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Is f < g?

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Is f < g?  [#permalink]

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New post 14 Aug 2016, 07:11
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

47% (01:59) correct 53% (01:56) wrong based on 161 sessions

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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 09:58
Bunuel wrote:
Is f < g?

(1) f < g + 1
(2) |f|/|g| < 1


Answer should be C.

Statement 1 : f < g+1.

Say f = -3 and g = -3, we will have -3 < -3+1 or -3<-2 BUT -3 is not less than -3. Hence, Insufficient.

For +ve , it will be satisfied.

Statement 2 : |f| < |g|

or say f = -2 and g = -3, we have |f| < |g| BUT f is not less than g.

For +ve , it will be satisfied.

Combining, now we need to take only +ve numbers for both Statement 1 and 2 be satisfied.

Hence, we will always get f<g. Sufficient.
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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 10:03
(1) the difference between f and g could be less than 1, thus we cannot determinate if f < g.
(2) abs of f is smaller than abs of g. f could be positive 2 and g could be -3 or 3. Ns.

(1+2)
not sufficient.
F = 10 g= 12 => fF=,3 g=-,6 f>g

Any other approaches for 1+2?

Ill Go with E
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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 10:16
Statement 1 : f<g+1
For this equation lets take following values of f & g and test if f < g
f = 2 g = 2 f is not less than g
f = 2 g = 3 f i s less than g

Insufficient



Statement 2:
|f|/|g| <1
this means |g|>|f|
In sufficient

1+2 statements;

f< g
sufficient
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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 10:22
MarkusKarl wrote:
(1) the difference between f and g could be less than 1, thus we cannot determinate if f < g.
(2) abs of f is smaller than abs of g. f could be positive 2 and g could be -3 or 3. Ns.

(1+2)
not sufficient.
F = 10 g= 12 => fF=,3 g=-,6 f>g

Any other approaches for 1+2?

Ill Go with E
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With your example,

F=,3 g=-,6 --> Statement 1 is not getting satisfied. Hence, this example is invalid.
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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 10:34
abhimahna wrote:
MarkusKarl wrote:
(1) the difference between f and g could be less than 1, thus we cannot determinate if f < g.
(2) abs of f is smaller than abs of g. f could be positive 2 and g could be -3 or 3. Ns.

(1+2)
not sufficient.
F = 10 g= 12 => fF=,3 g=-,6 f>g

Any other approaches for 1+2?

Ill Go with E
Posted from my mobile device

Posted from my mobile device


With your example,

F=,3 g=-,6 --> Statement 1 is not getting satisfied. Hence, this example is invalid.


-,6+1=,4>,3

What am I missing? :/
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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 10:53
MarkusKarl wrote:
abhimahna wrote:
MarkusKarl wrote:
(1) the difference between f and g could be less than 1, thus we cannot determinate if f < g.
(2) abs of f is smaller than abs of g. f could be positive 2 and g could be -3 or 3. Ns.

(1+2)
not sufficient.
F = 10 g= 12 => fF=,3 g=-,6 f>g

Any other approaches for 1+2?

Ill Go with E
Posted from my mobile device

Posted from my mobile device


With your example,

F=,3 g=-,6 --> Statement 1 is not getting satisfied. Hence, this example is invalid.


-,6+1=,4>,3

What am I missing? :/


You have taken f = 3 and g =-6.

Statement 1 says f < g+1.

as per your example, we are getting 3 < -6+1

or 3<-5 which is NOT true. hence, its an invalid example.
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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 10:57
I could have been clearer.
F=0.3
G=-0.6

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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 11:03
MarkusKarl wrote:
I could have been clearer.
F=0.3
G=-0.6

Posted from my mobile device


Yes, I believe I missed this condition. yes, you are right. Answer should be E.

Thanks Mate. :)
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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 11:16
abhimahna wrote:
MarkusKarl wrote:
I could have been clearer.
F=0.3
G=-0.6

Posted from my mobile device


Yes, I believe I missed this condition. yes, you are right. Answer should be E.

Thanks Mate. :)


My bad for not writing the numbers properly :)

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Re: Is f < g?  [#permalink]

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New post 14 Aug 2016, 12:39
Bunuel wrote:
Is f < g?

(1) f < g + 1
(2) |f|/|g| < 1


Q: is f-g < 0

1) f < g+1 => f-g<1, not sufficient.

2) |f| / |g| < 1, that means |g| > |f|, not sufficient as g can be -4 and f can -3 => f>g

1 + 2,

f < g +1
|f| < |g|

that means g has to be +2 > f so that above two statments work

answer is C
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Re: Is f < g?  [#permalink]

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New post 09 Feb 2017, 07:05
Bunuel, Sir, Please guide. According to me the answer is C...I donot understand how is OA E?
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Re: Is f < g?  [#permalink]

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New post 09 Feb 2017, 11:04
startjumprun wrote:
Is f < g?

(1) f < g + 1
(2) |f|/|g| < 1

Bunuel, Sir, Please guide. According to me the answer is C...I donot understand how is OA E?


Why do you think the answer is C? How did you get it? If you check the discussion above you'll find the examples proving that the answer is indeed E.

If f = 0.3 and g = -0.6, then the answer is NO.
If f = 1 and g = 2, then the answer is YES.
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Re: Is f < g?  [#permalink]

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New post 10 Feb 2017, 07:19
Thanks a lot Bunuel Sir for your time and explanation :) I missed to take into account decimals.. Will be careful next time.
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Re: Is f < g?  [#permalink]

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New post 13 Feb 2017, 17:13
Bunuel wrote:
Is f < g?

(1) f < g + 1
(2) |f|/|g| < 1


We need to determine whether f < g.

Statement One Alone:

f < g + 1

Statement one alone is not sufficient to answer the question. For instance, if f = 2 and g = 2, then f IS NOT less than g. However, if f = 1 and g = 2, then f IS less than g.

Statement Two Alone:

|f|/|g| < 1

We can multiply both sides of the inequality by |g| to get:

|f| < |g|

We can also determine that statement two alone is not sufficient to answer the question.

For instance, if f = -1 and g = -2, then f IS NOT less than g. However, if f = 1 and g = 2, then f IS less than g.

Statements One and Two Together:

Using the information from statements one and two, we know that f < g + 1 and that |f| < |g|. This is still not enough information to answer the question.

For instance, if f = -1 and g = -1.75, then f IS NOT less than g. However, if f = 1 and g = 2, then f IS less than g.

Answer: E
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Re: Is f < g?  [#permalink]

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