1. sqrt x - x > y - x
y -x < sqrt x - x


y - x < 0 when 0 < x < 1
y -x < some positive number when x > 1

Insuffcient.

2. x^3 > y
x^3 - x > y - x
y - x < x^3 - x

y - x < 0 when x is in (-inf, -1) U (0, 1)
y - x < some positive when x is in (-1, 0) U (1, +inf)

Insufficient

Combined. together.
y -x < 0 when x is in (0,1)
y - x < some positive, when (1, +inf)

Insufficient.

E is the answer

I marked E as well but that is wrong. Both are sufficient. Anyone else?

okigbo wrote:Is x > y?

(1) sqrt x > y

(2) x^3 > y



can someone pls lay out in detail how to combine both statements and test? i always stumble at this stage of DS questions. many thanks

The answer is C.

When we have exponents, we should always think about positive fractions, which behave weirdly.

(1) if we let x = 1/4 and y = 1/3, we validate the statement (since 1/2 > 1/3). Is 1/4 > 1/3? NO

we can also pick x = 100 and y = 1 (since 10 > 1). Is 100 > 1? YES

(2) we can pick x = 3 and y = 4 (since 27 > 4). Is 3 > 4? NO

we can also pick x = 100 and y = 1 (since 1000000 > 1). Is 100 > 1? YES

When we look at the statements together, we know that:

sqrtx > y and x^3 > y

Together, it's still easy to generate a "yes" answer (x=100 and y=1 worked for both). The question is can we still get a "no".

We could only generate a "no" from statement (1) when when we used positive fractions, so let's see if we can pick positive fractions in accord with (2) as well.

Well, if x is a postive fraction, x > x^3. We know that x^3 is greater than y, and if we put those two inequalities together we get:

x > x^3 > y

which clearly shows that, in this case, x > y.

Once we eliminate positive fractions, to satisfy (1) we can only pick values of x that ARE greater than y, so we're guaranteed a "yes" answer: choose C.

Try plugging in numbers:

1. If x=4, y=1, sqrt(4)>1, x > y
If x=1/4, y=1/3, sqrt*(1/4) > 1/3, x < y
Insufficient

2. If x = 2, y = 1, 2^3 > 1, x > y
If x = 2, y = 3, 2^3 > 3, x < y
Insufficient

Combine 1 and 2
If x = 4, y = 1, 4^3 AND sqrt(4) > 1, x > y
No set of numbers that satisfies stem1 and 2 to disprove x > y, I'd pick C. However, this took awhile...

Can someone please suggest a faster method?

1. sqrt(x) > sqrt(y) ; X,Y +ve result vary if it a fraction

2. x3 > y => result vary depending on +/-ve numbers ( integers/fraction)

Combined to gether X, Y must be +ve Integer . => we can determine whether X > Y.