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If m is an even integer, v is an odd integer, and m > v> 0, which of
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24 Jul 2017, 13:05
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If m is an even integer, v is an odd integer, and m > v> 0, which of
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24 Jul 2017, 15:27
carcass wrote: If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v ?
A. \(\frac{mv}{2} 1\)
B. \(\frac{mv1}{2}\)
C. \(\frac{mv}{2}\)
D. \(mv1\)
E. \(mv\) Let m = 10 & v = 1...........There are 4 Even number less than 10 and greater than 1 A. \(\frac{101}{2} 1\)= Fraction number.... Eliminate B. \(\frac{1011}{2}\) = 4......................... KeepC. \(\frac{101}{2}\) = Fraction................... EliminateD. \(1011\) = 8........................................ EliminateE. \(101\)=9............................................. EliminateAnswer: B




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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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26 Jul 2017, 02:10
The fastest way is plugging in some numbers.
Can someone show how to solve the problem algebraically?



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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26 Jul 2017, 16:15
carcass wrote: If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v ?
A. \(\frac{mv}{2} 1\)
B. \(\frac{mv1}{2}\)
C. \(\frac{mv}{2}\)
D. \(mv1\)
E. \(mv\) We can let m = 6 and v = 5. Since there are no (or zero) even integers less than m but greater than v, we see that the answer can be either B, (m  v  1)/2, or D, m  v  1, since either choice will produce 0 when we substitute m = 6 and v = 5. Now, let’s let m = 8 and v = 5. Since there is 1 even integer (namely, 6) less than m but greater than v, we see that the answer must be choice B, since (8  5  1)/2 = 1 whereas choice D will yield 8  5  1 = 2. Thus, the correct answer choice must be B. Answer: B
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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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27 Jul 2017, 01:28
Devbek wrote: The fastest way is plugging in some numbers.
Can someone show how to solve the problem algebraically? Hi Devbek, I'll try to explain. Number of even numbers between 2 different even numbers (eg, a,b) can be given by ((ab)/2)+1 (1 is added to include the first even number in the series) In the current question we need to find number of even numbers "greater than v" and "less than m". The even number after v(which is odd) is v+1 and the even number before m(which is even) is m2. (This step is done to exclude both v and m from the total number of even integers) Hence the number of even numbers between v and m (excluding v and m) should be ((m2v1)/2) + 1) (1 added to include the number v+1) Upon solving you will get (mv1)/2 Hope you got it.



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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27 Jul 2017, 03:06
Dkingdom wrote: Devbek wrote: The fastest way is plugging in some numbers.
Can someone show how to solve the problem algebraically? Hi Devbek, I'll try to explain. Number of even numbers between 2 different even numbers (eg, a,b) can be given by ((ab)/2)+1 (1 is added to include the first even number in the series) In the current question we need to find number of even numbers "greater than v" and "less than m". The even number after v(which is odd) is v+1 and the even number before m(which is even) is m2. (This step is done to exclude both v and m from the total number of even integers) Hence the number of even numbers between v and m (excluding v and m) should be ((m2v1)/2) + 1) (1 added to include the number v+1) Upon solving you will get (mv1)/2 Hope you got it. Still confusing, but thanks anyway



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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27 Jul 2017, 03:18
Devbek wrote: Dkingdom wrote: Devbek wrote: The fastest way is plugging in some numbers.
Can someone show how to solve the problem algebraically? Hi Devbek, I'll try to explain. Number of even numbers between 2 different even numbers (eg, a,b) can be given by ((ab)/2)+1 (1 is added to include the first even number in the series) In the current question we need to find number of even numbers "greater than v" and "less than m". The even number after v(which is odd) is v+1 and the even number before m(which is even) is m2. (This step is done to exclude both v and m from the total number of even integers) Hence the number of even numbers between v and m (excluding v and m) should be ((m2v1)/2) + 1) (1 added to include the number v+1) Upon solving you will get (mv1)/2 Hope you got it. Still confusing, but thanks anyway Formula to find count of even numbers in a consecutive series between two numbers is ((ab)/2) + 1. In this formula both a and b are included in the count. let m = 10 (even), v = 3(odd) What the question is asking : find the number of even numbers greater than v (3) and less than m (10). Breaking the question down :1) We need to exclude m and v from the total count of even numbers. v obviously is out because it is odd. To exclude m we need to use an even number less than m. Hence in order to apply the above stated formula we will use v+1 (4) and m2 (8). (Since we need to find only the count of even numbers that's why it is easier to keep the starting and ending numbers as even digits. Applying the formula : ((84)/2)+1) = 3 (i.e. 4,6,8) Remember :numbers are more than 3 and less than 10 and even. The above equation can be written as ((m2v1)/2)+1 = (mv1)/2



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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10 Sep 2017, 04:52
Devbek wrote: The fastest way is plugging in some numbers.
Can someone show how to solve the problem algebraically? You are better off just plugging in numbers, come the real test.



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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18 Sep 2017, 10:04
m is even and v is odd > (mv) is odd > (mv)/2 is not an interger => elimate A, C imidiately
SOLVE: Between m and v, we have [(m1)  (v+1) + 1] = (mv1) numbers, include both odd and even > the number of even is half of (mv1) => The answer is B



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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15 Oct 2017, 18:34
Hi, when we use number picking for method for this problem, it does not give a consistent result. For ex. if m=2; v=1 then non of the answers stand true.
please let me know what am i missing here?



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If m is an even integer, v is an odd integer, and m > v> 0, which of
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15 Oct 2017, 20:54
If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v? We are looking for even integers x that m< x <v > these three (m, v, x) are integers that means (m  v) cannot less than or equal to 1 (x cannot be an integer if it happens). So, m = 2 and v=1 are not the right numbers to pick. Hope it helps!



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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01 Dec 2017, 06:04
Hi all, I still don't understand. M>v>0 and we have to find a number x which is a even integer and m>x>v>0. Because x is a even integer, x can only be (B) or (D). If m=8, v=3: (mv1)/2=2 mv1= 4 2<v=3; 4>v=3 So (B) is wrong. If m=10, v =3 (mv1)/2=3 is an odd integer mv1=6 So (B) is wrong Please help me to explain these situations. Thank you



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If m is an even integer, v is an odd integer, and m > v> 0, which of
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02 Dec 2017, 13:41
carcass wrote: If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v ?
A. \(\frac{mv}{2} 1\)
B. \(\frac{mv1}{2}\)
C. \(\frac{mv}{2}\)
D. \(mv1\)
E. \(mv\) Greyfield wrote: Hi all, I still don't understand. M>v>0 and we have to find a number x which is a even integer and m>x>v>0. Because x is a even integer, x can only be (B) or (D). If m=8, v=3: (mv1)/2=2 mv1= 4 2<v=3; 4>v=3 So (B) is wrong. If m=10, v =3 (mv1)/2=3 is an odd integer mv1=6 So (B) is wrong Please help me to explain these situations. Thank you Greyfield , it looks as if you are interpreting one little part incorrectly. The wording is terse. And to use \(x\) can be confusing here. We are looking for a specially defined kind of number, defined by the prompt. The answers are supposed to tell us how many of those specially defined numbers there are  not the actual values of those specially defined numbers. Try rewriting this: . . .which of the following represents the number of even integers less than m and greater than vto THIS: . . .which of the following represents how many even integers there are which are less than m and greater than vYou wrote Quote: find a number x which is a even integer and m>x>v>0. Because x is a even integer, x can only be (B) or (D). Mmm... no. \(x\) is how many even integers there are. Let's use your numbers, m = 8, v = 3 Answer B) \(\frac{(mv1)}{2} = 2\) On the number line, between 8 and 3, there are two(2) even numbers: 4 and 6. Both are less than m=8 and greater than v=3. Answer B does not tell you "4 and 6." You have to discern that yourself. Your numbers: m = 10, v = 3 Between 10 and 3, there are, just as you calculated from B, three (3) even integers: 8, 6, and 4. All are less than 10 and greater than 3. Once you have figured out how many of those special integers there are, you cannot "go again" and plug your choices into (B) to find out which even integers they are. Hope that helps.
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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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10 Dec 2017, 19:22
Hi,
Can you explain why are you adding 1 to v and why are you subtracting 2 to m please?
Hi Devbek, I'll try to explain.
Number of even numbers between 2 different even numbers (eg, a,b) can be given by ((ab)/2)+1 (1 is added to include the first even number in the series) In the current question we need to find number of even numbers "greater than v" and "less than m". The even number after v(which is odd) is v+1 and the even number before m(which is even) is m2. (This step is done to exclude both v and m from the total number of even integers) Hence the number of even numbers between v and m (excluding v and m) should be ((m2v1)/2) + 1) (1 added to include the number v+1) Upon solving you will get (mv1)/2
Hope you got it.[/quote][/quote]



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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20 Dec 2017, 07:11
Mo2men wrote: carcass wrote: If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v ?
A. \(\frac{mv}{2} 1\)
B. \(\frac{mv1}{2}\)
C. \(\frac{mv}{2}\)
D. \(mv1\)
E. \(mv\) Let m = 10 & v = 1...........There are 4 Even number less than 10 and greater than 1 A. \(\frac{101}{2} 1\)= Fraction number.... Eliminate B. \(\frac{1011}{2}\) = 4......................... KeepC. \(\frac{101}{2}\) = Fraction................... EliminateD. \(1011\) = 8........................................ EliminateE. \(101\)=9............................................. EliminateAnswer: B Hi, I am not really sure why are you canceling out Choice D. I thought the question require us to find an EVEN integer that is LESS than m but GREATER than v? If you use m =10, v=1, isn't 8 fulfill this condition as well? since 8 is an even integer <10 but >1?



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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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20 Dec 2017, 07:30
xlgoh1992 wrote: Mo2men wrote: carcass wrote: If m is an even integer, v is an odd integer, and m > v > 0, which of the following represents the number of even integers less than m and greater than v ?
A. \(\frac{mv}{2} 1\)
B. \(\frac{mv1}{2}\)
C. \(\frac{mv}{2}\)
D. \(mv1\)
E. \(mv\) Let m = 10 & v = 1...........There are 4 Even number less than 10 and greater than 1 A. \(\frac{101}{2} 1\)= Fraction number.... Eliminate B. \(\frac{1011}{2}\) = 4......................... KeepC. \(\frac{101}{2}\) = Fraction................... EliminateD. \(1011\) = 8........................................ EliminateE. \(101\)=9............................................. EliminateAnswer: B Hi, I am not really sure why are you canceling out Choice D. I thought the question require us to find an EVEN integer that is LESS than m but GREATER than v? If you use m =10, v=1, isn't 8 fulfill this condition as well? since 8 is an even integer <10 but >1? HI.. we rae looking for NUMBERS of even numbers between v and m.. if v=1 and m=10, even integers are 2,4,6,8.. HOW many 4 so our answer should be 4 8 is ONE of the even integer BUT we are looking for NUMBERS of even integers between v and m
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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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11 Jan 2018, 10:50
Bunuel, Sir Can you please help me with this one ; if i am not able to get the answer after plugging the numbers. Thanks in advance.



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If m is an even integer, v is an odd integer, and m > v> 0, which of
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02 Jul 2018, 04:07
Bunuel niks18 pushpitkc chetan2u VeritasPrepKarishmaUsually in a PS if I take numbers satisfying the property given in question stem (here even/odd) I am usually left with only one answer choice. Here as suggested by ScottTargetTestPrep I had to further narrow down my options from 2 to 1 by taking more set of numbers. Is there any other time efficient way to come to OA?
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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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02 Jul 2018, 06:03
adkikani wrote: Bunuel niks18 pushpitkc chetan2u VeritasPrepKarishmaUsually in a PS if I take numbers satisfying the property given in question stem (here even/odd) I am usually left with only one answer choice. Here as suggested by ScottTargetTestPrep I had to further narrow down my options from 2 to 1 by taking more set of numbers. Is there any other time efficient way to come to OA? to find the number of items between x and y inclusive, it is xy+1.. say between 2 and 7... 72+1=6..... 2,3,4,5,6,7 so if we have consecutive numbers starting with O/E and finishing with E/O... half of them will be odd and half even so here m is even so m1 is odd and v is odd so v+1 will be even.. numbers between m1 and v+1 = m1(v+1)+1 = m1v1+1=mv1.. half will be even and half odd so \(\frac{mv1}{2}\)
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Re: If m is an even integer, v is an odd integer, and m > v> 0, which of
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02 Aug 2018, 05:01
adkikani wrote: Bunuel niks18 pushpitkc chetan2u VeritasPrepKarishmaUsually in a PS if I take numbers satisfying the property given in question stem (here even/odd) I am usually left with only one answer choice. Here as suggested by ScottTargetTestPrep I had to further narrow down my options from 2 to 1 by taking more set of numbers. Is there any other time efficient way to come to OA? Hi adkikaniActually, this comes with lot of training and number sense. In some questions, picking easy number could lead you with 23 potential solutions. So you will try another set to narrow down the the right choice. In my example above, I opened the range of m & v because Scott's first data set most probably will yield number =1 which GMAT test maker know how people will rush to choose small numbers to generate easy numbers. I hope it helps




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