Inequalities- Old Paper Tests

[kanha81]

Although, I got this question right while I was solving the problem, it took me about 4 mins to reach the answer. Any faster way of reaching the solution to problems such as listed below will be REALLY helpful:

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

Approach:
d>0: Let d=1/2, 1, 2

c>0 (?) True
d=1/2: 0 < 1-2c < 1
=> 0 < 1-2c => c < 1/2

1-2c < 1 => -2c < 0 => 2c > 0 => c > 0 . True

d=1: 0 < 1-c < 1
0 < 1-c => c < 1
1-c < 1 => -c < 0 => c > 0. True

[II] (c/d) > 1 (?) True
d=1/2: 2c > 1 => c > 1/2
c/d = 1/1/2 = 2 > 1. True

d=1: c > 1
c/d = 2/1 = 2 > 1. True

[III] c^2 + d^2 > 1 (?) False
d=1/2 and c > 0: 1 + (1/4) = 5/4 > 1. True

d=1 and c > 0: (1/4) + (1/4) = 1/2 < 1. False

Hence [spoiler] and [II] --> [C][/spoiler]

[Ian Stewart]

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

You might begin by separating the given three-part inequality into two inequalities, and then simplifying:

0 < 1 - c/d
c/d < 1

and
1 - c/d < 1
c/d > 0

So the given inequality just tells us that 0 < c/d < 1. Since d > 0, c must be greater than zero, and I is true. We have also shown that c/d < 1, so II must be true. There's no reason for III to be true; c and d could be 1/1000 and 1/100, respectively, or they could be 100 and 1000.

[DeepakR]

Since we know that 1-(c/d) should lie between 0 and 1 c/d should be <1 and should not <0 eg.) c/d can be 1/2 ,1/4 etc but it cannot be any negative values as 1-(c/d) will become >1 and wouldn't satisfy the equation.

Hence c cannot be c<0 or c cannot be c=0. Therefore C has to be c>0

I satisfies

The II one is obvious as 1-(c/d) lies between 0 and 1 hence c/d has to be less than 1 and >0.

Hence II satisfies

For III plug in values put c=1 and d=2 it will work. However if you put c=1/4 and d=1/2 then equation would be 1/16 + 1/4 <1 hence III will not work.

-Deepak

[kanha81]

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

You might begin by separating the given three-part inequality into two inequalities, and then simplifying:

0 < 1 - c/d
c/d < 1

and
1 - c/d < 1
c/d > 0

So the given inequality just tells us that 0 < c/d < 1. Since d > 0, c must be greater than zero, and I is true. We have also shown that c/d < 1, so II must be true. There's no reason for III to be true; c and d could be 1/1000 and 1/100, respectively, or they could be 100 and 1000.

Ian,
You make the problem look classically easy by piercing the arrows through the heart of the problem An excellent technique.

Much appreciated. Thanks a bunch.

[kanha81]

Since we know that 1-(c/d) should lie between 0 and 1 c/d should be <1 and should not <0 eg.) c/d can be 1/2 ,1/4 etc but it cannot be any negative values as 1-(c/d) will become >1 and wouldn't satisfy the equation.

Hence c cannot be c<0 or c cannot be c=0. Therefore C has to be c>0

I satisfies

The II one is obvious as 1-(c/d) lies between 0 and 1 hence c/d has to be less than 1 and >0.

Hence II satisfies

For III plug in values put c=1 and d=2 it will work. However if you put c=1/4 and d=1/2 then equation would be 1/16 + 1/4 <1 hence III will not work.

-Deepak

Thanks Deepak.