Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to pre-think assumptions and solve the most challenging questions in less than 2 minutes.

Sign up or for Target Test Prep’s weekly Quant webinar series. The free weekly webinar covers sophisticated, yet easy-to-deploy, tactics and strategies for handling commonly misunderstood, high-value GMAT quant problems.

Want to solve 700+ level Algebra questions within 2 minutes? Attend this free webinar to learn how to master the most challenging Inequalities and Absolute Values questions in GMAT

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

Show Tags

27 Apr 2019, 10:10

12

Top Contributor

1

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 2Y Since we can answer the REPHRASED target questionwith 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 questionwith certainty, statement 2 is SUFFICIENT

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

Show Tags

26 Apr 2019, 04:45

1

is 3(2.00X) > 2(3.00Y) ?

From S1:

3X < 2Y 2*3 and 3*2 are same. Now, the decimal value is only dependent on whether 3X > 2Y. As the statement says that 3X < 2Y We can say that 3(2.00X) is always lesser than 2(3.00Y) Sufficient.

From S2:

x+3 < y So, Y is > X. Then 3(2.00X) is always lesser than 2(3.00Y) Sufficient.

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

Show Tags

09 May 2019, 20:25

1

Totally agree with Brent’s approach to statement (1) – simplifying the question to “Is 3X > 2Y?” is the way to go.

For statement (2), an alternative to testing several cases is to use the fact that X and Y are digits, then apply a min/max approach.

Statement (2) tells us X < Y – 3, so since X and Y are digits, the most Y could be is 9, and therefore the most X could be is 5.

Now, test the question by assuming 3X > 2Y (i.e. 3X IS greater than 2Y), and then see what that tells us about X in combination with statement (2). To combine the two inequalities, we can use Y as the bridge.

\(\frac{3}{2}X > Y\)

Next, X < Y – 3 means X + 3 < Y and so

X + 3 < Y < \(\frac{3}{2}X\) then multiplying everything by 2 yields

2X + 6 < 2Y < 3X and so subtracting 2X from both ends yields

6 < X or X > 6, which contradicts the fact that X = 5 at most. Therefore, the assumption that 3X > 2Y must be false, and we get a definitive NO, which is sufficient.
_________________

SimplyBrilliantPrep.com - Harvard Grad GMAT Instructor with 99th Percentile Scores + Clients Admitted at All Top Business Schools

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

Show Tags

12 May 2019, 10:04

2

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

DS13841.01 OG2020 NEW QUESTION

The rephrased question:

Is \(3(2+0.00X)>2(3+0.00Y)\)

\(3\cdot 0.00X>2\cdot 0.00Y\)

\(3X>2Y\) ?

1) We know that \(3X<2Y\). Thus, the answer to the rephrased question is a definite No. \(\implies\) Sufficient

2) We know that \(X<Y-3\) and can use this statement information to further rephrase the question.

\(3X<3Y-9 \implies\) Is \(3Y-9>3X>2Y\) ?

Since \(Y\) is a digit, it cannot be true that \(3Y-9>2Y \implies Y>9\). Thus, the answer to the further rephrased question is a definite No. \(\implies\) Sufficient

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

Show Tags

14 May 2019, 19:16

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

Hence, we need to determine whether 3X is greater than 2Y or not. With this understanding, let us now analyse the individual statements.

Step 3: Analyse Statement 1 As per the information given in statement 1, 3X < 2Y.

• From this statement, we can definitely conclude that 3X is not greater than 2Y.

Hence, statement 1 is sufficient to answer the question.

Step 4: Analyse Statement 2 As per the information given in statement 2, X < Y – 3.

Or, X + 3 < Y.

Now, if 3X > 2Y, then

• Y < 3/2 X Or, X + 3 < Y < 3/2 X Or, X + 3 < 3/2 X Or, 2X + 6 < 3X Or, X > 6

Now, if X > 6, then from the relation X < Y – 3, we can say Y > 9.

• However, Y cannot be greater than 9, as Y is a digit. • Therefore, we can say 3X is not greater than 2Y.

Hence, statement 2 is sufficient to answer the question.

Step 5: Combine Both Statements Together (If Needed) Since we can determine the answer from either of the statements individually, this step is not required.