Bunuel wrote:
If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digits X and Y, is 3(2.00X) > 2(3.00Y) ?
(1) 3X < 2Y
(2) X < Y − 3
Given: 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digits X and Y Target question: Is 3(2.00X) > 2(3.00Y)?This is a good candidate for
rephrasing the target question.
Since X is the thousandths digit, we can write: 2.00X = 2 + X/1000
Since Y is the thousandths digit, we can write: 3.00Y = 3 + Y/1000
So, the target question becomes:
Is 3(2 + X/1000) > 2(3 + Y/1000)?Expand both sides:
Is 6 + 3X/1000 > 6 + 2Y/1000)?Subtract 6 from both sides:
Is 3X/1000 > 2Y/1000)?Multiply both sides by 1000 to get:
Is 3X > 2Y ?REPHRASED target question: Is 3X > 2Y?Aside: the video below has tips on rephrasing the target question Statement 1:3X < 2Y PERFECT!!
The answer to the REPHRASED target question is
NO, 3X is NOT greater than 2YSince we can answer the
REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: X < Y − 3Add 3 to both sides to get: X + 3 < Y
This one is TRICKY!!
The solution relies on the fact that X and Y are DIGITS (0, 1, 2, 3, 4, 5, 6, 7, 8 or 9)
Let's examine all possible DIGIT solutions to the inequality X + 3 < Y
case a: X = 0, and Y = 4,5,6,7,8 or 9. In all possible cases,
3X < 2Ycase b: X = 1, and Y = 5,6,7,8 or 9. In all possible cases,
3X < 2Ycase c: X = 2, and Y = 6,7,8 or 9. In all possible cases,
3X < 2Ycase d: X = 3, and Y = 7,8 or 9. In all possible cases,
3X < 2Ycase e: X = 4, and Y = 8 or 9. In all possible cases,
3X < 2Ycase f: X = 5, and Y = 9. In all possible cases,
3X < 2YNow that we've examine all possible values of X and Y, we can see that the answer to the REPHRASED target question is always the same:
NO, 3X is NOT greater than 2YSince we can answer the
REPHRASED target question with certainty, statement 2 is SUFFICIENT
Answer: D
Cheers,
Brent
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