## If $$2.00X$$ and $$3.00Y$$ are $$2$$ numbers in decimal form with thousandths digits $$X$$ and $$Y,$$ is

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### If $$2.00X$$ and $$3.00Y$$ are $$2$$ numbers in decimal form with thousandths digits $$X$$ and $$Y,$$ is

by VJesus12 » Thu Sep 16, 2021 11:49 am

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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) $$X < Y - 3$$

Source: Official Guide

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### Re: If $$2.00X$$ and $$3.00Y$$ are $$2$$ numbers in decimal form with thousandths digits $$X$$ and $$Y,$$ is

by [email protected] » Thu Sep 16, 2021 3:26 pm

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## Global Stats

VJesus12 wrote:
Thu Sep 16, 2021 11:49 am
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) $$X < Y - 3$$

Source: Official Guide
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 2Y
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: X < Y − 3
Add 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 < 2Y
case b: X = 1, and Y = 5,6,7,8 or 9. In all possible cases, 3X < 2Y
case c: X = 2, and Y = 6,7,8 or 9. In all possible cases, 3X < 2Y
case d: X = 3, and Y = 7,8 or 9. In all possible cases, 3X < 2Y
case e: X = 4, and Y = 8 or 9. In all possible cases, 3X < 2Y
case f: X = 5, and Y = 9. In all possible cases, 3X < 2Y

Now 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 2Y
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT