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If x and y are positive, is3x>7y? (1) x>y+4 (2) [#permalink]
 16 Jun 2011, 01:45
Question Stats:
8% (00:00) correct
91% (01:35) wrong based on 12 sessions
If x and y are positive, is3x>7y? (1) x>y+4 (2) -5x<-14y How to solve algebraically?
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Re: Inequality Approach [#permalink]
16 Jun 2011, 03:21
jamifahad wrote: If x and y are positive, is3x>7y?
(1) x>y+4
(2) -5x<-14y
How to solve algebraically? I think the OA is B the question can be rephrases as x/y>7/3 X>y+4 pick up number y = 1 x = 5 5/1>7/3 yes now Y= 100 X = 106 X/Y is not greater than 7/3 hence insufficient st2: -5x<-14y questions can be rephrases as x>2.33Y -5x<-14y( multiply -ve on both sides and flip the sign) 5x>14Y X>14/5Y X>2.8Y.. hence sufficient
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Re: Inequality Approach [#permalink]
16 Jun 2011, 03:44
the OA should be B to solve it graphically just draw all the three lines. you would notice that the 3x=7y and 5x=14y would intersect at 0. in the first quadrant the 5x=14y would be more steeper. since 5x>14 y => the area between x axis and 5x=14y. to check whether it satisfies 3x>7y : since the 5x=14y lies below 3x=7y => 3x>7y would always be true. Hence 2nd statement is sufficient. 1st statement is not sufficient. x>y+4 intersects at 7,3 with 3x=7y....after the intersection it the answer is NO. before the intersection the answer is Yes. Hence not sufficient. Answer = B.
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Re: Inequality Approach [#permalink]
16 Jun 2011, 03:58
If this was a problem solving question, I would have gotten a wrong reply, having said that since this is DS we don't need to solve we just need to know we can solve. Let's look at the question now We know x>7y/3 Statement 1 Sufficient because we put in a value of x and get a value of y. Let's assume from an understanding point of view x=4 x>y+4 y<0 hence x>0 Sufficient Statement 2 x>14y/5 Assume x=5 y=25/14 In this case 3x not greater than 7y Sufficient D  Posted from GMAT ToolKit Posted from GMAT ToolKit
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Re: Inequality Approach [#permalink]
16 Jun 2011, 04:08
3x>7y critical points are x = 7 and y = 3. Considering first quadrant. for 0< > x<7 and 0<y<3 is the region for which 3x > 7y. a x=6,y= 1 gives 3x > 7y and for x=10, y= 5 the sign changes. Not sufficient. b 5x > 14y 2<x < 14 and 0<y<5 satisfies. checking for region 2<x<7 and 0<y<3 gives 3x > 7y always. Hence B.
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Re: Inequality Approach [#permalink]
28 Jun 2011, 09:45
1. not sufficient.
2. 5x > 14y => 2.5x > 7y => 3x > 7y, hence sufficient.
B.
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Re: Inequality Approach [#permalink]
11 Jul 2011, 18:11
Can somebody please solve this algebrically?
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Re: Inequality Approach [#permalink]
11 Jul 2011, 20:20
1. try following values for (x,Y) (6, 1) -----------Yes (106, 100)--------No so 1 is not sufficient. B is the answer.
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Re: Inequality Approach [#permalink]
11 Jul 2011, 20:57
From 3x>7y
x > 7/3y
From (1) x > y+4
To prove (1) is enough, we must prove y+4 is always greater than or equal to 7/3y
To find out about this, we solve inequality
y+4 >= 7/3y,
we get y<= 3
Therefore, only when y<=3 that y+4 >= 7/3y
Not every value of y make y+4 >= 7/3y
Therefore (1) is insufficient
Doing the same thing with (2), we know (2) is sufficient
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Re: Inequality Approach [#permalink]
14 Jul 2011, 06:26
OA must be B. Given: x, y >0 Is 3x>7y? Rephrase Is x/y > 7/3? (since both x and y are +ve) St 1: x > (y+4) => x/y > 1+4/y. Insuff. as dependent on value of y. if y =1, Yes and if y=4, No. St 2: -5x < -14y => x/y >14/5. Suff. B. took me more than 2 mins 
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Re: Inequality Approach [#permalink]
31 Jul 2011, 11:06
There are more ways to solve stat 1: either multiple both sides by 3: you will get 3x>3y+12. combining both - you will notice that you need to check whether 3y+12>7y you can see immediately that you cannot know. plugging 1 and 10 showing that quickly. another way: x>y+4 we can also say x=y+4+c so the new question can be: 3(y+4+c)>7y? again - we cannot know. we can play now with c and y for the result we want.
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Re: Inequality Approach
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31 Jul 2011, 11:06
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