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
mgmat inequality
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Stuart@KaplanGMAT
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The answer is C.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
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.
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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. 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?
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GmatVerbal
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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.
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.
