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Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the

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Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 00:53
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 01:17
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Is y greater than y^2?

(1) y is greater than -1.

y= 2 yes
y=0.5 no

(2) y^2 is greater than 1.

y<-1 y>1

option B is sufficient

since only the values between -1<y<1; y is greater than y^2
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 01:38
(1) y is greater than -1.
(2) y^2 is greater than 1.

Clearly both individually cannot help in knowing whether y > y^2


Now by combining both
1) y>-1.....-0.5,0,0.5,1,2....
2) y^2>1..... -1>y>1....-3,-2.5,-2,-1.5,1.5,2...

Both y>y^2 and y<y^2 are possible


OA:E

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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 02:18
Is y greater than y^2?
q: y>y^2?
y-y^2>0
Solving for y :
is 0<y<1?


(1) y is greater than -1.
Yes. if y is between 0 and 1. No is outside this range. Insufficient


(2) y^2 is greater than 1.
y^2>1. Therefore, range is y>1, y<-1. Again Insufficient.

Combining, y>1. Ans C
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Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post Updated on: 06 Nov 2019, 02:11
to check
y>y^2
#1
y is greater than -1.
y>-1
y can be -1/2 , 0 we get no
or y = +1/2 , 2we get yes
insufficeint
#2
y^2>1
y can be + or -ve integer
sufficient
as in that case y>y^2 ; not possible
IMO B


(1) y is greater than -1.
(2) y^2 is greater than 1.

Originally posted by Archit3110 on 05 Nov 2019, 03:40.
Last edited by Archit3110 on 06 Nov 2019, 02:11, edited 1 time in total.
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 05:36
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Is y greater than \(y^2\)?
\(y > y^2\)
\(y - y^2 > 0\)
\(y(1 - y) > 0\)
\(0 < y < 1\) ?

(1) y is greater than -1.
\(y = \frac{-1}{2}\) NO
\(y = \frac{1}{2}\)

INSUFFICIENT.

(2) y^2 is greater than 1.
\(y^2 > 1\)
\(y^2 - 1 > 0\)
\((y - 1)(y + 1) > 0\)
y > 1, y > -1 SO y > 1
y = 2 NO
y = 3 NO

SUFFICIENT.

Answer B.
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 06:31
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Quote:
Is y greater than y^2?

(1) y is greater than -1.
(2) y^2 is greater than 1.


y>y^2 when 0<y<1 (proper fraction)

(1) y is greater than -1. insufic.

(2) y^2 is greater than 1. sufic.
y^2>1 then y>1 or y<-1; so y is not a proper fraction.

Answer (B)
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 06:32
1
Quote:
Is y greater than y^2?

(1) y is greater than -1.
(2) y^2 is greater than 1.


y>y^2 when 0<y<1 (proper fraction)

(1) y is greater than -1. insufic.

(2) y^2 is greater than 1. sufic.
y^2>1 then y>1 or y<-1; therefore y is not a proper fraction.

Answer (B)
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 07:41
1
Is \(y>y^{2}\)

(1) y is greater than -1.
y>-1 i.e y can be \(\frac{-1}{2},0,\frac{1}{2}\),1 etc
When Y=\(-\frac{1}{2}\) , is \(y>y^{2}\) ? i,e is \(\frac{-1}{2}>\frac{1}{4}\)? NO
When Y is 0, is \(y>y^{2}\) ? ie is 0<0 ? NO
When y is \(\frac{1}{2}\) , is \(\frac{1}{2}>\frac{1}{4}\), YES

Therefore Not Sufficient

(2) y^{2}>1
y can be -infinity<y<-1 or 1<y<infinity
If y is \(-\frac{3}{2}\), is \(-\frac{3}{2}\)>\(\frac{9}{4}?\) , NO
If Y is 2, Is 2>4, NO
If Y is \(\frac{3}{2}\) , Is \(\frac{3}{2}>\frac{9}{4}\)?NO

Therfore B is sufficient.

Answer is B
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 08:08
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Is y greater than y^2?

Is y> y^2?

(1) y is greater than -1. If y = -1/2, then y> y^2 will not be true and if y = 1/2, then y> y^2 will be true. Hence insufficient.

(2) y^2 is greater than 1. - It means y can be -ve or +ve. e.g. y = -2, then y> y^2 will not be true and y = 2, then y> y^2 will not be true. If we take lowest value of y^2 = 1.02 having y as + or - 1.01, then the answer will be confirm no.
Hence sufficient.

Imo. B
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 09:07
1
We are to determine if y>y^2.

1. y>-1
Not sufficient. This is because if y=1/2, then y^2=14 and y(1/2)>y^2 (1/4). But if y=2, then y^2=4 and y<y^2.

2. y^2>1
Sufficient.
y^2>1
y^2-1>0
(y-1)(y+1)>0
This is only true when y<-1 or y>1
with y<-1, y^2 (a positive number) is always greater than y (a negative number).
with y>1, y^2 is always greater than y.
Statement 2 is therefore sufficient.

The answer is therefore B.
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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 09:50
1
(1) y is greater than -1.
If 0 < y < 1, y is greater than y^2
If y > 1, y is less than y^2
—> Two different outcomes —> Insufficient

(2) y^2 is greater than 1.
—> y is either less than -1 or y is greater than 1
Case 1: y < -1, y is always less than y^2, since y^2 is always positive

Case 2: y > 1, y is always less than y^2,

—> y is never greater than y^2. A definite NO
—> Sufficient

IMO Option B

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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the  [#permalink]

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New post 05 Nov 2019, 13:28
1
Is y greater than y^2?

—>\(y > y^{2}\)
y*(y —1)< 0
—> 0 < y < 1?


(Statement1): y > —1
If y= 0.5, then YES
If y= —0.5, then NO
Insufficient

(Statement2): \(y^{2}> 1\)
(y—1)*(y+1) >0
—> y <—1 and y >1
(Always NO)
Sufficient

The answer is B

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Re: Is y greater than y^2? (1) y is greater than -1 (2) y^2 is greater the   [#permalink] 05 Nov 2019, 13:28
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