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can I solve like this?
n^3>n^2
n>0
so def n^2<n^3
so A/D

n^3<1
so n^3-1<0
cannot say
so A

however, answer is D
so Bunuel is it n^3 < 1 OR
n^3 > 1
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paidlukkha
can I solve like this?
n^3>n^2
n>0
so def n^2<n^3
so A/D

n^3<1
so n^3-1<0
cannot say
so A

however, answer is D
so Bunuel is it n^3 < 1 OR
n^3 > 1

Is n^2 < n^3 ?

Notice that the question basically asks whether n > 1.

(1) n^2 < 1 --> -1 < n < 1. The answer to the question is NO. Sufficient.

(2) n^3 < 1 --> n < 1. The answer to the question is NO. Sufficient.

Answer: D.
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Bunuel
paidlukkha
can I solve like this?
n^3>n^2
n>0
so def n^2<n^3
so A/D

n^3<1
so n^3-1<0
cannot say
so A

however, answer is D
so Bunuel is it n^3 < 1 OR
n^3 > 1

Is n^2 < n^3 ?

Notice that the question basically asks whether n > 1.

(1) n^2 < 1 --> -1 < n < 1. The answer to the question is NO. Sufficient.

(2) n^3 < 1 --> n < 1. The answer to the question is NO. Sufficient.

Answer: D.

Why have we not considered 0 as a possible value for n?? If n=0, then we do not have a clear answer..!!!

Please suggest.
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MeghaP
Bunuel
paidlukkha
can I solve like this?
n^3>n^2
n>0
so def n^2<n^3
so A/D

n^3<1
so n^3-1<0
cannot say
so A

however, answer is D
so Bunuel is it n^3 < 1 OR
n^3 > 1

Is n^2 < n^3 ?

Notice that the question basically asks whether n > 1.

(1) n^2 < 1 --> -1 < n < 1. The answer to the question is NO. Sufficient.

(2) n^3 < 1 --> n < 1. The answer to the question is NO. Sufficient.

Answer: D.

Why have we not considered 0 as a possible value for n?? If n=0, then we do not have a clear answer..!!!

Please suggest.

Even if n = 0 we still have the same NO answer to the question whether n^2 < N^3.
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Bunuel
paidlukkha
can I solve like this?
n^3>n^2
n>0
so def n^2<n^3
so A/D

n^3<1
so n^3-1<0
cannot say
so A

however, answer is D
so Bunuel is it n^3 < 1 OR
n^3 > 1

Is n^2 < n^3 ?

Notice that the question basically asks whether n > 1.

(1) n^2 < 1 --> -1 < n < 1. The answer to the question is NO. Sufficient.

(2) n^3 < 1 --> n < 1. The answer to the question is NO. Sufficient.

Answer: D.

Bunuel can you elaborate as to how you arrived at n>1?
Also are we assuming n to be an integer here?
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gmat1220
Is n^2 < n^3 ?

(1) n^2 < 1

(2) n^3 < 1

\(n^2<n^3\)

As \(n^2\) is non negative, we can divide the two sides by \(n^2\).

\(\frac{n^3}{n^2}>1………n>1\)

Thus n>1 will give a yes, while \(n\leq 1\)will give a no. \(n\leq 1\) will include n=0 as well.

(1) \(n^2 < 1\)
This means \(|n|<1………..-1<n<1\)
So the answer for ‘Is n>1?’ is NO.
Sufficient

(2) \(n^3 < 1\)
Take cube root on both sides
n<1. Again answer is NO.
Sufficient


D
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