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Re: Tough Inequality Challange [#permalink]
15 Jan 2010, 12:47

Hussain15 wrote:

If (x/y)>2, is 3x+2y<18?

(1) x-y is less than 2 (2) y-x is less than 2

It will be great to see how do you guys approach this lethal one.

Given x > 2y. Have to substantiate if 3x + 2y < 18.

Stmt-1: x < 2 + y. Keep substituting different values for y, we get ranges for x based on the stimulus condition and this statement, substitute these different values and we notice that certain values are applicable while many others aren't applicable to substantiate the posed question. Therefore, NS.

Stmt-2: can be rephrased as x > y - 2. Do the same method as above, same situation, no definitive answer. Therefore, NS.

combining both the statements, still substituting all possible values for y and deriving ranges for x, we can't really substantiate the given equation.

My answer is E. I wonder if there is a simpler way of solving problems of this kind. I used the brute force approach of substituting valid numbers for y and ended up getting wierder ranges for x and again, choose something which accidentally would substantiate the equation and mostly certain other numbers that do not. Took me more than a 10 mins handling work simultaneously, and if such questions appear on the real deal, I might as well give up on GMAT and pursue a PhD in Pure Math. _________________

Re: Tough Inequality Challange [#permalink]
15 Jan 2010, 13:02

7

This post received KUDOS

Expert's post

Hussain15 wrote:

If (x/y)>2, is 3x+2y<18?

(1) x-y is less than 2 (2) y-x is less than 2

It will be great to see how do you guys approach this lethal one.

I would solve this question with graphic approach, by drawing the lines. With this approach you will "see" that the answer is A. But we can do it with algebra as well.

\frac{x}{y}>2 tells us that x and y are either both positive or both negative, which means that all points (x,y) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality 3x+2y<18 is always true, so we should check only for I quadrant, or when both x and y are positive.

In I quadrant, as x and y are both positive, we can rewrite \frac{x}{y}>2 as x>2y>0 (remember x>0 and y>0).

So basically question becomes: If x>0 and y>0 and x>2y>0, is 3x+2y<18?

(1) x-y<2.

Subtract inequalities x>2y and x-y<2 (we can do this as signs are in opposite direction) --> x-(x-y)>2y-2 --> y<2.

Now add inequalities x-y<2 and y<2 (we can do this as signs are in the same direction) --> x-y+y<2+2 --> x<4.

We got y<2 and x<4. If we take maximum values x=4 and y=2 and substitute in 3x+2y<18, we'll get 12+4=16<18.

Sufficient.

(2) y-x<2 and x>2y: x=3 and y=1 --> 3x+2y=11<18 true. x=11 and y=5 --> 3x+2y=43<18 false.

Re: Tough Inequality Challange [#permalink]
15 Jan 2010, 21:35

Bunuel wrote:

Hussain15 wrote:

If (x/y)>2, is 3x+2y<18?

(1) x-y is less than 2 (2) y-x is less than 2

It will be great to see how do you guys approach this lethal one.

I would solve this question with graphic approach, by drawing the lines. With this approach you will "see" that the answer is A. But we can do it with algebra as well.

x/y>2 tells us that x and y are either both positive or both negative, which means that all points (x,y) satisfying given inequality are in I or III quadrants. When they are both negative (in III quadrant) inequality 3x+2y<18 is always true, so we should check only for I quadrant.

In I quadrant x and y are both positive and we can rewrite x/y>2 as x>2y>0 (remember x>0 and y>0).

(1) x-y<2.

Subtract inequalities x>2y and x-y<2 (we can do this as signs are in opposite direction) --> x-(x-y)>2y-2 --> y<2.

Now add inequalities x-y<2 and y<2 (we can do this as signs are in the same direction) --> x-y+y<2+2 --> x<4.

We got y<2 and x<4. If we take maximum values x=4 and y=2 and substitute in 3x+2y<18, we'll get 12+4=12<18.

Sufficient.

(2) y-x<2 and x>2y: x=3 and y=1 --> 3x+2y=11<18 true. x=11 and y=5 --> 3x+2y=43<18 false. Not sufficient.

Answer: A.

OA is "A". Thanks for detailed answer.

You have plugged the numbers in option 2, can it be done algeberically?? _________________

Re: Tough Inequality Challange [#permalink]
17 Jan 2010, 00:55

1

This post received KUDOS

Expert's post

Hussain15 wrote:

OA is "A". Thanks for detailed answer.

You have plugged the numbers in option 2, can it be done algeberically??

For (2) we have: y-2<x and 0<2y<x.

We'll be able to find the pair of (x,y) when 3x+2y<18 holds true and also when 3x+2y<18 doesn't hold true. As the lower limits for (x,y) is zero (x and y can take very small values ensuring 3x+2y<18 to hold true) and there is no upper limit for this pair (x and y can take huge values ensuring 3x+2y<18 not to hold true).

This question is quite hard and I really think that the best way to solve it is by drawing the lines OR by number plugging. _________________

Re: Tough Inequality Challange [#permalink]
21 May 2010, 08:10

1

This post received KUDOS

I spent 5 min for this question with incorrect ans .. There was no way I could have solved this question .. Very nice explanation Brunel ..

But I failed to understand the theory of addition and substraction for equalities with same sign and opposite signs respective .. Can you pls throw some light ..

Re: Tough Inequality Challange [#permalink]
21 May 2010, 08:28

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

sandeepuc wrote:

I spent 5 min for this question with incorrect ans .. There was no way I could have solved this question .. Very nice explanation Brunel ..

But I failed to understand the theory of addition and substraction for equalities with same sign and opposite signs respective .. Can you pls throw some light ..

You can only add inequalities when their signs are in the same direction:

If a>b and c>d (signs in same direction: > and >) --> a+c>b+d. Example: 3<4 and 2<5 --> 3+2<4+5.

You can only apply subtraction when their signs are in the opposite directions:

If a>b and c<d (signs in opposite direction: > and <) --> a-c>b-d (take the sign of the inequality you subtract from). Example: 3<4 and 5>1 --> 3-5<4-1.

Re: Tough Inequality Challange [#permalink]
01 Jun 2010, 14:26

Bunuel wrote:

Hussain15 wrote:

If (x/y)>2, is 3x+2y<18?

(1) x-y is less than 2 (2) y-x is less than 2

It will be great to see how do you guys approach this lethal one.

I would solve this question with graphic approach, by drawing the lines. With this approach you will "see" that the answer is A. But we can do it with algebra as well.

\frac{x}{y}>2 tells us that x and y are either both positive or both negative, which means that all points (x,y) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality 3x+2y<18 is always true, so we should check only for I quadrant, or when both x and y are positive.

In I quadrant, as x and y are both positive, we can rewrite \frac{x}{y}>2 as x>2y>0 (remember x>0 and y>0).

So basically question becomes: If x>0 and y>0 and x>2y>0, is 3x+2y<18?

(1) x-y<2.

Subtract inequalities x>2y and x-y<2 (we can do this as signs are in opposite direction) --> x-(x-y)>2y-2 --> y<2.

Now add inequalities x-y<2 and y<2 (we can do this as signs are in the same direction) --> x-y+y<2+2 --> x<4.

We got y<2 and x<4. If we take maximum values x=4 and y=2 and substitute in 3x+2y<18, we'll get 12+4=16<18.

Sufficient.

(2) y-x<2 and x>2y: x=3 and y=1 --> 3x+2y=11<18 true. x=11 and y=5 --> 3x+2y=43<18 false.

Not sufficient.

Answer: A.

+1 already for a great explanation.

Follow-up question: Would you mind detailing a graphical approach to this problem? I haven't taken a math course in 7 years so am a little rusty. Knowing how to solve such problems with a graph seems like it would be very useful. _________________

Re: Tough Inequality Challange [#permalink]
01 Jun 2010, 14:47

1

This post received KUDOS

Expert's post

alphastrike wrote:

+1 already for a great explanation.

Follow-up question: Would you mind detailing a graphical approach to this problem? I haven't taken a math course in 7 years so am a little rusty. Knowing how to solve such problems with a graph seems like it would be very useful.

Re: Tough Inequality Challange [#permalink]
26 Jun 2010, 01:16

Hi Bunuel, Thanks for the wonderful explanation. But in the actual exam when you have the clock ticking, how do we decide whether to try plugging in numbers or try solving it using algebra.I am not sure if there is any definite strategy for this but any inputs from your experience will help.

Re: Tough Inequality Challange [#permalink]
27 Jun 2010, 23:19

Given x/y > 2. i. x-y < 2: for this to be possible x and y have to be negative. Now since x and y are both negative, the equation in question will always result in a negative number. Hence, SUFFICIENT.

ii. y-x <2: For this to be possible x and y have to be positive. Now since x and y both are positive and x-y>-2, multiple solutions exist. Hence, NOT SUFFICIENT.

Re: Tough Inequality Challange [#permalink]
28 Jun 2010, 04:54

Expert's post

sunnyarora wrote:

Given x/y > 2. i. x-y < 2: for this to be possible x and y have to be negative. Now since x and y are both negative, the equation in question will always result in a negative number. Hence, SUFFICIENT.

ii. y-x <2: For this to be possible x and y have to be positive. Now since x and y both are positive and x-y>-2, multiple solutions exist. Hence, NOT SUFFICIENT.

Therefore, answer is A.

OA for this question is A, but your reasoning is not correct:

For (1) x=2>0 and y=0.5>0 satisfy both \frac{x}{y}>2 and x-y<2, so x and y can be positive as well.

For (2) x=-2<0 and y=-0.5>0 satisfy both \frac{x}{y}>2 and y-x<2, so x and y can be negative as well.

Solution for this problem is in earlier posts. _________________

Re: Tough Inequality Challange [#permalink]
28 Jun 2010, 07:08

we can solve just drawing th elines in geometric terms ,... for condition 1...draw lines of x>2y and x-y<2 and check the area whether it is lying below the line 3x+2y<18

same can be done for second condition

Bunuel wrote:

sunnyarora wrote:

Given x/y > 2. i. x-y < 2: for this to be possible x and y have to be negative. Now since x and y are both negative, the equation in question will always result in a negative number. Hence, SUFFICIENT.

ii. y-x <2: For this to be possible x and y have to be positive. Now since x and y both are positive and x-y>-2, multiple solutions exist. Hence, NOT SUFFICIENT.

Therefore, answer is A.

OA for this question is A, but your reasoning is not correct:

For (1) x=2>0 and y=0.5>0 satisfy both \frac{x}{y}>2 and x-y<2, so x and y can be positive as well.

For (2) x=-2<0 and y=-0.5>0 satisfy both \frac{x}{y}>2 and y-x<2, so x and y can be negative as well.

Re: If (x/y)>2, is 3x+2y<18? (1) x-y is less than 2 (2) [#permalink]
06 Apr 2012, 14:47

i followed the exact same procedure except at the very end. i tend to have problems switching back/forth between doing algebra, then stepping back and using logic, or stepping back and plugging numbers. in fact, I think a lot of the mistakes I make when I take the exam is trying to switch between techniques.

my problem stem 3x+2y<18 ==> is x < 18/3 -2y/3 ==> is x < 6 - 2y/3?

for (1). negative values of x/y are already sufficient, so these are for positive x and positive y

- first i try to isolate x because i took the problem stem and isolated x

x/y > 2 ==> x > 2*y

x -y < 2 ==> x < 2 + y

therefore

2* y < x < 2 + y

in order for the bolded problem stem to be true then the below must be true. x < 2+ y but is x < 6 - 2y/3?

2 + y <= 6 - 2y/3

y + 2y/3 <= 4 5y/3 <= 4 y <= 12/5 y <= 2 and 2/5 if the above is true then we know we are OK

then take that inequality with x in the middle and relate the y part only

Re: If (x/y)>2, is 3x+2y<18? (1) x-y is less than 2 (2) [#permalink]
23 Oct 2013, 21:09

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Re: If (x/y)>2, is 3x+2y<18? (1) x-y is less than 2 (2) [#permalink]
19 Jun 2014, 09:19

Hi Bunuel,

Thank you for the great solution. With regards to using graphs to solving the problem, do we get a grid kind of pad to be able to plot accurately and with ease?

Re: If (x/y)>2, is 3x+2y<18? (1) x-y is less than 2 (2) [#permalink]
20 Jun 2014, 04:48

Expert's post

startafresh wrote:

Hi Bunuel,

Thank you for the great solution. With regards to using graphs to solving the problem, do we get a grid kind of pad to be able to plot accurately and with ease?

This is what the scratchpad and pen in the test center will look like:

Attachment:

GMAT_Scratchpad.jpg [ 480.53 KiB | Viewed 568 times ]

Remember, in the GMAT testing center the scratch paper that is provided is laminated and you are given a sharp-tip erasable marker to use. _________________

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