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If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi
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Updated on: 17 May 2021, 06:17
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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 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 questionwith certainty, statement 2 is SUFFICIENT
Re: If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi
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12 May 2019, 09:04
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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
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26 Apr 2019, 03:45
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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
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09 May 2019, 19:25
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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. _________________
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Re: If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi
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14 May 2019, 18:16
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Expert Reply
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.
If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi
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20 Sep 2020, 00:05
ZoltanBP wrote:
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
Answer: D
Just expanding on Zoltan's explanation for Stmt (2)
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Re: If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi
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30 Sep 2020, 03:07
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Top Contributor
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 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 [it is given] if x =1 , then y>4, The minimum value of y will be 5 [ Remember one thing x, y must be single digit] 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.
Re: If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi
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14 May 2021, 05:55
Expert Reply
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
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.
Re: If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi
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22 May 2021, 23:52
Expert Reply
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
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Answer: Option D
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Re: If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi
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15 Sep 2021, 09:07
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
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
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Re: If 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digi [#permalink]
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