Data Sufficiency

AbeNeedsAnswers 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

D

Source: Official Guide 2020

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: Here's a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

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 a: X = 2, and Y = 6,7,8 or 9. In all possible cases, 3X < 2Y
case a: X = 3, and Y = 7,8 or 9. In all possible cases, 3X < 2Y
case a: X = 4, and Y = 8 or 9. In all possible cases, 3X < 2Y
case a: 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

Answer: D

Cheers,
Brent

Hi All,

We're told that 2.00X and 3.00Y are 2 numbers in DECIMAL form with THOUSANDTHS digits of X and Y, respectively. We're asked if 3(2.00X) > 2(3.00Y). This is a YES/NO question and can be solved with a mix of Arithmetic and TESTing VALUES. There's a great built-in 'shortcut' that we can take advantage of if we take the time to 'rewrite' the question that's asked...

To start, we can distribute the multiplication a bit in the question, which turns the question into: "Is 6 + 3(.00X) greater than 6 + 2(.00Y)?"
We can then cancel out the 6s: "Is 3(.00X) greater than 2(.00Y)?"
.... and then we can multiply both values by 1,000: "Is 3X greater than 2Y?" This is a far easier question to answer. It's also worth noting that since X and Y are both DIGITS, their values can only be integers from 0 - 9, inclusive.

(1) 3X < 2Y

Fact 1 tells us that 3X is LESS than 2Y, so since the question asks "is 3X GREATER than 2Y?", the answer is clearly NO.
Fact 1 is SUFFICIENT

(2) X < Y - 3

The inequality in Fact 2 can be rewritten as X + 3 < Y. Since X and Y are both DIGITS, this means that Y will always be AT LEAST 4 greater than X. For example...
IF...
X = 1, then Y must be 5 or greater
X = 2, then Y must be 6 or greater
Etc.
In all possible situations, 3X will be LESS than 2Y (by TESTing just the lowest possible value for Y in each situation, you can prove that this is the case. For example...
IF...
X=1, then (3)(1) = 3 and Y will be 5 or greater, so 2(5... or greater) will always be AT LEAST 10, so 3X will NEVER be greater than 2Y in this situation. The answer to the question is ALWAYS NO.
Fact 2 is SUFFICIENT

Final Answer: D

GMAT assassins aren't born, they're made,
Rich