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

Question

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

BrentGMATPrepNow · Accepted Answer

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 Answer: D Cheers, Brent RELATED VIDEO https://www.youtube.com/watch?v=MyG1K3ee69w

ZoltanBP · Answer

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 X3X>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 Answer: D

Chethan92 · Answer

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.

D is the answer.

Archit3110 · Answer

#1
3(2.00X) > 2(3.00Y
3X < 2Y
y=3 and x = 1
sufficient 
#2
x+3<y
agains sufficient to say that 3X < 2Y
IMO D

SimplyBrilliant · Answer

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.

SimplyBrilliant · Answer

For statement (2), even better, use 2Y as the bridge:

3X > 2Y > 2X + 6 then subtract 2X from both ends to get

X > 6, which again, is a contradiction, since X = 5 at most.

AkshdeepS · Answer

Question : Is 3(2.00X) > 2(3.00Y) ?

(1) 3X < 2Y

If X = 0, Y = 2 or more upto 9 ---- 3(2.000) < 2(3.002) ----- 6.000 < 6.004 ---- Answer to the question is NO.

If X = 1, Y = 3 or more upto 9 -----3(2.001) < 2(3.003) ------ 6.003 < 6.006 -----Answer to the Question is NO

If X = 2, Y = 4 or more upto 9 ----- 3(2.002) < 2(3.004) ------ 6.006 < 6.009------Answer to the Question is NO

By doing so we can say answer will always be NO.

Hence Sufficient.

(2) X < Y − 3

X - Y < − 3

We can use all the three values mentioned in Statement 1 to prove Sufficiency

Hence (D)

EMPOWERgmatRichC · Answer

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

EgmatQuantExpert · Answer

Solution

Steps 1 & 2: Understand Question and Draw Inferences
In this question, we are given

• 2.00X and 3.00Y are two numbers in decimal dorm with thousandths digits X and Y.
We need to determine

• Whether 3(2.00X) > 2(3.00Y) or not.
We can simplify the given expression as

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

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

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.

Hence, the correct answer choice is option D.

nick1816 · Answer

Question stem is basically whether 3X>2Y Statment 1- 3X<2Y Sufficient Statement 2 X3X Sufficient

reynaldreni · Answer

Just expanding on Zoltan's explanation for Stmt (2) Statement(2) X < Y-9 We get, 3Y - 9 > 3X Is 3X > 2Y ? X and Y can take values from 0...9 Min(3Y-9) = -9 (when Y=0) Max(3Y-9) = 18 (when Y=9) Min(2Y) = 0 Max(2Y) = 18 From the above we can say 3Y-9 <= 2Y and 3X < 3Y-9 Therefore, 3X > 2Y is False.

MHIKER · Answer

The question can be written as 600*3x>600*2y Is 3x>2y? Dividing both sides by 600. 1) Says 3x < 2y just opposite to the derived question stem. The answer is a definite NO. Sufficient. 2) x < y -3 Or, y > x+3 if x =1 , then y>4, The minimum value of y will be 5 x = 9, then y>12, the minimum value of y will be 13 It can be the value of y it exceeds one digit condition of y. so, the maximum value of x can be 6, let's see x =6, y = 9 Now, 3*1>2*5 NO 3*6>2*9 NO Statement two is also sufficient. Ans. D

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=Vot4eTHttC4

ScottTargetTestPrep · Answer

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

GMATinsight · Answer

Answer: Option D

Video solution by  GMATinsight

https://www.youtube.com/watch?v=E3dkuXh63bI

Archit3110 · Answer

target is 3(2.00X) > 2(3.00Y) X,Y can be 0 to 9 #1 3X < 2Y least value of x =1 and y = 2 sufficient to say that 3(2.001) > 2(3.002) would not be possible #2 X < Y − 3 least x= 0 and y =4 3(2.000) > 2(3.004) ; would not be possible sufficient option D

G-Man · Answer

Simplifying the question statement-----> 3* 200X/1000 > 2* 300Y/1000 = 3*(2000 + X) > 2*(3000+Y) = 3X > 2Y or 3X-2Y >0 (Positive) This is what we have to check. a) 3X < 2Y - Sufficient b) X < Y - 3 multiplying both sides by 3 3X < 3Y - 9 subtracting 2Y from both sides 3X - 2Y < Y-9 Y is a value between 0 and 9, which means Y-9 will be between -9 and 0 in either case, 3X - 2Y <=0 - Sufficient D is the correct answer

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