Answer

k = \frac{6m+2z}{2m+z} = \frac{2m+z}{2m+z}+\frac{2m+z}{2m+z}+\frac{2m}{2m+z}=2+\frac{2m}{2m+z} 
For k to equal 5, 6, or 7,  \frac{2m}{2m+z}  = 3, 4 or 5.

Case 1:  \frac{2m}{2m+z} = 3 , with the result that k=5

2m = 6m+3z 
 -3z = 4m 
 z = -\frac{4}{3}m

In this case, z will be an integer only if m is a multiple of 3.
Given that m>3, we get:
If m=6, then z=-8.
If m=12, then z=-16.
If m=18, then z=-24.
None of these results is viable, since the prompt requires than z>-7.
Implication:
Case 1 is not possible.
Thus, k≠5.
Eliminate B, C and E.

Case 2:  \frac{2m}{2m+z} = 5 , with the result that k=7

2m = 10m+5z 
 -5z = 8m 
 z = -\frac{8}{5}m

In this case, z will be an integer only if m is a multiple of 5.
Given that m>3, we get:
If m=5, then z=-8.
If m=10, then z=-16.
If m=15, then z=-24.
None of these results is viable, since the prompt requires than z>-7.
Implication:
Case 2 is not possible.
Thus, k≠7.
Eliminate D.

A

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

z and m are integers such that z > -7 and m > 3

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

z and m are integers such that z > -7 and m > 3

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards