Data Sufficiency

Given: f not equal to 1, is f^2/(f-1) > f ?

1) f is not an integer
2) f > 0

Please list your work and conclusion, I came to two answers.

cyrwr1 wrote:Given: f not equal to 1, is f^2/(f-1) > f ?

1) f is not an integer
2) f > 0

Please list your work and conclusion, I came to two answers.

i simplified the equation given in question and came to the question asking me is f>0?

2 solves that.. Hence
B it is!
OS pls!

Can you elaborate on what you mentioned?

I don't think the equation simplifies to "Is f>0?"


1) f is not an integer:
If we try f = 2.5 then f^2/(f-1) = 2.5^2/1.5 = 4.2, so it is >f
If we try f = 0.5 then f^2/(f-1) = 0.5^2/-0.5 = -0.5, so it is <f
INSUFFICIENT

2) f > 0
If we try f = 2 then f^2/(f-1) = 2^2/1 = 4, so it is >f
If we try f = 0.5 then f^2/(f-1) = 0.5^2/-0.5 = -0.5, so it is <f
INSUFFICIENT

Combining both statements doesn't give us anything new (see the equations in 1)

So IMO E

I got C

f^2/(f-1)>f

f^2/(f-1)-f>0

f^2/(f-1)-f(f-1)/(f-1)>0 (combining the left side over a common denominator)

(f^2-f^2+f)/(f-1)>0

f/(f-1)>0

This will be true for f<0 or f>1

1. f is not an integer, so it could be f=0.5 which is not in the solution set or f=1.5 which is in the solution set. INSUFFICIENT.

2. f>0. Same examples above provide two answers. INSUFFICIENT

1&2. Same examples above provide two answers. INSUFFICIENT

Ans: E

Hey Gokhra,
I had gotten E too but checked it a different way with rewriting it as:
is f^2> f^2-f ?

When you put in a positive >1, you get the result yes
when you put in a positive <1, you get the result yes too

say you use a 1/2, you would get is 1/4 > (1/4-1/2) ? and then have 1/4 > -1/4, that too is true.

so I am so confused about the solution. Unless my logic is incorrect, I cannot decide between choice B and E

You can't cross multiply as the denominator could be -ve.
GmatMathPro's solution is perfect

cyrwr1 wrote:Hey Gokhra,
I had gotten E too but checked it a different way with rewriting it as:
is f^2> f^2-f ?

When you put in a positive >1, you get the result yes
when you put in a positive <1, you get the result yes too

say you use a 1/2, you would get is 1/4 > (1/4-1/2) ? and then have 1/4 > -1/4, that too is true.

so I am so confused about the solution. Unless my logic is incorrect, I cannot decide between choice B and E

cyrwr1 wrote:Hey Gokhra,
I had gotten E too but checked it a different way with rewriting it as:
is f^2> f^2-f ?
Unless my logic is incorrect, I cannot decide between choice B and E

Your logic is incorrect. This is not a valid way to rewrite the inequality. I assume you got this by multiplying both sides of the inequality by (f-1)? This is only valid for positive values of f-1. Remember that when you multiply both sides of an inequality by a negative number you have to flip the direction of the inequality sign. f-1 could represent a negative value also, so when you rewrite it this way, you are essentially assuming f-1 is positive, and then you lose information about what happens when f-1 is negative. You should generally avoid multiplying or dividing both sides of an inequality by a variable for this reason. Notice that in my solution, I avoided doing any multiplication or division across the inequality sign when I rewrote it. If you do that, you can be sure you are preserving all of the original solutions (and non-solutions).

You CAN do it by multiplying both sides by f-1, but you have to be careful. To do it, you would have to break it down into cases. First, assume that f-1 is positive, or f-1>0 or f>1, and multiply to get f^2>f^2-f, which simplifies to f>0, but we are assuming that f>1, so we cannot include the numbers from 0 to 1 in the solution set. For this case, the solution is f>1.

Now, let's assume f-1<0 or f<1. Now when we multiply by f-1 we have to reverse the inequality sign because we are multiplying both sides by a negative: f^2<f^2-f which simplifies to f<0. We are also assuming that f<1, so all f that satisfy f<0 AND f<1 will be in the solution set, which is just f<0.

Combining, we get f<0 or f>1 for our solution set.

Hey GmatMathPro,

Thanks for catching my little error there.
I initially just tried #'s to test it out and in the end, though at an ambivalence, I chose E as the correct answer.



Thanks to everyone on this thread!

cyrwr1 wrote:Hey Gokhra,
I had gotten E too but checked it a different way with rewriting it as:
is f^2> f^2-f ?

When you put in a positive >1, you get the result yes
when you put in a positive <1, you get the result yes too

say you use a 1/2, you would get is 1/4 > (1/4-1/2) ? and then have 1/4 > -1/4, that too is true.

so I am so confused about the solution. Unless my logic is incorrect, I cannot decide between choice B and E

As you can see from the above posts, there is a compelling reason for not multiplying both sides with a variable whose sign you don't know. This is important to remember with GMAT inequality problems. Adding and subtracting is ok though.

In a problem like this, where the equation is pretty simple to start with, I find that plugging in a couple of numbers intuitively works best and fastest for me. Or you could simplify the equation as GMATMathPro has done above. That is probably the best and safest way to do it.

GmatMathPro wrote:

cyrwr1 wrote:Hey Gokhra,
I had gotten E too but checked it a different way with rewriting it as:
is f^2> f^2-f ?
Unless my logic is incorrect, I cannot decide between choice B and E

Your logic is incorrect. This is not a valid way to rewrite the inequality. I assume you got this by multiplying both sides of the inequality by (f-1)? This is only valid for positive values of f-1. Remember that when you multiply both sides of an inequality by a negative number you have to flip the direction of the inequality sign. f-1 could represent a negative value also, so when you rewrite it this way, you are essentially assuming f-1 is positive, and then you lose information about what happens when f-1 is negative. You should generally avoid multiplying or dividing both sides of an inequality by a variable for this reason. Notice that in my solution, I avoided doing any multiplication or division across the inequality sign when I rewrote it. If you do that, you can be sure you are preserving all of the original solutions (and non-solutions).

You CAN do it by multiplying both sides by f-1, but you have to be careful. To do it, you would have to break it down into cases. First, assume that f-1 is positive, or f-1>0 or f>1, and multiply to get f^2>f^2-f, which simplifies to f>0, but we are assuming that f>1, so we cannot include the numbers from 0 to 1 in the solution set. For this case, the solution is f>1.

Now, let's assume f-1<0 or f<1. Now when we multiply by f-1 we have to reverse the inequality sign because we are multiplying both sides by a negative: f^2<f^2-f which simplifies to f<0. We are also assuming that f<1, so all f that satisfy f<0 AND f<1 will be in the solution set, which is just f<0.

Combining, we get f<0 or f>1 for our solution set.

As always, a cool explanation. Thanks.

Can someone explain this by substituting values!!

TIA