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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82
When a, b and c are consecutive positive even integers such that a>b>c  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 63% (01:28) correct 37% (01:41) wrong based on 135 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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Math Expert V
Joined: 02 Aug 2009
Posts: 8283
Re: When a, b and c are consecutive positive even integers such that a>b>c  [#permalink]

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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

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

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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.

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Re: When a, b and c are consecutive positive even integers such that a>b>c  [#permalink]

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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!)

Cheers,
Brent
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Posts: 3158
Re: When a, b and c are consecutive positive even integers such that a>b>c  [#permalink]

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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.

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Re: When a, b and c are consecutive positive even integers such that a>b>c  [#permalink]

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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.

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Math Revolution GMAT Instructor V
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Posts: 8235
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]

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=>

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.
_________________ Re: When a, b and c are consecutive positive even integers such that a>b>c   [#permalink] 24 Jun 2018, 19:17
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