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If d > 0 and 0 < 1 - c/d < 1, which of the following must be true?

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If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   06 Mar 2014, 03:05
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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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C
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Bunuel
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   06 Mar 2014, 03:05
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SOLUTION

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

\(0<1-\frac{c}{d}<1\) --> add \(-1\) to all three parts of this inequality --> \(-1<-\frac{c}{d}<0\) --> multiply 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, hence this one is not always true.

Answer: C (I and II only).
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   06 Mar 2014, 03:58
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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

Sol: Given d>0 and 0 < 1 - c/d < 1---------> Eq a

Subtract from Eq a we get -1<-c/d<0

Mulitply by each term by -1 we get 1>c/d>0

Since d>0 and c/d>0 this implies c>0

also we can clearly see c/d> 1. At this stage options A,B and D can be rejected

Consider c^2+d^2> 1

If c=3 and d=5 then c^2+d^2>1
But if c=1/2 and d=3/4 then we have 1/4+9/25 or (25+36)/100 or 61 /100 which is less than 1

Ans is C
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   30 Aug 2017, 08:59
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Answer is C:

0<1-c/d<1
1-c/d>0
-c/d>-1
c/d<1 ________________________- ii point is proved

1-c/d<1
-c/d<0
c/d>0 , d>0 given so c/d to be >0 , implies c>0 ______i point proved

0<c/d<1
c=1 d=2______________c^2+d^2 =5 wow
c=0.1 d=0.2__________c^2 + d^2 = 0.05 ohh no.

hence only 1 and 2 are correct....
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   03 Oct 2017, 07:13
0<1-c/d<1

0<d-c/d<1

0<d-c<d

Ie d is >d-c and both d and c are positive (if c is -ve then d-c cannot be <d) and d is>c.

c>0 proved above

c/d<1 a smaller postive number divided by a larger +ve no will result in a decimal-Possible

c^2+d^2>1 if c and d are decimals then this inequality is not true.

Only 1 and II are a 'must' case.
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   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 true? [#permalink] New post   25 Jan 2018, 20:22
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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 true? [#permalink] New post   29 Jan 2018, 11:06
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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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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   25 Sep 2020, 00:04
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Bunuel 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



\(0 < 1 - \frac{c}{d} < 1\)........Since d>0, we can multiply the entire inequality by d, without changing the signs..
\(0*d <d*( 1 - \frac{c}{d}) < 1*d............0<d-c<d\)

Thus..
I) d-c<d or c>0
II) d-c>0 or d>c or c/d<1
Thus I and II are true..

Now III
III. \(c^2 + d^2 > 1\)
We know d>c>0, but we do not know the exact values as if both are very less for example 0.5>0.1>0.....0.5^2+0.1^2 = 0.25+0.01=0.26<1...answer is NO
But if d>1, answer is surely YES....1^2+0.1^2=1+0.01=1.01>1...answer is YES

So I and II are surely true..

C
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   01 Oct 2020, 16:54
for I, what if c>0 but c<d. like c=100 & d=99???
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   01 Oct 2020, 20:02
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AbhishekDhanraJ72 wrote:
for I, what if c>0 but c<d. like c=100 & d=99???


Then 0<1-c/d<1 will fail.
1-100/99=-1/99, so 1-c/d<0
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   17 Apr 2021, 23:28
I waited patiently on that trap from the last statement. Caught a sea bass.

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

Upon looking at this, it should be clear that c/d MUST be less than 1 to maintain the inequality: 0 < 1 - c/d < 1

Therefore, d > c and c > 0.

i) KEEP
ii) c/d < 1 is necessary KEEP.
iii) Don't forget to try fractions.

c = 1/11, d = 1/10 .

DROP

C is the answer.
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If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   30 Apr 2021, 00:29
JeffTargetTestPrep

What about if 1 > -c/-d > 0 here d is not > 0 but -c/-d is > 0

what I am missing? many thanks
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   02 May 2021, 00:58
Bunuel wrote:
SOLUTION

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

\(0<1-\frac{c}{d}<1\) --> add \(-1\) to all three parts of this inequality --> \(-1<-\frac{c}{d}<0\) --> multiply 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, hence this one is not always true.

Answer: C (I and II only).



why cant we have -1/-2 as c/d in option 1? where c is not greater than 0?
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   02 May 2021, 01:51
Expert Reply
Jaya6 wrote:
Bunuel wrote:
SOLUTION

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

\(0<1-\frac{c}{d}<1\) --> add \(-1\) to all three parts of this inequality --> \(-1<-\frac{c}{d}<0\) --> multiply 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, hence this one is not always true.

Answer: C (I and II only).



why cant we have -1/-2 as c/d in option 1? where c is not greater than 0?


Notice that the stem says that d > 0, so it cannot be -2 as in your example.
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   20 Sep 2021, 06:20
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Bunuel 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. c² + d² > 1

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II and III


Take the inequality: 0 < 1 - c/d < 1
Subtract 1 from all sides: -1 < -c/d < 0
Multiply all sides by -1 to get: 1 > c/d > 0 [when we multiply an inequality by a NEGATIVE value, we must REVERSE the direction of the inequality symbols]
Rearrange to get: 0 < c/d < 1

We can immediately see that statement II (c/d < 1) is true

Also, since 0 < c/d, we know that c/d is positive.
Since we're told d is positive, we can conclude that c must also be positive
In other words, c > 0, which means statement I is true.

Now let's analyze statement III
Notice that the values c = 0.1 and d = 0.2 satisfy all of the given information.
Now recognize that c² + d² = 0.1² + 0.2² = 0.01 + 0.04 = 0.05
So, it's not necessarily the case that c² + d² > 1
In other words, statement III need not be true.

Answer: C
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   14 Jan 2023, 21:26
Bunuel wrote:
SOLUTION

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

\(0<1-\frac{c}{d}<1\) --> add \(-1\) to all three parts of this inequality --> \(-1<-\frac{c}{d}<0\) --> multiply 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, hence this one is not always true.

Answer: C (I and II only).


Bunuel

Can you please explain why II must be true? It only states that c/d < 1 but not that 0 < c/d < 1. So c/d < 1 can have the possibility of c/d = -10.
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   14 Jan 2023, 21:54
Expert Reply
Sahil2208 wrote:
Bunuel wrote:
SOLUTION

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

\(0<1-\frac{c}{d}<1\) --> add \(-1\) to all three parts of this inequality --> \(-1<-\frac{c}{d}<0\) --> multiply 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, hence this one is not always true.

Answer: C (I and II only).


Bunuel

Can you please explain why II must be true? It only states that c/d < 1 but not that 0 < c/d < 1. So c/d < 1 can have the possibility of c/d = -10.


If c/d<0, then 1-(c/d) will become greater than 1.
But it is given that 0<1-c/d<1.
From 1-c/d<1, we get c/d>0.
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   14 Jan 2023, 22:18
Asked: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true?

Quote:
I. c > 0

1 - c/d < 1
c/d > 0
Since d>0
c > 0
MUST BE TRUE

Quote:
II. c/d < 1

0 < 1 - c/d < 1
c/d <1
MUST BE TRUE

Quote:
III. c^2 + d^2 > 1

c^2 + d^2 = c^2/d^2 * d^2 + d^2 = d^2 (c^2/d^2 + 1)
If c=.25; d=.5; c^2 + d^2<1
But if c=4; d=8; c^2 + d^2 > 1
NOT NECESSARILY TRUE

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II and III

IMO C
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink] New post   14 Feb 2023, 01:42
chetan2u wrote:
Sahil2208 wrote:
Bunuel wrote:
SOLUTION

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

\(0<1-\frac{c}{d}<1\) --> add \(-1\) to all three parts of this inequality --> \(-1<-\frac{c}{d}<0\) --> multiply 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, hence this one is not always true.

Answer: C (I and II only).


Bunuel

Can you please explain why II must be true? It only states that c/d < 1 but not that 0 < c/d < 1. So c/d < 1 can have the possibility of c/d = -10.


If c/d<0, then 1-(c/d) will become greater than 1.
But it is given that 0<1-c/d<1.
From 1-c/d<1, we get c/d>0.


Hi Chetan,
I have the same confusion as Sahil. Isnt what you just said makes II must be true if II would have said 0<c/d<1.
Statement II just asks if c/d >1 must be true for 0 < 1-(c/d) < 1
I didnt understand how statement II simply assumed that <1 is >0.
Had statement II stated that c/d>0, would we assume that >0 means <1 ?
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Re: If d > 0 and 0 < 1 - c/d < 1, which of the following must be true? [#permalink]
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