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

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If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink] New post 12 Jul 2009, 11:45
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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 bb on 12 Feb 2013, 22:11, edited 2 times in total.
OA added.
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Re: 3x > 7y? [#permalink] New post 12 Jul 2009, 16:22
simplify the question to "is x/y > 7/3"

stmt 1: insufficient.

rule - say p/q is a positive fraction [ie p<q], and you add a positive number a to both the numerator and denominator

then p/q < (p+a)/(q+a)

similarly, say p/q is such that p>q, and you add positive number a to both p and q, then p/q > (p+a)/(q+a) always

Here if we consider the fraction 7/3 as p/q, then as the values of x and y change (since they are variables), we will keep getting different values of x/y. It can be either greater or less than 7/3

You can also do it in the longer way by picking numbers for x and y, such that x > y+4; and will find that they can be either less or more than 7/3 for different values.
for x,y = 6,1;...... x/y = 6;
for x,y = 10, 5 ......x/y = 2

stmt 2: sufficient.
-5x < -14 y
or x/y > 14/5

Now 7/3 = 14/6 which is less than 14/5, so x/y is surely > 7/3

Last edited by rashminet84 on 12 Jul 2009, 20:18, edited 1 time in total.
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Re: 3x > 7y? [#permalink] New post 12 Jul 2009, 20:16
rashminet84 wrote:
simplify the question to "is x/y > 7/3"

stmt 1: insufficient.

rule - say p/q is a positive fraction [ie p<q], and you add a positive number a to both the numerator and denominator

then p/q < (p+a)/(q+a)

similarly, say p/q is such that p>q, and you add positive number a to both p and q, then p/q > (p+a)/(q+a) always

Here if we consider the fraction 7/3 as p/q, then as the values of x and y change (since they are variables), we will keep getting different values of x/y. It can be either greater or less than 7/3

You can also do it in the longer way by picking numbers for x and y, such that x > y+4; and will find that they can be either less or more than 7/3 for different values.
for x,y = 6,1;...... x/y = 6;
for x,y = 5,10 ......x/y = 2

stmt 2: sufficient.
-5x < -14 y
or x/y > 14/5

Now 7/3 = 14/6 which is less than 14/5, so x/y is surely > 7/3



If y=10, how can x be 5?
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Re: 3x > 7y? [#permalink] New post 12 Jul 2009, 20:18
oops, thanks for pointing out. Ill correct it. sorry for the inconvenience :oops:
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Re: 3x > 7y? [#permalink] New post 12 Jul 2009, 20:23
No inconvenience, at all. We are all helping each other! :)
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Re: 3x > 7y? [#permalink] New post 12 Jul 2009, 20:31
Jivana wrote:
No inconvenience, at all. We are all helping each other! :)


Oh yes, ofcourse :)

but sometimes people try a lot to make sense of a wrong post, thereby wasting time and effort. It might lead to frustration also when one comes to know that it was the mistake of the one who posted. :-D
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Re: 3x > 7y? [#permalink] New post 12 Jul 2009, 23:55
Jivana wrote:
ankur55 wrote:
If x and y are positive, is 3x > 7y?
(1) x > y + 4

y=1; x=5.1
3x=15.3; 7y=7.

3x<7y ? (Yes)


(2) -5x < -14y

y=1; x=5.1
-5x=-25.5; -14y = -14

-5x < -14y ? (Yes)


D


Thanks for the explanation but I still not get the same.
Can you pls elaborate how is the first statement sufficient?
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Re: 3x > 7y? [#permalink] New post 13 Jul 2009, 06:36
This is my approach..

3x > 7y to be proved...

given in 1: x > y+4
substitue this in the to be proved
3y+12 > 7y
12 > 4y
3 > y

ie y<3. since x and y are positive y max is 2.9

3x > 7(2.9)
3x> 20.3

hence i concluded 3x min is 20.3 and 7y max is 20.3 hence A is sufficient


second statement

5x > 14y
divide by 2
2.5x > 7y
so 3x > 7y

Hence 'D' is the answer.
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Re: 3x > 7y? [#permalink] New post 13 Jul 2009, 11:37
Ankur55,

I think I made a mistake, st.1 is insufficient. Only st.2 is.

If say, y=10, x=15, then 3x =45; 7y = 70. In this case 3x == 7y. So st.1 is insufficient.

Sorry about that!
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 01:29
skpMatcha wrote:
This is my approach..

3x > 7y to be proved...

given in 1: x > y+4
substitue this in the to be proved
3y+12 > 7y
12 > 4y
3 > y

ie y<3. since x and y are positive y max is 2.9

3x > 7(2.9)
3x> 20.3

hence i concluded 3x min is 20.3 and 7y max is 20.3 hence A is sufficient


second statement

5x > 14y
divide by 2
2.5x > 7y
so 3x > 7y

Hence 'D' is the answer.


I'm sorry but I'm still not clear skpMatcha, how is 1st statement sufficient.
Is there any better approach?
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 04:06
I think ans is D.What is the OA?
Both stmnts are sufficient.

Ankur,for proving the sufficiency of stmnt1 you can plug in the below values
y=1 since x>y+4 ,x can be 6 or greater than 5 .Thus 3x>7y(for x=6 and y=1) holds true

You can also plug in fractions to double check the ans.
Put y=1/2 since x>y+4, x has to be greater than 9/2.Take x=4.6
This time also 3x>7y
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 05:03
shrutisingh wrote:
I think ans is D.What is the OA?
Both stmnts are sufficient.

Ankur,for proving the sufficiency of stmnt1 you can plug in the below values
y=1 since x>y+4 ,x can be 6 or greater than 5 .Thus 3x>7y(for x=6 and y=1) holds true

You can also plug in fractions to double check the ans.
Put y=1/2 since x>y+4, x has to be greater than 9/2.Take x=4.6
This time also 3x>7y


Yes Shruti, OA is D.
Somehow plugging doesnt work for me sometimes in DS, hence I was trying to solve stmt 1 to check whether it is sufficient.
Anyway thanks a lot for explanation.
Plugging will work for this problem. Thanks again.
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 05:51
akur, i think the OA is incorrect.
Stmt 1 cannot be sufficient. Simply plugging values is telling you this.

take y to be anything 0,1,10,20

Now take x so that x is always > y+4

Y: 0, 1, 10, 20
X: 5, 6, 15, 25
7y: 0, 7, 70, 140
3x: 15, 21, 45, 75

For first two cases above (columns), 3x > 7y
but for 3rd and 4th cases, 3x < 7y

the statement that x > y+4 is not sufficient. Im pretty sure the OA is wrong. It would be better if you verify it from other sources.
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 06:06
Quote:
If x and y are positive, is 3x > 7y?
(1) x > y + 4
(2) -5x < -14y


(1) x > y + 4

Let x be 7 and y be 1
7>1+4
7>5

3(7)>7(1)
21>7

Let x be 10 and y be 5
10> 5+4
10>9

3(10)> 7(5)
30>!35

Hence A is insuff
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 06:33
Now I am in doubt..

to be proved is 3x > 7y

given x >y + 4 . Can any one verify if we can substitute x > y+4 in the to be proved inequation ??

can we say this ? 3(y+4) > 7(y) --> stmt 1

if yes then A makes sense..because stmt 1 will end up with y < 3 and for all values of 0 < y < 3 , 3x > 7y.

or is it wrong to approach this problem that way ? Please help !
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 07:48
skpMatcha wrote:
Now I am in doubt..

to be proved is 3x > 7y

given x >y + 4 . Can any one verify if we can substitute x > y+4 in the to be proved inequation ??

can we say this ? 3(y+4) > 7(y) --> stmt 1

if yes then A makes sense..because stmt 1 will end up with y < 3 and for all values of 0 < y < 3 , 3x > 7y.

or is it wrong to approach this problem that way ? Please help !


The problem with your approach is that it is already assuming both conditions to be true
ie, 3x>7y and that x > y+4

and after doing this, you get that these two conditions will be satisfied only when y<3, which is true.
As long as y<3, 3x >7y; but the moment y>3, then the condition reverses to 3x<7Y.
But for this problem you need to check whether 3x is always >7y or not, so you cannot assume the condition to be always true.
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 08:21
Quote:
The problem with your approach is that it is already assuming both conditions to be true
ie, 3x>7y and that x > y+4

and after doing this, you get that these two conditions will be satisfied only when y<3, which is true.
As long as y<3, 3x >7y; but the moment y>3, then the condition reverses to 3x<7Y.
But for this problem you need to check whether 3x is always >7y or not, so you cannot assume the condition to be always true.


Great ! Thanks for verifying.
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Re: 3x > 7y? [#permalink] New post 14 Jul 2009, 09:36
Hi All,

OA is D but I'm getting B after solving both the statements :(
Thats why I logged this question in the forum :)
So far I havnt got the conclusive solution why statement 1 is sufficient.
Or is it the OA is wrong?
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Re: 3x > 7y? [#permalink] New post 15 Jul 2009, 02:15
I think put it in simple

To define statement (1) is enough or not -- > it can say yes or no or Both. If it can say both --> then it is not sufficient right?

First if y = 1 and x =5 then 15 > 7 yes
Second if y = 10 and x = 15 then 45 > 70 no

So for the positive number x and y, we cannot sure that this Q 3x>7y or not? So statement (1) alone is not enough

For statement (2)
-5X < -14Y --- > Can rewrite to get 5X>14Y or 2.5X>7Y. Since x and y are positive and 2.5x>7y, so 3x>7y for sure.

So I prefer B krub.

Please correct me if I am worng krub.
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Re: 3x > 7y? [#permalink] New post 17 Jul 2009, 12:35
ankur55 wrote:
If x and y are positive, is 3x > 7y?
(1) x > y + 4
(2) -5x < -14y


is x/y>7/3


from 1

from one assuming x = y+4

is 3y+12>7y ie is 4y>12 ie is y>3........impossible to tell....insuff

from 2

5x>14y

x> 14/5y ie x> 2.8y assuming x = 2.8y

is 3(2.8y) > 7y

8.4y>7y.......suff

my answer is B
Re: 3x > 7y?   [#permalink] 17 Jul 2009, 12:35
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