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Director  V
Joined: 18 Feb 2019
Posts: 580
Location: India
GMAT 1: 460 Q42 V13 GPA: 3.6
If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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6 00:00

Difficulty:   45% (medium)

Question Stats: 61% (01:28) correct 39% (01:26) wrong based on 92 sessions

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If f(x) = 100 - x^2, is f(a) > f(b)?

(1) a^2 > b^2
(2) a/b > 1

Originally posted by kiran120680 on 06 Mar 2019, 20:56.
Last edited by Bunuel on 06 Mar 2019, 21:34, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Aug 2009
Posts: 7957
If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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2
kiran120680 wrote:
If f(x) = 100 - x^2, is f(a) > f(b)?

(1) a^2 > b^2
(2) a/b > 1

Let us modify the original condition...
$$f(x) = 100 - x^2....... f(a)=100-a^2$$ and $$f(b)=100-b^2$$
Thus f(a) > f(b) means $$100-a^2>100-b^2.......b^2>a^2$$...

(1) $$a^2 > b^2$$
Exactly opposite of what we were looking for. So, answer is No

(2) a/b > 1
This means $$\frac{a}{b} > 1........\frac{a}{b} - 1>0..........\frac{a-b}{b} >0$$
So two cases
(a) b<0.... a-b will also be<0... a-b<0.....a<b... So a<b<0, thus $$a^2>b^2$$, ans is NO
(b) b>0.... a-b will also be>0... a-b>0.....a>b... So a>b>0, thus $$a^2>b^2$$, ans is NO
Thus, sufficient

D
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Re: If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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kiran120680 wrote:
If f(x) = 100 - x^2, is f(a) > f(b)?

(1) a^2 > b^2
(2) a/b > 1

#1
a^2>b^2
a & b being + /-ve wont make any difference to say that f(a) <f(b)

sufficeint
#2
a/b>1
means a>b
if a >b then fa <fb
sufficient
IMO D
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Joined: 04 Jan 2017
Posts: 9
Re: If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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chetan2u wrote:
kiran120680 wrote:
If f(x) = 100 - x^2, is f(a) > f(b)?

(1) a^2 > b^2
(2) a/b > 1

Let us modify the original condition...
$$f(x) = 100 - x^2....... f(a)=100-a^2$$ and $$f(b)=100-b^2$$
Thus f(a) > f(b) means $$100-a^2>100-b^2.......b^2>a^2$$...

(1) $$a^2 > b^2$$
Exactly opposite of what we were looking for. So, answer is No

(2) a/b > 1
This means $$\frac{a}{b} > 1........\frac{a}{b} - 1>0..........\frac{a-b}{b} >0$$
So two cases
(a) b<0.... a-b will also be<0... a-b<0.....a<b... So a<b<0, thus $$a^2>b^2$$, ans is NO
(b) b>0.... a-b will also be>0... a-b>0.....a>b... So a>b>0, thus $$a^2>b^2$$, ans is NO
Thus, sufficient

D

The two cases youre assuming :
first when b<0 , it is not always that a-b will always be <0 . for eg. a could be -2 and b could be -3 . so a-b>0.
Intern  B
Joined: 03 Mar 2015
Posts: 12
Re: If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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Archit3110 wrote:
kiran120680 wrote:
If f(x) = 100 - x^2, is f(a) > f(b)?

(1) a^2 > b^2
(2) a/b > 1

#1
a^2>b^2
a & b being + /-ve wont make any difference to say that f(a) <f(b)

sufficeint
#2
a/b>1
means a>b
if a >b then fa <fb
sufficient
IMO D

[quote="Archit3110"]

does the statement 2 as explained by you holds true as "b" can be either positive or negative so considering a>b is correct ?? as a<b can also be correct .

Kindly shed some light.

and give kudos if i pointed out right.

regards,
Mahi..
GMAT Club Legend  D
Joined: 18 Aug 2017
Posts: 4987
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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1
mahipal wrote:
Archit3110 wrote:
kiran120680 wrote:
If f(x) = 100 - x^2, is f(a) > f(b)?

(1) a^2 > b^2
(2) a/b > 1

#1
a^2>b^2
a & b being + /-ve wont make any difference to say that f(a) <f(b)

sufficeint
#2
a/b>1
means a>b
if a >b then fa <fb
sufficient
IMO D

Archit3110 wrote:

does the statement 2 as explained by you holds true as "b" can be either positive or negative so considering a>b is correct ?? as a<b can also be correct .

Kindly shed some light.

and give kudos if i pointed out right.

regards,
Mahi..

mahipal
#2 given a/b > 1
so both a & b have to be of same sign and a >b then only #2 relation will stand true
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Intern  B
Joined: 03 Mar 2015
Posts: 12
If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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Archit3110 wrote:
mahipal wrote:
Archit3110 wrote:
If f(x) = 100 - x^2, is f(a) > f(b)?

(1) a^2 > b^2
(2) a/b > 1

#1
a^2>b^2
a & b being + /-ve wont make any difference to say that f(a) <f(b)

sufficeint
#2
a/b>1
means a>b
if a >b then fa <fb
sufficient
IMO D

Archit3110 wrote:

does the statement 2 as explained by you holds true as "b" can be either positive or negative so considering a>b is correct ?? as a<b can also be correct .

Kindly shed some light.

and give kudos if i pointed out right.

regards,
Mahi..

mahipal
#2 given a/b > 1
so both a & b have to be of same sign and a >b then only #2 relation will stand true[/quote]

Hi,

I feel you didn't get me.

a/b>1 does not imply a>b

suppose a=-3, b=-2

a/b>1 holds true ,but does a>b holds true ? ==> No.

If a/b>1 ==> we can imply that a^2>b^2 after squaring both sides.

Kindly correct me ,if i am wrong.

Mahi.
GMAT Club Legend  D
Joined: 18 Aug 2017
Posts: 4987
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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(1) a^2 > b^2
(2) a/b > 1[/quote]

#1
a^2>b^2
a & b being + /-ve wont make any difference to say that f(a) <f(b)

sufficeint
#2
a/b>1
means a>b
if a >b then fa <fb
sufficient
IMO D[/quote]

Archit3110 wrote:

does the statement 2 as explained by you holds true as "b" can be either positive or negative so considering a>b is correct ?? as a<b can also be correct .

Kindly shed some light.

and give kudos if i pointed out right.

regards,
Mahi..
[/quote]

mahipal
#2 given a/b > 1
so both a & b have to be of same sign and a >b then only #2 relation will stand true[/quote]

Hi,

I feel you didn't get me.

a/b>1 does not imply a>b

suppose a=-3, b=-2

a/b>1 holds true ,but does a>b holds true ? ==> No.

If a/b>1 ==> we can imply that a^2>b^2 after squaring both sides.

Kindly correct me ,if i am wrong.

Mahi.[/quote]
mahipal
please understand that in DS questions we need to use the given statements/ relations as the valid and true in #2 it says a/b > 1
so a/b>1 will be true in cases when a>b with both signs as same.. we cannot have a & b in opposite sign as it will break #2
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Intern  B
Joined: 03 Mar 2015
Posts: 12
Re: If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1  [#permalink]

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Hi,

I feel you didn't get me.

a/b>1 does not imply a>b

suppose a=-3, b=-2

a/b>1 holds true ,but does a>b holds true ? ==> No.

If a/b>1 ==> we can imply that a^2>b^2 after squaring both sides.

Kindly correct me ,if i am wrong.

Mahi.[/quote]
mahipal
please understand that in DS questions we need to use the given statements/ relations as the valid and true in #2 it says a/b > 1
so a/b>1 will be true in cases when a>b with both signs as same.. we cannot have a & b in opposite sign as it will break #2[/quote]

Hi,

Yes i agree ,completely with you that the given statements are considered to be right and then in light of these statements we have to consider the original question.

But statement #2 says

a/b>1

which you have inferred as a>b.

but here i am not considering opposite signs of a and b infant i am saying they both have same signs but sign can be both negative as well as both positive.

if we consider the case of both positive then what you are saying holds true.i.e if a/b>1 ==> a>b
But

if we consider the case of both negative then a/b>1 does not imply a>b

suppose a=-3 b=-2

then a/b>1 will hold true , but a>b will not hold true.

This is what i am saying.

If i am not clear now also ,do let me know .
As it is important for me to know what i am missing.

Regards,
Mahi. Re: If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1   [#permalink] 28 Apr 2019, 02:52
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