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x/y > 2 would mean that either both x and y are positive or both x and y are negative. If both x and y are negative, then the answer to the question stem is YES. So only the positive values of x and y need to be considered in trying to prove 3x + 2y<17 to be false.

Statement 1:
x-y < 2

For both x and y being positive, they need to be 'close' enough to zero in order to preserve x/y>2 and x-y<2. Therefore if x and y are small, 3x + 2y will be less than 17.

For eg. x = 3, y=1.1, 3x + 2y = 11.2 < 17. Other possible values of x and y will not be too far from those in this example and will satisfy 3x+2y<17. Therefore, statement 1 is sufficient.

Statement 2:
y-x < 2

If x=10, y=1, then 3x+2y will be larger than 17. Therefore insufficient.

Choose A.

P.S. My solution is a bit crude, especially in solving for statement 1. Would be interested in seeing a more analytical approach.

-BM-

I got B

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X/Y >2 -- > both X and Y must be +ve or both -ve

3X + 2Y <17 always true when X and Y are -ve..


1) X - Y < 2 --> eqn(1)

X/Y >2 --> X-2Y>0 --> 2y-x<0 eqn(2)
eqn(1)+eqn(2)
--> y<2 --> X<4

3X + 2Y < 3*4+2*2 <16

Statement 1 is sufficient

2) Y-X <2

WHEN X=2.1 Y=1

3X + 2Y <17 True

WHEN X=30.1 Y=15
3X + 2Y <17 False


not sufficient


A

can we really add inequalities to "solve" like equation ?

I solved this the following way :

3x+2y
=x+2(x+y)
=y{x/y+2(1+x/y)}

x/y >2
x/y = 2+ d (delta) where d >0

3x+2y = y(8+3d)


From (1)

x-y <2
x-y = 2-m (m>0)
y(x/y -1) = 2- m
y(2+d-1) = 2 -m

y = (2-m)/(d+1) <2 ---------(a)

y(d+1) = 2-m < 2
5y < 10 ..............................(b)

now 3x+2y = y(8+3d)
=y {5 + 3(d+1)}
= 5y + 3(2-m)
<10 + 6 -3m
<16 -3m

since m >0 so 3x+2y has to be less than 16

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Folks, any take on this....I am looking for more solutions (lucid) before I move away from this....

x/y < 2 From stem

1 x- y < 2

For this to be true both x and y have to be negative. If that's the case then 3x+2y is always < 17 - Sufficient.

2. y-x < 2 ; x<2y : For this to be true x is +ve and y is -ve always .. hence insufficient.

Choose A.

I doubt this is correct - but gave it a shot.

- pradeep

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I think you were too quick in concluding the bold statements above..Isn't it?

So x-y < 2 and x > 2y or x< y+2 and x > 2y

Try x +ve and y +ve - doesn't hold true
Try x -Ve and y +ve - doesn't hold true
x +ve and y -ve - doesn't not true

x -ve and y -ve - Holds true. Unless am missing something here.

Hence the conclusion.

I had to edit the post multiple times .. darn this is what happens when work and GMAT prep clash.

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GMATPREP1 - 550 (Oct 08)
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I somehow managed to get the answer as A. But not confident enough... sureshbala, i hope you are coming up with the explanation

Given x/y>2…..implies both x and y can be positive or both of them are negative
If both of them are negative, it is clear that 3x+2Y<0.
If both of them are positive we have x > 2y…......(1)

Statement1: x-y < 2 ...(2)
From (1) we have x-y > y .....(3)
From (2) and (3), y<x-y<2, so y<2.
From (2), we have x<y+2, so x<4.
Since y<2 and x<4, 3x+2y<16.
Hence statement (1) alone is sufficient

Coming to statement 2, we can think along the same lines and easily find two contradicting examples.
Let x=100 and y=1
Clearly x/y > 2, and y-x < 2 and the value of 3x+2y=302 (>18)
Let x=3 and y=1
Clearly x/y>2 and y-x<2 and the value of 3x+2y=11(<18)
So statement(2) is not sufficient
Hence the answer is Choice A.

Here is the next one...

The roots of a quadratic equation ax^2+bx+c=0 are integers and a+b+c>0 and a>0. Are both the roots of the equation positive?

1. Sum of the roots is positive.

2. Product of the roots is positive.

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Dear Suesh,
Thanks for a good Qs and a nice answer.

Dear Stuart,
I made this question right by making 1st condition by inequalities calculations, but in condition 2 I did not had an option but for to choose nos.

Can you or Suresh, please explain us wha all we need to consider before choosing smart nos. for cheking the answers, if we are choosing not to go by algebric methods.

Thanks.

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Basic rules:
a>0, this simply means the curve ax^2+bx+c=0 is 'U' shaped, having a single minimum point.
The sign of -b/2a can be used to identify if this single minimum point occurs in the positive x region (-b/2a > 0) or in the negative x region (-b/2a < 0).
The sign of c determines whether the curve intersects the y-axis in the positive y region (c>0) or the negative y region(c<0).

Statement 1: Sum of the roots is positive
This could mean either:
(i) both roots are positive (with c>0), or
(ii) one positive root and one negative root (with c<0, and the positive root having the larger absolute value)

For both these cases, the minimum point is in the positive x region. So we have:

-b/2a > 0
Since we know, a>0, therefore b must be less than 0.

a + b + c>0
We know that a>0 and b<0, but c could be either positive or negative. Therefore, we are unable to determine which of these two cases (i and ii) is true. Insufficient.

Statement 2: Product of the roots is positive
This could mean either:
(iii) both roots are positive (with -b/2a >0), or
(iv) both roots are negative (with -b/2a <0).

For both these cases, the y-intercept must be positive (c>0).

a + b + c>0
We know that a>0 and c>0, but b could be either positive or negative. Therefore, we are unable to determine which of these two cases (iii and iv) is true. Insufficient.

Both statements together:
From statement 1, we have established that b<0. This produces -b/2a > 0 (minimum point occuring in the positive x region)
From statement 2, we have established that c>0. The y-intercept in positive.

We can conclude now that both roots are positive.

Choose C.

-BM-