Is y^3 ≤ |y|? (1) y < 1 (2) y < 0

Is y^3 ≤ |y|?

(1) y < 1
(2) y < 0

Is y^3 ≤ |y|?

If y≥0 then:
y^3 ≤ |y|
y^3 ≤ y
y^3 - y ≤ 0

If y≤0 then:
y^3 ≤ |y|
y^3 ≤ -y
y^3 + y ≤ 0

1.) y < 1

For either case, positive or negative it holds true. For example:

for the positive case y<1.
If y=1/2 then 1/2^3 - 1/2 ≤ 0.
we only take cases between 0 and 1

for the negative case y<1
if y = -1/2 then -1/2^3 + -1/2 ≤ 0
we only take cases less than zero

2.) y < 0

We don't consider the positive case because we are only considering the negative case of y.

for the negative case y<0
if y = -1/2 then -1/2^3 + -1/2 ≤ 0
we only take cases less than zero
(Same as above)

Is this a correct way to solve the problem? It seems like others solved it in a different fashion.

Yes,it's correct. You can solve it correctly in many ways

stmt 1 :
y<1

test numbers:

y=1/2 , 1/8<1/2 - yes
y=0 0=0 , yes
y= any negative value

-ve <= +ve , always yes

sufficient.

stmt 2:
y<0

for any negative value, y3 will always be negative
-ve <= +ve ( always yes)

sufficient
Ans D

please give kudos if you like the solution

If we square it on both the sides then how do we get y<=1 ?

y^6 <= y^2
y^2 (y^4-1) <=0
As per + - + - method, since y^2, we do not consider it.
Y^4 - 1 gives us +1 and -1.

-1 +1
+ - +

So -1<= y <= 1. How did we get y<=1 ??

Is y^3 ≤ |y|?

(1) y < 1

Test Values with integers and fraction with positive and negative sign.

Let y = 1/2........Answer to question is yes

Let y = 0........Answer to question is yes

Let y = -1/2........Answer to question is yes

Let y =-2........Answer to question is yes

Sufficient

(2) y < 0

Use the same third and fourth examples above....Answer is Yes

Sufficient

Answer: D

Actually I wanted to know if the range is correct or not...Have i made any error in the algebriac expression ???

Nikhil - squaring both sides is not a good approach here, because there is no guarantee that both sides are positive. If y is negative, the squaring could flip the inequality.

Example take y=-2

(-2)^3 <= |-2| --> correct?

squaring both sides (without flipping inequality) gives
(-2)^6 <= (-2)^2 ==> This is obviously wrong!

I hope this helps and you see the point.

y is negative integer or negative fraction from both statement (1) and (2).

\(|y|\) will give only the positive value of y.

\(y ^3\) Cube of any negative integer or negative fraction will be negative and less than the positive value of y.

I is Sufficient. II is Sufficient. Answer (D)...

\(y^3 ≤ |y|\)

(1) \(y < 1\)

Consider \(y = \frac{1}{2}\)

\({\frac{1}{2}}^3 ≤ {|\frac{1}{2}|}\)

\({\frac{1}{8}} ≤ {\frac{1}{2}}\) ==> TRUE

Lets check for ZERO as well

\({0}^3 ≤ |0| = 0 = 0\) ==> TRUE

Now, lets check for Negative values as well, as we know that mod/absolute function will always give us positive values and cube of negative will always give us negative values, our L.H.S. Should always be < R.H.S. Lets test

\({\frac{-1}{2}}^3 ≤ {|\frac{-1}{2}|}\)

\({\frac{-1}{8}} ≤ {\frac{1}{2}}\) ==> TRUE

Hence, Eq. (1) is SUFFICIENT

2) \(y < 0\)

As y is negative, we know for Negative values as well, as we know that mod/absolute function will always give us positive values and cube of negative will always give us negative values, our L.H.S. Should always be < R.H.S. Lets test

\({\frac{-1}{2}}^3 ≤ {|\frac{-1}{2}|}\)

\({\frac{-1}{8}} ≤ {\frac{1}{2}}\) ==> TRUE

Hence, Eq. (2) is SUFFICIENT

As both (1) and (2) are SUFFICIENT

Answer is D

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Is y^3 ≤ |y|?

(1) y < 1
(2) y < 0

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