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

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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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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.

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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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Bunuel
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

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

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)
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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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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

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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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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
• 3(2.00X) > 2(3.00Y)
Or, 3(2 + 0.00X) > 2(3 + 0.00Y)
Or, 6 + 3X/1000 > 6 + 2Y/1000
Or, 3X > 2Y

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.

Hence, the correct answer choice is option D.

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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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Question stem is basically whether 3X>2Y

Statment 1-
3X<2Y
Sufficient

Statement 2
X<Y-3
Multiply both sides by 3

3X<3Y-9
3X<2Y+Y-9
3X-2Y<Y-9

Maximum value of Y is 9; hence 3X-2Y is always less than 0, or 2Y>3X

Sufficient

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

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.
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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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Bunuel
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.

Ans. D
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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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Video solution from Quant Reasoning:
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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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Bunuel
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.

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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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Bunuel
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]
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
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]
Bunuel
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

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

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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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