Is x|y| > y^2? (1) x > y (2) y > 0

Simplifying, is x|y| > |y|*|y| i.e is x > |y|
You can simplify, or more precisely, divide through by |y| if you know that y is non-zero.

The question "Is x|y| > y^2?" is equivalent to "Is y non-zero and x > |y|?"

(1) y can be 0. Also, x can be greater than |y|.
Not sufficient.
(2) Not sufficient, as nothing is known about x.

(1) and (2) together: y > 0 and x > y = |y|.
Sufficient.

Answer C.

I 'd like to know if my approach is correct

x | y | > y^2 ------> x < - y or x > y

1) x > y but we do not know if x is less of - y insuff

2) y> 0 we know only that y is positive but nothing about x

1) + 2) y is positive and x must be greater than y. suff.

is correct ??

x | y | > y^2 ------> x < - y or x > y NO

First of all, y must be non-zero. Then dividing through by |y|, we obtain x > |y| > 0.
From the given inequality, \(x|y| > y^2>0\) we deduce that x > 0, x cannot be negative.
And you cannot automatically assume that y is positive.

Can we take

\(Y^2\) = \([y][y]\) and cancel [y] from both sides if y is not equal to zero.

If \(y\neq{0}\) and \(x|y|>y^2\) we CAN reduce by \(|y|\) and get: \(x>|y|\).

Hope it's clear.

Is x|y|>y^2?

(1) x>y

Though we know that x>y, we don't know anything about the signs of x and y. For example:

x=2, y=1
x|y|>y^2
2|1|>1^2
2>1 Valid

x=2, y=-3
x|y|>y^2
2|-3|>-3^2
6>9 Invalid

INSUFFICIENT

(2) y>0
This tells us nothing about x. For example:

x=10, y=1
x|y|>y^2
10|1|>1^2
10>1 Valid

x=1, y=10
x|y|>y^2
1|10|>10^2
10>100 Invalid

INSUFFICIENT

1+2) x>y, y>0 ===> x>y>0
If y>0 then x is greater than zero (and y)
x=10, y=9
x|y|>y^2
10|9|>9^2
90>81

Think of it like this: The RHS is the smaller number times the smaller number. The LHS is the smaller number times a number larger than the smaller number.
SUFFICIENT

(C)

We have x|y|> y2

Now, to solve this, we need to understand, what do we actually want to know.

y2 = |y|2

x|y|>|y|2
x|y| - |y|2 > 0
or (x-|y|)|y|>0

Now, we know that |y| is always greater than 0, Thus, our inequality yield x-|y|>0

So Is x|y|>y2 translates to Is x> |y|

Now, lets come to the statements,

Statement 1 says, x>y; which is not sufficient. In case y is negative, we can't say anything about x being greater than absolute value of y

Statement 2 say, y>0, which in itself is insufficient as we don't know anything about x here.

But when we combine these two statements, we would get the answer. As we know when y > 0; |y| = y and statement 1 says x > y, and if y > 0 we can write x > |y|

Hence, the answer is C.

Hope it helps!!!

Kudos if it helped!!!

Everything is correct except that |y| is more than or equal to 0. So, x*|y| > y^2 after reducing by |y| can be translated: is \(x > |y|>{0}\)?

Hi Experts,

Is the following solution correct?
Question: Is x·|y| > y^2?
True when:
Case 1:
y>0 and x > y
Case 2:
y < 0 and x < -y

(1) NS; we don't know whether y > 0
(2) NS; we don't know anything about x

(1) + (2)
Case 1, so sufficient

For \(x*|y| > y^2\) to be true either \(0 < y < x\) or \(-x < y < 0\) (notice that in both cases x must be positive).

Can we rephrase the question as IxI>IyI??

reason.. if we square both side xIyI>y^2
we get x^2Y^2>Y^4

X^2>Y^2 or IXI>IYI...????

No, we cannot do this. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Check the following post for more: How to manipulate inequalities (adding, subtracting, squaring etc.).

Hope it helps.

Hi,
Bunuel too will be able to give a approach, but i will give you one--

\(x·|y| > y^2.....\)
what does this mean...
\(y^2-x|y|<0.............\)
since y^2 and |y| will be positive, we have to answer if x>|y|......OR x>0 and x>y...

lets see the statements-

\((1) x > y\)
say y is -ive and x is also -ive...... ans will be NO
if x is +ive, ans is YES
Insuff

\((2) y > 0\)
nothing about relation between x and y..
Insuff...

Combined-
x>y and y>0, so x>0..
ans is YES
Suff
C

Is x·|y| > y^2?

(1) x > y. If y=0, then no matter what x is x*|y| = y^2 = 0, and we'll have a NO answer to the question but if y=1 and x=2, then x*|y| > y^2, and we'll have an YES answer to the question. Not sufficient.

(2) y > 0. This implies that |y| = y, thus the question becomes is x·y > y^2. Reduce by y (we can safely do this since we know that y is positive): is x > y. We don't know that. Not sufficient.

(1)+(2) From (2) the question became whether x > y and (1) confirms that. Sufficient.

Answer: C.

Hope it's clear.

Y^2 is always positive, and |Y| is always positive. Y^2 could be written as Y^2= |Y||Y|

So we can safely divide both sides by |Y| without problem, then the question becomes: is x>|Y|???

(1) x > y.

Put x=2 & Y=1 .. so it is yes.

Put x=2 & Y=-3.. so it is NO

Insuff

2) Y>0. clearly nkthing about X.
Insuff

Combining 1 & 2

X>Y>0. We can take x=2 & Y=1


Answer: C

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.

There 2 variables and 0 equation. Thus we need 2 equations to solve for the variables; the conditions provide 2 equations, so there is high chance that (C) will be the answer.

The question \(x|y| > y^2\) is equivalent to \(x|y| > |y|^2\) or \(|y| ( x - |y| ) > 0\).
The equivalent question is if \(|y| ( x - |y| ) > 0\).

Condition 1)
\(x = 2\), \(y = 1\) : It is true.
\(x = 2\), \(y = 0\) : It is false.
This condtion is not true.

Condition 2)
This is not sufficient, since we don't know anything about \(x\).

Condition 1) & 2)
Since \(y > 0\), we have \(|y| > 0\).
Since \(x > y = |y| > 0\), \(x - |y| > 0\).
Thus \(|y| ( x - |y| ) > 0\).
Both conditions together are sufficient.

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.

Bunuel

Can we assume that since y^2 will always be positive and |y| will also be positive, the question is asking if x is positive? We will still get C as the answer.