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Re: If x, y, and z are distinct positive integers, is x(y – z) less than y [#permalink]
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22 Jul 2016, 13:56
AbdurRakib wrote:
If x, y, and z are distinct positive integers, is x(y – z) less than y(x – z) ?
(1) y > x
(2) x > z
Given to find whether x(y – z) less than y(x – z) ?
x(y – z) < y(x – z). => xy - xz < xy - yz. => -xz < - yz. ( adding -xz on both sides) => xz > yz (multiplied by -1 on both sides so sign value changes). => xz - yz > 0. => z(x-y) > 0....This is the required question stem.
Stat 1: y > x => y - x > 0 => -(x - y) > 0 = > x -y < 0 and this value is less than zero means negative.
sub x -y < 0 i.e. -ve value in question stem then z(-ve) is always < 0 ....Sufficient.
Stat 2: x > z => x - z > 0.. this form is not existing in the question stem...Hence Insufficient.
If x, y, and z are distinct positive integers, is x(y – z) less than y(x – z) ?
(1) y > x
(2) x > z
Target question:Is x(y – z) < y(x – z) ? This is a great candidate for REPHRASING the target question. Target question: Is x(y – z) < y(x – z) ? Expand both sides of inequality: xy - xz < xy - yz Subtract xy from both sides: -xz < -yz Add yz to both sides to get: yz - xz < 0 Factor the left side: z(y - x) < 0 Since we know z is POSITIVE, let's divide both sides by z to get: y - x < 0 ..... REPHRASED target question:Is y - x < 0 ?
Statement 1: y > x Subtract x from both sides to get: y - x > 0 PERFECT!! Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: x > z This does not help us answer the REPHRASED target question because they're no information about y Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Re: If x, y, and z are distinct positive integers, is x(y – z) less than y [#permalink]
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23 Jul 2016, 12:12
Q. x(y-z) < y(x-z)? or, -xz < -yz or, (since z is a distinct +ve integer, we can cancel z from both sides) or, y<x? -> this is the question we need to answer.
If x, y, and z are distinct positive integers, is x(y – z) less than y(x – z) ?
(1) y > x
(2) x > z
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations. We can modify the original condition and question as follows.
The question is asks x(y-z) < y(x-z) or x(y-z) - y(x-z) < 0. It is equivalent to -xz +yz < 0 or z(y-x) < 0 Since z > 0, the question asks if y - x < 0.
Thus, the condition 1) is sufficent, since the answer is always no.
Therefore, the answer is A.
For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.
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