## If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths

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### If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths

by BTGModeratorVI » Tue Mar 03, 2020 6:48 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
Answer: D
Source: Official Guide

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

by [email protected] » Wed Mar 04, 2020 6:26 am

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BTGModeratorVI wrote:
Tue Mar 03, 2020 6:48 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
Answer: D
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 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

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

by [email protected] » Fri May 14, 2021 5:56 am

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## Global Stats

BTGModeratorVI wrote:
Tue Mar 03, 2020 6:48 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
Answer: D
Source: Official Guide
Solution:

Question Stem Analysis:

We need to determine whether 3(2.00X) > 2(3.00Y) given that X and Y are the thousandths digits of 2.00X and 3.00Y, respectively. Notice that 3(2.00X) = 6 + 3X/1000 and 2(3.00Y) = 6 + 2Y/1000. Therefore, the question becomes whether 3X > 2Y.

Statement One Alone:

Since 3X < 2Y, we can say 3X is not greater than 2Y and hence the answer to the question is No. Statement one alone is sufficient.

Statement Two Alone:

Since X < Y - 3, we know that Y is at least 4 and 3X < 3Y - 9. If 3X > 2Y, we have:

2Y < 3X < 3Y - 9

If Y = 4, we have 8 < 3x < 3.

If Y = 5, we have 10 < 3x < 6.

If Y = 6, we have 12 < 3X < 9.

If Y = 7, we have 14 < 3x < 12.

If Y = 8, we have 16 < 3x < 15.

If Y = 9, we have 18 < 3x < 18.

As we can see, none of these double inequalities is a correct inequality. Therefore, we can say that 3X is not greater than 2Y, as we did in statement one. Statement two alone is sufficient.

Answer: D

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