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If \(d >0\) and \(0 < 1 - \frac{c}{d} < 1\), which of the following must be true?

I. \(c > 0\) II. \(\frac{c}{d} < 1\) III. \(c^2 + d^2 > 1\)

A. I only B. II only C. I and II only D. II and III only E. I, II, and III[/quote]

\(0<1-\frac{c}{d}<1\) --> add \(-1\) to all three parts of this inequality --> \(-1<-\frac{c}{d}<0\) --> mutliply by \(-1\) and as multiplying by negative flip signs --> \(1>\frac{c}{d}>0\).

So we have that: \(1>\frac{c}{d}>0\)

I. \(c>0\) --> as \(\frac{c}{d}>0\) and \(d>0\), then \(c>0\). Always true.

II. \(\frac{c}{d}<1\) --> directly given as true.

III. \(c^2 + d^2 > 1\) --> if \(c=1\) and \(d=2\), then YES, but if \(c=0.1\) and \(d=0.2\), then No, hences this one is not always true.

Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be [#permalink]

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02 Apr 2012, 18:18

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kashishh wrote:

Inequalities !! is it possible to get some tips and tools to solve these "sign" (< = > ) questions?

i dunno if my method can help you but i solved it in two ways.

0 < 1 - (c/d) < 1

the way i looked at this problem was to first simply the equation by subtracting 1 in the middle to both sides: -1 < -(c/d) < 0

and the given statement is that d is greater than zero, so d must be positive. if we know a variable is positive, you can safely multiply the variable without changing the signs. so i multiplied d to isolate c. -1(d) < -c < 0(d)

so we get -d < -c < 0

and then i divided the negative out and switched the signs. d > c > 0

the above answers statement I. c > 0 is true. for statement II, i manipulated the above equation by dividing d and i get 1 > c/d > 0 this answers statement II. c/d < 1

the last statement i just tested numbers using the d > c > 0 by using whole numbers and decimals.

the second way i solved this was using bunuel's way.

If d > 0 and 0 < 1 - c/d < 1, which of the following must be [#permalink]

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08 Nov 2014, 17:16

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This post was BOOKMARKED

Hey Guys

May I raise a small question regarding the third statement? If it would be INSTEAD of c^2 + d^2 > 1 be c^2 + d^2 > 0 , would the third statement be considered sufficient? My thinking is as follows:

1. as we know (out of proven first statement) that c>0: multiply the IE with itself leads to: c^2 >0 2. as we know (out of the question stem) that d>0: multiply the IE with itself leads to: d^2 >0 3. Combining the two IE from above by adding them up: c^2 + d^2 >0 4. The MODIFIED Statement 3 is considered sufficient

Would highly appreciate your inputs on my thought...

May I raise a small question regarding the third statement? If it would be INSTEAD of c^2 + d^2 > 1 be c^2 + d^2 > 0 , would the third statement be considered sufficient? My thinking is as follows:

1. as we know (out of proven first statement) that c>0: multiply the IE with itself leads to: c^2 >0 2. as we know (out of the question stem) that d>0: multiply the IE with itself leads to: d^2 >0 3. Combining the two IE from above by adding them up: c^2 + d^2 >0 4. The MODIFIED Statement 3 is considered sufficient

Would highly appreciate your inputs on my thought...

Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be [#permalink]

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21 Nov 2014, 08:42

Bunuel I have just one confusion here, 1> c/d > 0, however st 2 says c/d <1, doesnt the statement also include the cases where c/d<0, where the inequality wont hold so it isnt a MUST BE TRUE statement. Thanks in advance

Bunuel I have just one confusion here, 1> c/d > 0, however st 2 says c/d <1, doesnt the statement also include the cases where c/d<0, where the inequality wont hold so it isnt a MUST BE TRUE statement. Thanks in advance

We got that \(0 < \frac{c}{d} < 1\). So, the answer to the question whether \(\frac{c}{d} < 1\) is true, is YES. How else?
_________________

Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be [#permalink]

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07 Sep 2015, 05:13

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Bunuel wrote:

sunaimshadmani wrote:

I have just one confusion here, 1> c/d > 0, however st 2 says c/d <1, doesnt the statement also include the cases where c/d<0, where the inequality wont hold so it isnt a MUST BE TRUE statement. Thanks in advance

We got that \(0 < \frac{c}{d} < 1\). So, the answer to the question whether \(\frac{c}{d} < 1\) is true, is YES. How else?

I think what he meant was a different thing.

He assumes that c/d can also be negative. This is not possible, because c has to be bigger than 0 and d is positive (given).

Therefore c/d has to be smaller than 1 (and is NEVER negative).

Hope this helps people who struggeld with this part (like I did )

Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be [#permalink]

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21 Sep 2016, 06:18

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