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z and m are integers such that z > 7 and m > 3
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Updated on: 24 May 2019, 07:22
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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
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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.




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Re: z and m are integers such that z > 7 and m > 3
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Updated on: 23 May 2019, 19:13
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 > 3Options 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.




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Re: z and m are integers such that z > 7 and m > 3
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23 May 2019, 11:56
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.



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Re: z and m are integers such that z > 7 and m > 3
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23 May 2019, 12:31
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 > 3Options 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.



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Re: z and m are integers such that z > 7 and m > 3
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23 May 2019, 19:13
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.



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Re: z and m are integers such that z > 7 and m > 3
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23 May 2019, 19:16
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 > 3Options 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.



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Re: z and m are integers such that z > 7 and m > 3
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24 May 2019, 08:42
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 Answer: A Cheers, Brent
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Re: z and m are integers such that z > 7 and m > 3
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18 Jul 2019, 19:36
GMATPrepNow 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 Let's examine each statement, starting with.... i) 5 Option A: Lets say k=5 Therefore expression becomes \frac{6m+2z}{2m+z}[/m] 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 < 4Now 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!!



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Re: z and m are integers such that z > 7 and m > 3
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18 Jul 2019, 21:34
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
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Re: z and m are integers such that z > 7 and m > 3
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19 Jul 2019, 00:25
vishalkazone wrote: 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 This approach looks real good!! It can definitely help save some time..



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Re: z and m are integers such that z > 7 and m > 3
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19 Jul 2019, 11:00
prabsahi wrote: GMATPrepNow 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 Let's examine each statement, starting with.... i) 5 Option A: Lets say k=5 Therefore expression becomes \frac{6m+2z}{2m+z}[/m] 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 < 4Now 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!! Once you conclude that z < 4, we can combine that with the given information: x > 7Since z is an INTEGER, this means that either z = 5 or z = 6 When plug both of these possible zvalues into your equation m = (3/4)z, we see that, in both cases, the value of m is a NONINTEGER, which breaks the condition that m is an INTEGER. Does that help? Cheers, Brent
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Re: z and m are integers such that z > 7 and m > 3
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19 Jul 2019, 11:46
GMATPrepNow wrote: GMATPrepNow 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 Let's examine each statement, starting with.... i) 5 Option A: Lets say k=5 Therefore expression becomes \frac{6m+2z}{2m+z}[/m] 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 < 4Now 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!! Once you conclude that z < 4, we can combine that with the given information: x > 7Since z is an INTEGER, this means that either z = 5 or z = 6 When plug both of these possible zvalues into your equation m = (3/4)z, we see that, in both cases, the value of m is a NONINTEGER, which breaks the condition that m is an INTEGER. Does that help? Cheers, Brent[/quote] 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 !!



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Re: z and m are integers such that z > 7 and m > 3
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19 Jul 2019, 11:51
prabsahi wrote: 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 !! 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
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z and m are integers such that z > 7 and m > 3
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19 Jul 2019, 11:57
GMATPrepNow wrote: 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 !! 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[/quote] 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



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Re: z and m are integers such that z > 7 and m > 3
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19 Jul 2019, 12:03
This approach is not correct. Even/Even can be odd 6/2=3 vishalkazone wrote: 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



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Re: z and m are integers such that z > 7 and m > 3
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19 Jul 2019, 12:12
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




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