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Re: If x, y, and z are consecutive integers and x<y<z, is y an even number [#permalink]

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28 Dec 2017, 22:31

1. Two possibilities a) XZ odd- 5,7,9- consecutive but y is odd b) XZ odd ,5,6,7- consecutive but y is even 2. Consider 4,,5,6- y is odd and xz multiple of 8. Another 4,6,8_ y is even still xz multiple of 8 Combine together- xz are off only when x is odd and z is odd but to make xyz multiple of y , y must be multiple of 8 i.e y is even So answer is C

1. Two possibilities a) XZ odd- 5,7,9- consecutive but y is odd b) XZ odd ,5,6,7- consecutive but y is even 2. Consider 4,,5,6- y is odd and xz multiple of 8. Another 4,6,8_ y is even still xz multiple of 8 Combine together- xz are off only when x is odd and z is odd but to make xyz multiple of y , y must be multiple of 8 i.e y is even So answer is C

study mode

consecutive means one after another 5,6,7 5,7,9 would be consecutive ODD integers..

Quote:

If x, y, and z are consecutive integers and x<y<z, is y an even number?

1) xz is an odd number ONLY possible when both x and z are odd.. y is between x and z, so it has to be even Suff

2) xyz is a multiple of 8 many ways possible 7,8,9 ...YES 2,3,4 or 8,9,10...NO insuff

Re: If x, y, and z are consecutive integers and x<y<z, is y an even number [#permalink]

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29 Dec 2017, 02:30

MathRevolution wrote:

[GMAT math practice question]

If \(x\), \(y\), and \(z\) are consecutive integers and \(x<y<z\), is \(y\) an even number?

1) \(xz\) is an odd number 2) \(xyz\) is a multiple of \(8\)

Consecutive integers are integers that follow one another from a given starting point without skipping any integers {-4,-3,-2,-1,0,1,2...}. Any three consecutive integers will have one term that is a multiple of three. Two terms are consecutive integers if their difference is equal to 1 (d = 1).

1) \(xz\) is an odd number. Then, \(x=odd\) and \(z=odd\) because only \(odd*odd=odd\); So, {x=odd,y=even,z=odd} because they are consecutive; sufficient.

2) \(xyz\) is a multiple of \(8\). Then, \(xyz=8*something\), but could be anything, eg. {x,y,z=2,3,4; 8,9,10; 24,25,26; etc...}; insufficient.

Re: If x, y, and z are consecutive integers and x[#permalink]

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29 Dec 2017, 18:54

chetan2u wrote:

Binggm14 wrote:

1. Two possibilities a) XZ odd- 5,7,9- consecutive but y is odd b) XZ odd ,5,6,7- consecutive but y is even 2. Consider 4,,5,6- y is odd and xz multiple of 8. Another 4,6,8_ y is even still xz multiple of 8 Combine together- xz are off only when x is odd and z is odd but to make xyz multiple of y , y must be multiple of 8 i.e y is even So answer is C

study mode

consecutive means one after another 5,6,7 5,7,9 would be consecutive ODD integers..

Quote:

If x, y, and z are consecutive integers and x<y<z, is y an even number?

1) xz is an odd number ONLY possible when both x and z are odd.. y is between x and z, so it has to be even Suff

2) xyz is a multiple of 8 many ways possible 7,8,9 ...YES 2,3,4 or 8,9,10...NO insuff

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x,y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first.

Conditions 1) and 2) Since xz is an odd number, both x and z are odd integers. Since xz is an odd number, it follows from condition 2) that y is a multiple of 8. Therefore, y is an even number. Conditions 1) and 2) are sufficient when taken together.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1) Since xz is an odd integer, x and z are odd integers. As the three integers are consecutive, y must be even. Thus, condition 1) is sufficient.

Condition 2) If x = 7, y = 8, z = 9, the answer is ‘yes’. If x = 2, y = 3, z = 4, the answer is ‘no’. Since we do not have a unique answer, condition 2) is NOT sufficient.

Therefore, A is the answer.

Normally, in problems which require 2 additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E. Answer: A
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