Must be true-Inequalities question: Experts pls help

[mathewmithun]

if 0<1-f/g<1 and g>0, which of the followiing must be true?
1. f^2+g^2>1
2. f>0
3. f/g<1

A. 1 only
B. 3 only
C. 1 and 3 only
D. 2 and 3 only
E. 1,2 and 3 only

OA is D but I choose E as the answer.

STEP-1: 0<1-f/g<1 can be re-written as -1<-f/g<0 by subtracting 1.
So 0<f/g<1.

Since g>0, and f/g is +ve therefore f>0 need is true.
f/g<1 is the condition we got from step-1.

f^2+g^2>: Plugging in data let f=1 and g be 100, then we get the condition true
plugging f=1/2 and g=1 so f/g=1/2 which is still less than 1 and so f^2+g^2>1

So E is the answer I choose. Am i Correct? Also is step-1 mentioned above correct?

[AbhiJ]

What if f = 1/4 and g = 1/2 ?.
I think this should answer your question.

[mathewmithun]

What if f = 1/4 and g = 1/2 ?.
I think this should answer your question.

Thanks, got it

[Brent@GMATPrepNow]

if 0<1-f/g<1 and g>0, which of the followiing must be true?
1. f^2+g^2>1
2. f>0
3. f/g<1

A. 1 only
B. 3 only
C. 1 and 3 only
D. 2 and 3 only
E. 1,2 and 3 only

Let's deal with this in parts:

Part 1: 0 < 1-(f/g)
Add f/g to both sides to get: f/g < 1
Since we're told that g is positive, we'll multiply both sides by g to get: f < g

Part 2: 1-(f/d) < 1
Add f/g to both sides to get: 1 < 1 + f/g
Subtract 1 from both sides to get: 0 < f/g
If g is positive and f/g is positive, then f must be positive.
In other words, 0 < f

Combining the two parts, we get 0 < f < g

What kinds of conclusions can we make from this information?

1. f^2+g^2 >1 (no, this need not be true. For example, we could have f=0.1 and g=0.2, in which case f^2 + g^2 is not greater than 1.)
2. f>0 (yes, this must be true)
3. f/g < 1 (since f < g, and since f and g are both positive, this must be true)

Answer = D

Cheers,
Brent