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by logitech » Fri Nov 28, 2008 2:03 am
If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab > 0

OA E
LGTCH
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by austin » Fri Nov 28, 2008 3:59 am
If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab >= 0

a · |b| < a – b?

(1) a < 0 => a · |b| < 0
a = -3, b = 2 => a · |b| = -6 < a - b (= -5) YES
a = -4, b = 0 => a · |b| = 0 < a - b (= -4) NO
Inconsistent (REMEMBER b can be 0 in this case)

(2) ab >= 0 => a and b are both positive or both negative ( or zero).
a = -3, b = -2 => a · |b| = -6 < a - b (= -5) YES
a = 3, b = 2 => a · |b| = 6 < a - b (= 1) NO
Inconsistent

Combine, a and b are negative. ( b can also be zero)
a = -3, b = -2 => a · |b| = -6 < a - b (= -1) YES
a = -6, b = 0 => a · |b| = 0 < a - b (= -6) NO

E

If the second statement is (2) ab > 0 then the answer is C
a = -6, b = -1 => a · |b| = -6 < a - b (= -5) YES
a = -20, b = -1 => a · |b| = -20 < a - b (= -19) YES
a = -2, b = -1 => a · |b| = -2 < a - b (= -1) YES
a = -20, b = -19 => a · |b| = -380 < a - b (= -1) YES
C
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by logitech » Fri Nov 28, 2008 9:52 am
austin wrote:If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab >= 0

a · |b| < a – b?

(1) a < 0 => a · |b| < 0
a = -3, b = 2 => a · |b| = -6 < a - b (= -5) YES
a = -4, b = 0 => a · |b| = 0 < a - b (= -4) NO
Inconsistent (REMEMBER b can be 0 in this case)

(2) ab >= 0 => a and b are both positive or both negative ( or zero).
a = -3, b = -2 => a · |b| = -6 < a - b (= -5) YES
a = 3, b = 2 => a · |b| = 6 < a - b (= 1) NO
Inconsistent

Combine, a and b are negative. ( b can also be zero)
a = -3, b = -2 => a · |b| = -6 < a - b (= -1) YES
a = -6, b = 0 => a · |b| = 0 < a - b (= -6) NO

E

If the second statement is (2) ab > 0 then the answer is C
a = -6, b = -1 => a · |b| = -6 < a - b (= -5) YES
a = -20, b = -1 => a · |b| = -20 < a - b (= -19) YES
a = -2, b = -1 => a · |b| = -2 < a - b (= -1) YES
a = -20, b = -19 => a · |b| = -380 < a - b (= -1) YES
C
Still time consuming.
LGTCH
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by austin » Fri Nov 28, 2008 9:58 am
this question can be done without even solving...

b = 0 is a case where the consistency will fail

This question is a bit like:

Is x^2 + y^2 > 4a?

(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

adding we get x^2 + y^2 = 5a.. always true except when x = y = a = 0.

Answer is E... zero can be used a test point straight away (if nothing is mentioned about these variables - like +ve integers etc..)
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by logitech » Fri Nov 28, 2008 10:12 am
austin wrote:this question can be done without even solving...

b = 0 is a case where the consistency will fail

This question is a bit like:

Is x^2 + y^2 > 4a?

(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

adding we get x^2 + y^2 = 5a.. always true except when x = y = a = 0.

Answer is E... zero can be used a test point straight away (if nothing is mentioned about these variables - like +ve integers etc..)
I like this one!!
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by Spring2009 » Tue Dec 02, 2008 5:18 am
austin wrote:this question can be done without even solving...

b = 0 is a case where the consistency will fail

This question is a bit like:

Is x^2 + y^2 > 4a?

(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

adding we get x^2 + y^2 = 5a.. always true except when x = y = a = 0.

Answer is E... zero can be used a test point straight away (if nothing is mentioned about these variables - like +ve integers etc..)
Sorry, I don't get how to use your solution to solve logitech's question. Please explain ! Thanks.
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by austin » Tue Dec 02, 2008 9:52 am
Apply b = 0 in statements 1 and 2, they will fail the test of consistency.

Combine, a < 0 and ab >= 0
b<=0. Put b = 0 for the combined statement, still no consistent answers...
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