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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?

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


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: https://www.gmatprepnow.com/articles/han ... -questions
------------------------------------------------

E) (a - c)/4 = [(c + 4 ) - (c)]/4
= 4/4
= 1 (ODD!)

Answer: E

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

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
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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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Expert Reply
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
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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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=>

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