Math Revolution and GMAT Club Contest! If |3x| > |4y|, is x > y?

1)Using this info , the stated equation becomes. 3x > |4y|
Now if y is negative then x > y
If y is positive , then 3x > 4y so still>y

Hence statement one is sufficient .

2) | 3x| > 4y

If x is positive then x > y
But if negative , then x < y
So two. Answers are possible . not sufficient

So Correct Answer is A

If |3x| > |4y|, is x > y?

the question tells us that |3x|>|4y|
this is true only when |x|>|y|

how can we find out if x>y?
well, if x is positive, then y definitely can't be greater than x, since the y would have to be -x<y<x
if y is smaller than -x, then the absolute value of y is greater than the absolute value of x, and this is not possible, since we are already given that |x|>|y|

(1) x > 0
ok, now we know that x is positive, we know for sure that y should be -x<y<x, which only means that y is smaller than x, and the answer to our main question is YES, x>y.

we can cross B,C, and E.

and we're left with A and D. - if we don't know how to solve if further, at least we try to make a strategic guess 50% of getting the correct answer is better than 20%.

(2) y > 0
this doesn't help us much. let's take two examples:
x=3 and y = 2.
x>y and |x|>|y| -> satisfies the condition, and the answer is yes

now, let's say x = -7 and y=2
x<y but |x|>|y| -> satisfies the condition, but the answer is no.

we can cross D, and select A as the correct answer!

My answer to this question is A.

If |3x| > |4y|, is x > y?

(1) x > 0

(2) y > 0

Given : |3x| > |4y|
Looking at this inequality, we can say that since 3>4, |x| > |y| for the given statement to hold.
If we can prove that x > |y|, i.e x is greater than the magnitude of y, we should be good to say "Yes" to the question at hand.

1) x > 0
implies 3x > |4y|
Hence, x > 4/3 |y| or x > |y|. Sufficient!

2. y > 0
Considering |3x| > |4y|, if y>0, x will be greater than y, only if its positive, i.e if x>0
|3x| > 4y
=> |x| > 4y/3. We cannot comment on the sign of X. Insufficient!

If |3x| > |4y|, is x > y?

To check whether x > y or not, accept values of x & y for which |3x| > |4y| holds true .
To check whether x > y or not, discard values of x & y for which |3x| > |4y| does not hold true.

(1) x > 0

Case-1: x > 0, y >= 0
x > y, x = 2, y = 1 then |3x| > |4y| holds true as |3*2| > |4 *1| i.e. 6 > 4. Accepted.
x > y, x = 2, y = 0 then |3x| > |4y| holds true as |3*2| > |4 *0| i.e. 6 > 0. Accepted.
x < y, x = 1, y = 2 then |3x| > |4y| does not hold true as |3*1| > |4*1| i.e. 3 > 4 is wrong. Discarded.

Hence, Answer to the question "is x > y?" is YES!

case-2: x > 0, y < 0
x > y, x = 2, y = -1 then |3x| > |4y| holds true as |3*2| > |4 * (-1)| = |6| > |-4| = 6 > 4. Accepted.
x > y, x = 2, y = -2 then |3x| > |4y| does not hold true as |3*2| > |4 *(-2)| = |6| > |-8| = 6 > 8 is wrong. Discarded.

Hence, Answer to the question "is x > y?" is YES!

In Both case-1 & 2 We get definite answer "Yes". Statement-1 Sufficient!

(2) y > 0
Case-1: y > 0, x >= 0
x > y, x = 2, y = 1 then |3x| > |4y| holds true as |3*2| > |4*1| i.e. 6 > 4. Accepted.
x < y, x = 1, y = 2 then |3x| > |4y| does not hold true as |3*1| > |4*2| i.e. 3 > 8 is wrong. Discarded.
x < y, x = 0, y = 2 then |3x| > |4y| does not hold true as |3*0| > |4*2| i.e. 0 > 8 is wrong. Discarded.

Hence, Answer to the question "is x > y?" is YES!

case-2: y > 0, x < 0
x < y, x = -2, y = 1 then |3x| > |4y| holds true as |3*(-2)| > |4*1| = |-6| > |4| = 6 > 4. Accepted.

Hence, Answer to the question "is x > y?" is NO!

From Case-1 we get "Yes" and from case-2 we get "NO". Answer is not definite. Statement-2 Insufficient!

Hence "A" is the final answer!

The equation |3x|>|4y|
the possibilities for [x:y] are [+:+]/[-:-] or [+:-]/[-:+]

Condition 1 says - x is +ve this implies 4y can be = +ve or -ve ; x > y shall always be true. Sufficient
Condition 2 says - y is +ve this implies 3x can be = +ve or -ve ; x > y or x < y both are possible. Insufficient

Solution - A

Alternatively same can be shown on the number line
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You know the importance of kudos, be generous.

The answer is

We can simplify the Given statement as follows |x| >|y|*4/3 (we can take 4 & 3 out of the modulus). In other words, |x| will be slightly (1.33 times) greater than |y|. Since Mod X is greater Mod Y, we cannot say for sure whether x or y is positive or negative.

1) x>0, applying a +ve number to x in the equation |x| > |y| *4/3, we cannot determine y, it can be both +ve or -ve. For instance if x is 1.33, y can be both +1 or -1 -> in both cases x will still be 1.33. NOT Sufficient

2) y>0, applying a +ve number to y in the equation |x| > |y| *4/3, we cannot determine x, it can be both +ve or -ve. For instance if y is 1, x can be both +1.33 or -1.33 -> in both cases y will still be 1. NOT Sufficient

But combining, both we can easily say x > y, since x = y*4/3.

1. X>0
Let x=1, |3*1| > |4y|, this is only possible for y=0, x > y
for x=2, |3*2| > |4y|, this is only possinle for y=1,0, -1, again x>y
Sufficient

2. y>0
For y=1, |3x| > |4*1|, x could be -2 or 2.. Insufficient.

Ans: A

If |3x| > |4y|, is x > y?

(1) x > 0

(2) y > 0

Pick numbers:
- Both positive: if x = 3 and y =2 ==> x> y : ok
- Negative/Positive: x=-3 and y =2, then x< y
- Positive/Negative: x= 3 and y= -2 then x > y : ok
- Negative: x=-3 and y =-2, then x < y

So for x to be greather than y, x must be positive while y can be either positive or negative. Statement 1 alone is sufficient
A

Answer : (A).
Solution: (I) is sufficient to solve x>y (II) is not sufficient to solve x>y. Hereby,answer choices B, C, D, E can be eliminated easily. (A) is the correct choice.

(1) x>0

Case i. \(y\geq{0}\), x>0
|3x| > |4y|
=> 3x > 4y
=> x>y

Case ii. y<0, x>0
=> x>y

So (1) is sufficient.

(2) y>0

Case i. \(x\geq{0}\), y>0
|3x| > |4y|
=> 3x > 4y
=> x>y

Case ii. x<0, y>0
=> x<y

So, (2) is not sufficient.


Hence, A.

My take is A

(1) x>0 ; Say x = 5, then |3x| = 15, given condition |3x|>|4y|, that means y could lie between -3 to 3. In any condition x>y. Hence sufficient.
(2) y>0; Say y =5, then |4y| = 20, given condition |3x|>|4y|, then say x= 7, then it satisfies and x>y but if x=-7, then it satisfies given condition but x<y. Hence not sufficient.

Hope it helps!!!

Answer is A

If x > 0, then regardless of the value or sign of y , x > y

If y > 0 , then x can either be positive or negative

The answer is a.

(i) Given, x>0
Since x is positive, we can rewrite the inequality as
3x >|4y|
3x > 4|y|
x > (4/3) |y|
x > 1.something*|y|
Clearly we can conclude that x>y.
Hence sufficient.


(ii) y>0
Given y is positive and hence the inequality becomes
|3x|> 4y
|x|> (4/3)*y
|x|> 1.something*y
This can be true for a negative value of x, where x<y and also for a positive value of x where x>y.
Hence insufficient.

|3x|>|4y|
So, to make this true, it has to be |x|>|y|

Now st1: x>0,
as 3 is multiplied with x, and 4 is with y, x has to be bigger than y. even x will be > |y|.

For St 2: y>0,
lets say y is 6(and positive)... so 4y is 24... to satisfy initial condition, |x| has to be > 8, so, x can be -9 or +9(etc)... So, insufficient.

Thus A

Ans:A
Since the question has mod | | - it is obvious that the actual value of x will always be greater than y for the inequality to work.
The trick is to figure out the sign.

1. if x>0 - Sufficient
x=1, 3x = 3. For the inequality to work - y has to be 0. any other value for y will make |4y| > 3.

2. y>0. Not sufficient.
for y=5, x can be -7 or 7. in both cases |3x| > 20. Hence not sufficient.

Statement 1- If x>0, 3x is positive. y could be either positive or negative. If y is positive, since 3x>4y, x>y. If y is negative, anyways x>y. Sufficient.

Statement 2- If y>0, 4y is positive. x could be either positive or negative. If x is positive, since 3x>4y, x>y. If x is negative, it is possible that |3x|>|4y| but not necessarily x>y. Not sufficient.

Answer A.

|3X| >|4y|

1) X>0

|3X| = 3X ..

X= 1.33 |Y|

SO, X IS ALWAYS GREATER IRRESPECTIVE OF Y SIGN

2) y>0 NO INFO ABT X NS

1+2
X>0,Y>0

SO, 3X>4Y
X>(4/3)Y.. S

A