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Praetorian
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[#50] 2 Min. Challenge: Real Numbers 15 May 2004, 08:10
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code:97.15

Time yourself and Please explain your solution. its helps others.

Suppose a, b and c are real numbers for which
a/b > 1 and a/c 0
(B) a > b
(C) (a − c)(b − c) > 0
(D) a + b + c > 0
(E) abc > 0



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Emmanuel
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[#50] 2 Min. Challenge: Real Numbers 15 May 2004, 08:23
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Praetorian
code:97.15

Time yourself and Please explain your solution. its helps others.

Suppose a, b and c are real numbers for which
a/b > 1 and a/c < 0. Which of the following must be true?

(A) a + b − c > 0
(B) a > b
(C) (a − c)(b − c) > 0
(D) a + b + c > 0
(E) abc > 0
Show more


NOT E, A (a = -10, b = -9, c = 1), B (a = -10, b = -9), D (sate example).

So, C is the answer :-)

C is the answer because [a/b]/[a/c] < 0 => c/b < 0 => c/b - 1 < 0 => (c-b)/b < 0, and a/c - 1 < 0 => (a-c)/c < 0 =>

(c-b)(a-c)/bc > 0. But c/b < 0 => bc < 0 => (c-b)(a-c) < 0 => (b-c)(a-c) > 0!!!
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[#50] 2 Min. Challenge: Real Numbers 15 May 2004, 18:07
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C for me tried plugging in numbers and eliminated all but C.
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mirhaque
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[#50] 2 Min. Challenge: Real Numbers 15 May 2004, 18:29
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plugging in number is too time consuming. is there any shortcut?
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[#50] 2 Min. Challenge: Real Numbers 16 May 2004, 14:12
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Praetorian
code:97.15

Time yourself and Please explain your solution. its helps others.

Suppose a, b and c are real numbers for which
a/b > 1 and a/c < 0. Which of the following must be true?

(A) a + b − c > 0
(B) a > b
(C) (a − c)(b − c) > 0
(D) a + b + c > 0
(E) abc > 0
Show more


a/b > 1 means a and b are the same sign and |a|>|b|
a/c < 0 means a and c have different signs

so the possible cases:
(1) a and b are +ve, and c is -ve or
(2) a and b are -ve, and c is +ve.

I was quickly able to elimate (B) and (E). (B) doesn't hold true if both a and b are -ve, and (E) doesn't hold true if a and b are +ve and c is -ve.

(C) looked like it would be the right answer to me because it was the only remaining one that dealt with multiplication, so I decided to try it first.
Case (1): (Postive minus negative)*(Postive minus negative)>?0 which means (Positve + Positive)*(Positive +Positive)>?0 ...True for all cases

Case (2): (Negative - Postive)*(Negative - Postive) >?0, which means (Negative)*(Negative)>?0...True for all cases

Therefore (C) is the answer.
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[#50] 2 Min. Challenge: Real Numbers 19 May 2004, 06:28
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The answer is C.

First, note that a and b agree in signs and that c has a different
sign. Thus, a, b and −c have the same sign. Therefore, a − c and
b − c have the same sign. Hence their product must be positive.
None of the other inequalities need be true.
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[#50] 2 Min. Challenge: Real Numbers 19 May 2004, 11:30
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I got C as well. (1 min 20 secs)

Its really not that bad, if you take two simple examples:

a = 2
b = 1
c = -1

AND

a = -2
b = -1
c = 1

Go through each option (a thru e) for both scenarios. You can knock 'em out pretty quick.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Thank you for understanding, and happy exploring!