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If x, y, and z are positive integers, x=?
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16 Aug 2018, 07:51
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[ Math Revolution GMAT math practice question] If \(x, y\), and \(z\) are positive integers, \(x=\)? 1) \(y=x+1\) and \(z=x+3\) 2) \(x, y\), and \(z\) are prime numbers
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Re: If x, y, and z are positive integers, x=?
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16 Aug 2018, 07:56
If \(x, y\), and \(z\) are positive integers, \(x=\)? 1) \(y=x+1\) and \(z=x+3\) Any number of values can be taken...x=1, y=2, and z=4 or 10,11,13 and so on Insufficient 2) \(x, y\), and \(z\) are prime numbers Can be any combination.. 2,3,5 or 5,7,11 and so on Insufficient Combined.. X and X+1 are consecutive integers. And only way this is possible is 2 and 3, so X=2 Sufficient C
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Re: If x, y, and z are positive integers, x=?
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16 Aug 2018, 07:57
Statement 1 is clearly insufficient. Statement 2 is also insufficient. Combining both gives Y = 3, Z = 5 and X = 2. C is the answer.
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Re: If x, y, and z are positive integers, x=?
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16 Aug 2018, 08:00
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(x, y\), and \(z\) are positive integers, \(x=\)? 1) \(y=x+1\) and \(z=x+3\) 2) \(x, y\), and \(z\) are prime numbers S1  y=x+1 and z=x+3 x could be any value from 1,2... Insufficient. S2  x,y,z are prime numbers Insufficient Combining S1 and S2 If x=2, y=3 and z=5 (from S1) all 3 are prime. Sufficient. Answer C.
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Re: If x, y, and z are positive integers, x=?
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17 Aug 2018, 06:04
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(x, y\), and \(z\) are positive integers, \(x=\)? 1) \(y=x+1\) and \(z=x+3\) 2) \(x, y\), and \(z\) are prime numbers Given: x, y, and z are positive integers Target question: What is the value of x? Statement 1: y = x + 1, and z = x + 3 There are infinitely many solutions to the above equations, which means there are infinitely many possible answers to the target question. Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: x, y, and z are prime numbersSo, x can be ANY prime number. Statement 2 is NOT SUFFICIENT Statements 1 and 2 combined Statement 1 tells us that y = x + 1, which means y is 1 greater than x. This means x and y are CONSECUTIVE integers, which also means one value is ODD and one is EVEN Statement 2 tells us that x and y are prime numbers Since 2 is the only EVEN prime number, it must be the case that x is 2, and y is 3 Since we can answer the target question with certainty, the combined statements are SUFFICIENT Answer: C Cheers, Brent
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Re: If x, y, and z are positive integers, x=?
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19 Aug 2018, 17:41
=> 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. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Since y = x + 1, z = x + 3 and x, y, z are prime numbers, the only possibility is that x = 2, y = 3, and z = 5 because if x is an odd number, then y and z are two different even numbers, which cannot both be prime numbers. Thus, both conditions together are sufficient. Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B. Condition 1) There are many different possible values for x, y and z, including x = 1, y = 2, z = 4 and x = 2, y = 3, z = 5. Therefore, we don’t have a unique solution, and condition 1) is not sufficient. Condition 2) There are many different possibly values for x, y and z, including x = 2, y = 3 and z = 5, and x = 3, y = 5 and z = 7. So, we don’t have a unique solution, and condition 2) is not sufficient. Therefore, C is the answer. Answer: C In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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