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If x and y are negative numbers, is x<y? Updated on: 21 May 2023, 01:25
Originally posted by nitin6305 on 13 Aug 2013, 11:44.
Last edited by Bunuel on 21 May 2023, 01:25, edited 3 times in total.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

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Bunuel, can you please explain the graphical method to solve this question?

Thanks

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If x and y are negative numbers, is x<y? 13 Aug 2013, 12:09
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nitin6305
If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Bunuel, can you please explain the graphical method to solve this question?

Thanks
Show more
Dear nitin6305,
I'm happy to help with this. :-)

Idea #1: to graph an inequality, solve for slope-intercept form (i.e. y = mx + b form). When we do that
Graph (1) becomes: y > 3x/2 + 1/2
Graph (2) becomes: y > 2x/3 + 1/3
We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line.

Idea #2: any question about whether x is > or = or < y is question about the line y = x. Read this post about the magical properties of the line y = x.
https://magoosh.com/gmat/2012/gmat-math- ... -line-y-x/

Here's the image:
Attachment:
inequalities with y = x.JPG
inequalities with y = x.JPG [ 82.84 KiB | Viewed 24006 times ]
The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x.

Does all this make sense?
Mike :-)
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If x and y are negative numbers, is x<y? 07 Oct 2015, 07:03
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mikemcgarry
nitin6305
If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Bunuel, can you please explain the graphical method to solve this question?

Thanks
Dear nitin6305,
I'm happy to help with this. :-)

Idea #1: to graph an inequality, solve for slope-intercept form (i.e. y = mx + b form). When we do that
Graph (1) becomes: y > 3x/2 + 1/2
Graph (2) becomes: y > 2x/3 + 1/3
We plot the exact lines (y = 3x/2 + 1/2 and y = 2x/3 + 1/3) to determine the boundaries of the regions. If the inequality is y > mx + b, then the region representing the inequality is above the line; if the inequality is y < mx + b, then the region representing the inequality is below the line.

Idea #2: any question about whether x is > or = or < y is question about the line y = x. Read this post about the magical properties of the line y = x.
https://magoosh.com/gmat/2012/gmat-math- ... -line-y-x/

Here's the image:
Attachment:
inequalities with y = x.JPG
The darker green region, above the line y > 3x/2 + 1/23, is the region representing the first inequality. The yellow region, above the line y = 2x/3 + 1/3, is the region representing the second inequality. The bright spring green region is their overlap, the region satisfied by both inequalities. Notice that everything in that bright spring green region is above the solid green line, which is y = x. Points above the line y = x always have y > x.

Does all this make sense?
Mike :-)
Show more

Hi mikemcgarry

Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.

per the graph2 y > 2x/3 + 1/3 ------> the yellow region in the graph

I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.

Please help, I think I have confused things here since the OA is B.

Thanks in advance !

Regards,
SR
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If x and y are negative numbers, is x<y? 07 Oct 2015, 11:21
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Hi mikemcgarry

Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.

per the graph2 y > 2x/3 + 1/3 ------> the yellow region in the graph

I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.

Please help, I think I have confused things here since the OA is B.

Thanks in advance !

Regards,
SR
Show more
Dear solitaryreaper,
I'm happy to help.

Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y.

For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?"

By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question.

We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient.

Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change.

Does this answer your questions?
Mike :-)
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If x and y are negative numbers, is x<y? 07 Oct 2015, 11:33
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solitaryreaper
Hi mikemcgarry

Thanks for sharing the succinct graphical approach. I am having trouble understanding how the graph 2 gives a definite answer.

per the graph2 y > 2x/3 + 1/3 ------> the yellow region in the graph

I can see in the yellow region we have one small region where x>y and one bigger region where y>x. If that's true then we can't have a definite answer.

Please help, I think I have confused things here since the OA is B.

Thanks in advance !

Regards,
SR
Dear solitaryreaper,
I'm happy to help.

Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y.

For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?"

By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question.

We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient.

Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change.

Does this answer your questions?
Mike :-)
Show more

Hi Mike,

The stem says that x and y are negative numbers, so the OA is correct.
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If x and y are negative numbers, is x<y? 06 Nov 2017, 23:36
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nitin6305
If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Bunuel, can you please explain the graphical method to solve this question?

Thanks
Show more


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Since we have 2 variables and 0 equation, C is most likely to be the answer.

Condition 1) & 2)
\(3x + 4 < 2y + 3\)
\(2x - 3 < 3y - 4\)
If we add both inequalities, we have \(5x + 1 < 5y - 1\).
\(5x + 1 < 5y - 1 < 5y + 1 ⇔ 5x < 5y or x < y\).
Both conditions together are sufficient.

Since this question is one of key questions, we need to check A or B by CMT(Common Mistake Type) 4 (A).

Condition 1)
\(3x + 4 < 2y + 3\)

\(x = -2\), \(y = -1\) : Yes
\(x = -20\), \(y = -21\) : No.

This is not sufficient.

Condition 2)

\(2x - 3 < 3y - 4 ⇒ 3x - 4 < 3x - 3 < 2x - 3 < 3y - 4\) since \(x < 0\).
We have \(3x < 3y\) or\(x < y\).

This is sufficient.

Therefore, the answer is B.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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If x and y are negative numbers, is x<y? 18 Mar 2020, 05:39
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nitin6305
If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Bunuel, can you please explain the graphical method to solve this question?

Thanks


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Since we have 2 variables and 0 equation, C is most likely to be the answer.

Condition 1) & 2)
\(3x + 4 < 2y + 3\)
\(2x - 3 < 3y - 4\)
If we add both inequalities, we have \(5x + 1 < 5y - 1\).
\(5x + 1 < 5y - 1 < 5y + 1 ⇔ 5x < 5y or x < y\).
Both conditions together are sufficient.

Since this question is one of key questions, we need to check A or B by CMT(Common Mistake Type) 4 (A).

Condition 1)
\(3x + 4 < 2y + 3\)

\(x = -2\), \(y = -1\) : Yes
\(x = -20\), \(y = -21\) : No.

This is not sufficient.

Condition 2)

\(2x - 3 < 3y - 4 ⇒ 3x - 4 < 3x - 3 < 2x - 3 < 3y - 4\) since \(x < 0\).
We have \(3x < 3y\) or\(x < y\).

This is sufficient.

Therefore, the answer is B.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Hi @mathsrevolution VeritasKarishma Bunuel

for first statement 1, I tested values and concluded that st 1 is enough. My question here would be how do we know when we have checked "sufficient" values. I checked extreme values (x=-100 y= -1) and close values (x= -4 and y =-3) both cases gave me a YES. therefore I concluded with st 1 as sufficient.
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If x and y are negative numbers, is x<y? 18 Mar 2020, 10:07
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nitin6305
If x and y are negative numbers, is x<y?

(1) 3x+4<2y+3

(2) 2x−3<3y−4

Bunuel, can you please explain the graphical method to solve this question?

Thanks


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Since we have 2 variables and 0 equation, C is most likely to be the answer.

Condition 1) & 2)
\(3x + 4 < 2y + 3\)
\(2x - 3 < 3y - 4\)
If we add both inequalities, we have \(5x + 1 < 5y - 1\).
\(5x + 1 < 5y - 1 < 5y + 1 ⇔ 5x < 5y or x < y\).
Both conditions together are sufficient.

Since this question is one of key questions, we need to check A or B by CMT(Common Mistake Type) 4 (A).

Condition 1)
\(3x + 4 < 2y + 3\)

\(x = -2\), \(y = -1\) : Yes
\(x = -20\), \(y = -21\) : No.

This is not sufficient.

Condition 2)

\(2x - 3 < 3y - 4 ⇒ 3x - 4 < 3x - 3 < 2x - 3 < 3y - 4\) since \(x < 0\).
We have \(3x < 3y\) or\(x < y\).

This is sufficient.

Therefore, the answer is B.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Hi @mathsrevolution VeritasKarishma Bunuel

for first statement 1, I tested values and concluded that st 1 is enough. My question here would be how do we know when we have checked "sufficient" values. I checked extreme values (x=-100 y= -1) and close values (x= -4 and y =-3) both cases gave me a YES. therefore I concluded with st 1 as sufficient.
Show more

I will not use testing numbers to establish sufficiency. To establish insufficiency, yes, that is fine. If I am quickly able to arrive at two values which give a yes and a no, I can say that the statement is not sufficient and move on. But if all values I try give the same answer, I will need to establish sufficiency using logic.


(1) 3x + 4 < 2y + 3
3x < 2y−1
x < 2y/3 - 1/3
Add and subtract y/3 from RHS
x < y - y/3 - 1/3

x < y - (y+1)/3

Now, (y+1)/3 can be a positive number or a negative number say
when y = -1/2, (y+1)/3 is positive
when y = -2, (y+1)/3 is negative

So this inequality could give us e.g.
x < y - 1/6 (we don't know if x is less than y)
or x < y + 1/3 (yes, x is less than y)

Hence, not sufficient.
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If x and y are negative numbers, is x<y? 21 May 2023, 02:57
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If \(x\) and \(y\) are negative numbers, is \(x < y\)?

(1) \(3x + 4 < 2y + 3\).

Re-arrange to: \(3x < 2y-1\).

Express \(2y\) as \(3y -y\) to get: \(3x < 3y - y - 1\), which can also be written as: \(3x < 3y - (y + 1)\).

Divide each term by 3: \(x < y -\frac{y+1}{3}\).

Now, \(\frac{y+1}{3}\) can be positive if \(y > -1\), and negative if \(y < -1\).

When \(\frac{y+1}{3}\) is positive, \(y -\frac{y+1}{3}\) will be less than \(y\), so we'd have: \(x < y -\frac{y+1}{3} < y\).

When \(\frac{y+1}{3}\) is negative, \(y -\frac{y+1}{3}\) will be greater than y, so we'd have: \(y < y -\frac{y+1}{3}\). In this case, \(x\) could be between \(y\) and \(y -\frac{y+1}{3}\), thus being more than \(y\): \(y < x < y -\frac{y+1}{3}\), as well as to the left of \(y\), thus being less than \(y\): \(x < y < y -\frac{y+1}{3}\).

Not sufficient.

Alternatively, after getting \(3x < 2y-1\), we could plug in numbers. If \(x\) is a very small number, for instance -100, and \(y\) is -1, then \(x < y\) and the answer is YES. However, if \(x=-2\) and \(y=-2\), then \(x=y\) and the answer is NO.

(2) \(2x - 3 < 3y - 4\).

Re-arrange to: \(2x < 3y - 1\).

Express \(3y\) as \(2y + y\) to get: \(2x < 2y + y - 1\).

Divide each term by 2: \(x < y + \frac{y-1}{2}\).

Since \(y\) is negative, \(\frac{y-1}{2}\) will also be negative, thus \(y + \frac{y-1}{2}\) will be less than \(y\). Therefore, we have \(x < y + \frac{y-1}{2} < y\).

Sufficient.


Answer: B
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