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# When a, b and c are consecutive positive even integers such that a>b>c

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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When a, b and c are consecutive positive even integers such that a>b>c  [#permalink]

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21 Jun 2018, 00:45
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35% (medium)

Question Stats:

63% (01:28) correct 37% (01:40) wrong based on 142 sessions

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[GMAT math practice question]

When $$a, b$$ and $$c$$ are consecutive positive even integers such that $$a>b>c$$, which of the following must be an odd integer?

$$A. \frac{(a-c)}{2}$$
$$B. \frac{(c-a)}{2}$$
$$C. \frac{(a+c)}{2}$$
$$D. \frac{(a+c)}{4}$$
$$E. \frac{(a-c)}{4}$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Expert Joined: 02 Aug 2009 Posts: 8264 Re: When a, b and c are consecutive positive even integers such that a>b>c [#permalink] ### Show Tags 21 Jun 2018, 01:49 MathRevolution wrote: [GMAT math practice question] When $$a, b$$ and $$c$$ are consecutive positive even integers such that $$a>b>c$$, which of the following must be an odd integer? $$A. \frac{(a-c)}{2}$$ $$B. \frac{(c-a)}{2}$$ $$C. \frac{(a+c)}{2}$$ $$D. \frac{(a+c)}{4}$$ $$E. \frac{(a-c)}{4}$$ Two operations.. 1) c-a.. Since they are consecutive positive even integers, and a>b>c, c-a or a-c will be |4| So it will always leave an EVEN integer when divided by 2.. So eliminate A and B But what about div by 4, (a-c)/4=4/4=1, so ODD..C is the answer But let's see the addition 2) a+c It is nothing but 2b.. 2b will be even when div by 2 or 4 when B is multiple of 4, so eliminate C and D Example.. a>b>c => 10>8>6, so 2*8/4=4, even number Ans C _________________ Director Joined: 04 Aug 2010 Posts: 532 Schools: Dartmouth College When a, b and c are consecutive positive even integers such that a>b>c [#permalink] ### Show Tags 21 Jun 2018, 02:27 MathRevolution wrote: [GMAT math practice question] When $$a, b$$ and $$c$$ are consecutive positive even integers such that $$a>b>c$$, which of the following must be an odd integer? $$A. \frac{(a-c)}{2}$$ $$B. \frac{(c-a)}{2}$$ $$C. \frac{(a+c)}{2}$$ $$D. \frac{(a+c)}{4}$$ $$E. \frac{(a-c)}{4}$$ Test a=6, b=4 and c=2 in the five answer choices. Eliminate any answer choice that does not yield an odd integer. A: $$\frac{a-c}{2} = \frac{6-2}{2} = 2$$ --> Eliminate A. B: $$\frac{a-c}{2} = \frac{2-6}{2} =- 2$$ --> Eliminate B. C: $$\frac{a+c}{2} = \frac{6+2}{2} = 4$$ --> Eliminate C. D: $$\frac{a+c}{4} = \frac{6+2}{4} = 2$$ --> Eliminate D. _________________ GMAT and GRE Tutor New York, NY Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. GMAT Club Legend Joined: 11 Sep 2015 Posts: 4320 Location: Canada Re: When a, b and c are consecutive positive even integers such that a>b>c [#permalink] ### Show Tags 21 Jun 2018, 05:05 Top Contributor MathRevolution wrote: [GMAT math practice question] When $$a, b$$ and $$c$$ are consecutive positive even integers such that $$a>b>c$$, which of the following must be an odd integer? $$A. \frac{(a-c)}{2}$$ $$B. \frac{(c-a)}{2}$$ $$C. \frac{(a+c)}{2}$$ $$D. \frac{(a+c)}{4}$$ $$E. \frac{(a-c)}{4}$$ If a, b and c are consecutive EVEN integers, and a>b>c, then we we know that b is 2 greater than c, and a is 2 greater than b So, we can write: b = c + 2 a = c + 4 Now let's check the answer choices from E to A -----ASIDE---------------- This is one of those questions that require us to check/test each answer choice. In these situations, always check the answer choices from E to A, because the correct answer is typically closer to the bottom than to the top. For more on this strategy, see my article: http://www.gmatprepnow.com/articles/han ... -questions ------------------------------------------------ E) (a - c)/4 = [(c + 4 ) - (c)]/4 = 4/4 = 1 (ODD!) Answer: E Cheers, Brent _________________ Test confidently with gmatprepnow.com e-GMAT Representative Joined: 04 Jan 2015 Posts: 3239 Re: When a, b and c are consecutive positive even integers such that a>b>c [#permalink] ### Show Tags 21 Jun 2018, 05:39 Solution Given: • a,b and c are consecutive positive even integers and a>b>c. To find: • Among the given options which one is an odd integer. Approach and Working: • Since, a,b and c are consecutive positive integer and a>b>c, a = c+4, b=c+2. Now, let us check every option. A. $$\frac{(a−c)}{2}= \frac{(c+4 - c)}{2}= 2$$ B. $$\frac{(c−a)}{2}= \frac{-(a-c)}{2}= -2$$ C. $$\frac{(a+c)}{2}= \frac{2c+4}{2}= c+2= b$$ D. $$\frac{(a+c)}{4}= \frac{2c+4}{4}= \frac{c}{2+1}$$ • If c= 2 then $$\frac{c}{2}+1= 2$$ and if c= any other number than 2 then $$\frac{c}{2}+1$$= odd • Hence, we can say if $$\frac{c}{2}+1$$ is odd or not. E. $$\frac{(a-c)}{4}= \frac{4}{4}= 1$$ Hence, the correct answer is option E. Answer: E _________________ Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 9426 Location: United States (CA) Re: When a, b and c are consecutive positive even integers such that a>b>c [#permalink] ### Show Tags 24 Jun 2018, 16:56 MathRevolution wrote: [GMAT math practice question] When $$a, b$$ and $$c$$ are consecutive positive even integers such that $$a>b>c$$, which of the following must be an odd integer? $$A. \frac{(a-c)}{2}$$ $$B. \frac{(c-a)}{2}$$ $$C. \frac{(a+c)}{2}$$ $$D. \frac{(a+c)}{4}$$ $$E. \frac{(a-c)}{4}$$ We can let c = x, so b = x + 2 and a = x + 4. We see that (a - c)/4 = [(x + 4) - x]/4 = 4/4 = 1, which is an odd integer. Answer: E _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8566 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: When a, b and c are consecutive positive even integers such that a>b>c [#permalink] ### Show Tags 24 Jun 2018, 18:17 => Write $$a = 2n + 2, b = 2n$$ and $$c = 2n – 2$$. We check each of the alternatives. $$A. \frac{( a – c )}{2} = \frac{( 2n + 2 – ( 2n – 2 ) )}{2} = \frac{4}{2} = 2.$$ $$B. \frac{( c – a )}{2} = \frac{( 2n – 2 – ( 2n + 2 ) )}{2} = \frac{-4}{2} = -2.$$ $$C. \frac{( a + c )}{2} = \frac{( 2n + 2 + 2n – 2 )}{2} = \frac{4n}{2} = 2n.$$ $$D. \frac{( a + c )}{4} = \frac{( 2n + 2 + 2n – 2 )}{4} = \frac{4n}{4} = n.$$ $$E. \frac{( a – c )}{4} = \frac{( 2n + 2 – ( 2n – 2 ) )}{4}= \frac{4}{4} = 1$$ Only option E is guaranteed to be odd. Therefore, the answer is E. Answer : E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: When a, b and c are consecutive positive even integers such that a>b>c   [#permalink] 24 Jun 2018, 18:17
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