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Fractions problem

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by Troika » Mon Jul 09, 2012 4:47 pm
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

[spoiler]OA: C[/spoiler]
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Source: — Problem Solving |

by tutorphd » Mon Jul 09, 2012 4:52 pm
I would classify this as inequality problem with number properties (positive/negative), it has no fraction manipulation in it.

The double inequality 0 < 1 - c/d < 1 simplifies to 1 > c/d > 0 which implies that II is true.
c/d>0 means c and d have the same sign, so if d>0, then c>0, which is I itself.
You can show III is not always true by plugging numbers: keep the ratio c/d<1 but choose very small or very big c and d.

Ans: C
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by GMATGuruNY » Mon Jul 09, 2012 7:23 pm
Troika 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

[spoiler]OA: C[/spoiler]
0 < 1 - c/d < 1 [Subtract 1 from each part]
-1 < -c/d < 0 [Multiply through by -1, remembering to switch the direction of the inequality signs]
1 > c/d > 0 [Multiply by d -- since d>0, no need to worry about switching the direction of the inequality signs]
d > c > 0.

The portions in red show that both II and I must be true.
Eliminate A, B and D.

If c and d are fractions between 0 and 1, then III does not have to be true.
To illustrate:
If d = 1/2 and c = 1/4, then c² + d² = (1/4)² + (1/2)² = 1/16 + 1/4 = 5/16, which is not greater than 1.
Eliminate E.

The correct answer is C.
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