Fraction 1/c is bigger than fraction 1/d as we have to add something to fraction 1/d to make it equal to fraction 1/c.

So, if 1/c > 1/d then c < d

Ans: B

Plug in +ve fraction values:

----0---1/4----1/4 +1
1/d 1/c

1/4 < 5/4 --< 1/d < 1/c
BUT
4> 4/5 ---> d> c
Qty B > Qty A

There is an inherent contradiction in the given information, leading me to get the problem wrong. Let me explain.

Let's simplify the equation given. First we multiply everything by c and get

1 = c + c/d
Next we multiply by d
d = cd + c
d-cd = c
d(1-c) = c
d = c/(1-c)

If c is positive, then the 1-c in the denominator will be negative. Thus, d is negative. Thereby leading to a contradiction.

If c is positive, the denominator isn't necessarily negative.
For example, if c = 1/2, then d = 1

Cheers,
Brent

Thank you! That must have been from stress, can't believe I didn't see that. It's silly that it slipped me, double major in physics and mathematics and still have things like that slip me haha. GRE really challenges you to think very carefully, I like that aspect.

Think of this

c and d are positive numbers

1/c > 1+1/d

this means that C is a positive fraction less than 1, since we can only get a number greater than 1 if 1 is divided by something smaller than 1.
if c is super small, like .01 then

100 > 1+1/d

d cant be less than or equal to .01.. if it is then

100 > 1+1/.01
becomes
100 > 1 + 100... which is not true so we know that if c is equal to .01, then d has to be greater than .01
thus, d is greater than c.

The answer is B

given:1/c=1+1/d
solving equation gives d=cd+c
This implies D is Cd times greater than C

Nice fast solution mate.

Answer: b) B is greater
1/c = 1/d + 1

Multiply by cd on both sides of the equation

d = c + cd

Since c and d are both positive => cd > 0
This means that d > c

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