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(1) x > y+4. The best way to deal with this statement will be to pick numbers. On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get an YES answer with one chosen number(s) and a NO with another. Well, we can get an YES answer very easily: just pick large enough \(x\) and small enough \(y\) (for example 10 and 1 respectively) for NO answer: if \(x=10\) and \(y=5\) then \(10>5+4\) but \(3*10=30<7*5=35\). So this statement is not sufficient.

(2) -5x < -14y --> multiple both side by -1/2 and flip the sign because we multiply by a negative value --> \(2.5x>7y\), as both \(x\) and \(y\) are positive then \(3x>2.5x>7y\). Sufficient.

(1) \(x>y+4\) --> if \(x=6\) and \(y=1\) then \(3x=18>7=7y\), so the answer is YES but if \(x=20\) and \(y=15\) then \(3x=60<105=7y\), so the answer is NO. Two different answers, not sufficient.

(2) \(-5x < -14y\) --> \(2.5x>7x\) and as both unknowns are positive then naturally \(3x\) would be more than \(7y\), as \(3x>2.5x>7y\). Sufficient.

Answer: B.

OA must be wrong here, answer should be B.
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Re: if x and y are positive...need some help
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14 Dec 2010, 18:46

I have a doubt ....say if we get 5x > 7y in the second statement so x > 2.8y then this statement fails. Can you please explain whether this is correct.

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15 Dec 2010, 02:04

ajit257 wrote:

I have a doubt ....say if we get 5x > 7y in the second statement so x > 2.8y then this statement fails. Can you please explain whether this is correct.

There is a mistake in your reasoning: if 5x>7y, x>1.4y (not x>2.8y) x>1.4y => 3x>4.2y, which is not sufficient

By picking numbers: x=2 / y=1: 10>7 and 6<7 X=5 / y=1: 25>7 and 15>7

On the other hand if we had x>2.8y => 3x>8.4y>7y, the statement would have been sufficient.

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15 Dec 2010, 02:28

2

ajit257 wrote:

I have a doubt ....say if we get 5x > 7y in the second statement so x > 2.8y then this statement fails. Can you please explain whether this is correct.

5x > 7y --> x>1.4y not x > 2.8y.

Note that the question asks is \(\frac{x}{y}>\frac{7}{3}=\frac{35}{15}\)?

So if statement (2) were \(5x>7y\) then we would have: \(\frac{x}{y}>\frac{7}{5}=\frac{21}{15}\) (we can divide by y as given that it's positive) so \(\frac{x}{y}\) is more than \(\frac{21}{15}\) but we can't say whether it's more than \(\frac{35}{15}\), so this statement wouldn't be sufficient.

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16 Dec 2010, 14:29

apologies...calculation mistake. But thanks guys, for clarifying... i got the concept that i was overlooking ... appreciate your responses. A special thanks to Bunuel who has been answering a lot of my doubts...kudos to you !

Option A: x > y + 4 the minimum value of x , as evident from this statement, y + 4 . Lets substitute x=y+4 in 'is 3x>7y ?' to get 'is y <3?' ; we know y>0 but we do not know whether it is greater than 3 . so this option is NS.

Option B: -5x < -14y 5x > 14y x>14/5 Y Minimum value of x from this statement is X=14/5 Y Lets substitute x=14/5 Y in 'is 3x>7y ?' to get 'is y >0 ?' ; we know y>0 so this option is SUFFICIENT.

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20 Feb 2016, 11:18

ajit257 wrote:

If x and y are positive is 3x > 7y ?

(1) x > y+4 (2) -5x < -14y

Statement 1: x>y+4 -> multiply both sides with 3 3x > 3y + 12

pick y = 1 so 3x > 15. so 3x could be 15.1, 16, 20...100...and higher Question asks if 3x > 7y or 3x > 7(1). The answer is a definite yes because 15.1, 16, 20...100 are all more than 7. now pick y = 100 so 3x > 3(100) + 12 or 3x > 312. So 3x could be 312.2, 400, 700, 1000 and so on Question asks if 3x > 7y or 3x > 7(100). Based on the statement 1, we cannot answer. Because 400 is less 700 but 1000 is more than 700. The answer is sometimes yes and sometimes no. Therefore, Statement 1 is Insufficient.

Statement 2: I solved the exact same way as Bunuel.

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13 Mar 2016, 06:49

1

Excellent Question .. Choose y and x as 0.1,5 and then 10, 105 in the first statement we can easily say that its insufficient Statement two tells us that x>2.8y so x>2.33y => sufficient Hence B
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26 Oct 2016, 13:22

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mehdiov wrote:

If x and y are positive, is 3x > 7y?

(1) x > y + 4 (2) -5x < -14y

Target question:Is 3x > 7y?

Given: x and y are positive

Statement 1: x > y + 4 This statement doesn't FEEL sufficient, so I'm going to TEST some values. There are several values of x and y that satisfy statement 1. Here are two: Case a: x = 6 and y = 1 (this satisfies the condition that X > y + 4). In this case 3x is GREATER THAN 7y Case b: x = 10 and y = 5 (this satisfies the condition that X > y + 4). In this case 3x is LESS THAN 7y Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: -5x < -14y Divide both sides by -1 to get 5x > 14y NOTE: we need to compare 3x and 7y. So, let's fiddle with the inequality 5x > 14y Divide both sides by 2 to get 2.5x > 7y

IMPORTANT: If x is positive (which we're told it is), then 3x > 2.5x. So, let's add this to our inequality to get... 3x > 2.5x > 7y From this, we can conclude that it MUST be the case that 3x > 7x Since we can answer the target question with certainty, statement 2 is SUFFICIENT

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