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If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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Updated on: 01 Apr 2012, 13:10
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If d > 0 and 0 < 1  c/d < 1, which of the following must be true? I. c > 0 II. 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
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Originally posted by rahulraao on 03 Sep 2005, 06:16.
Last edited by Bunuel on 01 Apr 2012, 13:10, edited 1 time in total.
Edited the question and added the OA



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0 < 1  c/d < 1 => 0<c/d<1
with d>0
we get: c>0 and c/d<1
not true for c^2 + d^2 > 1
example: c=0.1 d=0.2
C



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If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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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. Answer: C (I and II only). Hope it's clear.
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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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01 Apr 2012, 14:27
Inequalities !! is it possible to get some tips and tools to solve these "sign" (< = > ) questions?



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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01 Apr 2012, 15:12
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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.



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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19 Jun 2013, 04:51



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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19 Jun 2013, 11:29
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0 < 1  c/d < 1
Multiply by d, 0 < d  c < d
Implies that c>0
Now, d  c > 0 implies that d>c and hence c/d < 1
information we now have is c>0, d>0 and d>c
c^2 + d^2 > 0 but it may or may not be greater than 1.
Hence I & II only.



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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21 Sep 2013, 16:58
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\(0<1\frac{c}{d}<1\)
First Part: \(0<1\frac{c}{d}\) \(\frac{c}{d}<1\) This satisfies Choice II
Second Part: \(1\frac{c}{d}<1\) \(0<\frac{c}{d}\) This satisfies Choice I as \(d>0\)
As for Choice III, if c and d are decimals then this gives us "False" but if they are integers, then "True"



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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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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...
Cheers, Rene



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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09 Nov 2014, 05:19



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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09 Nov 2014, 06:18
Bunuel, thanks very much for the feedback



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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



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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21 Nov 2014, 09:50



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



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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25 Jan 2018, 16:39
Bunuel, can we plug in smart numbers for statement #1? When I plug in a "2" for "d", it turns out to be false.
0 < 1  c/2 < 1
According to I, C is supposed to be a positive number. This statement works when C = 1, but it doesn't work when C = 2.
Plugging in numbers works with #2, making it true.



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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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25 Jan 2018, 20:22
OCDianaOC wrote: Bunuel, can we plug in smart numbers for statement #1? When I plug in a "2" for "d", it turns out to be false.
0 < 1  c/2 < 1
According to I, C is supposed to be a positive number. This statement works when C = 1, but it doesn't work when C = 2.
Plugging in numbers works with #2, making it true. It's the other way around. We are given that \(d >0\) and \(0 < 1  \frac{c}{d} < 1\) are true. If these two inequalities are true, then c will be positive. So you should consider only those numbers which satisfy \(d >0\) and \(0 < 1  \frac{c}{d} < 1\). For c and d, which satisfy \(d >0\) and \(0 < 1  \frac{c}{d} < 1\), c will turn out to be positive.
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Re: If d > 0 and 0 < 1  c/d < 1, which of the following must be [#permalink]
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29 Jan 2018, 11:06
rahulraao wrote: If d > 0 and 0 < 1  c/d < 1, which of the following must be true?
I. c > 0 II. 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 Simplifying the given inequality we have: 0 < 1  c/d < 1 1 < c/d < 0 d < c < 0 d > c > 0 So we see that c must be positive but less than d. Thus, I and II must be true. III does not have to be true since c and d could be some proper fractions, say both less than 1/2. Answer: C
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