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If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1

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If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1  [#permalink]

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New post 13 Apr 2018, 01:10
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

73% (01:11) correct 27% (01:34) wrong based on 106 sessions

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If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1  [#permalink]

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New post Updated on: 13 Apr 2018, 02:19
IMO B

St1 y>1 or y<-1
insufficient

St2 y^2+y>y^2 (sq quantity is +ve)
y>0
sufficient

Originally posted by NiharikaR on 13 Apr 2018, 01:13.
Last edited by NiharikaR on 13 Apr 2018, 02:19, edited 1 time in total.
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Re: If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1  [#permalink]

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New post 13 Apr 2018, 02:15
Bunuel wrote:
If y ≠ 0, is y > 0?


(1) \(|y| > 1\)

(2) \(\frac{y^2 + y}{y^2} > 1\)



NiharikaR, ans should be B..

(1) \(|y| > 1\)
tells us that y<-1 or y>1
insuff

(2) \(\frac{y^2 + y}{y^2} > 1\)..
Cross multiply as y^2 is POSITIVE
\(y^2 + y > y^2.............y>y^2-y^2.......y>0\)..
suff

B
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If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1  [#permalink]

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New post 13 Apr 2018, 02:27
Thanks. I read the question wrong. I thought its asking y>1.
Though I got y>0, I wrote insufficient.

I have to avoid such silly mistakes during exam
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Re: If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1  [#permalink]

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New post 13 Apr 2018, 05:12
I ) |y| > 1

y>1
y <-1

not sufficient

II)
\(y^2+y/y^2\) > 1

y^2/y^2 + y/y^2 > 1

1 + 1/y >1 ==> 1/y >0 ==> y>0

sufficient

Therefore B
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Re: If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1  [#permalink]

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New post 28 Apr 2018, 05:29
I) y could be 2 or -2
Insufficient

II) rewriting eqn
1+(1/y)>1

1/y>0
Y>0

So b is ANS

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Re: If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1  [#permalink]

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New post 30 Apr 2018, 11:39
Bunuel wrote:
If y ≠ 0, is y > 0?


(1) \(|y| > 1\)

(2) \(\frac{y^2 + y}{y^2} > 1\)



Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (y) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
If \(y = 1\), then \(y > 0\). The answer is "yes".
If \(y = -1\), then \(y < 0\). The answer is "no".
Since we don't have a unique solution, the condition 1) is not sufficient.

Condition 2)

\(\frac{y^2 + y}{y^2} > 1\)
\(⇔ y^2 + y > y^2\)
\(⇔ y > 1\)
Then we have \(y > 0\).

Condition 2) is sufficient.

Therefore, B is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: If y ≠ 0, is y > 0? (1) |y| > 1 (2) y^2 + y/y^2 > 1 &nbs [#permalink] 30 Apr 2018, 11:39
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