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# If x and y are positive, is 3x > 7y? (1) x > y + 4 (2)

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Manager
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If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink]  30 Aug 2007, 05:15
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65% (hard)

Question Stats:

50% (02:35) correct 50% (02:52) wrong based on 6 sessions
If x and y are positive, is 3x > 7y?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-is-3x-7y-106248.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jan 2012, 09:40, edited 2 times in total.
Topic is locked
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Manager
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[#permalink]  30 Aug 2007, 05:20
If w and c are integers, is w > 0?

(1) w + c > 50
(2) c > 48

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-w-and-c-are-integers-is-w-0-1-w-c-106355.html

[Reveal] Spoiler:
Answer: E.
Director
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Re: DS tricky! [#permalink]  30 Aug 2007, 05:58
for the 1 question

1st- suff,
says that x-4>y, so x>y

2- suff,
simply saying that y/x<1, so y<x

D must be the answer

Please, let know if I am missing something

this is good one
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Re: DS tricky! [#permalink]  30 Aug 2007, 20:01
solidcolor wrote:
IrinaOK wrote:
for the 1 question

1st- suff,
says that x-4>y, so x>y

this is good one

I didn't get it... The question was "is 3x > 7y?", not "is x>y".

Agree that 2nd is sufficient. But not the 1st.

Please tell me why this reasoning is wrong?

3x>7y-----3/7>y/x-------since 3/7 is less than 1, y/x must be less than 1,
this would mean y>x. So can we say that question is basicly asking if y>x? No? please, explain where is my mistake?
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Re: DS tricky! [#permalink]  30 Aug 2007, 20:16
Chill out.

Let's take an example.

x = 6, y = 1, then 3x = 18, 7y = 7, so 3x > 7y.
x = 10, y = 5, then 3x = 30, 7y = 35, so 3x < 7y.

I learnt from OG that sometimes we just pick up some easy values to check, rather than try to find a proper way to prove it. Sometimes it saves our time a lot.
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Re: DS tricky! [#permalink]  30 Aug 2007, 20:20
GMATBLACKBELT wrote:
Just use numbers to find that S1 is false.
I actually got E on this one...

BlackBelt, S2 is sufficient:

-5x <14y> 14y

(dividing both side by a negative number (-1), convert the sign from <to>)

Given that x and y are positive, 6x > 5x

Then we have 6x > 5x > 14y

or 6x > 14y

or 3x > 7y.
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Re: DS tricky! [#permalink]  30 Aug 2007, 20:30
Yah?

Hey I found something wrong with the appearance of my previous post, it doesn't appear as it is supposed to be, with the inequalities and stuff... Anyway good that you understand what I mean.
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Re: DS tricky! [#permalink]  30 Aug 2007, 21:26
Irina,

x>y+4

So x>y. Or x/y > 1.

But x/y > 1 doesn't mean x/y > 7/3 or 3x>7y.

Remember the reasoning line should go from bottom to top, i.e. "if x>y+4 then is 3x>7y", rather than top to bottom "if 3x>7y then x>y+4".

"x>y+4" is given, and we are to check if it is so, then whether 3x>7y. Not the other way round.
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[#permalink]  30 Aug 2007, 21:47
St1:
If y = 1, x = 6, then 3x > 7y.
If y = 10, x = 15, then 3x <7y> 14y/5
3x > 42y/5
Since 42y/5 > 7y, so 3x > 7y. Sufficient.

Ans B
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Re: DS tricky! [#permalink]  31 Aug 2007, 07:09
solidcolor wrote:
If x and y are positive, is 3x > 7y?

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

OK lets look at (2)

-5X<14y>42y

divide by 5

3X>42/5 y

42/5 >7

therefore 3x>7y

sufficient..
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Re: DS tricky! [#permalink]  31 Aug 2007, 08:47
solidcolor wrote:
If x and y are positive, is 3x > 7y?

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

Got B.

(1) Plug in x=6, y=1, 18>7 Yes!
x=15, y=10, 45>70 No!
INSUFFICIENT

(2) -5x < -14y => 5x > 14y
x/y > 14/5
x/y > 2.8
The question is asking if x/y > 7/3 which is same as x/y > 2.33
This is sufficient since 2.8 will always be greater than 2.33
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[#permalink]  31 Aug 2007, 08:51
solidcolor wrote:
If w and c are integers, is w > 0?

(1) w + c > 50
(2) c > 48

Got E

(1) if c=1, then w can be 50, which is positive
if c=52, then w can be -1, which is negative
INSUFFICIENT

(2) don't know anything about w, INSUFFICIENT

Together,
Say if c=49, w can be 2, positive
c=52, w can be -1, negative
INSUFFICIENT
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Re: DS tricky! [#permalink]  31 Aug 2007, 09:04
IrinaOK wrote:
solidcolor wrote:
IrinaOK wrote:
something wrong with HTML!!!

I will post this problem and my solution to it again under topic

"Inequality 3x, 7y"

if I can

I got the same problem. Anyway I understand what you mean.

Solid,

I just have one yes/no question.

does 3x>7y mean 3/7>y/x ?

....

In a nutshell, the equation 3x>7y can only be interpreted as x/y > 7/3 when y is positive.
Remember that when you divide or multiply a negative number in the inequality environment, the sign change...so let's consider out cases:

Case 1: x>0, y>0
3x>7y
Divide x by y, you get
x/y > 7/3

Case 2: x>0, y<0
3x>7y
Divide x by y, you get
x/y < 7/3 <- Notice the sign change because y is negative

Case 3: x<0, y>0
3x>7y
Divide x by y, you get
x/y > 7/3

Case 4: x<0, y<0
3x>7y
Divide x by y, you get
x/y < 7/3 <- Notice the sign change because y is negative
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Re: DS tricky! [#permalink]  16 Sep 2007, 12:15
solidcolor wrote:
If x and y are positive, is 3x > 7y?

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

ST1:

Let's take y = 1

Then, x > 5

We can take x = 6

Now 3x = 18
And 7y= 7

So Sufficient. You can verify by putting any other +ve values of y.

ST2:

-5x < -14y

Multiply both sides by (-1)
We get,
5x > 14y

Multiply both sides by (3/5)
We get,
3x > (42/5) y
=> 3x > 8.2 y

If 3x > 8.2y then 3x will always greater than 7y

Hence, 3x>7y Sufficient.

Thus, D should be the answer.

- Brajesh
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[#permalink]  17 Sep 2007, 13:46
solidcolor wrote:
If w and c are integers, is w > 0?

(1) w + c > 50
(2) c > 48

E.
According to Combination w can be any -ve value. e.g. c=52, w=-1 false.
or

c=49, w=1, true

for Q 1.

B is sufficient. If x> 14/5 *y
14/5> 7/3
--> sufficient.
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[#permalink]  17 Sep 2007, 15:13
Ans should be B

3x > 7y

St1: x > y+4 insufficient if you take different values of x & y. You can do this calculation in mind. For example: if you consider x=13 & y=8 then 3x=39 <7y> 7y=14

St2: -5x <14y> (42/5)y --------------- Multiply both sides by -3/5 to get 3x on leftside

i.e. 3x > 8(2/5)y > 7y

Hence B

Btw Irina one cannot multiple and divide inequalities with variables unless you know for sure that the variables are positive because multiplying or dividing by negative values changes the inequality sign.

In other words one cannot conclude that 3x>7y is equal to x/y > 7/3
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[#permalink]  17 Sep 2007, 17:58
If w and c are integers, is w > 0?

(1) w + c > 50
(2) c > 48

Simply E

Q2 - If x and y are positive, is 3x > 7y?

(1) x > y + 4
(2) -5x <14y>7/3

If we rephrase the Q, Is 3X>7Y or we can say X/Y > 1

St1- since X >Y , it got to be X/Y>1
St2- 5X>14Y or X/Y>14/5 means X/Y>1

So Both sufficient !! makes sense?????
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Re: If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink]  30 Jan 2012, 07:07
First question:
B
Statement (1) is not suff.
For example, x = 6, y =1, 3(6) > 7(1), YES
x = 9, y = 4, 3(9) < 7(4) NO
INSUFFICIENT.

Second question.
E
C can be 151 and W could be -100.
(-100) + 151 = 51 > 50
INSUFFICIENT, ANSWER E
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Re: If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink]  30 Jan 2012, 10:07
Expert's post
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 the numbers. On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. Well, we can get an YES answer very easily: just pick very large $$x$$ and very small $$y$$ (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 --> $$2.5x>7y$$, as both $$x$$ and $$y$$ are positive then $$3x>2.5x>7y$$. Sufficient.

Answer: B.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-is-3x-7y-106248.html
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Re: If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink]  30 Jan 2012, 10:09
Expert's post
If w and c are integers is w > 0 ?

(1) w + c > 50
(2) c > 48

Obviously each statement alone is not sufficient.

When taken together we can still have both Yes and No answers: if c=100 and w=10 then the answer is Yes but if c=100 and w=-10 then the answer is No.

Answer: E.

I feel that it should be w+c<50 for (1). Then as $$c$$ is an integer then the least value of it according to (2) will be 49, so max value of $$w$$ in order $$c+w$$ to be less than 50 is 0 (as $$w$$ is also an integer), which means that the answer to the question is $$w>0$$ is No. So if it were $$w + c < 50$$ the answer would be C and also the question would be much more interesting.

Hope it's clear.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-w-and-c-are-integers-is-w-0-1-w-c-106355.html
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Re: If x and y are positive, is 3x > 7y? (1) x > y + 4 (2)   [#permalink] 30 Jan 2012, 10:09
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