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If 0<2x+3y<10 and -10<3x+2y<0, then which of the

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If 0<2x+3y<10 and -10<3x+2y<0, then which of the

by Max@Math Revolution » Thu Jan 18, 2018 12:35 am
[GMAT math practice question]

$$If\ \ 0<2x+3y<0\ and\ -10<3x+2y<0$$ , then which of the following must be true?

I. x<0
II. y<0
III. x<y

A. I only
B . II only
C. I & II
D.I & III
E. I, II, &III
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by GMATGuruNY » Thu Jan 18, 2018 5:18 am
Max@Math Revolution wrote:[GMAT math practice question]

$$If\ \ 0<2x+3y<10\ and\ -10<3x+2y<0$$ , then which of the following must be true?

I. x<0
II. y<0
III. x < y

A. I only
B . II only
C. I & II
D.I & III
E. I, II, &III
Since 3x+2y < 0 and 0 < 2x+3y, we get:
3x+2y < 0 < 2x+3y
3x+2y < 2x+3y
x < y.
Thus, Statement III must be true.
Eliminate A, B and C.

The remaining answer choices both include Statements I and III.
Test whether Statement II must be true.
If we substitute y=1 into -10 < 3x+2y < 0 and 0 < 2x+3y < 10, we get:
-10 < 3x + (2*1) < 0
-10 < 3x+2 < 0
-12 < 3x < -2.

0 < 2x + (3*1) < 10
0 < 2x+3 < 10
-3 < 2x < 7.

The inequalities in blue are both satisfied by x=-1.
Since it's possible that y=1 and x=-1, Statement II does not have to be true.
Eliminate E.

The correct answer is D.
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by Scott@TargetTestPrep » Sat Jan 20, 2018 6:09 am
Max@Math Revolution wrote:[GMAT math practice question]

$$If\ \ 0<2x+3y<0\ and\ -10<3x+2y<0$$ , then which of the following must be true?

I. x<0
II. y<0
III. x<y

A. I only
B . II only
C. I & II
D.I & III
E. I, II, &III
We see that x and y can't be both positive or both negative. If x and y are both positive, then the second inequality will not hold. Similarly, if x and y are both negative, then the first inequality will not hold. Therefore, we must consider two separate cases: (1) x is negative and y is positive and (2) x is positive and y is negative.

Case 1. Let's assume that x is negative and y is positive.
For example, If x = -5 and y = 5, we see that we do have 0 < 2x + 3y < 10 and -10 < 3x + 2y < 0.

Case 2. Now let's assume that x is positive and y is negative. We see that the absolute value of y must be greater than the absolute value of x in order for the second inequality to hold.
For example, if x = 2, y has to be less than -3 in order to have -10 < 3x + 2y < 0. However, in that case, the first inequality will never hold since 2x + 3y will be negative. Thus we can't have x as positive and y as negative.
Thus it must be true that x is negative and y is positive and in that case we also have x < y. Thus, Statements I and III must be true.

Answer: D

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by Mo2men » Sun Jan 21, 2018 10:38 am
GMATGuruNY wrote:
Max@Math Revolution wrote:[GMAT math practice question]

$$If\ \ 0<2x+3y<10\ and\ -10<3x+2y<0$$ , then which of the following must be true?

I. x<0
II. y<0
III. x < y

A. I only
B . II only
C. I & II
D.I & III
E. I, II, &III
Since 3x+2y < 0 and 0 < 2x+3y, we get:
3x+2y < 0 < 2x+3y
3x+2y < 2x+3y
x < y.
Thus, Statement III must be true.
Eliminate A, B and C.

The remaining answer choices both include Statements I and III.
Test whether Statement II must be true.
If we substitute y=1 into -10 < 3x+2y < 0 and 0 < 2x+3y < 10, we get:
-10 < 3x + (2*1) < 0
-10 < 3x+2 < 0
-12 < 3x < -2.

0 < 2x + (3*1) < 10
0 < 2x+3 < 10
-3 < 2x < 7.

The inequalities in blue are both satisfied by x=-1.
Since it's possible that y=1 and x=-1, Statement II does not have to be true.
Eliminate E.

The correct answer is D.
Dear Micth,

Can we sum up the 2 inequalities? it will be as follows:

-10 < 5x + 5y < 10...........-2 < x + y < 2

In that case, x could be 1 & y could be 0 or vice versa
so I & II are invalid.

Where did I go wrong?

Thanks
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by GMATGuruNY » Sun Jan 21, 2018 2:28 pm
The converse of If A, then B is If B, then A.
The converse of an if-then statement is not necessarily true.
Consider the following case:
If John is in San Francisco, then John is in the United States.
The if-then statement above is true.
Converse:
If John is in the United States, then John is in San Francisco.
The converse in red is not necessarily true.
Mo2men wrote:
Max@Math Revolution wrote:[GMAT math practice question]

$$If\ \ 0<2x+3y<10\ and\ -10<3x+2y<0$$ , then which of the following must be true?

I. x<0
II. y<0
III. x < y

A. I only
B . II only
C. I & II
D.I & III
E. I, II, &III
Can we sum up the 2 inequalities? it will be as follows:

-10 < 5x + 5y < 10...........-2 < x + y < 2

In that case, x could be 1 & y could be 0 or vice versa
so I & II are invalid.

Where did I go wrong?

Thanks
Your algebra is correct.
The inequality in blue implies the following if-then statement:
If x and y satisfy the two inequalities, then x and y have a sum between -2 and 2.
The if-then statement above is true.
Converse:
If x and y have a sum between -2 and 2, then x and y satisfy the two inequalities.
The converse in red is not necessarily true.
Thus, we cannot conclude that ANY x-y combination with a sum between -2 and 2 will satisfy the two original inequalities.
Neither of your cases in red satisfies -10 < 3x+2y < 0.

An analogous example:
If we add together x>0 and y>0, we get x+y > 0.
Implied if-then statement:
If x>0 and y>0, then x+y > 0.
The if-then statement above is true.
Converse:
If x+y > 0, then x>0 and y>0.
The converse in red is not necessarily true:
If x=-1 and y=2, then x+y > 0 but x<0.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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EDIT

by Max@Math Revolution » Sun Jan 21, 2018 5:32 pm
=>
Label the inequalities as follows:
0<2x+3y<10 --- (1)
-10<3x+2y<0 --- (2)
We consider each statement individually.

Statement I:
Multiplying (1) by -2 yields -20 < -4x - 6y < 0, and multiplying (2) by 3 yields -30 < 9x + 6y < 0. Adding these inequalities gives -50 < 5x < 0 or -10 < x < 0.
This statement is true.

Statement II:
Multiplying (1) by 3 yields 0 < 6x + 9y < 30, and multiplying (2) by -3 yields 0 < -6x - 4y < 20. Adding these inequalities gives 0 < 5y < 50 or 0 < y < 10.
This statement is false.

Statement III:
Multiplying (1) by - 1 yields -10 < -2x - 3y < 0. Adding this to inequality (2) yields -20 < x - y < 0.
This implies that x < y, and statement III is true.

Therefore, the answer is D.
Answer: D
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