1st question :

Statement 1: x^3<x^2

If x^3 is lesser than x^2; there are 2 possibilities

x<0 (or) 0<x<1 (Check for values such as -2 (or) 1/2) Hence we cannot say x<0; INSUFFICIENT

Statement 2: x^3<x^4

Again, here there are 2 possibilities;

x>0 (or) x<0 (Check for values 2 and -2)

INSUFFICIENT

Together;
x^3<x^2
x^3<x^4.

Here we can rule out the 0<x<1 condition from Statement 1 (in bold)

Hence x<0 for sure. C IMO

2nd Question: Given ; -2x>3y

(1) y>0.

If y is positive RHS in the equality becomes positive. For the equality to hold good, LHS should be greater then RHS. Since there is a '-' sign in LHS
x should definitely be negative for the equality to hold good. SUFFICIENT

(2) 2x +5y =20

2x = 20-5y

Hence

-20+5y>3y (Sub above in bolded eq)

therefore, we can say y>10 (or) y is positive which is same as Statement 1(SUFFICIENT)

Hence D

tracyyahoo wrote:1.
Is x<0?

(1) x^3<x^2
(2) x^3<x^4

Statement 1:
x³ - x² < 0.
x²(x-1) < 0.
The critical points are x=0 and x=1.
These are the only values of x where x³ - x² = 0.
When x is any other value, x³ - x² < 0 or x³ - x² > 0.
To determine the range of x, test one value to the left and right of each critical point.

x<0:
Plug x=-1 into x³ - x² < 0:
(-1)³ - (-1)² < 0.
-2<0.
This works.
x<0 is part of the range.

0<x<1:
Plug x= 1/2 into x³ - x² < 0:
(1/2)³ - (1/2)² < 0.
-1/8 < 0.
This works.
O<x<1 is part of the range.

x>1:
Plug x=2 into x³ - x² < 0:
(2)³ - (2)² < 0.
6<0.
Doesn't work.
x>1 is not part of the range.

Two ranges satisfy statement 1: x<0 and 0<x<1.
Thus, we cannot determine whether x<0.
Insufficient.

Statement 2:
x^3 - x^4 < 0.
x³(x-1) < 0.
The critical points are x=0 and x=1.
To determine the range of x, test one value to the left and right of each critical point.

x<0:
Plug x=-1 into x³ - x^4 < 0:
(-1)³ - (-1)^4 < 0.
-2<0.
This works.
x<0 is part of the range.

0<x<1:
Plug x= 1/2 into x³ - x^4 < 0:
(1/2)³ - (1/2)^4 < 0.
1/16 < 0.
Doesn't work.
0<x<1 is not part of the range.

x>1:
Plug x=2 into x³ - x^4 < 0:
(2)³ - (2)^4 < 0.
-8<0.
This works.
x>1 is part of the range.

Two ranges satisfy statement 2: x<0 and x>1.
Thus, we cannot determine whether x<0.
Insufficient.

Statements 1 and 2 combined:
Only one range satisfies BOTH statements: x<0.
Sufficient.

The correct answer is C.