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TeHCM
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TeHCM
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This one seems to be more difficult.

1. (a,b) = (1,4)
a/(b-1) = 1/3 (< 1/2) & a/b = 1/4 < 1/2 ==> true

(a,b) = (2,0.5)
a/(b-1) = 2/(0.5-1) = -4 (< 1/2) & a/b = 2/(0.5) = 4 > 1/2 ==> False

INSUFF.

2. (a,b) = (1,4)
(a-1)/b = 0 < 1/2 & a/b = 1/4 < 1/2 ==> true

(a,b) = (2,4)
(a-1)/b = 1/4 < 1/2 & a/b = 1/2 ==> false.

INSUFF.

Combining (1) & (2),....

I am stuck! :stupid

(1,4) and (-1,-5) work for both equations and also a/b < 1/2

I am not able to come up with numbers to prove the eqn false. On my exam I would choose C!
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ywilfred
a/b < 1/2 -> 2a < b

1) a/(b-1) < 1/2 -> 2a < b-1 -> 2a-1 < b. Sufficient. 1 less than 2a will be even smaller than 2a.

2) 2a-2 < b. Sufficient. 2 less than 2a wil be even smaller than 2a.

Ans D



u say (1 < 2a) << 2a ? how? i dont get it , how do u correlate this to the given equation
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2a<b?

(1) a/(b-1) < 1/2

2a < b-1
Insufficient. 2a < b-1 doesn't nescessarily mean it's smaller than b. It could be a value between b-1 and b

(2) 2a-2 < b

a b
+ + --> 2a<b
+ - --> Not possible as it violates 2a-2 < b
- + --> 2a < b
- - --> 2a<b

Sufficient. For all possibilites where 2a-2<b, 2a<b.

Ans B
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Is it E?

Is a/b<1/2?

S1:
a/(b-1)<1/2

it means
=> b <> 1
case1: a = 0 => ans = yes.
case 2: a > 0, b = 0 => ans = undefined
case 3: a = -2, b = -4 => ans = no
INSUFFICIENT

S2: (a-1)/b<1/2
it means:
=> b <> 0
case 1: a = 1, = 3 => ans = yes
case 2: a = 1, b = 1 => ans = no
INSUFFICIENT

Together:
we know:
b <> 0, b <> 1

case 1: a = 0, b > 1 => ans = yes
case 2: a = 1, b = 1/2 => ans = no
(only case where ans = no is when 0 < b < 1)
INSUFFICIENT

PS: <> : not equal to
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business
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I think its A
2a<b???

case 1 : 2a<b-1

if this is true, (b-1) is more than 2a.
since b is always > b-1 we can say
b>(b-1)>2a sufficient - So A

case 2 :
2(a-1)<b
2a<b+2
if this is true, (b+2) is more than 2a
since (b+2) is always > b, we can say

(b+2)>2a>b or (b+2)>b>2a
so not sufficient!
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business
I think its A
2a<b???

case 1 : 2a<b-1

if this is true, (b-1) is more than 2a.
since b is always > b-1 we can say
b>(b-1)>2a sufficient - So A

case 2 :
2(a-1)<b
2a<b+2
if this is true, (b+2) is more than 2a
since (b+2) is always > b, we can say

(b+2)>2a>b or (b+2)>b>2a
so not sufficient!

How did you get bold part?
As we do not know sign of a,b (+ve, -ve or 0) we can not possibly simplify this way. For example:
a/b < 1/2
multiple both side by 2 gives
2a/b < 1
now, we cannot simplify this to:
2a < b as we do not know property of b.
if a > 0 and b < 0 we will have 2a > b
if a > 0 and b > 0 we will have 2a < b
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TeHCM
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OA is E, thanks Duttsit!!

I really was trying to avoid picking numbers (takes too long) and use some kind of logic....but my mind is too muddled :crying
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TeHCM
OA is E, thanks Duttsit!!

I really was trying to avoid picking numbers (takes too long) and use some kind of logic....but my mind is too muddled :crying


Hey TeCHM, I still think analytical way of solving problems is always good. This way you can almost be sure about your answer. It also gives you hint about numbers to pick.
However, for this problem, the analytical method I tried ran too lengthy to explain here. I would certainly think this is hard bin question.

Cheer up!! :beer
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duttsit
TeHCM
OA is E, thanks Duttsit!!

I really was trying to avoid picking numbers (takes too long) and use some kind of logic....but my mind is too muddled :crying

Hey TeCHM, I still think analytical way of solving problems is always good. This way you can almost be sure about your answer. It also gives you hint about numbers to pick.
However, for this problem, the analytical method I tried ran too lengthy to explain here. I would certainly think this is hard bin question.

Cheer up!! :beer


Thanks for the solution and nice advice duttsit!



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