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

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.

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

(1) x > 0

(2) y > 0

Stmt 1:

X>0
We have 2 scenarios for y
i) y>0 -> 3x>4y
-> x/y > 3/4
if x/y = 4/5 => x<y
if x/y = 2 => x>y
ii)if y<0 => x>y

Since we cannot say for sure => insuff

Stmt 2:

Y>0
We have 2 scenarios for x
i) x>0 -> 3x>4y
-> x/y > 3/4
if x/y = 4/5 => x<y
if x/y = 2 => x>y
ii)if x<0 => x<y

Since we cannot say for sure => insuff

Stmt 1 & 2: Still we have x/y>3/4 => Still we cannot say if x>y. Hence Insuff

Ans: E

|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

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

(1) x > 0

(2) y > 0


1)x > 0

==> X is not a negative number.
==> x > (4/3) y and x > -(4/3)y (Assuming we don't know if y>0)
The factor 4/3 is greater than 1
Test cases 1)Assuming integers
if y = 1; x > 1.33 ==> x >y
2)Assuming them to be decimals

if y = 0.5; x >0.66 ==>x >y

Hence sufficient.

2) y > 0

==> y is not a negative number.
y < (3/4)x .Since y > 0, y cannot be less than -(3/4)x.

Test cases
1)Assuming integers
if x = 1; y < 0.75 ==> x >y


2)Assuming them to be decimals
if x = 0.2; y < 0.15 ==>x >y

Hence sufficient.

Hence
(D)Both alone are sufficient.

QUESTION #1:

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

(1) x > 0

(2) y > 0


Soln: |3x| > |4y|, or 3*|x| > 4*|y| (since |A*B|=|A|*|B|; |3|=3, and |4|=4)
or \(\frac{3}{4}\)|x| > |y|, or |x| > |y| (if 3/4th of a is greater than b; a must also be greater than b, when a & b both +ve).

1. x>0. suppose x = 5. So, |y| < 5, or -5 < y < 5.
So, we see all the value of y must be less than x=5. or x>y. SUFFICIENT.

2. y>0. Suppose y = 5. So, |x| > 5. The range of x is thus: x > 5, or x < -5. In other words, x > y or x < -y. NOT SUFFICIENT.

Answer: A.

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

(1) x > 0
(2) y > 0

Solution :
Statement (1) : x > 0
suppose x = 0.1,
given |3x| > |4y|
0.3 > |4y|
y < 0.075
so x > y (Sufficient)

Statement (2) : y > 0
suppose y = 1,
given |3x| > |4y|
|3x| > 4
x > 4/3 ......... x > y
or x < -4/3 ....... x < y
we have two answers .. so Not Sufficient

Answer (A)

Answer : A
Explanation :
given = |3x|>|4y|

S1 : X>0, we know multiple of x is 3 and multiple of y is 4. So in order to have 3x more than 4y , y has to be less than x
i.e. if x=3, then |3x|=9 so 9>|4y|. Therefore y shall be less than 2.25. Which is x>y.
So S1 is sufficient.

S2 : y>0, we know |3x|>|4y|. x could be negative and still value of |3x| could be more than |4y|
i.e. x=-3, |3x| =9,
y=2, |4y| = 8
or
x=3, |3x| =9,
y=2, |4y| = 8
So S2 is not sufficient.

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

(1) x > 0

If x>0 then |x|= x
So, 3x >|4y|

y can be positive or negative.
If y is positive, 3x > 4y and x>4y/3 . Therefore x>y
If y is negative, since x is positive, x >y.
Therefore x is always greater than y, when x is positive. Choice A is sufficient.

(2) y > 0
If y is positive, then |3x| > 4y
Here x can be either positive or negative.
When x is positive, 3x > 4y and x>4y/3. Therefore x>y
When x is negative, -3x >4y and x<-4y/3. Therefore x<y.
This choice leads to one yes and one no answer. Hence not sufficient.

Therefore answer is choice A.

(1) x > 0 => 3x > |4y| => x > 4|y|/3 > |y| => x > y : Sufficient
(2) y > 0 is insufficient because if x > 0 then x > y, if x < 0 then x < y.

Answer A

since |3x|>|4y| we can take out 3 and 4 outside mode as both are positive (Absolute number). now we have |x|>4/3|y|. It imply absolute value of x is greater than 1.33times absolute value of Y. Now look at options
1. x>0 -->> If X is positive , then y can be positive or negative provided absolute value of y is less than . If Y >0, it has to be less than X, if Y <0, then x is always greater than y. In either case X>Y. Sufficient
2. Y>0 --> X can be > or <0. If X>0, then X>Y, if X<0( absolute value of X is greater than Y), then X <Y. Not Sufficient.

Ans :A

MATH REVOLUTION OFFICIAL SOLUTION:

This type of question appears within a score range of 45 to 48. Under the original condition, there are 2 variables (x and y) and 1 equation (|3x|>|4y|). In order to have the same number of equations, we need 1 more equation. In that sense, D seems highly correct answer because there is 1 equation each for both 1) and 2).

In case of 1), if x>0, then |3x|>|4y|→ 3x>|4y|≥|3y|≥3y→ 3x>3y → x>y, which means yes, this is sufficient.
In case of 2), if y>0, then |3x|>|4y|=4y, if x=2 and y=1, which means yes. However, if x=-2 and y=1, which means no. Because both yes and no are there, this is not sufficient. Therefore, the correct answer is A.

You can solve this question in less than 2 minutes while conventional way of solving takes 4-5 minutes.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

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

(1) x > 0

Let x = 2 & y= 1.......Answer to question is yes

Let x = 1/3 & y=1/6..Answer to question is yes

Let x =1 & y =0.........Answer to question is yes

As x =0, y =any negative value...Answer to question is yes

Sufficient

(2) y > 0

Let x = 2 & y= 1.......Answer to question is yes

Let x = -2 & y= 1.......Answer to question is No

Insufficient

Answer: A

A
From initial expression you know that the absolute value of X is greater than that of Y.
From that we can derive that if X is positive number, than X is greater than Y. If X is a negative number, than X is smaller than Y irrespective of the sign of Y.

Statement 1: Now we now that X is positive, hence X is greater than Y irrespective of the the sign of Y

Statement 2: Not sufficient. We cannot make implication. Two possible scenarios: X is negative -> smaller, X is positive -> larger

Hope that helped


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1)
If x>0, for |3x|>|4y|, 3x must be > 4y or > -4y. Divide both sides by 3 and we have x>4y/3 or x>-4y/3. This is true for all possible values of y so condition 1 is sufficient.

2)
If y>0, for |3x|>|4y|, 3x or -3x must be > 4y. Divide both sides by 3 and we have x>4y/3 or divide both sides by -3 and we have x<-4y/3. The former case means it's possible for x to be positive and greater than y and the latter case means it's possible for x to be negative while y is positive, therefore x<y. Insufficient.