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

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Joined: 22 Jun 2010
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If x and y are positive, is 3x > 7y? [#permalink]

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Updated on: 02 Jul 2013, 14:00
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Question Stats:

61% (01:07) correct 39% (01:39) wrong based on 365 sessions

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

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

Originally posted by mehdiov on 09 Sep 2010, 15:43.
Last edited by Bunuel on 02 Jul 2013, 14:00, edited 1 time in total.
Edited the OA.
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Posts: 46991
If x and y are positive is 3x > 7y ? [#permalink]

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14 Dec 2010, 06:39
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2
ajit257 wrote:
Q. if x and y are positive is 3x > 7y ?

(i) x > y+4
(ii) -5X < -14Y

I am not sure about the ans

No, D is not correct, answer should be B.

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

(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.

Answer: B.
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Re: please solve [#permalink]

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09 Sep 2010, 15:57
1
mehdiov wrote:
f x and y are positive, is 3x > 7y?

(1) x > y + 4

(2) -5x < -14y

I'm finding B

but accoding to me source the OA is

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

(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: please solve [#permalink]

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21 Sep 2010, 10:20
x^2 .y^4.y^4.zy from statement ..
Statement clearly insufficient as it does not explains the sign ..

Statement 2 show that either y is less than 0 or z is less than 0 . Hence substituting in question we know that statement is is less than 0
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If x and y are positive is 3x > 7y ? [#permalink]

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Updated on: 22 Feb 2014, 13:28
4
If x and y are positive is 3x > 7y ?

(1) x > y+4
(2) -5x < -14y
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Originally posted by ajit257 on 14 Dec 2010, 06:09.
Last edited by Bunuel on 22 Feb 2014, 13:28, edited 7 times in total.
Edited the stem and OA
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Re: if x and y are positive...need some help [#permalink]

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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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Re: if x and y are positive...need some help [#permalink]

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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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Re: if x and y are positive...need some help [#permalink]

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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.

Hope it's clear.
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Re: if x and y are positive...need some help [#permalink]

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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 !
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Re: Inequalities [#permalink]

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19 Feb 2011, 14:09
1
1
Is 3x>7y?

(1) x > y+4
x-y> 4

x=6, y=1
18>7. Yes

x=10
y=5

30<35. No

Not sufficient.

(2) -5x < -14y

5x>14y
x>(14/5)y
x>2.8y
3x>8.4y

So; 3x must be greater than 7y.
Sufficient.

Ans: "B"
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Re: If x and y are positive is 3x > 7y ? [#permalink]

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31 Jul 2014, 13:54
ajit257 wrote:
If x and y are positive is 3x > 7y ?

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

Interesting question. I got wrong answer for the first try (C). But (2) is sufficient.

-5x < - 14y -> 5x > 14y -> 2.5x > 7y -> 3x > 7y (x,y posisitve)
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Re: If x and y are positive is 3x > 7y ? [#permalink]

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17 Feb 2015, 21:43
mulitplying 2nd statement by -1/2 we get 2.5x>7y and since bothe x and y are positive so 3x >7y hence B is the answer.
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If x and y are positive is 3x > 7y ? [#permalink]

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10 Mar 2015, 07:36
ajit257 wrote:
If x and y are positive is 3x > 7y ?

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

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.

Answer B.

VeritasPrepKarishma could you please certify my approach ? thanks .

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Re: If x and y are positive is 3x > 7y ? [#permalink]

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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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Re: If x and y are positive is 3x > 7y ? [#permalink]

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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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Re: If x and y are positive, is 3x > 7y? [#permalink]

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26 Oct 2016, 13:22
Top Contributor
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

Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: http://www.gmatprepnow.com/articles/dat ... lug-values

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

Answer:

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Re: If x and y are positive is 3x > 7y ? [#permalink]

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Re: If x and y are positive is 3x > 7y ?   [#permalink] 16 May 2017, 02:27
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