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c and d are positive

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Carcass
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c and d are positive [#permalink] New post   25 Apr 2017, 02:58
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c and d are positive

\(\frac{1}{c} = 1 + \frac{1}{d}\)

Quantity A
Quantity B
c
d


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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B
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GreenlightTestPrep
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c and d are positive [#permalink] New post   26 May 2017, 12:00
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Carcass wrote:


c and d are positive

\(\frac{1}{c} = 1 + \frac{1}{d}\)

Quantity A
Quantity B
c
d


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Given: \(\frac{1}{c} = 1 + \frac{1}{d}\)

Multiply both sides of the equation by \(cd\) to get: \(d = cd + c\)

In Quantity B, replace \(d\) with \(cd + c\) to get:
Quantity A: \(c\)
Quantity B: \(cd + c\)

Subtract \(c\) from both sides of the equation to get:
Quantity A: \(0\)
Quantity B: \(cd\)

Since we're told \(c\) and \(d\) are both positive, we know that the product \(cd\) must be positive.

Answer: B

Cheers,
Brent
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Fatemeh
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Re: c and d are positive [#permalink] New post   01 Apr 2018, 12:51
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Answer: B
1/c equals 1 + 1/d so 1/c is something bigger than 1/d,
1/c > 1/ d so d > c

EVEN for negative values this correlation is true.
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Re: c and d are positive [#permalink] New post   04 Apr 2018, 07:46
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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
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Re: c and d are positive [#permalink] New post   16 Sep 2018, 23:10
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
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Re: c and d are positive [#permalink] New post   14 Dec 2018, 19:45
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



c and d are positive

\(\frac{1}{c} = 1 + \frac{1}{d}\)

Quantity A
Quantity B
c
d


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


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.
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Re: c and d are positive [#permalink] New post   15 Dec 2018, 06:42
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QuantumWonder wrote:
Carcass wrote:


c and d are positive

\(\frac{1}{c} = 1 + \frac{1}{d}\)

Quantity A
Quantity B
c
d


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


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
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QuantumWonder
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Re: c and d are positive [#permalink] New post   15 Dec 2018, 17:10
GreenlightTestPrep wrote:
QuantumWonder wrote:
Carcass wrote:


c and d are positive

\(\frac{1}{c} = 1 + \frac{1}{d}\)

Quantity A
Quantity B
c
d


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


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.
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HappyMathtutor
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Re: c and d are positive [#permalink] New post   15 Dec 2018, 21:44
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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
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Re: c and d are positive [#permalink] New post   17 Dec 2018, 03:39
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given:1/c=1+1/d
solving equation gives d=cd+c
This implies D is Cd times greater than C
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QuantumWonder
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Re: c and d are positive [#permalink] New post   17 Dec 2018, 07:17
vigjna wrote:
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.
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Siddhaarth
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Re: c and d are positive [#permalink] New post   19 Oct 2020, 17:55
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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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pihu25
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Re: c and d are positive [#permalink] New post   16 Jun 2026, 23:40
1
we can rewrite it as 1/c - 1/d = 1
d - c / cd = 1
Since, the RHS is +ve and both variables are +ve
therefore, d > c
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Re: c and d are positive [#permalink]
16 Jun 2026, 23:40
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